Complex Variables and
Analytic Functions
Complex Variables and
Analytic Functions
An Illustrated Introduction
Bengt Fornberg
University of Colorado
Boulder, Colorado
Cécile Piret
Michigan Technological University
Houghton, Michigan
Society for Industrial and Applied Mathematics
Philadelphia
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Library of Congress Cataloging-in-Publication Data
Names: Fornberg, Bengt, author. | Piret, Cécile, author.
Title: Complex variables and analytic functions : an illustrated
introduction / Bengt Fornberg (University of Colorado, Boulder,
Colorado), Cécile Piret (Michigan Technological University, Houghton,
Michigan).
Description: Philadelphia : Society for Industrial and Applied Mathematics,
[2020] | Series: Other titles in applied mathematics ; 165 | Includes
bibliographical references and index. | Summary: “This book is the first
primary introductory textbook on complex variables and analytic
functions to use predominantly functional illustrations”-- Provided by
publisher.
Identifiers: LCCN 2019030487 (print) | LCCN 2019030488 (ebook) | ISBN
9781611975970 (paperback) | ISBN 9781611975987 (ebook)
Subjects: LCSH: Functions of complex variables--Textbooks. | Analytic
functions--Textbooks.
Classification: LCC QA331.7 .F67 2020 (print) | LCC QA331.7 (ebook) | DDC
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LC record available at https://lccn.loc.gov/2019030487
LC ebook record available at https://lccn.loc.gov/2019030488
is a registered trademark.
Contents
Preface
1
2
ix
Complex Numbers
1.1
How to think about different types of numbers . . . . . .
1.2
Definition of complex numbers . . . . . . . . . . . . . .
1.3
The complex number plane as a tool for planar geometry
1.4
Stereographic projection . . . . . . . . . . . . . . . . . .
1.5
Supplementary materials . . . . . . . . . . . . . . . . .
1.6
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .
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1
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Functions of a Complex Variable
2.1
Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2
Some elementary functions generalized to complex argument by means
of their Taylor expansion . . . . . . . . . . . . . . . . . . . . . . . .
2.3
Additional observations on Taylor expansions of analytic functions . .
2.4
Singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5
Multivalued functions—Branch cuts and Riemann sheets . . . . . . .
2.6
Sequences of analytic functions . . . . . . . . . . . . . . . . . . . . .
2.7
Functions defined by integrals . . . . . . . . . . . . . . . . . . . . . .
2.8
Supplementary materials . . . . . . . . . . . . . . . . . . . . . . . .
2.9
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
19
22
33
38
43
51
54
56
62
3
Analytic Continuation
71
3.1
Introductory examples . . . . . . . . . . . . . . . . . . . . . . . . . . 71
3.2
Some methods for analytic continuation . . . . . . . . . . . . . . . . 73
3.3
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
4
Introduction to Complex Integration
4.1
Integration when a primitive function F (z) is available
4.2
Contour integration . . . . . . . . . . . . . . . . . . .
4.3
Laurent series . . . . . . . . . . . . . . . . . . . . . .
4.4
Supplementary materials . . . . . . . . . . . . . . . .
4.5
Exercises . . . . . . . . . . . . . . . . . . . . . . . . .
5
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93
95
97
106
112
114
Residue Calculus
119
5.1
Residue calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
5.2
Infinite sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
v
vi
Contents
5.3
5.4
5.5
5.6
6
7
8
9
10
11
Analytic continuation with use of contour integration
Weierstrass products and Mittag–Leffler expansions .
Supplementary materials . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . .
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152
160
165
166
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173
173
176
181
184
186
Elliptic Functions
7.1
Some introductory remarks on simply periodic functions .
7.2
Some basic properties of doubly periodic functions . . . .
7.3
The Weierstrass ℘-function . . . . . . . . . . . . . . . .
7.4
The Jacobi elliptic functions . . . . . . . . . . . . . . . .
7.5
Supplementary materials . . . . . . . . . . . . . . . . .
7.6
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .
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191
191
191
194
197
203
207
Conformal Mappings
8.1
Relations between conformal mappings and analytic functions .
8.2
Mappings provided by bilinear functions . . . . . . . . . . . .
8.3
Riemann’s mapping theorem . . . . . . . . . . . . . . . . . .
8.4
Mappings of polygonal regions . . . . . . . . . . . . . . . . .
8.5
Some applications of conformal mappings . . . . . . . . . . .
8.6
Revisiting the Jacobi elliptic function sn(z, k) . . . . . . . . . .
8.7
Supplementary materials . . . . . . . . . . . . . . . . . . . .
8.8
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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209
211
212
214
215
219
222
226
227
Transforms
9.1
Fourier transform . . . . . . . . . . . . . . .
9.2
Laplace transform . . . . . . . . . . . . . . .
9.3
Mellin transform . . . . . . . . . . . . . . . .
9.4
Hilbert transform . . . . . . . . . . . . . . .
9.5
z-transform . . . . . . . . . . . . . . . . . . .
9.6
Three additional transforms related to rotations
9.7
Supplementary materials . . . . . . . . . . .
9.8
Exercises . . . . . . . . . . . . . . . . . . . .
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231
231
245
255
259
264
264
266
268
Wiener–Hopf and Riemann–Hilbert Methods
10.1
The Wiener–Hopf method . . . . . . . . . .
10.2
A brief primer on Riemann–Hilbert methods
10.3
Supplementary materials . . . . . . . . . .
10.4
Exercises . . . . . . . . . . . . . . . . . . .
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273
273
286
288
289
Gamma, Zeta, and Related Functions
6.1
The gamma function . . . . . .
6.2
The zeta function . . . . . . .
6.3
The Lambert W-function . . .
6.4
Supplementary materials . . .
6.5
Exercises . . . . . . . . . . . .
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Special Functions Defined by ODEs
291
11.1
Airy’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292
11.2
Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293
Contents
11.3
11.4
11.5
11.6
12
vii
Hypergeometric functions . . . . .
Converting linear ODEs to integrals
The Painlevé equations . . . . . .
Exercises . . . . . . . . . . . . . .
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299
303
307
314
Steepest Descent for Approximating Integrals
12.1
Asymptotic vs. convergent expansions .
12.2
Euler–Maclaurin formula . . . . . . . .
12.3
Laplace integrals . . . . . . . . . . . . .
12.4
Steepest descent . . . . . . . . . . . . .
12.5
Supplementary materials . . . . . . . .
12.6
Exercises . . . . . . . . . . . . . . . . .
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317
317
319
322
329
346
350
Bibliography
355
Index
359
Preface
The topics of Complex Variables and Analytic Functions are of fundamental importance not only in pure mathematics, but also throughout applied mathematics, physics, and
engineering. Their theorems and formulas also simplify many results from calculus and
for functions of real variables. There is already a wide choice of textbooks available, ranging from timeless classics such as by Whittaker and Watson [41], Copson [8] and Ahlfors
[2],1 to many later ones, raising the question of why anyone would want to see yet another
one. Our main motivation lies in the evolution that has occurred in other fields. For the
last half century or so, it has been unthinkable to use introductory text books for calcu2
lus that do
√ not graphically illustrate the basic elementary functions, such as f (x) = x ,
f (x) = x, f (x) = sin x, f (x) = log x, etc. When we first taught a course on complex
variables and analytic functions, we became puzzled about why complex variables texts
should not also visually build on the real cases, familiar to all students, and then liberally
illustrate how these same functions extend away from the real axis. While formulas alone
for some students might provide a feasible alternative to visual impressions, we are aiming
this text at students that find the latter to be helpful for gaining at least their initial intuitive feeling for the subject. Although we have included an abundance of illustrations (and
give brief code templates for displaying analytic functions with MATLAB and Mathematica), this book is an introduction to the classical theory of complex variables and analytic
functions. It contains enough materials to support a two-semester course, but has been
structured to make it easy to omit chapters or sections as needed for a one-semester course
(offering a lot of flexibility in course emphasis). SIAM’s website for this book, available
from www.siam.org/books/ot165, contains “Notes to Instructors,” with suggestions for different one- and two-semester syllabi, ideas for supplementary student projects, etc. Once
a solution manual for all the exercises in the text has been developed, it will also provide
information for how instructors can get access to it.
Textbooks often differ with regard to the order in which topics are covered. One strategy
is to make sure each step follows rigorously from previous steps. While that can have
some appeal, it might not necessarily be the best order for developing an initial conceptual
understanding, and it also does not correspond to how mathematical problem solving and
research is carried out.2 Introducing key ideas early on might require certain issues to be
revisited later, once additional tools have fallen in place. In either case, the end knowledge
1 Supplementing these texts, Jahnke and Emde’s “Tables of Functions” [28] has excellent (precomputer era)
illustrations, but lacks textbook-type materials.
2 The eminent mathematician Paul Halmos writes in his autobiography [24, page 321]: “Mathematics is not a
deductive science - that’s a cliché. When you try to prove a theorem, you don’t just list the hypotheses, and then
start to reason. What you do is trial and error, experimentation, guesswork. You want to find out what the facts
are, and what you do is in that respect similar to what a laboratory technician does.”
ix
x
Preface
will be similar, but it is our belief that the latter approach is better suited for an introductory
text. A case in point concerns singularities, where we here bring these up early on, before
having developed all the tools of contour integration that are needed for a certain (Laurentexpansion based) singularity characterization.
We have in this book never sacrificed correctness for simplicity. However, there inevitably arise situations where it becomes unavoidable to choose a compromise path between strictest mathematical rigor and more heuristic arguments. Believing that the former
is better suited for finalizing proofs than for gaining insights and for solving problems, we
have not shied away from the latter when that is more appropriate (but have then made this
choice clear in the text).
The emphasis that is given here to separate topics also differs somewhat from many
other texts. For example, we view analytic continuation as fundamental to a good understanding of the nature of analytic functions. Instead of just listing one approach for this
task (the circle-chain theorem, which incidentally is quite impractical in many contexts),
we have included numerous approaches. We also discuss and illustrate multivalued functions and their Riemann sheets more extensively than what is customary in introductory
texts. The main applications of analytic functions have changed quite significantly during
the last decades. For example, conformal mapping was more central to applied mathematics before recent advances in scientific computing (which have vastly extended the range
of equations that can be addressed effectively). On the other hand, analytic function techniques are gaining in significance in other areas (such as when used in conjunction with
numerical methods). Rather than discussing different applications (say, why scientists and
engineers routinely use Fourier or Laplace transforms, complex exponentials, etc.), our focus has been on providing the analytic function based tools that are needed across a wide
range of fields.
The present material originated from lecture notes first developed by Bengt Fornberg
in 2006 and then used jointly by us in 2008 (at the University of Colorado, Boulder). After
that, they were mostly put aside until Cécile Piret revitalized them when resuming teaching
the topic in 2015 at Michigan Technological University.
As noted above, a main purpose of extending from the real axis to the complex plane
is to greatly simplify a vast number of tasks in applied mathematics, engineering, etc. The
renowned mathematician Jacques Hadamard expressed this succinctly (paraphrasing an
earlier statement from 1900 by Paul Painlevé):
“The shortest path between two truths in the real domain passes through the complex
domain.”
However, complex numbers go far beyond being just a mathematical trick that, once
having done its magic, ought to quickly and gracefully disappear. It is our hope that this
book will make its readers not only appreciate their utility, but also come to regard them as
equally “natural” as, say, negative integers.
Acknowledgments: While developing the present book, we received extensive and
thorough comments from numerous experts, both on educational and on technical aspects.
We want to extend special thanks to Dr. Tom Bogdan and Professors Willy Hereman, Paul
Martin, Nick Trefethen, and Grady Wright for many suggestions, as well as much encouragement. From the first concept suggestion to the final product, it has been a delight to
work with Elizabeth Greenspan at SIAM. Personally, for love and support, B.F. thanks
Natasha and C.P. thanks Erin, Maximilien, and Juliette.
Chapter 1
Complex Numbers
1.1 How to think about different types of numbers
1.1.1 Integers, rational numbers, and real numbers
The origins of counting are lost in prehistory. For a very long time, the only numbers used
were positive integers, in modern notation 1,2,3, . . . . Such numbers can be used to count
apples, oranges, days, frying pans, etc. The earliest number system extension was to allow
for rational numbers, such as 3/7 or 11/8. These are clearly useful in measuring things like
distances and weights. They can still be associated with counting (like pieces of a whole
pie), and they simplify division by allowing all cases (apart from division by zero). An
important point to note is that it is usually best not to think of a rational number as a pair of
two integers. Even when represented as a ratio of two integers, the combination is usually
best viewed as a single number (and we use it as such in, say, representing a single position
on the real axis), not as some sort of a 2-component vector.
Zero and negative numbers originally made little sense. However, much algebra—such
as subtraction—became much easier if these were allowed. Else there would need to be
extra rules about when subtraction is permitted. Also, there are many applications for
which both the input data and the final answer are positive, but when nevertheless going
via negative numbers during intermediate steps makes it easier to reach the final answer.
We have now given up on the idea that numbers need to directly correspond to counting
objects.
The Greeks, over 2000 years ago, noted that even the rational numbers were insufficient
to measure all distances, such as the length of the diagonal of a square with side length one.
This discovery was deeply agonizing, since it caused fear that the Gods would avenge this
human exposure of their imperfection in creation. The discovery of the irrational numbers
which, together with rational numbers, form the set of real numbers was kept secret for
quite some time.
1.1.2 Complex numbers
The historical perspective above may be useful as a background when we now extend the
real number system to complex numbers. There are no 3+5i apples, just as there are no −7
apples, so we are already used to numbers being generalized past being merely counters for
1
2
Chapter 1. Complex Numbers
objects. Much like for rational numbers, a complex number can be represented by means
of a pair of numbers (and we graphically use these two real numbers as coordinates in
picturing a complex number’s location in the complex plane). Nevertheless, it is often best
to think about z = 3 + 5i as a single number, just as 3/7 is a single number. This book is
largely about what happens
´ to our usual functions, such as y = sin x, and to our calculus
tools, such as df (x)/dx, f (x)dx, etc., when we use z complex instead of x real. Any
other way of thinking of z than as a single number would for such work be very awkward.3
It is natural to wonder whether one next should continue generalizing complex numbers
to some kind of hypercomplex ones, with more than two components. It can be shown that
such attempts (of which there have been many) will require very severe sacrifices—rules
such as a·b = b·a and/or others that we want to take for granted will have to be abandoned,
typically causing much greater losses than gains. Complex numbers in many ways is THE
most natural number system possible. Making excursions into the complex plane often
provides the simplest solution strategy also for problems where all ingredients as well as
the final answers are all real. Also, analytic functions (meaning functions w = f (z) for
which df /dz exists, the primary topic of this book) are in many ways much simpler than
real functions f (x). A few aspects which we will encounter early on in this book are the
following:
1. If a function f (z) can be differentiated once, it can be differentiated infinitely many
times.
2. If an analytic function f (z) is defined uniquely on any interval, no matter how short,
it is automatically defined uniquely away from that line segment as well.
3. If the magnitude of an analytic function is everywhere bounded by some constant,
the only possibility is that the function is identically constant.
4. A polynomial equation of degree n will always have exactly n roots.
The term complex is unfortunate, since generalizing to complex numbers simplifies much
of calculus and algebra. The complex number system is the most natural system in which
to do most mathematics. Other cases, such as integers, rational numbers, real numbers,
etc., are just restrictive subclasses, with often more difficult rules.
1.2 Definition of complex numbers
We recall that negative numbers were introduced in order to always make subtraction possible, with rather immediate practical applications. In contrast, when complex numbers
were first conceived (to have some formal notation for all square roots and for solutions
to all quadratic equations, such as x2 + 1 = 0), their practical utility was at first very
unclear. A key step in advancing these complex numbers from being mostly meaningless
notational abstractions occurred when Girolamo Cardano (1501–1576) described a for3
mula for solving a general cubic equation. One
√ case he considered was z − 15z − 4 = 0,
with the three
√ roots z1 = 4, z2,3√= −2 ± 3. However, his procedure gave one root as
z1 = (2 + −121)1/3 + (2 − −121)1/3 . On observing that, in our modern notation,
3 Roger Penrose, in his bestseller The Road to Reality [37] writes: “When we get used to playing with these
complex numbers, we cease to think of a + i b as a pair of things, namely the two real numbers a and b, but think
instead of a + i b as an entire thing on its own, and we could use a single letter, say z, to denote the whole complex
number z = a + i b.”
1.2. Definition of complex numbers
3
(2+i)3 = 2+11i and (2−i)3 = 2−11i, z1 can be simplified to z1 = (2+i)+(2−i) = 4.
For a purely real-valued problem, a temporary excursion into the world of complex numbers had produced a real-valued solution that no previously available systematic approach
had been able to reach. This is likely the first known example of the (much later formulated) quote in the Preface: “The shortest path between two truths in the real domain passes
through the complex domain.”
√
In 1777, Euler assigned the symbol i to the imaginary units satisfying i = −1. Not
until in the early 19th century did the complex number plane become recognized as a
natural extension of the number line. The real axis becomes then just a special case.
1.2.1 Complex number system: Algebraic introduction
Let i have the property that i2 = −1. Assume that all the standard rules of algebra continue
to hold:
1. z1 + z2 = z2 + z1 and z1 z2 = z2 z1 (commutativity in addition and multiplication),
2. (z1 + z2 ) + z3 = z1 + (z2 + z3 ) and (z1 z2 ) z3 = z1 (z2 z3 ) (associativity in addition
and multiplication),
3. (z1 + z2 ) z3 = z1 z3 + z2 z3 and z1 (z2 + z3 ) = z1 z2 + z1 z3 (distributivity).
Next, let z1 = a + ib and z2 = c + id, where a, b, c, d are real. Then the following hold:
1. z1 + z2 = (a + c) + i(b + d) and z1 − z2 = (a − c) + i(b − d),
2. z1 z2 = (a + ib)(c + id) = (ac − bd) + i(ad + bc),
z1
a + ib
ac + bd
a + ib
c − id
(ac + bd) + i(bc − ad)
3.
=
= 2
+
=
=
z2
c + id
c + id
c − id
c2 + d2
c + d2
bc − ad
.
i 2
c + d2
The results are in these cases complex numbers. While the steps in cases 1 and 2 are
straightforward, the idea in Case 3 (eliminating a complex factor c+id from a denominator
by multiplying by c−id
c−id ) is also very often applicable.
1.2.2 Complex number system: Graphical representation
A complex number z = x + iy is made up of a real part x, denoted by Re z, and of an
imaginary part,4 y = Im z. The complex z-plane was originally devised by Caspar Wessel,
a Danish-Norwegian cartographer, in a paper published in 1797, but which would receive
little initial attention outside of Scandinavia, maybe because it was written in Danish.5 This
plane can be viewed as a two-dimensional Cartesian coordinate system whose horizontal
axis corresponds to the real part of the numbers and whose vertical axis corresponds to the
imaginary part of the numbers. We can thus display z on the complex plane as a point with
abscissa x and ordinate y, as in Figure 1.1.
We can also display operations on the numbers in the complex plane. Addition and subtraction are illustrated in Figure 1.2. One can add and subtract by forming parallelograms,
4 Note
that y = Im z in itself is a real-valued number.
a result, the complex plane is sometimes referred to as the Argand plane, after an independent description
of it by J.-R. Argand in 1813.
5 As
4
Chapter 1. Complex Numbers
Im(z)
z = x +iy
iy
x
Re(z)
Figure 1.1. Representation of a number z = x + iy in the complex plane.
Im(z)
z1 + z2
z1
z2
Re(z)
Im(z)
z1
z1 − z2
z2
Re(z)
Figure 1.2. Addition and subtraction of two complex numbers.
1.2. Definition of complex numbers
5
in the same way that one would perform addition or subtraction of two vectors. In order to
display multiplication and division of two complex numbers, we need first to represent z1
and z2 in polar form. A complex number has a magnitude and an
√argument (or angle); cf.
Figure 1.3. The point z1 = a + ib has a magnitude r1 = |z1 | = a2 + b2 and an argument
θ1 . The point z2 has a magnitude r2 and an argument θ2 . Then, z1 = r1 (cos(θ1 )+i sin(θ1 ))
and z2 = r2 (cos(θ2 ) + i sin(θ2 )). We obtain
z1 z2 = r1 (cos(θ1 ) + i sin(θ1 )) · r2 (cos(θ2 ) + i sin(θ2 ))
= r1 r2 ((cos(θ1 ) cos(θ2 ) − sin(θ1 ) sin(θ2 )) + i(cos(θ1 ) sin(θ2 ) + sin(θ1 ) cos(θ2 )))
= r1 r2 (cos(θ1 + θ2 ) + i sin(θ1 + θ2 ))
and similarly
z1
r1
= (cos(θ1 − θ2 ) + i sin(θ1 − θ2 )).
z2
r2
The magnitude of a product is r1 r2 , the product of the magnitudes, and the argument
of a product is θ1 + θ2 , the sum of the arguments. Similarly, the magnitude of a quotient
is r1 /r2 and the argument of a quotient is θ1 − θ2 . The product rule can be expressed as
follows:6
|z1 · z2 | = |z1 | · |z2 | and arg(z1 · z2 ) = arg(z1 ) + arg(z2 ),
|z1 /z2 | = |z1 |/|z2 | and arg(z1 /z2 ) = arg(z1 ) − arg(z2 ).
The complex conjugate of a complex number z = x + iy is z̄ = x + iy = x − iy (i.e.,
swapping the sign of the imaginary part). A useful formula for finding the magnitude of a
complex number is r2 = |z|2 = z̄z. One can easily check that
• z1 ± z2 = z1 ± z2 ,
• z1 · z2 = z1 · z2 ,
• z1 /z2 = z1 /z2 .
If p(z) is a polynomial with real-valued coefficients, it follows from these rules that p(z) =
p(z). This is a special case of the Schwarz reflection principle described in Section 3.2.2.
1.2.3 Examples of polar rules
1. i2 = −1. This makes sense, since arg(i) = π/2 and |i| = 1. If we square i, we
square the magnitude and double the argument; see the illustration in Figure 1.3.
This had better be true, since it is the property that initiated the whole present topic
of complex variables.
2. We can graphically find solutions to z k + 1 = 0, k = 1, 2, 3, by alternatively considering z k = −1. We find all the points on the unit circle (the circle of radius 1
centered at the origin) for which k arg(z) = π + 2nπ, where n is an integer. For
example, if k = 2, arg(z) = π/2 + nπ when n = 0, 1. The k = 3 case is displayed
in Figure 1.4(a).
6 The
dot “·” indicating multiplication will usually later be omitted unless it is helpful for clarity.
6
Chapter 1. Complex Numbers
Im(z)
z1
r1
z2
θ1
r2
θ2
Re(z)
(a) Polar form of complex numbers.
Im(z)
i
π/2
π
1 Re(z)
−1
(b) Display of i2 = −1.
Figure 1.3. If we square i, we square the magnitude (here equal to one) and double the
argument (here π/2), i.e., the result ends up at −1.
3. We can similarly find solutions to z k − 1 = 0 by considering that z k = 1. These k
solutions are called the kth roots of unity. They are also equispaced around the unit
circle. The third roots of unity are shown in Figure 1.4(b).
4. The product rule implies de Moivre’s formula
n
(cos θ + i sin θ) = cos nθ + i sin nθ
(1.1)
in the special case of n integer.
After we, in the next chapter, have introduced functions of complex variables, such as
f (z) = ez , we can simplify (and generalize) many polar rule results. One such case, noted
at the end of Section 2.2.2, removes the restriction on n in (1.1).
Example 1.1. Use the polar rules to simplify z = √ 1+i√ .
1+i 3
1.2. Definition of complex numbers
7
(a) z 3 + 1 = 0.
(b) z 3 − 1 = 0.
Figure 1.4. Illustrations of the solutions to two single cubic equations.
√
For the numerator, we get |1 + i| = 2 and √
arg(1 + i) = arctan
1=
√
√ π/4. Regarding
3,
i.e.,
|1
+
i
3|
=
1 + 3 = 2, and
the denominator,
we
start
by
considering
1
+
i
√
√
√ √
π
arg(1 + i 3) = arctan 3 = π/3. Thus, |z| = 2/ 2 = 1 and arg z = π4 − 12 · π3 = 12
.
π
π 7
Hence z = cos 12 + i sin 12 .
7 Since both (+z)2 and (−z)2 evaluate to z 2 , the reverse process of taking a square root will give two answers,
√
of the opposite sign. It is common to let z denote the value in the right half-plane, but to include both values
when instead writing z 1/2 . Multivalued functions are discussed further in Section 2.5.
8
Chapter 1. Complex Numbers
We should note that the function arctan x (with x real) always returns a value in the
range [− π2 , π2 ] while the argument θ of a complex number has no corresponding limitation.
If z is in the 1st or 4th quadrants (counting these counterclockwise starting from x > 0,
Im(z)
, but we need to either add π (if in 2nd quadrant) or
y ≥ 0), we can use θ = arctan Re(z)
subtract π (if in 3rd quadrant) to this result.
Example 1.2. Express z = −1 − i in polar form.
p
√
We obtain r = Re(z)2 + Im(z)2 = 2 and, since z is in the 3rd quadrant, θ =
−1
8
(arctan −1
) − π = π4 − π = − 3π
4 .
1.3 The complex number plane as a tool for planar
geometry
Most geometric problems in planar geometry can be solved by means of analytic geometry
(which has little or nothing to do with analytic functions). It is a branch of algebra that is
used to model geometric objects by representing points, lines, etc., by formulas in Cartesian
coordinates x, y in 2-D (and x, y, z in 3-D). There are certain cases where the complex
plane z = x + iy provides convenient algebraic shortcuts over more general formulations
in x and y. We give here just a single classical example, with several further ones in the
Exercises.
George Gamow’s delightful book One, Two, Three . . . Infinity [21] describes the following treasure hunt problem:
Example 1.3. A young adventurer found among his great-grandfather’s belongings a piece
of parchment with instructions to find a hidden treasure. It described the location of the
island, with the following further instructions (here translated to contemporary English):
“Find the meadow with a single oak tree and a single pine tree. Start from the gallows,
walk to the oak, counting your steps. Turn at the oak right and walk the same distance
again. Put there a stake in the ground. Return to the gallows, walk to the pine (counting
your steps), turn this time left, walk the same distance, and place another stake. Dig halfway between the stakes, and the treasure is there.” The adventurer found the island, the
meadow, and the two trees, but all traces of the gallows were lost. He dug desperately in
many places, and finally returned home, ruined. Could he have done better?
Figure 1.5(a) shows the adventurer just having landed on the island. Had he known
the basics of complex arithmetic, he would have placed a complex plane over the island,
in such way that the oak was located at −1 and the pine at +1, and denoted the unknown
location of the gallows by Γ. Part (b) of the figure shows this same complex plane, reduced
to its basic mathematical essentials, and also two arbitrarily chosen guesses Γ and Γ0 for
the location of the gallows. The instructions for the two walks starting from Γ give stake
positions A and B satisfying
A − (−1) = i (Γ − (−1)),
B − 1 = i (1 − Γ),
8 It
is common to use the range (−π, π] for θ, but z = r (cos θ + i sin θ) still holds if we to a value for θ add
any integer multiple of 2π.
1.3. The complex number plane as a tool for planar geometry
9
(a) Sketch of island
A'
2
A
1.5
1
treasure
B
0.5
B'
oak
-2
-1.5
-1
0.5 pine1
-0.5
1.5
2
-0.5
'
-1
(b) Superposed complex plane
Figure 1.5. Illustration of the treasure hunt in Example 1.3: (a) Sketch of island (reproduced from [21], with permission from Dover, Inc.), (b) Simplified illustration showing only the
mathematical essentials (in case of two different locations Γ and Γ0 of the gallows; A, B and A0 , B 0
are the stake locations).
= i .9 To find the treasure, it was therefore completely
from which it follows that A+B
2
unnecessary to know the starting point of the walks. He could have found the treasure
immediately at the location +i.
9 We utilized here that adding a constant to a complex number amounts to a translation in the complex plane,
and multiplication by i a rotation of 90◦ around the origin, leaving magnitudes intact.
10
Chapter 1. Complex Numbers
Figure 1.6. The stereographic projection of the lines Re z = 0 and Im z = 0 (in bold) and
two arbitrary circles (viewed from a location above the complex plane, with Im z large and positive).
The North Pole (denoted by a black dot) maps to infinity, and the South Pole maps to zero.
1.4 Stereographic projection
Although addition, subtraction, multiplication, and division are all well defined in the numbering system defined so far, there still persists a notable limitation. We cannot divide by
zero. In terms of complex numbers, this becomes naturally handled if we project the complex number plane to a sphere of radius one, placed on top of the origin in the plane, in
the way shown in Figure 1.6. We clearly get a one-to-one map between the plane and the
sphere. However, ∞ (infinity, in all directions) in the plane maps to the North Pole, a point
much like any other point on the sphere. There is a trade-off, however, in defining complex
numbers as points on the sphere instead of on the plane. By doing so, we lose the nice immediate illustrations of the arithmetic operations +, −, ∗, /, but instead gain a better way to
think about divisions by zero. Maybe more importantly, the sphere helps in thinking about
singularities (such as poles and branch points, introduced later); these may be located in
the finite part of the complex plane and/or at infinity.
Examples
1
Consider the function f1 (z) = z−(1+i)
. It diverges to ∞ at z = 1 + i, a point on the
plane with a matching point on the sphere, where it is said to have a pole singularity. Next,
consider the function f2 (z) = z. This function goes to ∞ in the limit of z → ∞ following
any direction in the complex plane. The limits in all of these directions map on the sphere
to its top (North Pole). Thus, on the sphere, the two functions f1 (z) and f2 (z) do not differ
in “character,” but only in where the singularity is located.
This stereographic projection has a number of nice properties, such as the following:
• Locally, angles between intersecting curves are preserved between the plane and the
sphere.
• A circle in the complex plane maps to a circle on the sphere.
• A circle on the sphere which contains the North Pole maps to a straight line in the
plane.
1.5. Supplementary materials
11
These properties are all discussed further in the Exercises in Section 1.6.
1.4.1 Formulas
It is easy to find the correspondence between points z = x + iy in the plane and points
(X, Y, Z) on the sphere (satisfying X 2 + Y 2 + (Z − 1)2 = 1):
Plane → Sphere
X=
4x
x2 +y 2 +4
Y =
x2 +y 2 +4
Z=
2(x2 +y 2 )
x2 +y 2 +4
Sphere → Plane
4y
x=
2X
2−Z
y=
2Y
2−Z
1.5 Supplementary materials
1.5.1 Constructing regular polygons with ruler and compass
Geometric constructions using just a ruler and a compass are possible only when the result
can be expressed algebraically as solving a sequence of linear and quadratic equations.
Constructing regular polygons is closely related to properties of roots of unity.
Example 1.4. Show that the regular pentagon can be constructed using ruler and com2π
pass.10 Derive algebraic expressions for cos 2π
5 and sin 5 .
The task is equivalent to finding the nontrivial roots to z 5 − 1 = (z − 1)(z 4 + z 3 + z 2 +
z + 1) = 0. Writing these roots in turn as z1 , z2 , z3 , z4 , their sum is −1 (by the relation
between roots and coefficients of a polynomial; −1 is the negative of the coefficient for z 3
in this quartic polynomial). Noting that z4 = z1 and z3 = z2 , we next group them in pairs
z + z + z2 + z3 = −1.
|1 {z }4
| {z }
y1
y2
(1.2)
Then y1 + y2 = −1 (directly from (1.2)) and also y1 · y2 = (z1 + z4 )(z2 + z3 ) = z1 z2 +
z1 z3 + z4 z2 + z4 z3 = z3 + z4 + z1 + z2 = −1. Again using the relations between
roots and√coefficients of a polynomial,
y1 and y2 are the roots of y 2 + y − 1 = 0, i.e.,
√
1
1
y1 = 2 ( 5 − 1) and y2 = 2 (− 5 − 1). Next, y1 = z1 + z4 = z1 + 1/z1 gives the
q
√
√
quadratic z12 − y1 z1 + 1 = 0, with solution z1 = 14 ( 5 − 1 ± 2i 52 + 25 ), from which
q
√
√
1
2π
we read off cos 2π
=
(
5
−
1)
and
sin
=
(5 + 5)/8.
5
4
5
10 In one of the most famous mathematical discoveries of all times, C.F. Gauss (in 1796) used a related argument (involving successive stages of grouping roots) to find that a regular polygon with N corners is possible
to construct with ruler and compass if and only if N = 2k p1 p2 · · · · · ps , where k = 0, 1, 2, . . . and pi ,
n
i = 1, 2, . . . , s, are distinct prime numbers of the form p = 2(2 ) + 1, n = 0, 1, 2, . . .. At present, it
is unknown whether there are any further such “Fermat primes” than what are obtained by n = 0, 1, 2, 3, 4.
The cases n = 5, 6, . . . , 32 are all known to give composite numbers; some evidence suggests that the
Fermat number sequence will never produce anyq
further prime beyond n = 4. The n = 2 case gives
p
p
√
√
√
√
2π
1
cos 17 = 16 (−1 + 17 + 34 − 2 17 + 2 17 + 3 17 − 170 + 38 17), with a similar formula
. This case is discussed in some detail in [11].
for sin 2π
17
12
Chapter 1. Complex Numbers
1.6 Exercises
Exercise 1.6.1. Find the magnitude and the argument of the following complex numbers:
(a) 1 − i.√
(b) 1 + i 3.
4
1+i
√
(c)
.
√ 2
(d)
3 + 4i.
(e) The√roots of z 7 + 128 = 0.
3
π
(f) 1+i
1+i . Compute from this cos 12 .
1
(g) (1−i)8 .
(h) √ 1+i√ .
1+i 3
4
1+i
(i)
.
1−i
(j) ii/2 .
Exercise 1.6.2. Express each of the following numbers in the form a + bi, where a and b
are real:
1
.
(a) 6+2i
(b)
(c)
(2+i)(3+2i)
.
1−i
2
3
1 + 1+i
.
√
(d) The roots of z 2 + 32i z − 6i = 0.
(e) The roots of z 5 = 2.
√
(f) The roots of z 2 − i 3z − 1 = 0.
4
1−i
.
(g)
1+i
(h) i1/2 .
√ 2
1−i 3
(i)
.
2+2i
Exercise 1.6.3. Show that if z0 is a root to the polynomial equation z n +a1 z n−1 +a2 z n−2 +
· · · + an = 0 with real coefficients, then so is z0 .
Exercise 1.6.4. Show that the only prime number of the form n4 + 4 with an n integer is 5,
obtained for n = ±1.
Hint: Find the roots of z 4 + 4 = 0 and thus split z 4 + 4 as a product of four linear factors.
Rearrange these four factors into two quadratic ones, with no imaginary parts remaining.
Exercise 1.6.5. Show that the product of two positive integers that each can be written as
the sum of two squares is itself the sum of two squares.
Hint: Solution options include to (i) follow the idea of Exercise 1.6.4 or to (ii) utilize the
relation |z1 | |z2 | = |z1 z2 |.
Exercise 1.6.6. Let z and w be any two complex numbers. Prove the following relations:
(a) z + z̄ = 2 Re(z).
(b) Re(z) ≤ |z|.
(c) |z − w| ≤ |z| + |w|.
1.6. Exercises
13
(d) |zw| = |z||w|.
(e) |wz + wz| ≤ 2|wz|.
(f) z − z̄ = 2 i Im(z).
Exercise 1.6.7. Draw the set of points that satisfy the following:
(a) Im(z + 2) = 3.
(d) |z − 1| + |z + 1| = 3.
(b) |z − i| < 2.
(e) |z − 1| − |z + 1| = 2.
(c) |z − i + 2| = |z + 2i − 1|.
Exercise 1.6.8. Show that the lines Re(a z +b) = 0 and Re(c z +d) = 0 are perpendicular
if and only if Re(a c) = 0.
Exercise 1.6.9. Determine which of the following sets of points form a triangle ∆αβγ with
a 90o corner:



 α=1−i
 α=1+i
 α=1−i
β = 5 + 5i ,
β = 5 + 5i ,
β = 4 − 2i .
(b)
(c)
(a)



γ = −2 + i
γ = −4 − 2i
γ = −1 − i
Exercise 1.6.10. Let z, w be complex numbers, and define E = zw + zw + 2|zw|. Show
that E is real and nonnegative.
Exercise 1.6.11.
h One can
i associate a complex number a + ib with a real antisymmetric
a −b
2 × 2 matrix b a . Show that all the complex arithmetic operations (+ − ×/) then
match the equivalent real-valued matrix operations (with “/” corresponding to a multiplication with the matrix inverse).
h
i
a + bi
c − di
Comment: Considering regular matrix algebra for 2×2 matrices −c
forms
− di a − ib
one entry point to the subject of quaternions (historically the best-known attempt to generalize complex numbers).
Exercise 1.6.12. Carry out the algebra Cardano used for finding all the roots to the (reduced) cubic equation z 3 − 15z − 4 = 0.11
Hint: The classic approach for solving a general cubic equation x3 + bx2 + cx + d = 0
proceeds as follows: Set x = y −b/3 to obtain a reduced cubic of the form y 3 +py +q = 0.
3
Next set y = z − p/(3z) to obtain z 3 − (p/3z)
0, which is a quadratic equation
q + q =√
3
3
in z , with two (out of six) solutions z1,2 = −q/2 ± R with R = (p/3)3 + (q/2)2 .
Straightforward algebra will then show that z1 z2 = −p/3, from which in turn follows that
y = z1 + z2 satisfies the reduced cubic. With one root known, it then suffices to solve a
quadratic equation for the remaining roots.
Exercise 1.6.13. Let α, β, γ be complex numbers representing the corners of a triangle.
Show that this triangle is equilateral (all sides the same length) if and only if
α2 + β 2 + γ 2 − αβ − αγ − βγ = 0.
(1.3)
11 G. Cardano (1501–1576) attributes in his 1540 publication the solution method to N.F. Tartaglia (1500–1557),
who likely was not the first one either to find it. Cardano’s student L. Ferrari (1522–1565) had by then already
found a way to reduce a general quartic equation to a cubic one. N.H. Abel showed in 1823 that no closed form
solution (using radicals, e.g., square root, cube root, etc.) is possible for the general quintic and higher degree
polynomial equations (sometimes known as the Abel–Ruffini theorem).
14
Chapter 1. Complex Numbers
Hints: The following are two solution options:
(i) If the triangle is equilateral, then
γ − β = (β − α) C,
α − γ = (γ − β) C,
β − α = (α − γ) C,
√
√
where C = − 21 + i 23 if α, β, γ are oriented clockwise, else C = − 12 − i 23 . Consider
ratios of the equations above. If (1.3) holds, reverse the argument to obtain ratios, such as
γ−β
β−α
α−γ = γ−β .
(ii) Note that rotating/scaling the corner set by b (complex, 6= 0) and translating by a
(complex) (i.e., α → a + bα, β → a + bβ, γ → a + bγ) just causes α2 + β 2 + γ 2 − αβ −
αγ − βγ to become multiplied by b2 , which does not affect the expression being zero, or
not. It thus suffices to look at a special case, such as choosing α = −1 and β = +1.
Exercise 1.6.14. Show that the three roots to the cubic equation z 3 + az 2 + bz + c = 0
form an equilateral triangle if and only if a2 = 3b.
Hint: Let the roots be α, β, γ. Expand (z − α)(z − β)(z − γ) and then use the result of
Exercise 1.6.13. A symbolic algebra package is helpful.
Exercise 1.6.15. Consider an arbitrary planar triangle ABC (cf. Figure 1.7(a)). Outside
each of its sides, equilateral triangles CBa, ACb, and BAc form. For each of these, denote
their centroids (average of its corner locations) by α, β, γ, respectively. Show that the
triangle αβγ is equilateral.
Hint: Following up on the example in Section 1.3, place a complex plane over the figure
such that A = −1, B = 1, and C = z. Similarly to previously walking equal distances
and turning 90◦ , turn 60◦ to obtain a, b, c and from these α, β, γ in terms of z. Apply the
result of Exercise 1.6.13 (again, a symbolic algebra package is helpful).
Exercise 1.6.16. Change the triangle ABC in Exercise 1.6.15 to a quadrilateral ABCD.
This time, α, β, γ, δ are the centroids of squares based on each side of the quadrilateral
(see Figure 1.7(b)). Show that the lines αγ and βδ joining these are of equal length and
orthogonal to each other.
Exercise 1.6.17. Show that, for the stereographic projection, a circle in the z-plane corresponds to a circle on the sphere.
Hint: Recall the correspondences between points z = x+iy in the regular complex z-plane
and points (X, Y, Z) on the sphere as given in Section 1.4.1. Furthermore, note that a circle on the sphere is given by the intersection of it with a plane AX + BY + CZ − D = 0.
Exercise 1.6.18. Given a point in the complex z-plane, show that the following steps produce the value w = 1/z: (i) Form 4z and project this value to the sphere, (ii) rotate the
sphere half a turn around a line through the sphere center, parallel to the X-axis, and (iii)
project the point back again from the sphere to the z-plane.
Note: In some descriptions of stereographic projection, the unit sphere is shifted down by
one, so its equator (rather than its south pole) lies in the (x, iy)-plane. In this case, the
factor 4 in step (i) should be omitted.
1.6. Exercises
15
a
C
α
β
b
A
B
γ
c
(a) Exercise 1.6.15.
γ
D
C
δ
β
A
B
α
(b) Exercise 1.6.16.
Figure 1.7. The polygons described in Exercises 1.6.15 and 1.6.16.
Exercise 1.6.19. Let z1 and z2 be two points in the complex plane, with counterparts Z1
and Z2 on the stereographic sphere. Show that their distance (in 3-space) becomes
|Z1 − Z2 | = q
4 |z1 − z2 |
q
.
2
2
4 + |z1 |
4 + |z2 |
Hint: See Figure 1.8. Apply traditional results from planar geometry (Pythagoras’ theorem,
similar triangles, etc.).
16
Chapter 1. Complex Numbers
N
2.5
Z1
2
Z2
1.5
1
S
z1
0.5
3
2
0
z2
1
0
-0.5
-2
-1
-1
0
-2
1
2
3
4
-3
Figure 1.8. The geometry and notation used in Exercise 1.6.19.
Exercise 1.6.20. Verify that angles between intersecting curves are preserved between the
plane and the stereographic sphere.
Hint: Consider an infinitesimal triangle in the plane, and apply the result of Exercise
1.6.19.
Exercise 1.6.21. Differentiating the equations for X, Y, Z in Section 1.4.1 with respect to
x and y gives, after some further algebra, that
∂X
∂X
∆x +
∆y
∂x
∂y
2
+
∂Y
∂Y
∆x +
∆y
∂x
∂y
2
+
∂Z
∂Z
∆x +
∆y
∂x
∂y
2
=
16 ( (∆x)2 + (∆y)2 )
.
(4 + x2 + y 2 )2
Deduce also from this result that the stereographic mapping is conformal (just scales and
rotates any infinitesimal triangle in the (x, y)-plane).
Chapter 2
Functions of a Complex
Variable
A real function of one variable has the form y = f (x). A complex function has similarly
the form w = f (z), where w and z now are complex quantities. Let z = x + iy and
w = u + iv, with x, y, u, v real. Thus, w = u(x, y) + iv(x, y), where u(x, y) and v(x, y)
are two real functions of two real variables.12 If we want to fully visualize w = f (z), we
can plot u(x, y) and v(x, y) separately. For example, in the case that f (z) = z 2 , we can
expand f (z) = z 2 = (x + iy)2 = (x2 − y 2 ) + i (2xy), i.e., Re(f (z)) = u(x, y) = x2 − y 2
and Im(f (z)) = v(x, y) = 2xy. Thus, to picture f (z), we can plot u(x, y) and v(x, y). It
is often helpful to also display |f (z)| and arg(f (z)), as shown in the rightmost column of
subplots in Figure 2.1.13 The top and bottom rows of subplots show similarly the functions
f (z) = z and f (z) = z 3 , respectively. Along the x-axis (marked by a thick red curve
in the left and center subplots and as a black curve in the right subplots), we recognize as
special cases the standard results when z = x is a real variable.
When displaying the magnitude of a function as a surface over the (x, y)-plane, we will
color it according to the displayed function’s phase angle (arg f (z)), choosing colors corresponding to the color wheel repeated identically in each plot of this kind (some different
color choices have been used in the literature for this type of displays; we follow here the
convention established, for example, in [39]).
We will soon come to consider much more complicated functions than pure monomials.
However, these types of graphical displays will continue to work very well for showing
very general functions over the complex plane. As a further example, Figure 2.2 illustrates
similarly the function f (z) = 12 (z + z1 ). Sometimes, it is helpful to also include a fourth
subplot, here showing what the phase portrait of the function looks like (same as the third
subplot, but viewed from straight above, making the magnitude information invisible unless
contour lines for this are included). It is critically important to know what the functions
“look like” to understand and best utilize them.
12 Like in the real-valued case, functions may be either single- or multi-valued (or undefined). For example, in
the real-valued case, f (x) = x1/2 is undefined for x < 0, single-valued for x = 0,
√and double-valued (with a
choice of ±) for x > 0. Somewhat arbitrary conventions may apply, such as writing x instead of x1/2 to imply
the positive choice.
√
13 We recall that |f (z)| =
u2 + v 2 and arg(f (z)) = arctan(v/u) if f (z) is in quadrant 1 or 4; else add or
subtract π to get a result in the range (−π, π].
17
2
y
-1
-2
-2
-1
x
0
1
-2
y
-1
0
1
2
2
1
2
3
z
z
1
0
y
-1
-2
-2
-1
x
0
0
x
-1
-2
-2
-1
y
0
1
z
1
0
y
-1
-2
-2
-1
x
0
1
2
4
2
0
-2
-4
4
2
0
-2
-4
2
2
1
2
4
2
0
-2
-4
4
2
0
-2
-4
1
0
y
-1
-2
-2
-1
x
0
1
2
0
2
2
4
-2
-1
x
0
1
-2
-2
-1
y
0
1
2
u(x, y) = Re f (z)
0
y
-1
-2
-2
-1
x
0
1
2
1
0
1
2
0
2
2
4
-1
x
0
v(x, y) = Im f (z)
1
2
0
2
2
2
2
4
4
2
0
-2
-4
1
0
y
-1
-2
-2
-1
x
0
1
2
Chapter 2. Functions of a Complex Variable
4
2
0
-2
-4
|f (z)| together with arg f (z)
18
Figure 2.1. Re f (z), Im f (z), and |f (z)| together with arg f (z) displayed for the three
functions f (z) = z, z 2 , z 3 . In the first two columns of plots, some positive contour lines
√ are shown
in blue and some negative ones in green. In the last column, the magnitude |f (z)| = u2 + v 2 is
displayed vertically, and the phase angle is displayed according to the “color wheel” at the bottom
left of each of these subplots (showing how colors are associated with phase angles).
2.1. Derivative
19
5
5
0
0
-5
2
-5
2
2
1
1
0
2
1
1
0
0
-1
y
0
-1
-1
-2
-2
x
y
-1
-2
-2
x
(a) Re f (z)
(b) Im f (z)
(c) |f (z)| (elevation) and
arg f (z) (color)
(d) As (c), but seen from above
(with contours for |f (z)|)
Figure 2.2. Graphical representations of f (z) = 21 (z + z1 ). The function values along the
real axis are marked by thick red or black lines. The colors in (c) and (d) are related to the function’s
phase angle (arg) according to the small color disk. The phase portrait (subplot (d)) will be omitted
in most further cases (unless it reveals some features not apparent from the first three subplots). In
this present case, it also includes contour lines for the values 1,2,3,4, . . . of the magnitude.
2.1 Derivative
(x)
, where we require the limit to
In the real-valued case, f 0 (x) = lim∆x→0 f (x+∆x)−f
∆x
be the same from the right and the left. In the case of a complex function, f 0 (z) =
(z)
lim∆z→0 f (z+∆z)−f
, where the limit now should be the same when ∆z → 0 from
∆z
any direction of the complex plane. In particular, the limits in two main directions must be
the same: the horizontal direction (when ∆z is purely real) and the vertical direction (when
∆z is purely imaginary). This suffices for obtaining the Cauchy–Riemann equations.
Definition 2.1. The function f is analytic at z0 if f (z) is differentiable in some neighborhood of z0 (no matter how small).14 The function f is analytic in a region if it is analytic at
all points in that region. If we describe a function as analytic, without specifying any point
14 Alternatively
expressed as an open region including z0 .
20
Chapter 2. Functions of a Complex Variable
or region, that means there is some region within which it is analytic.15 The function f is
holomorphic if it is analytic. The terms are synonyms. An analytic function is entire if its
region of analyticity includes all points in C, the finite complex plane, excluding infinity.
2.1.1 Cauchy–Riemann equations
We defined f 0 (z) = lim∆z→0
do the following:
f (z+∆z)−f (z)
,
∆z
where f (z) = u(x, y) + iv(x, y). Next, we
1. Choose ∆z = ∆x real. Then
f 0 (z) = lim
∆x→0
u(x + ∆x, y) − u(x, y)
∆x
+i
v(x + ∆x, y) − v(x, y)
∆x
2. Choose ∆z = i∆y purely imaginary. Then f 0 (z) = lim∆y→0
∂v
i v(x,y+∆y)−v(x,y)
= −i ∂u
i∆y
∂y + ∂y .
=
∂v
∂u
+i .
∂x
∂x
u(x,y+∆y)−u(x,y) +
i∆y
For the two results to be the same, u(x, y) and v(x, y) must therefore be related by the
Cauchy–Riemann (C-R) equations:







∂u
∂v
=
,
∂x
∂y
∂v
∂u
=− .
∂x
∂y
(2.1)
Theorem 2.2. If the function f (z) is differentiable, the C-R equations hold.16
We have already shown this result just above. For the reverse direction, some minor
extra conditions are needed. Theorem 2.40 shows that the C-R equations together with
ux , uy , vx , vy being continuous at a point suffices for f 0 (z) to exist at that point. If the C-R
equations are valid in a full neighborhood of a point, the extra requirements get reduced to
u and v, themselves being continuous (i.e., we no longer need to verify continuity also of
the partial derivatives).17
There are many consequences to the C-R equations:
1. A purely real function f (z) = u(x, y) is not analytic unless identically constant.
2. If f (z) is differentiable once, it is differentiable an infinite amount of times (Theorem
4.11). There is no counterpart of this for real-valued functions.
3. The functions u(x, y) and v(x, y) each satisfy Laplace’s equation
uxx + uyy = (ux )x + (uy )y = (vy )x + (−vx )y = 0
and, similarly,
vxx + vyy = 0.
15 For example, we describe sin z, log z, and 1 all as analytic functions, although the last two have a singular
1−z
(exceptional) point at z = 0 and z = 1, respectively.
16 When there is little danger of misunderstandings, we leave out certain formalities, here such as z belonging
to an open region (meaning a region that does not include its boundary points).
17 The continuity requirements on u(x, y) and v(x, y) are needed to exclude cases such as f (z) = e−1/z 4 ,
for which the C-R equations “happen” to hold also at the point of singularity z = 0. A more detailed discussion
can be found in [23].
2.1. Derivative
21
Thus, neither u nor v can have a local minimum or a local maximum. Typically, at a
local maximum, both uxx < 0 and uyy < 0, which is incompatible with uxx +uyy =
0. A strict proof follows from Theorem 4.16, which states that the value at any point
of a Laplace equation solution is the average of the values around the periphery of
any circle centered at the point.
4. Given u, we can compute v up to a constant, and vice versa. Each, satisfying
Laplace’s equation, is called a harmonic function; the two are called the harmonic
conjugates of each other.
5. The gradient vectors for u(x, y) and v(x, y) satisfy ∇u · ∇v = (ux , uy ) · (vx , vy ) =
ux vx + uy vy = −vy uy + uy vy = 0; i.e., they are orthogonal to each other (or
we are at a location with f 0 (z) = 0). The same holds for level curves to the two
surfaces u(x, y) and v(x, y), since level curves are orthogonal to gradient vectors;
see Figure 2.3.
1
0
-1
-1
0
1
Figure 2.3. This plot shows for f (z) = z 2 − 3 the level curves of u(x, y) as solid curves
and of v(x, y) as dashed curves. They are orthogonal except at z = 0 where f 0 (0) = 0.
2.1.2 Taylor series verification of analyticity
In order to test whether a function is analytic, we can thus check whether u(x, y) and
v(x, y) are continuous and satisfy the C-R equations in some open region. Another procedure will also turn out to be very useful. If a function has a Taylor expansion, it is analytic
at least within its radius of convergence (radius of the largest disk in the complex plane surrounding the expansion point within which the series converges, to be discussed in much
more detail shortly). It is sufficient to find a Taylor expansion in x and then replace x by z.
For example, we have the following:
• f (z) = log(1 + z) is analytic (at least inside the unit circle |z| = 1). One can easily
d
1
find a series expansion for log(1 + x), using the fact that dx
log(1 + x) = 1+x
and
22
Chapter 2. Functions of a Complex Variable
the geometric series identity
x−
2
x
2
+
3
x
3
1
1+x
= 1 − x + x2 − x3 + − · · · . Thus, log(1 + x) =
− +···.
• f (z) = z is analytic. It is its own Taylor series.
• f (z) = z̄ = x − iy does not have a Taylor series in z. Checking with the C-R
∂v
equations, we see that ∂u
∂x = 1 and ∂y = −1, so (2.1) does not hold. Thus, this
function is nowhere analytic.
• If g(z) is analytic, then f (z) = g(z̄) is analytic. Proof: Let z = x + iy and g(z) =
u(x, y) + iv(x, y). Then, g(z̄) = u(x, −y) − iv(x, −y) = a(x, y) + ib(x, y). So
ax = ux , ay = −uy , bx = −vx , by = vy , which implies that, since g is analytic,
ax = ux = vy = by and that ay = −uy = vx = −bx . Since the C-R equations are
satisfied, f (z) = g(z̄) is analytic.
2.2 Some elementary functions generalized to complex
argument by means of their Taylor expansion
P∞
Consider a function f (x) that for x real has a Taylor series expansion f (x) = k=0 ak xk
(where we, for simplicity,
have expanded about x = 0).18 We can then substitute z for x
P∞
k
and let f (z) = k=0 ak z be the generalization of the function to a complex argument.
If the original series converged for −R < x < R, we will show (Theorem 4.21) that the
complex version will then converge for all z satisfying |z| < R. Typically, the function
f (z), even if known only through its Taylor coefficients, is completely defined outside this
disk as well. How to then find its values for |z| ≥ R is the topic of analytic continuation,
which will be addressed in Chapter 3.
Example 2.3. Extend
f (x) = ex = 1 + x +
x3
x2
+
+ ···
2!
3!
to a complex argument.
Since the functional definition is in the form of a Taylor series, we just replace x with
z and thus obtain
z3
z2
+
+ ··· .
f (z) = ez = 1 + z +
2!
3!
It is now easy to verify that all the standard relations for the exponential function hold also
in the complex case. For example, when x1 , x2 are real, it holds that ex1 · ex2 = ex1 +x2 .
This implies that if we Taylor expand in its two variables (using standard procedures from
real-valued calculus) the function
f (x1 , x2 ) = ex1 · ex2 − ex1 +x2
x21
x31
x22
x32
= 1 + x1 +
+
+ ···
1 + x2 +
+
+ ···
2!
3!
2!
3!
(x1 + x2 )2
(x1 + x2 )3
− 1 + (x1 + x2 ) +
+
+ ···
2!
3!
= a0,0 + (a1,0 x1 + a0,1 x2 ) + (a2,0 x21 + a1,1 x1 x2 + a0,2 x22 ) + · · · ,
18 A
Taylor series expanded about z = 0 is also known as a Maclaurin expansion.
2.2. Elementary functions generalized to complex argument by Taylor expansions 23
then every Taylor coefficient ai,j must be zero. These Taylor coefficients do not at all
depend on what values we substitute for x1 and x2 , so the result will be zero also if we
substitute in complex numbers z1 and z2 . Consequently, the relation ez1 · ez2 = ez1 +z2
must hold also for complex arguments z1 and z2 .
In the same way, we can conclude that any functional relation that holds when x is real
(and the function(s) involved are Taylor expandable) will again hold when x is replaced by
z complex.
2.2.1 Relations between exponential and trigonometric functions
If z = x + iy, we get from the result above that
ez = ex+iy = ex · eiy ,
where furthermore
y2
y3
y4
y5
eiy = 1 + iy −
−i +
+ i ···
2!
3! 4! 5!
y2
y4
y3
y5
= 1−
+
··· + i y −
+
· · · = cos y + i sin y.
2!
4!
3!
5!
(2.2)
The generalization of the exponential function to complex arguments can therefore also be
written as
ez = ex (cos y + i sin y).
(2.3)
An interesting special case is Euler’s identity, eπi + 1 = 0, connecting the five fundamental
numbers 0, 1, e, π, and i.
Figure 2.4 shows what the exponential function looks like in the complex plane (numerous similar illustrations for other standard analytic functions will soon be central to
our description of them). This figure was produced by the MATLAB statements shown in
Section 2.8.4. We can recognize the real-valued function y = ex along the real axis in the
upper plot, along the real axis, drawn in red. Using Mathematica, similar plots are obtained
by the code shown in Section 2.8.4.
We saw for the monomials f (z) = z, z 2 , z 3 in Figure 2.1 and now again for the
exponential function that a Taylor expandable function, previously defined only along the
real axis, will have a unique extension to the complex plane. A strong statement about this
will soon be given as Theorem 2.12.
2.2.2 Trigonometric functions represented in terms of the
exponential function
Since the relation eiy = cos y + i sin y holds for y real, it must according to our discussion
above also hold when y is complex; i.e., we obtain what are known as Euler’s relations19
eiz = cos z + i sin z,
(2.4)
e−iz = cos z − i sin z.
(2.5)
2
19 Euler observed in a letter to Johann Bernoulli, dated October 18, 1740, that the unique solution to d y + y =
dx2
0, y(0) = 2, y 0 (0) = 0 can be written either as y(x) = 2 cos x or as y(x) = eix + e−ix , also noting that
similarly 2i sin x = eix − e−ix . His first publication on this (in 1748) includes (2.2). However, Roger Cotes
wrote (in Phil. Trans. Royal Soc. 29, 1714, 5–45) what, in modern notation, amounts to i φ = log(cos φ +
i sin φ); see Exercise 2.9.18.
24
Chapter 2. Functions of a Complex Variable
60
40
20
0
-20
-40
-60
10
5
4
2
0
0
-5
y
-2
-10
x
-4
(a) Re ez .
60
40
20
0
-20
-40
-60
10
5
4
2
0
0
-5
y
-2
-10
-4
x
(b) Im ez .
(c) |ez | and arg ez .
Figure 2.4. Graphical representations of f (z) = ez .
2.2. Elementary functions generalized to complex argument by Taylor expansions 25
Adding and subtracting these equations give
cos z =
eiz + e−iz
,
2
(2.6)
eiz − e−iz
.
(2.7)
2i
Graphically, we can see what this becomes for cos z (Figure 2.5) and for sin z (Figure 2.6). The two functions differ only by a phase shift (translation) in the real direction
(cos z = sin(z + π2 )), and they grow exponentially fast in the imaginary direction (away
from the real axis). These relations (2.6) and (2.7) motivate how the functions cosh x and
sinh x are defined for real x and thus extended to complex z as
sin z =
ez + e−z
= cos(iz),
2
ez − e−z
1
sinh z =
= sin(iz).
2
i
Many trigonometric identities (at first sight not having anything to do with complex
numbers) can be derived very easily with the use of Euler’s formulas.
cosh z =
Example 2.4. Derive the addition theorems for sin(α + β) and cos(α + β).
We have (from (2.4))
ei(α+β) = cos(α + β) + i sin(α + β)
and also
ei(α+β) = eiα eiβ
= (cos α + i sin α)(cos β + i sin β)
= (cos α cos β − sin α sin β) + i(cos α sin β + sin α cos β).
Equating the real and imaginary parts now give the standard addition theorems.
PN
Example 2.5. Evaluate k=−N cos kx.
N
X
N
X
cos kx =
k=−N
eikx = e−N ix 1 + eix + e2ix + · · · + e2N ix
k=−N
|
Imaginary part vanishes, since
= e−N ix
|
PN
1 − e(2N +1)ix
1 − eix
k=−N
{z
}
sin kx = 0; now a finite geometric progression.
=
{z
e−(N +1/2)ix − e(N +1/2)ix
e−ix/2 − eix/2
}
Routine way to simplify denominator of this type; here multiply numerator and denominator by e−ix/2
=
−2i sin N + 12 x
sin N + 21 x
=
−2i sin 12 x
sin 12 x
|
{z
}
This is a purely real-valued result (when x is real), as to be expected
.
26
Chapter 2. Functions of a Complex Variable
30
20
10
0
-10
-20
-30
4
2
10
5
0
0
-2
y
-5
-4
x
-10
(a) Re cos z.
30
20
10
0
-10
-20
-30
4
2
10
5
0
0
-2
y
-5
-4
-10
x
(b) Im cos z.
(c) | cos z| and arg cos z.
Figure 2.5. Graphical representations of f (z) = cos z.
2.2. Elementary functions generalized to complex argument by Taylor expansions 27
30
20
10
0
-10
-20
-30
4
2
10
5
0
0
-2
y
-5
-4
x
-10
(a) Re sin z.
30
20
10
0
-10
-20
-30
4
2
10
5
0
0
-2
y
-5
-4
-10
x
(b) Im sin z.
(c) | sin z| and arg sin z.
Figure 2.6. Graphical representations of f (z) = sin z.
28
Chapter 2. Functions of a Complex Variable
While this result can also be proven by induction over N , the present derivation is more
practical, since it does not require the answer to be known in advance.
n
We also obtain de Moivre’s formula (1.1) from the identity eix = einx .
2.2.3 Logarithm function
From the (real-valued) Taylor expansion for log(1 + x) it follows that we can define the
extension to complex z (with |z| < 1) as
1
1
1
log(1 + z) = z − z 2 + z 3 − z 4 ± · · · .
2
3
4
(2.8)
By the principle that functional relations that hold in real cases must carry over to complex
variables, it will hold that log(z1 z2 ) = log(z1 ) + log(z2 ). Writing z = reiθ gives
log z = log r + iθ = log |z| + i arg z,
(2.9)
a counterpart of (2.3) in terms of extending an elementary function from the real axis to the
complex plane (and here also outside the radius of convergence R = 1 for (2.8)). Figure 2.7
displays the log z function over the domain −4 ≤ Re z ≤ 4, −4 ≤ Im z ≤ 4. Something
we have not seen before is here visible along the negative real axis (leading us in Section
2.5 to introduce the concepts of branch cuts and Riemann surfaces).20
In the displays of different functions f (z), we typically show surface plots for Re f (z),
Im f (z), and |f (z)|. For the first two of these, we noted in Section 2.1.1 that they can
never have any local (finite) maxima or minima, since they obey Laplace’s equation. From
(2.9) follows a similar result for |f (z)|. This quantity has a local minimum (with value
zero) wherever f (z) is zero, but cannot otherwise have any local finite extreme points. If
it did, so would log |f (z)| = Re log f (z), since log is a monotonic function (for positive
arguments), but this is not allowed for the real part of an analytic function (this is later
expressed again in Theorem 4.18).
2.2.4 Inverse hyperbolic and trigonometric functions
Solving for z in the RHSs of (2.6) and (2.7) (and their hyperbolic counterparts) gives expressions for their inverse functions, for example
1+z
1
,
(2.10)
arctanh z = log
2
1−z
arcsin z = −i log(iz + (1 − z 2 )1/2 ),
(2.11)
arccos z = −i log(z + (z 2 − 1)1/2 ),
i
1 − iz
arctan z = log
.
2
1 + iz
(2.12)
(2.13)
Figures 2.8–2.10 illustrate the first three of these.
20 We will there also note that the branch cut (line of discontinuity) can be placed in different ways; the negative
real axis is merely an arbitrary choice that often is convenient.
2.2. Elementary functions generalized to complex argument by Taylor expansions 29
(a) Re log z.
(b) Im log z.
(c) | log z| and arg log z.
(d) arg log z and contour lines
for | log z|.
Figure 2.7. Graphical representations of f (z) = log z.
Example 2.6. Derive (2.10), arctanh z =
for z real.
1
2
log
1+z
1−z
, and compare it to its counterpart
We first note that w = arctanh z corresponds to z = tanh w =
1+z
1+z
from which follows e2w = 1−z
and thus (2.10): w = 12 log 1−z
.
x
−x
ew −e−w
ew +e−w
=
e2w −1
e2w +1 ,
For x real, tanh x = eex −e
+e−x asymptotes to ±1 when x → ±∞, and we obtain the
inverse function by reflecting its graph across the line y = x, cf; Figures 2.11(a)–(b).
The curve in Figure 2.11(b) matches the central (red) section −1 < z < 1 in Figure
2.8(a). Nothing else in these Figures 2.8(a)–(b) is apparent from the case of a real argument. Regarding the imaginary part in Figure 2.8(b), we see two shifted copies of the
logarithm function from Figure 2.7(b), as to be expected from writing the arctanh formula
as arctanh z = 12 log(1 + z) − 12 log(1 − z).
30
Chapter 2. Functions of a Complex Variable
(a) Re arctanh z.
(b) Im arctanh z.
(c) |arctanh z| and arg arctanh z.
Figure 2.8. Graphical representations of f (z) = arctanh z. Notice the lines of discontinuity, like for the logarithm function highly visible on the imaginary part of the function, but not
present for the real part.
2.2. Elementary functions generalized to complex argument by Taylor expansions 31
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
5
10
5
0
0
-5
y
-5
x
-10
(a) Re arcsin z.
4
3
2
1
0
-1
-2
-3
-4
5
10
5
0
0
-5
y
-5
-10
x
(b) Im arcsin z.
(c) | arcsin z| and arg arcsin z.
Figure 2.9. Graphical representations of f (z) = arcsin z.
32
Chapter 2. Functions of a Complex Variable
3
2.5
2
1.5
1
5
0.5
0
0
-10
-8
-6
-4
-2
0
2
4
6
y
-5
8
10
x
(a) Re arccos z.
4
3
2
1
0
-1
-2
-3
-4
-5
0
y
5
10
8
6
2
4
0
-2
-4
-6
-8
-10
x
(b) Im arccos z.
(c) | arccos z| and arg arccos z.
Figure 2.10. Graphical representations of f (z) = arccos z.
2.3. Additional observations on Taylor expansions of analytic functions
y
33
y
1.5
1.5
1
1
0.5
0.5
x
-1.5
-1
-0.5
0.5
1
x
1.5
-1.5
-1
-0.5
0.5
-0.5
-0.5
-1
-1
-1.5
-1.5
(a) The function y = tanh x.
1
1.5
(b) The inverse function y = arctanh x.
Figure 2.11. Graphical representation of the functions y = tanh x and y = arctanh x.
The two curves are identical apart from a reflection in the line y = x (exchanging the roles of x
and y).
2.3 Additional observations on Taylor expansions of
analytic functions
We introduced above the standard analytic functions through their Taylor expansions. Before using them much further, below are some observations of their domains of convergence.
Theorem 2.7. The following are two key results on the region in which Taylor expansions
converge (assuming for notational simplicity expansions about z0 = 0; else replace z by
z − z0 below):
1. Convergence occurs for all z within a circular domain |z| < R with divergence
for all z with |z| > R, where R is some value from 0 to ∞ (including 0 and ∞).
Convergence on the boundary |z| = R can vary between functions and boundary
locations.
2. The radius R is the largest possible, such that f (z) is differentiable (i.e., analytic) at
every point with |z| < R.
These statements will be proven later in Section 4.2.8. A collection of various convergence tests can be found in Section 2.8.3. We will use these tests in the following examples.
Example 2.8. Determine R for f (z) = ez = 1 +
1
1!
z+
1
2!
z2 +
1
3!
z3 + · · · .
Since f 0 (z) = ez , the derivative exists and is finite for all finite z. By part 2 above, the
radius of convergence is R = ∞ (meaning that the series converges for all |z| < R = ∞).
The result also follows immediately also from the root and ratio tests (Theorems 2.35 and
2.36).
Example 2.9. Determine R for f (z) = log(1 + z) = z − 21 z 2 + 13 z 3 − + · · · .
34
Chapter 2. Functions of a Complex Variable
1
Since f 0 (z) = 1+z
, the only singularity is located at z = −1, implying that the radius
of convergence is R = 1. Alternatively, we can note that R = 1/ lim|1/n|1/n = 1 or that
n|
R = lim n→∞ |a|an+1
| = 1.
Example 2.10. Determine R for
1
1+z 2
= 1 − z2 + z4 − z6 + − · · · .
The radius of convergence is again R = 1, since the function is differentiable everyn|
where apart from at z = ±i. Also R = 1/ lim|an |1/n = 1, but lim n→∞ |a|an+1
| does not
exist, so the root test is applicable but the ratio test is not.21
Example 2.11. Determine R for
´∞
0
e−t
1+zt dt
= 1 − 1!z + 2!z 2 − 3!z 3 + 4!z 4 − + · · · .
This function will be considered in more detail in Example 3.18. The radius of convern|
gence is R = 0, easiest seen from R = lim n→∞ |a|an+1
| = 0, but it follows also from the
other tests.
P∞
n
It is easy to verify that if fP
(z) =
the disk of radius R, then
n=0 an z is valid
Pin
∞
∞
0
n−1
00
we can differentiate (f (z) = n=0 nan z
, f (z) = n=0 n(n − 1)an z n−2 , . . .) or
integrate however many times we want and obtain expansions that will still converge in the
disk of radius exactly R.
Theorem 2.12. If two analytic functions coincide on any curve segment, no matter how
small (or any point set with a limit point), then the two functions are identically the same.22
Proof. P
Let the limit point be z = 0 and the difference between the two functions be
∞
f (z) = n=0 an z n . From f (z) being zero at all the points it follows (by continuity) that
P∞
= n=1 an z n−1 , also zero at all the points, implying that
a0 = 0. Consider next f (z)
z
a1 = 0, etc. All the Taylor coefficients, and thus also f (z), must vanish.
This last result raises fundamental questions about different possible representations of
1
a single function. As illustrated in Figure 2.12, f (z) = 1−z
is defined everywhere except
at z = 1, where the function has a pole. Next, define g(z) = 1 + z + z 2 + z 3 + · · · .
This function g(z) converges only in the disk |z| < 1 and diverges for |z| > 1. Do
these functions represent the same function everywhere? This will be the topic of analytic
continuation (Chapter 3). Although the function itself exists everywhere, its particular
representation as a Taylor series fails outside |z| = 1.
Given, at a point a, values for f (a), f 0 (a), f 00 (a), . . . , we can write down the Taylor
expansion at that point
f (z) = f (a) + (z − a)f 0 (a) +
(z − a)2 00
f (a) + · · · .
2!
(2.14)
This relation is theoretically important, but it is usually impractical to differentiate a nontrivial function many times. We will next describe two methods (beyond (2.14)) to obtain
21 For
the definition of lim, see Theorem 2.35.
result extends as far away as the functions remain analytic; “continuation” of analytic functions is discussed in Chapter 3.
22 The
2.3. Additional observations on Taylor expansions of analytic functions
Im(z)
35
Im(z)
1
1
Re(z)
Re(z)
(a) f (z) is defined for all z except z = 1.
(b) g(z) converges only in |z| < 1.
1
Figure 2.12. Although the function f (z) = 1−z
exists everywhere (except at z = 1), its
particular representation g(z) as a Taylor series fails on and outside |z| = 1
Taylor expansions of given functions. To go in the reverse direction and obtain a function
from its Taylor expansion, we will need analytic continuation.
2.3.1 Get the expansion from an already known expansion
Example 2.13. Derive Taylor expansions for log(1 + z) and
1
series 1+z
= 1 − z + z2 − z3 + z4 − + · · · .
1
1+z 2
based on the geometric
Integrating the geometric series gives log(1 + z) = z − 21 z 2 + 13 z 3 − 14 z 4 + 15 z 5 −
+ · · · + C. To find C, let z = 0, leading to C = 0. Replacing z by z 2 in the geometric
1
2
4
6
8
series gives 1+z
2 = 1 − z + z − z + z − +··· .
Example 2.14. Determine the Taylor expansion for f (z) = arcsin z.
We first note that f 0 (z) =
√ 1
.
1−z 2
The binomial theorem, generalized to noninteger
s(s−1)(s−2) 3
2
powers, states that (1 + z) = 1 + sz + s(s−1)
z + · · · . Using this
2! z +
3!
1 2
2
0
z4 −
with s = −1/2 and swapping z for −z gives f (z) = 1 + 2 z + (−1/2)(−3/2)
2!
(−1/2)(−3/2)(−5/2) 6
1 3
3 5
5 7
z + · · · and therefore f (z) = arcsin z = z + 6 z + 40 z + 112 z + · · ·
3!
(with the integration constant vanishing, since arcsin 0 = 0 or, equivalently, since arcsin z
is an odd function of z).
s
2.3.2 Method of undetermined coefficients
This method applies, for example, when one wants to Taylor expand a rational function.
Other cases include finding expansions to solutions of ODEs. See also Section 2.8.1 for
inverting a Taylor expansion.
Example 2.15. Find the Taylor expansion of f (z) =
3−4z
1+2z .
Let the expansion be
3 − 4z
= a0 + a1 z + a2 z 2 + a3 z 3 + · · · .
1 + 2z
36
Chapter 2. Functions of a Complex Variable
Multiplying up the denominator gives
3 − 4z = a0 + (a1 + 2a0 ) z + (a2 + 2a1 ) z 2 + (a3 + 2a2 ) z 3 + · · · .
Since this is an identity, the coefficients on the two sides must match for all powers of z.
We thus obtain
3 = a0 ,
−4 = a1 + 2a0 ,
0 = a2 + 2a1 ,
(2.15)
0 = a3 + 2a2 ,
..
.
from which all the coefficients follow recursively: a0 = 3, a1 = −10, an+2 = −2an+1 ,
n = 0, 1, 2, . . . or evaluated: f (z) = 3 − 10 z + 20 z 2 − 40 z 3 + 80 z 4 + − · · · .
5
An alternative approach is to write f (z) = 3−4z
1+2z = −2 + 1+2z and then note that the
second term is the sum of an infinite geometric progression.
With the power of modern computers, many results relating to analytic functions are
nowadays first discovered numerically, and only later (if at all) verified analytically. This
brings up the intriguing issue of the amount of numerical evidence that is needed for it
to become plausible that an observed result will always remain true. The book The Computer as a Crucible, subtitled An Introduction to Experimental Mathematics [6], contains
a fascinating discussion of this subject. The following example, again concerning Taylor
expanding a rational function, can be found in it. The example is highly exceptional, but it
can still serve as a useful warning to be careful!
Example 2.16. Find some leading terms in the Taylor expansion of
8 + 7z − 7z 2 − 7z 3
.
1 − 6z − 7z 2 + 5z 3 + 6z 4
Does the recursion a0 = 8, a1 = 55, an+2 = 1 + a2n+1 /an , n = 0, 1, 2, . . . (with the
brackets b·c meaning rounding down to nearest integer), produce the exact result for all the
Taylor coefficients of f (z)?
f (z) =
For a problem involving quite a lot of algebra, computational tools greatly simplify
tedious labor (as well as greatly reduce the risk of algebraic errors). For Mathematica, note
in this context its command “Series” and for MATLAB the functions “conv” and “deconv.”
The sequence of Taylor coefficients (all integers) begins
a0 = 8,
a1 = 55,
a2 = 379,
a3 = 2612,
a4 = 18002,
a5 = 124071,
a6 = 855106,
a7 = 5893451,
a8 = 40618081,
a9 = 279942687,
..
.
The singularity of f (z) located nearest to the origin occurs around z = 0.145094, implying
2.3. Additional observations on Taylor expansions of analytic functions
37
that the coefficients will forever grow each step roughly by a factor of 7. It has been found
that the proposed recursion formula for the coefficients an is exact all the way up through
n = 11, 055, but it fails for n = 11, 056.
g(z)
z
√
Example 2.17. Determine the Taylor expansion for f (z) = arcsin
= h(z)
based on the
1−z 2
3 5
5 7
known expansions g(z) = z + 61 z 3 + 40
z + 112
z + · · · and h(z) = 1 − 12 z 2 − 81 z 4 −
5 8
1 6
16 z − 128 z − · · · .
This problem generalizes the rational function case above in that the degree is infinite
in both the numerator and denominator. That makes, however, no difference in how one
g(z)
proceeds. Noting that the ratio will be an odd function, we let the expansion for h(z)
be
g(z)
h(z)
= a1 z + a3 z 3 + a5 z 5 + · · · . Multiplying up the denominator and equating coefficients gives again a set of recursive equations that can be solved explicitly:
1
=
1
6
= − 12 a1 + a3 ,
3
40
5
112
and thus f (z) =
arcsin
z
√
1−z 2
a1 ,
= − 18 a1 − 12 a3 + a5 ,
=
..
.
1
− 16
a1
= z + 23 z 3 +
8 5
15 z
−
1
8 a3
+
16 7
35 z
−
1
2 a5
(2.16)
+ a7 ,
+ ··· .
Another common case of using the method of undetermined coefficients arises if the
function f (z) satisfies some simple ODE (ordinary differential equation).
Example 2.18. Determine the Taylor expansion for f (z) =
it satisfies
arcsin
z
√
1−z 2
based on an ODE that
Once we differentiate and simplify, we notice that f (z) satisfies the linear ODE (1 −
z 2 )f 0 (z) = 1 + z f (z) with f (0) = 0. Assume that f (z) = a0 + a1 z + a2 z 2 + · · · ,
implying f 0 (z) = a1 + 2a2 z + 3a3 z 2 + · · · . After substituting these expansions into the
ODE, we can equate the coefficients and get any number of coefficients recursively. Noting
z
√
that f (z) = arcsin
is an odd function of z again simplifies the algebra (since that implies
1−z 2
all even terms in the Taylor expansion must vanish). In either case, starting with a0 = 0
z
8 5
16 7
√
= z + 32 z 3 + 15
z + 35
z + ··· .
gives again f (z) = arcsin
1−z 2
Especially for nonlinear ODEs of the general form f 0 (z) = F (z, f (z)), where F is analytic in both its arguments, even the method of undetermined coefficients often becomes
cumbersome if one needs manyP
terms. Assume we have already obtained the beginning of
n
the Taylor expansion, fn (z) = k=0 ck z k (at first just knowing c0 from the initial condition), substituting this truncated series into the ODE and then re-expanding and integrating
once gives one more term. This is relatively straightforward to carry out (or implement
computationally), and can be repeated indefinitely to obtain any number of terms.
Example 2.19. Find the leading coefficients c0 , c1 , . . . , c5 in the Taylor expansion of the
(z)
solution to dfdz
= 1 + sin f (z), f (0) = 0.
38
Chapter 2. Functions of a Complex Variable
Following the recipe just above, the iterations proceed as follows (going down row-byrow):
−
−
−
f0 (z) = 0
df1 (z)
dz
∼ 1 + sin f0 (z) ∼ 1
f1 (z) = z
df2 (z)
dz
∼ 1 + sin f1 (z) ∼ 1 + z
f2 (z) = z + 21 z 2
df3 (z)
dz
∼ 1 + sin f2 (z) ∼ 1 + z + 21 z 2
f3 (z) = z + 12 z 2 + 61 z 3
df4 (z)
dz
∼ 1 + sin f3 (z) ∼ 1 + z +
1 2
z
2
∼ 1 + sin f4 (z) ∼ 1 + z +
1 2
z
2
df5 (z)
dz
+ 0z
3
3
+ 0z −
..
.
1 4
z
4
f4 (z) = z +
1 2
z
2
f5 (z) = z +
..
.
1 2
z
2
−
+
1 3
z
6
4
+ 0z
+
1 3
z
6
+ 0z 4 −
1 5
z
20
2.4 Singularities
Functions f (x), x ∈ R, with which we are familiar can usually be extended to become
analytic functions w = f (z). However, no matter how smooth these are along the real axis,
they can become infinite (or nondifferentiable) at certain points (or regions) in the complex
plane. These are called singularities. We have encountered some examples already. For
1
instance, f (x) = 1+x
2 is everywhere infinitely differentiable along the real axis, but, when
1
generalized to a complex argument, f (z) = 1+z
2 (shown in Figure 2.13) has singularities
(because of division by zero) at z = ±i.
There is a difficulty in introducing the concept of singularities because a good understanding of the different forms they can take requires the use of Laurent expansions, which
in turn are best derived by means of integration along contours in the complex plane. However, singularities play a key role in developing integration techniques. Hence, we follow
here a two-step approach for discussing singularities. In this present Section 2.4, we limit
ourselves to a quite heuristic (nonrigorous) overview, sufficient for developing complex integration in Chapter 4. Once we have Laurent expansions available, we revisit singularities
for a more in-depth treatment in Section 4.3.2.
We next want to describe the different ways in which an analytic function can be singular. There turns out to be six types of singularities. The first four refer to isolated
singularities, meaning that the function is analytic in a full neighborhood of the singular
point.
2.4.1 Removable singularities
This type of “singularity” is a problem of a particular representation of the function, and
not of the function itself. It should therefore not be counted as a type of singularity, but
it nevertheless traditionally is, thus being a misnomer. In contrast to genuine singularities,
removable ones do not limit the radius of convergence for a Taylor series.
Example 2.20. The function f (z) = sinz z has a singularity at z = 0, at which point
1 3
1 5
it is undefined. However, sin z = z − 3!
z + 5!
z − + · · · . The expansion sinz z =
1 2
1 4
1 − 3! z + 5! z − + · · · converges for all z and is thus analytic for all z. It is an entire
function that uniquely provides the missing value of f (0) = 1. Wherever we center a
Taylor expansion of this function, it will have R = ∞ (i.e., a removable singularity does
2.4. Singularities
39
3
2
1
0
-1
-2
-3
2
1
2
1
0
0
-1
y
-1
-2
x
-2
(a) Re
1
1+z 2 .
3
2
1
0
-1
-2
-3
2
1
2
1
0
0
-1
y
-1
-2
x
-2
(b) Im
1
1+z 2 .
1
(c) | 1+z
2 | and arg
1
1+z 2 .
1
Figure 2.13. Graphical representations of f (z) = 1+z
2 . The Runge function f (z) =
has two first order pole singularities in the complex plane but is analytic everywhere on the
real axis.
1
1+z 2
40
Chapter 2. Functions of a Complex Variable
not influence a Taylor series’ radius of convergence; it is not a feature of a function but
only of a particular way to represent it).
A more formal characterization of the next two types of singularities has to await our
description of Laurent expansions in Section 4.3. We can, however, already now describe
their key features and note how they differ both in character and when displayed visually.
2.4.2 Poles
These singularities typically arise through division by zero. In the neighborhood of such
a point, |f (z)| gets uniformly large. One can recognize a pole by its signature “chimney
look” in the graph of |f (z)|, together with a simple phase angle (color) pattern on the
chimney (e.g., Figures 2.2 and 2.13, to be compared with the infinity of color strips we
soon will see for essential singularities in Figure 2.14). A function that has poles as its only
singularities in the finite complex plane is called a meromorphic function.
Example 2.21. The function f (z) = z1 has a first order pole singularity23 at z = 0, f (z) =
1
1
z 2 has a second order pole singularity at z = 0, and f (z) = sin z has first order pole
singularities at z = πk, k = 0, ±1, ±2, . . . , in all cases caused by divisions by zero.
Near a pole singularity, the phase angle coloring shows great simplicity and regularity.
Returning to Figure 2.2 with f (z) = z + z1 , we see a first order pole (from the term z1 ) at
z = 0 and first order zeros at z = ±i. In the phase portrait (subplot (d)), we note that the
color pattern near the pole is identical to that surrounding the zeros, apart from a reversal
in the order of the colors. For zeros and poles of higher orders (such as z 2 , 1/z 2 , etc.), the
colors (phase angles) would be traversed the same number of times as the order.
2.4.3 Essential singularities
The structure of a function near an essential singularity is far more complicated. In fact (as
will be shown later in Theorem 4.30), the function comes arbitrarily close to every single
finite value infinitely many times in any neighborhood of the singular point (no matter how
small).
Example 2.22. The function f (z) = e1/z has an essential singularity at z = 0.
1
x−iy
x
y
y
This function f (z) = e1/z = e x+iy = e x2 +y2 = e x2 +y2 (cos( x2 +y
2 ) − i sin( x2 +y 2 ))
converges to 0 as z → 0 from the negative real axis, converges to +∞ as z → 0 from the
positive real axis, and is rapidly oscillatory as z approaches 0 from any other direction. Due
to unbound oscillations of infinite density near the origin, the function cannot be well represented there by our usual mesh plots of its real and imaginary parts. The magnitude/phase
plots in Figure 2.14 show a totally different pattern than for poles: well structured but with
infinitely “dense” phase angle stripes near the singularity.
For all the singularity types above, we return to the same value if we follow the function
along a small path that goes around the singular point.
23 Also
known as a simple pole.
2.4. Singularities
41
(a) |e1/z | and arg e1/z .
(b) arg e1/z .
Figure 2.14. Graphical representations of f (z) = e1/z . Illustration of the essential singularity.
2.4.4 Branch points
Example 2.23. Consider the function f (z) = log z = log reiθ = log r + iθ.
Moving around the origin once in the positive direction increases θ by 2π, as we saw in
Figure 2.7(b).
How to “deal with” such functions will be discussed in Section 2.5. Very briefly, a
branch point is any point such that, moving around it once, one does not return to the same
function value as one started with. A branch cut is a “barrier curve” that one can draw in the
complex plane, from one branch point to another one, if one wishes to make a multivalued
function appear as if it was single-valued. The path of a branch cut is quite arbitrary and
will mostly depend on conventions. For example, if one wants log z to be single-valued for
42
Chapter 2. Functions of a Complex Variable
θ ∈ (−π, π], one would place the branch cut (barrier not to be crossed) along the negative
real axis. If one instead wants to work within θ ∈ [0, 2π), one would place it along the
positive real axis. While the shape of this artificial barrier (branch cut) is arbitrary, its end
points (branch points) are not. For the log function, it needs to go along some non-self–
intersecting path from z = 0 out to infinity (which, in view of the stereographic projection,
can also be thought of as a single point).
Example 2.24. If not imposing any branch cut restrictions, then Im(log 1) = iθ, where
√ ) = i( π + 2kπ), k ∈ Z (integer). With the common
θ = 0, ±2π, ±4π, . . . and log( 1+i
4
2
choice of placing the branch cut along the negative real axis, only the θ = 0 and k = 0
cases are relevant (i.e., the result is then single-valued).
Example 2.25. In the case of the square root function f (z) = z 1/2 , Figure 2.17 shows two
common branch cut choices (discussed further in Section 2.5).
Note that we can encounter somewhat similar cases for real variables. Apart from
needing to restrict the x-values of y = arcsin x to −1 ≤ x ≤ 1 to keep f (x) defined
as a real-valued quantity, single-valuedness is typically obtained by the further restriction
−π/2 ≤ f (x) ≤ π/2.
Back to complex variables: Software such as Mathematica or MATLAB typically impose branch cuts following common conventions, for instance along the negative real axis
in cases where this is appropriate. Another much less “crude” approach for achieving
uniqueness is to extend the complex plane through Riemann sheets (or surfaces), making
branch cuts unnecessary (see Section 2.5 below). When using Riemann sheets, also multivalued analytic functions can be considered in their entirety, rather than just partially.
2.4.5 Cluster points
An example would be a function with poles located at an infinite set of points with a finite
limit point.
1
Example 2.26. The function f (z) = sin1 1 is singular at z = πk
, k = 0, ±1, . . . . These
z
are all poles, apart from the cluster point at z = 0. The magnitude and phase plots in Figure
2.15 show the string of poles along the real axis and the intricate structure of the phase near
the cluster point at the origin. This point should not be viewed as an essential singularity,
since it is not isolated (it contains no full neighborhood that is singularity-free).
2.4.6 Natural boundary
A natural boundary is a genuine barrier (or boundary) past which a function no longer exists (i.e., it is not just that a certain functional form fails to work, such as a Taylor expansion
beyond its radius of convergence). Typically, every point on this boundary is a cluster point
of singularities in the sense described just above for a single cluster point (in Section 2.4.5).
We will discuss this later in the context of analytic continuation (including an illustration
in Figure 3.2).
2.5. Multivalued functions—Branch cuts and Riemann sheets
43
(a) |1/ sin z1 | and arg (1/ sin z1 ).
(b) arg (1/ sin z1 ).
Figure 2.15. Graphical representations of f (z) =
1
1
sin z
. Illustration of the cluster point singularity.
2.5 Multivalued functions—Branch cuts and Riemann
sheets
The simplest example of multivalued functions is provided by the square root function
f (z) = z 1/2 . When using only real numbers, the function y = x2 can be inverted graphically by reflection in the line y = x (cf. Figures 2.16(a)–(b)). For x positive, there are √
two
choices of numbers√that, when squared, return the value x. If one denotes these by ± x,
it is common to let x be the positive choice.
With instead a complex argument z, Figure 2.18 reveals a much richer functional behavior. We recognize again
√ the red curve from Figure 2.16(b) in Figure 2.18(a), but nothing
else. The two choices ± x are merely curves on smooth connected surfaces. The function
f (z) = z 1/2 is double-valued for all z, apart from an anomaly (branch point) at z = 0 (and
a counterpart branch point at z = ∞).
Chapter 2. Functions of a Complex Variable
2
2
1.5
1.5
1
1
0.5
0.5
0
0
y
y
44
-0.5
-0.5
-1
-1
-1.5
-1.5
-2
-2
-1.5
-1
-0.5
0
0.5
1
x
(a) y = x2 near the origin.
1.5
2
-2
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
x
√
(b) y = ± x near the origin.
√
Figure 2.16. The real-valued function y = x2 , and its inverse y = ± x, as obtained by
reflection in the line y = x (shown in blue).
To analyze f (z) = z 1/2 in more detail, consider z in its polar form representation
z = reiθ , where we
term 2πik, k integer, to θ in the exponent without changing
√ can add
√ any
iθ
the result. Thus, z = re 2 +ikπ , where it makes
sense to limit k to k = 0, 1 as further
√
choices
become
just
repetitions.
For
instance,
1
=
ekiπ , for k = 0, 1, recovering the real
√
case 1 = ±1; squaring either +1 or −1 gives the result +1.
There are two main ways to proceed when representing a multivalued function:
1. Impose some “barrier” (branch cut) in the complex plane such that, if we do not
allow z to move across it, the result will be single-valued.
2. Accept double (or, more generally, multivalued) surfaces of function values, such as
those displayed here in Figures 2.18(a)–(b) and also in several further figures.
We will next discuss these two choices in some more detail.
2.5.1 Branch cuts
√ iθ
√
Staying for now with the f (z) = z 1/2 example, we can limit ourselves to z = re 2 +ikπ ,
θ ∈ (−π, π], and the single case of k = 0. This limits z to the region shown in Figure
2.17(a), with an artificial “barrier” (branch cut) placed along the negative real axis. With
this restriction, all we will be able to “see” of the function is what is at or above height
zero in Figure 2.18(a), and on the surface sheet that tilts upwards for increasing imaginary
part in Figure 2.18(b). These parts are single-valued and in many cases suffice for what
may be needed. This k = 0 case is often called the function’s principal value or primary
Riemann sheet. Instead choosing k = 1 will give us the rest of the function, the parts that
were omitted when choosing k = 0.
Even when considering both cases k = 0 and k = 1, the results will depend on our
quite arbitrary choice of locating the branch cut in the z-plane along the negative real
2.5. Multivalued functions—Branch cuts and Riemann sheets
Im(z)
p
(1 + i)/ 2
Im(z)
p
(1 + i)/ 2
1 Re(z)
i
−i
1 Re(z)
−1
i
p
(−1 + i)/ 2
p
(1 − i)/ 2
(a)
45
√
z with θ ∈ (−π, π).
(b)
√
z with θ ∈ (0, 2π).
Figure 2.17. Two branch cut choices for f (z) = z 1/2 . The numbers shown give the
function values at the respective locations.
axis. Figure 2.17(b) illustrates the branch cut instead placed along (immediately below) the
positive real axis. If we thus define θ ∈ [0, 2π), we get a jump discontinuity
at the branch
√
i
is
the
same
in either
cut on the positive real
axis.
The
figure
shows
that
the
value
of
√
case, but the value of −i differs. In Figure 2.18(a), the primary sheet would now be what
we encounter when starting from the top red curve and moving one full turn in the positive
(counterclockwise) direction until immediately before we reach the bottom red curve. In
part (b), we would follow the top surface all the way around once. It will depend on the
application what choice of branch cut is the most convenient to use.
√
How can we test whether a point is a branch point? For example, let f (z) = z − z0 .
We can test z0 by√centering a small circle at that point: let z − z0 = reiθ . The function
becomes f (z) = reiθ/2 . We then monitor the function values when θ increases by 2π. If
the starting value differs from the end value, it means that the point that is being tested is a
branch point, and that we need to introduce a branch cut if we want to make the function
appear single-valued. For example, if we choose θ ∈ [0, 2π), let’s test the point z =√z0 .
Let’s follow a small circle around z0 , starting
0. At the starting point, f (z) = r.
√ at θ = √
After one revolution, θ = 2π and f (z) = reiπ = − r. Thus z = z0 is a branch point.
√ A branch cut always joins two branch points. We know that z = z0 is one (for f (z) =
z − z0 ), but where is the other one? There is no other one for finite z0 . Let’s simplify the
notation by setting z0 = 0 and then test z = ∞. Considering the stereographic projection,
z = ∞ is just one point like any other. To test it, let’s consider z = 1t around t = 0. Then
f (t) = √1t , and we see that t = 0 (and thus z = ∞) is also a branch point. The branch cut
will need to link those two points.
Example 2.27. Discuss the different branch cut options for
f (z) = (z − 1)1/2 (z + 1)1/2 .
Figure 2.19 shows the real and imaginary parts of f (z). The function clearly has two
solution sheets and branch points at z = 1 and z = −1.
√ To explore whether infinity also
(1−t)(1+t)
is a branch point, we set z = 1t and obtain f (t) =
. Thus, there’s a pole at
t
t = 0, but not a branch point. The cut will thus need to link z = −1 and z = 1 together.
Following the convention to place branch cuts along the real axis, if possible, there are
two natural choices: (i) directly along [−1, +1], and (ii) via infinity, i.e., along [−∞, −1]
46
Chapter 2. Functions of a Complex Variable
(a) Re z 1/2 .
(b) Im z 1/2 .
(c) |z 1/2 | .
Figure 2.18. Visualization of f (z) = z 1/2 . The phase angle coloring corresponds to the
primary branch (upper surface) in Parts (a) and (b) (phase angle θ in the range − π2 < θ ≤ π2 ).
2.5. Multivalued functions—Branch cuts and Riemann sheets
47
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
1
0.5
0
-0.5
-1
y
-3
-2
-1
0
1
2
3
x
(a) Re f (z).
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
1
0.5
0
-0.5
y
-1
-3
-2
-1
0
1
2
3
x
(b) Im f (z).
Figure 2.19. Real and imaginary parts of f (z) = (z − 1)1/2 (z + 1)1/2 . Each consists of
two sheets.
and [1, ∞]. Figure 2.20 focuses on Im f (z) and illustrates the corresponding single-valued
solutions that arise from the two branch cut strategies. For each choice of branch cuts, it
is clear that the two single-valued functions can be combined to form the multivalued one
shown in Figure 2.19(b).
Without relying on the figures above, we can directly verify that the cut presented in
Figure 2.21 makes the function single-valued. Let’s follow the contour around the cut to
make sure that we end up with the same function value as the one with which we started.
1/2 1/2 θ1 +θ2
Let z + 1 = R1 eiθ1 and z − 1 = R2 eiθ2 . Thus, (z + 1)1/2 (z − 1)1/2 = R1 R2 ei 2 .
2
Let Θ = θ1 +θ
2 . We want to make sure that the function’s argument Θ has changed by
an integer multiple of 2π after having gone around the contour. Table 2.1 shows that after
going around the cut, the function’s argument has increased by 2π. Thus, the value of the
function remains unchanged. The function is therefore single-valued with this cut.
48
Chapter 2. Functions of a Complex Variable
2
2
1.5
1.5
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
-2
1
-2
1
0.5
0
-0.5
-1
y
-3
-1
-2
0
1
2
3
0.5
0
-0.5
-1
y
x
(a) Cut [−1, 1], Primary sheet.
-3
-2
-1
0
2
1
3
x
(b) Cut [−1, 1], Secondary sheet.
2
2
1.5
1.5
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
-2
1
-2
1
0.5
0
-0.5
y
-1
-3
-1
-2
0
1
2
3
0.5
0
-0.5
y
x
(c) Cuts [−∞, −1] and [1, ∞],
Primary sheet.
-1
-3
-2
-1
0
2
1
3
x
(d) Cuts [−∞, −1] and [1, ∞],
Secondary sheet.
Figure 2.20. Imaginary parts of f (z) = (z − 1)1/2 (z + 1)1/2 . Parts (a), (b): the two
solutions when the branch cut is placed along [−1, 1]; parts (c), (d): the two solutions when branch
cuts are placed along [−∞, −1] and [1, ∞].
Im(z)
z
θ2
IV
θ1
I
Re(z)
1
−1
III
II
Figure 2.21. Branch cut for f (z) =
p
(z − 1)(z + 1).
2.5.2 Riemann sheets
If we don’t want to restrict the domain by a branch cut, we obtain a multivalued function
with multiple Riemann sheets. Figure 2.18 illustrated this for the function f (z) = z 1/2 .
Figure 2.22 shows similar illustrations for f (z) = z 1/3 , which is a function with three
Riemann sheets. We note along the entire real axis one option that matches the real-valued
2.5. Multivalued functions—Branch cuts and Riemann sheets
49
Table 2.1. Changes in the angle Θ as one follows the path shown in Figure 2.21.
Step
I
II
III
IV
Total
Changes in Θ
θ1 gains 2π
No change in the angle
θ2 gains 2π
No change in the angle
2
Θ = θ1 +θ
has increased by 2π
2
√
case of y = 3 x (with imaginary part zero), but we see here all the three options for all
complex values z.
While one can think of Figures 2.22(a)–(b) as three solution sheets over a single complex “base plane,” an interesting alternative is to think of the base plane replaced by a
triple layered complex plane, with layers connected as if one just “flattened down” either
of the triple surfaces, to get three sheets that lie together without any vertical separations.
From the perspective of this triply connected complex base plane, the function has become
entirely single-valued (and with no branch cuts needed).
The function f (z) = (z − 1)1/2 (z + 1)1/2 just considered in Example 2.27 displayed
also how the full set of function values can be viewed either as Riemann sheets or as
decomposed into pieces with two different branch cut strategies.
Figure 2.23 displays in an equivalent way f (z) = log z, with an infinity of Riemann
sheets (which all coincide for the real part, reminiscent of Figure 2.7, but now with additional sheets displayed). As noted above, it will depend on the application whether it is best
to consider the infinity of sheets, or to make a branch cut (for example along the negative
real axis), and then only have a small part of the function available at a time.
Example 2.28. Illustrate the function f (z) = log
z+1
z−1
.
Apart from a trivial factor of 21 , we recognize in Figure 2.24(b) the function Re arctanh z
from Figure 2.8(a). As in Example 2.27, both z = 1 and z = −1 are branch points. Like
for the function f (z) = log z, the multivaluedness is not visible in the real part (only in
the imaginary part and in the magnitude). We see a function with an infinity of Riemann
sheets. However, if we choose a path that takes us around both branch points (and does not
cross the real axis between −1 and +1), we return to the same function value. The point
z = ∞ is, in this case, not a branch point.24 It is nevertheless entirely possibly to place
branch cuts along [−∞, −1] and [1, +∞] (i.e., letting the cut between −1 and +1 along
its way pass through infinity), giving a function that is analytic and single-valued within
that cut domain. For the imaginary part, that choice would correspond to using the upper
half-plane (towards the lower left in Figure 2.24(c)) from one level and the lower half-plane
from the level below.
24 It is visually clear from Figures 2.24(b)–(c) that, following a circular path very far out, will return to the
same function value after going around once. Analytically, the same thing is most easily seen by writing f (z)
as f (z) = log
regular point.
1
1+ z
1
1− z
. Values of z near infinity keep the argument of the log function near one, which for it is a
50
Chapter 2. Functions of a Complex Variable
(a) Re z 1/3 .
(b) Im z 1/3 .
(c) |z 1/3 | .
Figure 2.22. Visualization of f (z) = z 1/3 . The coloring corresponds to the primary
branch (phase angle θ in the range − π3 < θ ≤ π3 ).
2.6. Sequences of analytic functions
51
(a) Re log z.
(b) Im log z (showing only three Riemann sheets).
Figure 2.23. Visualization of f (z) = log z.
2.6 Sequences of analytic functions
We will in several contexts come across sequences of analytic functions and, if they are
converging, their limits. This turns out to be yet another context in which the results for
analytic functions turn out simpler than for, say, infinitely differentiable functions on the
real axis. The next three examples illustrate some issues in the real-valued case. These are
followed by Vitali’s theorem for the analytic function case.
Consider real-valued functions. Let fn (x) be continuous on [a, b], and let the sequence of functions fn (x) converge at every point x as n → ∞. This does not imply
that limn→∞ fn (x) = f (x) is continuous.
Example 2.29. Discuss the convergence of the function sequence
0
, 0 ≤ x < 1 − n1 ,
fn (x) =
n (x − 1) + 1 , 1 − n1 ≤ x ≤ 1,
as n → ∞.
52
Chapter 2. Functions of a Complex Variable
3
2
f(x)
1
0
-1
-2
-3
-4
-2
0
2
4
x
(a) f (x) = log
x+1
x−1
for x real.
3
2
1
0
-1
-2
-3
4
2
0
-2
x
-6
-4
-2
0
2
-44
y
(b) Re log
z+1
z−1
.
15
10
5
0
-5
-10
-15
4
2
0
-2
-45
x
(c) Im log
z+1
z−1
4
3
1
2
0
-1
-2
-3
-4
-5
y
(showing three Riemann sheets).
z+1
Figure 2.24. Visualization of f (z) = log z−1
. In order to see the imaginary part better,
the viewpoint for parts (b) and (c) is located in the third quadrant.
2.6. Sequences of analytic functions
53
This sequence converges pointwise (at every x-location on [0,1]) to zero for 0 ≤ x < 1
and to one for x = 1; see Figure 2.25(a). The limit function f (x) (in green) is obviously
not continuous. The convergence is not uniform, since one can find an ε > 0 such that, no
matter how large n is, one can then choose an x such that |fn (x) − f (x)| > ε.
Example 2.30. Discuss the convergence of the function sequence
fn (x) = 2
n
X
(−1)k+1
k=1
sin kx
k
as n → ∞.
Each function fn (x) is 2π-periodic, converges pointwise to x for x ∈ (−π, π), and
equals zero for x = ±π. Convergence is again not uniform, also for an additional reason
(other than approximating a discontinuous function by a sequence of continuous ones). Due
to the Gibbs’ phenomenon, there arises a narrow “overshoot” next to the jumps (occurring
at x = ±π, ±3π, . . .), reaching in the limit approximately 9% of the jump height; see
Figure 2.25(b).25
It can be useful to consider limits of functions that converge pointwise and then be able
to tell something about the limit. For real functions, a small step in that direction is the
concept of uniform convergence, which can be defined more formally as
Given ε > 0, there exists an N (ε) such that |fn (x) − f (x)| < ε for all x ∈ [a, b]
whenever n > N ().
The two examples above were not uniformly convergent. With uniform convergence, the
fact that the functions fn (x) are continuous implies that the limiting function f (x) is cond
tinuous as well. Still this does not guarantee that the derivative(s) dx
fn (x) will converge
d
toward dx
f (x), the derivative of the limiting function.
Example 2.31. The sequence fn (x) = sinnnx converges uniformly to zero. In fact, given
d
fn (x) = cos nx
ε > 0, then |fn (x) − f (x)| < ε whenever n > N (ε) = 1ε . However, dx
d
does not converge towards dx 0 = 0.
Back to complex variables. Life is much easier for analytic functions.
Theorem 2.32 (Vitali’s convergence theorem). If fn (z) are analytic and bounded (|fn (z)|
< M constant) in a domain D (open; containing a full neighborhood of some point), and
if limn→∞ fn (z) = f (z) pointwise in D, then f (z) is also analytic there, and hence
dk
dk
limn→∞ dz
k fn (z) = dz k f (z) for all k = 1, 2, 3, . . .. Furthermore, it suffices if we know
pointwise convergence only on a point set with a finite limit point in D.
Note that Examples 2.30 and 2.31 do not contradict Vitali’s theorem, since the functions
are bounded only along the real axis, and not in a full complex neighborhood of any point.
25 See
also Exercise 9.8.11.
54
Chapter 2. Functions of a Complex Variable
1.2
1
0.8
y
0.6
0.4
n=2
4
10
0.8
0.9
0.2
0
-0.2
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
1
x
(a) fn (x) =
0
n (x − 1) + 1
,
,
0 ≤ x < 1 − n1
1 − n1 ≤ x ≤ 1
4
n = 20
3
n=3
2
y
1
0
-1
-2
-3
-4
-6
-4
-2
0
2
4
6
x
(b) fn (x) = 2
Pn
k+1 sin kx
k=1 (−1)
k
Figure 2.25. The functions fn (x) in Examples 2.29 and 2.30, respectively, shown for some
different n-values. Both cases feature pointwise convergence of a sequence of continuous functions.
However, in both cases, the convergence is nonuniform, and the limit functions are discontinuous.
2.7 Functions defined by integrals
Consider the function f (z) defined as follows:
ˆ
f (z) =
b
g(z, t)dt ,
a
assuming if nothing else is specified a straight line integration path from a to b. Let g(z, t)
be an analytic function of z and a continuous function in t. We want to know whether f (z)
2.7. Functions defined by integrals
55
´b
df
is analytic. If |a|, |b| < ∞, then dz
= f 0 (z) = a ∂g
∂z (z, t)dt must exist. Thus, f (z) has
one derivative and is
therefore
analytic.
If
b
=
∞,
we
can redefine f (z) as the limit of the
´n
sequence fn (z) = a g(z, t)dt and then use the convergence of sequence results (Vitali’s
convergence theorem). If the integral exists and is finite, then f (z) is again analytic. The
process is identical if a = −∞.
Let’s now examine two first examples of functions defined by integrals:
• Consider the function
ˆ
f (z) =
0
∞
e−t
dt.
1 + zt
(2.17)
If z is a real negative number, the integral doesn’t converge as the denominator becomes 0
at t = −1/z. Else, it is an analytic function.
• Another example is given by the Γ-function
ˆ ∞
Γ(z) =
e−t tz−1 dt.
(2.18)
0
Let’s analyze it piece by piece, by observing the potential difficulties of the integral:
1. The infinite interval can be a problem. However, since e−t makes the integrand decay very rapidly, the limit of the sequences of functions with the upper
limit of integration b → ∞ exists and is finite. The sequence is thus always
convergent.
2. At first glance, we note that there are values of z for which the integrand becomes divergent at the origin if the lower limit of integration a → 0. However,
having a divergent integrand along the integration path doesn’t always mean
´1
that the integral will diverge (e.g., 0 t1p dt for p < 1). We want to show that
the integral nevertheless converges if Re z > 0. As t & 0, the Taylor expansion of e−t becomes extremely accurate, and we can express e−t t z−1 =
1 z
1 z+1
t z−1 − 1!
t +´ 2!
t
− +· · · .The first term is the most singular one at the
origin; we get 0 tz−1 dt = z1 t z t=0 . Writing t z = ez log t , we see that t z for
t & 0 is well defined only for Re z > 0 (then goes to zero; further terms in the
expansion impose the weaker restrictions, Re z > −1, Re z > −2, etc.)
By means of (2.18), Γ(z) is thus defined when Re z > 0 and it is analytic because of the
observations above for a → 0 and for b → ∞. We have by this defined Γ(z) in the right
half-plane.
Integrating by parts reveals the interesting property
ˆ ∞
ˆ ∞
ˆ ∞
−t z−1
z−1
−t
−t z−1 ∞
Γ(z) =
e t dt = −
t d(e ) = −e t |0 +
e−t tz−2 (z − 1)dt
0
= (z − 1)Γ(z − 1)
0
0
if Re z > 1.
Functional equations such as this one arise often for analytic equations. This one can
alternatively be written as
Γ(z + 1) = zΓ(z) if Re z > 0.
(2.19)
56
Chapter 2. Functions of a Complex Variable
´∞
Noticing that Γ(1) = 0 e−t dt = 1, and then Γ(2) = 1Q
· 1, Γ(3) = 2 · 1, Γ(4) = 3 · 2 · 1,
n
etc., we have generalized the factorial function n! = k=1 k from positive integers to
complex arguments, as z! = Γ(z + 1) if Re z > 0.
The natural follow-up question to having generalized Γ(z) to an analytic function in
the complex plane becomes the following: Is the boundary limiting the Γ-function to the
right-hand side of the complex plane (Re z > 0) an actual limitation of the function or
just an artifact of the functional representation used in (2.18)? It will soon turn out that we
can analytically continue Γ(z) to the left half-plane by several methods (all, by necessity,
giving the same result).
2.8 Supplementary materials
2.8.1 Taylor series for the inverse of an analytic function
Especially in Chapter 12, we will repeatedly come across the task of inverting a Taylor
series, i.e., given the coefficients a0 , a1 , a2 , . . . in
w = f (z) = a0 + a1 (z − z0 ) + a2 (z − z0 )2 + · · ·
obtain the coefficients b0 , b1 , b2 , . . . for the inverse function
z = g(w) = b0 + b1 (w − a0 ) + b2 (w − a0 )2 + · · ·
(with both expansions centered at the matching locations z = z0 , w = a0 = f (z0 )).
Before describing two approaches, let us note that the task is quite laborious if done by
hand and that symbolic algebra packages handle it very effectively.
Lagrange’s inversion formula
In the case of f 0 (z0 ) 6= 0 (i.e., a1 6= 0), Lagrange gave an explicit expression for the
bk -coefficients in terms of derivatives of f (z) (rather than more immediately from the ak coefficients):26
k !
1
dk−1
z − z0
b0 = z0 ,
bk =
lim
, k = 1, 2, 3, . . . .
k! z→z0 dz k−1 f (z) − f (z0 )
While of theoretical interest, the following approach is in most cases more practical.
Inversion by equating coefficients
Without loss of generality, we can assume both z0 = 0 and a0 = 0. The two expansions
are then
w = f (z) = a1 z + a2 z 2 + · · · ,
2
z = g(w) = b1 w + b2 w + · · · .
26 This
5.1.1.
(2.20)
(2.21)
can be proved, for example, by combining (4.12), the result in Exercise 8.8.1, and Method 3 in Section
2.8. Supplementary materials
57
Case with a1 6= 0: Direct substitution of (2.21) into (2.20) gives w = a1 (b1 w + b2 w2 +
· · · ) + a2 (b1 w + b2 w2 + · · · )2 + · · · and, after equating coefficients in w,
1 = a1 b1 ,
0 = a2 b21 + a1 b2 ,
0 = a3 b31 + 2a2 b1 b2 + a1 b3 ,
0 = a4 b41 + 3a3 b21 b2 + a2 b22 + 2a2 b1 b3 + a1 b4 ,
..
..
.
.
The pattern that emerges here shows that the successive coefficients b1 , b2 , b3 , . . . become
uniquely determined as long as a1 6= 0.
Case with a1 = 0, a2 6= 0: Also this case arises frequently in Chapter 12.27 The expansion (2.20) of specified coefficients now lacks its leading term a1 z, and becomes
w = f (z) = a2 z 2 + a3 z 3 + a4 z 4 + · · · ,
where we this time have assumed that a2 6= 0. We will in this case need to modify (2.21)
to include half-integer powers as well:
z = g(w) = b1/2 w1/2 + b1 w1 + b3/2 w3/2 + b2 w2 + · · · .
The same process of substituting the latter expansion into the former gives now
1 = a2 b21/2 ,
0 = a3 b31/2 + 2a2 b1/2 b1 ,
0 = a4 b41/2 + 3a3 b21/2 b1 + a2 b21 + 2a2 b1/2 b3/2 ,
0 = a5 b51/2 + 4a4 b31/2 b1 + 3a3 b1/2 b21 + 3a3 b21/2 b3/2 + 2a2 b1 b3/2 + 2a2 b1/2 b2 ,
..
.
..
.
In the first equation, there is a choice of sign for b1/2 . Once that choice has been made, the
subsequent coefficients b1 , b3/2 , b2 , . . . follow uniquely.
2.8.2 Exchanging the order of summation and integration
In the following, we will on several occasions change the order of operators, in particular
between infinite sums and integration, raising the question of the validity of
!
ˆ X
X ˆ
?
fn (x) dx =
fn (x)dx
n
n
(assuming that each sum and integral converges). Simple examples show that some caution
is needed when changing orders of operators.
27 The
approach in this case generalizes quite straightforwardly to when a1 = a2 = · · · = ap−1 = 0, ap 6=
0. We don’t give any details here, since applications then become quite rare.
58
Chapter 2. Functions of a Complex Variable
Regarding double sums, we can consider the infinite sized matrix A with nonzero elements only along its main diagonal and first subdiagonal:


1

 −1 1




−1 1
A=
.


−1 1


.. ..
.
.
P∞
P∞
Clearly,
row sums n=1 am,nP
and column
sums m=1 am,n converge, but nevertheP∞bothP
P
∞
∞
∞
less m=1 ( n=1 am,n ) = 1 and n=1 ( m=1 am,n ) = 0.
Turning to double integrals (with infinitely differentiable integrands), the situation is
no different. For example,
ˆ ∞ ˆ ∞
ˆ ∞ ˆ ∞
x2 − y 2
π
π
x2 − y 2
dy dx = , while
dx dy = −
2 + y 2 )2
2 + y 2 )2
(x
4
(x
4
1
1
1
1
(2.22)
2
−y 2
y
∂2
(as obtained, for example, by noting that (xx2 +y
2 )2 = ∂x∂y arctan x ).
Exceptional cases (as both above) typically boil down to cancellations between infinite
quantities.28 If applying magnitudes to all functions/terms involved still leads to finite
results, the exchange is correct. As a consequence of Fubini’s theorem, the following will
hold.
Theorem 2.33. Consider an infinite sequence of functions fn (x). If either
!
ˆ X
X ˆ
|fn (x)| dx < ∞ or
|fn (x)|dx < ∞,
n
n
then the exchange of order
ˆ X
n
!
fn (x) dx =
X ˆ
fn (x)dx
n
is valid.
The most severe restriction in the following theorem is that the integration path cannot
be infinite.29
Theorem
2.34. In the case of (i) integration along an interval of finite length, and (ii)
P
f
(x)
being a uniformly
n
n
sum of continuous functions, the exchange of order
´ P
P ´ convergent
( n fn (x)) dx = n
fn (x)dx is valid.30
28 If we divide the domain for (2.22) in two halves by the line y = x, then the double integrals over these halves
both become infinite, and with opposite signs.
29 When verifying that restrictions in theorem formulations are necessary (rather than reflecting shortcomings in proofs), specific examples can be very useful. Following up on the idea in (2.22), let fn (x) =
2 P∞
n2 −x2
πx
. By techniques from Chapter 5, we obtain
1 − sinh
/ 2x2 (unin=1 fn (x) =
πx
(n2 +x2 )2
´ ∞ P∞
´
1
formly in x) and 1
fn (x) dx = 12 (1 + π − π coth π), whereas 1∞ fn (x)dx = − 1+n
2 and
´∞
n=1
P∞
1
π
n=1
1 fn (x)dx = 2 (1 − π coth π), a difference of 2 .
30 We will in Chapter 4 generalize integrals from intervals along the real axis to paths in the complex plane.
This theorem remains valid when these paths are piecewise smooth and of finite length.
2.8. Supplementary materials
59
Neither of the two theorems requires any analyticity properties of fn (x), and proofs
for them can be found in calculus texts. In the following, when exchanges of order can be
justified by either of these theorems, we make no explicit comments about this.
2.8.3 Convergence of infinite series
The following theorems provide techniques to test whether infinite series converge or not.
We refer the reader to classical calculus textbooks for details and proofs.
P∞
Theorem 2.35. The root test: If a power series takes the form f (z) = n=0 an z n , then
the radius of convergence R = 1/ limn→∞ |an |1/n , where lim denotes the least upper
bound (also known as limit superior, or upper limit) of the sequence and is defined below.
This formula for the radius of convergence of a power series is often referred to as the
Cauchy–Hadamard theorem.
As defined just above, lim denotes the least upper bound. The following are three
equivalent ways to define it. A = limn→∞ Sn of a sequence {Sn }∞
n=1 if (i) the sequence
is bounded from above and has A as its largest limit point, (ii) A = limN →∞ supn>N Sn ,
and (iii) A is the smallest number such that for ε arbitrarily small, Sn > A + ε only a finite
number of times.
P∞
Theorem 2.36. The ratio test: The series n=1 an converges absolutely if L < 1, and
|
. If the series is a power series of the form
diverges if L > 1, where L = lim n→∞ |a|an+1
n|
P∞
n|
n
f (z) = n=0 an z , it will converge in case that lim n→∞ |a|an+1
| exists. In that case, the
radius of convergence R is equal to this limit.
Theorem 2.37. The limit comparison test: If the terms of the sequences {ak } and {bk }
are real and positive, and if limn→∞ abnn = c with 0 < c < ∞, then either both series
P∞
P∞
n=1 an and
n=1 bn converge or both diverge.
Theorem 2.38. The integral
´ ∞ {ak } are ak = f (k), with f
P∞ test: If the terms of the sequence
continuous, the series n=N an converges if and only if N f (x)dx < ∞, for N > 0.
Theorem 2.39. Weierstrass M-test: Let |bj (z)| ≤ Mj , an
P upper bound for eachPanalytic
function bj (z) for j = 0, 1, 2, 3, . . . in some region R. If j Mj converges, then j bj (z)
converges uniformly in R.
2.8.4 Codes for visualization of analytic functions
Most illustrations in this book were generated using either MATLAB31 or Mathematica.32
A sample code using each of the systems is shown below.
MATLAB
The following MATLAB script together with function display_function produce the three
subplots of Figure 2.4.
31 The
MathWorks, Natick, MA.
Research, Champaign, IL.
32 Wolfram
60
Chapter 2. Functions of a Complex Variable
%
Test code for displaying an analytic function ; here exp(z)
clear ; close all;
bx = [ -4 ,4 , -10 ,10]; %
Domain (box) surrounding origin to be displayed
nx = 41; %
Number of nodes along real axis for re & im plots
ny = 41; %
Number of nodes along imag axis for re & im plots
bounds = [ -60 ,60; -60 ,60; 0 ,60]; %
Lower and upper in the three displays
f = @(z) exp(z); %
Function to display
display_function (f,bx ,nx ,ny , bounds ); %
Create the three subplots
function display_function (f,bx ,nx ,ny , bounds )
%
The input parameters described in the script that calls this routine
lw = 4;%
Set LineWidth for highlighting values along real axis
x = linspace (bx(1),bx(2),nx); y = linspace (bx (3) , bx (4) , ny );
[xr ,xi] = meshgrid (x,y); z = complex (xr ,xi);
figure (1) %
Plot the real part using mesh
mesh(xr ,xi ,real(f(z))); colormap ([0 0 0]); hold on;
xlabel (’\itx ’); ylabel (’\ity ’); title(’Real part ’)
xlim(bx (1:2)); ylim(bx (3:4)); zlim( bounds (1 ,:));
plot3 (x, zeros (size(x)), real(f(x)),’r’,’LineWidth ’,lw ); %
Highlight real axis
figure (2) %
Plot the imaginary part using mesh
mesh(xr ,xi ,imag(f(z))); colormap ([0 0 0]); hold on;
xlabel (’\itx ’); ylabel (’\ity ’); title(’ Imaginary part ’)
xlim(bx (1:2)); ylim(bx (3:4)); zlim( bounds (2 ,:));
plot3 (x, zeros (size(x)), imag(f(x)),’r’,’LineWidth ’,lw ); %
Highlight real axis
figure (3) %
Plot the magnitude as a surface
%
Increase resolution for smoother display
x = linspace (bx(1),bx(2),nx *4); y = linspace (bx (3) , bx (4) , ny *4);
[xr ,xi] = meshgrid (x,y); z = complex (xr ,xi);
p = surf(xr ,xi ,abs(f(z)), angle(-f(z))); % Display the surface
set (p,’EdgeColor ’,’none ’); colormap hsv (600); hold on;
xlabel (’\itx ’); ylabel (’\ity ’); title(’Magnitude , with phase plot ’)
xlim(bx (1:2)); ylim(bx (3:4)); zlim( bounds (3 ,:));
plot3 (x, zeros (size(x)), abs(f(x)),’k’,’LineWidth ’,lw ); %
Highlight real axis
axes(’Position ’ ,[0.05 0.05 .17 .17]) %
Add color wheel for phase information
[th ,r] = meshgrid ( linspace (-pi ,pi), linspace (0 ,1));
[X,Y] = pol2cart (th+pi ,r);
contourf (X,Y,th ,100,’ linestyle ’,’none ’); hold on %
Show wheel colors
plot ([ -1 1] ,[0,0],’k ’); plot ([0 0],[-1,1],’k ’); %
Show Re and Im axes
plot(cos (0:0.01:2* pi),sin (0:0.01:2* pi),’k ’); colormap hsv (600);
axis equal ; axis off
Mathematica
If one wants to quickly get surface plots of a complex function with Mathematica, just
a few line statements suffice for each plot. The first line below specifies the exponential
function f (z) = ez , and the following lines produce plots of its real and imaginary parts,
and of its magnitude (with phase angle coloring).
f[z_] = E^z;
(* bounding box *)
{x1 , x2 , y1 , y2 , z1 , z2} = {-4, 4, -10, 10, -60, 60};
(* Curves on the Real Axis *)
{re , im , abs} =
ReleaseHold [
Hold[ ParametricPlot3D [{x, 0, a[f[x]]} , {x, x1 , x2},
PlotStyle -> Directive [ Thickness [0.01] , b]]] /. {{a -> Re , b -> Red},
{a -> Im , b -> Red}, {a -> Abs , b -> Black }}];
(* Real Part *)
Show[ Plot3D [Re[f[x + I*y]], {x, x1 , x2}, {y, y1 , y2},
PlotRange -> {z1 , z2}, AxesLabel -> {" Re(z)", "Im(z)"} ,
PlotLabel -> Re[f[z]], ViewPoint -> {-1.3, -2.4, 2}] , re]
(* Imaginary Part *)
Show[ Plot3D [Im[f[x + I*y]], {x, x1 , x2}, {y, y1 , y2},
2.8. Supplementary materials
61
PlotRange -> {z1 , z2}, AxesLabel -> {" Re(z)", "Im(z)"} ,
PlotLabel -> Im[f[z]], ViewPoint -> {-1.3 , -2.4, 2}] , im]
(* Magnitude and Argument *)
Show[ ComplexPlot3D [f[z], {z, x1 + I y1 , x2 + I y2}, PlotRange -> {0, z2},
Mesh -> False , PlotPoints -> 100, ViewPoint -> { -1.3 , -2.4, 2},
Epilog -> Inset[ ComplexPlot [z, {z, -1 - 1 I, 1 + 1 I},
RegionFunction -> Function [{z}, Abs[z] < 1], Axes -> True , Ticks ->None ,
BoundaryStyle -> Directive [Black , Thin], Mesh -> None , Frame -> None ,
ImageSize -> 50, PlotPoints -> 100, ColorFunction -> None], {0.05 , 0.1}] ,
AxesLabel -> {" Re(z)", "Im(z)"}, PlotLabel -> Abs[f[z]]] , abs]
2.8.5 Select proofs
Theorem 2.40. If the C-R equations (2.1) hold and the partial derivatives ux , uy , vx , vy
are continuous functions of x and y, then f (z) is differentiable.33
Proof. Let f (z) = u(x, y) + iv(x, y) and introduce the infinitesimal quantities dz =
dy
1
= dx−i
dx + i dy ⇒ dz
|dz|2 . Then
(dx − i dy)
f (z + dz) − f (z)
=
[{u(x + dx, y + dy) − u(x, y)}
dz
|dz|2
+ i {v(x + dx, y + dy) − v(x, y)}]
(dx − i dy)
=
[{ux dx + uy dy} + i {vx dx + vy dy}] ,
|dz|2
where we, in the last line, replaced u(x + dx, y + dy) (and likewise for v) with the leading
terms of their 2-D Taylor expansions about (x, y). Expanding the product, some terms drop
out due to the C-R equations, and we are left with
f (z + dz) − f (z)
1 =
(dx)2 (ux + ivx ) + (dy)2 (vy − iuy ) ,
dz
|dz|2
which, after using the C-R equations again, further simplifies to
ux + ivx , thus showing that f 0 (z) exists.
(dx)2 +(dy)2
(ux
|dz|2
+ ivx ) =
2.8.6 Some apparent paradoxes due to multivaluedness
Certain areas of mathematics are quite rich in seemingly counterintuitive results (such as
set theory and statistics). In the context of analytic functions, the common theme in such
examples is usually having neglected to properly account for multivaluedness.
p
√
Example 2.41. 1 = 1 = (−1)2 = −1.
In this simple case, we clearly neglected the multivaluedness of the square root in the
last equality. Remembering the convention that we, by the square root of a positive number,
imply the positive choice would have eliminated the error here.
33 From
the continuity of the partial derivatives, it is understood that u, v, and thus f exist in a neighborhood of
z = x + iy.
62
Chapter 2. Functions of a Complex Variable
Example
2.42. Let z = −1 + i. From the logarithm definition (2.9) for −π <2 arg z ≤ π,
log z 2 = log (−2i) = log 2 − i π2 . However, 2 log z = log 2 + i 3π
2 6= log z .
As in the previous example, the function is restricted to its primary sheet. Giving up
this restriction would restore the functional property that log z 2 = 2 log z.
Example 2.43.
1
−1
=
−1
1
⇒
q
1
−1
=
q
−1
1
⇒
√
√ 1
−1
=
√
√−1
1
⇒
1
i
=
i
1
⇒ 1 = −1.
The argument −1 lies exactly on the typical branch cut for the square
√ root function,
which confuses
the
branch
choices.
Although
one
casually
may
write
i
=
−1, it is safest
√
to think of −1 as a multivalued expression.
Example 2.44. e2πi = 1 ⇒ e2πi+1 = e. If we substitute this expression for e into the last
2πi+1
2
2
relation’s LHS, we get e2πi+1
= e ⇒ e(2πi+1) = e ⇒ e−4π +4πi+1 = e and
2
e−4π = 1.
Although there are no multivalued functions in immediate sight, we need to recall that
expressions of the form az have to be interpreted as az = ez log a . In the a = e case, ez
is defined as a single-valued function, and the same is the case when a > 0, under typical
conventions. However, there is no similar convention for the case when a is either negative
or complex.
2.9 Exercises
Exercise 2.9.1. Write as a + ib the following complex quantities. If there are several
solutions, give them all:
i
(a) e(e )
e2+2πi
(b)
e1+πi/3
(c) log 1
(d) log e
(e) cos i
1/3
1
(f)
3
(g) i1/3 √
(h) (1 + 3 i)i
1/2
i
(i)
1+i
Exercise 2.9.2. Verify that the contour lines for magnitude equal to one in Figure 2.2 are
perfect circles centered at ±i and passing through ±1.
Exercise 2.9.3. Derive (2.10) and (2.11) for (a) arctanh z and (b) arcsin z, respectively.
Exercise 2.9.4. Write in the form f (z) = u(x, y) + i v(x, y) the following function, and
verify in each case the C-R equations:
2.9. Exercises
63
(a) z 3
1
z+i
(c) cos z
(b)
Exercise 2.9.5. Consider the function f (x, y) = 1 + x − 3(x2 − y 2 ) + i y(1 − 6x).
a. Verify by the C-R equations that f (x, y) is analytic.
1
(z − z) to (i) again verify analyticity, and
b. Use the substitutions x = 12 (z + z), y = 2i
(ii) obtain a simple expression for f (z).
Exercise 2.9.6. Are the following functions analytic? Check using the C-R equations. If
they are, also write them as functions of z:
(a) f (x, y) = y 3 − 3x2 y + i(x3 − 3xy 2 + 2)
(b) f (x, y) = ex−iy
Exercise 2.9.7. An arbitrary triangle can be scaled and rotated to have two corners at
0 + 0i, 1 + 0i and the third at z = x + iy. Notable points for a triangle include
N (z) intersection of the normals from a corner to the opposite side,
M (z) intersection of the medians, going from a corner to the midpoint of the opposite
side, and
C(z) center of the circumscribed circle (passing through all the corners)
(see Figure 2.26). Straightforward geometry will show that
x(1 − x)
1
y
1
i
x(1 − x)
N (z) = x + i
, M (z) = (x + 1) + i , C(z) = +
y−
:
y
3
3
2 2
y
(a) Determine which (if any) of the three functions N (z), M (z), C(z) are analytic
functions of z.
N (z)−C(z)
(b) Verify that M
(z)−C(z) = 3, independent of the value of z.
Comment: The result in (b) tells that, for an arbitrary triangle, the points N , M , C always
lie on a straight line, with M located 2/3 of the way from N to C. This result is due to
Leonhard Euler; the line is known as the “Euler line” of the triangle.34
Exercise 2.9.8. Let f (x, y) = u(x, y) + iv(x, y) be an analytic function, where u(x, y) is
given below. Find v(x, y), the harmonic conjugate of u(x, y), and express the function in
terms of z:
y
(a) u(x, y) = x2 +y
2
(b) u(x, y) = cos x cosh y
Exercise 2.9.9. Derive the expressions for the C-R equations in terms of polar coordinates.
Exercise 2.9.10. Use Euler’s formulas to obtain easy proofs for the trigonometric identities
cos 3θ = 4 cos3 θ − 3 cos θ
.
sin 3θ = 3 sin θ − 4 sin3 θ
34 The Euler line features in several contexts, for example, the points where the lines from one corner intersect
the opposite side in Figures 2.26(a)–(b) and the midpoints of the lines from the corners to N (z) (nine points in
all) lie on a single circle, with center exactly halfway between N (z) and C(z).
64
Chapter 2. Functions of a Complex Variable
0.6
z = x + iy
0.5
0.4
0.4
0.3
0.3
0.2
0.1
z = x + iy
0.6
0.5
0.2
N(z)
0.2
M(z)
0.1
0.4
0.6
0.8
1
0.2
-0.1
-0.1
-0.2
-0.2
-0.3
-0.3
(a) Intersection of the normals
0.4
0.6
0.8
1
(b) Intersection of the medians
z = x + iy
0.6
0.5
0.4
0.3
C(z)
0.2
0.1
0.2
0.4
0.6
0.8
1
-0.1
-0.2
-0.3
(c) Center of circumscribed circle
Figure 2.26. The three points N , M , C for a triangle, shown with red markers, and
described in Exercise 2.9.7.
Exercise 2.9.11. Show that
Pn
n
X
k=1
k=0
cos kθ =
sin kθ =
sin
sin( 12 (n+1)θ ) cos
sin
1
2 (n
nθ
2
θ
2
+ 1)θ sin nθ
2
sin θ2
and
.
Exercise 2.9.12. With z = x + iy, show that | sinh x| ≤ | cosh z| ≤ | cosh x|.
Exercise 2.9.13. Using a calculator, one quickly sees that π π ≈ 36.4622 and ee ≈
15.1543. Can you spot what similarly evaluates approximately to 0.20788? Explain!
Exercise 2.9.14. Calculate all values of arctan z when (a) z = 0, (b) z = 2i, (c) z = 1 + i.
Exercise 2.9.15. Show that all solutions to sin z =
k integer.
5
3
are given by zk = (2k+ 12 )π±i log 3,
Exercise 2.9.16. Use, for example, the MATLAB code given in Section 2.8.4 (or some
other software) to plot the real part, imaginary part, magnitude, and phase portrait of the
following functions:
2.9. Exercises
65
(a) f (z) = z 6
(b) f (z) = tan z
(c) f (z) = arctan z
Exercise 2.9.17. Derive the following formulas:
(a) arcsin z = −i log(iz + (1 − z 2 )1/2 )
d
arcsin z = (1−z12 )1/2
(b) dz
Exercise 2.9.18. Derive Cote’s relation i φ = log(cos φ + i sin φ) by evaluating
p
d
log
1 + z2 + z ,
dz
and then substitute z = i sin φ and integrate with respect to φ.
ExerciseP
2.9.19. Find the radius of convergence of the following series:
∞
(a)
zk
Pk=1
∞
zk
(b)
k=1 (k+1)!
P∞ k! k
(c)
k=1 kk z
1+z
(d) The Taylor series of 2−z
about z0 = i
Exercise 2.9.20. Find the Taylor series expansions about z = 0 of the following functions,
valid in the given regions:
1
(a) 1−z
2 , |z| < 1
(b) cosh z, |z| < ∞
2
(c)
ez −1−z 2
,
z3
0 < |z| < ∞
Exercise 2.9.21. Find the leading coefficients c0 , c1 , . . . , c4 in the function
y(x) =
∞
X
ck x k
k=0
that solves the ODE
dy(x)
dx
= x + sin y(x), y(0) = 0.
Exercise 2.9.22. If a series is not of Taylor form, it may converge to different functions in
different regions. Show that
∞
X
z k−1
1/(1 − z)2 if |z| < 1
=
.
k
k+1
1/ z(1 − z)2
if |z| > 1
(1 − z )(1 − z
)
k=1
Hint: Show first that
PN
k=1
z k−1
(1−z k )(1−z k+1 )
=
1
z(1−z)
1
1−z
−
1
1−z N +1
.
Exercise 2.9.23. ThePBernoulli numbers {Bk }∞
the Euler numbers {Ek }∞
k=0
k=0 are
Pand
∞ Bk k
∞ Ek k
z
1
defined by ez −1 =
k=0 k! z and cosh(z) =
k=0 k! z , respectively, named after
Jacob Bernoulli (1654–1705) and Leonhard Euler (1707–1783).
(a) Find the first three nonzero Bernoulli numbers and the first three nonzero Euler
numbers.
(b) Find the radius of convergence of both series.
66
Chapter 2. Functions of a Complex Variable
Exercise 2.9.24. Give explicit formulas for u(x, y) and v(x, y) in the case of f (z) =
Simplify the result as far as you can.
√
z.
√
√
Exercise 2.9.25. In the expression 3 2 + 2i + 3 2 − 2i, each cube root can take three
different values. The nine possible values for the sum turn out to be
−2, √
1 ± √3,
1 ± i 3,
√
√
(four combinations).
− 12 (1 ± 3)(1 ± i 3)
√
Verify the three real-valued cases (i.e., −2√and 1 ± 3).
Hints: (i) Note
that the three options for 3 2 − 2i are the complex conjugates of the three
√
3
π
= π3 − π4 .
options for 2 + 2i, and (ii) 12
Exercise 2.9.26. Discuss the singularities of the following functions:
(a) f (z) = coszz−1
4
(b) f (z) = z 3 e1/(z−1)
(c) f (z) = ez1−1
P∞
Exercise 2.9.27.
to Example 2.5, show that k=1 k1 cos kx = − log |2 sin x2 |.
P∞As 1a follow-up
Hint: Since k=1 k eikx diverges for all x if differentiated term-by-term, consider temP∞ rk ikx
d
porarily dx
with 0 < r < 1 and, at some later stage of the algebra, let r
k=1 k e
increase to 1.
Exercise 2.9.28.
(a) Show that (z − eiθ )(z − e−iθ ) = z 2 − 2z cos θ + 1.
(b) Write down the (2n)th roots of −1 in the form eiθ , and deduce
z
2n
+1=
n Y
2
z − 2z cos
k=1
(2k − 1)π
2n
+1 .
(2.23)
(c) Substitute z = i into (2.23). Show that for n even
3π
(2n − 1)π
1π
· cos
· · · · · cos
= (−1)n/2 21−n
cos
2n
2n
2n
and for n odd
cos2
1π
2n
· cos2
3π
2n
· · · · · cos2
(n − 2)π
2n
= n · 21−n .
Exercise 2.9.29. Consider Exercise 1.6.15. Solve it again, noting that α(z), β(z), γ(z) will
all be linear functions of z, so f (z) = α2 + β 2 + γ 2 − αβ − αγ − βγ will be a quadratic
polynomial. If this evaluates to zero for three different z-values, it must be identically zero.
Exercise 2.9.30. Prove Ptolemy’s theorem: With a quadrilateral inscribed in a circle, as
shown in Figure 2.27, it holds that a · c + b · d = α · β.
Note: If the quadrilateral is a rectangle, the theorem reduces to Pythagoras’ theorem.
2.9. Exercises
67
b
α
c
β
a
d
Figure 2.27. Illustration of Ptolemy’s theorem a · c + b · d = α · β for a quadrilateral
inscribed in a circle.
Hint: Show first the relations
1
|eiβ − eiα | = 2 sin (β − α) if 0 < α < β < 2π,
2
and then
|eiβ − eiα | · |eiγ − eiδ | + |eiβ − eiγ | · |eiα − eiδ | = |eiα − eiγ | · |eiβ − eiδ |
if 0 < α < β < γ < δ < 2π.
Exercise 2.9.31. Figure 2.28(a) shows the stereographic projection of the phase portrait of
f (z) = z. Plots (b) and (c) show, respectively, the southern and the northern hemispheres
viewed from inside the sphere. Explain why the only difference between plots (b) and (c) is
the direction of the color scheme.
Exercise 2.9.32. The rows of pictures in Figure 2.29 are labeled (a) to (f). The real axis
is drawn in bold. These are illustrations of the functions f (z) that are listed below. Match
each row with its corresponding function and explain your choice:
(i) cos(z)
√
(ii)
z3 + 1
(iii) ez
(iv) ez z 6
(v) z21−1
(vi) 103 z + z 6
Exercise 2.9.33. Consider f (z) = log z 2 − 1 :
(a) Find the types and locations of all the singularities of f (z).
(b) Test whether f (z) has a singularity at z = ∞.
(c) Make this function single-valued by introducing a branch cut. Carry on the procedure introduced in Table 2.1 to verify the single-valued nature of the function.
68
Chapter 2. Functions of a Complex Variable
(a) Stereographic projection of the phase portrait of f (z).
(b) Southern hemisphere.
(c) Northern hemisphere.
Figure 2.28. Graphical representations of f (z) = z, in (b) and (c) as seen from the sphere center.
Exercise 2.9.34. Discuss the options for placing branch cuts to make the following functions single-valued (a, b, c are finite and distinct complex constants).
(a) f (z) = log z−a
z−b
(b) f (z) = log
(z−a)(z−c)
z−b
1 2
1 3
Exercise 2.9.35. Consider w = f (z) = ez − 1 = z + 2!
z + 3!
z + · · · . Obtain the first
2
3
three terms in the inverse expansion z = b1 w + b2 w + b3 w + · · · by means of
(a) Lagrange’s inversion formula, and (b) the method of undetermined coefficients.
Verify your answers against the direct expansion of z = log(1 + w).
2.9. Exercises
Stereographic projection
69
Southern hemisphere
Northern hemisphere
(a)
(b)
(c)
(d)
(e)
(f)
Figure 2.29. Stereographic projection of the phase portraits for Exercise 2.9.32. The
hemispheres are shown as seen from the sphere center.
Chapter 3
Analytic Continuation
A particular representation of a function might be valid only in a limited region of the
complex plane, although the function itself may exist well past that region. For example,
a function defined by its Taylor series can only be evaluated (by immediate summation)
inside this expansion’s circular domain of convergence, which is limited in size by the
distance to the nearest singularity. The topic of analytic continuation is about changing
the way a function is formulated so that it can be evaluated also in a larger domain. All
extensions of this type give identical results (although their functional forms and domains
of validity may differ). This chapter gives examples of a number of continuation methods.
This topic provides yet another example of mathematics becoming greatly simplified
when functions are analytic. There is no counterpart for nonanalytic functions (no matter
how many times these are differentiable).
3.1 Introductory examples
Example 3.1. Consider the following three functions:
P∞
1. f1 (z) = k=0 z k = 1 + z + z 2 + z 3 + · · · . The radius of convergence is R = 1.
This follows from either of Theorems 2.7, 2.35, or 2.36.
´∞
2. f2 (z) = 0 e−t(1−z) dt. If the integral had been over a finite interval rather than
over [0, ∞], it would have defined an entire function. With the upper limit ∞, it will
converge for Re z < 1 but diverge for Re z ≥ 1.
1
3. f3 (z) = 1−z
. This expression represents an analytic function in the whole complex
plane, with the exception of a single first order pole located z = 1.
The key point in this example is that all three functions above are identically the same.
The boundaries for f1 (z) and f2 (z) are not natural ones, but just artifacts of the particular
way we have used for representing the function f3 (z). Figure 3.1 illustrates this function
over the domains corresponding to the three different representations.
The goal of analytic continuation is to reformulate functional expressions so they can
be used in larger regions than was originally the case. It can happen that continuation is not
possible. For instance, in the following example, the unit circle forms a natural boundary,
and it is not possible to continue the function past it.
71
72
Chapter 3. Analytic Continuation
(a) f1 (z)
(b) f2 (z)
(c) f3 (z)
Figure 3.1. Illustration of Example 3.1. The regions of validity for the three functional
representations f1 (z), f2 (z), and f3 (z).
3.2. Some methods for analytic continuation
Example 3.2. Consider the function f (z) =
73
P∞
k=1
z k! = z + z 2 + z 6 + z 24 + · · · .
Theorem 2.35 again gives R = 1, and the domain of convergence will be the same as
for the function f1 (z) in Example 3.1. Consider a point z = e2πik/m where k and m are
arbitrary integers. Then z n! = e2πin!k/m = 1 for all n ≥ m. This means that f (z) → +∞
whenever z is of the form z = re2πik/m and r → 1. The function f (z) must therefore have
singularities located densely all the way around the unit circle, making |z| = 1 a natural
boundary. It cannot be continued past it.
What does a function like this look like graphically? Figures 3.2(a)–(b) show its magnitude and phase angles when displayed within circles of radii 0.99 and 0.999999, respectively. There is of course no difference between the two displays throughout the inner
region, but we see more fine structure entering near the edge in the second case. Moving
even closer to the unit circle, the fine structure becomes infinitely dense. We can also note
that |f (z)| does not increase monotonically as the unit circle is approached. In the phase
angle color patterns, we can see tell-tale signs of increasing numbers of zeros. Near the
origin, this function f (z) becomes of course very close to f (z) = z, as displayed in the
top right subplot of Figure 2.1.
P∞
As a follow-up to this example, we can consider g(z) = k=1 z k! /k!. Since g 0 (z) =
f (z)/z, this function g(z) will also have |z| = 1 as a natural boundary. Nevertheless,
counterparts to Figures 3.2(a)–(b) would seem visually to be perfectly smooth also along
the unit circle.
Whenever a function is given in a form which only permits it to be evaluated in some
limited region of the complex plane, it becomes of great interest to see whether it can be
rewritten in some other form that permits the domain to be extended, preferably to the
whole complex plane. There is no similar opportunity for functions of a real variable, even
if these are known to be infinitely differentiable, as can be seen from considering functions
such as
−1/x
e
, x>0
f (x) =
.
0
, x≤0
It is quite common in this context that the initial representation is a Taylor series. The
remainder of this chapter describes a number of ways to carry out analytic continuation.
This chapter is placed in its present position in the book, since it requires no more preliminaries than what are already available. However, it can also be treated as an appendix to
the book, to be used as a reference resource when needed.
3.2 Some methods for analytic continuation
Each of the following nine subsections describes a different method for analytic continuation:
1. Circle-chain method
2. Schwarz reflection principle
3. Use of a functional equation
4. Partitioning of an integration interval
5. Replace Taylor coefficients by integrals or sums
74
Chapter 3. Analytic Continuation
(a) Display of |f (z)| over |z| ≤ 1 − 10−2 .
(b) Display of |f (z)| over |z| ≤ 1 − 10−6 .
Figure 3.2. Illustration of |f (z)| (with phase coloring) for f (z) =
slightly different disks around the origin. Illustration of a natural boundary.
P∞
k=1
z k! over two
6. Subtraction of a similar series or integral
7. Borel summation
8. Ramanujan’s formula
9. Padé approximations
The Padé approach differs from the other ones in that it only rarely can be used for mathematically exact analytic continuation. We have nevertheless included it here, since many
applications can utilize its ability to provide quite accurate continuations even from highly
incomplete information (such as from a relatively low number of leading terms in truncated
3.2. Some methods for analytic continuation
75
y
x
Figure 3.3. Schematic illustration of a circle-chain continuation.
Taylor expansions). Still further examples of continuation methods (which require the use
of complex integration) are given in Section 5.3.
3.2.1 Circle-chain method
This approach is rarely practical for actual applications, but it is of interest in that it gives
insight into what analytic continuation amounts to. Suppose that we have a function with
point-type singularities (poles, branch points, or essential singularities) in the (a priori unknown) locations marked by red crosses in Figure 3.3. Assume that we are given initially a
Taylor expansion centered at the origin of the complex plane, and the task is to follow how
the function changes along the green (curved) path in the figure. The Taylor expansion will
converge only in the domain bounded by the solid blue circle, extending out to the nearest
singularity. Within this circle, the expansion completely describes the function. Hence, we
can choose any other point inside it, on the continuation path, and create a new expansion
centered at this new location. Its domain of convergence will likely reach outside the original one. The process can in theory be repeated indefinitely, allowing us to proceed, for
example, along the indicated chain of circles.
From a practical (as opposed to a theoretical) perspective, this approach is seldom usable. Even when the complete (infinite) set of Taylor coefficients is known analytically, it
is rare that one can find convenient closed forms also for the subsequent expansions (however, one such example is given in Example 3.3 and another one in Exercise 3.3.3). If the
initial Taylor coefficients can only be found numerically, this continuation process becomes
extremely ill conditioned.
This circle-chain method (as well as the mathematical concept of analytical continuation) requires the initial Taylor expansion’s radius of convergence R to be nonzero. Nevertheless, some of the approaches we describe later can be applied with apparent success
76
Chapter 3. Analytic Continuation
even in some cases of power series expansions with R = 0, as seen in Example 3.18 and
Exercises 3.3.5(b)–(c).
An interesting situation arises if one tries to continue all the way around a singularity.
Will the results agree when the circle-chain continuation comes back to the start point? The
answer will be yes when going around poles and essential singularities, but no in the case
of a branch point.
P∞ n
Example 3.3. Continue the function f (z) = n=1 zn by means of a chain of circles once
in the clockwise direction around the singularity at z = 1.
We can start by noting (but will not use further) the fact that f (z) = − log(1 − z), so
the end result after having continued step-by-step around the singularity should return the
same function as we started with, apart from an increase in its imaginary part by 2πi.
The original expansion is centered at z0 = 0. Let the subsequent expansions be centered
at z1 , z2 , . . . , zp = z0 , where zk = 1 − e−2iπk/p , k = 1, 2, . . . , p (see Figure 3.4). From
the Taylor expansion of f (z) it follows that
f 0 (z) =
∞
X
z n−1 =
n=1
and therefore f (m) (z) =
(m−1)!
(1−z)m .
1
1−z
(3.1)
The Taylor expansion centered at z1 therefore becomes
n
∞
∞
X
X
1 (n)
1 z − z1
z − z1
f (z1 )(z − z1 )n = f (z1 ) +
= f (z1 ) + f
.
n!
n 1 − z1
1 − z1
n=0
n=1
(3.2)
Differentiating (3.2), we obtain from its first and third expressions
n−1
∞ 1 X z − z1
1
0
,
f (z) =
=
1 − z1 n=1 1 − z1
1−z
f (z) =
confirming that (3.1) again holds when now expanding around z = z1 (and its different domain of convergence). Looking at the first and last expressions in (3.2), a pattern becomes
clear. For the next expansion, around the point z2 , we get similarly
z2 − z1
z − z2
f (z) = f (z1 ) + f
+f
,
1 − z1
1 − z2
etc. After p steps, we arrive at
f (z) = f (z1 ) + f
|{z}
|
new f (z)
z2 − z1
1 − z1
+f
z3 − z2
1 − z2
{z
+ ··· + f
additive constant
zp − zp−1
1 − zp−1
+ f
}
|
z − zp
.
1 − zp
{z
}
old f (z) since zp = 0
Are we back to the function value we started with? Noting that zk = 1 − e−2iπk/p , all
the p terms in the “additive constant” become equal to f (z1 ) with z1 = 1 − e−2iπ/p .
Furthermore, the value of this additive constant must be a fixed number, independent of p.
To determine its value, we can let p → ∞. This gives
f (z )
z1
z2
1
p f (z1 ) = p 1 − e−2πi/p
= p 1 − e−2πi/p
1+
+ 1 + ··· ,
2
3
{z
} z1
|
{z
}|
|
{z
}
z1
→2πi as p→∞
→1 as p→∞
3.2. Some methods for analytic continuation
77
y
z2
z1
z0 =zp
1
x
(a) The centers of each consecutive Taylor
expansions are shown as blue dots.
(b) The imaginary part of each expansion,
drawn within its own domain of convergence.
Figure 3.4. Illustrations of the sequence of Taylor expansion centers in Example 3.3.
which shows that it must evaluate to 2πi. Continuation with the circle-chain method all the
way around the singularity at z = 1 thus did not come back to the same expansion as we
started with, revealing that the singularity was a branch point. Every time we continue in
the clockwise direction around z = 1, this function f (z) increases in value by 2πi.
3.2.2 Schwarz reflection principle
Suppose that we are given a function f (z) that is
1. known to be analytic in some region D (see Figure 3.5), and
2. real on the real axis (or on some part of it).
We showed in Section 2.1.2, last bullet item, that the function g(z) = f (z̄) will then also
be analytic. However, this g(z) is not defined on D but on D (the reflection of D in the
78
Chapter 3. Analytic Continuation
Im z
D
Re z
D̄
Figure 3.5. Schematic illustration of the Schwarz reflection principle.
real axis). Since g(z) agrees with f (z) along a section of the real axis, the two functions
must be the same (Theorem 2.12). The function f (z) has thus been continued from D to
D̄. Another way to formulate the result is as follows.
Theorem 3.4. An analytic function f (z) that is real on any segment of the real axis takes
complex conjugated values at complex conjugated locations in the z-plane.
The result can be generalized significantly. For any analytic function, there will be
some curve(s) along which Im f (z) = 0. If we can find some analytic change of variable
z = z(w) such that the segment corresponds to real values of w, we can apply the reflection
principle to f (z(w)).
3.2.3 Use of a functional equation
We will illustrate this approach with three examples. The first two extend
ˆ ∞
Γ(z) =
e−t tz−1 dt,
0
convergent for Re z > 0, to the whole complex plane.
Example 3.5. Continue the gamma function by utilizing the functional equation
Γ(z + 1) = zΓ(z).
We noted in Section 2.7 that the functional equation follows from integration by parts
in the defining integral and is valid (like the integral for Γ(z)) throughout Re z > 0. If we
3.2. Some methods for analytic continuation
(a) Γ(z) =
´∞
0
79
e−t tz−1 dt converges for Re z > 0.
(b) Using the functional equation Γ(z + 1) = zΓ(z) allows
(after one step) the computation of Γ(z) for Re z > −1.
Figure 3.6. Analytic continuation of Γ(z) using the functional equation given in Example 3.5.
introduce H(z) = Γ(z+1)
, it will thus hold that H(z) = Γ (z) when Re z > 0. Recalling
z
Theorem 2.12, the functions H(z) and Γ (z) are identical. However, with Γ(z) defined for
Re z > 0, H(z) = Γ(z+1)
becomes defined for Re z > −1, and thus so now is Γ(z).
z
This has continued Γ(z) through the strip −1 < Re z ≤ 0 (and we can note that Γ(z) will
have a pole at z = 0); see Figure 3.6. This process can be repeated indefinitely, showing
that Γ(z) exists uniquely, is defined throughout the entire complex plane, and has poles at
z = 0, −1, −2, . . . as its only singularities.
Example 3.6. Continue the gamma function by utilizing the functional equation
Γ(z)Γ(1 − z) =
π
.
sin πz
(3.3)
80
Chapter 3. Analytic Continuation
Different proofs for this equation will be given later (Theorem 5.15 and in the comments
following Theorem 6.2; see also Example 3.14). This relation gives the value of the gamma
function at the location z whenever it is available at a location 1 − z. Whenever Re z < 0
(i.e., we are outside where the integral definition holds), Re (1 − z) ≥ 1, so Γ(1 − z) can
be evaluated. In this case, Γ(z) has in a single step been continued over the whole complex
plane (omitting its poles).
The next example requires a convergence theorem for Dirichlet series.
P∞
Theorem 3.7. The regions of convergence and divergence for a Dirichlet series n=1 annz
are half-planes Re z > α and Re z < α, respectively, with α the least value that makes the
right half-plane singularity-free.
Example 3.8. Continue the Riemann zeta function
ζ(z) =
∞
X
1
z
n
n=1
(3.4)
with use of the functional equation
ζ(z) = 2z π z−1 sin
1
πz Γ(1 − z)ζ(1 − z) .
2
(3.5)
The functional equation (cf. Theorem 6.6) will immediately give the continuation everywhere that Re z ≤ 12 if values are available for Re z ≥ 12 (just as was the case for the
gamma function in the above example). By Theorem 3.7 above, we can find the
Pconver∞
gence boundary by inspecting the sum (3.4) for z = x real. Comparing the sum n=1 n1x
´ ∞ dt
1
shows that (3.4) converges for Re z > 1. What remains for
to the integral 1 tx = x−1
us is therefore to somehow continue (3.4) from Re z > 1 down to Re z ≥ 1/2, after which
(3.5) completes the task of continuation to the whole complex plane.
One idea for this is to turn the sum in (3.4) into an alternating one. From
ζ(z) =
1
1
1
1
1
1
+ z + z + z + z + z + ···
z
1
2
3
4
5
6
follows
2
2
2
2
ζ(z) = z + z + z + · · · .
2z
2
4
6
Subtracting this second series from the first one gives
1
1
1
1
1
1
1 − 21−z ζ(z) = z − z + z − z + z − z + − · · · .
1
2
3
4
5
6
(3.6)
This new sum is again of Dirichlet type, and it therefore suffices to test convergence for
z = x real. Since the sum in this case alternates and the magnitudes of the terms decay for
all x > 0, it will, by Theorem 3.7, also converge for all z with Re z > 0. The continuation
of (3.4) is complete.
As a sideline, we can note that
1 − 21−z
1 − e(1−z) log 2
= lim
= − log 2
z→1 1 − z
z→1
1−z
lim
3.2. Some methods for analytic continuation
81
and
lim
z→1
1
1
1
1
1
1
− z + z − z + z − z + −···
1z
2
3
4
5
6
= log 2,
from which (3.6) gives that limz→1 (z − 1)ζ(z) = 1. One can thus deduce that the singularity of ζ(z) at z = 1 is a first order pole, and then from (3.5) that this pole is the
only singularity of the zeta function in the complex plane (and also that ζ(0) P
= − 12 ). We
∞
can further note a few other values of the zeta function. Evaluating ζ(2) = n=1 n12 is
2
known as the Basel problem, shown by Euler in 1734 to be equal to π6 . Using contour integration, we will obtain this result as a special case of evaluating ζ(2m), m = 1, 2, 3, . . . ,
1
in Example 5.25. With this value for ζ(2), the functional equation gives ζ(−1) = − 12
.
Finding similar closed-form expressions for ζ(k) for k = 3, 5, 7, . . . remains an unsolved
problem.
3.2.4 Partitioning of an integration interval
Example 3.9. Extend the gamma function (again) from its integral definition (2.18), but
without use of a functional relation.
´∞
In the integral definition Γ(z) = 0 e−t tz−1 dt, the upper integration limit causes no
problem (thanks to the fast decay of the exponential function). The limitation Re z > 0
comes from what happens at the origin. Thus, we split the interval in two parts:
ˆ
ˆ
1
−t z−1
Γ(z) =
0
|
e t
{z
dt
}
∞
e−t tz−1 dt
{z
}
+
1
Analytic for Re z > 0; Continue this part
|
.
Analytic for all z; Entire function.
On the short interval [0, 1], the Taylor expansion of e−t converges rapidly, giving
ˆ
ˆ
1
z−1 −t
t
e dt =
0
1
t
0
z−1
∞
X
(−t)n
n!
n=0
!
dt =
ˆ 1
∞
X
(−1)n
tz−1+n dt .
n!
0
n=0
|
{z
}
=
1
z+n
if Re z > −n
So it might not look as if we have achieved anything new. The first term in the sum has
n = 0, so the restriction would still seem to be Re z > 0. However, let us ignore this and
consider the function
η(z) =
∞
X
(−1)n 1
n! z + n
n=0
|
{z
}
ˆ
∞
+
Converges for all z. Poles at z = 0, −1, −2, . . .
1
|
e−t tz−1 dt .
{z
}
(3.7)
Entire function.
Since η(z) = Γ(z) for Re z > 0, we have achieved (again by Theorem 2.12) the continuation of Γ(z) to the whole complex plane.
3.2.5 Replace Taylor coefficients by integrals or sums
Again, we describe the approach by means of an example.
82
Chapter 3. Analytic Continuation
Example 3.10. Continue the function
f (z) =
∞
X
zn
√ .
n
n=1
The Taylor series converges for |z| < 1 and
diverges for |z| > 1. We start by noting that
√
´∞
´∞
2
2
a simple change of variables in the relation 2π = 0 e−t dt gives √1n = √2π 0 e−nt dt.
Therefore
!
ˆ ∞ X
ˆ ∞
ˆ
∞
∞
2 X n ∞ −nt2
2z
dt
2
n −nt2
f (z) = √
z e
dt = √
.
z
e
dt = √
t2 − z
π n=1
π
π
e
0
0
0
n=1
In the last equality, we made use of the fact that the sum inside the integral is a geometric series and therefore can easily be summed in closed form. The function f (z) is now
uniquely defined everywhere away from along z ≥ 1 (where the integral becomes singular,
making it natural to place a branch cut along the real axis from z = 1 to z = ∞).
Section 5.3.2 describes how one can further continue this function in both directions
across the branch cut. Different (related) options are available for generalizing the √1n
coefficients in this example to n1α ; see (3.22) and Exercise 3.3.9.
3.2.6 Subtraction of a similar series or integral
Example 3.11. Continue the function f (z) =
zk
k=0 1+2−k .
P∞
The original series
|z| > 1. For large k, the terms become very close to
P∞ diverges for
1
those of the series k=0 z k = 1−z
. This makes it natural to look at the difference between
these two sums. After minor simplifications,
∞
f (z) −
X (z/2)k
1
=−
= −f (z/2).
1−z
1 + 2−k
(3.8)
k=0
Hence, we stumbled upon a functional equation
f (z) −
1
= −f (z/2).
1−z
(3.9)
Since the original sum converges for |z| < 1, equation (3.9) allows us to compute f (z) for
|z| < 2 and then repeating again for |z| < 4, etc. The function is thereby continued to the
whole complex plane. It will have poles at z = 1, 2, 4, 8, . . . .
1
It is tempting to write (3.9) as f (z) = 1−z
− f (z/2) and then apply this relation
1
1
1
repeatedly, to obtain f (z) = 1−z − 1−z/2 + 1−z/4
− + · · · . However, this sum diverges
for all z. A much better idea is to repeat the original subtraction idea, first on the sum in
(3.8), then on the sum to which this gives rise, etc. After some manipulations, this gives
1
1
1/2
1/4
1/8
−
+
−
+ −··· ,
(3.10)
f (z) = + z
2
1−z
1 − z/2 1 − z/4 1 − z/8
which converges rapidly for all values of z.
3.2. Some methods for analytic continuation
83
y
x
Figure 3.7. Schematic illustration of the region of convergence of a Borel sum.
3.2.7 Borel summation
Assume that we are given a function f (z) defined by a Taylor series f (z) =
with positive radius of convergence R. Then
φ(z) =
P∞
k=0
ak z k
∞
X
ak z k
k=0
k!
is an entire function. Next, we form the function
ˆ ∞
ˆ ∞
∞
X
ak z k tk
dt
η(z) =
e−t φ(zt)dt =
e−t
k!
0
0
k=0
ˆ ∞ −t k
∞
∞
X
X
e t
k
=
ak z
dt =
ak z k = f (z).
k!
0
k=0
{z
} k=0
|
= 1 for all k
This new function η(z) clearly agrees with f (z) within the domain of convergence |z| < R.
But is it a continuation of f (z)? It can be shown that η(z) will converge in the smallest
polygon that can be constructed from the singularities of f (z) in the way that is indicated
in Figure 3.7. In this schematic figure, the red ×’s mark the singularities of f (z) and the
circle shows where the Taylor expansion around the origin will converge. Through each
singularity, we draw a line orthogonal to the direction to the origin. The function η(z) will
converge inside the resulting polygon (marked by solid lines in the figure).
P∞ k
Example 3.12. Apply Borel summation to f (z) =
= 1 + z + z2 + z3
k=0 z
+ ··· .
84
Chapter 3. Analytic Continuation
P∞ k
We get immediately φ(z) = k=0 zk! = ez , and therefore the Borel extension takes
´ ∞ −t
´∞
the form f (z) = 0 e φ(zt)dt = 0 e−t(1−z) dt. This has brought us from f1 (z) to
f2 (z) in Example 3.1. As we noted there, the new form converges for Re z < 1, entirely
agreeing with the description above of where integrals obtained through Borel summation
will converge.
3.2.8 Ramanujan’s formula
Analytic continuation of a Taylor expansion around z = 0 can be viewed as reconstructing
a function globally based on knowing f (0), f 0 (0), f 00 (0), f 000 (0), . . .. If we instead know
the function values f (z) at an infinite sequence of points zk , k = 1, 2, 3, . . . , which has a
finite limit point, there is in theory enough information available to extract all derivatives
at this point.35 If the infinite sequence of points does not have a finite limit point, there will
in general not be a unique solution. However, the following formula by Ramanujan may
nevertheless be applicable if certain additional conditions are met. In case that the values
for f (0), f (1), f (2), f (3), . . . are known, it states that
π
f (−z) =
sin πz
ˆ
∞
tz−1 f (0) − t f (1) + t2 f (2) − + · · · dt
(3.11)
0
(where we may need analytic continuation to interpret the sum f (0) − t f (1) + t2 f (2) −
+ · · · over the full interval 0 ≤ t < ∞).
It is trivial to see that both f (z) ≡ 0 and f (z) = sin πz satisfy f (0) = f (1) = f (2) =
· · · = 0. However, if it is also known that |f (z)| < CeA|z| with A < π holding when
Re z > 0, it can be shown that (3.11) uniquely determines f (z).
In the two examples below, we use the formula somewhat differently: enter known
functions and obtain nontrivial integral formulas. Both cases provide analytic continuations
of integrals that converge only for Re z > 0.
Example 3.13. Substitute f (k) = 1, k = 0, 1, 2, . . . into Ramanujan’s formula
(3.11).
Obviously, f (z) ≡ 1 will satisfy the given data (and also the growth condition for the
right half-plane). Equation (3.11) then gives the relation
ˆ
∞
0
Example 3.14. Substitute f (k) =
tz−1
π
dt =
.
1+t
sin πz
1
k! ,
k = 0, 1, 2, . . . , into (3.11).
We arrive similarly at
ˆ
∞
e−t tz−1 dt =
0
π
.
Γ(1 − z) sin πz
Since we recognize the integral as Γ(z), we have obtained (3.3).
35 cf.
Theorem 2.12.
3.2. Some methods for analytic continuation
85
An alternative version of Ramanujan’s formula36 (with more rapid convergence of the
f (z)
series inside the integral) is obtained by redefining f (z) → Γ(z+1)
:
ˆ
Γ(z)f (−z) =
0
∞
t2
t
tz−1 f (0) − f (1) + f (2) − + · · · dt .
1!
2!
(3.12)
Later in this text, Example 9.32 and Exercise 12.6.6 outline two different (somewhat heuristic) arguments that lead to (3.11) and (3.12).
3.2.9 Padé approximations
This approach frequently becomes an exact continuation procedure if it is applied to the
Taylor expansion of a rational function. However (and in contrast to our previous methods),
it can also be extremely powerful as a computational tool if all that is known of a Taylor
expansion are good approximations for a finite number of leading coefficients (which is
insufficient information for exact analytic continuation). Precise convergence results are
only partially known. A good review of the topic, as well as on its connection to continued
fractions, can be found in Sections 8.3–8.6 of [4].
Taylor expansions (and often also asymptotic expansions; cf. Chapter 12) can often be
accelerated quite dramatically (or turned from divergent to convergent) by being rearranged
into a ratio of two such expansions.
A Padé approximation,
PN
an z n
N
PM
(z) = Pn=0
(3.13)
M
n
n=0 bn z
(normalized by b0 = 1) generalizes the Taylor expansion with equally many degrees of
freedom,
M
+N
X
TM +N (z) =
cn z n ,
(3.14)
n=0
with the two being the same in case M = 0. The Padé coefficients are normally found by
starting from a Taylor expansion
c0 + c1 z + c2 z 2 + · · · =
a0 + a1 z + a2 z 2 + · · ·
1 + b1 z + b2 z 2 + · · ·
and then requiring that both sides match to the highest degree possible at the origin. Multiplying by the denominator gives the following equivalent set of coefficient relations:
a0 = c0 ,
a1 = c1 + c0 b1 ,
a2 = c2 + c1 b1 + c0 b2 ,
a3 = c3 + c2 b1 + c1 b2 + c0 b3 ,
...
(3.15)
With the ci given, each new line introduces two new unknowns, ai and bi . The system
would appear to be severely underdetermined. However, if we specify the degree of the
36 Also
known as Ramanujan’s master theorem.
86
Chapter 3. Analytic Continuation
numerator to be N , of the denominator to be M , and of the truncated Taylor expansion to
be M + N , there will be just as many equations as unknowns (ignoring all terms that are
O(z M +N +1 )). We next solve for all the unknown coefficients, as the following example
shows.
Example 3.15. Given T5 (z), determine P32 (z).
In this case of M = 3, N = 2, M + N = 5, the system (3.15) becomes truncated as
follows:
a0 = c0 ,
a1 = c1 + c0 b1 ,
 a2 = c2 + c1 b1 + c0 b2 ,
 0 = c3 + c2 b1 + c1 b2 + c0 b3 ,
0 = c4 + c3 b1 + c2 b2 + c1 b3 ,
No more a’s available

0 = c5 + c4 b1 + c3 b2 + c2 b3 ,
past limit O(z 2+3+1 ).
No more b’s available
After solving the bottom three equations for b1 , b2 , b3 , the top three explicitly give a0 , a1 ,
a2 . This same idea generalizes to other values of M and N .37
As noted above, a key use of Padé approximations is to extract improved information
from power series expansions with only a limited number of known terms. Transformation
to Padé form usually accelerates convergence, and often allows good approximations to be
found even well outside a Taylor expansion’s radius of convergence (which might even be
R = 0).
Example 3.16. Find the increasing order Padé approximations for f (z) = 1 − z + z 2 −
z3 + − · · · .
The Padé expansions based on the truncated Taylor sums are shown in Table 3.1. The
main diagonal (and the diagonal below it) usually gives the best results. This example is atypical in that f (z) is a rational function, implying that it will produce the exact
result.
Table 3.1. Beginning of Padé table for f (z) = 1 − z + z 2 − z 3 + − · · · .
M - order of
denominator
0
1
2
3
...
0
1
1
1+z
1
1+z
1
1+z
N - order of numerator
1
2
3
1 − z 1 − z + z2 1 − z + z2 − z3
1
1
...
1+z
1+z
1
.
.
.
1+z
...
...
...
...
Example 3.17. Approximate f (2) when we only know the first few terms in the Taylor
, but convergent only if
expansion f (z) = 1 − 12 z + 31 z 2 − 14 z 3 + 15 z 4 − + · · · (= ln(1+z)
z
|z| < 1).
37 It should, however, be noted that this (conceptually simplest) implementation of the Padé method can be
numerically “fragile.” Robust codes need to take precautions for the event of singular (or near-singular) systems.
3.2. Some methods for analytic continuation
87
The Padé table of Table 3.2 is laid out like Table 3.1, but includes only its M = 0
and M = N entries, and for these shows only the numerical values for z = 2 and, in
parentheses, the errors in these compared to 21 log 3 ≈ 0.5493. In spite of z = 2 being well
outside the domain of convergence for the Taylor series of f (z) (as is again visible from
the top M = 0 row in the table), the Padé method allows fast and accurate calculation of
the analytically continued result at z = 2.
Table 3.2. Values for f (2) from Padé approximations; in parentheses is their difference to
0
0
1
(0.4507)
1
M - order of
denominator
2
3
N - order of numerator
1
2
3
0
1.3333
−0.6667
(−0.5493) (0.7840) (−1.2160)
0.5714
(0.0221)
0.5507
(0.0014)
0.5494
(0.0001)
log 3.
...
0.5493
(0.0000)
4
...
4
2.5333
(1.9840)
1
2
...
...
Example 3.18. Compare Taylor- and Padé approximations for the Stieltjes function f (z) =
´ ∞ e−t
dt.
0 1+zt
The integral defining f (z) is singular for z < 0 (real and negative), but it is well defined
for other values of z in the complex plane. Figures 3.8(a)–(b) show the result of a direct
evaluation of f (z). The presence of a branch discontinuity along the negative real axis is
obvious.
We next Taylor expand f (z) around z = 0 (for example, by repeated integration by
parts, or by noting that f (z) satisfies z 2 f 0 (z) + (1 + z)f (z) − 1 = 0, f (0) = 1, and then
equating coefficients). The resulting expansion becomes
f (z) ≈
∞
X
(−z)k k! .
k=0
This diverges for all nonzero values of z (its radius of convergence is R = 0). Truncation
after the sixth power gives
T6 (z) ≈ 1 − z + 2z 2 − 6z 3 + 24z 4 − 120z 5 + 720z 6 .
As seen in Figure 3.9(a), T6 (z) resembles the true Stieltjes function at most in a very small
region close to the origin. However, after converting T6 (z) to its P33 (z) Padé counterpart
P33 (z) =
1 + 11z + 26z 2 + 6z 3
,
1 + 12z + 36z 2 + 24z 3
(3.16)
we see in Figure 3.9(b) a somewhat respectable approximation of the original function.
Being a rational function, and therefore single-valued, the Padé approximation can never
88
Chapter 3. Analytic Continuation
(a) Real part of f (z) =
´∞
0
e−t
1+zt dt.
(b) Imaginary part of f (z) =
´∞
0
e−t
1+zt dt.
Figure 3.8. The real and imaginary parts of the Stieltjes function.
(a) Im T6 (z).
(b) Im P33 (z).
(c) Im T30 (z).
15
(d) Im P15
(z).
Figure 3.9. Displays of the imaginary parts of truncated Taylor and corresponding Padé
approximations of the Stieltjes function, starting from Taylor expansions using terms up through
degrees 6 and 30, respectively.
3.3. Exercises
89
reproduce the branch discontinuity. It has, however, placed its pole singularities along the
negative real axis in an attempt to mimic the branch discontinuity there.38 Finally, Figures
15
3.9(c)–(d) compare T30 (z) with P15
(z). In the case of the Stieltjes function, it can be
proven that the Padé approximations along the M = N diagonal in the Padé table will
converge to the true function exponentially fast (as higher degrees are used) everywhere in
the complex plane away from the negative real axis. Somehow, the everywhere divergent
Taylor expansion (R = 0) does contain complete information about the function it was
based on, and the Padé approach allows this information to be recovered.
3.3 Exercises
Exercise 3.3.1. The Taylor expansions of a function f (z) around z = α and z = β,
respectively, take the forms
f (z) = a0 + a1 (z − α) + a2 (z − α)2 + a3 (z − α)3 + · · ·
= b0 + b1 (z − β) + b2 (z − β)2 + b3 (z − β)3 + · · · .
If we define δ as δ = β−α, show that we can obtain the b-coefficients from the a-coefficients
by



 
a0
b0
1 1δ 1δ 2 1δ 3 1δ 4 · · ·
 b1   0 1 2δ 3δ 2 4δ 3 · · ·   a1 



 

 b2   0 0

1
3δ 6δ 2 · · · 
  a2 

 
,


 b3  =  0 0
0
1
4δ · · ·   a3 


 


 b4   0 0

0
0
1 · · ·   a4 

 
.. ..
..
..
..
..
..
..
.
.
. .
.
.
.
.
where the coefficients for the powers of δ are given by Pascal’s triangle.
Hint: Let t = z − β = z − α − δ, i.e., z − α = t + δ. Equate coefficients for powers of t.
Exercise 3.3.2. Given a Taylor series c1 z + c2 z 2 + c3 z 3 + · · · , setting z =
rearranging it in powers of t gives a1 t + a2 t2 + a3 t3 + · · · , where
a1
a2
a3
a4
t
1+t
and
= c1 ,
= c2 − c1 ,
= c3 − 2c2 + c1 ,
= c4 − 3c3 + 3c2 − c1 ,
..
.
etc. (with coefficients from Pascal’s triangle).
(a) Verify the result above.
(b) Already, Newton used the result above to rewrite the expansion
arctan z = z −
z5
z7
z3
+
−
+ −···
3
5
7
(3.17)
38 In cases of branch points (such as in this example), “quadratic” Padé approximations [15] are generally
preferable over the “linear” Padé approximations described here.
90
Chapter 3. Analytic Continuation
into
z
arctan z =
1 + z2
2
1+
3
z2
1 + z2
2·4
+
3·5
z2
1 + z2
2
2·4·6
+
3·5·7
z2
1 + z2
!
3
+ ···
(3.18)
Compare the regions in which (3.17) and (3.18) converge. Did this rewrite amount to an
analytic continuation of the original series?
P∞
Exercise 3.3.3. Consider the Taylor expansion f (z) = n=0 (−z)n .
(a) Determine the radius of convergence R for the Taylor series for f (z).
(b) Sum the Taylor series in closed form.
(c) Based on the result in part (b), show that the Taylor expansion of f (z) around
n+1
n
P∞
z = 21 becomes f (z) = n=0 (−1)n 23
z − 21 .
(d) From the expansion coefficients obtained in part (c), determine the radius of convergence for the new expansion. Draw where the two expansions converge in the complex
plane, and mark the point z = −1.
For parts (e) and (f) below, use only the original Taylor series coefficients for f (z) (i.e.,
not the closed-form expression for f (z)) to re-expand f (z) around the point z = 12 . You
m+1
P∞
1
k!
= (−1)m 23
can assume it is known that k=m (−1)k m!(k−m)!
.
2k−m
n
d
(e) Obtain the new Taylor coefficients by working out the values for dz
n f (z), n =
1
0, 1, 2, . . . , at z = 2 .
(f) Substitute z → ζ + 12 in the original Taylor series, rearrange the result into powers
of ζ, and then change the variable back ζ → z − 12 .
P∞
k
Exercise 3.3.4. Consider the function f (z) = k=0 z (2 ) :
(a) Determine the radius of convergence of the Taylor expansion for f (z).
(b) Show that f (z) satisfies the functional equation
f (z) = z + f (z 2 ).
(c) By means of the functional equation, show that the function cannot be continued
outside the Taylor series radius of convergence (i.e., its boundary forms a natural boundary
for the function itself).
Pm−1 k
m
Hint: From the functional equation, first deduce that f (z) = f (z (2 ) )+ k=0 z (2 ) holds
for m = 1, 2, 3, . . . . The result then follows by a similar argument to the one used in Example 3.2.
Comment: In the previous example and in this exercise, the ratio between successive exponents are k!/(k − 1)! = k and 2k /2k−1 = 2, respectively. The Ostrowski–Hadamard gap
theorem tells one that if R = 1 and the ratio of successive (integer) exponents is bound from
below by a constant greater than 1, the unit circle always becomes a natural boundary.
Exercise 3.3.5. This exercise supplements Example 3.18.
(a) Determine the radius of convergence of the Taylor series
f (z) =
∞
X
(−1)k k!z k .
k=0
(3.19)
.
3.3. Exercises
91
(b) Use Borel summation to obtain from (3.19) the formula
ˆ ∞ −t
e
f (z) =
dt,
1
+ zt
0
(3.20)
and determine the range of validity for this integral representation of f (z).
(c) Use instead the method from Section 3.2.5 to´ obtain (3.20) from (3.19). Hint: Note
∞
from the definition of the gamma function that k! = 0 e−t tk dt.
(d) Starting from (3.20), derive (3.19) by repeated integration by parts.
(e) Again, start from (3.20) and arrive at (3.19) by the different method of first showing that (3.20) implies that f (z) satisfies the ODE
z 2 f 0 (z) + (1 + z)f (z) − 1 = 0
(3.21)
with initial condition f (0) = 1. Then assume that f (z) has a Taylor expansion
f (z) = a0 + a1 z + a2 z 2 + a3 z 3 + · · ·
and use (3.21) to determine the unknown coefficients.
Exercise 3.3.6.
(a) By a change of variables in the integral definition of the gamma function, derive
ˆ ∞
1
1
=
tz−1 e−nt dt.
(3.22)
nz
Γ(z) 0
(b) Following one of the analytic continuation methods, show that the Riemann zeta
function
∞
X
1
ζ(z) =
z
n
n=1
can be rewritten as
ζ(z) =
1
Γ(z)
ˆ
0
∞
tz−1
dt.
et − 1
(3.23)
(c) Determine for what values of z the integral in part (b) converges.
Exercise 3.3.7. By combining the results in (3.23) and (3.12) with the definition of Bernoulli
numbers in Exercise 2.9.23, make a plausible argument that arrives at the formula
ζ(−k) = (−1)k
Bk+1
, k = 0, 1, 2, . . . ,
k+1
giving the values of the zeta function for nonpositive integer arguments.
Comment: Combining equations with little regard for convergence, etc., is not mathematically rigorous, but is nevertheless often extremely helpful in terms of “spotting” relations.
We will later in this text (Section 6.2) arrive rigorously at this same relation.
P∞
Exercise 3.3.8. Let f (z) = k=0 (k + 1) z k .
1
(a) Show by some direct means that f (z) = (1−z)
2.
Apply Borel summation to the Taylor series, to obtain φ(z) = (1 + z) ez and η(z) =
´ ∞ (b)
−t (1−z)
e
(1 + zt)dt.
0
92
Chapter 3. Analytic Continuation
(c) Show that this integral converges for Re z < 1 and diverges for Re z ≥ 1.
1
(d) Evaluate η(z) for Re z < 1, to then obtain η(z) = (1−z)
2.
Exercise 3.3.9.
(a) Show that the identity in Example 3.10 can be generalized as
α−1
ˆ 1
1
1
1
=
log
xn−1 dx.
nα
Γ (α) 0
x
α−1 dt
´1
P∞
n
z
(b) Use part (a) to show that n=1 nz α = Γ (α)
log 1t
1−z t .
0
Exercise 3.3.10.
P∞ √
(a) Derive for f (z) = n=1 n z n the analytic continuation
ˆ ∞
2
2z
ex
f (z) = √
dx.
π 0 (ex2 − z)2
Hint: Start with the result in Example 3.10.
(b) Verify the result by Taylor expanding the integrand around z = 0, and then integrate term-by-term.
Exercise 3.3.11. Carry out the algebra steps that produce (3.10).
Exercise 3.3.12. Padé approximations are somewhat “fragile” in that they sometimes cannot give the expected order of accuracy. Let T (z) = c0 + c1 z + c2 z 2 + c3 z 3 + · · · .
Determine the condition(s) we need to impose on these coefficients in order to obtain the
expected accuracy in the cases of
(a) P11 (z),
(b) P12 (z),
(c) P21 (z).
(d) Show that the most accurate P12 (z) approximation in the case of c2 = 0 is P12 (z) =
c0 + c1 z. Is that approximation unique?
Exercise 3.3.13. In some cases, closed-form expressions are available for Padé approxiPn
(z)
(2n−k)!n!
, where pn (z) = k=0 (2n)!k!(n−k)!
z k . Verify by
mations. For example, Pnn (ez ) = ppnn(−z)
direct evaluation that this indeed matches the Taylor expansion to the expected number of
coefficients in the cases of n = 0, 1, 2.
Exercise 3.3.14. In one of the most famous letters in the history of mathematics (dated
January 16, 1913), the Indian mathematical genius Srinivasa Ramanujan (1887–1920) introduces himself to the English mathematician G. H. Hardy (1877–1947) and then gives
some stunning examples of original formulas. The letter also contains the seemingly whimsical statements
1 − 1! + 2! − 3! + 4! − + · · · ≈ 0.596
and
1
.
12
Based on the examples in this chapter, explain what lies behind both of these results.
1 + 2 + 3 + 4 + ··· = −
Chapter 4
Introduction to Complex
Integration
Real-valued calculus centers around differentiation and integration. In the complex case,
ability to differentiate led to the concept of analytic functions. Such functions also have
very special properties when it comes to integration. Various integration theorems will
become key tools for establishing a wide range of general results for analytic functions.
This chapter provides an introduction to the topic of complex integration, leaving the key
technique of residue calculus to Chapter 5.
Real-valued case:
In real variable calculus, a definite integral is defined by
ˆ
b
f (x)dx = lim
X
∆xk →0
a
f (xk )∆xk ,
(4.1)
k
as illustrated in Figure 4.1(a).39 The key theorem states that if there is a function F (x)
´b
d
such that dx
F (x) = f (x), then a f (x)dx = F (b) − F (a). Finding such a function F (x)
(which may not exist in terms of elementary functions) is then often the only means available for evaluating integrals analytically.
Complex-valued case:
In the complex plane, we consider instead
ˆ
zb
f (z)dz = lim
za
∆zk →0
X
f (zk )∆zk ;
(4.2)
k
´z
cf. Figure 4.1(b). This integral can similarly be evaluated as zab f (z)dz = F (zb ) − F (za ),
d
F (z) = f (z) (as shown in Theorem 4.1 below). Although the definition (4.2)
where dz
39 In (4.1) and similarly later in (4.2), the numerical values for ∆x and ∆z are the differences between the
k
k
end and the start values for each of the pieces that the integration path has been divided into. These pieces contain
the values xk and zk , respectively.
93
94
Chapter 4. Introduction to Complex Integration
y
y = f (x)
∆x k
a
b
x
(a) Integration in a real variable.
Im z
zb
Γ
Re z
za
∆zk
(b) Integration in a complex variable.
Figure 4.1. Conceptual difference between integrating in a real and a complex variable.
does not require f (z) to be analytic, two major novelties arise if this is the case:
(i)
As long as F (z) is single-valued, it is unnecessary to specify the integration
path Γ (also known as contour) that was followed from za to zb .40
(ii)
By utilizing this path independence, it will turn out that many definite integrals can be evaluated exactly, without us needing to obtain any closed-form
expression for F (z) (the topic of residue calculus in Section 5.1).
40 Continuously
deforming one path into another, it is required that f (z) remains analytic also along the intermediate ones. This can be expressed as the domain of analyticity being simply connected.
4.1. Integration when a primitive function F (z) is available
95
4.1 Integration when a primitive function F (z) is
available
Theorem 4.1. If there exists a function F (z) such that F 0 (z) = f (z), then
F (zb ) − F (za ).
´ zb
za
f (z) dz =
Proof.
d
dz
ˆ
´ z+∆z
z
za
f (ξ)dξ = lim
f (ξ)dξ −
za
f (ξ)dξ
∆z
∆z→0
za
´z
´ z+∆z
f (ξ)dξ
∆z
z
= lim
∆z→0
= lim
∆z→0
f (z)∆z
∆z
= f (z)
=
d
F (z).
dz
Therefore,
d
dz
ˆ
z
F (z) −
f (ξ)dξ
= 0,
za
which implies
ˆ
z
F (z) −
f (ξ)dξ = C,
za
where C is a constant. Setting z = za , we get C = F (za ), and setting z = zb , we obtain
ˆ
zb
f (ξ)dξ = F (zb ) − F (za ).
za
Next, consider a closed loop as in Figure 4.2, where za = zb . If F (z) exists and is
single-valued
inside and on Γ, it will follow from za = zb that F (zb ) − F (za ) = 0, i.e.,
¸
that Γ f (ξ)dξ = 0.
We made above the assumption that F (z) exists. This can be justified by considering a
circle-chain continuation of f (z) along the integration path. We can then integrate each of
these Taylor expansions (with one free constant, as to be expected). Recall that the radius
of convergence never changes if we integrate or differentiate a Taylor series. So F (z) exists
along a path if f (z) exists along it.
We have not yet proven that a Taylor series converges
up to the nearest singularity. In
¸
fact, the proof for that will use the properties of f (z)dz. Hence, we would like to arrive
at the result
˛
f (z)dz = 0,
Γ
96
Chapter 4. Introduction to Complex Integration
Im z
Γ
Re z
za = z b
Figure 4.2. Integration along a closed contour Γ is denoted by
¸
Γ
f (z)dz.
where Γ is a closed loop containing no singularity, without having to invoke the existence
of F (z). This is accomplished in the next theorem.
4.1.1 Cauchy’s and Morera’s theorems
¸
Theorem 4.2 (Cauchy). Γ f (z)dz = 0 when f (z) is an analytic function41 that is
singularity-free on and inside the contour Γ.
Proof.
˛
˛
f (z)dz =
Γ
Γ
˛
(u + iv) (dx + idy)
| {z } | {z }
f (z)
dz
˛
(udx − vdy) + i
=
Γ
(udy + vdx).
Γ
Recall from multivariable calculus Green’s theorem the following: Let g(x, y) and h(x, y)
and their first partial derivatives be continuous in a domain D bounded by a closed loop Γ.
Then,
˛
¨ ∂h ∂g
(gdx + hdy) =
−
dxdy.
∂x ∂y
Γ
D
41 For
the proof, we need that f (z) satisfies the C-R equations and that f 0 (z) is continuous.
4.2. Contour integration
In our case,
˛
97
˛
˛
(udx − vdy) + i (udy + vdx)
Γ
Γ
¨ ¨ ∂u
∂u ∂v
∂v
+
dxdy + i
−
dxdy
=−
∂x ∂y
∂x ∂y
D
D
|
{z
}
|
{z
}
f (z)dz =
Γ
=0 by C-R
(4.3)
=0 by C-R
=0
Morera’s theorem is the reverse to Cauchy’s theorem.
¸
Theorem 4.3 (Morera). If f is continuous and if Γ f (z)dz = 0 for any contour Γ in the
domain, then f (z) is an analytic function within the domain.
The main themes in this and the next chapter will be the following:
¸
• Since evaluating Γ f (z)dz along certain contours Γ might be problematic, we will
learn how to change paths to make the evaluations easier.
• We will see what happens when the closed loop Γ actually does contain a singularity.
´b
• If we want to evaluate a real integral a f (x)dx, we will see how to modify the path
to utilize properties in the complex plane.
Combinations of these techniques will lead to very powerful methods to evaluate many
definite integrals also in cases where we cannot find F (z). These methods also lead to
accurate estimates of integrals in various singular limit situations, as discussed in Chapter
12: Steepest Descent for Approximating Integrals.
4.2 Contour integration
Example 4.4. Evaluate
ˆ
z̄ dz,
Γ
where Γ is the path shown in Figure 4.3.
We first note that f (z) is NOT an analytic function, so Morera’s theorem tells us that
the result will be path dependent. Also, the contour is not a closed loop, so Green’s theorem
(as used in the proof of Theorem 4.2) is not available to us. We thus need to work out this
integral one line segment at a time. Here,
f (z) = z̄
= x − iy,
so
ˆ
ˆ
z̄dz =
Γ
ˆ
1
xdx +
0
0
1
x2
(1 − iy)(idy) =
2
1
1
y2
+i y−i
= 1 + i.
2 0
0
If we do continue the path from 1+i first to i and then back to the origin (thus going around
a unit square), we pick up an additional i − 1, giving for the full path around the square
98
Chapter 4. Introduction to Complex Integration
Im z
1+i
0
Re z
1
Figure 4.3. Path for the integral in Example 4.4.
the value 2i. For this closed loop, we could also have used (4.3) (skipping the last step of
applying the C-R equations), giving
ˆ
¨
¨
z̄ dz = −
(0 + 0)dxdy + i
(1 + 1)dxdy = 2i.
2
2
2
Example 4.5. Evaluate
ˆ
z n dz,
Γ
where n is an integer and Γ is any closed contour that goes around the origin once in the
positive (counterclockwise) direction.
We solve this problem in two ways:
1. by changing the path and evaluating the integral explicitly around a simplified path,
and
2. by using the primitive form F 0 (z) = f (z).
If n ≥ 0, then f (z) is analytic inside Γ, which
leads to Γ z n dz = 0. However, if n < 0, then there is a pole at the origin. In that case, one
can consider a different contour, such as the one shown in Figure 4.4(b) (note the positive
direction around the origin for both Γ and γ). This new contour excludes the origin and
thus the pole. It will integrate to zero:
ˆ
˛
ˆ
ˆ
ˆ f (z)dz =
+
− +
f (z)dz
(4.4)
Method´1: Modifying the contour:
Γ
←
γ
→
=0.
´Since one can choose the paths → and ← arbitrarily close to each other,
f (z)dz cancel out. Therefore,
→
˛
˛
f (z)dz =
f (z)dz,
γ
Γ
´
←
f (z)dz and
4.2. Contour integration
99
Γ
Γ
γ
(a) Path for the integral in Example 4.5.
(b) Modified contour, with a red dot marking
the pole at the origin.
Figure 4.4. Different integration paths considered in Example 4.5. Note in part (b) that we
have marked the path direction for the small circle in the positive direction (causing the minus sign
in (4.4)).
where both are followed in the positive direction. The argument shows that we, in general,
can move the contour as we want without changes to the value of the integral, as long
we don’t move it across any singular point. Instead of following Γ, we can thus integrate
around the circle γ centered at the origin and with radius R. Let z = Reiθ , and therefore
dz = iReiθ dθ. Then
˛
˛
n
z dz =
z n dz
Γ
γ
ˆ
2π
Rn einθ iReiθ dθ
=
0
ˆ
2π
= iRn+1
ei(n+1)θ dθ
0
h
i2π
1
ei(n+1)θ
i(n + 1)
0
= 0 if n 6= −1.
= iR
n+1
However, if n 6= −1, then
˛
˛
z −1 dz =
Γ
z −1 dz
γ
ˆ
2π
R−1 e−iθ iReiθ dθ
=
0
ˆ
2π
=
idθ
0
= 2πi.
100
Chapter 4. Introduction to Complex Integration
Modifying contours in this fashion is an often used technique, which generalizes straightforwardly to cases with many singularities, as will be illustrated later in Figure 5.1.
Let f (z) = z n . If n 6= −1, then
F (z) =
Therefore, after having followed the closed contour Γ, F (zend ) = F (zstart ),
since F (z) is single-valued. Therefore, in the n 6= −1 case,
˛
z n dz = 0.
Method 2: Finding the primitive function F (z):
z n+1
n+1 .
Γ
However, if n = −1, then F (z) = log(z). After having gone through one cycle of the
contour,42 the value of F (z) must have increased by 2πi. Therefore,
˛
z n dz = 2πi.
Γ
We arrive at the same conclusion as before.
The above result can alternatively be formulated as follows.
Theorem 4.6. For n integer,
1
2πi
˛
Γ

 0
dz
0
=

(z − z0 )n
1
if z0 is outside Γ,
if z0 is inside Γ and n 6= 1,
if z0 is inside Γ and n = 1.
The case of n = 1 and z0 right on the curve Γ will be discussed in Section 5.1.4.
4.2.1 Cauchy’s argument principle
Example 4.7. Let p(z) = A(z − a1 )(z − a2 ) · · · · · (z − an ) be a polynomial of degree
n.43 Evaluate the following integral:
˛ 0
1
p (z)
I=
dz .
2πi Γ p(z)
We obtain
p0 (z)
d
=
log (p(z))
p(z)
dz
d
=
(log A + log(z − a1 ) + log(z − a2 ) + · · · + log(z − an ))
dz
1
1
1
=
+
+ ··· +
.
z − a1
z − a2
z − an
(4.5)
This implies that the value of I equals the number of roots lying inside Γ. We can tell the
number of roots inside a contour by inspecting the polynomial only along the contour.
This result generalizes in various ways, to Cauchy’s argument principle (next) and to
Rouché’s theorem (Theorem 5.21); see also the Gauss–Lucas theorem (Exercise 5.6.38).
42 Without
43 We
having placed any branch cut in its way.
will soon see (Section 4.2.4) that any polynomial of degree n indeed can be factored in this way.
4.2. Contour integration
101
Theorem 4.8 (Cauchy’s argument principle). If f (z) is a meromorphic function (its
only singularities are poles), and if it has N zeros and, as only singularities, P poles
(accounting for their multiplicities) inside the contour Γ, then
˛ 0
1
f (z)
N −P =
dz.
2πi Γ f (z)
Proof. We write f (z) as
QN
(z − ak )
g(z),
f (z) = Qk=1
P
k=1 (z − bk )
where g(z) has neither zeros nor poles within Γ. Then
f 0 (z)
d
=
log (f (z))
f (z)
dz
d
=
dz
=
N
X
k=1
N
X
log(z − ak ) −
k=1
P
X
!
log(z − bk ) + log g(z)
k=1
P
X 1
1
−
+
z − ak
z − bk
function that is
singularity-free within Γ
k=1
,
and the result follows when integrating around Γ.
We will revisit this result in Section 5.1.6 describing Rouché’s theorem and winding
numbers.
4.2.2 Cauchy’s integral formula
As a preliminary to Cauchy’s integral formula, let us note the following estimate, which
often turns out to be useful.
Theorem 4.9. The integral along a path Γ can be bounded from above by
ˆ
f (z)dz ≤ M · L,
Γ
where M is the upper bound of |f (z)| on Γ and where L is the length of Γ.
Proof.
ˆ
f (z)dz =
Γ
lim
X
∆zk →0
≤ lim
∆zk →0
X
f (zk )∆zk
|f (zk )| |∆zk |
| {z }
≤M
X
≤ M lim
|∆zk |
∆zk →0
|
{z
}
=L
= M · L.
102
Chapter 4. Introduction to Complex Integration
Theorem 4.10 (Cauchy’s integral formula). Let f (z) be analytic inside Γ. Then, for z
inside Γ,
˛
f (ξ)
1
dξ.
(4.6)
f (z) =
2πi Γ ξ − z
Significance
1. We have mentioned before that f (z) is completely determined by its values along
any curve segment (Theorem 2.12). So the existence of a formula of this kind is not
altogether surprising. If we know its values along a whole closed contour, Cauchy’s
integral formula provides its values at points inside the contour in a remarkably simple way.
2. We will soon deduce from (4.6) that if f (z) is analytic, it is not just once but infinitely
many times differentiable.
3. This formula forms the foundation for a much more solid understanding of Taylor
expansions, and their generalization to Laurent expansions.
Proof. As in Figure 4.4(b), we modify our contour Γ to become a circle centered at z with
radius δ, and call this new contour γ. Thus
˛
˛
1
f (ξ)
1
f (ξ)
dξ =
dξ
2πi Γ ξ − z
2πi γ ξ − z
˛
1
(f (ξ) − f (z)) + f (z)
=
dξ
2πi γ
ξ−z
˛
˛
1
(f (ξ) − f (z))
1
dξ
=
.
dξ +f (z)
2πi γ
ξ−z
2πi γ ξ − z
{z
}
{z
}
|
|
=1
=Iδ
Consider the first of the integrals Iδ , and let δ become arbitrarily small. Then, the continuity
of f implies that |f (ξ) − f (z)| < ε whenever |z − ξ| < δ. Thus (by Theorem 4.9)
ε
· 2πδ
|Iδ | ≤
δ |{z}
|{z}
L
M
= ε2π
→ 0 when δ → 0.
Therefore,
1
2πi
˛
Γ
f (ξ)
dξ = f (z).
ξ−z
Theorem 4.11. If f (z) is analytic (once complex differentiable) in some region, it can then
be differentiated any number of times.
Proof. The RHS of (4.6) clearly possesses any number of derivatives, and then so must the
LHS. In fact, we obtain the following formula for the kth derivative:
˛
k!
f (ξ)
(4.7)
f (k) (z) =
dξ.
2πi Γ (ξ − z)k+1
4.2. Contour integration
103
4.2.3 Liouville’s theorem
Theorem 4.12. If f (z) is entire (no singularity for finite z) and bounded (|f (z)| ≤ M ),
then f (z) ≡ C for C constant.
Proof. Equation (4.7) becomes for the first derivative (k = 1)
1
f (z) =
2πi
˛
0
Γ
f (ξ)
dξ
(ξ − z)2
and we choose as the contour a circle of radius R centered at the point z. Then,
˛
f (ξ)
1
dξ
2π Γ (ξ − z)2
1 M
M
≤
2πR =
.
2
2π R
R
|f 0 (z)| =
Since M is a constant and R can be arbitrarily large, this inequality is a contradiction unless
f 0 (z) = 0. From this it follows then that f (z) must be a constant.
By considering higher k-values (than k = 1) in (4.7), we obtain the following generalized result.
Theorem 4.13. If f (z) is entire and satisfies a bound |f (z)| ≤ A |z|n for n a nonnegative
integer, then f (z) is a polynomial of degree at most n.
4.2.4 The fundamental theorem of algebra
Theorem 4.14. A polynomial p(z) of degree n ≥ 1 has at least one root.
Proof. Suppose that p(z) has no root and that n ≥ 1. We wish to construct a contradiction.
1
. It satisfies the two conditions needed to apply
To this end, we consider f (z) = p(z)
Liouville’s theorem:
1.
1
p(z)
is an entire function (since p(z) has no root).
2.
1
p(z)
is bounded since |p(z)| → ∞ as |z| → ∞.
1
By Liouville’s theorem, p(z)
and therefore also p(z) must be constant. This violates the
assumption that p(z) is a polynomial of degree n ≥ 1. The hypothesis that p(z) has no root
must therefore have been false.
After dividing out the root we now know must have existed, the argument can be repeated.
Theorem 4.15. A polynomial pn (z) of degree n > 0 will always have exactly n roots (not
necessarily at different locations from each other).
104
Chapter 4. Introduction to Complex Integration
4.2.5 Mean value theorem
Theorem 4.16. The mean value theorem:
ˆ 2π
1
f (z) =
f (z + Reiθ )dθ
2π 0
(4.8)
(assuming no singularity inside the circle).
Proof. Use Cauchy’s integral formula
f (z) =
1
2πi
˛
Γ
f (ξ)
dξ
(ξ − z)
and let the contour be a circle centered at z with radius R. A change of variables yields
ξ − z = Reiθ and dξ = iReiθ dθ. Thus
ˆ 2π
1
f (z) =
f (z + Reiθ )dθ.
2π 0
This last theorem states that the value of an analytic function at any interior point z is
the average of the function integrated around a circle centered at z. One can separate the
real from the imaginary part in the above equation, so the theorem holds also for Re f (z)
and for Im f (z) separately.
4.2.6 Revisiting max/min theorems
Theorem 4.17 (max/min theorems). u(x, y) = Re f (z) and v(x, y) = Im f (z) cannot
have a local (finite) maximum or minimum.
Proof. Assume that the real part u(x, y) reaches a local maximum at some point z. However, the value of u at that point is the average of the values of u around a circle centered
at z. This is a contradiction.
Theorem 4.18. The magnitude |f (z)| of an analytic function cannot attain a local minimum or maximum except at a z-value for which |f (z)| = 0 and at a singular point.
Proof. Assume that |f (z)| attains a local min/max that it is neither 0 nor ∞. Then
log(|f (z)|) also attains a local min/max, since log (for a real-valued argument) is a monotonic function. However, log(|f (z)|) is the real part of the function log(f (z)), which is
analytic. By the theorem above, this is impossible.
Somewhat related to this theorem is whether anything can be said if it only is known
that a function f (z) is analytic (without singularities) in a half-plane and bounded along its
edge.
Theorem 4.19 (Phragmén–Lindelöf). Let f (z) be analytic (singularity-free) in the right
half-plane Re z ≥ 0 and |f (iy)| < M (constant) along the imaginary axis. Then, if
β
|f (z)| < K e|z| for some constant K and β < 1, in the half-plane, it will hold everywhere
across it that |f (z)| < M .
4.2. Contour integration
105
γ
Proof. Define F (z) = f (z) e−ε z , where ε > 0 is arbitrarily small, and β < γ < 1.
The maximum principle tells us that |F (z)| < M throughout the half-plane, and the result
follows by letting ε → 0.
Roughly speaking, the Phragmén–Lindelöf theorem tells us that the function, bounded
along the imaginary axis, must either grow exponentially fast or stay bounded in the halfplane—nothing in between can happen. There are many theorems of this general character,
for example for functions that are bounded along two intersecting lines z = r eiθ1 and
z = r eiθ2 , r ∈ [0, ∞].
4.2.7 Poisson’s integral formula
Theorem 4.20. If f (z) is analytic inside a circle C, which we here take to be the unit circle,
its value at an arbitrary point z = r eiθ , 0 ≤ r < 1, is given from the values along C by
ˆ 2π
1
1 − r2
iθ
f (r e ) =
dt.
(4.9)
f (eit )
2π 0
1 + r2 − 2r cos(t − θ)
We note the following:
1−r 2
(i) The Poisson kernel 1+r2 −2r
cos(t−θ) is purely real-valued for all real values of r, t, θ.
(ii) If we write f = u + iv with u and v real, then (4.9) will hold separately for the
two harmonic functions u(x, y) and v(x, y). Similarly, if a real-valued function u(x, y) is
2
2
harmonic, i.e., continuous44 and obeying ∂∂xu2 + ∂∂yu2 = 0, it will satisfy (4.9).
(iii) Setting r = 0, equation (4.9) reduces to the mean value theorem (Theorem 4.16)
(see (4.8)).
A proof for Theorem 4.20 is given in Section 4.4.
4.2.8 Revisiting Taylor expansions
Radius of convergence
In the following two theorems, we consider for notational simplicity expansions centered
at the origin.
Theorem 4.21. If f (z) =
|z| < |z0 |.
P∞
j=0 bj z
j
converges for z0 6= 0, then it also converges for all
Significance: The domain of convergence for a Taylor series must be a circular disc.
Anything else would become a contradiction.
P∞
Proof. Assume that f (z) = j=0 bj z j converges for z0 6= 0. Then, from some index J
and onwards (j ≥ J), it will hold that |bj z0j | < 1. Also from this index,
|bj z j | = |bj z0j |
<
z
z0
= Mj
44 Recall
4
f (z) = e−1/z from Section 2.1.1.
j
z
z0
j
106
Chapter 4. Introduction to Complex Integration
P
with |M | < 1, since |z| < |z0 |. Since PMj converges, by the Weierstrass M-test theorem
∞
(Theorem 2.39), we can conclude that j=0 bj z j converges.
Theorem 4.22. If f (z) is analytic for |z| ≤ R, then its Taylor series f (z) =
with bj =
f (j) (0)
j!
P∞
j=0 bj z
j
converges to f (z) at each point |z| < R.
Significance: This theorem implies that a Taylor series always converges in the greatest
circle up to the nearest singularity.
Proof. Let Γ be the circle of radius R, and let |z| < R. By Cauchy’s integral formula,
˛
1
f (ξ)
f (z) =
dξ
2πi Γ ξ − z
!
˛
1
1
1
=
f (ξ)
dξ.
2πi Γ
ξ 1 − zξ
z
ξ
Since |ξ| = R and |z| < R, it holds that
1
ξ
1
1−
< 1, which implies
!
z
ξ
=
∞
X
zj
j+1
ξ
j=0
is a convergent series. Thus,
˛
∞ ∞
X
X
1
f (j) (0) j
f (ξ)dξ
j
f (z) =
z
=
z
2πi Γ ξ j+1
j!
j=0 |
j=0
{z
}
=bj =
(4.10)
f (j) (0)
j!
is also convergent.
The largest number R for which the power series converges inside the disk |z| < R is
called the radius of convergence.
4.3 Laurent series
A
series is a natural extension of a Taylor series. The Taylor series takesP
the form
PLaurent
∞
∞ aj
1
j
a
z
and
converges
when
|z|
<
R.
If
we
define
t
=
,
we
obtain
a
series
j=0 j
j=0 tj
z
P
∞
which converges for |t| > R1 . If we are given a sum j=−∞ cj z j , it would be natural to
split it:




∞
∞
−1
X
X
X
cj z j = 
cj z j  + c0 + 
cj z j  ,
j=−∞
j=−∞
j=1
for which the region of convergence could be expected to be a circular annulus centered at
the origin, with a singularity on each of its bounding circles; cf. Figure 4.5.
Theorem 4.23. If f (z) is analytic in an annulus, then there is a Laurent series corresponding to it, converging everywhere inside the annulus.
4.3. Laurent series
107
Im z
z2
z1
I
II
III
IV
z0
z3
Re z
Figure 4.5. Regions of validity for the Taylor (I) and Laurent (II, III, and IV) expansions
centered at z0 with singular points z1 , z2 , and z3 .
Proof. Consider the annulus shaded in gray in Figure 4.6, and the integration path C =
C1 + C2 , oriented as shown, and with the radial segments adjacent to each other, i.e., with
canceling integrals. Let z be a point inside the annulus, and choose radii r1 and r2 such
that R1 < r1 < |z| and R2 > r2 > |z|. By Cauchy’s integral theorem,
ˆ
1
f (ξ)
f (z) =
dξ
2πi C ξ − z
ˆ
ˆ
f (ξ)
1
f (ξ)
1
dξ +
dξ.
=
2πi C1 ξ − z
2πi C2 ξ − z
For the first integral, |z| > |ξ|, and
1
1
=−
ξ−z
z
1
1−
!
ξ
z
1
ξ
ξ2
= − − 2 − 3 − ···
z
z
z
converges. Likewise, for the second integral, |ξ| > |z|, and
1
1
=
ξ−z
ξ
1
1−
!
z
ξ
=
z
1
z2
+ 2 + 3 + ···
ξ
ξ
ξ
converges. Substituting these into (4.11) gives
f (z) = −
ˆ
ˆ
−1 +∞ X
X
1
f (ξ)
1
f (ξ)
n
dξ
z
+
dξ
zn.
n−1
n−1
2πi
ξ
2πi
ξ
C1
C2
n=−∞
n=0
(4.11)
108
Chapter 4. Introduction to Complex Integration
Im z
R2
r2
C1
C2
r1
R1
Re z
z
Figure 4.6. Paths used in the derivation of the Laurent series theorem (Theorem 4.23).
Here C1 was followed in the negative (clockwise) direction. If we follow both contours in
the positive direction, no further change occurs if we change both paths to become a single
path going around once counterclockwise within the annulus. We rename it C.
4.3.1 Summary of Taylor versus Laurent series around a point z0
Taylor series
A Taylor series is valid out to the nearest singularity
∞
X
1
f (z) =
an (z − z0 ) with an =
2πi
n=0
ˆ
n
C
f (ξ)
f (n) (z0 )
dξ
=
(ξ − z0 )n+1
n!
(4.12)
for any path C going around z0 once counterclockwise, inside the region of analyticity.
Laurent series
A Laurent series is valid in an annulus between singularities. Each annulus is associated
with a separate Laurent expansion:
∞
X
1
f (z) =
an (z − z0 ) with (again) an =
2πi
n=−∞
ˆ
n
C
f (ξ)
dξ,
(ξ − z0 )n+1
(4.13)
now for −∞ < n < +∞. The contour C loops around z0 once, counterclockwise, within
the annulus in question. In this Laurent case, there is no longer any counterpart to the
(n)
closed-form expression an = f n!(z0 ) from the Taylor case.
4.3. Laurent series
109
Example 4.24. Give the three Taylor/Laurent expansions centered at the origin for the
following function:
1
f (z) =
.
(z − 1)(z − 2)
Since there are two distinct poles, the three expansions will be as follows (corresponding to the regions marked by I, II, and III in Figure 4.7):
1. A Taylor series converging inside the disk of radius 1.
2. A Laurent series converging in the annulus between the circles of radius 1 and radius 2.
3. A Laurent series converging outside the circle of radius 2.
1
=
The partial fraction expansion f (z) = (z−1)(z−2)
to manipulating geometric progressions:
−1
(z−1)
+
1
(z−2)
simplifies the algebra
1. The first expansion is a Taylor series converging in a disk centered
P∞at the origin
1
with radius 1. The first function (1−z)
can be directly expanded as k=0 z k , which
1
converges for |z| < 1 as required. The second term, (z−2)
, can be rewritten as
P∞ z k
−1
1
1
.
k=0 2
(z−2) = 2 (1− z2 ) . When |z| < 2, this term can be expressed as
Adding the expansions for both terms gives
∞ X
1−
f (z) =
k=0
1
2(k+1)
zk .
2. The second expansion is a Laurent series converging in an annulus centered at the
origin, between the circles of radii 1 and 2. The first term of the function needs
1
to be expanded outside of the unit circle. We thus write it as (1−z)
= z1 (1−1 1 ) .
z
The expansion that we found above for the second term is still valid in the present
annulus. The resulting series becomes
∞
X
f (z) =
ak z k ,
k=−∞
where
ak =
−1
1
2k+1
,
,
k ≤ −1,
k ≥ 0.
3. Finally, the third expansion is a Laurent series converging outside the circle of radius
2. The first term of the function can be expanded as above. The second term can be
1
rewritten as (z−2)
= z1 (1−1 2 ) to obtain a convergence in the requested region. The
z
resulting series is
k+1
∞
X
1
k
f (z) =
(2 − 1)
.
z
k=0
z− z1
Example 4.25. Expand f (z) = e
around z = 0.
110
Chapter 4. Introduction to Complex Integration
Im z
III
II
I
1
2
Re z
Figure 4.7. Regions of validity for the expansions in Example 4.24.
This function has essential singularities at z = 0 and at z = ∞. We can thus find a
Laurent expansion that will converge between them (thus for all finite z except at z = 0).
2
3
1
1
Direct Taylor expansion f (z) = ez− z = 1 + z − z1 + 21 z − z1 + 3!
z − z1 + · · ·
suggests that, when expanding out the different powers in the sum, we will end up with a
Laurent-type expansion.
It is in this case somewhat tricky to obtain the Laurent coefficients explicitly. However,
P∞ zr P∞ (−1)s z−s 1
1
one can proceed as follows: f (z) = ez− z = ez · e− z =
=
r=0 r!
s=0
s!
P∞ P∞ (−1)s zr−s
. If we now set r = n + s, i.e., r − s = n, the double sum can
r=0
s=0
r!s!
s
P∞
P∞
be rearranged into n=−∞ cn z n , where cn = s=0 s!(−1)
(n+s)! . This last sum turns out to
appear also in the context of Bessel functions (cf. Section 11.2) and can in terms of these
be given in closed form as cn = Jn (2).
4.3.2 Revisiting singularities
We have now better tools available to characterize singularities. The lowest index in a
Laurent sum, centered at a singularity, will give us information about the type of singularity
with which we are faced. Let’s revisit the first four types:
1. Removable singularities: We saw earlier that a removable singularity is only an
artifact of the function representation. Any trace of this “singularity” completely
disappears once the function has been expanded into a power series centered at that
point.45 Since the singularity has vanished, the resulting expansion around this point
is strictly a Taylor series.
2. Poles: A function exhibiting an N th-order pole at a point z0 can be expanded into a
45 Or
centered anywhere, but with the point included within the radius of convergence.
4.3. Laurent series
111
Laurent series centered at z0 . It will have the following form:
∞
X
cj (z − z0 )j .
j=−N
We call c−N the pole’s strength, the coefficient associated with the lowest index
(−N ). Another coefficient of interest is c−1 , which we call the residue. We will in
the following often denote the residue of a function f (z) at z = z0 as Res(f (z), z =
z0 ), or more briefly, Res(f (z), z0 ).
3. Essential singularities: An essential singularity is characterized by a Laurent series
for which there are cj 6= 0 for arbitrarily large negative indices j. The Laurent series
will thus take the form
∞
X
cj z j .
j=−∞
Once again, we shall soon give particular attention to the residue c−1 .
4. Branch points: Like a Taylor series, a Laurent series can only produce single-valued
results. In the Laurent case, there may, however, be branch points inside the inner
circle, as long as these cancel in the sense that the result becomes single-valued in
the annulus of convergence.
Example 4.26. Describe the possible Taylor and Laurent expansions around the origin for
the function f (z) = (z 2 − 1)1/2 .
This function was illustrated in Figures 2.19 and 2.20. The branch points are located
at z = ±1. If we are interested in Taylor expanding around z = 0, we orient the branch
cuts so they become directed outwards, for example, [−∞, −1] and [+1, +∞]. Each of
the two solution sheets now has its own Taylor expansion, i.e. (by the binomial expansion
s(s−1)(s−2) 3
2
(1 + z)s = 1 + sz + s(s−1)
s + · · · ),
2! z +
3!
f (z) = ±i 1 − z 2
1/2
= ±i
1
1
1
1 − z2 − z4 − z6 − · · · .
2
8
16
Regarding Laurent expansions for |z| > 1, we need to instead place the branch cut between
+1 and −1 inside the unit circle (which is possible, since z = ∞ is not a branch point).
There are then again two solution sheets, each with a separate Laurent expansion. We write
1/2
now f (z) instead as f (z) = ±z 1 − z12
, and the same binomial expansion formula
gives similarly
1
f (z) = ±z 1 − 2
z
1/2
= ±z
1
1
1
− ···
1− 2 − 4 −
2z
8z
16z 6
.
As can be inferred from Figure 2.20, there is no one-to-one correspondence between the
Taylor and the Laurent versions. The Taylor expansion holds within the unit circle, and
assumes the branch cut is away from this region, as in parts (c) and (d) of the figure,
whereas the Laurent expansions assume regularity outside of the unit circle, as depicted in
parts (a) and (b) of the figure.
112
Chapter 4. Introduction to Complex Integration
Example 4.27. Describe singularities of f (z) =
(z−1)2
z(z+1)3
.
There are two pole singularities (at z = 0 and at z = −1):
• Consider the Laurent series centered at z = 0. The second factor
2
1 (z−1)
z (z+1)3
(z−1)2
(z+1)3
2
in f (z) =
is analytic at z = 0 and can be expanded as 1 − 5z + 13z − · · · . The
coefficient c−1 = 1 is both the strength of the pole at z = 0, of order 1, and the
residue.
• Consider next the Laurent series centered at z = −1. We now write f (z) as f (z) =
2
(z−1)2 1
. The function (z−1)
is analytic around z = −1 and thus has a
(z+1)3
z
z
2
= −4 − (z + 1)2 − (z + 1)3 − · · · . Thus the
Taylor series centered there: (z−1)
z
function exhibits a third order pole, with a strength of −4.
Example 4.28. Describe the singularities of f (z) =
z 1/2 −1
z−1 .
Due to the z 1/2 term in the numerator, there will be a branch point at z = 0, giving
rise to two solution sheets. At z = 1, there occurs a divide by zero. On one solution sheet,
11/2 = +1 and the singularity becomes removable, whereas on the other sheet 11/2 = −1
and we get a first order pole with residue −2. Figure 4.8 displays this function. In part
(a), we see the real parts of the two solutions coinciding for x < 0 and the imaginary parts
coinciding for x > 0. For the real part, we see the pole at z = 1 affecting only one of
the two solution branches. Part (b) illustrates Im f (z) in the complex plane, showing again
how one solution sheet is unaffected by the pole that is present on the other one.
For the examples we have shown previously of multivalued functions, there were great
similarities between the solutions on different sheets. This latest example illustrates that
this by no means needs to be the case.
The following is an interesting result for the character of an analytic function near an
essential singularity.
Theorem 4.29 (Picard). Take any arbitrarily small neighborhood of an essential singularity. The function f (z) then takes every complex value with at most one exception infinitely
many times within this neighborhood.
The following theorem is slightly weaker, but much easier to prove; see Section 4.4.
Theorem 4.30 (Caserati–Weierstrass). Take any arbitrarily small neighborhood of an
essential singularity. The function f (z) then comes arbitrarily close to every complex
value infinitely many times within this neighborhood.
4.4 Supplementary materials
Proof of Poisson’s integral formula: Theorem 4.20.
Note: This proof below is somewhat “tricky,” but it relies only on contour integration.
By means of a Fourier series expansion (Section 9.1.1), the formula can be proven more
directly; see Exercise 9.8.6.
4.4. Supplementary materials
113
3
2
1
0
-1
-2
-3
-3
-2
-1
0
1
2
3
1
2
3
x
1
0.5
0
-0.5
-1
-3
-2
-1
0
x
(a) Re f (x) (top) and Im f (x) (bottom), both along the x-axis.
(b) Im f (z) in the complex plane near the origin.
1/2
Figure 4.8. The function f (z) = z z−1−1 , displayed along the real axis, and its imaginary
part in the complex plane (with a part of the surface near the pole cut out to better display the
pole-free solution sheet).
Denoting the evaluation point z = r eiθ , and changing variable ξ = eit (i.e., dξ =
i e dt), equation (4.6) becomes
it
f (r eiθ ) =
1
2π
ˆ
2π
f (eit )
0
eit
eit − r eiθ
dt.
(4.14)
We next separate the “kernel” inside the large parentheses into its real and imaginary parts:
eit
eit
[1 − r cos(t − θ)] − i [r sin(t − θ)]
=
.
− r eiθ
1 + r2 − 2r cos(t − θ)
114
Chapter 4. Introduction to Complex Integration
We can now spot that changing r → 1r leaves the imaginary part unchanged. Furthermore,
since r < 1, it will hold that 1r > 1, and (4.14) becomes after this change r → 1r a closed
loop integration without any singularity inside the path, implying that it evaluates to zero.
Subtracting this integral from (4.14) eliminates the imaginary parts and gives
f (r eiθ ) − 0 =
1
2π
ˆ

1
cos(t
−
θ)]
 [1 − r cos(t − θ)]

r
f (eit ) 
−
 dt .
1
1
1 + r2 − 2r cos(t − θ)
1 + 2 − 2 cos(t − θ)
r
r

2π
0
[1 −
The proof is now finished by simplifying the expression inside the large parentheses to
1−r 2
1+r 2 −2r cos(t−θ) .
The Caserati–Weierstrass theorem (Theorem 4.30) stated that if a function f (z) has
an essential singularity at a point z0 , then it comes arbitrarily close to every complex value
infinitely many times within this neighborhood.
Proof. This is a proof by contradiction: Say that f (z) never comes within ε of some value
1
is then bounded in magnitude by
f0 whenever |z − z0 | < δ. The function g(z) = f (z)−f
0
´
g(ξ)
1
1
ε in this neighborhood. By (4.13) its Laurent coefficients are an = 2πi C (ξ−z0 )n+1 dξ, so
by Theorem 4.9,
1 −n 1
|an | <
δ
2π .
2π
ε
With ε > 0 fixed and freedom to choose δ arbitrarily small, this tells us that an = 0 for all
1
can then
negative n, i.e., z = z0 is a regular point for g(z). The function f (z) = f0 + g(z)
have at most a pole at z = z0 , contradicting that this was an essential singularity.
4.5 Exercises
¸
Exercise 4.5.1. Let Γ be the unit square, with corners at ±1 ± i. Evaluate Γ z 2 dz in the
following three ways:
(a) Give the answer directly by referring to some general theorem.
(b) Evaluate explicitly the line integrals along the four sides, and add together the
results.
(c) Use Green’s theorem to convert the contour integral over to a double integral, and
then work out this double integral (i.e., do NOT refer to the C-R equations).
¸
Exercise 4.5.2. Let Γ again be the unit square, with corners at ±1 ± i. Evaluate Γ z̄ 2 dz
in the following two ways:
(a) Evaluate explicitly the line integrals along the four sides, and add together the
results.
(b) Use Green’s theorem to convert the contour integral over to a double integral, and
then work out this double integral.
Exercise 4.5.3. Evaluate
´
Γ
sin z dz, where Γ goes from i to π.
4.5. Exercises
115
Exercise 4.5.4. Let f (z) be an entire function with |f (z)| ≤ A|z| for all z ∈ C, where A
is a constant. Show that f (z) = B z, where B is a constant.
Exercise 4.5.5. Show that the roots of a polynomial are continuous functions of its coefficients.
Hint: Assume that a root jumps discontinuously. Consider the result of Example 4.7, choose
a suitable contour Γ, and obtain a contradiction.
2
Exercise 4.5.6. It is obvious from the form of f (z) = (z−1)
(z−i)3 that this function has N = 2
zeros and P = 3 poles. Nevertheless, show that N −P = −1 by directly applying Cauchy’s
argument principle (Theorem 4.8) to f (z).
Exercise 4.5.7. Prove the fundamental theorem of algebra directly from Theorem 4.8. Let
f (z) = pn (z) be a polynomial of degree n ≥ 1 and choose as the integration path a circle
|z| = R where R is very large.
Note: Since the task is to prove that pn (z) has root(s), your argument must not assume that
pn (z) can be factorized.
Exercise 4.5.8. Prove Cauchy’s estimate: If f (z) is differentiable inside Γ, if the circle
|z − z0 | = R, and if |f (ζ)| < M with |ζ − z0 | = R, then |f (k) (z0 )| ≤ k!M
for k =
Rk
0, 1, 2, . . . .
Exercise 4.5.9. Prove that the only entire functions f (z) that satisfy |f (z)| < 1 + |z|1/2
are constants.
Exercise 4.5.10. In Example 2.15, we Taylor expanded f (z) = 3−4z
1+2z around the origin and
obtained (using undetermined coefficients) the leading coefficients a0 = 3, a1 = −10 and
then the recursion an+2 = −2an+1 , n = 0, 1, 2, . . . . Derive exactly the same information
f (z)
1
instead from applying (4.12), written as an = 2πi
dz.
Γ z n+1
1
Exercise 4.5.11. Consider the function f (z) = z(z−2)(z
2 +1) :
(a) Give a closed-form expression (as simplified as you can) for the Laurent coefficients
when f (z) is expanded in a form valid near the origin.
(b) Give the coefficients c−1 , c0 , c1 in a Laurent expansion for f (z) around the origin,
valid far out from the origin.
(c) Give the similar three coefficients c−1 , c0 , c1 for a Laurent expansion valid in the
vicinity of z = i.
Figure 4.9 illustrates the function f (z) and the regions of validity for the three expansions.
Exercise 4.5.12. Describe all the singularities (none, removable, pole (if so, what order?),
essential, etc.) in the finite plane (i.e., no need to consider z = ∞) of the following
functions:
z)−1
(a)
f (z) = (cos
sin z .
(b)
f (z) =
z 1/2 +1
z+1 .
z 1/2 −1
z−1 .
(c)
f (z) =
Hint for (b) and (c): Recall that, for z 6= 0, the quantity z 1/2 can take two different values.
116
Chapter 4. Introduction to Complex Integration
(a) Original function from
Exercise 4.5.11.
(b) Validity of the expansion
in problem part (a).
(c) Validity of the expansion
in problem part (b).
(d) Validity of the expansion
in problem part (c).
1
Figure 4.9. The function f (z) = z(z−2)(z
2 +1) in Exercise 4.5.11, and the regions of
validity (colored) for the Laurent expansions in the three cases (with figures produced using 30 terms
in the respective expansions).
Exercise 4.5.13. Identify the types of all singularities to the following functions, and give
for these (if applicable) order, strength, and residue:
(a)
f (z) = 1/ 1 − z1 .
2
(b)
f (z) = 1/ 1 − z12 .
(c)
f (z) = 1/ sin(ez ) .
(d)
f (z) = (tan z)1/2 .
Exercise 4.5.14. Consider the function f (z) = (z + 1)1/2 (z − 1)1/2 . Does this function
have zero, one, or two Taylor expansions around the origin? If any exist, give the first three
nonzero coefficients.
2 s(s−1)(s−2) 3
Hints: (i) Recall the binomial expansion (1+z)s = 1+sz+ s(s−1)
z +· · · ,
2! z +
3!
and (ii) compare with Figures 2.19 and 2.20.
4.5. Exercises
117
Im z
Im z
br anch cut
br anch point
Re z
Re z
br anch point
br anch point
(a) Branch cuts
(b) Considered integration path.
Figure 4.10. Branch cuts and integration paths considered in Exercise 4.5.15.
Exercise 4.5.15. Consider the function f (z) = ((z + 1)(z − 1)(z − i))1/3 . It is proposed
to make this function single-valued by means of making a branch cut as shown in Figure
4.10.
Determine whether this branch cut is sufficient by means of the following two approaches:
(a) Inspect whether z = ∞ is a branch point (by making the change of variable z =
1/t).
(b) Follow the function around the path outlined in Figure 4.10(b) and see whether we
get back to the same function value as we started with.
Exercise 4.5.16. Let t be a parameter, and consider the analytic function
1
f (z, t) = et(z− z )/2 .
(4.15)
(a) Tell where in the complex z-plane f (z, t) has singularities.
Comment: The Laurent expansion of f (z, t) (in terms of z) around the origin will naturally
have coefficients that depend on the parameter t:
1
et(z− z )/2 =
∞
X
Jn (t)z n .
(4.16)
n=−∞
These coefficients are known as the J-Bessel functions, and they are of great importance in
mathematical physics, partly because y(t) = J±n (t) provide solutions to the differential
equation
dy
d2 y
+ (t2 − n2 )y = 0,
(4.17)
t2 2 + t
dt
dt
which has a great tendency to arise in many situations, one being separation of variables
in polar coordinates.
(b) Show that y(t) = J±n (t) indeed satisfy the differential equation (4.17).
Hint: By direct differentiation, show that f (z, t) satisfies
2
∂f
∂
∂f
2
2∂ f
t
+t
+t f =z
z
.
∂t2
∂t
∂z
∂z
Then use the fact that the coefficients in a Laurent expansion are uniquely determined.
118
Chapter 4. Introduction to Complex Integration
(c) From the integral formula for Laurent coefficients, show that
ˆ π
ˆ
1
1 π
−i(nθ−t sin θ)
Jn (t) =
e
dθ =
cos(nθ − t sin θ)dθ.
2π −π
π 0
Comment: It is possible to extract much more information about the J-Bessel functions
from (4.15) than what is illustrated in this example; see Sections 11.2.1 and 11.2.2. For
example, Taylor expanding the exponential on the LHS of (4.16) quickly leads to the Taylor
expansions of the Jn (t) functions, showing that they are entire functions of t (i.e., analytic,
and with no singularity in the finite complex t-plane).
Exercise 4.5.17. If f (z) is analytic and has no zeros for |z| ≤ R, show that
1
log |f (0)| =
2π
ˆ
2π
log |f (R eiθ )|dθ.
(4.18)
0
Note: This result is generalized in Exercise 5.6.3 to the case where f (z) has zeros inside
the circle (Jensen’s formula).
Exercise 4.5.18. Verify the following alternate way to express Poisson’s integral formula
(4.9) (with z = r eiθ , |z| ≤ 1):
f (z) =
1
2π
ˆ
2π
f (eit )
0
1 − |z|2
dt.
|eit − z|2
Exercise 4.5.19. In Poisson’s integral formula (4.9), f (z) is specified around the periphery
of the unit circle, and the harmonic function satisfying this inside the circle is obtained.
If we instead specify f (z) along the real axis, a counterpart formula giving the harmonic
extension to the upper half-plane is
ˆ ∞
1
y
f (x + i y) =
f (t)Py (x − t)dt, where Py (x) =
.
(4.19)
2
π x + y2
−∞
(a) Verify that (4.19) is correct.
(b) With use of (4.19), derive an explicit form for a harmonic function over the upper
half-plane that, along the real axis, takes the values
1 if − 1 ≤ x ≤ 1,
f (x) =
0
otherwise.
(c) Solve the same problem as in (b), but instead by considering Figure 2.8.
Chapter 5
Residue Calculus
The previous chapter described how contour integrals of analytic functions become path
independent (as long as paths do not move across singularities). From this followed a number of important theorems, as well as an improved understanding of Taylor (and Laurent)
expansions. This chapter follows up on this by introducing residue calculus, which can be
used to evaluate a wide variety of definite integrals (many of which, at first thought, may
not appear to have anything to do with complex variables). Applications of residue calculus also include evaluating many infinite sums in closed form, additional opportunities for
analytic continuation, etc.
5.1 Residue calculus
Let z1 , z2 , z3 , . . . , zn be the singularities (poles or essential singularities) of a function
f (z) inside contour Γ, as shown in Figure 5.1. Choosing as integration path Γ, furthermore
going around each singularity in the direction shown by arrows, and including cancelling
integrations up/down each vertical line segment results in an overall path that is closed and
encloses no singularity. Therefore
˛
ˆ
+
Γ
ˆ
+
z1
ˆ
ˆ
+··· +
+
z2
z3
= 0.
(5.1)
zn
Changing direction when going around the singularities changes the sign of their contribution, giving
˛
ˆ
ˆ
ˆ
ˆ
=
+
+
+··· +
.
(5.2)
Γ
z1
z2
z3
zn
We can thus change the original contour Γ to just go around each singular point separately,
all in the same positive direction. Recall that
1
2πi
˛
Γ

 0
dz
0
=

(z − z0 )n
1
if z0 is outside Γ,
if z0 is inside Γ and n 6= 1,
if z0 is inside Γ and n = 1
119
120
Chapter 5. Residue Calculus
z2
z3
z1
z4
z5
Γ
Figure 5.1. Change of contour in residue calculus.
for n integer. Locally to each singularity, there is a Laurent expansion
f (z) =
∞
X
cn (z − zj )n .
n=−∞
Only the term associated with n = −1, i.e., the residue c−1 , will contribute to the integral.
Hence, we have the following.
Theorem 5.1.
˛
N
X
f (z)dz = 2πi
Γ
c−1
j=1
{z
|
zj
}
Sum of the residues at the zj
= 2πi

N
X

j=1


Res(f, zj ) ,

where the zj are all the singularities inside Γ.
5.1.1 Shortcuts to calculate residues
We wish to compute the residue Res(f (z), z0 ). However, doing this by means of the
Laurent expansion around z = z0 may require more work than necessary. In case the
singularity is a pole of order m, the following gives three alternative options:
N (z)
D(z) , where N (z0 ) 6= 0 and D(z) has a simple root
N (z0 )
D 0 (z0 ) . Note that what one decides to include in N (z)
1. In case of m = 1, write f (z) =
at z0 . Then, Res(f (z), z0 ) =
and in D(z) is quite arbitrary; making D(z) as simple as possible will make it easier
to take the derivative.
Justification: Taylor expanding around z0 gives f (z) =
N (z0 )
D 0 (z0 )
·
1
(z−z0 )
+ O(1).
N (z0 )+N 0 (z0 )(z−z0 )+···
0+D 0 (z0 )(z−z0 )+···
=
5.1. Residue calculus
121
2. If m is arbitrary and f (z) can conveniently be written in the form f (z) =
then Res(f (z), z0 ) =
φ
φ(z)
(z−z0 )m ,
(m−1)
(z0 )
(m−1)! .
Justification: Expanding φ(z) in a Taylor series centered at z0 gives φ(z) = φ(z0 ) +
2
0)
(z − z0 )φ0 (z0 ) + (z−z
φ00 (z0 ) + · · · , and therefore
2!
φ(z0 )
φ0 (z0 )
φ00 (z0 )
φ(m−1) (z0 )
φ(z)
+···
=
+
+
+
·
·
·
+
(z − z0 )m (z − z0 )m (z − z0 )m−1 2!(z − z0 )m−2
(m − 1)!(z − z0 )
with the last displayed term above giving the result.
3. If m is arbitrary and the rewrite described above is not convenient, we can instead
1
dm−1
m
use Res(f (z), z0 ) = (m−1)!
limz→z0 dz
f (z)), where the limit usum−1 ((z − z0 )
ally is best handled by l’Hôpital’s rule.
Justification: Around the pole,
f (z) =
c−m+1
c−1
c−m
+
+ ... +
+ ··· ,
(z − z0 )m
(z − z0 )m−1
(z − z0 )1
and therefore
(z − z0 )m f (z) = c−m + c−m+1 (z − z0 )1 + . . . + c−1 (z − z0 )m−1 + · · ·
and
dm−1
((z − z0 )m f (z)) = (m − 1)! c−1 + m! c0 (z − z0 ) + · · · ,
dz m−1
from which the result follows.
5.1.2 Examples
Example 5.2. Evaluate
˛
ze1/z dz.
I=
|z|=1
The only singularity inside the unit circle is the essential singularity at z = 0. We thus
expand the integrand at that point to find the function’s residue there:
1
1
1/z
ze
=z 1+ +
+ ···
z
2!z 2
1
+ ··· .
=z+1+
2!z
Thus, Res(ze1/z , z = 0) = 1/2, and
˛
1
I=
ze1/z dz = 2πi · = πi .
2
|z|=1
Figure 5.2 shows that the integrand is of extreme complexity around the singularity point.
It is remarkable how that did not at all complicate the calculation of the integral going
around it—all that mattered was the value of the residue.
122
Chapter 5. Residue Calculus
2
0
-2
0.2
0.1
0.2
0
0.1
-0.1
0
-0.1
-0.2
-0.2
y
-0.3
x
-0.3
(a) Re z e1/z
(b) |z e1/z |, with phase angle coloring
Figure 5.2. Illustrations of the integrand in Example 5.2, f (z) = z e1/z , near the origin.
Example 5.3. Compute Res
ez
z cos z , z
=0 .
We can let N (z) = ez / cos z and D(z) = z. Then Res
ez / cos z
,z
z
=0 =
1. Alternatively, we can choose N (z) = ez and D(z) = z cos z, which gives
0
z = 0 = cose0−0 = 1.
Example 5.4. Evaluate
ˆ
∞
I=
−∞
x2
dx.
1 + x4
e0 / cos 0
=
1
ez
Res z cos z ,
5.1. Residue calculus
123
Im z
Γ
R
z2
z1
Re z
z4
z3
Figure 5.3. Contour in Example 5.4.
2
z
4
To compute this integral with residue calculus, we consider f (z) = 1+z
4 . Since z + 1
1+i
−1+i
1−i
−1−i
has four simple zeros, z1 = √2 , z2 = √2 , z3 = √2 , z4 = √2 , these are simple poles
of f (z).46 Consider the open contour Γ, a semicircle of radius R, as shown in Figure 5.3.
´ z2
≤ M · L, where L is the length
Recall that the magnitude of the integral Γ 1+z
4 dz
of the contour (πR) and where M is the upper bound of the function on the contour. In
order to find this bound, recall the following triangle inequalities: |z1 + z2 | ≤ |z1 | + |z2 |,
2
z2
≤ RR
|z1 + z2 | ≥ |z1 | − |z2 |, but also |z1 + z2 | ≥ |z2 | − |z1 |. Therefore, 1+z
4
4 −1 . Thus,
´ z2
´
3
z2
dz ≤ RπR
dz → 0 as R increases. The
4 −1 → 0 as R → ∞. Therefore,
Γ 1+z 4
Γ 1+z 4
contour Γ was open. Next, consider the closed contour, say Λ, which is made of the contour
Γ followed by the section of the real axis from −R to R. Since two poles are located inside
of this new contour,
˛
Λ
ˆ R
z2
x2
dz
+
dx
4
4
Γ 1+z
−R 1 + x
ˆ ∞
x2
=0+
dx as R → ∞
4
−∞ 1 + x
= 2πi{Res(f, z1 ) + Res(f, z2 )}.
z2
dz =
1 + z4
ˆ
Using the formulas above, we next determine Res(f, zk ) =
N (zk )
D 0 (zk ) ,
1
4z1 =
k = 1, 2, where
√
N (z) = z and D(z) = z + 1. This gives Res(f, z1 ) =
2(1 + i)/8 and
√
1
Res(f, z2 ) = 4z2 = 2(−1 − i)/8. Thus, 2πi{Res(f, z1 ) + Res(f, z2 )} = √π2 , and
2
4
ˆ
∞
−∞
x2
π
dx = √ .
4
1+x
2
46 If one tries to evaluate the integral using partial fractions (rather than residue calculus), the first step towards
x2
1
x
√
√
arriving at 1+x
− 2 √x
requires a factorization of x4 + 1, which is easiest done
4 =
2
2 2
x − 2x+1
x + 2x+1
by complex arithmetic, writing z 4 + 1 as [(z − z1 )(z − z3 )] [(z − z2 )(z − z4 )].
124
Chapter 5. Residue Calculus
´∞
Note: When the integral to compute is of the type −∞ f (x)dx, the open half-circle contour
Γ shown above will often ´be used, with R → ∞. We showed that, for our particular
integrand in Example 5.4, Γ f (z)dz → 0. This will always be true in the case where both
the numerator and the denominator are polynomials, and the degree of the denominator is
at least two more than the degree of the numerator.
Example 5.5. Evaluate
ˆ
∞
I=
−∞
cos kx
dx
(x + b)2 + a2
with k, a > 0 and b ∈ R.
cos kz
The integrand f (z) = (z+b)
2 +a2 has two simple poles, located at z1 = ia − b and
z2 = −ia − b. We would like to bound the magnitude of the integral as we did in the
previous example. However, cos kz oscillates with amplitudes that grow exponentially
fast once we consider its values far away from the real axis (as follows from cos kz =
eikz +e−ikz
; cf. Figure 2.5). In order to bypass this issue, let’s instead consider the function
2
eikz
f (z) = (z+b)
2 +a2 . In this case,
ˆ
∞
−∞
eikx
dx =
(x + b)2 + a2
ˆ
∞
−∞
|
ˆ ∞
cos kx
sin kx
dx +i
dx .
(x + b)2 + a2
(x
+
b)2 + a2
−∞
{z
} |
{z
}
Imaginary part
Real part
´∞
eikx
Thus, we can evaluate −∞ (x+b)
2 +a2 dx and retain the real part only as the answer. As for
the boundedness of the integrand, eikz = e−ky eikx , which implies that |eikz | = e−ky → 0
as y → ∞ (had k been negative, we could similarly have used a contour in the lower halfplane). Let x = R cos θ and y = R sin θ be the coordinates of z = x + iy, a point on the
open semicircle contour Γ of radius R as illustrated in Figure 5.4. Let’s next find an upper
Im z
ia − b
Γ
R
Re z
−ia − b
Figure 5.4. Contour in Example 5.5.
5.1. Residue calculus
125
bound for the magnitude of the integral. We have
ˆ
eikz
eikz
≤
dz
(πR)
2
2
(z + b)2 + a2
Γ (z + b) + a
e−Rk sin θ
(πR)
|(z + b)2 + a2 |
πe−Rk sin θ
≤
a 2
R |(cos θ + i sin θ + Rb )2 | − | R
|
→ 0 as R → ∞.
=
To obtain a closed contour (so that the residue theorem can be applied), consider the contour
Λ, which is made of the half-circle Γ and which is closed by also including the real axis
from −R to R (and letting R → ∞):
ˆ
ˆ R
˛
eikz
eikx
eikz
dz
=
dz
+
dx
2
2
2
2
2
2
Γ (z + b) + a
−R (x + b) + a
Λ (z + b) + a
ˆ ∞
eikx
=0+
dx
2
2
−∞ (x + b) + a
= 2πi Res(f, ia − b).
Calculating the residue at z = ia − b, we obtain
Res(f, ia − b) =
e−ka (−i cos kb − sin kb)
2a
(here using the assumption that a > 0; for a < 0 the relevant pole would have been the one
at z = −ia − b). Thus,
ˆ ∞
πe−ka (cos kb − i sin kb)
eikx
dx
=
.
2
2
a
−∞ (x + b) + a
Finally, by only retaining the real part, we obtain the result
ˆ ∞
cos kx
πe−ka cos kb
I=
dx
=
.
2
2
a
−∞ (x + b) + a
As a concluding observation, we can note from the form of the original integral that the
answer must be an even function in both a and k. We can therefore drop the requirements
that these parameters are positive and give the final answer as
ˆ ∞
cos kx
πe−|ka| cos kb
dx
=
I=
2
2
|a|
−∞ (x + b) + a
for a, b, k all real (with divergence if a = 0).
´∞
Integrals of the type −∞ f (x)eikx dx are very common. The semicircle contour Γ
is
in such cases. The following result gives an easy way to ensure that
´ used routinely
ikz
f
(z)e
dz
→
0 as R → ∞.
Γ
Theorem 5.6 (Jordan’s lemma). If f (z) → 0´uniformly as R → ∞, Γ is a half-circle as
shown in Figures 5.3 and 5.4, and k > 0, then Γ f (z)eikz dz → 0.
126
Chapter 5. Residue Calculus
Example 5.7. Evaluate
ˆ
2π
I=
0
dθ
,
A + B sin θ
where A > 0 and where |B| < A (in order to avoid divisions by zero on the integration
path).
We often encounter the more general form
trates a couple of general approaches.
´ 2π
0
g(sin θ, cos θ)dθ. The following illus-
Method 1: Let
1
.
A + B sin z
Considering the contour sketched in Figure 5.5(a), we find that for the integrand f (z) =
f (z) =
1
A+B sin z
ˆ
f (z)dz → 0,
Γ3
since sin z grows exponentially
we go up the complex plane. We also notice
´ fast the further
´
that f (z) is 2π-periodic, so Γ2 f (z)dz + Γ4 f (z)dz = 0. The three extra paths Γ1 , Γ2 , Γ3
will therefore together not contribute anything to the integral (in the limit of Γ3 shifted
up towards +i∞). The next step is to find the singularities inside the contour, illustrated
again together with f (z) in Figure 5.5(b). The single pole inside the contour occurs where
A
A + B sin z0 = 0, i.e., at z0 = − arcsin B
. The residue becomes
Res(f, z0 ) =
N (z0 )
1
=
D0 (z0 )
B cos z
=
z=− arcsin
A
B
B
1
p
1 − (sin z)2
z=− arcsin
A
B
1
= √
.
i A2 − B 2
Therefore,
2π
I=√
.
2
A − B2
Method 2: The second “routine” approach for trigonometric integrals over a full period is
to let z = eiθ . Then
1 iθ
1
1
cos θ =
e + e−iθ =
z+
,
2
2
z
1 iθ
1
1
sin θ =
e − e−iθ =
z−
,
2i
2i
z
and
1
dz .
iz
For 0 ≤ θ ≤ 2π (or equivalently −π ≤ θ ≤ π), the path for z goes once around the unit
circle in the positive direction. With this variable change,
˛
1
dz
I=
1
1
iz
|z|=1 A + B 2i z − z
˛
2
dz
=
.
A
B |z|=1 z 2 + 2i B
z−1
dθ =
5.1. Residue calculus
127
Im z
Γ3
Γ4
Γ2
0
Γ1
2π Re z
(a) Schematic integration path
(b) Magnitude/phase plot with path in case A = 3, B = 1.
Figure 5.5. Contour and function in Example 5.7.
A
The denominator factorizes as z 2 + 2i B
z − 1 = (z − z1 )(z − z2 ) with
z1,2 = i
−A ±
√
A2 − B 2
.
B
√
2
2
Since A > |B|, the second root (pole) z2 = i −A− BA −B will fall outside the unit circle,
and from z1 · z2 = −1 it follows then that z1 falls inside it. From
1
1
Res
, z = z1 =
,
(z − z1 )(z − z2 )
z1 − z2
128
Chapter 5. Residue Calculus
it follows then that the relevant residue evaluates to
I=
2i
√B
A2 −B 2
and, with that,
2π
2
B
=√
.
(2πi) √
B
2i A2 − B 2
A2 − B 2
Example 5.8. Evaluate
ˆ π/2
dθ
, a, b > 0, for n = 0, 1, 2, . . . .
In (a, b) =
2
(a cos θ + b sin2 θ)n+1
0
A common technique to manipulate integrals is to differentiate with respect to its parameters, to see whether anything simpler happens to arise (sometimes known as “Feynman’s trick,” although used since the beginning days of calculus; one might even introduce
artificial parameters if the integral does not contain any). In the present case, we will stumble onto the recursion relation
∂
∂
1
+
In (a, b) = −
In+1 (a, b),
∂a ∂b
n+1
´ π/2
dθ
showing that it is sufficient to evaluate I0 (a, b) = 0 a cos2 θ+b
. The routine varisin2 θ
iθ
able change z = e just introduced
(Example
5.7,
Method
2)
gives,
after
a few simplifi¸
i
z dz
cations, I0 (a, b) = − a−b
,
with
the
integration
path
once
around
the unit
a+b
z 4 +2 a−b z 2 +1
¸
dt
i
2
circle. Changing variable t = z , dt = 2z dz gives next I0 (a, b) = − a−b t2 +2 a+b
t+1
a−b
(where the factor 2 in dt = 2z dz has canceled against making t go around the unit circle once rather than twice). Since the product√of the
roots in the denominator polynomial
√
a−√b
√ . We thus obtain
is 1, the only relevant root becomes t1 = − √a+ b , with residue 4a−b
ab
a−b i
π
I0 (a, b) = 2πi − a−b 4√ab = 2√ab . From the recursion relation then follows
π
(a + b),
4(ab)3/2
π
(3a2 + 2ab + 3b2 ),
I2 (a, b) =
16(ab)5/2
···
I1 (a, b) =
With some more effort, one can from the recursion relation also deduce an expression for
general n:
n X
π
2k
2(n − k) k n−k
In (a, b) =
a b
.
k
n−k
(4ab)n+1/2
k=0
Example 5.9. Evaluate
˛
I=
Γ
z2
dz
+ a2
around the contour shown in Figure 5.6.
We find that
Res
1
1
,
−ia
=−
z 2 + a2
2ai
5.1. Residue calculus
129
Im z
ia
Re z
Γ
−ia
Figure 5.6. Contour in Example 5.9.
and, similarly,
Res
1
, +ia
2
z + a2
=
1
.
2ai
Since the contour goes around the pole z = ia twice, we shall count the function’s residue
at that point twice and the residue at z = −ia only once. We thus find
1
1
π
I = 2πi 2
−
= .
2ia 2ia
a
The winding number w(zj ) of the curve Γ around a point is the number of times that Γ
winds around zj (with positive means counterclockwise). We have in the example above
w(ia) = 2 and w(−ia) = 1.
Example 5.10. Evaluate
ˆ
I=
0
∞
x3
dx
, a>0.
+ a3
One might again consider using as contour a circle
´ of radius R centered at the origin,
since, far out in any direction in the complex plane, x f (z)dz → 0 as R → ∞. However,
the interval of integration goes from 0 to ∞ and not from −∞ to ∞. Before describing a
more general approach next in Example 5.11, we consider first a more specialized idea.
Since z only appears as z 3 , we note that both z = r and z = re2πi/3 give the same
result when raised to the third power. We thus choose the contour as in Figure 5.7 and
perform the substitutions z = r and z = re2πi/3 along the respective radial path segment.
130
Chapter 5. Residue Calculus
Im z
z1 = ae iπ/3
2π/3
z2 = −a
Re z
Γ
z3 = ae5iπ/3
Figure 5.7. Contour in Example 5.10.
This gives
ˆ
0
∞
dr
+
3
r + a3
´
Since x
obtain that
dz
z 3 +a3
ˆ
x
dz
+
3
z + a3
ˆ
0
∞
ei2π/3 dr
= 2πi Res
r3 + a3
1
πi/3
,
z
=
ae
.
1
z 3 + a3
1
πi/3
→ 0 as R → ∞, and since Res( z3 +a
)=
3 , z1 = ae
ˆ ∞
dr
2πi
(1 − e2πi/3 ) = 2 2πi/3 ,
3 + a3
r
3a e
0
which simplifies to
ˆ
∞
0
1
,
3a2 e2πi/3
we
dr
2π
= √
.
r3 + a3
3 3a2
It was in this case quite lucky that the integrand itself provided a return path from
infinity back to the origin, along which it took the same values as along the real axis. As
we will see next, this type of “luck” is unnecessary. The particular behavior of Im log z
(recall Figure 2.7(b)) offers a more general approach for creating a suitable return path.
5.1.3 Special contours
Example 5.11. Evaluate
ˆ
I=
0
∞
dx
, a, b > 0.
(x + a)(x + b)
5.1. Residue calculus
131
This example can be easily solved using partial fractions (and no complex variables),
but we will ignore that for now. We will
instead use an idea that can be applied for the
´∞
much more general integrals of the type 0 f (x)dx. We only need the integrand to decay
fast enough for the contributions from a path distant from the origin to be insignificant.
Instead of tackling the integral exactly as it stands, consider in the present case
˛
(log z)dz
J=
, a, b > 0,
(z
+ a)(z + b)
Γ
along the keyhole contour shown in Figure 5.8.
Γ
Figure 5.8. Keyhole contour in Example 5.11.
´∞
¸
In general, to compute 0 f (x)dx, we will consider Γ (log z)f (z)dz. Why is this a
good idea? If we follow the contour, the contributions to the integral become the following:
´
´∞
1. → (log z)f (z)dz = 0 log(z)f (z)dz .
´
2. " (log z)f (z)dz → 0 as outer radius R → ∞ .
R
´
´∞
´∞
3. ← ((log z) + 2πi)f (z)dz = − 0 (log z)f (z)dz − 2πi 0 f (z)dz .
´
4. ! (log z)f (z)dz → 0 as inner radius → 0 .
Thus
ˆ
ˆ
+
→
"R
ˆ
ˆ
+
+
←
!
ˆ
!
∞
(log z)f (z)dz = −2πi
f (x)dx
0
= 2πi
X
Res((log z)f (z), zj )
j
and
ˆ
∞
f (x)dx = −
0
X
j
Res((log z)f (z), zj ).
132
Chapter 5. Residue Calculus
In our example, we obtain
I = − (Res((log z)f (z), −a) + Res((log z)f (z), −b)) = −
=
log(−a) log(−b)
+
−a + b
−b + a
log(b/a)
.
b−a
The extra factor log z canceled out for the integrand, but still provided us with a closed
contour for the integration over the semi-infinite interval [0, +∞]. If the function we want
to integrate already contains a factor of log z in the numerator, we can still multiply with
an extra log z; see Exercises 5.6.19 and 5.6.20.
Example 5.12. Evaluate
ˆ
1
I=
−1
√
1 − x2
dx.
1 + x2
Figure 5.9 shows the integrand over
√ the integration interval x ∈ [−1, 1] (assuming the
conventional sign choice such that 1 = +1). As usual, displaying the function over
the complex plane reveals a much richer picture. The function has two branch points, at
z = ±1 (but none at z = ∞), making it natural to introduce a branch cut between −1 and
+1. The real and imaginary parts of the integrand f (z) then become as shown in Figure
5.10. We note in particular the (first order) poles at z = ±i.
√
In order to determine the residues at the poles, we look first at the numerator 1 − z 2
(Figure 5.11). The value of this function at z = 0, just above the branch cut, is +1. Moving
z straight up along the imaginary
axis from 0 to i, as illustrated in green in Figure 5.11, will
√
just change this +1 to + 2. Since no branch cuts were passed along this way, no sign was
swapped. For the pole at z = −i, we have a choice in thinking. The value at z = 0 should
this time evaluate to −1, and then
√ reason
√ moving z down from 0 to −i will for the same
smoothly
change
that
−1
to
−
2.
Alternatively,
if
we
start
at
z
=
+i
with
the
1 − z2 =
√
+ 2 and move z in a big swing, illustrated in blue, around to −i (in either direction, well
2
outside the [−1, 1] cut),
√ 1 − z will have gone one full turn around the origin, and (having
√
2 will have made half a turn, i.e., swapped sign from + 2
not crossed any
cut)
1
−
z
√
N (z0 )
to become − 2. We can now evaluate the residue using the Res(f (z), z0 ) = D
0 (z ) rule
0
1
f(x)
0.8
0.6
0.4
0.2
0
-0.2
-1
-0.5
0
0.5
1
x
√
Figure 5.9. The integrand f (x) =
1−x2
1+x2
displayed over −1 ≤ x ≤ +1.
5.1. Residue calculus
133
3
2
1
0
-1
3
-2
2
-3
1
-3
0
-2
-1
-1
0
-2
1
2
x
3
(a) Real part of f (z) =
y
-3
√
1−z 2
1+z 2 .
3
2
1
0
-1
3
-2
2
-3
1
-3
0
-2
-1
-1
0
-2
1
2
x
3
y
-3
(b) Imaginary part of f (z) =
√
1−z 2
1+z 2 .
√
1−z 2
Figure 5.10. Real and imaginary parts of f (z) = 1+z2 , using a branch cut along
[−1, +1]. The sign is chosen so the function is positive along the cut and smooth above it.
(Section 5.1.1):
√
+
1 − z2
2z
z=+i
√
1 − z2
=−
2z
√
2
2i .
z=−i
Both poles have thus the same residue,
We give next three different methods for proceeding.
√
2
=
.
2i
134
Chapter 5. Residue Calculus
1.5
1
0.5
0
-0.5
-1
-1.5
1
0.5
0
-0.5
-1
y
-2
-3
x
(a) Real part of f (z) =
√
3
2
1
0
-1
1 − z2.
2
1
0
-1
-2
1
-3
0.5
-2
-1
0
0
-0.5
1
x
2
-1
3
y
√
(b) Imaginary part of f (z) = 1 − z 2 .
Figure 5.11. Real and imaginary parts of the integrand’s numerator, f (z) =
using a branch cut along [−1, +1]. The branch is chosen to satisfy f (0) = +1.
√
1 − z2,
Method 1: A simple fact that could quite easily escape attention if it weren’t for Figure
5.10 is that Re f (z) is identically zero over [−∞, −1] and [+1, +∞] and Im f (z) is antisymmetric along the real axis, meaning that there will be no difference in value if we
change the integration interval from [−1, +1] to [−R, +R] for any R > 1. Doing that, we
then close the loop via a large (radius R) half-circle Γ in the upper half-plane. It will then
hold that
ˆ
√
I + f (z)dz = 2πi · {residue of pole at z = +i} = π 2 .
(5.3)
Γ
5.1. Residue calculus
135
´
It remains to evaluate Γ f (z)dz. Since f (z) is singularity-free for |z| > 1, it will there
possess a Laurent expansion:
q
√
±iz 1 − z12
i
1 − z2
3
=±
f (z) =
= 2
1 − 2 + ··· .
1 + z2
z
2z
z 1 + z12
Taking the integral over the semicircle, only the leading term ± zi will contribute in the
limit of R → ∞. We need, however, to choose the sign so it corresponds to the function
branch we are interested in, i.e., with f (0) = +1, as shown in Figure 5.10(b). As z moves
up the imaginary axis, the value should (beyond the pole)
be negative
´
´ and increase towards
zero. That requires the negative sign. We then find Γ f (z)dz = Γ (− zi )dz, which, after
´
changing variables {z = Reiθ , dz = Ri eiθ dθ}, evaluates to Γ (− zi )dz = π. From (5.3)
√
now follows I = π( 2 − 1).
Method 2: When one wants to integrate along one side of a branch cut, it is often advantageous to extend the contour to also run along the other side of the cut, and in the opposite
direction. Hence, we consider the contour illustrated in Figure
5.12. On´ the return path,
´
below the cut, both f (z) and dz have changed ´sign, so ← f (z)dz = → f (z)dz = I.
Since the integrals around the two circular paths f (z)dz → 0 as their radii ε → 0, the
integral around the closed loop will be 2 I. Had there been a residue associated with the
branch cut, this would have provided a value for I. The alternate strategy in cases such as
this is to just expand the contour into a very large circle, radius R, surrounding the origin.
Expanding the contour in this way will (i) cause the integral to change according to the two
poles that now have been included (cf. Figure 5.13), and (ii) allow an exact evaluation for
R → ∞.
Im z
i
Γ
1
−1
Γ
Re z
−i
Figure 5.12. First contour in Example 5.12; the second contour will be to expand it to a
large circle, radius R, surrounding the origin.
√
From the discussion above, each of the poles has residue 2i2 , and the integral around
the full circle will be twice what we evaluated for the half-circle under Method 1. Putting
this together,
ˆ
√
2I +
f (z)dz = 2πi {sum of residues}, i.e., 2I + 2π = 2 2π,
"
136
Chapter 5. Residue Calculus
Im z
i
Γ
1
−1
Re z
−i
Figure 5.13. Illustration of the “first” and “second” contours (as shown in Figure 5.12
and the large circle, respectively) and why the integrals around these two contours add up to the
contributions from the two poles.
and we therefore again obtain
I=π
√
2−1 .
Method 3: Let’s perform the change of variable x = sin θ. Then,
dx = cos θdθ. Also, 1 + x2 = 1 + sin2 θ. Therefore,
ˆ π/2
cos2 θ
I=
2 dθ
−π/2 1 + sin θ
ˆ
1 π cos2 θ
dθ
=
2 −π 1 + sin2 θ
√
1 − x2 = cos θ and
because the integrand is π-periodic. A further change of variable t = tan θ allows (with
some effort) closed-form integration. However, with the limits now being ±π, contour
integration is easier.
The standard further change of variables when dealing with trigonometric functions
over a full period (cf. Example 5.7, Method 2) is to set z = eiθ . Then cos θ = 21 z + z1 ,
1
sin θ = 2i
z − z1 , and ieiθ dθ = dz, implying dθ = −i
z dz. Therefore,
˛
1
1 2
1
−i
4 z+ z
I=
dz
2 |z|=1 1 + −1 z − 1 2 z
4
z
2
˛
z2 + 1
−i
dz
=
2 |z|=1 z 4z 2 − (z 2 − 1)2
2
˛
z2 + 1
i
=
dz,
(5.4)
2 |z|=1 z (z − z1 ) (z − z2 ) (z − z3 ) (z − z4 )
5.1. Residue calculus
137
Figure 5.14. Magnitude/phase portrait of the integrand in (5.4), with the integration path
marked by a dashed black circle.
√
that are located inside the unit circle
where zj = ±1 ± 2, j =√1, 2, 3, 4. The only roots√
are at z = 0, z = z1 = 1 − 2, and z = z2 = −1 + 2; see Figure 5.14. We find that
Res(f, 0) = 1,
−1
Res(f, z1 ) = √ ,
2
−1
Res(f, z2 ) = √
2
and therefore
√
i
2πi − 2 + 1
2
√
=π
2−1 .
I=
Example 5.13. Evaluate
ˆ
∞
I=
−∞
eαx
dx.
1 + ex
In order for the integral to converge at x = −∞ and at x = +∞, we need Re α > 0 and
eαz
Re α < 1, respectively. In the complex z-plane, f (z) = 1+e
z has poles at z = ±iπ, ±3iπ,
±5iπ, etc.; see Figures 5.15 and 5.16. This pole pattern, and even more how the phase angle
for f (z) seems to be constant along lines parallel to the x-axis, suggests the integration
path that is shown, following the real axis from −R to +R, and returning from R + 2πi
to −R + 2πi , with also two vertical sides along which the contributions go to zero when
´ eα(z+2πi)
´ eαz
2πiα
R → ∞. Along the upper return path, ← 1+e
dz = −e2πiα I,
(z+2πi) dz = e
← 1+ez
implying that for the full loop, we get
αz ˆ
ˆ
e
+
= 1 − e2πiα I = 2πi Res (f, z = πi) = 2πi
= 2πi e(α−1)πi ,
z
e
→
←
z=πi
138
Chapter 5. Residue Calculus
Im z
3iπ
Pole
iπ
Re z
−iπ
Figure 5.15. Contour for Example 5.13.
Figure 5.16. Magnitude/phase portrait of the integrand in Example 5.13 (shown in the
case of α = 1/2), with the same integration path as shown schematically in Figure 5.15 here marked
by a dashed rectangle.
which simplifies to
I=
π
.
sin πα
Example 5.14. Evaluate
ˆ
I=
0
∞
xa−1
dx, a ∈ R.
1+x
Convergence requires a > 0 (x = 0) and a < 1 (x = ∞). A notable feature of the
a−1
integrand, which we write as f (z) = z1+z , is its branch point at z = 0. That suggests a
couple of opportunities. First, we have seen before that, when integrating from a branch
point, it can be convenient to locate the branch cut along it and choose a closed loop that
returns on the opposite side of the same cut (a “keyhole” contour). This will become our
5.1. Residue calculus
139
Method 1. Another possibility is to start by rewriting xa−1 = e(a−1) log x and change
variable t = log x, after which the integrand becomes free of branch points. Just possibly,
the new integral will provide additional solution opportunities.
Method 1: Placing the branch cut immediately below the outgoing integration path and
creating a closed loop that returns just below it suggests the contour shown in Figure 5.17.
Im z
Γ
Re z
x = −1
Figure 5.17. Keyhole contour for Example 5.14.
Let’s consider all the contributions to the closed-loop integral:
1. Res(f, −1) =
2.
3.
´
→
´
←
f (z)dz =
(−1)a−1
1
´∞
0
f (z)dz = −
= eiπ(a−1) .
xa−1
1+x dx
´∞
0
= I.
e(a−1)(log x+i2π)
dx
1+x
= e2πi (a−1) I.
´
4. ! f (z)dz → 0 as ε → 0 .
´
5. " f (z)dz → 0 as R → ∞ .
R
Putting everything together, we obtain
2πi eiπ(a−1)
1 − e2πi (a−1)
π
=
.
sin πa
I=
Method 2: Alternatively, one can consider some analytic modifications (partial integration, change of variables, etc.) before applying contour integration. Given that xa−1 =
140
Chapter 5. Residue Calculus
e(a−1) log x , the variable change t = log x would seem natural. Then x = et , dx = et dt,
and the integral becomes
ˆ ∞ t(a−1) t
ˆ ∞ a−1
e
e
x
dt
dx =
t
1+x
−∞ 1 + e
0
ˆ ∞
eta
=
dt
(5.5)
t
−∞ 1 + e
π
=
,
sin πa
where the last equality follows from the preceding Example 5.13.47
Some follow-up observations to this example include the following:
1. We recognize (5.5) as the result of Example 3.13.
2. A simple variable change generalizes (5.5) to
ˆ ∞ a−1
x
π
.
dx =
b
1+x
b sin π ab
0
3. The integral (5.5) provides one of several ways to prove a key functional identity for
the Γ(z)-function.
Theorem 5.15. The gamma function satisfies the functional equation
π
Γ(z)Γ(1 − z) =
.
sin(πz)
(5.6)
Proof. If 0 < Re z < 1, then
∞
ˆ
Γ(z)Γ(1 − z) =
0
|
=
ˆ ∞
tz−1 e−t dt
sz e−s ds
{z
} 0
Change of variable: t→su
ˆ ∞
z−1 −su
(su)
e
ˆ
sdu
0
∞
sz e−s ds .
0
Combining into a double integral, let the parameter s inside the first integral be the same as
the s inside the second one, so s−z cancels sz . Then
ˆ ∞ˆ ∞
Γ(z)Γ(1 − z) =
sz−1 uz−1 e−su ss−z e−s duds.
0
Now split up again:
ˆ
∞
uz−1
ˆ
0
0
|
∞
0
ˆ ∞ z−1
u
e−su e−s ds du =
du.
1+u
0
{z
}
1
1+u
=
π
.
sin πz
´
cos 2zt
1
following two integrals are sometimes attributed to Ramanujan: 0∞ cosh
dt = 2 cosh
and
πt
z
1
π
π
=
(valid
for
|Im
z|
<
and
|Re
z|
<
,
respectively).
These
also
follow
from
a
0
2 cos z
2
2
variable change in the result of Example 5.13.
47 The
´∞
cosh 2zt
dt
cosh πt
5.1. Residue calculus
141
We have just shown that
f (z) = Γ(z)Γ(1 − z) −
π
≡ 0 when 0 < Re z < 1.
sin πz
Recalling the idea of analytic continuation (and Theorem 2.12), this relation will hold for
all complex z.
5.1.4 Principal value integrals
A natural follow-up to discussing the effect on an integral when a singularity is located
inside vs. outside of a contour is what happens if it is located exactly on a contour (cf.
´1
Figure 5.18). For example, does a concept such as −1 dx
x make sense? Will both left and
right halves diverging to infinity make the sum meaningless? Figure 5.19(a) illustrates the
´ −ε
´ 1 dx
case of −1 dx
x + ε x , with ε small, and the contour closed in the upper half-plane. Parts
(b) and (c) give the integral values for two choices of connecting z = −ε to z = +ε.
Going around this small inner circle one full revolution in the positive direction contributes
Γ
(a)
´
Γ
f (z)dz = 0
(c)
(b)
´
Γ
´
Γ
f (z)dz = 2πiRes(f, z0 )
f (z)dz = πiRes(f, z0 )
Figure 5.18. The principal value integral can be thought of as the average of integrals with
a pole inside vs. outside of the contour.
142
Chapter 5. Residue Calculus
Γ
Γ
ε
−ε
(a) Principal value integral contour
(b)
ε
´−ε
f
(z)dz
=0
Γ
Γ
ε
−ε
´
(c)
Γ
f (z)dz = 2πiRes(f, z0 )
Figure 5.19. Contours considered for deriving the key principal value integral result.
2πi Res(f, 0), and going around it half a revolution similarly contributes ±1πi Res(f, 0),
with the ± sign according to direction. Starting from either the case in part (b) or part (c)
of the figure, we see that the integral value in case (a) (as ε → 0) should be 1πi Res(f, 0).
We have thus arrived at the following.
Key result.
As is illustrated in Figure 5.18(c) with a singularity on the boundary,
ˆ
f (z)dz = πi Res(f, z0 )
Γ
(the average of the integrals in the situations for which the singularity is inside and outside
of the contour Γ). For this result, the singularity must be on a curve segment that is locally
smooth (in particular, it cannot be at a corner).
Example 5.16. Show that
ˆ
I=
0
∞
sin x
1
dx =
x
2
ˆ
∞
−∞
sin x
π
dx = .
x
2
´ ∞ iz
Following the idea from Example 5.5, we compute 12 −∞ ez dz and extract its imaginary part (turning an ordinary integral into one of principal value type). Closing the contour by a half-circle in the upper half-plane (cf. Figure 5.20) does not, by Jordan’s lemma,
change the value (as R → ∞). Since the pole at z = 0 is located right on the integration
´ ∞ iz
iz
path, we will use half the residue Res( ez , z = 0) = 1. Thus, −∞ ez dz = 12 · 2πi = πi.
´ ∞ cos z
´∞
Separating real and imaginary parts gives −∞ z dz = 0 and −∞ sinz z dz = π, i.e.,
´ ∞ sin x
I = 0 x dx = 12 π.
5.1. Residue calculus
143
Im z
Γ
R
Re z
Figure 5.20. Contour in Example 5.16.
From this example follows two ways to express the discontinuous sign-function in
terms of continuous functions (with the latter integral being of principal value type):48

 −1
0
sign α =

+1
ˆ
if α < 0
1 ∞ sin αx
1
if α = 0 =
dx =
π −∞ x
πi
if α > 0
∞
−∞
eiαx
dx .
x
(5.7)
Various notations
´ can be
´ fflused for emphasizing that an integral is of principal value
type,49 including P , P V , , etc.
5.1.5 Half-plane splitting
A problem that arises quite frequently (and which we will revisit in Sections 9.1.8, 9.4, and
10.1) is to split a function f (z) into a sum of two functions
f (z) = f − (z) + f + (z)
(5.8)
that are singularity-free and go to zero in the upper and lower half-planes, respectively
(with here the “+” and “−” superscripts denoting the half-plane to which all growth and
singularities (if any) are confined).50 We have already utilized splittings of this form (in
1 iz
1 −iz
Examples 5.5 and 5.16): cos z = 12 eiz + 12 e−iz and sin z = 2i
e − 2i
e . The purpose
in these examples was to obtain a half-plane in which we could close integration contours.
´ ∞ sin 4x
curious extension of (5.7), implying as a special case −∞
dx = π, is that
x
sin 4x
Qn
x
cos
dx
=
π
holds
exactly
for
n
=
1,
2,
.
.
.
,
30,
but
fails
for
n = 31 (as noted in
k=1
−∞
k
x
the Preface of [31]).
49 Meaning that the path goes right through a first order pole, contributing πi times its residue.
50 In many texts, “+” and “−” are used in the opposite way as here (and/or placed as subscripts instead of
superscripts). We will later (in Chapter 10) use subscripts “+” and “−” to split functions in their parts along the
positive and negative real axes, respectively. With the present definitions, the sign will in all cases indicate the
direction of nontrivial behavior.
48 A
´∞
144
Chapter 5. Residue Calculus
To obtain the split, it suffices to know f (x) only along the real x-axis and then form
ˆ ∞
1
f (x)
f − (z) =
dx for Im(z) > 0,
(5.9)
2πi −∞ x − z
ˆ ∞
f (x)
1
dx for Im(z) < 0.
(5.10)
f + (z) = −
2πi −∞ x − z
These two functions have the required decay properties (Exercise 10.4.2) and, by the
principal value idea, return the f (z) as their sum when z is real. Each of these two
functions (assuming there are no issues with convergence) is then analytically continued
into the opposite half-planes. This can be done trivially, since (5.8) for Im z ≥ 0 gives
f + (z) = f (z) − f − (z) and for Im z ≤ 0 gives f − (z) = f (z) − f + (z). In case f (z) is
real along the real axis, it will also hold that f − (z) = f + (z) and f + (z) = f − (z).
Example 5.17. Split f (z) =
1
1+z 2
into f (z) = f − (z) + f + (z).
1
i
i
The partial fraction decomposition f (z) = 1+z
2 = 2(i+z) + 2(i−z) gives immediately
1
−
(z) =
the answer. However, as an exercise, we apply (5.9) to f (x) = 1+x
2 , giving f
´∞
1
1
dx.
We
can
close
the
contour
in
either
half-plane.
Choosing
the
upper
2πi −∞ (x−z)(1+x2 )
1
i
one, there are two poles inside the contour, with residues adding up to 1+z2 + 2(z−i) =
i
i
−
+
2(i+z) . The 2πi factors cancel, giving again f (z) = 2(i+z) . Then follows f (z) =
i
f − (z) = 2(i−z)
. Choosing instead to close the contour in the lower half-plane, the only
i
. Changing the sign (since
pole to be concerned with is at x = −i, with residue − 2(i+z)
i
−
this contour goes around the pole clockwise) gives again f (z) = 2(i+z)
. The original
function and its two half-plane components are shown in Figure 5.21.
Example 5.18. Find the f (z) = f − (z) + f + (z) decomposition in the case of f (x) =
2
e−x .
Equation (5.9) becomes
ˆ ∞ −x2
2
1
1
e
dx = e−z (1 + erf (iz)) ,
(5.11)
f (z) =
2πi −∞ x − z
2
´z
2
where erf(z) = √2π 0 e−t dt. As this integral stands in (5.11), it is difficult to evaluate
by contour integration. However, different arguments in Examples 9.21 and 9.42 both give
2
this same result f − (z) = 12 e−z (1 + erf(iz)). The bottom row of subplots in Figure 9.5
shows f (z) and its two half-plane components. These functions are visualized again (in a
slightly different context) in Figure 9.12. Parts (a) and (b) of that figure show the real and
2
imaginary parts of f (z) = e−z , and parts (e) and (f) show 2f − (z).
−
5.1.6 Winding numbers and Rouché’s theorem
We start by recalling Theorem 4.8 (Cauchy’s argument principle).
Lemma 5.19. Consider a function f (z) and a contour Γ, as schematically illustrated in
Figure 5.22. If the function f (z) has N zeros and P poles inside Γ, and there are no
¸ f 0 (z)
1
singularities other than these poles inside of Γ, then 2πi
f (z) dz = N − P .
5.1. Residue calculus
145
(a) f − (z)
(b) f + (z)
(c) f (z)
Figure 5.21. The function f (z) =
1
1+z 2
and its half-plane components f − (z) and f + (x).
Here is a slightly different argument that also shows the result. Let us modify the
contour Γ to go around each zero and each pole separately in small circles. Say z0 is a zero
of order m. Then f (z) = (z − z0 )m g(z), where g(z) is nonzero and analytic inside the
circle, and
f 0 (z)
m(z − z0 )m−1 g(z) + (z − z0 )m g 0 (z)
=
f (z)
(z − z0 )m g(z)
m
g 0 (z)
=
+
.
z − z0
g(z)
| {z }
| {z }
contributes m2iπ
contributes nothing
146
Chapter 5. Residue Calculus
Γ
Figure 5.22. Contour in Lemma 5.19, where we identify the poles as × and the zeros as •.
We can use the same reasoning for poles. The pole of order n contributes −2πin to the
integral.
From this result, we can introduce the concept of winding number:
˛ 0
1
f (z)
N −P =
dz
2πi
f (z)
1
[log f (z)]end
=
start
2πi
1
=
[ log |f (z)| +i
arg(f (z))
]end
start
| {z }
2πi | {z }
Same at start and end
changes by 2π each time f (z) goes around the origin
= {winding number of f (z) with respect to Γ}.
The color plots we have frequently used to illustrate phase angles of functions also
show the concept of winding number “in action.” As an example, Figure 2.2(d) illustrates
the phase angles for the function f (z) = z+ z1 , which has one first order pole (at z = 0) and
two simple zeros (at z = ±i). If we follow any curve that circles close to one of the zeros,
the color changes along it correspond to once around the color wheel, while going around
the pole near to it goes through the colors in reverse order. For a completely arbitrary
closed path in the z-plane, we can follow the net circulation it has produced with regard
to the color wheel. For example, going around the origin once in the positive direction far
out from it, the net effect in this case matches once around the color wheel in the positive
direction, telling us that N − P for that path equals 1. The number of times we have gone
around the color wheel when following a path equals the winding number for the path.
Example 5.20. Say f (z) has no poles inside Γ (P = 0). If in this case the winding number
is 2 (cf. Figure 5.23), then N = 2, i.e., f (z) has two zeros inside Γ.
A consequence of this observation is the following theorem.
Theorem 5.21 (Rouché’s theorem). Let f (z) and g(z) be analytic (no singularities) inside
Γ (closed). If |f (z)| > |g(z)| on Γ, then f (z) and f (z) + g(z) have the same number of
zeros inside Γ.
Proof. Let z go around Γ. If |f (z)| > |g(z)| on Γ, then |(f (z) + g(z)) − f (z)| < |f (z)|.
Thus f (z) + g(z) is always closer to f (z) than f (z) is the origin. Think of walking a dog
5.1. Residue calculus
147
Im w
Γ
w = f (z)
st ar t/end
Re w
Im z
Re z
Figure 5.23. Illustration of Example 5.20.
Im w
Re w
Figure 5.24. Illustration of the w-plane in Rouché’s theorem (Theorem 5.21). We see
w = f (z) as a solid curve and w = f (z) + g(z) as a dashed curve.
a number of times around a tree, located at the origin of the complex plane. If a (variable
length) leash at any moment is shorter than the person’s distance to the tree, the dog will
have to move around the tree equally many times as the person. In this illustration (cf.
Figure 5.24), think of the person’s position as w = f (z) and the dog’s as w = f (z) + g(z).
The leash information is |f (z)| > |g(z)|, and we have then concluded that f (z) + g(z) and
f (z) have the same winding number. By the result just above, the two functions f (z) and
f (z) + g(z) must have the same number of zeros inside Γ.
As a consequence of Rouché’s theorem, we obtain another proof of the fundamental
theorem of algebra.
Theorem 5.22. A polynomial pn (x) of degree n has exactly n roots.
Proof. Write the polynomial as
pn (z) = |{z}
z n + an−1 z n−1 + an−2 z n−2 + · · · + a1 z + a0 .
|
{z
}
f (z)
g(z)
148
Chapter 5. Residue Calculus
Let z go around a circle of radius R, centered at the origin. The function f (z) = z n clearly
has n roots inside it (all at z = 0). Around the periphery of the circle, |f (z)| = Rn and
|g(z)| < n·max |aj | Rn−1 . If we choose R > n·max |aj |, it will hold that |f (z)| > |g(z)|
on the periphery, and therefore pn (z) will also have n roots inside the circle.
Example 5.23. How many zeros does p(z) = z 4 +6z +3 have in the complex plane? Also
determine how many of these satisfy |z| < 2.
The degree of the polynomial is 4. Therefore, there are four zeros. Now, let f (z) = z 4
and let g(z) = 6z + 3. When |z| = 2, |f (z)| = 24 = 16 and |g(z)| = |6z + 3| < 6|z| + 3 =
15. Since |f (z)| > |g(z)| around the contour, f (z) and p(z) = f (z) + g(z) have the same
number of zeros inside it. The function f (z) there has four zeros, and thus so does p(z);
i.e., all four zeros of p(z) satisfy |z| < 2.
5.2 Infinite sums
An important
´ sums. When trying to
P∞ application of residue calculus is to evaluate infinite
evaluate n=−∞ f (n), the most direct approach is to consider Γ f (z)π cot πzdz around
πz
a large contour Γ . This idea is based on the fact that π cot πz = π cos
sin πz (see Figure 5.25)
has a simple pole with residue 1 at each integer value, so the residue of the integrand at
each of these poles becomes Res(f (z) π cot πz, z = n) = f (n).
sum of the residues
PThe
∞
associated with all the integer values becomes the original sum n=−∞ f (n). We choose
a large square as contour Γ, which encloses the integers −N, −N + 1, . . . , N − 1, N along
the real axis, and consider the limit of N → ∞. As also seen from Figure 5.25, the
function π cot πz converges to −πi away from the real axis in the positive imaginary side
of the complex plane and to πi away from the real axis in the negative imaginary side of the
complex plane. We further let the vertical sides of our contour cross the real axis halfway
between two consecutive poles. The function π cot πz is then bounded along the whole
contour.
Example 5.24. Evaluate for α real and nonzero the sum S =
We first rewrite S as
S=
P∞
1
n=1 n4 +α4 .
∞
1 X
1
1 1
−
.
4
4
2 n=−∞ n + α
2 α4
Following the outline above, we consider the following integral:
˛
π cot πz
dz.
I=
4
4
Γ z +α
1
1
Further, we note that z4 +α
≤ N 4 −α
4
4 as N → ∞. Thus, we can deduce that the integral
¸ π cot
along the contour vanishes: Γ z4 +απz
4 dz = 0, which implies that
N
X
n=−N
|
Res
X
3
π cot πz
π cot πz
,z = n +
Res
, z = zj = 0,
z 4 + α4
z 4 + α4
j=0
{z
}
P∞
n=−∞
f (n)
5.2. Infinite sums
149
10
5
0
-5
-10
3
2
1
-3
0
-2
-1
-1
0
x
-2
1
-3
2
3
y
(a) Re π cot πz.
10
5
0
-5
-10
3
2
1
-3
0
-2
-1
-1
0
x
-2
1
-3
2
3
y
(b) Im π cot πz.
(c) |π cot πz|.
Figure 5.25. Graphical representation of the function π cot πz, as seen from a viewpoint
in the third quadrant (i.e., the positive real axis is directed upwards and to the left).
150
Chapter 5. Residue Calculus
where zj are the poles of the function f (z) =
∞
X
f (n) = −
n=−∞
X
1
z 4 +α4 .
Therefore, we obtain
Res (f (z)π cot πz, z = zj ) .
j
iπ
ijπ
In the present case, zj = αe 4 + 2 for j = 0, 1, 2, 3. Thus Res(f (z)π cot πz, z = zj ) =
π cot πzj
. After some simplification, we obtain
4zj3
√
√ !
∞
X
√
1 1
1
sinh πα 2 + sin πα 2
3
√
√
=−
+ 4 2πα
S=
.
n4 + α 4
2 α4
cosh πα 2 − cos πα 2
n=1
Im z
Γ
Re z
Figure 5.26. Contour in Example 5.24.
Example 5.25. Evaluate
S(m) =
∞
X
1
, m = 1, 2, 3, . . .
k 2m
k=1
(generalizing the Basel problem
P∞
1
k=1 k2
=
π2
6 ).
´
πz
Following the same idea, we let f (z) = π cot
and note that f (z)dz will vanish
z 2m
when integrated around the same contour as shown in Figure 5.26, i.e.,
2S(m) + Res (f (z), z = 0) = 0 .
The residue at z = 0 follows most easily from the Taylor expansion
πz cot πz = 1 −
π 2 2 π 4 4 2π 6 6
z −
z −
z − ··· ,
3
45
945
5.2. Infinite sums
151
i.e.,
π2
π4
π6
, S(4) =
, S(6) =
,....
6
90
945
The Taylor coefficients above are closely related to the Bernoulli numbers Bk , k = 0, 1, 2,
. . . , defined by
∞
X
z
Bk k
=
z ,
(5.12)
z
e −1
k!
S(2) =
k=0
since this relation can be rearranged into πz cot πz =
general formula becomes therefore
S(m) =
P∞
2k
k B2k
k=0 (−1) (2k)! (2πz) .
∞
X
1
(−1)m+1 (2π)2m
=
B2m , m = 1, 2, 3, . . . .
k 2m
2(2m)!
The
(5.13)
k=1
Example 5.26. Evaluate
S=
∞
X
n=−∞
a2
1
.
− n2
Consider the integral
˛
I=
Γ
π cot πz
dz.
a2 − z 2
It vanishes along the same contour as shown in Figure 5.26 (as N → ∞). The function
1
π cot πa
f (z) = a2 −z
2 has simple poles at z = ±a. Thus, Res(f (z)π cot πz, z = ±a) =
−2a ,
and therefore
∞
X
1
π cot πa
=
.
S=
2
2
a −n
a
n=−∞
With a slightly different notation, this can be expressed as
∞
X
1
π cot πz
=
.
2 − n2
z
z
n=−∞
(5.14)
For alternating series, we will use the function sinππz instead of π cot πz, with the same
πf (z)
n
contour as before. The residue of sin
πz at an integer value n is (−1) f (n).
In case an infinite sum contains a binomial factor, one might be able to use the relation
ˆ
n
1
(1 + z)n
= {coeff. of z k in (1 + z)n } =
dz,
(5.15)
k
2πi C z k+1
where C is a contour around the origin.
Example 5.27. Evaluate
S=
∞ X
2n
n=0
n
·
1
.
5n
152
Chapter 5. Residue Calculus
Substituting the integral (5.15) for the binomial factor gives
∞ X
2n
n=0
n
ˆ
∞ ˆ
∞
X
1
1 X
(1 + z)2n
(1 + z)2n dz
1
· n =
·
=
n
5
2πi n=0 C (5z)
z
2πi C n=0 (5z)n
ˆ
5
dz
.
=
2πi C 3z − 1 − z 2
!
·
dz
z
It was here necessary to assume that the geometric progression is convergent, which fails
2
4
both when |z| is small and when it is large. However, for |z| = 1, it holds that (1+z)
5z ò 5 ,
1
so we choose C as the unit circle. The integrand has first order poles at z1 = 2 (3 − 5) ≈
√
0.38 and at z2 = 12 (3 + 5) ≈ 2.62. Only the first one is inside the unit circle, giving
√
√
1
1
S = 5 · Res( 3z−1−z
5)) = 5.
2 , z = 2 (3 −
5.3 Analytic continuation with use of contour
integration
Contour integration provides a powerful tool also for analytic continuation, adding opportunities beyond those discussed in Chapter 3. The approaches highlighted below apply
to functions that are defined by integrals, and they utilize contour changes, but in quite
different ways.
5.3.1 Change of contour allowing exact evaluation
Example 5.28. Continue the function f (z) =
´∞
ez t
dt.
−∞ 1+et
As it stands, this integral defines f (z) only for 0 < Re z < 1, since it otherwise
diverges at t = −∞ and at t = ∞, respectively. However, we showed in Example 5.13
that the integral (for such z-values) evaluates to f (z) = sinππz . Therefore, this amounts
to an analytic continuation of the f (z)-function to the full complex plane, as illustrated in
Figure 5.27.
5.3.2 Change of contour to extend a function across a branch cut
The following two examples illustrate this approach. In the first case, we have an analytic
solution available with which to verify the result.
´∞
1
Example 5.29. Continue the function f (z) = 0 (s+z)(s+1)
ds across Re z < 0.
1
1
1
1
From Example 5.11 (or by using partial fractions: (s+z)(s+1)
= z−1
s+1 − s+z ), it
follows that
log z
f (z) =
.
(5.16)
z−1
This is a multivalued function, with a branch cut most naturally placed along the negative
real z-axis. From the closed form (5.16), we would expect the infinitely many solution
sheets to be given by
f (z) =
log z + 2kπi
,
z−1
k ∈ Z.
(5.17)
5.3. Analytic continuation with use of contour integration
(a) f (z) =
´∞
ez t
dt
−∞ 1+et
153
and its region of convergence.
(b) f (z) =
π
sin πz .
Figure 5.27. Illustration of Example 5.28.
Since it is rare that integrals are available in this type of closed form, we will now
find its next sheet, continuing from below the negative real, utilizing contour changes and
residue calculus. We need to consider two complex planes:
• z-plane, where f (z) is defined (as used in Figure 5.28),
´∞
1
ds.
• s-plane, in which we do the contour integration 0 (s+z)(s+1)
Consider first a z-value z = z1 just below the negative z-axis, as indicated by an orange
154
Chapter 5. Residue Calculus
1.5
1
0.5
0
-0.5
4
2
4
2
0
0
-2
-2
y
-4
x
-4
(a) Re
log z
z−1 .
2
1
0
-1
-2
-3
4
2
4
2
0
0
-2
-2
y
-4
x
-4
(b) Im
log z
z−1 .
5
4
3
2
1
0
4
4
2
2
0
0
-2
-2
y
-4
(c)
log z
z−1
-4
and arg
x
log z
z−1 .
Figure 5.28. The primary solution sheet for f (z) =
log z
.
z−1
5.3. Analytic continuation with use of contour integration
155
z2
Br anch
cut
s1
Re z
0
I nt e g r at ion
path
0
z1
s2
(a) Complex z-plane.
Re s
(b) Complex s-plane.
Figure 5.29. Figures for Example 5.29.
dot in Figure 5.29(a). This causes a pole of the integrand in the s-plane at the location
s1 = −z1 (orange dot in Figure 5.29(b)).
We next carry out the following steps:
• Deform the contour in the s-plane to follow the dashed path. This makes no difference in the value of the integral.
• In the z-plane, slide the evaluation point up from z = z1 to z = z2 . In the s-plane,
the pole moves from s1 to s2 . The integral value changes smoothly, so it must now
represent f (z) on a secondary sheet.
• Change the contour in the s-plane back to the straight line. This path change crosses
1
, meaning that the integral has
over the pole at s = s2 = −z2 , with residue 1−z
2
2πi
changed in value by − 1−z2 .
We finally note that this straight line integral is exactly how f (z2 ) is evaluated on the
2πi
primary sheet. These two sheets therefore differ in value by − 1−z
, in perfect agreement
with (5.17).
´∞
Example 5.30. Continue the function f (z) = √2zπ 0 esds
across its branch cut.
2
−z
In Example 3.10, we continued the function f (z) =
P∞
n=1
zn
√
from the unit disk
n´
∞ dx
2z
√
. Figures
π 0 ex2 −z
to the full complex plane, cut along [1, +∞], obtaining f (z) =
5.30(a)–(b) illustrate this extension.
This function f (z) can be continued further, across this cut, following the same idea
as in Example 5.29. To use the same notation, we rename x to s, and then consider two
(a) Re f (z).
(b) Im f (z).
Figure 5.30. Real and imaginary parts of the function f (z) =
2z
√
π
´∞
0
dx
.
2
ex −z
156
Chapter 5. Residue Calculus
(a) Re f (z).
(b) Im f (z).
Figure 5.31. Real and imaginary parts of the function f (z) =
ing a second solution sheet.
2z
√
π
´∞
0
dx
,
2
ex −z
also includ-
separate complex planes, z = x+iy
´ ∞ for the function f (z) and s = σ +it for the integration
. If we, for example, move z from below to above
path, thus writing f (z) = √2zπ 0 esds
2
−z
the cut, we can at the same time deform the integration contour (from the straight line path
from 0 to +∞ in s) to still go from 0 to +∞, but taking a curved path in the complex
2
s-plane such that es goes above also the new value for z. The integral will then evaluate
to a value of f (z) that is smoothly continued from the primary sheet, i.e., located on the
second sheet. When we next change the path in the s-plane back to follow the straight
line from s = 0 to s = +∞, the path moves across the first order pole (in the s-plane)
of es21−z , which is located at s = (log z)1/2 . Calculating its residue as usual, the effect
q
becomes that the value of f (z) has changed by 2i logπ z . However, this integration (z
in the upper half-plane and the straight line integration path from s = 0 to s = +∞) is
exactly how f (z) is evaluated
on the primary z-plane sheet. The values of f (z) on the two
q
sheets thus differ by 2i logπ z . In the upper half-plane, we subtract this quantity from the
values of f (z) on the primary sheet, and we similarly add in the lower half-plane to obtain
the extension of f (z) to a second sheet. We recognize in Figures 5.31(a)–(b) the primary
z-plane sheet (from Figure
5.30) but see also the secondary one. Due to the difference
q
√
π
between the sheets (2i log z , with log z in the denominator), the second sheet has a new
branch point at z = 0 (barely visible as a small spike in Figure 5.31(a)).
5.3.3 Change to a Hankel-type contour
Example 5.31. Continue the gamma function Γ(z) =
´∞
0
e−t tz−1 dt.
5.3. Analytic continuation with use of contour integration
157
C
H
Figure 5.32. The contours C and H in Example 5.31.
The definition, as it stands, works only for Re z > 0, since it otherwise diverges at the
lower integration limit
´ t = 0.
Let us consider C et tz−1 dt (note the sign in the exponent) over the path marked C
in Figure 5.32 (entering from −∞ just below the branch cut along the negative real axis,
going around the origin, and exiting back to −∞ just above the cut).
Under the assumption that Re z > 0, the contribution when going around the origin
will be vanishingly small. With t = reiθ , we obtain
tz−1 = e(z−1) log t = e(z−1)(log r+iθ) = rz−1 eiθ(z−1) ,
i.e.,
tz−1 =
Thus
rz−1 e−iπ(z−1)
rz−1 eiπ(z−1)
ˆ
: on path in towards the origin (θ = −π),
: on path out from the origin (θ = π).
ˆ
C
ˆ
∞
et tz−1 dt = e−iπ(z−1)
e−r rz−1 dr − eiπ(z−1)
0
∞
e−r rz−1 dr
0
= −2i sin π(z − 1) Γ(z)
= 2i sin(πz) Γ(z).
Hence,
Γ(z) =
1
2i sin πz
ˆ
et tz−1 dt.
(5.18)
C
Up to this point, we have not achieved any analytic continuation (as convergence near the
origin in t still requires Re z > 0). However, we now change the contour to the one that is
158
Chapter 5. Residue Calculus
marked H (for Hankel) in the figure. This will not change the value of the integral, but the
fact that the contour now keeps some distance away from the origin means that the integral
will converge for all values of z, i.e., Γ(z) has been continued as a meromorphic function
to the complete complex plane.
By the identity (5.6), we can write (5.18) as
1
1
=
Γ(z)
2πi
ˆ
et t−s dt.
(5.19)
H
Example 5.32. Continue the Riemann zeta function ζ(z) =
P∞
1
n=1 nz
The definition, as it stands, works only for Re z > 1. In a first stage, we rewrite the
sum into an integral, which features the same domain restriction. However, the integral
formulation can then be continued by changing to a Hankel-type contour.
Following Exercise 3.3.6,
the integral definition of the
´ ∞ the change of variable t →
´nt∞in z−1
1
gamma function Γ(z) = 0 e−t tz−1 dt gives n1z = Γ(z)
t
e−nt dt. Therefore
0
ζ(z) =
∞
X
1
z
n
n=1
∞ ˆ ∞
1 X
tz−1 e−nt dt
Γ(z) n=1 0
!
ˆ ∞
∞
X
1
=
tz−1
e−nt dt
Γ(z) 0
n=1
ˆ ∞ z−1
1
t
=
dt.
Γ(z) 0 et − 1
=
(5.20)
For the same reason as with the original integral definition of the gamma function, this
integral representation is limited to Re z > 0. The remedy is now exactly the same as in
´ z−1
the previous example. After considering C ett −1 dt, where C is the same path as in Figure
5.32, and using the relation Γ(z)Γ(1 − z) = π/ sin(πz) to further simplify the final result,
we obtain
ˆ
Γ(1 − z)
tz−1
ζ(z) =
dt.
(5.21)
−t − 1
2πi
C e
We next change the contour from C to H where H is the same Hankel contour as previously. This new representation for the zeta function is defined and finite for all z 6= 1 (the
additional poles of the gamma function gets canceled by the fact that the integral turns out
to vanish when z = 1, 2, 3, . . .).
Apart from the branch singularity at t = 0, the integrand also features poles at t =
±2πi, ±4πi, . . . . It is assumed that the contour H is not deformed so much that any of
these fall inside it. However, in case that H is so deformed that all these poles become
enclosed by H, it is shown in Section 6.4 how this formulation (5.21) leads to a proof of
the functional equation for the zeta function:
ζ(z) = 2z π z−1 sin
1
πz Γ(1 − z)ζ(1 − z).
2
5.3. Analytic continuation with use of contour integration
159
5.3.4 Change to a Pochhammer-type contour
The primary application for this method is to functions defined by an integral over a finite
interval and where the domain restrictions come from singularities at both ends.
´1
Example 5.33. Continue the beta function B(p, q) = 0 tp−1 (1 − t)q−1 dt.
The beta function is analytic in both of its variables p and q. The integral definition
works only when Re p > 0 and Re q > 0 (since it otherwise diverges at t = 0 and at t = 1,
respectively).
One way to continue the beta function is to derive the closed-form expression B(p, q) =
Γ(p)Γ(q)/Γ(p + q) (cf. Section 6.1.2) and then rely on an analytically continued representation of the gamma function.
For the Pochhammer approach, we note that the integrand g(t) = tp−1 (1 − t)q−1 is
analytic in a complex t-plane, apart from branch point singularities at t = 0 and t = 1
(unless p and/or q are integers). We start at the origin in a complex t-plane, and move
along the path marked 1 → 2 → 3 → 4 in Figure 5.33(a), taking us back to the origin.
4
1
t =0
2
3
t =1
t
t =1
t
(a) Contour C.
t =0
(b) Contour P.
Figure 5.33. The contours C and P in Example 5.33. For the contour C, the paths between
t = 0 and t = 1 all coincide with the real line segment [0, 1], and the loops at t = 0 and t = 1 are
infinitesimally small.
160
Chapter 5. Residue Calculus
The magnitude of g(t) will be single-valued along this path, while the argument of g(t)
will change according to the sequence 0 → −2π(q − 1) → −2π(q − 1) − 2π(p − 1) →
−2π(p−1) → 0; i.e., g(t) is back to its original value when the full trajectory is completed.
We note that each of the two singularities t = 0 and t = 1 has been gone around twice, but
for each, the two times have been in opposite (canceling) directions.
With Re p > 0 and Re q > 0, the contributions around the small near-circular parts will
be vanishingly small. The integral around the full loop will thus add up to
ˆ
= B − e−2π(q−1)i B + e−2π(q−1)i e−2π(p−1)i B − e−2π(p−1)i B
C
= −4e−iπ(p+q) sin(pπ) sin(qπ)B,
i.e.,
−eiπ(p+q)
B(p, q) =
4 sin(pπ) sin(qπ)
ˆ
tp−1 (1 − t)q−1 dt.
C
We next change the contour from C to the one denoted by P (for Pochhammer) in
Figure 5.33(b). This will not change the value of the integral, but the fact that the contour
now keeps some distance away from the singularities implies that the integral will converge
for all values of p and q.
We can note that it would not have been helpful to introduce branch cuts from the two
branch points (at t = 0 and t = 1) in order to create a single-valued function. No matter
how such cuts would have been placed, the contour would have needed to move across
these artificially introduced barriers. With a Riemann surface approach to multivaluedness,
there is no difficulty in this regard.
5.4 Weierstrass products and Mittag–Leffler
expansions
Applies to
Describes a functions through its
Weierstrass Products
entire functions
(no singularities)
zeros
Mittag–Leffler Expansions
meromorphic functions
(poles are only singularities)
poles
The general treatment of these expansions is quite involved, so we will limit ourselves
to illustrative examples.
5.4.1 Weierstrass products
Let f (z) be an entire function with a finite number of zeros (all simple), located at z =
a1 , a2 , a3 , . . . , an , and consider pn (z) = 1 − az1 1 − az2 · · · · · 1 − azn . Then pfn(z)
(z)
is a zero-free entire function, and h(z) = log
h(z)
f (z)
pn (z)
will be entire. We can thus write f (z)
as f (z) = pn (z)e
. Knowing the zeros of f (z) is enough to determine pn (z), but not
enough to exactly determine f (z).
Generalizing this to the case of infinitely many zeros a1 , a2 , . . . suggests forming
∞ Y
z
p(z) =
1−
,
an
n=1
5.4. Weierstrass products and Mittag–Leffler expansions
161
bringing up the issue of
when infinite products converge. This requires investigating the
Q∞
tail end of the product n=M (1 − azn ) for M 1, to see whether this converges to a limit
that is neither 0 nor ±∞.
Example 5.34. Find a product expansion for sin πz.
The function sin πz has zeros at 0, ±1, ±2, . . . , so sinzπz has zeros at ±1, ±2, . . . ,
making it natural to form
n
z
z o n
z
z
z o
z
1−
1−
... · 1 +
1+
1+
... .
p(z) =
1−
1
2
3
1
2
3
While both of the products diverge, we can note that
p1 (z) =
∞ n
Y
z z/k o
1−
e
k
k=1
and
p2 (z) =
∞ n
Y
z −z/k o
e
1+
k
k=1
both converge by Theorem 5.40. To see why, note that the terms in p1 (z) become
1−
2 3 z
z
z
z2
z
1+ +O
=
1
−
+
O
,
k
k
k2
k2
k3
P∞
P∞
3 2
and similarly for p2 (z). The sums n=1 kz + O kz 3
= |z|2 n=1 k12 + O kz3
converge. In the present example, we can either rewrite p(z) as p(z) = p1 (z) · p2 (z) or,
simpler still, group the factors to directly give the convergent product
p(z) =
∞ Y
k=1
z2
1− 2
k
.
We will from this know that
sin πz
=
z
(
∞ Y
k=1
z2
1− 2
k
)
eh(z) ,
(5.22)
where h(z) is an entire function, still to be determined. It will turn out that eh(z) ≡ π. A
good way to see this starts by taking the log on both sides, giving
log(sin πz) = log z + h(z) +
∞
X
k=1
z2
log 1 − 2
k
Differentiating gives
∞
X
1
1
π cot πz = + h0 (z) + 2z
.
z
z 2 − k2
k=1
.
162
Chapter 5. Residue Calculus
P∞
1
0
Recalling (5.14), z k=−∞ z2 −k
2 = π cot πz, it follows that h (z) = 0 and thus h(z) = h
is a constant. Taking the limit z → 0 on both sides of (5.22) gives that eh = π. Therefore
∞ Y
sin πz
z2
=
1− 2 .
πz
k
(5.23)
k=1
A classical special case was computed by Wallis in 1670, who arrived at it in a different
way. If z = 1/2, then
π
2·2 4·4 6·6
=
·
·
· ....
2
1·3 3·5 5·7
5.4.2 Mittag–Leffler expansions
These expansions represent a function in terms of its poles.
Example 5.35. Find a closed-form expression for
f (z) =
∞
X
k=−∞
1
.
(z − k)2
Clearly the series converges and it has period 1 (indeed, f (z + 1) = f (z)). Because it
is 1-periodic, it has the same Laurent expansion at every pole. We check this expansion at
z = 0:
X
1
1
+
2
z
k 2 (1 − z/k)2
k6=0
2
X 1 1
z
z2
= 2+
·
1
+
+
+
·
·
·
z
k2
k z2
k6=0
!
∞
X
1
0
1
= 2 + +2
+ 0 z + (· · · )z 2 + · · · .
z
z
k2
f (z) =
k=1
Next consider the function
get the Laurent expansion
π2
.
sin2 πz
It is also 1-periodic. Expanding around the origin, we
π2
1
0 π2
+
+ 0z + ··· .
=
+
z2
z
3
sin2 πz
P∞
2
1
Therefore, h(z) = sinπ2 πz − k=−∞ (z−k)
2 is an entire function that also is 1-periodic.
It is easy to show that h(z) → 0 up and down the period
in Figure 5.34 (for
P∞strip shown
1
1
example by noting that |z−k|
,
2 for each k, and thus also
2
k=−∞ |z−k| are monotonically
decreasing if Im |z| increases). Thus, by Liouville’s theorem, h(z) ≡ constant. Since
both functions tend to 0 far out in the period strip, this constant must be h = 0.
We have thus obtained
∞
X
π2
1
=
,
sin2 πz k=−∞ (z − k)2
(5.24)
5.4. Weierstrass products and Mittag–Leffler expansions
163
Period Strip
Im w
Re w
Figure 5.34. Period strip for both
π2
sin2 πz
and
P∞
1
k=1 (z−k)2 .
which is a typical example of a Mittag–Leffler expansion (expressing here the function
π2
in terms of its poles).
sin2 πz
An
side result is the following alternative way to arrive at the sums I(m) =
P∞ interesting
1
k=1 k2m , m = 1, 2, 3, . . . (cf. Example 5.25). Equating the Laurent series of both functions around z = 0 gives the following sequence of relations:
∞
X
π2
1
=
,
k2
6
k=1
∞
X
1
π4
=
,
4
k
90
k=1
···
As commented on again in Section 6.2, finding corresponding closed-form expressions for
odd positive integers 3,5,7, . . . is a longstanding open problem.
Example 5.36. Find a closed-form expression for
f (z) =
∞
X
k=−∞
1
.
z−k
(5.25)
As it stands, the series is only conditionally convergent. The following are two ideas to
bypass this issue:
1. Add a term which will cancel out. Notice that
∞
X
1
1 X
f (z) =
= +
z−k
z
k=−∞
k6=0
1
1
+
z−k k
|
{z
}
size O
1
k2
, so converges
.
164
Chapter 5. Residue Calculus
2. Pair up the terms k and −k. We can then rewrite f (z) as
m
∞ X
1
1 X
1
1
f (z) = lim
= +
+
m→∞
z−k
z
z+k z−k
k=−m
k=1
∞
X
1
1
.
= + 2z
2
z
z − k2
k=1
We next show two ways to evaluate this sum.
Method 1: We notice from the last equation above that this function has the same pole
structure as π cot πz. Furthermore, both functions have period 1 and are bounded up and
down the period strip. By P
Liouville’s theorem, they can differ only by a constant. Since
∞
1
both π cot πz and z1 + 2z k=1 z2 −k
2 are odd functions, the constant has to be zero (as
there is no other odd constant). Thus
∞
X
k=−∞
1
= π cot πz .
z−k
P∞
´Method 2: The standard approach that we use for evaluating k=−∞ f (k) is to consider
f (z)π cot πzdz around the large rectangular path in Figure 5.26. As usual (although
slightly more subtle to show in this particular case), the integral vanishes as the path is
moved increasingly far out. To avoid a confusion of notation, we write (5.25) as f (ξ) =
P
∞
1
π cot πz
k=−∞ ξ−k and thus consider the integral of
ξ−z . Apart from at z integer, there is
one more pole, at z = ξ, with residue −π cot πξ. Since all residues should add up to zero,
we arrive again at f (ξ) = π cot πξ.
Note that we can similarly find that
∞
X
(−1)k
π
=
.
z−k
sin πz
k=−∞
Mittag–Leffler expansions provide a supplementary approach to the one described in
Section 5.2 for evaluating many infinite sums.
Example 5.37. Find a closed-form expression for
∞
X
S=
k=−∞
1
.
1 + k2
(5.26)
The present idea is to form the 1-periodic analytic function
G(z) =
∞
X
k=−∞
1
,
1 + (z + k)2
for which we want the value G(0). Considering G(z) in the complex plane, its only
N
=
singularities will be first order poles at z = −k ± i, k integer, with residues D
0
1
i
=
∓
.
Recalling
that
π
cot
πz
has
poles
with
residues
+1
at
all
integers,
it
2(z+k) z=−k±i
2
is clear that G(z) has exactly the same pole structure (locations and residue) as the function
H(z) =
πi
[cot(π(z + i)) − cot(π(z − i))] ;
2
5.5. Supplementary materials
165
Figure 5.35. Magnitude/phase plot of the function G(z) = H(z) in Example 5.37.
cf. Figure 5.35. The difference G(z) − H(z) is therefore an entire, 1-periodic function
which furthermore is easily seen to approach zero when |Im z| → ∞. By Liouville’s theorem, the difference is identically zero, and we obtain S = G(0) = H(0) = π coth π.
5.5 Supplementary materials
5.5.1 Convergence of infinite products
Theorem 5.38.
If the terms of the sequence {ak } are real and positive,
P∞
if and only if k=1 log (ak ) converges.
= 1. Since limn→∞ an = 0, limn→∞
Proof. limx→0 log(1+x)
x
comparison test (Theorem 2.37), the result follows.
Q∞
log(1+an )
an
k=1
ak converges
= 1. By the limit
We give without proofs two further results regarding infinite products.
P∞
Theorem P
5.39. If the terms of the sequence
{ak } are real and k=1 a2k converges, then
Q
∞
∞
the series k=1 ak and the product k=1 (1 + ak ) converge or diverge together.
Note: Convergence
Qn of the product to zero occurs only if it has a zero factor (i.e., an ak =
−1). If limQ
n→∞
k=1 (1
+ ak ) = 0 but no termQank = −1, then
the product diverges to 0.
n
Example: k=2 1 − k1 diverges, yet limn→∞ k=2 1 − k1 = limn→∞ 21 23 43 . . . n−1
n =
limn→∞ n1 = 0.
Theorem 5.40.
(uniformly).
Q∞
n=1
(1 + un (z)) converges (uniformly) in z if
P∞
n=1
|un (z)| converges
166
Chapter 5. Residue Calculus
5.6 Exercises
Exercise 5.6.1. Evaluate
¸
Γ
e1/z dz, where Γis the unit circle.
1
Exercise 5.6.2. If |ζ| < R, show that 2π
ζ).
Hint: Let z = R eiθ , write the integral as
of this with respect to ζ.
´ 2π
0
log |R eiθ − ζ| dθ = log R (independent of
1
2π Im
¸
log(z−ζ)
dz,
z
and consider the derivative
Exercise 5.6.3. Assume f (z) is analytic for |z| ≤ R and that it has there zeros at
z = z1 , z2 , . . . , zn , none of which are at z = 0 and |z| = R. Show Jensen’s formula
(generalizing (4.18)):
Rn
1
log |f (0)| + log
=
z1 · z2 · · · · · zn
2π
ˆ
f (z)
(z−z1 )(z−z2 )·····(z−zn )
Hint: Apply (4.18) to g(z) =
5.6.2.
2π
log |f (R eiθ )|dθ .
0
and then use the result in Exercise
´∞
π
Exercise 5.6.4. Show that 0 x4 +2x2dx
cos 2α+1 =| 4 cos α | if cos α 6= 0.
Hint: Key solution steps include showing that the four poles are at the locations {i eiα ,
i e−iα , −i eiα , −i e−iα } and have residues
1
1
i
1
1
i
1
1
i
1
1
i
−
,
−
−
,
−
+
,
+
,
8 sin α
cos α
8
sin α
cos α
8
sin α
cos α
8 sin α
cos α
respectively.
Exercise 5.6.5. Show that
Exercise 5.6.6. Evaluate
´∞
0
e−x −cos x
x
´∞
Exercise 5.6.7. Show that
x sin x
dx,
−∞ a2 +x2
´∞
dx
−∞ 1+x2p
=
dx = 0.
where a is a real-valued nonzero constant.
π
,
π
p sin( 2p
)
where p is a positive integer.
Exercise 5.6.8. If α and β are real and positive constants, show that
ˆ
0
∞
cos αx
dx =
x+β
ˆ
0
∞
x e−αβx
dx.
1 + x2
If you were to numerically evaluate the result for some different values of α and β,
which of the two versions do you think would be best to use?
´t
Exercise 5.6.9. We want to evaluate I = 0 a+bdθcos θ , where a >| b |, and t lies in the
interval (0, 2π). By the change of variable z = eiθ , derive
ˆ 1
1
1
I= √
−
dz,
(5.27)
z − z2
i a2 − b2 Γ z − z1
5.6. Exercises
167
where z1 and z2 are the pole locations of the integrand in the z-plane, and Γ is the arc of
the unit circle from z = 1 to z = eit .
Note: Although we don’t have a closed loop (and thus cannot readily use the residue
method), we have nevertheless arrived at an integral (5.27) that we can evaluate directly,
giving rise to two logarithms. Since we know that the result will be real, we furthermore
need only to consider the imaginary √parts of these logarithms. After some simplifications,
2 −b2 sin t . You do not need to carry out any of the
one arrives at I = √a21−b2 arctan ab+a
cos t
steps beyond arriving at (5.27).
Exercise 5.6.10. Show that
not equal to zero.
´∞
cos ax
dx
−∞ b2 +x2
´∞
π −|a||b|
,
|b| e
=
where a and b are real, and b is
log x
1+x2 dx
= 0 in the following two ways:
´1 ´∞
´∞
(a) Without contour integration, split 0 = 0 + 1 and in one of these integrals,
change variable x → 1t .
(b) Use contour integration.
Note: Part (a) is a reminder that, even for integrals that look natural for contour integration, it is worthwhile to also look for simplifying reformulations. The same idea as here
gives, for example, the surprising result that
ˆ ∞
dx
π
= ,
2 )(1 + xa )
(1
+
x
4
0
Exercise 5.6.11. Show that
0
independently of the value for a.
´∞
= 2√π2 a3 .
´∞
Hint: The variable change t − 1/t = x gives I(a) = −∞ 21 1 +
Exercise 5.6.12. Show that I(a) =
0
dt
a4 +(t−1/t)4
√ x
4+x2
dx
a4 +x4 .
Exercise 5.6.13. Assume a > 0. Utilize the result in Exercise 5.6.11 to show that
ˆ ∞
π log a
log x
dx =
.
2 + a2
x
2a
0
Exercise 5.6.14. Show that (a)
´∞
0
log2 x
1+x2 dx
=
π3
8 ,
(b)
´∞
0
log x
(1+x2 )2 dx
= − π4 .
´ π/2
Exercise 5.6.15. Evaluate 0 log(sin x)dx by
(a) the result in Exercise 2.9.27,
(b) utilizing the relation sin x = 2 sin x2 cos x2 ,
(c) direct contour integration.
Exercise 5.6.16. Show that
Exercise 5.6.17. Show that
´∞
0
´∞
x1/3
1+x3 dx
=
x4
dx
−∞ 1+x8
π
3 cos
=
π
2
π
18
.
q
1−
√1 .
2
Exercise 5.6.18. Evaluate the same integral as in Example 5.10: I =
but use instead the keyhole contour shown in Figure 5.8.
´∞
0
dx
x3 +a3 ,
a > 0,
168
Chapter 5. Residue Calculus
Exercise 5.6.19. Show that
f (z) =
´∞
dx
(x+a)2 +b2
0
1
b
=
arctan
b
a
(a ≥ 0, b > 0) by considering
log z
(z+a)2 +b2 .
´∞
log x dx
0 (x+a)2 +b2
2
(log z)
(z+a)2 +b2 .
Exercise 5.6.20. Show that
by considering f (z) =
=
1
b
arctan
b
a
log
√
a2 + b2 (a ≥ 0, b > 0)
´∞
cos γx dx
π
−βγ
Exercise 5.6.21. Show that 0 (x2 +α
− β e−αγ ), where
2 )(x2 +β 2 ) = 2αβ(α2 −β 2 ) (α e
α, β, γ are all positive. Do you need to consider separately the case of α = β?
Note: The result generalizes immediately to α, β, γ real, since the integrand is an even
function of its three parameters.
Exercise 5.6.22. Assuming that 4c − b2 > 0, show that
Exercise 5.6.23. Show that
ˆ
∞
0
´∞
dx
−∞ (x2 +bx+c)2
=
4π
.
(4c−b2 )3/2
1 ea + 1
1
sin ax
dx
=
−
.
2πx
a
e
−1
4 e − 1 2a
(5.28)
´ ∞ ez log t
Exercise 5.6.24. Consider the function f (z) = 0 (1+t
2 )2 dt:
(a) Determine the region in the complex z-plane where the integral converges.
(b) Analytically continue f (z) by means of re-expressing it as f (z) = 4π(1−z)
.
cos 1 πz
2
´ ∞ cos ax − cos bx
dx = π(b − a).
−∞
x2
bz
is an entire
Hint: Correct what is wrong with the following argument: f (z) = cos az−cos
z´2
´∞
∞ eiaz −eibz
dz.
function, so any closed-loop integral is zero. Also −∞ f (z) dz = Re −∞
z2
Now closing
the
loop
far
out
in
the
upper
half-plane
makes
a
negligible
change
due
to
the
´
´∞
estimate x · · · dz ≤ M · L ≤ R22 · πR → 0 as R → ∞. Hence −∞ · · · dx = 0.
Exercise 5.6.25. Show that for a, b ≥ 0,
Exercise 5.6.26. If A + B + · · · + C = 0 and a, b, . . . , c are positive, show that
ˆ ∞
A cos ax + B cos bx + · · · + C cos cx
dx = −A log a − B log b − · · · − C log c .
x
0
Note: Later, in connection with Laplace transforms, we obtain a quite different approach
(to contour integration) for solving this problem; cf. Exercise 9.8.19.
ffl ∞
π
= − 3√
.
´ 0 3 dx
√ .
Hint: It follows from Example 5.10 that −∞ x3 −1 = − 32π
3
Exercise 5.6.27. Show that
0
dx
x3 −1
ffl ∞ a−1
Exercise 5.6.28. Show that 0 xx−1 dx = −π cot πa, 0 < a < 1.
´0
a−1
Hint: The value for −∞ xx−1 dx follows from Example 5.14.
Exercise 5.6.29. Show that
´ 2π
0
2−sin θ
2−cos θ dθ
=
4π
√
.
3
5.6. Exercises
169
Exercise 5.6.30. Show the following:
´ 2π
(a) 0 5+3dθcos θ = π2 .
´ 2π
5π
(b) 0 (5+3dθ
cos θ)2 = 32 .
Exercise 5.6.31. The integrals in Exercise 5.6.30 can be generalized significantly. Assume
a > b > 0. Then show the following:
´ 2π
2π
(a) 0 a+bdθcos θ = (a2 −b
2 )1/2 .
´ 2π
dθ
2πa
(b) 0 (a+b cos θ)2 = (a2 −b
2 )3/2 .
´ 2π
π(2a+b)
dθ
(c) 0 (a+b cos2 θ)2 = a3/2 (a+b)3/2 .
´ 2π
´ 2π
2m−1
π
Exercise 5.6.32. Show that 0 (cos θ)2m dθ = 0 (sin θ)2m dθ = 4m−1
m−1 , m =
1, 2, 3, . . . .
´ 2π
´ 2π
Note: Trivially, 0 (cos θ)2m−1 dθ = 0 (sin θ)2m−1 dθ = 0, m = 1, 2, 3, . . . .
Exercise 5.6.33. Evaluate
´ 2π
0
cot
θ−a−i b
2
dθ, where a, b are real nonzero constants.
√
√
4
Exercise 5.6.34. The function f (x) = e− x sin 4 x has the strange property of being
orthogonal to every polynomial over the interval [0, ∞] (something which is impossible for
a´ not identically zero continuous function over a finite interval). In particular, it holds that
∞ n
x f (x)dx = 0 for n = 0, 1, 2, . . . . Prove this.
0
´∞
√
2
Exercise 5.6.35. Several ways are available to show that −∞ e−x dx = π (see Exercise
6.5.8). It is, however, surprisingly tricky to do this with residue calculus. One way this
2
can be achieved is by integrating f (z) = eiπz tan πz (shown in Figure 5.36(a)) over the
contour shown in its part (b) and letting R → ∞. Carry this out.
Exercise 5.6.36. Solve the ODE w0 (z) = zw(z) + z p , with w(0) = w0 using the series
method, for p = 1 and p = 2.
Exercise 5.6.37. Knowing that
´∞
−∞
2
e−x dx =
√
π , contour integration of the function
−z 2
f (z) = e
leads´ to a number of´related relations:p
∞
∞
(a) Show that 0 cos x2 dx = 0 sin x2 dx = 12 π2 .
´ ∞ −x2
´∞
2√
2
(b) Show that 0 e
cos(2ax)dx = 21 e−a π and that 0 e−x sin(2ax)dx =
´
2
2
a
e−a 0 ex dx.
Note: We will come across the latter expression (known as Dawson’s function D(z) =
2 ´z
2
e−z 0 et dt) again as the Hilbert transform of a Gaussian; cf. Example 9.42. Its elementary properties include that it satisfies the ODE dD(z)
dz + 2zD(z) = 1, D(0) = 0.
Exercise 5.6.38. One striking feature in Figure 5.37 is that the roots of the derivative of a
polynomial seem to fall within the convex hull (smallest enclosing convex polygon) of the
polynomial’s roots. Prove the Gauss–Lucas theorem, which states that this will always be
the case.
170
Chapter 5. Residue Calculus
2
(a) Magnitude and phase of f (z) = eiπz tan πz.
(R,iR) (R+1,iR)
4
2
-6
-4
-2
2
4
-2
-4
(-R,-iR) (-R+1,-iR)
(b) Integration contour
Figure 5.36. Integrand and contour in Exercise 5.6.35.
6
5.6. Exercises
171
2.5
1
2
1.5
0.5
1
0
0.5
0
-0.5
-0.5
-1
-1
-1.5
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2
-1
0
1
2
3
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
Figure 5.37. Three cases of randomly chosen degree 10 polynomials p10 (z) with their
roots shown in red and the nine roots of p010 (z) shown in green.
Hints: Note that (i) it suffices to show that if all roots of p(z) fall in the right half-plane,
so do those of p0 (z), (ii) the result is true for a root of p0 (z) that P
is also a root of p(z), and
1
(iii) equation (4.5) tells that if z is a simple root of p0 (z), then
z−ak = 0. Rearrange
P ak
P 1
this into z
|z−ak |2 =
|z−ak |2 , and deduce the result.
Exercise 5.6.39. Carry out the half-plane splitting f (z) = f − (z)+f + (z) for f (z) =
Exercise 5.6.40. Evaluate S(m) =
Exercise 5.6.41. Show that
(−1)k
k=1 k2m ,
P∞
(−1)k
k=0 (2k+1)3
P∞
=
π3
32
m = 1, 2, 3, . . . .
.
Exercise 5.6.42. Find a closed-form expression for
x < π.
(−1)k
k=−∞ 1+k2
P∞
cos kx, where −π <
P∞
PN
Exercise 5.6.43. Interpreting k=−∞ as limN →∞ k=−N , show the following:
(a)
∞
X
(−1)k eikx
π/α −i x/α
=
e
for − π < x < π.
1 + αk
sin π/α
k=−∞
sin z
z .
172
Chapter 5. Residue Calculus
(a) Re f (z)
(b) Im f (z)
´ ∞ e−t
Figure 5.38. Real and imaginary parts of the Stieltjes function f (z) = 0 1+zt
dt, when
continued from the primary sheet in the lower half-plane across the negative real axis. Note the
complicated structure of the second sheet near the origin.
(b) By means of a variable change in the result in (a), show that
( π/α
∞
−i(x+π)/α
X
for − π < x < 0,
eikx
sin π/α e
=
π/α
−i(x−π)/α
1 + αk
for 0 < x < π.
sin π/α e
k=−∞
(c) Derive the result in (b) directly by contour integration.
Hints for (c): (i) Consider first 0 < x < π; the result for −π < x < 0 follows then by
complex conjugation. (ii) Just like π cot πz has poles with residue 1 at z integer, so does
π(cot πz − i), which furthermore goes to zero in the lower half-plane.
P∞
Exercise 5.6.44. Show that k=1 z12 = −1, where the zk are the (infinitely many) soluk
tions to the equation z = ez .
z
−1
has a pole of residue one at each of these zk -locations. Therefore
Hint: The function eez −z
´
z
e −1
consider C z2 (ez −z) dz around a large rectangle.
Exercise 5.6.45.
(a) Show that ez − 4z 2 + 1 = 0 has exactly two roots for | z | < 1.
´z
(b) Show that |ez −1| ≤ e+1 for |z| ≤ 1, for example by considering ez −1 = 0 ew dw.
Use this result to establish that ez − (2z + 1) = 0 has exactly one root for | z | < 1.
´ ∞ e−t
Exercise 5.6.46. The Stieltjes function f (z) = 0 1+zt
dt was illustrated in Figure 3.8. If
we place the branch cut along [0, ∞] instead of along [−∞, 0], the continuation from the
lower to the upper half-plane will instead look as shown in Figure 5.38. The origin is still
the only singular point, but that now has a more complicated structure. Give a closed-form
expression for how the upper half-plane now differs from the primary sheet version of it.
Chapter 6
Gamma, Zeta, and
Related Functions
The gamma function Γ(z) is ubiquitous in the mathematical and applied sciences. All its
key properties have been well known for centuries. The zeta function ζ(z) (for complex
arguments z) became prominent in a pioneering study of the distribution of primes, presented in 1859 by Bernhard Riemann. Although many formulasPrelate the two functions,
∞
and the zeta function is easily defined (for example as ζ(z) = n=1 n1z if Re z > 1), it
still contains some of the most outstanding mathematical challenges of our time, and, as
such, has provided inspiration to generations of professional mathematicians and amateurs
alike.
6.1 The gamma function
We introduced the gamma function in Section 2.7 as
ˆ
∞
e−t tz−1 dt
Γ(z) =
0
and showed that it satisfies the functional equations Γ(z + 1) = zΓ(z) (see (2.19)) and
π
Γ(z)Γ(1 − z) = sin(πz)
(Theorem 5.15 and (5.6)). We also noted that it generalizes the
factorial function both to noninteger and to complex arguments, since n! = 1·2·3·· · ··n =
Γ(n+1). It can be evaluated rapidly for all complex z by combining either of the functional
equations with the asymptotic expansion (12.23), which is highly accurate when Re z is
large. We see immediately from (3.7) that the only singularities of the gamma function are
poles at the nonpositive integers, with residues
Res(Γ(z), z = −n) =
(−1)n
, n = 0, 1, 2, . . . .
n!
(6.1)
As shown in Figure 6.1, the rapid growth in magnitude for increasing Re(z) is associated
with oscillations in the real part (and likewise in the imaginary part) as we move away from
the real axis. The function Γ(z) approaches zero very rapidly as we move out in the left
half-plane (except for the string of poles along the negative real axis).
173
174
Chapter 6. Gamma, Zeta, and Related Functions
15
10
5
x
-4
-3
-2
-1
1
2
3
4
-5
-10
-15
(a) Γ(x) for x real.
(b) Re Γ(z).
(c) |Γ(z)| with phase coloring.
(d) |1/Γ(z)| with phase coloring.
Figure 6.1. Graphical representations of Γ(z) and 1/Γ(z).
6.1.1 Product representations of the gamma function
There are two important product representations of the Γ(z) function.
Theorem 6.1.
Γ(z) = lim
n! nz
.
n→∞ z(z + 1) · · · · · (z + n)
(6.2)
∞ h
Y
1
z −z/n i
= z eγz
1+
e
,
Γ(z)
n
n=1
(6.3)
Theorem 6.2.
where
γ = lim
n→∞
1+
1 1
1
+ + · · · + − log n
2 3
n
≈ 0.5772156649 .
(6.4)
6.1. The gamma function
175
Proofs for both are given in Section 6.4. From these relations, several further results
follow. For example, (6.3) shows immediately that 1/Γ(z) is an entire function, and in
1
particular that Γ(z) will never take the value of zero. Forming Γ(z)Γ(−z)
leads in a few steps
π
(utilizing (5.23)) to the functional relation Γ(z)Γ(1 − z) = sin(πz) (cf. (5.6)). Substituting
Γ(1 − z) = −zΓ(−z) leads to an exact formula for the magnitude of the gamma function
up/down the imaginary axis. For y real,
|Γ(iy)| =
π
y sinh πy
1/2
.
6.1.2 The beta function
The beta function B(p, q) is defined by
ˆ
1
tp−1 (1 − t)q−1 dt
B(p, q) =
(6.5)
0
(choosing the principal values for the powers) and is related to the gamma function through
the relation
Γ(p)Γ(q)
B(p, q) =
,
(6.6)
Γ(p + q)
proved below in Section 6.4. These equations (6.5) and (6.6) provide one (of several) ways
to prove the duplication formula51 for the gamma function.
Theorem 6.3 (Legendre).
1
Γ(2z) = π −1/2 22z−1 Γ(z)Γ z +
.
2
(6.7)
Three different ways to prove this relation are outlined in Exercises 6.5.9(a), 6.5.9(b),
and 12.6.9.
While the definition (6.5) only works for Re p > 0 and Re q > 0 (as the integral otherwise diverges), we saw in Example 5.33 how it could be continued to all p and q by means
of a Pochhammer integral. The explicit formula (6.6) provides also such a continuation.
Replacing t by 1 − τ shows that B(p, q) = B(q, p). Other substitutions that lead to often
useful identities are
ˆ π/2
2
t = sin θ gives B(p, q) = 2
(sin θ)2p−1 (cos θ)2q−1 dθ
(6.8)
0
ˆ
and
τ
t=
1+τ
∞
gives B(p, q) =
0
tp−1
dt.
(1 + t)p+q
(6.9)
In Section 11.3.3, we will come across the incomplete beta function, which
´ z has one
additional argument z replacing 1 as the end point of the integral: B(p, q; z) = 0 tp−1 (1−
t)q−1 dt.
51 This
formula is the m = 2 special case of Γ(mz) = (2π)
1, 2, 3, . . . (due to Gauss).
1−m
2
1
mmz− 2
Qm−1
k=0
Γ z+
k
m
, m =
176
Chapter 6. Gamma, Zeta, and Related Functions
Figure 6.2. The beta function B(p, q) displayed over the domain −2.5 ≤ p, q ≤ 6.2.
Since B(p, q) depends on two independent complex parameters, it is difficult to display it graphically. However, in the special case that both p and q are real, it can be
illustrated as in Figure 6.2. It decays smoothly to zero inside the first quadrant in the
(p, q)-plane. However, it follows from (6.6) that it flips between ±∞ when either p or q
crosses 0, −1, −2, . . . . Furthermore, p + q = 0, −1, −2, . . . are zero contour lines.
6.2 The zeta function
In Section 3.2.3,
P∞ we analytically continued the Riemann ζ-function, which we introduced
as ζ(z) = n=1 n1z , from Re z > 1 down to Re z > 0 using the reformulation ζ(z) =
P∞ (−1)n+1
1
and finally to the entire complex plane by means of the functional
n=1
(1−21−z )
nz
z z−1
equation ζ(z) = 2 π
sin 21 πz Γ(1 − z)ζ(1 − z). We also found that ζ(z)’s only
singularity is a simple pole at z = 1, with residue 1.
Regarding values of the zeta function at integer arguments, we deduced from Example
5.25 that
ζ(2k) =
(−1)k+1 B2k (2π)2k
, k = 1, 2, 3, . . . ,
2(2k)!
(6.10)
and then, via the functional equation, that
ζ(−k) = (−1)k
Bk+1
, k = 0, 1, 2, . . . ,
k+1
(6.11)
P∞
where the B-coefficients stand for the Bernoulli numbers defined by ezz−1 = k=0 Bk!k z k
1
1
(see (5.12)), i.e., B0 = 1, B1 = − 21 , B2 = 16 , B3 = 0, B4 = − 30
, B5 = 0, B6 = 42
,....
In particular, ζ(k) vanishes at all the negative even integers, forming the trivial zeros of the
zeta function. Missing from this summary of values at integer locations are the odd positive
integers 3, 5, 7, . . . . Surprisingly little is known about these, with the main exception that
ζ(3) is known to be irrational.
6.2. The zeta function
177
(b) |ζ(z)|
(a) Re ζ(z)
(c)
1
ζ(z)
Figure 6.3. Plots from left to right of ζ(z) in the strip [−5, 5] × [−50, 50] surrounding the
imaginary axis: (a) Re ζ(z), (b) |ζ(z)|, and (c) |1/ζ(z)|. In the last subplot, the first nontrivial zeros
along the line Re z = 12 are seen as narrow spikes.
6.2.1 The critical strip
The vertical strip 0 ≤ Re z ≤ +1 is for the zeta function called its critical strip. Figure
6.3 shows the function in the somewhat wider strip −5 ≤ Re z ≤ +5 (and −50 ≤ Im z ≤
+50) surrounding the imaginary axis, and Figure 6.4 shows ζ(x) along the real x-axis,
with the first few of the trivial zeros, located at the negative even integers, highlighted with
red dots. Figure 6.5 shows |ζ(z)| along two sections of the critical line Re z = 12 . The
nontrivial zeros are seen to be quite irregularly distributed.
In the context of the zeta function, it is common in the literature to denote the complex
plane by s = σ + it rather than by z = x + iy. Hence, we use t in place of y in some cases
below.
(x)
2
1
x
-20
-15
-10
-5
5
10
-1
-2
Figure 6.4. ζ(x) plotted along the real axis. We notice the pole at x = 1. The trivial zeros
at x = −2, −4, −6, . . . are marked with red dots.
178
Chapter 6. Gamma, Zeta, and Related Functions
3.0
2.5
2.0
1.5
1.0
0.5
10
20
30
40
(a) |ζ( 12 + it)| shown for 0 ≤ t ≤ 40.
5
4
3
2
1
110
120
130
140
(b) |ζ( 12 + it)| shown for 100 ≤ t ≤ 140.
Figure 6.5. Plots of |ζ( 12 + it)|.
The Riemann hypothesis is one of the most famous unsolved problems in all of mathematics. It states that all of the (infinitely many) nontrivial zeros lie exactly on the line
Re z = 12 , by now known to be true for the first 1013 of them. Either proving or disproving
the Riemann hypothesis would have extensive consequences in mathematics. At the time
this book is being written, there is a one million dollar prize for solving this problem.52
6.2.2 Relation to prime numbers
Already, the ancient Greeks knew that there are infinitely many prime numbers. As a minor
illustration of the extensive connections between primes and the ζ- function, we will show
below this same result based on the fact that ζ(z) has a pole at z = 1.
Theorem 6.4. There are infinitely many primes.
52 One
of the seven Millennium Problems, as presented by the Clay Mathematics Institute.
6.2. The zeta function
179
Proof. Let {p1 , p2 , p3 , . . .} = {2, 3, 5, . . .} be the set of all the prime numbers. Because
every integer can be written in exactly one way as a product of primes,
1
1
1
1
1
1
1 + z + 2z + · · ·
1 + z + 2z + · · ·
1 + z + 2z + · · · · · ·
p1
p1
p2
p2
p3
p3
1
1
1
1
1
= z + z + z + z + z + · · · = ζ(z).
1
2
3
4
5
Each of the factors in the product is an infinite geometric progression that can be summed
in closed form, giving the relation
∞ Y
1
ζ(z) = 1/
1 − z , Re z > 1.
(6.12)
pk
k=1
If there were only finitely many primes, this would be a finite and not an infinite product.
For z = 1, it would then evaluate to a finite rational number, contradicting the fact that
ζ(1) = ∞.
The function π(x) denotes the number of primes not exceeding x. Viewed as a function
of x real (rather than just of x integer), it is a step function that increases by 1 every time x
reaches a prime. Some manipulations of (6.12)53 lead to
ˆ ∞
π(x)
log(ζ(z)) = z
dx.
(6.13)
z − 1)
x
(x
0
Because π(x) = 0 for x < 2 (since 2 is the first prime), the lower integration limit can be
set to 2 instead of to 0.
A key goal is to invert relations of this type, to obtain better insights about the primes.
Instead of having ζ(z) as an explicit function of π(x), it would be preferable to have π(x)
as an explicit function of ζ(z). A somewhat
Pmore effective starting point than the function
π(x)
(which
we
can
define
as
π(x)
=
p<x 1) turns out to be the function ψ(x) =
P
log
p
(which
increases
by
log
p
each
time x passes an integer power of a prime).
pn <x
54
This function can be shown (H. von Mangoldt, 1895) to satisfy
X xρ
1
x2
ψ(x) = x −
+ log
− log 2π,
(6.14)
ρ
2
x2 − 1
ρ
where ρ runs over all the nontrivial zeros of ζ(z).
The exact distributions of the primes can in theory, by relations such as (6.14), be
deduced from the locations of the ζ-function’s nontrivial zeros, which, however, still (and
for the foreseeable future) contain many unsolved questions. The independent discoveries
in 1896 by J. Hadamard and C.-J. de la Vallée Poussin that these zeros must lie strictly
inside the critical strip sufficed for them to prove the prime number theorem:
π(x)
= 1.
x→∞ x/ log(x)
lim
(6.15)
P
P∞
1
(6.12), with Re z > 1, it follows that log ζ(z) =
p log 1−p−z =
n=0 π(n) − π(n −
P∞
P∞
1
1
1
1) log 1−n
=
π(n)
log
−
=
π(n)
log
1 − (n + 1)−z −
−z
n=0
n=0
1−n−z
1−(n+1)−z
´
P
n+1
d
z
log 1 − n−z . Since dx
log(1 − x−z ) = x(xzz−1) , this gives log ζ(z) = ∞
dx
n=0 π(n) n
x(xz −1)
´ ∞ π(x)
P∞ ´ n+1 π(x)
= z n=0 n
dx
=
z
dx.
z
z
0 x(x −1)
x(x −1)
´
54 The counterpart to (6.13) becomes d log(ζ(z)) = −z ∞ ψ(x)x−z−1 dx, which is in the form of a Mellin
0
dz
transform (cf. Section 9.3).
53 From
180
Chapter 6. Gamma, Zeta, and Related Functions
180
160
140
120
100
80
60
40
20
0
0
100
200
300
400
500
600
700
800
900
1000
x
Figure 6.6. The function π(x) with two approximations. The lower curve always underestimates π(x). In contrast, the function π(x) will intersect the upper curve infinitely many times as
x → ∞. These two curves get somewhat close around x = 110. The first intersection has not yet
been accurately located, but is known to occur somewhere in the interval 1019 < x < 1.4 · 10316 .
In terms of the ψ-function, this corresponds to limx→∞ ψ(x)
x = 1. Figure 6.6 compares up
x
to x = 1, 000 the true π(x) against the approximation indicated by (6.15) π(x) ≈ log(x)
´x 1
and the (for very large x) much more accurate π(x) ≈ 2 log(t) dt.55
The pioneering paper on the topic of primes and the zeta function was written by B. Riemann in 1859. Excellent survey books (ranging from elementary to increasingly advanced)
include [10, 13, 38].
While the ζ-function provides the most striking connection between analytic functions
and number theory (properties of integers), other analytic functions can also provide such
insights. Exercise 6.5.15 gives a classical illustration of this.
6.2.3 The critical line
Much attention has been given to the ζ-function not just in the critical strip 0 ≤ Re z ≤ +1
but also on the critical line itself, Re z = 21 .56 The functional equation (3.5) can be rearranged into the form
1
+ i t = e−i ϑ(t) Z(t),
ζ
(6.16)
2
where both ϑ(t) and Z(t) are analytic functions in t, odd and even, respectively, that are
purely real-valued for t real. The Riemann–Siegel theta function ϑ(t)
1 it
1 it
log π
i
log Γ
+
− log Γ
−
−
t
ϑ(t) = −
2
4
2
4
2
2
∞ X
γ + e log π
t
2t
=−
t − arctan 2t +
− arctan
2
n
4n + 1
n=1
describes the phase angle of ζ 12 + it . It becomes increasingly smooth far out both to the
right and to the left in the complex t-plane; cf. Figure 6.7(a) for its values along a section
55 The
convergence in (6.15) is very slow. At the largest x-value for which π(x) is presently known exactly,
π(x)
x = 1027 , x/ log(x) ≈ 0.984.
56 A key tool for this is the Riemann–Siegel formula, described in Section 12.5.1.
6.3. The Lambert W-function
181
20
15
10
5
0
-5
0
10
20
30
40
30
40
t
(a) Function ϑ(t)
3
2
1
0
-1
-2
-3
0
10
20
t
(b) Functions Z(t) (solid) and ζ
1
2
+ it
(dotted)
Figure 6.7. Displays of the two functions ϑ(t) and Z(t) introduced in (6.16).
of the positive real axis. The function Z(t), displayed similarly in part (b), retains all the
information about the nontrivial zeros. Each time it crosses zero for t real corresponds
to a zero of the ζ-function exactly on the critical line. In view of (6.16), |Z(t)| (here
shown as a dotted curve) matches perfectly the function ζ 12 + it that was displayed
before in Figure 6.5(a). Riemann’s hypothesis would fail if any oscillation of the Z(t)function (beyond its first zero) would not quite reach up/down to the level zero. Figure
6.5(b) indicates a slightly “close call” around t = 112. Higher up along the critical line,
numerous extremely close calls have been documented, but, as of yet, no exception has
been found.
6.3 The Lambert W-function
Part (a) of Figure 6.8 shows the elementary function f (x) = x ex and part (b) its inverse
function, commonly denoted as the Lambert function W(x) (graphically a reflection of
182
Chapter 6. Gamma, Zeta, and Related Functions
x exp(x)
0.5
0.4
0.3
0.2
0.1
x
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
-0.1
-0.2
-0.3
(a) f (x) = x ex
W(x)
-0.2
0.2
0.4
x
-0.5
-1
-1.5
-2
-2.5
-3
-3.5
(b) W(x)
Figure 6.8. The function f (x) = x ex and its inverse function W(x).
f (x) in the line y = x and satisfying x = W(x)eW(x) ). It shows a primary branch valid for
−1/e < x < ∞ and a secondary (lower) one for −1/e < x < 0. This function W(x) also
appears in a large number of applications (which are beyond the scope of this summary).
Interesting mathematical properties include the following:
P∞
• On the primary sheet, W(z) = n=1
by the ratio test in Theorem 2.7),
(−n)n−1 n
z
n!
(converging for |z| < 1/e, as seen
• It provides closed-form solutions in x for equations such as px = ax + b (p > 0 and
log a
a 6= 0), and xx = a (a > 0, for which x = W(log
a) ),
• When converging, z z
z.
.
=
W(− log z)
− log z ,
6.3. The Lambert W-function
183
Real part
0.5
0
-0.5
-1
0.4
0.4
0.2
0.2
0
0
-0.2
-0.2
-0.4
-0.4
y
-0.6
x
-0.6
(a) Re W(z)
1
Imag part
0.5
0
-0.5
-1
0.5
0.5
0
0
-0.5
y
-0.5
x
(b) Im W(z)
Figure 6.9. Real and imaginary parts of the primary solution branch of W(z).
• It solves certain nonlinear ODEs, such as
d
dz W(z)
=
W(z)
z(1+W(z)) ,
• It provides solutions to certain delay-differential equations, such as dy
dt = αy(t) +
βy(t − τ ) with, as initial condition, y(t) specified throughout some interval t ∈
[−τ, 0].
Figure 6.9 shows how the primary branch of W(x) extends over the complex plane,
single-valued when placing a branch cut along −∞ ≤ z ≤ −1/e. These extensions are
somewhat reminiscent of the upper solution sheets for the square root function f (z) =
z 1/2 , as shown in Figure 2.18. Figure 6.10 shows the W(z)-function when also including
the next sheet in each direction across the branch cut. The lower solution branch from
Figure 6.8(b) is now apparent. The sheet structure has become more reminiscent of that
for the log z function (cf. Figure 2.23), not entirely surprising, since that was the inverse
function to ez and W(z) is the inverse function to z ez . The function W(z) has branch
184
Chapter 6. Gamma, Zeta, and Related Functions
1
0
-1
Real part
-2
-3
-4
-5
-6
-7
-8
0.5
0
-0.5
y
-0.4
-0.6
-0.2
0
0.2
0.4
0.6
x
(a) Re W(z)
(b) Im W(z)
Figure 6.10. Real and imaginary parts of the three sheets W−1 (z), W0 (z), W1 (z) of the
Lambert W-function.
points at different locations on different sheets: at z = −1/e for the primary sheet, also at
z = 0 for the adjacent two sheets, and only at z = 0 for all the (infinitely many) further
sheets.
6.4 Supplementary materials
Theorem 6.1 (stated). Γ(z) = limn→∞
n! nz
z(z+1)·····(z+n) .
n
´n
Proof. Since Γ(z) = limn→∞ 0 e−t tz−1 dt and e−t = limn→∞ 1 − nt , we can write
n
ˆ n
ˆ 1
t
(1 − τ )n τ z−1 dτ.
Γ(z) = lim
1−
tz−1 dt = lim nz
n→∞ 0
n→∞
n
0
6.4. Supplementary materials
185
For n integer, repeated integration by parts gives
1
ˆ 1
ˆ
1 z
n 1
(1 − τ )n τ z−1 dτ =
(1 − τ )n−1 τ z dτ
τ (1 − τ )n +
z
z
0
0
0
n(n − 1) · · · · · (1)
= ··· =
.
z(z + 1) · · · · · (z + n)
Q∞ 1
Theorem 6.2 (stated). Γ(z)
= z eγz n=1 1 + nz e−z/n .
Proof. Equation (6.2) can be written as
h 1
z
z
z −z log n i
= lim z 1 +
1+
· ··· · 1 +
e
Γ(z) n→∞
1
2
n
h z −z
z −z
z − z [1+ 12 +···+ n1 −log n]z i
= lim z 1 +
e 1 1+
e 2 · ··· · 1 +
e n ·e
n→∞
1
2
n
∞ h
Y
z −z/n i
= z eγz
1+
e
.
n
n=1
Theorem 6.5. The beta function B(p, q) =
´1
B(p, q) =
0
tp−1 (1 − t)q−1 dt satisfies
Γ(p)Γ(q)
.
Γ(p + q)
Proof. Starting with the integrals for Γ(p) and Γ(q), we obtain
ˆ ∞
ˆ ∞
Γ(p)Γ(q) =
e−t tp−1 dt
e−t tq−1 dt
0
0
ˆ ∞
ˆ ∞
−t p−1
q
−tx q−1
=
e t dt t
e x dx
ˆ0 ∞
ˆ ∞ 0
=
xeq−1 dx
e−t(x+1) tp+q−1 dt
0
0
ˆ ∞
ˆ ∞
xq−1
dx
e−τ τ p+q−1 dτ
=
(1 + x)p+q
0
0
and the result follows from using (6.9).
Theorem 6.6. The ζ-function satisfies the functional equation
1
ζ(z) = 2z π z−1 sin
πz Γ(1 − z)ζ(1 − z).
2
(6.17)
Proof. (Outline) The following is a sketch of the key steps in P
one of the proofs provided
∞
by Riemann. Example 5.32 started from the definition ζ(z) = n=1 n1z and then showed
both
ˆ ∞ z−1
1
t
dt,
(6.18)
ζ(z) =
Γ(z) 0 et − 1
for which the contour Γ1 (the interval [0, +∞]) is indicated in part (a) of Figure 6.11, and
ˆ
Γ(1 − z)
tz−1
ζ(z) =
dt
(6.19)
−t
2πi
−1
Γ2 e
186
Chapter 6. Gamma, Zeta, and Related Functions
←
+(2N-1)πi
↓
→
-N
←
→
Γ1
↑
+N
Γ2
Γ3
↓
↑
→
-(2N-1)πi
Figure 6.11. The three integration paths Γ1 , Γ2 , Γ3 considered in the proof of Theorem
6.6, shown in parts (a)–(c), respectively. There is in all cases a branch cut along the negative real
axis.
(now valid for all z), where Γ2 is the Hankel contour shown in Figure 6.11(b) (well separated from the negative real axis). This last integrand has poles at t = ±2πi, ±4πi, . . . (red
1
±πi(z−1)/2
dots in Figure 6.11(c)), with residues − (2πn)
, n = 1, 2, . . . . A direct esti1−z e
mate shows that the integral around the path Γ3 (shown in green in Figure 6.11(c)) will vanish if N → ∞ (meaning that the path is moved increasingly far out while passing halfway
between poles on the imaginary axis). Hence, ζ(z) (based on (6.19)) will equal Γ(1−z)
·2πi·
P∞2πi 1
{sum of residues}, where the sum of the residues becomes (2π)z−1 2 sin π2 z n=1 n1−z
=
(2π)z−1 2 sin π2 z ζ(1 − z). The last equality holds for Re z < 0, but the resulting identity will then by continuation hold for all z (note a sign change due to the direction of
integration along Γ2 ).
6.5 Exercises
Exercise 6.5.1. Starting from the product representation (6.3) of Γ(z),
∞ Y
z −z/n
1
= z eγz
1+
e
,
Γ(z)
n
n=1
deduce the functional equation Γ(z + 1) = z Γ(z) together with Γ(1) = 1.
Hint: The following sequence of intermediate steps lead to the result:
0
P∞
(z)
1
(a) ΓΓ(z)
= − z1 − γ − n=1 z+n
− n1 .
0
0
(z+1)
(z)
(b) ΓΓ(z+1)
− ΓΓ(z)
− z1 = 0.
(c) Γ(z + 1) = α z Γ(z), where α is a constant that needs to be determined.
(d) limz→0 z Γ(z) = 1.
(e) Γ(1) = 1. In this step, you need that γ = limn→∞ 1 + 21 + 13 + · · · + n1 − log n .
6.5. Exercises
187
Exercise 6.5.2. Starting from the result in part (a) of Exercise 6.5.1, show that
∞
X
1
d2
log
Γ(z)
=
.
dz 2
(z
+
n)2
n=0
Note: This function is a special (m = 1) case of the polygamma function ψ (m) (z) =
P∞
dm+1
1
(m)
(z) = (−1)m+1 m! n=0 (z+n)
=
m+1
dz m+1 log Γ(z), with properties including ψ
´
´
m −zt
z−1
∞
1
t e
t
m+1
m
(−1)
1−e−t dt = − 0 1−t (log t) dt.
0
Exercise 6.5.3. Show that the Taylor expansion of log Γ(z + 1) around z = 0 is
log Γ(z + 1) = −γz +
∞
X
(−1)k ζ(k)
k=2
k
zk .
(6.20)
Hint: From (6.2) it follows that
"
log Γ(z + 1) = lim
n→∞
log n! + (1 + z) log n −
n
X
#
log(1 + z + k) .
k=0
Note also that log(1 + z + k) = log(1 + k) + log(1 +
z
1+k ).
Exercise 6.5.4. Show that Γ0 (1) = −γ by (a) using (6.20) and (b) using (6.3).
Exercise 6.5.5. Use, for example, the result of Exercise 6.5.4 to show that
ˆ ∞
γ=−
e−t log t dt.
0
´ a+1
Exercise 6.5.6. Show that, for a ≥ 0, a log Γ(x)dx = a log a − a + 12 log 2π.
Hint: Differentiate with respect to a, simplify, integrate back, and use (12.24) (in the x →
∞ limit) to determine the integration constant. Another option for determining the constant
is to consider the case a = 1, take the logarithm of (5.6), and integrate over [0, 1] (recalling
the result of Exercise 5.6.15).
Exercise 6.5.7. Show that the beta function satisfies the identity B(p, q+1)+B(p+1, q) =
B(p, q).
´∞
√
2
Exercise 6.5.8. The definite integral −∞ e−x dx = π is most easily evaluated by con´∞ ´∞
2
2
sidering the double integral −∞ −∞ e−(x +y ) dxdy and then changing to polar coordinates:
(a) Carry out the details of the derivation just outlined.
(b)
Change the variable
in the integral definition
of the gamma function to obtain
´ ∞ −x
2
e
dx = Γ 12 , and then evaluate Γ 21 from the functional equation (5.6):
−∞
π
Γ(z)Γ(1 − z) = sin(πz)
.
´1
1
(c) Evaluate Γ 2 from the fact that B 12 , 21 = 0 √ dt
= {set t = sin2 x} = π.
t (1−t)
188
Chapter 6. Gamma, Zeta, and Related Functions
Exercise 6.5.9. Prove Theorem 6.3, Γ(2z) = π −1/2 22z−1 Γ(z)Γ z + 21 , in the following
ways:
´1
(a) Expand on the sequence of steps Γ(z)·Γ(z)
=B(z, z) = 0 tz−1 (1 − t)z−1 dt =
Γ(2z)
´1
√
1−2z
2 0 1 − x2 z−1 dx = {set x = u} = 21−2z B 12 , z =
{set t = 1+x
2 } = 2
Γ( 1 )·Γ(z)
21−2z Γ 2z+ 1 .
( 2)
1/2
(n!)2 22n+2
. Since the
(b) Using (6.2), obtain Γ(z)Γ z + 21 = 2−2z Γ(2z) limn→∞ n (2n+1)!
1
limit is a constant, its value can be obtained by setting z = 2 .
Note: Still another proof, reminiscent of part (b) but not utilizing (6.2), is outlined in
Exercise 12.6.9.
Exercise 6.5.10. Show that
´ π/2 √
´ π/2 √
Γ( 34 )Γ( 12 )
sin x dx = 0
cos x dx = 2Γ
,
(a) 0
( 54 )
2
´ π/2 dx
´ π/2 dx
[Γ( 1 )]
(b) 0 √sin
= 2√42π .
= 0 √cos
x
x
Hint: Make the variable change t = sin2 x and then use (6.5) and (6.6).
Exercise 6.5.11. Show that the even order (nontrivial) Bernoulli numbers satisfy
ˆ
B2k = (−1)k+1 4k
0
∞
x2k−1
dx
−1
(6.21)
e2πx
by Taylor expanding both sides of (5.28) in powers of a and then equating coefficients
(using (5.12)).
´ ∞ xm
Note: In view of (6.10), this result can also be obtained as a special case of 0 e2πx
−1 dx =
ζ(m+1) m!
(2π)m+1 ,
m = 1, 2, 3, . . . , which follows from (6.18).
Exercise 6.5.12. Another way to continue the zeta function based on (6.18) other than
changing to a Hankel contour is to apply the “partitioning of an integration interval,”
method from Section 3.2.4. Carry this out, and use this to show that the zeta function’s only
singularity is a first order pole at z = 1 with residue 1.
Exercise 6.5.13. Show that the primes pk , k = 1, 2, 3, . . . , in the long run are more densely
distributed than the
sequence k 1+ε , k = 1, 2,
matter how small we choose ε > 0.
P∞
P3,∞. . ., no
1
Hint: Show that k=1 p1k diverges, while k=1 k1+ε
converges.57 Equation (6.12) and
Theorem 5.40 (from basic calculus) are helpful.
Exercise 6.5.14. Show that the nontrivial zeros of ζ(z) are limited to the critical band
0 ≤ Re z ≤ 1.
Hint: For z with Re z > 1, consider the logarithm of (6.12). Then, for z with Re z < 0,
consider the functional equation (6.17).
57 Reminiscent
M =
P
of Euler’s constant γ (see (6.4) and Example 12.3), theP
Meissel–Mertens constant is given by
1
− log(log(n)) ≈ 0.2614972128. The divergence of ∞
k=1 p is thus extraordinarily slow.
1
pk ≤n pk
k
6.5. Exercises
189
P∞
(n2 ) = 1 +
Exercise 6.5.15. The Jacobi theta function, defined by ϑ(z) =
n=−∞ z
P∞
2
2 n=1 z (n ) , can be shown to satisfy
3z 3
4z 4
5z 5
2z 2
z
4
+
+
+
+
·
·
·
.
(6.22)
+
[ϑ(z)] = 1 + 8
1−z
1 + z2
1 − z3
1 + z4
1 − z5
Taylor expanding the RHS will further reveal that all its Taylor coefficients are positive.
Deduce from this that every positive integer can be written as a sum of no more than four
integer squares.
Notes: (i) This result is a special case of the still partly unsolved Waring problem, where
squares are generalized to other integer powers. (ii) This function ϑ(z) provides an additional example (cf. Example 3.2 and Exercise 3.3.4) of a function with a natural boundary
(here also located at |z| = 1).
Exercise 6.5.16. Show that the Lambert W-function satisfies the ODE
W(z)
d
dz W (z) = z (1+W(z)) .
Exercise 6.5.17. The equation 2x = 5x has two real-valued solutions, x1 ≈ 0.2355 and
x2 ≈ 4.488. Show that they are given by x = -W(− log5 2 )/ log 2, where we choose either
the upper or lower branch shown in Figure 6.8(b).
Chapter 7
Elliptic Functions
Functions such as sin z or ez have one direction of periodicity (along lines parallel to the
real and the imaginary axes, respectively). With the complex plane being 2-dimensional,
it turns out that one can also create analytic functions that are periodic in two separate directions. The most important such class is known as elliptic functions, and we focus here
on two such families, introduced by Weierstrass and Jacobi, respectively. Some of their
key applications may at first appear unrelated to the double periodicity property, including
for example, providing analytic solutions to certain integrals and nonlinear ODEs (ordinary
differential equations). The name elliptic functions originates from the fact that one of the
integral cases gives the circumference of an ellipse explicitly as a function of its major
and minor axes (not possible to express in terms of “standard” functions). The monograph
Elliptic Functions [3] contains much more information than is given here, including connections to planar geometry.
7.1 Some introductory remarks on simply periodic
functions
Consider first a function f (z) with a single period ω, satisfying f (z + ω) = f (z). For
example, f (z) = sin z, with period ω = 2π, was illustrated previously in Figure 2.6 and is
shown again here in Figure 7.1(a). The values it takes in its period strip, shaded in Figure
7.1(b), repeat in each equally sized strip that is parallel to it. The basic strip, as shown, is not
uniquely determined, as it can be shifted any amount sideways, say to α ≤ Re z < α + 2π,
where α is any real number. For more complicated periodic functions, with singularities, it
can be convenient to choose α so that the edges of the period strip become singularity-free.
If a function is, say, 2π-periodic, it has of course also the periods −2π, ±4π, ±6π, . . . .
The ratio between these different choices is real-valued, and we will in general focus on
the smallest one, choosing here ω = 2π.
7.2 Some basic properties of doubly periodic functions
With z being a complex variable, it becomes possible for analytic functions to have two
genuinely different periods ω1 and ω2 (with the ratio τ = ω1 /ω2 not real-valued). We
will soon see that such functions (unless constant) cannot be singularity-free. If they are
191
192
Chapter 7. Elliptic Functions
(a) Real part of f (z) = sin z.
Im z
−2π
0
2π
4π
Re z
(b) Period strip for f (z) = sin z.
Figure 7.1. A period strip (here 0 ≤ Re z < 2π) for the function f (z) = sin z.
meromorphic (only poles as singularities), they are known as elliptic functions. Figure 7.2
illustrates the concept of a 2-D period box, here shown with the distinct periods ω1 = 1
and ω2 = i. Apart from translating the period box (convenient in case we wish to avoid
singularities on the box sides), we can also define the periods genuinely differently, as
shown with the dashed parallelogram.
Theorem 7.1. A nonconstant doubly periodic function must have at least one singularity
inside its period box.
7.2. Some basic properties of doubly periodic functions
193
Im z
Re z
Figure 7.2. Period box(es) for a function with primary periods ω1 = 1 and ω2 = i.
Proof. If it does not have any singularity inside the period box, it will be bounded in
magnitude there and therefore, by the periodicities, everywhere across the complex plane.
By Liouville’s theorem (Section 4.2.3), it becomes a constant.
Theorem 7.2. A (nonconstant) elliptic function has equally many poles as it has zeros
inside the period box.
Proof. Consider the difference between the number of zeros and number of poles in the
´ f 0 (z)
´
1
dz = 0 (Theorem 4.8, here with denoting integraperiod box: N − P = 2πi
f (z)
tion around the four sides of the period box). The last equality follows from the fact that
f 0 (z) must have the same double periodicity, so the contributions to the integral from the
opposite box sides will cancel. The number N = P is called the order of the elliptic
function.
Corollary 7.3. An elliptic function takes every value equally many times.
Theorem 7.4. There does not exist any elliptic function of first order.
Proof. If there was one, then
residue at the
simple pole
=
1
2πi
ˆ
f (z)dz = 0
(by cancellation between the sides of the period box), i.e., the pole does in fact not
exist.
194
Chapter 7. Elliptic Functions
7.3 The Weierstrass ℘-function
7.3.1 Construction
Although we have identified some properties that elliptic functions must possess if they
exist, we have not yet demonstrated that they in fact do exist. The easiest demonstration
is to explicitly construct one. Our first example will be the Weierstrass ℘-function. This
function has a double pole at the origin, with zero residue, and we denote its two periods
by ω1 and ω2 (with τ = ω1 /ω2 not real).
First attempt at construction: Let us place a double pole of the type 1/z 2 at the origin
and also at each periodic repetition of this point, i.e., at ω = n1 ω1 + n2 ω2 , where n1 and
n2 run through all integers (zero, positive, and negative). The simplest attempt would be to
consider
X
1
.
(7.1)
P (z) =
(z
−
ω)2
ω=n ω +n ω
1
1
2
2
However, the convergence properties of this sum are not obvious. It is not absolutely convergent, as we can see by comparing with
ˆ
∞
−∞
ˆ
∞
−∞
1
dxdy =
r2
ˆ
0
2π
ˆ
∞
0
ˆ ∞
1
1
r
dr
dθ
=
2π
dr = ∞,
r2
r
0
(7.2)
where we in all the integrals have excluded some region around the origin and only focused
on behaviors towards infinity.
Second attempt at construction: The idea will be to change the function P (z) from
(7.1) in concept as little as possible (so that it still features unit strength double poles at the
same locations). So here is another attempt:
X
1
1
1
℘(z) = 2 +
− 2 .
(7.3)
z
(z − ω)2
ω
ω6=0
With the 1/z 2 -term separated out in front, and then having inserted −1/ω 2 inside the sum
(which no longer includes the case of ω = 0), each term becomes (for z fixed and ω large)
of size O(1/ω 3 ), instead of O(1/ω 2 ) as before. That extra power of ω means that we get
one more power of r in the denominators of (7.2), showing that (7.3) will converge for z
fixed. With convergence thus taken care of, it remains to verify that we did not lose the
double periodicity when modifying (7.1) to (7.3).
Theorem 7.5. The Weierstrass ℘-function, as defined by (7.3), is doubly periodic.
P
1
Proof. From (7.3) it follows, by differentiation, that ℘0 (z) = −2 ω (z−ω)
3 , which
3
converges (since its terms are again of size O(1/ω )). This sum is clearly doubly periodic,
implying that ℘0 (z + ωi ) − ℘0 (z) = 0, i = 1, 2. Integration gives
℘(z + ωi ) − ℘(z) = C.
(7.4)
Substituting z = −ωi /2 into (7.4) gives ℘(ωi /2) − ℘(−ωi /2) = C. Since (7.3) shows that
℘(z) is an even function of z, the constant C must be zero. Equation (7.4) then confirms
that ℘(z) indeed is doubly periodic.
7.3. The Weierstrass ℘-function
195
40
20
0
-20
-40
1.5
1
0.5
1.5
0
1
0.5
-0.5
y
0
-0.5
-1
-1.5
-1
x
-1.5
(a) Re ℘(z)
40
20
0
-20
-40
1.5
1
0.5
1.5
0
1
0.5
-0.5
y
0
-0.5
-1
-1.5
-1
-1.5
x
(b) Im ℘(z)
Figure 7.3. Real and imaginary parts of the Weierstrass ℘-function in the case of periods
ω1 = 1 and ω2 = i.
Figures 7.3 and 7.4 show the ℘-function in the special case when the periods have been
chosen as ω1 = 1 and ω2 = i. The pattern of equispaced double poles extends over the full
complex plane.
7.3.2 Some nonlinear ODEs that are satisfied by the ℘-function
Elliptic functions satisfy various different nonlinear ODEs, many of which arise in applications. We continue to use the ℘-function as an illustrative example.
The Laurent expansion of ℘(z) around the origin will take the form
1
+ 0 + a2 z 2 + a4 z 4 + · · · ,
(7.5)
z2
P
where the coefficients depend on the periods: ak = (k+1) ω6=0 1/ω k+2 , k = 2, 4, 6, . . . .
Since ℘(z) is even, all coefficients for odd powers vanish. That is also the case for a0 , seen
℘(z) =
196
Chapter 7. Elliptic Functions
Figure 7.4. Magnitude and phase plot for the same case as in Figure 7.3.
from letting z → 0 in (7.3). From (7.5) follows
℘0 (z) = −
2
+ 2a2 z + 4a4 z 3 + · · · ,
z3
℘0 (z)2 =
4
8a2
− 2 − 16a4 + · · · .
z6
z
and therefore
We obtain the same leading term
out by subtraction:
4
z6
3
if we form 4[℘(z)] , implying that we can cancel it
℘0 (z)2 − 4℘(z)3 = −
20a2
− 28a4 + · · · .
z2
Adding 20a2 ℘(z) to this will additionally cancel out the first term in the RHS, leading to
℘0 (z)2 − 4℘(z)3 + 20a2 ℘(z) = −28a4 + · · · .
(7.6)
The LHS of (7.6) is doubly periodic with its only possible singularity location (in the period
box) at z = 0. However, the RHS is singularity-free at this point. By Theorem 7.1, the
LHS function must be a constant. All further terms in the RHS of (7.6) (beyond −28a4 )
must therefore also have vanished.
It is conventional to introduce the new parameters g2 = 20a2 and g3 = 28a4 . We can
then write the ODE satisfied by ℘(z) as
℘0 (z)2 = 4℘(z)3 − g2 ℘(z) − g3 .
(7.7)
If g2 and g3 are specified, one can compute matching periods ω1 and ω2 from the formula
for ak . The same procedure as just employed will give many additional ODEs satisfied by
℘(z), such as
1
(7.8)
℘00 (z) = 6℘(z)2 − g2 ,
2
℘000 (z) = 12℘(z)℘0 (z),
etc. We will return to (7.8) in Section 11.5.1.
7.4. The Jacobi elliptic functions
197
7.3.3 Closed-form expressions for certain integrals based on the
℘-function
p
dz
Equation (7.7) can be written as 1 = ℘0 (z)/ 4℘(z)3 − g2 ℘(z) − g3 . Using d℘
dz = 1/ d℘ ,
p
we obtain for the inverse function z(℘) the relation z 0 (℘) = 1/ 4℘3 − g2 ℘ − g3 . At this
point, ℘ has become the name of an independent variable, which we just as well can call,
say, x. By means of the inverse function z(x), we have thus obtained an analytic expression
for the indefinite integral
ˆ
dx
p
(7.9)
3
4x − g2 x − g3
for arbitrary constants g2 and g3 . By choosing other elliptic functions, (7.9) can be generalized to
ˆ
dx
√
.
arbitrary quartic polynomial in x
7.4 The Jacobi elliptic functions
This family of elliptic functions differs from the Weierstrass ℘-functions primarily in that
the period box is assumed to be rectangular, and containing two first order poles (of opposite residues) instead of a single pole (of order 2 and residue zero).58 Many of the notations
for these functions come from their connection with conformal mappings; see Section 8.6.
For example, the Jacobi sn(z, k) function arises there in the implicit form
ˆ
sn(z,k)
z=
du
[(1 −
0
u2 )(1
1/2
− k 2 u2 )]
,
(7.10)
where k is a parameter in the range 0 ≤ k ≤ 1.59 However, we start here with a more direct
approach to the Jacobi elliptic functions by defining a periodic box and then just placing
two poles inside it.
7.4.1 The Jacobi sn(z,k ) function
We first create a period box of size 0 ≤ Re z ≤ 4K, 0 ≤ Im z ≤ 2K 0 , where both K and
K 0 depend on a single free parameter k ∈ [0, 1] through the relations
ˆ
K(k) =
0
1
du
1/2
[(1 − u2 )(1 − k 2 u2 )]
and K 0 (k) = K(k 0 ), with k 0 =
p
1 − k2 .
(7.11)
(We follow here again notational conventions: K and K 0 are related functions, and k and k 0
are related parameter values, with the primes here having nothing to do with derivatives.)
Section 7.4.3 describes a simpler representation of this function K(k) (and thereby also
K 0 (k)). Figure 7.5 shows the two functions. In the period box, we next place a first order
58 There is nevertheless a close connection: Equation (7.7) tells us that ℘(z) will not have a double zero unless
the parameter g3 = 0. The inverse 1/℘(z) will therefore in general be another elliptic function of order 2, but
with two separate first order poles (of opposite residue).
59 It is in the literature also common to parameterize the Jacobi elliptic functions by m = k 2 instead of by the
“elliptic modulus” k directly, as we do here (for example, both MATLAB and Mathematica use m).
198
Chapter 7. Elliptic Functions
6
K
K'
/2
5
4
3
2
/2
1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
k
Figure 7.5. The two functions K(k) and K 0 (k). K(0) = K 0 (1) =
2
√
intersect at k = √12 , with K( √12 ) = K 0 ( √12 ) = Γ( 14 ) /(4 π).
π
.
2
The two curves
pole with residue +1/k at z = iK 0 and one with residue −1/k at z = 2K +iK 0 . Repeating
this period box across the full complex plane, and summing the contributions from all
the resulting infinitely many poles, gives rise, like for our first attempt at constructing
the Weierstrass ℘-function, to a divergent double sum. However, this divergence can be
avoided similarly to the “second attempt” (also in Section 7.3.1). The result is known as
the Jacobi sn(z, k) function, illustrated over one period box in Figure 7.6.
If we let the parameter k approach zero, the period box approaches 2π in width, while
its height increases towards infinity. The sn(z, k) function then approaches sin z (which
we can see indications of along the real axis already in Figure 7.6(a)). In the opposite limit
of k approaching 1, the period box increases towards infinity in width while shrinking in
height to π, and sn(z, k) then approaches tanh z. The red curves in Figure 7.7 show these
variations in sn(x, k) over one period [−2K, +2K] in the real variable x for some different
values of k.
7.4.2 Some other Jacobi elliptic functions
Similar to how we from sin z obtain cos z, we can from sn(z, k) create numerous additional
elliptic functions, with the most commonly used ones being cn(z, k) and dn(z, k), defined
through the relations
cn2 (z, k) + sn2 (z, k) = 1
and
dn2 (z, k) + k 2 sn2 (z, k) = 1.
The green and blue curves in Figure 7.7 show counterparts to the red sn-curves for the functions cn(x, k) and dn(x, k), respectively. For k increasing from 0 to 1, both cn(z, k) and
dn(z, k) transition to 1/ cosh(z), starting from cos(z) and 1, respectively. These elliptic
functions solve numerous nonlinear ODEs in closed form. For example,
sn(x, k) solves
cn(x, k) solves
dn(x, k) solves
d2 y
dx2
+ (1 + k2 )y − 2k2 y 3 = 0
and
2
+ (1 − 2k2 )y + 2k2 y 3 = 0
and
2
− (2 − k2 )y + 2y 3 = 0
and
d y
dx2
d y
dx2
dy 2
dx
dy 2
dx
dy 2
dx
= (1 − y 2 )(1 − k2 y 2 ),
= (1 − y 2 )(1 − k2 + k2 y 2 ),
= (y 2 − 1)(1 − k2 − y 2 ).
There is also a wealth of closed-form integrals and other relations satisfied by the Jacobi
elliptic functions. The NIST Handbook of Mathematical Functions [35, Chapter 22] contains a summary. As noted above, one of the integrals provides a closed-form expression
for the circumference of an ellipse.
7.4. The Jacobi elliptic functions
199
5
0
-5
4
3
2
6
5
4
1
3
y
2
0
1
0
x
(a) Re sn(z, k) for k = 12 .
5
0
-5
4
3
2
6
5
4
1
3
y
2
0
1
0
x
(b) Im sn(z, k) for k = 12 .
(c) |sn(z, k)| for k = 12 .
Figure 7.6. Visualization across one period box of sn(z, k) in the case of k = 12 . For
this k-value, K ≈ 1.6857503548 and K 0 ≈ 2.1565156475; i.e., the period box is roughly of size
[0, 6.7430] × [0, 4.3130]. In parts (a) and (b), some positive contour lines are highlighted in blue
and negative ones in green. As usual, values along the real axis are shown in red for (a) and (b), and
in black for (c).
200
Chapter 7. Elliptic Functions
k = 0.999
1
0
-1
-10
-8
-6
-4
-2
0
2
4
6
8
10
2
4
6
8
10
2
4
6
8
10
k = 0.99
1
0
-1
-10
-8
-6
-4
-2
0
k = 0.9
1
0
-1
-10
-8
-6
-4
-2
0
k=0
1
sn
cn
dn
0
-1
-10
-8
-6
-4
-2
0
2
4
6
8
10
Figure 7.7. The variations of the sn(x, k), cn(x, k), and dn(s, k)-functions along the real
axis over one spatial period in the cases of four different k-values.
7.4.3 The arithmetic-geometric mean
We have in previous sections often defined analytic functions as the limit of a convergent
sequence of simpler functions, such as any time we define a function through its Taylor series, Weierstrass product, or Mittag–Leffler sum, or by means of an integral. The
arithmetic-geometric mean (AGM) is yet another process that generates convergent sequences. We describe it first as a function AGM(x, y) of two real variables, x ≥ 0 and
y ≥ 0. One then sets x0 = x, y0 = y and iterates
xn+1 =
xn + yn
,
2
yn+1 =
√
xn yn ,
n = 0, 1, 2, . . . .
(7.12)
The two sequences {xn } and {yn } approach each other, converging to a joint limit we
denote by AGM(x, y). Apart from the trivial cases of either x = 0 or y = 0, the convergence rate is extraordinarily high, with the number of digits in common between xn
and yn roughly doubling with each iteration (see Exercise 7.6.5). Figure 7.8 illustrates this
function AGM(x, y). It is zero along both the x- and the y-axes, and equals x when y = x
(green straight line; the red curve is explained below).
In 1799, C.F. Gauss, by numerical hand calculations, “stumbled onto” the relation
K(k) =
π
2 AGM(1, k 0 )
(7.13)
7.4. The Jacobi elliptic functions
201
3
2
1
0
3
2
1
y
0
0.5
0
1.5
1
2
3
2.5
x
Figure 7.8. The function AGM(x, y) in the first quadrant of the (real) (x, y)-plane. The
red and green curves are explained in the text.
π
(and thus K 0 (k) = 2 AGM(1,k)
) in the notation of (7.11).60 Gauss followed this up by
noting61 that, for x, y ≥ 0,
ˆ
π/2
0
ˆ
dθ
p
x2
cos2
θ+
y2
2
=
sin θ
0
π/2
dϕ
q
x+y 2
2
cos2 ϕ +
√
xy
2
.
(7.14)
sin2 ϕ
From this it follows directly that
ˆ
π/2
0
π
dθ
p
=
.
2 AGM(x, y)
x2 cos2 θ + y 2 sin2 θ
(7.15)
As outlined in Exercise 7.6.6, this in turn leads to a proof of (7.13). Another consequence
of (7.15) (cf. Exercise 7.6.8) is
ˆ
1
1 π
dθ
√
=
AGM(1 + x, 1 − x)
π 0
1 − x2 cos2 θ
2
2
2
1
1·3
1·3·5
2
4
=1+
x +
x +
x6 + · · · . (7.16)
2
2·4
2·4·6
The red curve in Figure 7.8 shows this function AGM(1 + x, 1 − x) over the range −1 ≤
x ≤ 1.
Since the recursion for AGM is rapidly converging, replacing x with z (complex) in
AGM(1+x, 1−x) will give rise to an analytic function. This function is displayed in Figure
7.9. Along the real axis, for −1 ≤ x ≤ 1, we recognize the red curve from Figure 7.8,
60 G
=
1 √
AGM(1, 2)
=
2
π
´1
0
√ dx
1−x4
=
1
B( 14 , 21 )
2π
constant.
61 By means of the ingenious change of variable sin θ =
≈ 0.8346268417 has become known as the Gauss
2x sin ϕ
.
(x+y)+(x−y) sin2 ϕ
202
Chapter 7. Elliptic Functions
2
1.5
1
0.5
2
0
-2
1
-1
0
0
-1
1
x
2
y
-2
(a) Re AGM(1 + z, 1 − z).
1
0.5
0
-0.5
2
-1
-2
1
-1
0
0
-1
1
x
2
-2
y
(b) Im AGM(1 + z, 1 − z).
(c) Magnitude and phase angle for AGM(1 + z, 1 − z).
Figure 7.9. Display of the analytic function AGM(1 + z, 1 − z).
7.5. Supplementary materials
203
but see now also that the function AGM(1 + z, 1 − z) is well defined across the complex
plane, however with branch points at z = ±1. The three expressions in (7.16) all represent
the same function, but have quite different characters. The AGM expression is superior for
numerical evaluations, while the integral arises in numerous application contexts.62 The
validity of the Taylor series representation is limited by its radius of convergence R = 1.
7.5 Supplementary materials
7.5.1 The modular function λ(τ )
The RHS in (7.7) is a cubic in ℘(z), which can be factored as
℘0 (z)2 = 4(℘(z) − e1 )(℘(z) − e2 )(℘(z) − e3 ),
(7.17)
where the ei , i = 1, 2, 3, are the three zeros of the order 3 elliptic function ℘0 (z). From
the basic properties of the ℘-function, it transpires that e1 = ℘(ω1 /2), e2 = ℘(ω2 /2),
e3 = ℘((ω1 + ω2 )/2) will satisfy this (cf. the proof in Section 7.5.4). These ei -values are
always different from each other. If one further notes that each of them becomes multiplied
by 1/t2 if the two ωi are both multiplied by t, one can conclude that the ratio (e3 −e2 )/(e1 −
e2 ) will become a single-valued analytic function of the ratio τ = ω1 /ω2 . This function
(e3 −e2 )
λ(τ ) = (e
is known as a modular function. It is of great interest in its own right,
1 −e2 )
independently of its definition based on an elliptic function. We can’t here go through its
properties in any detail (for that, see more advanced books, such as Ahlfors [2] or Copson
[8]), but let us mention that w = λ(z) maps the boundary of the region Ω, shown in Figure
7.10 on the left, to the real axis and the inside one-to-one to the upper half-plane (Figure
7.10 (right)). Given this, it is natural to display this function graphically over the region Ω
rather than, as we usually do, over the region surrounding origin in the complex plane; see
Figure 7.11.
In Figure 7.12 (left), we show also the region Ω reflected across the imaginary axis,
denoting this by Ω. By Schwarz’s reflection principle, the function λ(z) will in this region
take complex conjugated values compared to in Ω, implying that its values will be as shown
in 7.12 (right), where we note the branch cuts along [−∞, 0] and [1, ∞]. We denote the
inverse function to w = λ(z) as z = λ−1 (w). It can be shown to have no other singularities
than these branch points at w = 0 and w = 1.
7.5.2 Picard’s theorem
This theorem provides intriguing insights into the possible behaviors of entire functions
(functions that are singularity-free in the finite complex plane). As a background, let us
first recall that a polynomial of degree d takes every finite value d times (when counting
the multiplicity of roots). We also know from Liouville’s theorem (Theorem 4.12) that, if
the magnitude of an entire function is bounded, it is a constant. The Caserati–Weierstrass
62 A
classic example concerns the period T of a pendulum of length l, with all its mass at its end. If
θ0 is its maximal deviation from vertical, k = sin θ20 , and g is the acceleration of gravity, then T =
q ´
q
q
2 gl 0π √ dθ
= 2π gl /AGM(1 + k, 1 − k). For θ0 small, so is k, and T ≈ 2π gl . If θ0
2
2
1−k sin θ
increases towards π, then T → ∞.
204
Chapter 7. Elliptic Functions
Figure 7.10. Illustration of the mapping w = λ(z). We use here colors, not to indicate the
phase, but as a texture, identifying matching points in the z-plane and the w-plane. In particular, we
see that λ(z) maps the edge of the domain Ω, shown on the left to the real w-axis. Letting z follow
the Ω-domain boundary along the path 1 → 1 + i∞(→ i∞) → 0 → 1, then w follows the real
axis from left to right: −∞ → 0(→ 0) → 1 → +∞. The colored regions (extending to infinity) are
mapped one-to-one to each other. The bottom boundary of Ω is a half-circle centered at z = 12 .
theorem (Theorem 4.30) tells us that an entire function comes arbitrary close to any finite
value infinitely many times. Some entire functions actually take (rather than just come
arbitrarily close to) every value infinitely many times (e.g., sin z), while others avoid one
value altogether (e.g., ez and Γ(z) never take the value zero for any finite z).
Theorem 7.6 (Picard, 1879). An entire function that avoids two values must be a constant.
Proof. (The following is a brief sketch only, outlining the key idea behind the original
proof by Picard.) If the function F (z) avoids the two (finite) values a and b, we consider
f (z) = F (z)−a
b−a and it remains to prove that an entire function that avoids both 0 and 1 must
be a constant. The essential idea is to consider the function Λ(z) = λ−1 (f (z)) and then
apply Morera’s theorem (Theorem 4.3). If we choose any closed path in the z-plane, we
know that f (z) along this path can never encircle either of the points 0 or 1 (as we otherwise
could shrink the path to a point at which this would conflict with the fact that f (z) cannot
take these values). This means that the integral also of Λ(z) around the path will be zero,
and thus (by Morera’s theorem) represent an analytic function of z. However, the values
of Λ(z) are all in the upper half-plane (cf. Figure 7.12 (left)). The function ei Λ(z) is then
a bounded entire function and according to Liouville’s theorem a constant. Then, so are
Λ(z), λ(z), f (z), and F (z).
Another related result (also by Picard) is that if a meromorphic function (poles as only
singularities, here also including at infinity) omits three values, it has to be a constant.
7.5. Supplementary materials
205
(a) Re λ(z)
(b) Im λ(z)
Figure 7.11. Real and imaginary parts of the modular function λ(z). For part (a) we note
that Re λ(z) increases monotonically as z follows the boundary 1 → ∞ → 0 → 1. For (b) we note
that Im λ(z) is zero along the whole boundary and everywhere positive inside it.
7.5.3 Conformal mapping of a doubly periodic region to a torus
Similar to how the stereographic projection (Section 1.4) mapped the whole complex plane
to the surface of a sphere, the doubly periodic regions that arise for elliptic functions can
be mapped conformally to tori.
Example 7.7. Consider the square doubly periodic region −π ≤ x ≤ π, −π ≤ y ≤ π and
the mapping
cos x
X=√
,
2 − cos y
Y =√
sin x
,
2 − cos y
Verify that this mapping indeed is conformal.
sin y
Z=√
.
2 − cos y
(7.18)
206
Chapter 7. Elliptic Functions
Figure 7.12. Once again, in this illustration, the colors are used as a texture rather than
to indicate the argument. The one-to-one mapping is given by λ(z) between a part of the upper
half-plane (composed of the regions Ω and Ω) and the entire complex plane. On the right, note the
branch cuts along [−∞, 0] and [1, ∞].
The mapping is illustrated in Figure 7.13, with the doubly periodic region in the complex z = x + iy plane to the left and the resulting surface in X, Y, Z space to the right.
Following Exercise 1.6.21, we evaluate
2 2 2
∂Y
∂Y
∂Z
∂Z
(∆x)2 + (∆y)2
+
∆x+
∆y +
∆x+
∆y = √
,
∂x
∂y
∂x
∂y
( 2−cos y)2
(7.19)
from which it follows that the mapping indeed is conformal.
∂X
∂X
∆x+
∆y
∂x
∂y
The torus in this example is the surface
of revolution generated by a circle of radius 1
√
in the (X, Z)-plane, centered
at
X
=
2,
Z
= 0 and rotated
√ around the Z-axis (described
√
as having major axis 2, minor axis 1, and aspect ratio 2).63
7.5.4 Select proofs
Theorem 7.8. The three zeros of ℘0 (z) in the period box are z1 = ω1 /2, z2 = ω2 /2, and
z3 = (ω1 + ω2 )/2.
Proof. Since ℘(z) is even and has a period ω1 , it holds that ℘(z) = ℘(−z) = ℘(ω1 − z),
and therefore ℘0 (z) = −℘0 (ω1 − z). For z = ω1 /2,we then get ℘0 (ω1 /2) = −℘0 (ω1 /2),
i.e., ℘0 (ω1 /2) = 0. The same argument can be utilized for the other two roots. Since
℘0 (z) is an elliptic function of order 3, it cannot have any further zeros (inside the period
box).
63 If
the (rectangular) period box is not square, two different tori aspect ratios become possible (by exchanging
the roles of x and y in a formula similar to (7.18)).
7.6. Exercises
207
3
2
0.5
y
Z
1
0
0
-0.5
-1
-1
-2
2
1
-3
-2
0
2
x
2
1
0
0
-1
-1
Y
-2
-2
X
Figure 7.13. The conformal mapping given by (7.18) from the periodic square −π ≤ x ≤
π, −π ≤ y ≤ π to the surface of a torus in X, Y, Z-space.
7.6 Exercises
Exercise 7.6.1. Consider the Weierstrass ℘(z) function in the special case of the periods
being ω1 = 1 and ω2 = i:
(a) Show that ℘(z) has a double zero at the center of the square period box, i.e., at
z = 12 (1 + i).
(b) Show that ℘(z) · ℘(z + 12 (1 + i)) is identically constant.
Hint for (a): Deduce first that (i) Re ℘(z) = 0 along the diagonals of the period box, and
(ii) Im ℘(z) = 0 along the sides of the box, as well as along Re z = 21 and Im z = 12 .
Exercise 7.6.2. Show that ℘(z) has a double zero if and only if g3 = 28a4 = 0.
Exercise 7.6.3. Consider once again the Weierstrass ℘(z) function in the special case of
the periods being ω1 = 1 and ω2 = i.
(a) Show that g3 = 0.
P
(b) Compute g2 . Using the formula ak = (k + 1) ω6=0 1/ω k+2 , carry out the procedure described in Section 5.2 to obtain
!
∞ X
2 8
1
2
4
4
+
− coth(nπ) + 2 coth(nπ)
.
g2 = 60π
45 n=1 3 3
Notice that the sum converges rapidly and that already two terms are enough to give an accurate approximation. For comparison, Mathematica’s WeierstrassInvariants[{0.5, 0.5I}]
(based on half-periods) gives the result {189.073, 0}.
(c) Show that g2 = −4 · ℘(z) · ℘(z + 12 (1 + i)) for all z ∈ C.
Exercise 7.6.4. For some k-values in the range k ∈ [0, 1], evaluate the corresponding
values for K(k) and K 0 (k) by means of (7.13). Verify that your results match what is
quoted in the captions of Figures 7.5 and 7.6.
208
Chapter 7. Elliptic Functions
Exercise 7.6.5. Prove the quadratic convergence of the AGM iterations.
Hint: Derive from (7.12) the relation xn+1 − yn+1 = 4(xn+11+yn+1 ) (xn − yn )2 .
Exercise 7.6.6. Make the change of variable u = sin θ in (7.11) to show that (7.13) follows
from (7.15).
Exercise 7.6.7. Show that the value for the two integrals in Exercise 6.5.10(a) can be
written as 1/G where G is the Gauss constant (cf. Section 7.4.3).
Exercise 7.6.8. Derive from (7.15) both the results in (7.16).
Exercise 7.6.9. The Jacobi elliptic function sn(z, k) is characterized by having first order
poles of opposite residue placed at the locations z = 0 + iK 0 and z = 2K + iK 0 within the
period box, of horizontal and vertical sides of lengths 4K and 2K 0 , respectively. Show that
immediate double summation over all repetitions of this period box leads to a not absolutely
convergent sum, and then show how a certain regrouping of the terms will provide this
(convergence even if we take the magnitude of all its terms).
Exercise 7.6.10. Verify that the surface in Example 7.7, as given by (7.18), indeed is the
surface
√ of revolution generated by a circle of radius 1 in the (X, Z)-plane, centered at
X = 2, Z = 0, and rotated around the Z-axis.
Chapter 8
Conformal Mappings
The previous chapters have contained many illustrations of analytic functions f (z), often
in the form of displaying Re f (z), Im f (z), and/or |f (z)| as surfaces over the complex
z-plane. A different type of illustration is obtained by choosing some region in the z-plane
and then showing in a complex w-plane where the corresponding values of w = f (z) end
up.64 Figure 8.1(a) shows an arbitrarily chosen region in the z-plane, and parts (b) to (i)
the matching regions in the w-plane for some different simple choices of f (z). All of these
cases illustrate one-to-one mappings for the displayed regions (i.e., over these regions, the
inverse function z = f −1 (w) is also single-valued). Only case (b) (a combination of a
translation and a rotation) would be single-valued in both directions for arbitrary regions.
Maybe the most striking feature in these figures is that the grid, shown in green in part
(a), will correspond to a curve set in the w-plane where again all the curves intersect at right
angles. This will turn out to be the case at any point where f (z) is analytic, with f 0 (z) 6= 0.
This, furthermore, is a special case of a more general property: Not only right angles but all
angles (including sign) are locally preserved. In reverse, only analytic functions w = f (z)
can have this conformal mapping property.65
Figure 8.1 is somewhat unsatisfactory in that neither the initial region in the z-plane, nor
the resulting regions in the w-plane, is of any special significance. What makes the topic of
conformal mappings important for certain applications is that one can specify both regions
independently, and there will then exist (a 3-parameter family66 of) analytic functions w =
f (z) that perform that mapping in a one-to-one manner. Although it is somewhat rare to
be able to find an appropriate f (z) in closed form, it can typically be computed accurately
(with algorithms that fall outside the scope of this book).
The rest of this chapter will highlight the following:
• Relations between conformal mappings and analytic functions (Section 8.1).
• Mappings provided by bilinear functions (Section 8.2).
64 That
is, for each point that is marked in the z-plane (for example, by color), evaluate w = f (z) and graphically mark the corresponding point in the w-plane identically.
65 Mappings between the complex plane and other types of surfaces can also be conformal, as we recall from
the stereographic projection (Section 1.4 and Exercises 1.6.20 and 1.6.21) and the mapping from a square to a
torus (in Section 7.5.3).
66 The number “3” here follows from the special case of mapping the inside of the unit circle to itself, as
discussed following Theorem 8.7.
209
210
Chapter 8. Conformal Mappings
y
2.5
y
2
6
2
1.5
y
1.5
1
4
1
2
0.5
0.5
0
0
0.5
1
1.5
0
2
0.5
1
(a) Initial domain in
complex z-plane.
1.5
2
-4
-2
0
2
4
x
x
(b) The mapping
√ z.
w = 1 + 1+i
2
(c) The mapping
w = z2.
x
y
2.5
2
y
6
1.5
4
0.6
1
0.4
0.5
y
2
0.2
0
0
0.2
0.4
0.6
0.8
1
-2
1.2
0
2
(d) The mapping
w = z 1/3 .
4
-1
6
0
1
x
x
(e) The mapping
w = ez .
(f) The mapping
w = arctan z.
x
3
0.6 y
2.5
0.4
2
0.2
1.5
1
y
0.5
y
0.5
x
1
1
-0.2
x
-0.2
-0.4
-0.4
-0.6
0.5
x
0
1
2
-0.5
(g) The mapping
w = z −1/3 .
(h) The mapping
w = 21 z + z1 .
(i) The mapping
w = sn(z, 0.7).
Figure 8.1. (a) An arbitrarily chosen domain in the first quadrant. (b)–(i) The images of
this domain (and of the green lattice) when mapped to a complex w-plane through some different
mapping functions w = f (z).
8.1. Relations between conformal mappings and analytic functions
211
• Riemann’s mapping theorem (Section 8.3).
• Mappings of polygonal regions (Section 8.4).
• Some applications of conformal mappings (Section 8.5).
8.1 Relations between conformal mappings and
analytic functions
A pair of functions
u = u(x, y)
v = v(x, y)
(u, v, x, y real) form a conformal (angle-preserving) map if any two intersecting curve segments in the (x, y)-plane become two intersecting curve segments in the (u, v)-plane, such
that the angle between the curves (at the intersection points) has stayed the same. The
following theorem shows that this property is unique to analytic functions.
Theorem 8.1. The mapping
u = u(x, y)
v = v(x, y)
with ux , uy , vx , vy continuous is conformal at a location z = x + iy if and only if f (z) =
u + iv is an analytic function, with f 0 (z) 6= 0.
Proof. If f (z) is analytic at z0 , its Taylor expansion there starts w = f (z) = f (z0 ) +
(z − z0 )f 0 (z0 ) + · · · ; i.e., it is locally around z0 a combination of a translation (given by
f (z0 )), a uniform scaling (given by |f 0 (z0 )|), and a rotation (given by arg f 0 (z)), all of
which preserve angles.
The proof for the reverse statement (local angles preserved implies analyticity) is more
lengthy and is given in Section 8.7 as Theorem 8.18.
The next theorem makes it relatively simple to check that a mapping function is oneto-one.
Theorem 8.2. Let D be a finite region in the z-plane, with boundary δD, corresponding to
E and δE in the w = f (z) plane, with neither boundary self-intersecting. Then
(a) w = f (z) provides a one-to-one map from D to E, and
(b) if z moves around δC in the positive direction, so does w around δE.
Proof. Let w0 be a point inside E, and consider
ˆ
ˆ
1
f 0 (z) dz
1
dw
I=
=
,
2πi δD f (z) − w0
2πi δE w − w0
where z moves around δD in the positive direction. By Theorem 4.8, I = N − P , where
N and P are the number of zeros and poles, respectively, of f (z) − w0 . With P = 0,
N ≥ 0, and the RHS = ±1 according to the direction w moves around δE, the theorem
follows.
212
Chapter 8. Conformal Mappings
In the following, we will reserve the expression “conformal mapping” for such ones
that have the one-to-one property. Let us note again that the functions w = f (z) and
z = g(w), mapping between the domains D and E in the z- and w-planes, respectively,
are assumed to be analytic (singularity-free) within their respective domains.
8.2 Mappings provided by bilinear functions
It turns out that bilinear functions67
az+b
,
(8.1)
cz+d
the ratio between two linear expressions in z, have a number of remarkable properties when
it comes to mapping simple regions.
w=
Theorem 8.3. Any straight line or circle in the z-plane becomes mapped by a bilinear
function into another circle or straight line.
Proof. If c = 0, the result is obvious, since (8.1) then just amounts to a sequence of (i)
scaling/rotation (forming a z), (ii) shift (adding b), and (iii) scaling/rotation (dividing by
d), none of which can change a circle/line away from still being a circle/line. For c 6= 0,
we rewrite (8.1) as
az+b
1
1
w=
=
(bc − ad)
+a ,
cz+d
c
cz + d
and it is again just a sequence of scaling/rotations and shifts, with the only novelty that it
also includes one inversion. Hence, it remains to show that if z is on a circle/line in the
complex plane, so is w = 1/z. However, this follows from Exercises 1.6.17 and 1.6.18. A
circle/line in the complex plane becomes a circle on the stereographic sphere, and inversion
is associated with a rotation of the sphere, i.e., preserves the circular shape. Then, it again
becomes a circle/line when projected back to the complex plane.
Theorem 8.4. Successive bilinear mappings become again a bilinear mapping.
z+b1
Proof. Straightforward algebra shows that w = ac11z+d
followed by ζ =
1
a3 z+b3
written as ζ = c3 z+d3 , where
a3 b3
a2 b2
a1 b1
=
.
c3 d 3
c2 d2
c1 d1
Theorem 8.5. The inverse of a bilinear mapping w =
w+b2
bilinear map z = ac22w+d
.
2
h
a1
c1
b1
d1
i
a1 z+b1
c1 z+d1
a2 w+b2
c2 w+d2
can be
(8.2)
takes also the form of a
z+b1
6= 0, since otherwise w = ac11z+d
reduces to w =
1
h
i h
i−1
constant (with no z-dependence). By (8.2), we should choose ac22 db22 as ac11 db11
,
since the combined mapping then becomes the identity; i.e., starting from z, we get back
to z.
Proof. We can assume det
67 Also
known as Möbius transforms or linear fractional transforms.
8.2. Mappings provided by bilinear functions
1
1
1
−1
213
1
1
1
−1
1
−1
−1
−1
1
−1
−1
(a) z-plane.
(b) w-plane.
(c) z-plane.
(d) w-plane.
Figure 8.2. Mapping by the bilinear function w = 1/z of the inside of the unit circle to its
outside and by a shifted unit circle to a half-plane. This function maps zero to infinity, and vice versa.
Among the family of bilinear mappings, the Blaschke factor,
z−a
,
w=λ
1 − az
(8.3)
where a is an arbitrary complex constant and where |λ| = 1, is particularly important due
to the following result.
Theorem 8.6. The bilinear mapping (8.3) maps z = a to 0 and |z| = 1 to |w| = 1.
Proof. The result
= a → 0 is obvious. The second part needs verification: |w|2 =
zz z−a
z−a
z−z a+a a
z−a
1−a z
1−a z = 1−a z−a z+a a z z = 1. In the last equality, we utilized twice that z is on
the unit circle, i.e., that z z = 1.
Figures 8.2 and 8.3 illustrate the mappings performed by two different bilinear functions. This can be contrasted to the mapping in Figure 8.4, which is not bilinear, and hence
can map a circle to a different type of curve, here to an ellipse. The function in this figure
is the same as the function shown earlier in Figure 2.2.
Theorem 8.7. There is no other way to conformally (one-to-one) map the inside of the
unit circle and its center to themselves than through the Blaschke factor (8.3) for a = 0,
w = λz, where λ is a complex constant of magnitude 1.
1
1
1
1
−1
1
−1
1
1
−1
−1
−1
(a) z-plane.
1
−1
−1
−1
(b) w-plane.
(c) z-plane.
(d) w-plane.
Figure 8.3. Two mappings by the bilinear function w = (z − 12 )/(1 − 12 z). Since this
function is of the form w = (z − a)/(1 − a z), the mapping takes the unit circle to itself (red) and
inside to inside and outside to outside. The blue circle becomes again a circle (since the mapping is
of bilinear form). This function maps z = 0 to w = − 21 , z = 2 to w = ∞, and z = ∞ to w = −2.
The inverse mapping is z = (w + 21 )/(1 + 12 w).
214
Chapter 8. Conformal Mappings
1
1
1
1
1
−1
1
−1
1
−1
1
−1
−1
−1
−1
(a) z-plane.
−1
(b) w-plane.
(c) z-plane.
(d) w-plane.
Figure 8.4. Two mappings by the function w = 12 (z + 1/z). This function is not bilinear,
and we note in the second part a circular arc becoming an elliptic arc. The mapping is singular at
the points z = ±1 (corresponding to w = ±1), and local angles are doubled at these points.
Proof. Let w = f (z) be another such mapping, and consider g(z) = f (z)/z. Since the
mapping is one-to-one, f (z) has a unique zero inside the unit disk. It must be located at the
origin, since z = 0 is mapped to w = 0, and hence the singularity of g(z) at z = 0 must
be removable, making g(z) and consequently log(g(z)) = log |g(z)| + i arg g(z) analytic
within the unit disk. By Theorem 4.18 (see also Section 2.2.3), as a harmonic function,
log |g(z)| can have neither a minimum nor a maximum within the disk. Since |g(z)| = 1,
and so log |f (z)| = 0, around |z| = 1, log |g(z)| must remain identically 0 within the disk;
i.e., |g(z)| has to be identically 1.
The theorem above can quite easily be strengthened to apply also when f (z) is known
to be analytic only for |z| < 1 rather than for |z| ≤ 1 by considering a limit process as
|z| % 1 (known as the Schwarz lemma).
Between Theorems 8.6 and 8.7, we can conclude that a mapping from the unit circle
to itself has precisely three free (real) parameters, of which two can be used to choose the
(complex) point a and the third to specify λ = eiθ (obeying |λ| = 1). Combining some of
the results above leads to the following.
Theorem 8.8. The most general mapping of the upper half-plane to itself can be written in
the form w = az+b
cz+d , where a, b, c, d are all real and satisfy ad > bc.
Theorem 8.9. The most general conformal mapping of the whole stereographic sphere to
itself takes the form of a bilinear function.
The next section tells how some of the results above can be greatly generalized, from
the unit circle (or upper half-plane) to arbitrarily shaped domains, assumed, however, to be
simply connected, i.e., neither the whole plane nor regions with nonconnected boundaries
(as were seen in the subplot of Figure 8.3).
8.3 Riemann’s mapping theorem
Theorem 8.10. Let D be a simply connected domain (excluding the entire complex plane).
There exists then a 3-parameter family of functions f (z) that maps the unit circle one-toone to D (or, in reverse, D to the unit circle).
8.4. Mappings of polygonal regions
215
This result was formulated by Riemann as a brief part of his 1851 Ph.D. thesis. Weierstrass later found the proof to be incomplete, and the first rigorous one was not given until
1912. While strict proofs remain technically difficult, we sketch in Section 8.5.2 a heuristic
argument that makes the theorem seem very plausible.
Current rigorous proofs are all of “existence type” and not “constructive” for actually
obtaining a mapping function from the domain shape D. There is, however, a rich literature
on how to achieve this numerically [17, 40]. In the particular case when D is a polygon,
there exists a systematic approach that we will describe next. However, even this will
require computational methods for all but the simplest cases [12].
Domains in the complex plane are described as being simply, doubly (triply, etc.) connected according to their character on the stereographic sphere (cf. Section 1.4). If a curve
between any two points can be continuously deformed (without exiting the domain) to any
other such curve, the domain is simply connected. If there can exist two (but no more)
families of curves that cannot be deformed to each other, it is doubly connected, etc. For
example, the shaded regions in Figures 8.2 and 8.4 are all simply connected, whereas the
ones in Figure 8.3 are doubly connected. With the exception that the entire complex plane
cannot be conformally mapped to any other region, conformal mappings are possible only
between equally connected domains.
8.4 Mappings of polygonal regions
The classical method for mapping to the inside of a polygon comes in two versions dependent on the region we map from: either from the upper half-plane or from the inside of
the unit circle (without an intermediate bilinear mapping going via the upper half-plane).
The resulting formulas turn out to be virtually identical in the two cases. We follow here
the first choice. For the second one, see Exercise 8.8.11 and Example 11.3. A different
derivation is given in Ahlfors [2, Chapter 6].
The following example of creating a mapping function from the upper half-plane to the
inside of a triangle generalizes immediately to the case when the target region instead is a
general polygon.
Example 8.11. Map conformally the upper half-plane from Figure 8.5(a) to the inside of
the triangle shown in Figure 8.5(b).
We simplify first the problem by rotating and translating the target triangle (and scaling
it if we so wish), so that one of its sides (here marked blue) falls along the w-axis; cf.
Figure 8.5(c). Let then z0 = −∞ correspond to w0 = w(−∞), z1 correspond to w1 , and
z2 correspond to w2 . Since the mapping has three free parameters, we can for convenience
choose z1 = −1 and z2 = +1. When a point z moves along the z-axis over [−∞, z1 ], the
mapping function w(z) needs to satisfy arg w0 (z) = 0 (so that w(z) follows the first side
of the triangle, shown in blue). Then for z ∈ [z1 , z2 ], it has to hold that arg w0 (z) = πα1
(to match the slope of the second triangle side, shown in red). Finally, for z ∈ [z2 , +∞],
arg w0 (z) needs to be increased with an additional πα2 (with side shown in green). In other
words, arg w0 (z) needs, along the real z-axis, to be a piecewise constant function which at
z = z1 and z = z2 jumps by the amounts πα1 and πα2 , respectively.
Recalling the behavior of the log z function (cf. Figure 2.7(b)), we recognize in this
function the perfect “building block” for creating functions that are piecewise constant
216
Chapter 8. Conformal Mappings
z0 = ∞
z1 = −1
z2 = 1
(a) z-plane.
(b) w-plane.
πα2
w2
πα1
w0
πα0
w1
(c) w-plane.
Figure 8.5. Steps in mapping the upper half-plane to the inside of a triangle, as described
in Example 8.11. In part (c), the corners are denoted by w0 , w1 , w2 and the exterior angles are
πα0 , πα1 , πα2 with here α0 = 17/20, α1 = 3/4, α2 = 2/5.
along the real axis. For example, the imaginary part of
g(z) = C − α1 log(z − z1 ) − α2 log(z − z2 )
(with C some constant) jumps by πα1 when z (real) increases past z1 and by πα2 when
it passes z2 . Displayed over the upper half-plane, it looks as shown in Figure 8.6 in the
present case of z1 = −1, α1 = 3/4 and z2 = +1, α2 = 2/5 (i.e., the jumps are of sizes
πα1 and πα2 , respectively, matching the angles in Figure 8.5).
What we need is a function w0 (z) such that arg w0 (z) (rather than Im w0 (z)) has this
step structure. However, the Euler identity ez = ex+i y = ex (cos y + i sin y) tells us that
the two functions arg and Im are closely related. For any complex z, it holds that
arg ez = Im z .
8.4. Mappings of polygonal regions
217
4
3
2
1
0
-1
1
0.8
0.6
0.4
0.2
y
0
-3
-2
-1
0
1
2
3
x
Figure 8.6. Imaginary part of the function g(z) = −α1 log(−1 − z) − α2 log(+1 − z),
constructed to be piecewise constant along the real axis, with jumps of sizes 43 π and 25 π at the
locations z1 = −1 and z2 = 1, respectively. The colors correspond to those of Figure 8.5.
Hence, we form
w0 (z) = eg(z) = eC (z − z1 )−α1 (z − z2 )−α2 ,
(8.4)
0
and arg w (z) now matches the requirements imposed by the angles and the straight sides
of the triangle.
The further generalization from a target triangle to a general target polygon with corners
at w0 , w1 , . . . , wn is now immediate. Also renaming eC = A, we have arrived at the
following.
Theorem 8.12 (Schwarz–Christoffel). A function w(z) that maps the upper half-plane
to the inside of a polygon, and [−∞, z1 , z2 , . . . , zn ] (on the real axis) to the polygon with
corners [w0 , w1 , w2 , . . . , wn ] and external angles [πα0 , πα1 , πα2 , . . . , παn ], satisfies
w0 (z) = A (z − z1 )−α1 (z − z2 )−α2 · · · · · (z − zn )−αn .
(8.5)
In practical use of these mappings obtained by the Schwarz–Christoffel theorem, one
should note the following:
• Only three (real-valued) parameters can be chosen arbitrarily. We have above used
one to enforce that z = −∞ maps to the polygon corner w0 . Given that the integral
typically is difficult to handle in closed form, further choosing z1 = −1 and z2 = +1
may simplify the algebra.
• The constants A and z3 , z4 , . . . , zn will then need to be obtained from enforcing that
(8.5) obeys the mapping requirements that z = [−∞, z1 , z2 , . . . , zn ] correspond to
w = [w0 , w1 , w2 , . . . , wn ]. The algebra in this step may become difficult or impossible unless approached numerically. The following two examples will be revisited
in Section 8.5.
218
Chapter 8. Conformal Mappings
8
8
6
6
4
4
2
2
0
0
-2
-2
-4
-10
-8
-6
-4
-2
0
2
(a) Upper half-plane near the z-axis.
4
-6
-4
-2
0
2
4
6
(b) Mapping of grid shown in part (a).
Figure 8.7. Mapping provided by the function w(z) in (8.6), Example 8.13.
Example 8.13. Map the upper half-plane (cf. Figure 8.7(a)) to the region indicated in part
(b) (with a step height of π).
We can use the three free parameters to ensure that the mapping function w(z) will
satisfy w(−∞) = −∞, w(−1) = πi, and w(1) = 0. With exterior angles at the two latter
points − π2 and + π2 , respectively, we have α1 = − 21 and α2 = 12 . By (8.5), the mapping
function w(z) will satisfy
1/2
z+1
0
w (z) = A
,
z−1
from which follows by integration
w(z) = A · (z 2 − 1)1/2 + log z + (z 2 − 1)1/2 + B.
(8.6)
The two conditions w(−1) = πi and w(1) = 0 determine the constants in (8.6) as A = π1
and B = 0. With some care in choosing the appropriate branch cut location for the square
root function (for example, implement (z 2 − 1)1/2 as (z + 1)1/2 (z − 1)1/2 ), the z-plane
grid in Figure 8.7(a) maps to the curve set shown in part (b).
Example 8.14. Map the upper half-plane (cf. Figure 8.8(a)) to the region indicated in part
(b).
The key idea in this case is to let the target region have three corners at w(−1) = −1,
w(0) = −i b, w(1) = 1 and then consider the limit of b → +∞. The matching exterior
angles become − π2 , π, − π2 , and we thus need to use the exponents 12 , −1, 21 , i.e.,
w0 (z) = A(z + 1)1/2 z −1 (z − 1)1/2 ,
which, when integrated, gives
1
2
1/2
w(z) = A (z − 1) + arcsin
+ B.
z
Requiring that w(−1) = −1 and w(1) = 1 gives A =
2
π
and B = 0.
(8.7)
8.5. Some applications of conformal mappings
219
1
3
2.5
0.5
2
0
1.5
1
-0.5
0.5
-1
0
-0.5
-1.5
-1
-2
-1.5
-3
-2
-1
0
1
2
3
(a) Upper half-plane near the z-axis
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
(b) Mapping of grid shown in part (a)
Figure 8.8. Mapping provided by the function w(z) in (8.7), Example 8.14.
8.5 Some applications of conformal mappings
Many applications of conformal mappings rely on the fact that the change of independent
variables leaves Laplace’s equation invariant.
Theorem 8.15. If ψ(u, v) satisfies
ψuu + ψvv = 0,
(8.8)
then, after a conformal mapping
u = u(x, y),
v = v(x, y),
ψ expressed in x, y will also satisfy Laplace’s equation
ψxx + ψyy = 0.
(8.9)
Proof. (Outlines of two approaches for proving the result).
1. If ψ(u, v) satisfies (8.8), then it is the real part of an analytic function Ψ(w) in a
complex (u, v)-plane. A further analytic variable change w = f (z) (w = u + iv, z =
x + iy) gives Ψ(f (z)), which now has to be analytic in z, implying (8.9).
2. In a less elegant manner, one arrives at the same result from using the chain rule
∂2
∂2
on ∂x
2 + ∂y 2 ψ(u(x, y), v(x, y)) and then simplifying by means of the C-R equations
ux = vy , uy = −vx .
If we want to solve (8.8) in some region of a (u, v)-plane, with values of ψ given on
a boundary, a conformal mapping preserves this boundary data while having changed the
shape of the region. Starting from a domain in which the Laplace solution is trivial, one
can thus obtain corresponding solutions also in domains of much more general shapes.
8.5.1 Ideal fluid flow in 2-D
In the context of 2-D fluid dynamics, it is customary to let u(x, y) and v(x, y) denote the
velocities of fluid particles in the x- and y-directions, respectively. The governing equations
220
Chapter 8. Conformal Mappings
for 2-D steady, ideal flow then become
Incompressible: ux + vy = 0,
(8.10)
uy − vx = 0
(8.11)
Irrotational:
(with a sign reversal in v compared to the C-R equations). It follows from (8.10) that we
can introduce a stream function ψ(x, y) such that
(
∂ψ
u=
∂y ,
v=
− ∂ψ
∂x .
From (8.11) it follows then that this function ψ(x, y) satisfies
ψxx + ψyy = 0 .
(8.12)
Level curves of ψ(x, y) are streamlines (paths that the fluid particles will follow). The following example introduces the mapping we will use in our first illustration of this concept.
Example 8.16. Map conformally the upper half-plane outside the unit circle to the complete upper half-plane (cf. Figure 8.9(b)–(c)).
Figure 8.9(a) illustrates the function f (x) = 12 x + x1 for a real argument x. It is
clearly monotonically increasing for x > 1 and decreasing for x < −1. Considering the
complex counterpart function
1
1
w = f (z) =
z+
,
(8.13)
2
z
it is therefore clear that the cyan and green line segments in Figure 8.9(b) correspond to
the similarly colored ones in part (c). The upper part of the unit circle in the z-plane is
given by z = eiθ , 0 ≤ θ ≤ π, and will correspond to w = 21 eiθ + e1iθ = cos θ, i.e., the
straight line between −1 and +1. The mapping will
be conformal everywhere apart from
where f 0 (z) = 0, which, with f 0 (z) = 12 1 − z12 , occurs at z = ±1 (already noted in
part (a) of the figure). We see this lack of conformity again when comparing parts (b) and
(c); the 90◦ boundary angles in the z-plane have become 180◦ in the w-plane. At all other
locations, the mapping will have to be conformal. Similarly, the lower half-plane outside
the unit circle in the z-plane will map to the lower half-plane in the w-plane. Note that this
mapping (8.13) is not a bilinear one: putting its RHS in common denominator form will
make the numerator quadratic.
Figure 8.10 illustrates the steps needed to obtain the flow past the central objects in the
figure’s parts (a), (c), and (d): the unit circle, and flat plates inclined at different angles
relative to free stream. We denote the planes in parts (a) and (b) as complex z- and wplanes, respectively. As we know from Example 8.16, the mapping
1
1
z+
(8.14)
w=
2
z
maps the outside of the circle in (a) to the outside of the flat line segment [−1, 1] in (b). We
write as usual z = x + iy and w = u + iv. In the w-plane, the flow field is trivial, described
8.5. Some applications of conformal mappings
221
f(x)
2
1
-3
-2
-1
1
2
3
x
-1
-2
(a) The function f (x) =
1
2
x+
1
x
.
2.5
2
1.5
1
0.5
-3
-2
-1
1
2
3
z-plane
(b) The domain we map from in the z-plane.
2.5
2
1.5
1
0.5
-3
-2
-1
1
2
3
w-plane
(c) Target domain in the w-plane.
Figure 8.9. Illustrations of the function f (z) =
mapping it provides for z complex.
1
2
z+
1
z
along the real axis and of the
222
Chapter 8. Conformal Mappings
by the stream function ψ(u, v) = v. We then get from (8.14)
1
1
u + iv =
x + iy +
2
x + iy
y
x
1
1
+
i
,
=
1+ 2
1
−
2
x + y2
2
x2 + y 2
from which itfollows (equating imaginary parts) that ψ(u, v) = v corresponds to ψ(x, y) =
y
1
2 1 − x2 +y 2 , which thus becomes the matching stream function in the z = (x + iy)-plane
(cf. Figure 8.10(a)).
Multiplying z by a complex number eiα will in (a) rotate the free stream, but has no
effect on the shape of the central circle. Doing this, and then mapping from the z-plane to
the w-plane using (8.14), brings the central circle again to the horizontal flat plate, but the
far-field flow direction has been changed by the angle α. This gives the streamlines that are
shown in parts (c) and (d), for 45◦ and 90◦ turns, respectively. Considerable amounts of
information can now be deduced. For example, fluid velocities are inversely proportional
to the separation of adjacent stream lines. Parts (c) and (d) thus reflect limitations of the
ideal flow assumptions, as it would suggest infinite velocities at the ends of the inclined
plates.
8.5.2 Static electrical fields in 2-D
Electrical fields in vacuum also satisfy Laplace’s equation. Just as edges of solid objects
must coincide with a streamline (i.e., no fluid passing through its surface), electric potentials are constant over the surface of a conductor. Looking again at Figure 8.10, we can
identify the solid curves with equipotential curves and the dashed curves with field lines.
Parts (c) and (d) show how these latter ones become very concentrated at “spikes” in the
shape of a conductor, and much less so along smooth surfaces (as in part (b)).
Figures 8.7 and 8.8 can now also be interpreted as illustrating both flow- and electrostatic fields in their respective geometries. As a concluding example, Figure 8.11 shows in
part (a) the field (fluid or electrostatic) between two infinite parallel plates, and in parts (b)
and (c) its counterpart after the mappings w = ez and w = z + ez , respectively. From part
(b), we can deduce that the logarithm function describes the stream function (or electrostatic field) around a point source/sink (or a point charge).While the fields in parts (a) and
(b) are conceptually trivial, that is not the case in part (c), showing flow exiting a 2-D pipe
or, equivalently, the field around the edge of a capacitor.
8.6 Revisiting the Jacobi elliptic function sn(z, k )
From our discussion of the Schwarz–Christoffel mapping, it becomes apparent that (7.10)
represents the conformal mapping illustrated in Figure 8.12: mapping the upper half-plane
shown in part (a) to the inside of the rectangle in part (b). The inverse function z(w) is
clearly real-valued whenever w lies on one of the straight sides shown in part (b). By the
Schwarz reflection principle (Section 3.2.2), reflecting this rectangle as indicated by going
from the red rectangle in Figure 8.13 to the composite of four rectangles shown here in four
black (dashed line) rectangles will have created a doubly periodic function. As such, we can
shift the period box as shown to arrive at the rectangle marked in green. By this, we have
8.6. Revisiting the Jacobi elliptic function sn(z, k )
223
5
2
4
1.5
3
2
1
1
0.5
0
0
-1
-0.5
-2
-1
-3
-1.5
-4
-2
-3
-5
-6
-4
-2
0
2
4
-2
-1
0
1
2
3
6
(b) w = (u + iv)-plane; w = 12 (z + z1 ).
(a) z = (x + iy)-plane.
3
2.5
2
2
1.5
1
1
0.5
0
0
-0.5
-1
-1
-1.5
-2
-2
-2.5
-3
-3
-2
-1
0
1
2
3
(c) Mapping to flat plate after 45◦ turn.
-2
-1
0
1
2
(d) Mapping to flat plate after 90◦ turn.
Figure 8.10. A trivial flow field (part (a)) and its counterparts after different conformal mappings.
recovered exactly the same Jacobi elliptic sn(z, k) function as we introduced previously in
Section 7.4. This confirms that (7.10) indeed represents a doubly periodic function over the
claimed period box.
Example 8.17. Heuristically motivate Riemann’s mapping theorem (Theorem 8.10).
Following Copson [8], we first check on what properties a function w = f (z) must
have if it maps the inside of a curve Γ (such as shown in Figure 8.14, part (a)) to the inside
of the unit circle, part (b), with f (a) = 0. Being a one-to-one mapping, f (z) can have no
other zero inside (or on) Γ than z = a, so we can write
f (z) = (z − a) eφ(z) .
Since |f (z)| = 1 on Γ, we obtain as the real part of the logarithm of (8.15)
log |z − a| + Re φ(z) = 0.
(8.15)
224
Chapter 8. Conformal Mappings
3
2
1
0
-1
-2
-3
-7
-6
-5
-4
-3
-2
-1
0
1
2
(a) Finite section of the area between two infinite plates,
located vertically at ±i in a complex z-plane.
6
4
2
0
-2
-4
-6
-8
-6
-4
-2
0
2
4
6
8
z
(b) The mapping w = e (or z = log w).
8
6
4
2
0
-2
-4
-6
-8
-10
-5
0
5
10
z
(c) The mapping w = z + e .
Figure 8.11. Two mappings from the infinite strip |Im z| ≤ π.
8.6. Revisiting the Jacobi elliptic function sn(z, k )
(a) z-plane.
225
(b) w-plane.
Figure 8.12. (a) Upper half-plane with the marked segments color-coded and (b) the resulting map in the w-plane through the relation (7.10), with matching color coding.
The term V = Re φ(z) is the real part of an analytic function, and will therefore satisfy
2
2
Laplace’s equation ∂∂xV2 + ∂∂yV2 = 0, where z = z + iy. If we can show there exists a
unique solution to Laplace’s equation which (i) vanishes on Γ, and (ii) stays finite within
Γ apart from behaving like log |z − a| (which itself satisfies Laplace’s equation away from
the singularity at z = a) near to z = a, then the real part of φ(z) can be obtained from
V (x, y) = log |z − a| + Re φ(z) .
We can then from this obtain a matching imaginary part of φ(z) from the C-R equations.
However, a function V (x, y) with the required properties will need to exist if we can
presume that nature will not come to a halt if subjected to the experiment of finding the
Figure 8.13. Illustrations of reflecting and shifting the rectangle shown in Figure 8.12(b)
to obtain the period box for the Jacobi elliptic function sn(z, k) .
226
Chapter 8. Conformal Mappings
(a) z-plane.
(b) w-plane.
Figure 8.14. (a) Region in the z-plane and (b) target region in the w-plane for a mapping
function w = f (z).
electrostatic potential in the configuration in Figure 8.14(a), where the outer boundary Γ is
an electric conductor and where a point charge has been placed at the location z = a.
8.7 Supplementary materials
Theorem 8.18. If the mapping
u = u(x, y)
v = v(x, y)
with ux , uy , vx , vy continuous is conformal in the vicinity of a location z = x + iy , then
f (z) = u + iv is an analytic function, with f 0 (z) 6= 0.
Proof. Consider infinitesimal steps dx and dy in the (x, y)-plane, with corresponding steps
du and dv in the (u, v)-plane. We have then du = ux dx + uy dy and dv = vx dx + vy dy.
Preserving angles locally means that all sides of any infinitesimal triangle scale with the
same factor h(x, y), i.e., du2 + dv 2 = h2 (dx2 + dy 2 ). With the expressions above for du
and dv, this becomes
(u2x + vx2 − h2 )dx2 + (u2y + vy2 − h2 )dy 2 + (ux uy + vx vy )dxdy = 0.
For this to hold identically for any combination of dx and dy, all three coefficients need to
vanish, i.e.,

u2x + vx2
= h2 ,



u2y + vy2


 u u +v v
x y
x y
= h2 ,
= 0.
We can satisfy the top equation with ux = h cos α, vx = h sin α and the second by uy =
h cos β, vy = h sin β, after which the third equation gives α − β = ± π2 , from which
8.8. Exercises
227
follows either {ux = vy , uy = −vx } or {ux = −vy , uy = vx }. We recognize the first
option as the C-R equations for u + iv = f (x + iy), implying that f (z) is analytic. The
second option similarly corresponds to u + iv = f (x + iy). Although the latter version
also preserves scales locally, and sizes of angles, it reverses directions and is hence not
what we accept as being conformal.
8.8 Exercises
Exercise 8.8.1. With the notation and the result of Theorem 8.2, w = f (z) provides a
one-to-one map between z ∈ D and w ∈ E. Show that the inverse mapping z = f −1 (w)
´
ζ f 0 (ζ)
1
is given by z = 2πi
dζ.
δD f (ζ)−w
Exercise 8.8.2. Construct a function w(z) that maps the inside of the unit circle to the
upper half-plane.
Hint: One approach is to apply suitable shifts and rotations to the mapping w = 1/z
illustrated in Figure 8.2 (note especially the rightmost two subplots).
Exercise 8.8.3. The exponential function w = ez and trigonometric functions (e.g., w =
sin z) are useful ingredients in many conformal mapping cases.
(a) Map the infinite strip |Im z| < π to the right half-plane.
(b) Map the semi-infinite strip Re z < 0, |Im z| < π to the upper half of the unit circle.
(c) Show that w = sin z maps the semi-infinite strip 0 < Re z < π2 , Im z > 0 to the
first quadrant.
(d) Show that w = sin z maps a rectangle with its two bottom corners at z = ± π2 to
the upper half of an ellipse.
Exercise 8.8.4.
(a) Show that w = z + z1 maps the upper half of the unit circle to the lower half-plane.
√
(b) Map the inside of the unit circle to its upper half (in particular, explain why w = z
is not a solution to this problem).
(c) Map the region in the first quadrant above the hyperbola y = 1/x to the full first
quadrant.
Exercise 8.8.5. Consider the upper half-plane with two nonoverlapping circular “holes.”
Either describe the sequence of steps you would take to map this region to the annulus
between two circles, both centered at the origin, or explain why it cannot be done.
Exercise 8.8.6. Prove Theorem 8.8.
Exercise 8.8.7. Give a direct proof (not via stereographic mappings) of the fact that if z is
located on a circle or straight line in the complex z-plane, so is w = 1/z in the complex
w-plane (cf. the proof of Theorem 8.3).
Exercise 8.8.8. Carry out the details of the second of the outlined proofs for Theorem 8.15.
Exercise 8.8.9. Let f (z) be an entire function that satisfies |f (z)| = 1 for |z| = 1. Show
that it then must be of the form f (z) = eiϑ · z n with θ real and n a nonnegative integer.
228
Chapter 8. Conformal Mappings
Hint: One way to show this uses these steps: (i) Show that f (z) has a zero for |z| < 1,
or else it is a constant, and (ii) if f (z) has zeros for |z| < 1, note that these can all be
“divided away” by functions of the form (8.3).
Exercise 8.8.10. For the cases in Figure 8.3, electrostatic field lines are trivial in the zplane. Create images of the corresponding field lines through the gray shaded areas in the
corresponding w-planes.
Exercise 8.8.11. Show the following counterpart to Theorem 8.12: The mapping function w(z) from the inside of the unit circle to the inside of a closed polygon with corners
[w0 , w1 , w2 , . . . , wn ] and external angles [πα0 , πα1 , πα2 , . . . , παn ] satisfies
w0 (z) = A (z − z0 )−α0 (z − z1 )−α1 (z − z2 )−α2 · · · · · (z − zn )−αn .
(8.16)
Note: The formula (8.16) is identical to (8.5) except that the factor (z − z0 )−α0 is now also
included (meaning that the pre-image to w0 is now a point z0 on the unit circle, rather than
−∞).
Hint: For the proof of Theorem 8.12, we relied on the fact that Im(− log(z0 − z)) features
a step but is otherwise piecewise constant when z0 is real and z moves along the real axis.
Figure 8.15(a) shows a corresponding result for when z0 is on the unit circle and z = eiθ
moves around the circle. The function
φ(z) = Im(− log(z0 − z))
with z0 = 1 satisfies
(8.17)
1
(8.18)
φ eiθ = (π − θ), 0 < θ < 2π,
2
for a reason that can be deduced from the geometry observation in Figure 8.15(b). Extending (8.17) to include [z0 , z1 , z2 , . . . , zn ] (all on the unit circle), deduce that when z moves
around the unit circle, the function w(z) defined by (8.16) will obey both of the required
key properties:
(i) w(z) becomes a straight line segment when z moves between any two adjacent zi
points, and
(ii) w(z) makes the specified changes in direction when z passes any of the zi points.
8.8. Exercises
229
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-1.5
-1
-0.5
0
0.5
-1
-1.5
1
-0.5
0.5
0
1
1.5
y
x
(a) φ(z) = Im(− log(z0 − z)) shown for z0 = 1.
1
0.8
0.6
θ/2
0.4
θ
y
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-1.5
-1
-0.5
0
0.5
1
x
(b) A theorem from plane geometry.
Figure 8.15. The imaginary part of the logarithm function when following a circle going
through its singular point (here, the unit circle and the singularity at z = 1).
Chapter 9
Transforms
We have often introduced analytic functions
´ ∞by means of infinite sums (e.g., Taylor expansions) or by integrals (such as Γ(z) = 0 e−t tz−1 dt in Section 2.7). A more general
approach in the case of integrals is to choose a path C and a kernel function K(z, w) and
then, for every function f (z), define a counterpart function g(w) as
ˆ
g(w) =
f (z) K(z, w) dz.
(9.1)
C
This procedure becomes particularly useful if
(i)
there is a similar integral formula available to return to f (z) if g(w) is specified
(with the two formulas then forming a transform pair), and
(ii)
some tasks become simpler when carried out in the transformed variable.
In the discussion below of the most frequently used transforms, the focus will be on their
mathematical properties rather than on their very wide range of applications. Even for
functions that are real-valued along the real axis, knowledge of analytic functions will play
a key role in simplifying and in utilizing the transforms. The chapter starts with the Fourier
transform. The further transforms are all related to the Fourier cases (via variable changes,
etc.). Common features include that taking derivatives of a function usually becomes a
purely algebraic operation in the transformed variable and that the transforms possess convolution theorems.
9.1 Fourier transform
One way to introduce the Fourier transform is through the Fourier series in the limit when
the period increases to infinity. Hence, we first briefly summarize the series case and how
its coefficients may be found.
9.1.1 Fourier series
For f (x) real-valued and periodic over [−π, π], the Fourier series of f (x) can be defined as
f (x) = a0 +
∞
X
ak cos(kx) +
k=1
∞
X
k=1
231
bk sin(kx),
(9.2)
232
Chapter 9. Transforms
where
ˆ π
1
f (x)dx,
2π −π
ˆ
1 π
ak =
f (x) cos(kx)dx, k = 1, 2, . . . ,
π −π
ˆ
1 π
f (x) sin(kx)dx, k = 1, 2, . . . .
bk =
π −π
a0 =
(9.3)
Thus f (x) can be seen as a superposition of Fourier modes (sine and cosine functions) and
the coefficients tell us how much of each mode is needed to recover the original function.
The assumptions that are needed for (9.2) (and equivalently (9.4)) to converge for all x are
quite subtle. Requiring the function f (x) to be continuous and periodic is not quite enough
(see, for example, [29, Chapter 18]), while requiring it to be continuously differentiable is
stronger than required.
Equations (9.4) and (9.5) provide yet another example of how algebra simplifies if we
replace trigonometric functions with complex exponentials.
Theorem 9.1. If f (x) can be represented as
∞
X
f (x) =
ck eikx ,
(9.4)
f (x) e−ikx dx.
(9.5)
k=−∞
then
1
ck =
2π
ˆ
π
−π
Proof. If (9.4) holds, then the RHS in (9.5) evaluates (after renaming the summation variable from k to n) to
1
2π
ˆ
π
−ikx
f (x)e
−π
dx =
∞
X
cn
n=−∞
|
1
2π
1
0
ˆ
π
i(n−k)x
−π
e
{z
dx
= ck ,
}
if n = k,
otherwise.
In the case when f (x) is real-valued, the complex coefficients ck will satisfy c−k = ck .
Their decay rate for increasing k will reflect how smooth f (x) is. If f (x) is m times
differentiable (after having been extended periodically), then |ck | = O(|k|m+1 ), and if it
is analytic in a strip surrounding the x-axis, the decay rate will be exponential, with a rate
that depends on the strip width.
Example 9.2. Estimate how fast the Fourier coefficients ck go to zero for the function
1
f (z) = 3+sin
z.
This function was displayed
in Figure
√ 5.5(b). Its poles are located where sin z = −3,
i.e., at zk = − π2 + 2πk ± i log(3 + 2 2), k ∈ Z. Because of the periodicity, the contour
√
in (9.5) can be shifted up/down with any amount less than log(3+2 2) without any change
9.1. Fourier transform
233
in the ck . Since the Fourier series must remain convergent (being the expansion of the analytic function along
the shifted
and the e±ikz factors have grown in magnitude with
√
√path)
1
|k| log(3+2 2)
|k|
up to e
= (3 + 2 2) , the ck will decay in magnitude like O (3+2√
.
2)|k|
The only information needed here for f (z) was the imaginary part of the singularity closest
to the real axis.
A key use of analytic functions in the context of Fourier series arises in evaluating
integrals of the form (9.5). In fact, it was shown in Example 5.7 that the poles in the
1
i
√
upper half-plane of f (z) = 3+sin
z have the residue − 2 2 . From this follows the exact
coefficients in Example 9.2: ck =
k
i√
2 2
1
√ k
2)
(3+2
for k ≥ 0 (and c−k = ck , since f (x) is
real for x real).
9.1.2 Fourier transform: Definition and inversion
Changing the period in x from [−π, π] to [−Lπ, Lπ] changes the frequencies present from
k
eikx to ei( L )x , where k runs through all integers. In the limit of L → ∞, there will be a
continuum of frequencies present, typically denoted by ω ∈ (−∞, ∞).
Theorem 9.3. Defining the Fourier transform of f (x) as
ˆ ∞
1
ˆ
f (ω) = √
f (x) e−iωx dx ,
2π −∞
the inversion formula becomes
1
f (x) = √
2π
ˆ
(9.6)
∞
fˆ(ω) eiωx dω.
(9.7)
−∞
Together, (9.6) and (9.7) form a transform pair (much like (9.4) and (9.5)).68 A proof
for this theorem, based on the limit argument just mentioned, is given in Section 9.7.
In some situations, a slightly more explicit notation is convenient:
fˆ(ω) = F {f (x)}(ω) and f (x) = F −1 {fˆ(ω)}(x).
The integrals that arise when using (9.6) and (9.7) are often perfectly suited for contour
integral treatment. Essential to the utility of the Fourier transform, as well as for all the
other transforms considered in this chapter, is that they are linear:
F {α f (x) + β g(x)}(ω) = α F {f (x)}(ω) + β F {g(x)}(ω) ,
generalizing to any number of terms.
Example 9.4. Determine the Fourier transform of a Gaussian
2
f (x) = e−αx ,
(9.8)
where α > 0.
68 In some literature, the Fourier transform is defined slightly differently than here. Inserting a factor 2π in each
of the exponents eliminates the factors √1 in front of each integral. Instead of having the factor √1 in front
2π
2π
1
in just one of the two places (then making the forms of
of both integrals, there is sometimes instead a factor 2π
(9.6) and (9.7) somewhat unsymmetric). It is also not uncommon to swap the sign in the two exponents.
234
Chapter 9. Transforms
We find from (9.6)
ˆ ∞
2
1
fb(ω) = √
e−αx e−iωx dx.
2π −∞
The integrand is an entire function. Since its magnitude grows large up and down the
complex plane, no matter the values for α > 0 and ω (cf. Figure 9.1), it does not seem
promising to try to close the integration path by some loop far out. Another option that
has been successful in some of our previous examples is to run a return path parallel to
the original one. That turns out to work again here. An algebraically elegant way to go
about that starts by completing the square in the exponent. We then obtain (denoting the
integration variable by z instead of by x)
ˆ ∞
ω2
iω 2
1
e−α(z+ 2α ) − 4α dz
fˆ(ω) = √
2π −∞
ˆ ∞
iω 2
ω2
1
= √ e− 4α
e−α(z+ 2α ) dz.
2π
−∞
This can be interpreted as having moved the integration contour vertically up by a distance
of ω/2α, which does not change its value compared to integrating along the real axis. Since
´∞
pπ
2
we for that case know the value to be −∞ e−αz dz = α
, we obtain
ω2
1
fˆ(ω) = √ e− 4α .
2α
(9.9)
In words, the Fourier transform of a Gaussian becomes again a Gaussian. Changing
α to make one of the Gaussians narrower will make the other one correspondingly wider.
One basic feature of Fourier transforms is that a “pulse” can never be narrow in both the
(usually physical) variable x and in the transform variable ω.69
Example 9.5. Let
f (x) = e−|x| .
Determine fˆ(ω), and then invert this result back to recover f (x).
We find from (9.6)
ˆ 0
ˆ ∞
ˆ ∞
1
1
−|x| −iωx
x −iωx
−x −iωx
ˆ
e
e
dx = √
e e
dx +
e e
dx
f (ω) = √
2π −∞
2π
−∞
0
r
1
i
i
2 1
=√
+
=
.
π ω2 + 1
2π i + ω i − ω
Inverting back requires us to evaluate
1
f (x) = √
2π
ˆ
∞
−∞
r
2 1
eiωx dk.
π ω2 + 1
This is not an integrand to which we can find a primitive function. It is, however, easy to
evaluate the definite integral with contour integration. We consider two cases:
√
√
as probability distributions, (9.8) and (9.9) have standard deviations σx = 1/ 2α and σω = 2α,
respectively, making σx σω = 1 for all Gaussians. This is the minimal possible value for the product of matching
σx and σω , forming a basis for Heisenberg’s uncertainty principle.
69 Viewed
9.1. Fourier transform
235
2
Re e−z
2
Figure 9.1. Magnitude and phase plot of the function f (z) = e−α z in the case of α = 1.
2
and Im e−z are illustrated separately in a different context in Figures 9.12(a)–(b).
1. If x > 0, consider a contour such as illustrated in Figure 5.20, in the upper half
of the complex ω-plane. By Jordan’s lemma, the contribution from the semicircle
vanishes, and using the residue at the only relevant pole (at ω = i) gives f (x) =
−x
√1 2πi √1 e
= e−x .
2π
2π i
2. If x > 0, extend the integration path similarly with the semicircle in the lower halfex
= ex .
plane. We similarly now obtain f (x) = − √12π 2πi √12π −i
Combining the two cases, the final result becomes f (x) = e−|x| .
We see in this example an instance of a general rule relating the smoothness of a function (not analytic along the x-axis) to the decay rate of its Fourier transform. Since f (x)
has a jump in its first derivative, |fˆ(ω)| = O(1/ω 2 ) for |ω| increasing. In Example 9.10,
we will similarly see jumps in the third derivative leading to |fˆ(ω)| = O(1/ω 4 ).
9.1.3 Some key features of the Fourier transform
The following two results are particularly useful in applications (both readily derived from
the Fourier transform definition).
Theorem 9.6. If a function f (x) has the Fourier transform fˆ(ω), then for n = 1, 2, 3, . . .
dn
dn
f (x) has the transform (iω)n fb(ω) and xn f (x) has the transform (i)n n fb(ω).
n
dx
dω
In our alternative notation,
n
d
dn
F
f
(x)
(ω) = (iω)n F {f (x)}(ω) and F {xn f (x)}(ω) = (in ) n F {f (x)}(ω).
n
dx
dω
236
Chapter 9. Transforms
Theorem 9.7.
F {f (x + a)}(ω) = eiωa F {f (x)}(ω) .
In words, shifting a function the distance a to the left multiplies its Fourier transform by a
factor of eiωa .
Example 9.8. Give a closed-form solution u(x, t) to the heat equation
∂u(x, t)
∂ 2 u(x, t)
=
∂t
∂x2
(9.10)
for t > 0, satisfying an initial condition u(x, 0), x ∈ (−∞, ∞). Then evaluate this expres2
sion for the special case of u(x, 0) = e−x .
Applying the Fourier transform in the x-direction to (9.10) gives
ˆ ∞
ˆ ∞ 2
1
1
∂u(x, t) −iωx
∂ u(x, t) −iωx
√
e
dx = √
e
dx ,
∂t
2π −∞
2π −∞ ∂x2
which simplifies to
ˆ ∞
ˆ ∞
∂
1
1
−iωx
2
−iωx
√
u(x, t) e
dx = (−ω ) √
u(x, t) e
dx ,
∂t
2π −∞
2π −∞
i.e.,
∂
û(ω, t) = −ω 2 û(ω, t) .
∂t
2
For each value of ω, this is an ODE in t, with the closed-form solution û(ω, t) = e−ω t û(ω, 0).
Inverting û(ω, t) back to physical x-space gives
ˆ ∞
2
1
u(x, t) = √
e−ω t û(ω, 0) eiωx dω .
2π −∞
2
In the special case of u(x, 0) = e−x , Example 9.4 provides all the ingredients needed for
a closed-form evaluation of the integral, giving
2
1
e−x /(1+4t) .
u(x, t) = √
1 + 4t
(9.11)
This solution is shown in Figure 9.2 (with the initial condition highlighted in red).
In this example above, we solved a PDE using the first part of Theorem 9.6. We will in
Section 11.1.1 base a transform solution of an ODE on the second part of this theorem.
Theorem 9.9. A function f (x) (not identically zero) and its Fourier transform fˆ(ω) cannot
both have compact support (be identically zero for sufficiently large |x| and |ω|, respectively).
´R
Proof. If f (x) has compact support, then (9.6) becomes fˆ(ω) = √12π −R f (x)e−iωx dx
for some finite R. This shows fˆ(ω) to be an entire function of ω. If it is identically zero on
any curve segment (consider a piece of the ω-axis sufficiently far out), it will thus have to
be identically zero.
9.1. Fourier transform
237
Figure 9.2. The solution (9.11) to the heat equation in Example 9.8, with the Gaussian
initial condition.
Example 9.10. Illustrate the Fourier transform of a cubic B-spline.
Given real-valued data yk at discrete 1-D locations xk , k = 1, 2, . . . , n, a cubic spline
interpolant s(x) is made up of a separate cubic polynomial in each subinterval, with the
extra requirement that, at the interior data locations xk , k = 2, 3, . . . , n − 1, where two
different cubics meet, s(x), s0 (x), and s00 (x) shall all remain continuous (for uniqueness,
two further conditions are needed, which we do not need to investigate here). The simplest “building block” to create such a cubic spline interpolant is known as a B-spline, the
narrowest cubic spline that connects identical zero nontrivially back to identically zero.
Figure 9.3(a) shows the cubic B-spline B(x) at unit-spaced data locations. It is nonzero
only over four subintervals and is here normalized such that its integral is 1. Writing it
down piece-by-piece becomes somewhat tedious:
[−∞, −2] → 0,
[−2, −1] → 43 + 2x + x2 + 61 x3 ,
[−1, 0]
→
2
3
− x2 − 21 x3 ,
[0, +1]
→
2
3
− x2 + 21 x3 ,
[+1, +2]
→
4
3
− 2x + x2 − 61 x3 ,
(9.12)
[+2, +∞] → 0.
However, its Fourier transform B̂(ω) becomes surprisingly simple (see Exercise 9.8.10):
1
B̂(ω) = √
2π
sin(ω/2)
ω/2
4
.
(9.13)
Figure 9.3(b) shows that also B̂(ω) looks very much like a single “pulse” with compact
support. However, Theorem 9.9 tells us this cannot be the case, and part (c) of the figure indeed displays low-amplitude irregularities for large ω. Since B(x) features discontinuities in its third derivative, these amplitudes decay like O(1/ω 4 ), as is apparent
from (9.13).
238
Chapter 9. Transforms
0.4
2/3
0.6
0.3
0.4
1/6
0.2
0.2
1/6
0
0
-4
-2
0.1
0
0
2
0
-10
4
-5
0
5
10
x
(a) The equispaced cubic B-spline B(x).
(b) Its Fourier transform B̂(ω).
10-4
6
4
2
0
10
15
20
25
30
(c) Detail of B̂(ω) for larger ω.
Figure 9.3. Illustrations of the cubic B-spline B(x) and its Fourier transform B̂(ω).
9.1.4 Parseval’s relations
For the Fourier transform, this relation takes the following form.
Theorem 9.11.
ˆ
ˆ
∞
∞
|fˆ(ω)|2 dω .
|f (x)|2 dx =
−∞
−∞
Proof. With use of (9.7) and (9.6), we obtain
ˆ
ˆ
∞
∞
|f (x)|2 dx =
−∞
f (x) f (x)dx
ˆ ∞
1
=
f (x) √
fˆ(ω) e−ixω dω dx
2π −∞
−∞
ˆ ∞
ˆ ∞
1
=
fˆ(ω) √
f (x)e−ixω dx dω
2π −∞
−∞
ˆ ∞
ˆ ∞
=
fˆ(ω) fˆ(ω) dω =
|fˆ(ω)|2 dω.
−∞
ˆ ∞
−∞
−∞
A generalization of this theorem is considered in Exercise 9.8.7.
(9.14)
9.1. Fourier transform
239
A counterpart result for the Fourier series (9.4) f (x) =
ˆ
∞
X
π
|f (x)|2 dx = 2π
−π
P∞
ikx
k=−∞ ck e
is that
|ck |2 .
(9.15)
k=−∞
Both (9.14) and (9.15) offer opportunities to evaluate various infinite integrals and sums,
respectively.
Example 9.12. Show that
´∞
−∞
sin x 2
x
dx = π.
If we happen to know the transform pair
f (x) =
1
0
if |x| < 1
if |x| > 1
r
⇔
fˆ(ω) =
2 sin ω
,
π ω
(9.16)
the result follows immediately
from (9.14); equation (9.16) is obtained through fˆ(ω) =
q
´
1
√1
1 · e−iωx dx = π2 sinω ω .
2π −1
If we do not know (9.16), we can still note that the integrand is the square of f (x) =
´
sin x
of fˆ(ω) (on the chance that fˆ(ω)2 dω might be simpler to
x , suggesting
´ a calculation
integrate than f (x)2 dx). Hence, we use (9.6) to obtain
ˆ ∞
sin x −iωx
1
fˆ(ω) = √
e
dx
2π −∞ x
1
1 ∞ ei (1−ω)x
1
=√
dx −
x
2i
2π 2i −∞
p
π/2 if |ω| < 1
=
,
0
if |ω| > 1
∞
−∞
ei (−1−ω)x
dx
x
where we for the last step used (5.7). Parseval’s relation (9.14) now gives
´ 1 p π 2
dω = π.
2
−1
´∞
−∞
sin x 2
x
dx =
Example 9.13. Find the infinite sum obtained by choosing f (x) = x in (9.15).
k
P∞
From (9.3) it follows (for −π < x < π) that x = f (x) = 2 k=1 (−1)
sin kx.
k
P
(−1)k+1 ikx
eikx −e−ikx
With sin kx =
, this becomes x = k6=0 ik e , and thus c0 = 0 and
2i
´π
P
ck = (−1)k+1 /(ik) for k 6= 0. Equation (9.15) now gives −π x2 dx = 2π k6=0 k12 ,
P∞
2
which simplifies to k=1 k12 = π6 .70
9.1.5 Convolution theorems
Most transforms have convolution theorems of one form or another. In the case of Fourier
expansions, they become as follows.
70 The
Basel problem we encountered earlier in Sections 3.2.3 and 5.2.
240
Chapter 9. Transforms
Fourier series case
Let A(x) and B(x) be 2π-periodic functions:
A(x) =
∞
X
αk eikx ,
αk =
k=−∞
B(x) =
∞
X
βk eikx ,
k=−∞
βk =
1
2π
1
2π
ˆ
π
A(x)e−ikx dx,
(9.17)
B(x)e−ikx dx.
(9.18)
−π
ˆ
π
−π
The convolutions of the two functions and of the two coefficient sequences, respectively,
are defined as
ˆ π
1
C(x) =
A(t)B(x − t) dt (physical space, again 2π-periodic),
(9.19)
2π −π
γk =
∞
X
αν βk−ν
(Fourier coefficient space, k ∈ Z).
(9.20)
ν=−∞
What amounts to a convolution in one of the spaces becomes a product in the other, as
shown in the following theorem.
Theorem 9.14. With the notation just above, it will hold that
ˆ π
∞
X
1
ikx
C(x) =
{αk βk } e ,
αk βk =
C(x) e−ikx dx,
2π −π
(9.21)
k=−∞
A(x)B(x) =
∞
X
γk e
ikx
,
k=−∞
1
γk =
2π
ˆ
π
A(x)B(x) e−ikx dx.
(9.22)
−π
The proofs for the two versions are similar. The first one is proven in Section 9.7.
Fourier transform case
This case is more symmetric in that both directions of the transform, (9.7) and (9.6), respectively, are integrals over (−∞, ∞). For two functions f (x) and g(x), with Fourier
transforms fˆ(ω) and ĝ(ω), respectively, we define the convolutions in the two spaces as
ˆ ∞
1
f (t)g(x − t) dt ,
(9.23)
h(x) = √
2π −∞
ˆ ∞
1
k̂(ω) = √
fˆ(ξ)ĝ(ω − ξ) dξ .
(9.24)
2π −∞
The counterpart results to those in Theorem 9.14 become now the following.
Theorem 9.15.
ˆ ∞
1
h(x) = √
fˆ(ω)ĝ(ω) eiωx dω,
2π −∞
ˆ ∞
1
f (x)g(x) = √
k̂(ω) eiωx dω,
2π −∞
ˆ ∞
1
fˆ(ω)ĝ(ω) = √
h(x) e−iωx dx,
2π −∞
(9.25)
ˆ ∞
1
−iωx
k̂(ω) = √
f (x)g(x) e
dx.
2π −∞
(9.26)
9.1. Fourier transform
241
9.1.6 Poisson’s summation formula
Theorem 9.16. Let f (x) have the Fourier transform fˆ(ω); then
∞
X
1
fˆ(k) eikx
f (x + 2πk) = √
2π
k=−∞
k=−∞
(9.27)
∞
X
1
ˆ
f (k) e−ikω .
f (ω + 2πk) = √
2π
k=−∞
k=−∞
(9.28)
∞
X
and, equivalently,
∞
X
The LHSs of each of these two equations are 2π-periodic in their arguments x and ω,
respectively. The RHSs give, in very simple explicit form, their Fourier series expansions.
The proofs for the two parts are essentially identical. The first part is shown in Section 9.7.
9.1.7 The generalized Fourier transform
One apparent limitation of Fourier transforms is that the defining integrals (9.6) and (9.7)
need to be convergent. An easy way around this restriction is illustrated in the following
example.
Example 9.17. Find the generalized Fourier transform of the function f (x) = |x|3 .
´∞
Although √12π −∞ |x|3 e−iωx dx diverges, it will converge if we insert a factor inside
the integral that decays to zero sufficiently fast as |x| → ∞. Choosing, for example, this
factor as e−α|x| , we get convergence for all α > 0 and can then take the limit as α & 0.
Thus,
ˆ ∞
1
fˆ(ω) = lim √
|x|3 e−α|x| e−iωx dx
α&0
2π −∞
12 1
12(ω 4 − 6ω 2 α2 + α4 )
√
= lim
=√
.
(9.29)
2
2
4
α&0
2π(ω + α )
2π ω 4
There is much arbitrariness in the choice of this extra factor. One naturally tries to find one
such that the resulting integral can be evaluated in closed form.
One example of the utility of the generalized Fourier transform is shown in Exercise
9.8.10. The example below illustrates both the utility of this generalized Fourier transform
and the results in the preceding sections on convolutions and Poisson’s summation formula.
Example 9.18. Given the function φ(x) = |x|3 , find coefficients λk , k ∈ Z, such that it
holds for all n ∈ Z that
∞
X
1, n = 0,
λk φ(n − k) =
(9.30)
0, n 6= 0.
k=−∞
242
Chapter 9. Transforms
Data of this type, taking the value one at one point and zero at all other points, is
known as cardinal data and arises frequently in approximation theory.71 Given how fast
φ(x) grows, convergence of this sum will require that the λk go quite rapidly to zero for
|k| increasing. The various functions that arise in the following solution are illustrated in
Figure 9.4.
We recognize the LHS of (9.30) as being of the same form as (9.20) and define therefore
the two 2π-periodic functions
Λ(x) =
Φ(x) =
∞
X
k=−∞
∞
X
λk eikx ,
(9.31)
φ(k)eikx .
(9.32)
k=−∞
The first equation in (9.22) tells us that Λ(x) · Φ(x) = 1 · e0x = 1, and therefore
Λ(x) = 1/Φ(x).
(9.33)
A solution plan emerges now: With φ(k) given, use (9.32) to find Φ(x), use then (9.33) to
get Λ(x). Equation (9.31) tells us then that Fourier expanding Λ(x) will give the coefficients we want.
The first step poses immediately a problem: The sum (9.32) diverges.
√ However,
P∞ its form
matches perfectly the RHS of (9.28), with the LHS then becoming 2π k=−∞ φ̂(ω +
1
2πk). By the previous Example 9.17, φ̂(ω) = √12
, and this last sum therefore con2π ω 4
verges just fine. It can be evaluated by the contour integration method in Section 5.2, giving
4 sin4 (ω/2)
2+cos ω
Φ(ω) = 4 sin
4 (ω/2) , from which it follows that Λ(ω) =
2+cos ω . By (9.31) we get the λk
´ π 4 sin4 (ω/2) −ikω
1
as the Fourier coefficients of Λ(ω), i.e., λk = 2π −π 2+cos ω e
dω. This integral,
for k ∈ Z, is of a standard form for contour integration. Carrying out the details of this
gives the result
√

−4 + 3√ 2
, k = 0,


 19 − 6 3
, |k| = 1,
2
√
λk =
k
(−1)
3
3



√
, |k| ≥ 2.
(2 + 3)|k|
These coefficients decay exponentially fast as |k| increases, ensuring that (9.30) indeed
converges for all values of n.
9.1.8 Half-plane splitting by means of the Fourier transform
The Fourier transform offers an alternative to the principal value approach in Section 5.1.5
for splitting an analytic function f (z) into f (z) = f − (z) + f + (z), with the two functions
f − (z) and f + (z) singularity-free and going to zero in the upper and lower half-planes, respectively (i.e., having all growth and singularities (if any) in the half-plane
corresponding
´∞
to the sign in the superscript). The key idea is to form fˆ(ω) = √12π −∞ f (x) e−iωx dx as
usual but, when inverting back, split the integration interval [−∞, ∞] for ω in two parts
71 Cases
with more general φ(x)-functions are discussed, for example, in Section 3.3.9 of [18].
9.1. Fourier transform
243
1
100
0.8
80
0.6
60
0.4
40
0.2
20
0
-1
-0.5
0
0.5
0
-10
1
-5
(a) φ(x) = |x|3 .
0
(b) φ̂(ω) =
100
5
80
4
60
3
40
2
20
1
0
5
10
√12 14 .
2π ω
0
-3
-2
-1
0
(c) Φ(ω) =
1
2
3
-3
-2
2+cos ω
.
4 sin2 (ω/2)
-1
(d) Λ(ω) =
0
1
2
3
2
4 sin (ω/2)
2+cos ω .
1.5
1
0.5
0
-0.5
-1
-20
-15
-10
-5
0
5
10
15
20
(e) Resulting coefficients λk .
Figure 9.4. Illustration of the sequence of functions arising in Example 9.18. The functions
in parts (a), (b), (e) are shown truncated sideways, whereas the ones in parts (c) and (d) are [−π, π]periodic.
(denoting the independent variable by z rather than by x):
ˆ 0
ˆ +∞
1
1
f (z) = √
fˆ(ω) eiωz dω + √
fˆ(ω) eiωz dω = f + (z) + f − (z). (9.34)
2π −∞
2π 0
Both formulations (see (9.34) and (5.9)–(5.10)) are equivalent; see Exercise 10.4.1.
Some additional notation is helpful in expressing relations, such as the one in (9.34), as
244
Chapter 9. Transforms
clearly and concisely as possible. We thus supplement the superscript notation − and + by
also introducing subscripts − and + , to denote half-line splits along the real axis.
Definition: If f (x) is defined for −∞ < x < ∞, we can write it as
f (x) = f− (x) + f+ (x),
(9.35)
where
f− (x) =
f (x) ,
0
,
x < 0,
x ≥ 0,
and f+ (x) =
0
, x < 0,
f (x) , x ≥ 0.
Both for superscripts and subscripts, we adhere to the convention that the sign indicates the direction of nontrivial functional behavior. Depending on the context (left/right),
real axis splits and (lower/upper) half-plane splits can be applied in “physical” space or in
Fourier space. Since there is a difference in sign in the exponent when transforming from
physical to Fourier space, we obtain the following.
Theorem 9.19.



F {f− (x)}(ω) = fˆ− (ω)
ˆ+

 F {f+ (x)}(ω) = f (ω)
{z
}
|
signs are the same
and
similarly



F {f − (x)}(ω) = fˆ+ (ω)
.
+
ˆ

 F {f (x)}(ω) = f− (ω)
|
{z
}
signs are opposite
(9.36)
In words, the Fourier transforms of the real axis-split functions f− (x) and f+ (x), when
evaluated for complex ω, provide the half-plane split versions of f (x)’s Fourier transform
fˆ(ω). Similarly, the Fourier transforms of the half-plane split functions f − (x) and f + (x)
become the real-axis splits of f (x)’s Fourier transform fˆ(ω).
Proof. Each of the pairs of equations in (9.36) add up to F {f (x)}(ω) = fˆ(ω), so what
remains to show (for the four equations, in turn) are
F −1 {fb− (ω)}(x) = 0
F −1 {fˆ+ (ω)}(x) = 0
F {f − (x)}(ω) = 0
F {f + (x)}(ω) = 0
for x > 0,
for x < 0,
for ω < 0,
for ω > 0.
Each LHS above is an integration along the real ω- or x-axis. Closing the contours far out
in the half-plane in which the integrand goes to zero (and is singularity-free) gives these
results (by Theorem 5.6, Jordan’s lemma).
The relations in Theorem 9.19 will play a key role in the Wiener–Hopf discussion in
Section 10.1.
The following two examples illustrate this procedure for obtaining half-plane splits in
the same two cases as before in Examples 5.17 and 5.18, respectively.
Example 9.20. Calculate f − (z) in case of f (z) =
1
1+z 2 .
9.2. Laplace transform
245
Since we want to arrive at f − (z), the relevant equation in (9.36) is the top right one:
p
F {f − (x)}(ω) = fˆ+ (ω). Following Example 9.5, we obtain fˆ(ω) = π2 e−|ω| , and therefore
ˆ +∞ r
π −ω iωz
i
1
e e dω =
,
f − (z) = √
2
2(i + z)
2π 0
as before.
2
Example 9.21. Calculate f − (z) in case of f (z) = e−z (Gaussian).
2
As in the previous example, the first step is to obtain fˆ(ω) = √12 e−ω /4 (from (9.9)),
and therefore
ˆ ∞
2
1
−
e−ω /4 eiωz dω .
f (z) = √
2 π 0
Following the idea in Example 9.4, we complete the squares in the exponents,
ˆ ∞
2
2
ω
1
f − (z) = √ e−z
e−( 2 −iz) dω ,
2 π
0
and then make the variable change ω2 − iz = t, i.e., 12 dω = dt, giving
ˆ ∞
ˆ ∞
ˆ −iz
2
2
2
2
2
1
1
f − (z) = √ e−z
e−t dt = √ e−z
e−t dt −
e−t dt
π
π
−iz
0
0
1 −z2
(1 + erf(iz)) .
= e
2
The successive steps in this derivation are illustrated in Figure 9.5.
9.2 Laplace transform
The Laplace transform is widely used in many fields. In the context of electrical engineering, it offers a particularly convenient approach for solving many linear ODE initial value
problems (often in time, over the semi-infinite range t ∈ [0, ∞)).
ˆ
Definition 9.22.
fˆ(s) =
∞
f (x) e−sx dx.
(9.37)
0
We will again use either the “hat” notation, or the more explicit version L{f (x)}(s) =
fˆ(s), to refer to the transform and mostly use s as the transform variable (to further help
distinguish the Laplace transform from the Fourier transform). While evaluating (9.37)
often is quite straightforward,72 it is much less obvious how to get back from fˆ(s) to f (x).
Without using complex variables, there are two main approaches:
1. Recognize the transform in a tabulated collection of known cases. This approach
can be combined with techniques to modify certain types of transforms fˆ(s) into
combinations of other functions that are more likely to be tabulated (such as splitting
rational functions into partial fraction form, etc.).
72 We assume the integral converges for some s-values, assured, for example, if f (x) is of exponential order,
i.e., there exist constants A and B such that |f (x)| ≤ A eBx for x ≥ 0. In this case, (9.37) converges for
Re s > B.
246
Chapter 9. Transforms
1
0.8
0.6
0.5
0.4
0.2
0
-3
-2
-1
0
1
2
0
-3
3
-2
-1
0
x
2
(b) fˆ(ω) =
(a) f (x) = e−x
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
-3
-2
-1
(c) fˆ− (ω) =
0
1
2
0
-3
3
fˆ(ω) , ω < 0
0
, ω≥0
-2
√1
2π
-1
(d) fˆ+ (ω) =
(e) f (z) = e−z
4
1
´∞
−∞
3
f (x)e−iωx dx
0
2
1
2
3
0
, ω≤0
ˆ
f (ω) , ω > 0
2
4
2
2
0
2
0
0
-2
0
-2
-2
0
x
(f) f − (z) =
2
√1
2π
2
´∞
−∞
-2
0
y
x
fˆ+ (ω)eiωx dω
2
(g) f + (z) =
√1
2π
2
´∞
−∞
y
fˆ− (ω)eiωx dω
Figure 9.5. f (x) = e−x and the steps in calculating f − (z) and f + (z).
9.2. Laplace transform
247
2. Use the (highly impractical) explicit formula
(−1)k
f (x) = lim
k→∞
k!
k+1
dk ˆ k
k
f
.
dxk
x
x
With use of complex variables, there is a vastly more practical general formula available
(obtained by recognizing the Laplace transform as a “disguised” case of the Fourier transform).73
9.2.1 Inversion formula
Theorem 9.23. The inversion formula for the Laplace transform is
f (x) =
ˆ
1
2πi
c+i∞
fˆ(s)esx ds,
(9.38)
c−i∞
where we will choose c so that all of the singularities of fˆ(s) have Re s < c.
Proof. We first extend f (x) to x < 0 by setting f (x) = 0 for x < 0. Then
ˆ
∞
fˆ(s) =
f (x)e−sx dx.
(9.39)
−∞
Let’s next make a change of variables to get what will look like a Fourier transform, which
we know how to invert. Thus, let s = c + ik, where c and k are real (it will soon become
apparent why the constant c is required). This gives
ˆ
∞
fˆ(s = c + ik) =
−∞
|
f (x)e−cx e−ikx dx
{z
}
Formula for the Fourier transform of
and therefore
f (x)e−cx =
1
2π
ˆ
,
√
2πf (x)e−cx
∞
fˆ(c + ik)eikx dk .
−∞
Next, we change the variables back to s = c + ik , ds = idk, and we multiply everything
by ecx .
Once again, we are in a case where the required integrals often are very well suited
for contour integration approaches. In this application, extending the contour to become closed (allowing residue calculus to be applied) is known as forming a Bromwich
contour.
inverting, one might want to verify that lims→∞ fˆ(s) = 0, since otherwise, f (x) needs to be a
“generalized function” (such as a delta function).
73 Before
248
Chapter 9. Transforms
Im s
Γ
i
R
c
Re s
−i
Figure 9.6. Contour in Example 9.24.
Example 9.24. Given
f (x) = sin x ,
determine fˆ(s) and then invert the result to arrive back at f (x).
We find
ˆ
fˆ(s) =
ˆ
∞
−sx
sin x e
∞
eix−sx dx = Im
dx = Im
0
0
eix−sx
i−s
∞
=
0
s2
1
.
+1
(9.40)
This can now be inverted by
f (x) =
1
2πi
ˆ
c+i∞
c−i∞
esx
1
ds =
2
1+s
2πi
ˆ
Γ
esx
ds,
1 + s2
where the contour Γ is a vertical line to the right of the singularities. If x ≥ 0, we can close
the contour to the left, as displayed in Figure 9.6. By Jordan’s lemma, the contribution
of the integral around the semicircle vanishes as its radius R → ∞, giving (from the two
poles)
eix − e−ix
1
2πi
= sin x.
f (x) =
2πi
2i
If, on the other hand, x < 0, the contour needs to be closed to the right, giving f (x) = 0.
In other words, a Laplace transform represents a function that can be nontrivial for x ≥ 0,
but is assumed to vanish identically for x < 0 (of course, nothing stops us from applying
the Laplace transform to sin(−x), obtaining a transform representation of the sine function
for negative arguments instead of for positive ones).
9.2. Laplace transform
249
Im s
Γ
i
R
c
ε
Re s
−i
Figure 9.7. Contour in Example 9.25.
Example 9.25. Find f (x) which has as its Laplace transform
log s
fˆ(s) =
.
1 + s2
Since the factor log s requires a branch cut from s = 0 to infinity, we cannot this
time follow the contour in Figure 9.6. However, we can place the cut along the negative
real axis and then use the keyhole contour illustrated in Figure 9.7. Listed below are the
contributions from the contour and from the residues:
1. Res(f, s = i) =
π ix
4e .
2. Res(f, s = −i) = π4 e−ix .
´
´
´
´∞
∞ log(r)+iπ −xr
1
3. → f (z)dz + ← f (z)dz = 2πi
dr − 0
1+r 2 e
0
´ ∞ e−xr
dr.
0 1+r 2
´
4. | f (z)dz| ≤ M · L ≈ 2π log → 0 as → 0 .
´
5.
f (z)dz → 0 as R → ∞ by Jordan’s lemma.
R
log(r)−iπ −xr
dr
1+r 2 e
=
Therefore, adding all the contributions, we obtain
ˆ ∞ −xr
π
e
f (x) = cos x −
dr,
2
1 + r2
0
where the integral cannot be expressed in terms of elementary functions in closed form.
One often useful feature of the Laplace transform (for example in the context of solving
ODEs, as in Section 9.2.4) is that also discontinuous functions f (x) transform into analytic
functions fˆ(s).
250
Chapter 9. Transforms
(a) f (x).
(b) fˆ(s).
Figure 9.8. The function f (x) in Example 9.26, and its Laplace transform fˆ(s).
Example 9.26. Find the Laplace transform fˆ(s) of the step function

 1 , 0≤x<1
−1 , 1 < x < 2 ,
f (x) =

0 , x>2
and then invert fˆ(s) to recover f (x).74
´∞
´1
´2
−s 2
We obtain directly fˆ(s) = 0 f (x) e−sx dx = 0 1 e−sx dx− 1 1 e−sx dx = (1−es ) .
The singularity at s = 0 is removable, so fˆ(s) is an entire function. Figure 9.8 shows the
two functions f (x) and fˆ(s).
The integrand in the Bromwich integral for inverting fˆ(s) becomes fˆ(s) esx . While the
first factor decays in the right half-plane, the second factor grows there, making it unclear
how to close an integration contour. Hence, we split fˆ(s) esx as
1
2
1
fˆ(s) esx = esx − es(x−1) + es(x−2) .
s
s
s
(9.41)
In contrast to fˆ(s) esx , which is an entire function, each of these terms has a pole at the
origin as its only singularity, with residues 1, −2, 1, respectively. The columns in Figure
9.9 display these three terms in the cases of x = 12 , 32 , 52 , respectively (representative for
the three subintervals 0 < x < 1, 1 < x < 2, and x > 2). Following a vertical path in the
right half-plane, we now close the contour in whichever half-plane the magnitude of the
integrand goes to zero. Just keeping track of whether this implies encircling a pole or not
recovers the step function f (x).
74 Values
assigned to f (x) at single points (such as here at the points of discontinuity x = 1 and x = 2) do not
influence fˆ(s). When inverting, (9.38) returns at such points the average from the two sides.
9.2. Laplace transform
2
− es(x−1)
s
1 s(x−2)
e
s
x = 5/2
x = 3/2
x = 1/2
1 sx
e
s
251
Figure 9.9. The three functions of s which, when added, form fˆ(s) esx in Example 9.26
(cf. (9.41)), shown for three representative values of x.
252
Chapter 9. Transforms
9.2.2 The inverse Laplace transform of rational functions
It follows from the definition of the gamma function Γ(z) =
´∞
0
tz−1 e−t dt that
L{xk }(s) = Γ(k + 1)/sk+1 .
(9.42)
Limiting k to the nonnegative integers and using that k! = Γ(k + 1) gives
1
2!
1
, L{x}(s) = 2 , L x2 (s) = 3 , . . . ,
s
s
s
k!
L xk (s) = k+1 , k = 0, 1, 2, . . . .
s
L{1}(s) =
(9.43)
From (9.37) it follows that
L {f (x) eax } = L{f (x)}(s − a)
(9.44)
for an arbitrary constant a. Together, these last two equations imply that
L xk eax =
k!
, k = 0, 1, 2, . . . .
(s − a)k+1
(9.45)
If f (x) is bounded at the origin and has the rational function fˆ(s) = p(s)
q(s) as its Laplace
transform, the degree of p(s) will be less than that of q(s). The partial fraction expansion
of fˆ(s) will only contain terms that are multiples of expressions of the same form as the
RHS in (9.45). This relation therefore provides a way to find f (x).
Example 9.27. Find the function f (x) with Laplace transform fˆ(s) =
s2 −5
s4 −1
.
Method 1: Given that the zeros of the denominator are s = {−1, +1, −i, +i} (all distinct),
1
we find by standard methods of calculus the partial fraction expansion to be fˆ(s) = s+1
−
3i 1
3i 1
1
s−1 + 2 s+i − 2 s−i . Each term is a k = 0 special case of (9.45), giving after a quick
simplification f (x) = e−x − ex + 3 sin x.
Method 2: Viewed as an analytic function of s, fˆ(s) has as its only singularities first order
poles at s = {−1, +1, −i, +i}. By the N/D0 rule (see “Shortcuts to calculate residues” in
−ix
Section 5.1.1), the residues of the integrand fˆ(s)esx in (9.38) become {e−x , −ex , 3i
,
2e
3i ix
− 2 e }. With a vertical path to the right of the singularities and closing it to the left, (9.38)
gives f (x) as the sum of the residues; i.e., again f (x) = e−x − ex + 3 sin x.
Had there been poles of second or higher order present in fˆ(s), Method 1 would generalize immediately. For Method 2, the N/D0 rule would no longer apply, so some more
effort would be needed to calculate the residue at the poles of higher multiplicity. The
function f (x) will then be the sum of the residues.
We can from this example spot an easy way to obtain the coefficients in a partial fraction
expansion of a rational function with first order poles only; each is given by applying the
N/D0 rule to the rational function.
The inverse Laplace transform of rational functions can be obtained also without using
complex variables. If a real-valued partial fraction expansion only contains terms of the
9.2. Laplace transform
253
forms
n!
= L {xn eax } ,
(s − a)n+1
s−a
= L {eax cos bx} ,
(s − a)2 + b2
b
= L {eax sin bx} ,
(s − a)2 + b2
(9.46)
(9.47)
(9.48)
the inversion is immediate. However, if the quadratic denominator in (9.47) or (9.48)
appears raised to higher powers, the corresponding formulas become more complicated.
When using complex variables, no similar complication arises, since partial fraction expansions will then never contain terms in forms different from what is given by (9.46).75
9.2.3 Some key features of the Laplace transform
dn
For the Fourier transform, Theorem 9.6 gave a formula for F dx
(ω) in terms of
n f (x)
F {f (x)} (ω). The corresponding formula for the Laplace transform becomes somewhat
more involved, since the interval is finite, x ∈ [0, ∞], and we can no longer assume functions to vanish at both ends.
Theorem 9.28.
n
d
n
n−1
n−2 0
(n−1)
f
(x)
(s)
=
s
L{f
(x)}(s)
−
s
f
(0)
+
s
f
(0)
+
·
·
·
+
f
(0)
.
L
dxn
(9.49)
In particular,
d
L
f (x) (s) = s L{f (x)}(s) − f (0) ,
(9.50)
dx
2
d
L
f
(x)
(s) = s2 L{f (x)}(s) − (s f (0) + f 0 (0)) .
(9.51)
dx2
Proof. Repeated integration by parts gives
n
ˆ ∞
d
L
f (x) (s) =
f (n) (x)e−sx dx
dxn
0
ˆ ∞
h
i∞
= f (n−1) (x)e−sx
+s
f (n−1) (x)e−sx dx
0
0
h
i∞
(n−1)
(n−2)
= −f
(0) + s
f
(x)e−sx
0
ˆ ∞
+s
f (n−2) (x)e−sx dx
0
=
···
= −f (n−1) (0) − s f (n−2) (0) − · · · − sn−1 f (0)
ˆ ∞
n
+s
f (x) e−sx dx.
0
75 A
simple case illustrating this arises in the solution to part (b) of Example 9.31.
254
Chapter 9. Transforms
Theorem 9.29.
L {xn f (x)} (s) = (−1)n
dn
L {f (x)} (s), n = 0, 1, 2, . . . .
dsn
This result follows from (repeatedly) differentiating the definition of the Laplace transform (9.37) with respect to s.
Like for the other transforms we consider, the Laplace transform possesses a convolution theorem.
Theorem 9.30. Defining the convolution of f (x) and g(x) as
ˆ x
h(x) =
f (ξ)g(x − ξ)dξ ,
0
it will hold that
ĥ(s) = fˆ(s) · ĝ(s).
(9.52)
The proof is given in Section 9.7.
9.2.4 Application of the Laplace transform to solving constant
coefficient ODEs
The following example illustrates the main ideas that are involved.
Example 9.31. Solve ODE initial value problem y 00 (x)+y(x) = g(x), y(0) = 1, y 0 (0) =
2 for the following three choices of RHS function g(x):
(a) g(x) = 0,
(b) g(x) = cos x,
(c) g(x) = tan x.
(a) From (9.49) it follows that s2 ŷ(s) − (s + 2) + ŷ(s) = 0, i.e., ŷ(s) = ss+2
2 +1 =
1
1
1
1
2 + i s+i + 2 − i s−i . Inverting, using the k = 0 case of (9.45), gives y(x) =
1
−ix
+ 12 − i e+ix = cos
2 +i e
x + 2 sin x. (b) Since L {cos x} (s) = L 12 eix + e−ix (s) = s2s+1 , the Laplace transform of
s
the ODE becomes s2 ŷ(s) − (s + 2) + ŷ(s) = s2s+1 , and ŷ(s) = ss+2
2 +1 + (s2 +1)2 . We
recognize the first term from part (a); its inverse Laplace transform is cos x + 2 sin x.
The second term, originating from the ODE’s RHS function, can be expanded in partial
i
1
i
1
i
s
−ix
−
fractions as (s2 +1)
2 = 4 (s+i)2 − 4 (s−i)2 , with inverse Laplace transform 4 x e
i
1
1
ix
76
x
e
=
x
sin
x.
The
ODE
solution
thus
becomes
y(x)
=
cos
x
+
2
sin
x
+
x
sin
x.
4
2
2
(c) The Laplace transform of the ODE becomes in this case s2 ŷ(s) − s − 2 + ŷ(s) =
1
L{tan x}(s), and ŷ(s) = ss+2
2 +1 + s2 +1 L{tan x}(s). We recognize again the first term
from part (a). Regarding the second term s21+1 L{tan x}(s), there is a complication in that
there does not exist any practical explicit expression for L{tan x}(s). However, noting
that s21+1 = L{sin x}(s), the task becomes finding the inverse Laplace transform of the
76 This case features resonance, with the solution amplitude growing unbounded. This happens when a forcing
function g(x) contains a component with a frequency that equals a natural frequency of the unforced (g(x) = 0)
and undamped equation.
9.3. Mellin transform
255
6
4
2
x
0
/2
3 /2
2
5 /2
3
7 /2
-2
-4
Part (a): Solution to homogeneous ODE
Part (b): Change in solution due to inhomogeneous term
Part (c): Change in solution due to inhomogeneous term
-6
Figure 9.10. The solution to the homogeneous ODE in Example 9.31(a), and the additional
terms that have to be added to this solution for the inhomogeneous cases in parts (b) and (c). Note
that the two latter curves do not affect the ODE’s initial conditions on y(0) and y 0 (0).
L{sin x}(s)·L{tan x}(s), By the convolution theorem (Theorem 9.30), this equals
´product
x
sin(x
− ξ) tan ξdξ. Although also not easy to evaluate, this integral can be found in
0
cos x
closed form: sin x + (cos x) log 1+sin
x .
Figure 9.10 shows the solution to the homogeneous ODE (part (a)), and the terms that
have to be added to this when including the inhomogeneous77 RHSs in parts (b) and (c).
9.3 Mellin transform
This transform resembles the Laplace transform, but with xs−1 replacing the factor e−sx
in the definition, i.e.,
ˆ ∞
ˆ
M {f (x}(s) = f (s) =
f (x) xs−1 dx.
(9.53)
0
The range of s-values for which fˆ(s) is defined (if not resorting to analytic continuation)
will depend on the properties of f (x) at x = 0 and x = ∞, and will take the form
of a vertical band in the complex s-plane. This fundamental band can be empty (e.g.,
for constants and all monomials xα ) or it can cover the whole s-plane (as for f (x) =
1
e−(x+ x ) ). As illustrated in Example 5.14, contour integration is often well suited for
evaluating integrals of the form (9.53).
9.3.1 Inversion formula and some transform properties
The inversion formula is also similar to the Laplace transform case,
ˆ γ+i∞
1
f (x) =
x−s fˆ(s)ds,
2πi γ−i∞
(9.54)
where the vertical path has to be chosen inside this band (for a proof, see Theorem 9.44 in
Section 9.7). Like for the other key transforms, large numbers of transform pairs are known
[34]. A few are listed in Table 9.1, and some general properties in Table 9.2.
77 “Inhomogeneous”
and “nonhomogeneous” are synonyms.
256
Chapter 9. Transforms
Table 9.1. Some examples of Mellin transforms. Note that fˆ(s) is the same in the second
and third cases, but that the fundamental band differs.
f (x)
0, x ≤ 1
xa , x > 1
fˆ(s)
1
−
s+a
Fundamental band
Re s < a
π
sin πs
π
sin πs
π
s sin πs
−1 < Re s < 0
e−x
Γ(s)
Re s > 0
1
ex − 1
ζ(s)Γ(s)
Re s > 1
1
1+x
x
−
1+x
log(1 + x)
sin x
sin
πs
2
0 < Re s < 1
−1 < Re s < 0
Γ(s)
−1 < Re s < 1
Table 9.2. Some analytic properties of the Mellin transform.
dk
M (log x)k f (x) (s) = k M {f (x)} (s)
ds
k
d
Γ(s)
M
f (x) (s) = (−1)k
M {f (x)} (s − k)
dxk
Γ(s − k)
´ x
1
M 0 f (ξ)dξ (s) = − M {f (x)} (s + 1)
s
M {f (ax)} (s) = a−s M {f (x)} (s)
Example 9.32. Use (9.54) to invert fˆ(s) = Γ(s) g(−s), where g(s) is analytic, and recover Ramanujan’s master theorem, as given in (3.12).
´ γ+i∞ −s
1
Equation (9.54) gives f (x) = 2πi
x Γ(s) g(−s) ds. Since Γ(s) has poles at
γ−i∞
each nonpositive integer, we choose γ = 0+ . Forming a Bromwich contour with a large
left semicircle CR of radius R leads to
1
f (x) + lim
R→∞ 2πi
ˆ
x
CR
−s
Γ(s) g(−s) ds =
∞
X
Res x−s Γ(s) g(−s), s = −k . (9.55)
k=0
The integral vanishes when R → ∞ due to the fast decay of the gamma function in the left
half-plane (cf. Figures 6.1(b)–(d)). Stirling’s formula (Example 12.8) provides the leading
asymptotic behavior of Γ(z) for large values of |z|. Equation (12.24) yields
∼ (R log R) cos θ − R θ sin θ − R cos θ
log Γ R eiθ = Re log Γ R eiθ
1
cos θ
+ ···
+ (log 2π − log R) +
2
12 R
9.3. Mellin transform
257
if one avoids θ = ±π (the negative real axis). The contribution of the contour integral
along the semicircle can be bounded from above as follows:78
´
CR
x−s Γ(s) g(−s) ds ≤ πR xR g −R eiθ
θ|
R−R|cos
√
eR
R
√
2π → 0 as R → ∞,
due to the rapid decay of the factor R−R|cos θ| .
With the integral absent in (9.55), we next use (6.1) to recover Ramanujan’s formula
ˆ ∞
x2
(9.56)
xs−1 g(0) − x g(1) +
g(2) − + · · · dx = Γ(s) g(−s).
2!
0
9.3.2 Application of the Mellin transform to harmonic sums
Applications of Fourier and Laplace transforms often utilize that (i) differentiation becomes
an algebraic operation, and (ii) the transforms possess convolution theorems. The Mellin
transform has the additional feature that a rescaling of the independent variable reduces to
a multiplication of the transform (noted as the last entry in Table 9.2).
Theorem 9.33. If f (x) has the Mellin transform fˆ(s), then f (ax) has the transform
a−s fˆ(s).
´∞
Proof. By (9.53), the transform of f (ax) is 0 f (ax) xs−1 dx. After the variable change
´∞
ax = t, this becomes 0 f (t) (a1−s ts−1 ) (a−1 dt) = a−s fˆ(s).
This scaling property provides an approach to solve functional equations of the type
f (x) = g(x) + f (ax),
with g(x) and the constant a given, since, on the transform side, this relation becomes
fˆ(s) = ĝ(s)/(1 − a−s ). Since the transform is linear, this result generalizes greatly, to
harmonic sums:
X
F (x) =
λk f (µk x).
(9.57)
k
Here f (x) is a “base function” and µk and λk represent frequencies and amplitudes, respectively. The Mellin transform of this sum factorizes as
!
X
−s
F̂ (s) =
λk (µk )
fˆ(s) .
(9.58)
k
Example 9.34. Find the Mellin transform of f (x) =
P∞
1
k=1 k
·
P∞
k=1
1
k
−
1
k+x
.
x/k
1+x/k ,
we recognize it as a harmonic sum (9.57)
with λk = µk =
Applying (9.58) and the third case in Table 9.1 gives fb(s) =
P∞ 1 1
ζ(1−s)
π
− sin πs k=1 k k−s = − π sin
πs for −1 < Re s < 0.
After rewriting the sum as
1
k.
78 Here assuming some mild bounds on any growth of g(−s) and also technical details about Γ(s) along the
negative real axis.
258
Chapter 9. Transforms
Table 9.3. Some integral equations that can be solved with the Mellin transform. In these
examples, y > 0.
Solution in terms of fˆ(s)
ĝ(s)
fˆ(s) =
K̂(s)
Integral equation
´∞
g(y) = 0 f (x) K(x y)dx
g(y) =
g(y) +
´∞
0
´∞
f (x) K(y/x)dx
ĝ(s − 1)
fˆ(s) =
K̂(s − 1)
f (x) K(x y)dx = f (y)
ĝ(s) + K̂(s) ĝ(1 − s)
fˆ(s) =
1 − K̂(s) K̂(1 − s)
0
9.3.3 The Mellin transform applied to some integral equations
The remarkably wide range of features of the Mellin transform has made it applicable in
several contexts, including analytic number theory79 (for example in proofs of the prime
number theorem), closed-form evaluation of certain integrals [16], and the solution of some
types of integral equations [34]. We limit ourselves here to quote a few examples of the
latter in Table 9.3.
9.3.4 Connection between the Mellin transform and properties of
f (x) at x = 0 and x = ∞
Although (9.53) only defines fˆ(s) within a fundamental strip α < Re s < β, it can typically be analytically continued far beyond this strip. For example, all the cases of fˆ(s)
given in Table 9.1 are meromorphic functions (i.e., their only singularities in the finite
complex s-plane are poles).80 These poles tell a lot about how f (x) behaves in the vicinities of x = 0 and x = ∞ (as can be deduced from closing the contour in (9.54) in the
λ
left and right half-planes, respectively). Each simple pole s+a
to the left of the fundamental strip (including on its edge) tells us that a local power series expansion at x = 0 will
λ
contain a term λxa . Poles of higher multiplicity, (s+a)
k+1 , indicate the presence also of
k
a
k
logarithmic terms as x & 0; (−1)
k! λx (log x) . Matching results for the right of the strip
λ
λ
are the following: A simple pole s+a implies a term − x1a as x → ∞, and (s−a)
k+1 a term
k
− (−1)
k!
λ
k
xa (log x) .
Example 9.35. Use the result just above to invert fˆ(s) =
π
sin πs .
As the second and third cases in Table 9.1 showed, the answer will also depend on what
the fundamental strip is. In the second case, the relevant poles (with Re s ≤ 0) are located
79 One
example is provided by the relation
0
x−s R(x)dx
=
Γ(2−s)
− ζ(2s−2) , where R(x)
=
n+1 n
(−1)
x
n=1 (n−1)!ζ(2n)
P∞
´∞
(following from (9.56)). R(x) is known as the Riesz function. Recalling the poles of
Γ(2−s)
Γ(z) and trivial zeros of ζ(z), ζ(2s−2) would be pole-free for 0 < Re s < 2 if it was not for the nontrivial zeros
of ζ(z). It can from this be deduced that the Riemann hypothesis is true if and only if the growth of R(x) for
x → +∞ satisfies |R(x)| ≤ C x1/4+ε , where C is a constant and ε > 0 is arbitrarily small.
80 If fˆ(s) is not meromorphic (e.g., fˆ(s) being an entire function in case of f (x) = e−(x+1/x) ), the results in
this section are not applicable.
9.4. Hilbert transform
259
at s = 0, −1, −2, . . . . and have residues 1, −1, 1, −1, . . . . The expansion of f (x) at x = 0
should thus be of the form f (x) = 1 − x + x2 − x3 + · · · , indeed matching the Taylor
1
expansion for f (x) = 1+x
. In the third case, the pole at s = 0 should be omitted, giving
x
. The fundamental strip can similarly be translated with
instead the result f (x) = − 1+x
any integer amount, giving different f (x).
The argument just above was not quite rigorous, since we used only partial information
about fˆ(s). In particular, f (x) could also have contained terms that for x & 0 become
smaller than any power of x (such as e−1/x ). Several options are available for ruling
out such additional functions, including (i) knowledge that f (x) is analytic at x = 0,
and (ii) simple estimates for |fˆ(s)| up/down the fundamental band. For precise results in
this context, see, for example, [36]. In certain contexts (such as for creating asymptotic
expansions; see Chapter 12), no additional information is needed.
Example 9.36. Find the function f (x) (analytic at x = 0) that has fˆ(s) =
its Mellin transform in the strip 0 < Re s < n.
k
The relevant poles (locations and residues) are given by (−1)
k
2, . . . . The function f (x) should thus, at x = 0, have an expansion
f (x) =
Γ(s)Γ(n−s)
Γ(n)
Γ(n+k) 1
Γ(n) s+k ,
as
k = 0, 1,
∞
X
(−1)k Γ(n + k) k
x .
k!
Γ(n)
k=0
−n
We can recognize the terms as k xk , i.e., the sum is the binomial expansion of f (x) =
(1 + x)−n .
1
In the special case of n = 1, this gives for the Mellin transform of f (x) = 1+x
the
π
result fˆ(s) = Γ(s)Γ(1 − s) = sin πs , in the band 0 < Re s < 1, matching Example
9.35.
9.4 Hilbert transform
The Hilbert transform has significant applications in signal processing (such as for digital
filters and for approximating instantaneous properties of oscillatory time series), and also
in physiology, spectroscopy, microscopy, geophysical imaging, etc. A key feature of it is
that it allows analytic function techniques to be applied to real-valued “signals” that might
not be differentiable (or even continuous). It also forms one (of several) entry points to the
topic of Riemann–Hilbert problems, briefly summarized in Chapter 10.
If a function f (z) = u(x, y) + iv(x, y) is analytic everywhere in the upper half-plane,
and |f (z)| → 0 there for |z| → ∞, then the values along the real axis for u(x, 0) and
v(x, 0) are related through the Hilbert transform. Knowing one of these two functions
u(x, 0) and v(x, 0) uniquely determines the other, without the need to make any calculations that extend outside the x-axis.
Example 9.37. Determine the Hilbert transforms of cos ωx and sin ωx, assuming ω
real.
If ω > 0, we consider f (z) = eiωz = cos ωz + i sin ωz, and if ω < 0, we consider
f (z) = e−iωz = cos ωz − i sin ωz (in both cases ensuring decay in the upper half-plane).
260
Chapter 9. Transforms
We thus obtain
if ω > 0 :
if ω < 0 :
H{cos ωx} = + sin ωx and H{sin ωx} = − cos ωx,
H{cos ωx} = − sin ωx and H{sin ωx} = + cos ωx.
(9.59)
In general it will hold (as we saw an example of in this special case) that H{H{u(x)}} =
−u(x), implying that there is no need to have any separate formula for the inverse of the
Hilbert transform.
In most cases, the Hilbert transform cannot be simply “read off” just by inspection
based on elementary functions. Theorems 9.38 and 9.39 provide general procedures to
obtain the transform.
Theorem 9.38. For f (z) = u(x, y) + iv(x, y) with the properties just described (f (z) =
u(x, y) + iv(x, y) analytic with |f (z)| → 0 for |z| → ∞ in the upper half-plane), it holds
that
1
π
1
v(x, 0) = +H{u(x, 0)} = −
π
u(x, 0) = −H{v(x, 0)} = +
∞
v(t, 0)
dt,
−∞ t − x
∞
u(t, 0)
dt.
−∞ t − x
(9.60)
(9.61)
Proof. If we close the integration with a large semicircle in the upper half-plane, the integral
along that part will vanish, and therefore
∞
−∞
f (t)
dt = πi f (x)
t−x
(using half the residue for the first order pole at t = x, since this lies on the integration
path). Writing f (t) = u(t, 0) + iv(t, 0), this becomes
u(x, 0) + iv(x, 0) =
1
πi
∞
−∞
u(t, 0)
i
dt +
t−x
πi
∞
−∞
v(t, 0)
dt,
t−x
and the result follows from separating the real and the imaginary parts.
A quite different approach for obtaining a Hilbert transform of a function, not requiring
any principal value integrals, is provided by its close relation to Fourier transforms.
Theorem 9.39. If a function u(x) has the Fourier transform û(ω), then H{u(x)} is obtained as the inverse Fourier transform of −i sign(ω) û(ω).
Proof. The relations (9.59) can be summarized in a single equation as
H{eiωx } = −i sign(ω) eiωx ,
(9.62)
where ω and x (both real) can be of either sign. With û(ω) denoting the Fourier transform
of u(x), it then follows that
ˆ ∞
ˆ ∞
1
1
iωx
H{u(x)} = H √
û(ω)e dω = √
û(ω)H(eiωx )dω
2π −∞
2π −∞
ˆ ∞
1
=√
−i sign(ω) û(ω)eiωx dω.
2π −∞
9.4. Hilbert transform
261
Theorem 9.40. The Hilbert transform commutes with the derivative operator, i.e.,
H
du(x)
dx
=
d
H{u(x)},
dx
which further implies
H
dk u(x)
dxk
=
dk
H{u(x)},
dxk
k = 1, 2, . . . .
(9.63)
This result follows from either of the two theorems just above. With these three theorems, the Hilbert transform can be calculated for a variety of functions. Some examples
beyond (9.59) include
a
H
a2 + x2
sin ax
H
x
n
o
2
H e−αx
H {Π(x)}
H {δ(x)}
x
,
a2 + x2
1 − cos ax
= sign(a)
,
x
ˆ x
√
2
2
2
= √ D( αx), where α > 0, D(x) = e−x
et dt,
π
0
x + 12
1 if |x| < 12 ,
1
, where Π(x) =
= log
π
0 if |x| > 12 ,
x − 12
ˆ ∞
1
, where δ(x) = 0 for x 6= 0,
δ(x) = 1.
=
πx
−∞
=
The last of these (together with (9.63)) opens up one approach towards interpreting concepts such as derivatives of delta functions. This relation (9.63) together with linearity
H {α1 u1 (x) + α2 u2 (x)} = α1 H {u1 (x)} + α2 H {u2 (x)} can supplement the results in
Theorems 9.38 and 9.39 for determining Hilbert transforms of a variety of functions.
In the following two examples, we determine the Hilbert transform directly from the
principle that f (z) = u(x, 0) + i H {u(x, 0)} will decay to zero in the upper half-plane.
Example 9.41. Find the Hilbert transform of the cubic B-spline function introduced in
Example 9.10.
A sometimes more convenient form for B(x) than (9.12) is
B(x) =
1
|x + 2|3 − 4 |x + 1|3 + 6 |x|3 − 4 |x − 1|3 + |x − 2|3
12
(9.64)
(since this also describes the unique piecewise cubic function, with continuous second
derivative and the same nodes, vanishing for x ≤ −2 and for x ≥ +2, and normalized the
same). This expression can alternatively be written as
B(x) =
1
sign(x + 2)(x + 2)3 − 4 sign(x + 1)(x + 1)3 + 6 sign(x)(x)3
12
− 4 sign(x − 1)(x − 1)3 + sign(x − 2)(x − 2)3 .
262
Chapter 9. Transforms
(a) Re f (z), equals B(x) for z real.
(b) Im f (z), equals H{B(x)} for z real.
Figure 9.11. Real and imaginary parts of the function f (z) given in (9.65), showing along
the real axis a cubic B-spline and its Hilbert transform, respectively.
With sign x a step function, as is Im log z for z real, we are thus led to consider the analytic
function
i
f (z) =
log(z + 2)(z + 2)3 − 4 log(z + 1)(z + 1)3 + 6 log(z)(z)3
(9.65)
6π
− 4 log(z − 1)(z − 1)3 + log(z − 2)(z − 2)3
and recover from this B(x) = Re f (x). Since f (z) decays to zero (due to cancellations
between its terms) for |z| → ∞ in the upper half-plane, we can pick up its Hilbert transform
immediately from H{B(x)} = Im f (x); cf. Figure 9.11. This function f (z) has, due to
its logarithms, branch points at z = −2, −1, 0, 1, 2. With the standard choice for branch
cuts, these will not get into the upper half-plane.
2
Example 9.42. Find the Hilbert transform of the Gaussian function f (x) = e−z .
2
We start in Figures 9.12(a)–(b) by illustrating the Gaussian function f (z) = e−z in
the complex plane. If we can alter this function to another analytic one such that (i) the
2
real part along the real axis remains unchanged e−x , and (ii) it decays to zero in the
upper half-plane, we can from its imaginary part along the real axis read off the Hilbert
transform. As ´a further background, we illustrate in parts (c) and (d) the error function
2
z
erf (z) = √2π 0 e−t dt. Not surprisingly, this function diverges in the same sectors as
2
does e−z , with the additional striking feature that it converges very rapidly to +1 in the
sector surrounding the positive real axis and to −1 in the opposite sector. If we thus form
2
erf (iz), we rotate these two images 90◦ clockwise. Then multiplying them by e−z and
2
adding to e−z , the large oscillations in the upper half-plane ought to cancel out (without
2
altering the real part of e−z along the real axis). Parts (e) and (f) of the figure shows this
indeed to be the case. We can now read off the Hilbert transform from the function along
the real axis in the last of these figures, obtaining
ˆ ix
ˆ x
n
o
2
2
2
2
2
2
2
2
H e−x = Im √ e−x
e−t dt = √ e−x
et dt = √ D(x),
π
π
π
0
0
where D(x) is known as Dawson’s function. This same result follows also from Example
9.21.
9.4. Hilbert transform
263
2
2
0
0
-2
-2
-4
-4
-3
-3
-2
-2
4
3
-1
4
0
1
1
0
0
-1
2
x
2
0
1
1
3
-1
2
-2
3
x
y
-3
4
-1
2
-2
3
2
y
-3
4
-4
-4
2
(a) Re e−z .
(b) Im e−z .
2
2
0
0
-2
-2
-4
-4
-3
-3
-2
-2
4
3
-1
4
1
1
0
0
-1
2
x
2
0
1
1
3
-1
2
0
-2
3
-4
(c) Re erf (z) = Re
x
y
-3
4
√2
π
-1
2
´z
0
-2
3
2
e−t dt.
-4
(d) Im erf (z) = Im
2
2
0
0
-2
y
-3
4
√2
π
´z
0
2
e−t dt.
-2
-4
-4
-3
-3
-2
4
3
-1
2
0
1
1
0
4
3
-1
2
0
1
1
0
-1
2
x
-2
-2
3
-3
4
y
-4
h
i
2
2
(e) Re e−z +e−z erf (iz) .
-1
2
x
-2
3
-3
4
y
-4
h
i
2
2
(f) Im e−z +e−z erf (iz) .
Figure 9.12. Illustration of the steps in obtaining the Hilbert transform of a Gaussian.
264
Chapter 9. Transforms
9.5 z-transform
This transform can either be seen as a discrete version of the Laplace transform or as a case
of a regular Taylor (or Laurent) expansion.81
If the x-variable in (9.37) is not continuous, 0 ≤ x ≤ ∞, but discrete, k = 0, 1, 2, . . .,
with corresponding function values fk , a counterpart to the Laplace transform can be written as
∞
X
fˆ(z) =
fk z −k .
(9.66)
k=0
This function fˆ(z) is known as the z-transform of the sequence {fk } and is widely used, for
example, in signal processing. After the variable change z → z1 , (9.66) takes the form of
a standard Taylor expansion. It can then be inverted by any means that we have for Taylor
expanding a function; cf. Sections 2.3 and 4.2.8. If the Taylor expansion has radius of
convergence R = 0, ideas from analytic continuation and asymptotic expansions (Chapters
3 and 12, respectively) may be applicable. Especially in geophysical applications, the
standard Taylor version is commonly used for the z-transform:
fˆ(z) =
∞
X
fk z k .
(9.67)
k=0
Example 9.43. Solve the linear inhomogeneous recursion relation
xk + 5xk−1 =
1
, k = 1, 2, 3, . . . ,
k
x0 given.
(9.68)
Let the z-transform of the sequence {xk } (using the definition (9.67)) be
fb(z) = x0 + x1 z + x2 z 2 + x3 z 3 + · · · , i.e.,
5z fb(z) =
5x0 z + 5x1 z 2 + 5x2 z 3 + · · · .
(9.69)
Adding these two sums and using (9.68) gives
1
1
1
1
fb(z) =
x0 + z + z 2 + z 3 + · · · .
1 + 5z
1
2
3
Resisting the temptation of simplifying 11 z + 12 z 2 + 13 z 3 + · · · = − log(1 − z), but instead
1
Taylor expanding 1+5z
= 1 + (−5)z + (−5)2 z 2 + · · · and expanding the RHS, followed
by equating coefficients with (9.69), gives
"
#
k
X
1
xk = (−5)k x0 +
.
n(−5)n
n=1
9.6 Three additional transforms related to rotations
The list of transform pairs is almost (if not literally) inexhaustible. We conclude by just
hastily mentioning three transform pairs that often become important in applications that
feature rotations or axial symmetries.
81 Additionally
known as generating functions.
9.6. Three additional transforms related to rotations
265
9.6.1 Abel transform
The Abel transform of a function f (r) is defined by
ˆ
∞
A {f (r)} (s) = 2
s
f (r) r
√
dr,
r2 − s2
(9.70)
with the inverse transform given by
f (r) = −
1
π
ˆ
∞
r
dA(s)
ds
√
ds
s2 − r2
(9.71)
(see Exercise 9.8.29).82 In 2-D radially symmetric cases, with r2 = x2 + y 2 , functions
b can be interpreted as the
f (x, y) will come to depend on r only, as f (r). The function A(s)
integral of f (x, y) along an infinite straight line that passes a distance of s from the origin
in the (x, y)-plane.
9.6.2 Radon transform
We again consider straight line integrals across the (x, y)-plane, as in the Abel transform
case, but remove the requirement of axial symmetry. The integrals will then depend not
only on the line’s distance s to the origin, but also on the angle θ of the line relative to
the x-axis. A function f (x, y) will be transformed to Fb(s, θ). Johann Radon discovered
in 1917 its inversion formula, which can again be expressed in integral form. In this case,
the exact inversion formula turns out to be impractical for actual use, but extremely effective computational algorithms for it were later found, leading to Nobel Prize winning
advances in medical imaging. X-ray absorption along straight lines can be recorded, and
the inverse transform provides then high resolution reconstructions / images of the interior
of the recorded object.
9.6.3 Hankel transform
The Hankel transform pair for a function f (r) is given by
ˆ
∞
H{f (r)}(s) =
ˆ
f (r) Jν (sr) r dr,
(9.72)
H(s) Jν (sr) s ds.
(9.73)
0
f (r) =
∞
0
Here, Jν (r) denotes the Bessel J-function of order ν, to be introduced later in Section
11.2 and illustrated in Figures 11.5 and 11.6. If a function f (x1 , x2 , . . . , xn ) is radially
symmetric in an n-dimensional space, so will its Fourier transform f (ω1 , ω2 , . . . , ωn ) be
in its n-dimensional transform space. The radial behaviors of these two functions are thus
both 1-D functions, and it transpires that they are the Hankel transforms of each other,
when choosing ν = n2 − 1. This transform is a case where the two transforms in the pair
are identical to each other.
82 Different
1
2
definitions exist in the literature, such as using
in the denominator to α with 0 < α < 1.
´s
0
instead of
´∞
s
, and also generalizing the power
266
Chapter 9. Transforms
9.7 Supplementary materials
Theorem 9.3. If f (x) =
√1
2π
´∞
−∞
fˆ(ω) eiωx dω, then fˆ(ω) =
√1
2π
´∞
−∞
f (x) e−iωx dx.
Proof. For the interval x ∈ [−π, π], we recall (9.4) and (9.5). The Fourier modes
present are given by eikx , k = 0, ±1, ±2, ±3, . . . . If we extend the period by a factor
of L, instead considering x ∈ [−Lπ, Lπ], the modes present increase, to ei ωk x , with
ωk = 0, ± L1 , ± L2 , ± L3 , . . . . The Fourier expansion can then be written as
ˆ Lπ
∞
1
1 X
f (x) e−i ωk x dx.
c(ωk )ei ωk x , with c(ωk ) =
f (x) =
L
2π −Lπ
k=−∞
In the limit of L → ∞, the sequence ωk → ω (continuous variable),
ˆ ∞
f (x) →
c(ω)ei ωx dω,
−∞
´∞
and c(ωk ) → c(ω) =
f (x) e−i ωx dx. It is arbitrary where the 2π factor is placed,
√−∞
so we can define fˆ(ω) as 2πc(ω), arriving at (9.7) and (9.6).
1
2π
Proof of the first part of Theorem 9.14. With the notation in (9.17)–(9.20), we obtain
ˆ π
ˆ π
1
1
−ikt
−iky
αk βk =
A(t) e
dt
B(y) e
dy
2π −π
2π −π
ˆ π ˆ π
1
=
A(t)B(y) e−ik(t+y) dtdy .
4π 2 −π −π
Replacing y by y = x − t, i.e., t + y = x and dy = dx, the double integral becomes
ˆ π ˆ π
ˆ π
1
1
1
A(t)B(x − t) dt e−ikx dx =
C(x) e−ikx dx.
2π −π 2π −π
2π −π
Proof of the first part of Theorem 9.16 (Poisson’s summationP
formula). The task is to
∞
determine the Fourier coefficient ck of the 2π-periodic function n=−∞ f (x + 2πn). By
(9.5), these are
!
ˆ π
∞
X
1
f (x + 2πn) e−ikx dx.
ck =
2π −π n=−∞
It makes
if we
the factor e´−ikx to e−ik(x+2πn) , since e−i2πkn = 1.
´ π no
Pdifference
Pchange
∞
∞ ´π
∞
Then −π −∞ · · · dx = −∞ −π · · · dx = −∞ · · · dx, and
ˆ ∞
1
1
ck =
f (x) e−ikx dx = √ fˆ(k).
2π −∞
2π
Proof of Theorem 9.30. With the notation in the theorem formulation,
ˆ x
ˆ ∞
−sx
ĥ(s) =
e
f (ξ)g(x − ξ)dξ dx
0
0
ˆ ∞ ˆ ∞
−sx
=
e
g(x − ξ)dx f (ξ)dξ
0
ξ
ˆ
ˆ ∞
∞
−sξ
−s(x−ξ)
=
e f (ξ)
e
g(x − ξ)dx dξ = fˆ(s) · ĝ(s).
0
ξ
9.7. Supplementary materials
267
0
x
0
Figure 9.13. The first octant in the (x, ξ)-plane (extending to infinity to the right).
For the change in integration order between the first and second lines, note that the domain
we integrate over is the first octant in the (x, ξ)-plane, as shown shaded in Figure 9.13.
Theorem 9.44. The inversion formula for the Mellin transform (9.53)
ˆ ∞
ˆ
f (s) =
f (x) xs−1 dx, α < Re s < β,
0
is given by
1
f (x) =
2πi
ˆ
γ+i∞
fˆ(s) xs ds,
(9.74)
γ−i∞
where α < γ < β.
Proof. As for the inversion formula for the Laplace transform in Section 9.2.1, we will
make variable changes that reduce the problem to the inversion of a Fourier transform.
Starting with the definition of fˆ(s), we get
ˆ ∞
ˆ
f (s) =
f (x) xs−1 dx
{set x = et ; dx = et dt}
0
ˆ ∞
=
f (et ) est dt
{set s = γ − iω},
−∞
ˆ ∞
ˆ ∞
t γt −iω t
fˆ(γ − iω) =
f (et ) e(γ−iω) t dt =
f (e )e
e
dt.
−∞
−∞
t
The RHS is now a standard Fourier transform of f (e ) eγt . Inverting it gives
ˆ ∞
1
f (et )eγt =
fˆ(γ − iω) eiω t dω.
2π −∞
We next undo the variable change s = γ − iω by setting ω = γ−s
i , dω = i ds, to obtain
ˆ
ˆ
γ+i∞
e−γt γ+i∞ ˆ
1
f (et ) =
f (s) e(γ−s) t ds =
fˆ(s) e−s t ds.
2πi γ−i∞
2πi γ−i∞
Undoing the first variable change, x = et , now gives (9.74).
268
Chapter 9. Transforms
9.8 Exercises
P∞
Exercise 9.8.1. Determine the coefficients in the Fourier series k=−∞ ck ei kx = e−ix/α ,
−π < x < π. Compare your result with that of Exercise 5.6.43(a).
Exercise 9.8.2. In the style of Example 9.13, determine what sum you get when applying
(9.15) to f (x) = x2 .
Exercise 9.8.3.
Show that the coefficients ak in the Fourier cosine expansion f (x) =
P∞
ecos x = a0 + k=1 ak cos kx converge to zero as ak ≈ 1/(k! 2k−1 ).
ix
−ix
Hint: One way to solve this problem starts from noting that f (x) = e(e +e )/2 =
1 eix +e−ix
1 eix +e−ix 2
1 + 1!
+ 2!
+ · · · . Find the coefficient for eikx (which is the same
2
2
as for eikx ) when this sum is fully expanded.
Comments: As to be expected from Example 9.2 and the fact that f (z) = ecos z is an entire
function, the coefficients ak go to zero faster than exponentially. A different way to solve
this problem is outlined in Exercise 12.6.21.
Exercise 9.8.4. The first part of Theorem 9.14 was proved in Section 9.7. Provide a proof
for the second part.
Exercise 9.8.5. Derive Poisson’s integral formula (Theorem 4.20) by considering the
Fourier series expansion of f (eiθ ).
Hints: (i) With z = x + iy = r eiθ , note that fn (z) = r|n| einθ satisfies Laplace’s equation
P∞
2
for all n ∈ Z, and (ii) n=−∞ r|n| einθ = 1+r21−r
−2r cos θ .
Exercise 9.8.6. Derive Parseval’s relation (4.20) for a Fourier series.
Exercise 9.8.7. Prove the following generalization to Parseval’s theorem 9.11:
ˆ
ˆ
∞
∞
f (x) g(x) dx =
−∞
−∞
fb(ω) gb(ω) dω .
Exercise 9.8.8. Determine the Fourier transform of
f (x) =
cos x,
0
− π2 ≤ x ≤ π2 ,
otherwise.
´∞
x)2
Exercise 9.8.9. Show that −∞ (1−cos
dx = π3 .
x4
Hint: You might start by verifying the Fourier transform pair

 1 + x,
1 − x,
f (x) =

0
−1 ≤ x ≤ 0,
0 ≤ x ≤ 1,
otherwise
r
⇔
fb(ω) =
2 1 − cos ω
.
π
ω2
Exercise 9.8.10. By combining (9.29) and (9.64), derive the Fourier transform for the cubic
B-spline, as given in (9.13).
9.8. Exercises
269
Exercise 9.8.11. Figure 2.25(b) illustrated Gibbs’ phenomenon, an overshoot in the Fourier
interpolant to a discontinuous function, amounting to about 9% of the height of the jump.
Derive a closed-form expression for this quantity.
Hint: A convenient case for deriving this is provided by the function
−1/2,
−π < x < 0,
f (x) =
1/2,
0 < x < π,
P∞
. Consider the derivative of the truncated
with Fourier series f (x) = π2 k=0 sin(2k+1)x
2k+1
P
n
2
0
series fn (x) = π k=0 cos(2k + 1)x, sum this exactly (cf. Example 2.5), find its first
positive zero, evaluate fn (x) at this point, and let n → ∞.
Exercise 9.8.12. Give the details for recovering the initial function in Example 9.26.
Exercise 9.8.13. Determine L−1
Exercise 9.8.14. Determine L−1
s e−2s
3
(s2 +1)(s2 +4)
s2 −1
(x).
(x).
Exercise 9.8.15. Use the Laplace transform to solve the ODE y 00 (x)−8y 0 (x)+7y(x) = 5x,
with initial conditions y(0) = 1, y 0 (0) = 2.
Exercise 9.8.16. Use the Laplace transform to solve the ODE y 000 (x) + y 0 (x) = ex , with
initial conditions y(0) = y 0 (0) = y 00 (0) = 0.
Exercise 9.8.17. The Laplace transform method generalizes immediately from scalar ODEs
to systems. Use it to solve
0
u (x) = 3u(x) − 3v(x) + 4,
u(0) = 1,
v 0 (x) = 2u(x) − 2v(x) − 1,
v(0) = 0.
Exercise
9.8.18. If f (x) has the period X (with X > 0), show that L{f (x)}(s) =
´
X
0
f (ξ)e−sξ dξ
.
1−e−sX
Exercise 9.8.19.
(a) Integrate the definition of the Laplace transform (9.37) to arrive at
ˆ ∞
ˆ ∞
f (x) −sx
b
f (σ)dσ =
e
dx .
x
s
0
´∞
(b) Letting s → 0 in (9.75) gives 0
an alternate solution to Exercise 5.6.26.
f (x)
x
dx =
´∞
0
(9.75)
fb(s) ds. Use this relation to obtain
Exercise 9.8.20. One of the (many) strengths of transforms
is how well they handle step
´∞
and delta functions. Let δ(x) = 0, x 6= 0, but with −∞ δ(x)dx = 1. Assume that a > 0.
(a) Show that L{δ(x − a)}(s) = e−as .
(b) Show by Theorem 9.23 that L−1 {e−as
´ ∞} (x) = 0 for x 6= a and (for example using
the result of Exercise 9.8.19, part (b)) that −∞ L−1 {e−as } (x) dx = 1.
270
Chapter 9. Transforms
Exercise 9.8.21. The moments of a function f (x) are defined by
ˆ
∞
xn f (x) dx , n = 0, 1, 2, . . . .
µn =
0
Assuming f (x) goes to zero sufficiently fast that there are no convergence issues, deduce
that the moments alternatively can be obtained from the Laplace transform of f (x) as
µn = (−1)n
dn b
f (s)
dsn
, n = 0, 1, 2, . . . .
s=0
Hint: Recall Theorem 9.29.
Exercise 9.8.22. Show that L{log x}(s) = − 1s (log s + γ).
Hint: Differentiate (9.42) with respect to k (cf. “Feynman’s trick”) and then use the result
in Exercise 6.5.4.
Exercise 9.8.23. The “sine integral” function is defined by Si(x) =
its Laplace transform becomes
L {Si(x)} (s) =
´x
0
sin t
t dt.
Show that
1
1
arctan .
s
s
Hint: Combine Theorem 9.29 with (9.40) and (9.50).
Exercise 9.8.24. Perform the following Mellin transforms (and determine their fundamental band):
(a)
M ex1−1 (s) = ζ(s) Γ(s),
P
∞
Hint: ex1−1 = n=1 e−nx for x > 0,
1
Γ(n−s)
.
(b)
M (1+x)n (s) = Γ(s)Γ(n)
Note: This problem is the inverse of the one in Example 9.36.
t
Hint: Change variable x = 1−t
and then use (6.5) and (6.6).
Exercise 9.8.25. Give the coefficients of the Taylor expansion about the origin of f (x) =
1
ex −1 in terms of ζ(z) and Γ(z), using Exercise 9.8.24(a):
(a) Using the connection between the Mellin transform and properties of f (x) at x = 0
and x = ∞.
(b) Using the Bernoulli numbers definition given in Exercise 2.9.23.
Hint for (b): Use the result in Exercise 3.3.7.
Exercise 9.8.26. Find the inverse Mellin transform of M (s) = ζ(s) Γ(s), with the fundamental band Re(s) ≥ 0.
Hint: Use Ramanujan’s master theorem (see (3.12)).
Exercise 9.8.27. Confirm the following alternative definition of the Hilbert transform:
H{u(x)} = −
1
lim
π ε→0
ˆ
ε
∞
u(x + t) − u(x − t)
dt .
t
9.8. Exercises
271
Exercise 9.8.28. A typical amplitude modulated signal is a product of a “message signal” fL (t) containing only Fourier modes in a low range, and a “carrier signal” fH (t)
that contains only high Fourier modes, with no overlap between the two ranges. Show
Bedrosian’s theorem for their Hilbert transforms: H{fL (t) fH (t)} = fL (t) H{fH (t)}.
Hint: (i) The linearity of the Hilbert transform makes it sufficient to consider a just single
low mode ωL and a single high mode ωH , (ii) note (9.62), and (iii) note that sign (ωL +
ωH ) = sign (ωH ).
Exercise 9.8.29. Verify the inversion formula (9.71) for the Abel transform (9.70) (assuming r f (r) → 0 and r f 0 (r) → 0 as r → ∞).
Hint: Integrate (9.70) by parts to obtain f 0 (r) inside the integral, and then evaluate dA(s)
ds .
´β
dt
The relation α √t−α√β−t = π (for α < β) may come in handy.
Exercise 9.8.30. Show that, like for the Fourier transform, the Abel transform of a Gaussian is again a Gaussian:
r
n
o
2
π −α s2
e
.
A e−α r (s) =
α
Note: In this case, the two Gaussians differ only by a scalar factor; otherwise their widths
are the same.
Chapter 10
Wiener–Hopf and
Riemann–Hilbert
Methods
These techniques can be seen as far-reaching generalizations of both the Laplace and the
Hilbert transforms. In the Wiener–Hopf case, analytic functions are sought which satisfy
different relations along the positive and negative real axes. In the Riemann–Hilbert generalization, information is instead provided along a curve C that can be either closed or open
(often in the form of linear relation between the function’s real and imaginary parts on the
two sides of C). If C has end points, certain additional issues arise.
We focus first, in Section 10.1, on the Wiener–Hopf method before turning to the
Riemann–Hilbert method in Section 10.2. The present goal is limited to give a general
“flavor” of what the techniques can do, and to their key underlying ideas. For more comprehensive treatments on the two topics, see, for example, [22, 33] and [1], respectively.
10.1 The Wiener–Hopf method
Among the most important features of the transforms that we studied in Chapter 9 were
their convolution theorems. Applying the Fourier and the Laplace transform, respectively,
we can, for example, immediately (by these theorems) solve integral equations of the forms
ˆ ∞
f (x) =
κ(x − y)f (y)dy + g(x), x ∈ (−∞, ∞),
−∞
and
ˆ
x
κ(x − y)f (y)dy + g(x),
f (x) =
x ∈ [0, ∞),
0
where the “kernel” κ and the function g(x) are given; the function f (x) is to be found.
Also important in applications, but not handled directly by any of these transforms, is the
case
ˆ ∞
f (x) =
κ(x − y)f (y)dy + g(x), x ∈ [0, ∞).
0
This is just one of several Wiener–Hopf applications we will describe below. Others will
involve solving certain types of ODEs and PDEs. All have in common that both formulations and answers involve no complex variables, but complex arithmetic is nevertheless
indispensable throughout the solution process.
273
274
Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
10.1.1 Preliminary notion: Half-plane splittings revisited
Previously, in Sections 5.1.5 and 9.1.8, we described two approaches for splitting an analytic function f (z), given along the real axis, into f (z) = f + (z) + f − (z), singularity-free
and decaying to zero in the lower and upper half-planes, respectively.83 Both this splitting
and its product/quotient counterparts f (z) = f + (z) · f − (z) and f (z) = f + (z)/f − (z) are
key components of the Wiener–Hopf method. They are commented on next.
Sum splitting
In the two sections where we described this previously, the primary focus was on functions
defined along the real x-axis. The leftmost two relations in Theorem 9.19,



F {f− (x)}(ω) = fˆ− (ω),
ˆ+
(10.1)

 F {f+ (x)}(ω) = f (ω),
{z
}
|
signs are the same
considered, however, the case where the function to be split resides in Fourier space: fˆ(ω)
instead of f (z). For the subscripts in (10.1), we recall their definition:
f (x) , x ≥ 0,
0
, x ≥ 0,
f+ (x) =
and f− (x) =
0
, x < 0,
f (x) , x < 0.
Product and quotient splittings
Given a function fˆ(ω) in the Fourier space, we can similarly split it into either a product
fˆ(ω) = fˆ+ (ω) · fˆ− (ω) or a quotient fˆ(ω) = fˆ+ (ω)/fˆ− (ω).
Example 10.1. Find product and quotient splits of fˆ(ω) =
1
1+ω 2 .
Factorizing the denominator gives us immediately the product decomposition, with
1
1
fˆ− (ω) = ω+i
and fˆ+ (ω) = ω−i
. These functions are free of both singularities and
zeros in Im(ω) ≥ 0 and Im(ω) ≤ 0, respectively. For a quotient split fˆ+ (ω)/fˆ− (ω),
1
we swap fˆ− (ω) = ω+i
to fˆ− (ω) = ω + i. We note that we usually have implied that
−
fˆ (ω) not only is singularity-free in the upper half-plane, but also that it vanishes in that
half-plane. In this instance, fˆ− (ω) no longer vanishes in Im ω ≥ 0.
√
Example 10.2. Find a product split of fˆ(ω) = ω 2 − λ2 , where Re(λ), Im(λ) > 0.
1/2
Again, immediate factorization gives the answer: fˆ− (ω) = (ω + λ)
and fˆ+ (ω) =
1/2
(ω − λ) . We can direct the branch cuts towards −i∞ and +i∞, to keep the respective
half-planes singularity- and zero-free.
For the more complicated cases, one can rewrite fˆ(ω) = fˆ+ (ω) · fˆ− (ω) as log fˆ(ω) =
log fˆ+ (ω) + log fˆ− (ω), use a sum split of log fˆ(ω), and then exponentiate the resulting
83 We recall again that the meaning of “+” and “−” in this context does not agree between all texts. Whether
used as superscripts for half-planes or as subscripts for splits along the real axis, we use the sign to denote the
direction of nontrivial behavior.
10.1. The Wiener–Hopf method
275
terms. A number of technical issues can arise (such as how to handle the multivaluedness that arises for poles and zeros of fˆ(ω)), and the algebra often becomes cumbersome.
Specialized texts, such as [7, 22], discuss this further.
10.1.2 The Wiener–Hopf procedure
The Wiener–Hopf method relies on three key steps, which the following examples are
organized around. These steps are the following:
Step 1: Wiener–Hopf equation
The first step is to recast the problem to be solved for an unknown function f (x) = f− (x)+
f+ (x) into a relation of the form
γ(ω) fˆ+ (ω) + fˆ− (ω) = h(ω) ,
(10.2)
where γ(ω) and h(ω) are known functions, and fˆ− (ω) and fˆ+ (ω) remain to be determined.
Step 2: Decomposition and analytic continuation
Further rearrange (10.2) into a form from which fˆ− (ω) and fˆ+ (ω) both can be deduced
by using Liouville’s theorem. The key rearrangement steps are (i) quotient split γ(ω) =
γ + (ω)/γ − (ω), (ii) multiply by γ − (ω), (iii) sum split γ − (ω)h(ω), (iv) collect all + terms
on one side and all − terms on the other side.
Step 3: Half-line inverse Fourier transform
Having found fˆ− (ω) and fˆ+ (ω), calculate f− (x) and f+ (x) by means of (10.1).
Case with convergent Fourier transforms
For the sake of simplicity, we will initially assume that all of the functions which we consider decay for x → ±∞.
Example 10.3. Determine a function f (x) which satisfies the integral equation
ˆ ∞
f (x) = −4
e−|x−t| f (t) dt + xe−2x ,
x ≥ 0.
(10.3)
0
Step 1: Wiener–Hopf equation
The integral equation (10.3) gives no information about f (x) for x < 0. To be able to use
Fourier transforms, we first generalize the RHS function xe−2x to
xe−2x , x ≥ 0,
g(x) = g+ (x) =
0
, x < 0.
We next write
f+ (x) =
f (x) , x ≥ 0,
0
, x < 0,
and, by (10.3),
f− (x) =
−4
´∞
0
0
e−|x−t| f (t) dt
, x ≥ 0,
, x < 0.
276
Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
It now holds for all real x values that
ˆ ∞
e−|x−t| f+ (t) dt = g+ (x),
f+ (x) + f− (x) + 4
x ∈ R.
−∞
We next take the Fourier transform of this equation. By (10.1) and the Fourier convolution
theorem (Section 9.1.5), this gives the Wiener–Hopf equation
√
fˆ+ (ω) + fˆ− (ω) + 4 2π κ̂(ω) fˆ+ (ω) = ĝ + (ω).
(10.4)
q
´∞
Here κ̂(ω) = √12π −∞ e−|x| e−ixω dx = π2 ω21+1 is the standard Fourier transform of
´ ∞ x(−2−iω)
xe
dx =
the kernel κ(x) = e−|x| . The functions fˆ+ (ω) and ĝ + (ω) = √1
√ −1
2π(ω−2i)2
ˆ−
2π
0
are singularity-free in the complex ω-plane’s lower half (Im(ω) ≤ 0), and
f (ω) is similarly singularity-free in the complex ω-plane’s upper half (Im(ω) ≥ 0). For
κ(x), κ̂(ω), g+ (x), and ĝ + (ω), see Figure 10.1. Since f (x) and g(x) decay as x → ±∞,
fˆ+ (ω), fˆ− (ω), ĝ + (ω) must all be analytic along the real axis. The objective is to solve
for fˆ+ (ω) and to then use the inversion formula to obtain f+ (x). However, fˆ− (ω) is also
unknown in (10.4). The beauty of the Wiener–Hopf method is that this sole equation (10.4)
provides enough information to solve for both fˆ+ (ω) and fˆ− (ω). The key ingredients for
doing this are analytic continuation and Liouville’s theorem.
Step 2: Decomposition and analytic continuation
Substituting the known quantities into (10.4) results in
2
1
ω + 9 ˆ+
= −fˆ− (ω),
f (ω) + √
2
ω +1
2π(ω − 2i)2
(10.5)
2
+
γ (ω)
ω +9
which is of the form (10.2). We therefore quotient split γ(ω) = ω
2 +1 = γ − (ω) with
ω+i
−
−
γ + (ω) = ω−3i
ω−i and γ (ω) = ω+3i . After multiplying both sides by γ (ω), we obtain
fˆ− (ω)(ω + i)
ω − 3i ˆ+
(ω + i)
.
(10.6)
f (ω) + √
= −
2
ω−i
(ω + 3i)
2π(ω − 2i) (ω + 3i)
{z
}
|
{z
}
|
Singularity-free in Im(ω)≥0
Singularity-free in Im(ω)≤0
The term
(ω+i)
2π(ω−2i)2 (ω+3i)
still has singularities in both regions, so the next step is to
√
√
3/(5 2π )
i 2/π
(ω+i)
−
sum split this function. By partial fractions, √2π(ω−2i)
=
2 (ω+3i)
(ω−2i)2
25(ω−2i) +
√
i 2/π
25(ω+3i) . After reordering the terms, we reach the goal of having obtained an identity
between two functions that are singularity-free and, also importantly, decay to zero in the
opposite half-planes. We call this function E(ω):
√
3
ω − 3i ˆ+
i 2
f (ω) + √
− √
ω−i
25 π(ω − 2i)
5 2π(ω − 2i)2
|
{z
}
√
Singularity-free in Im(ω)≤0
√
fˆ− (ω)(ω + i)
i 2
=−
− √
= E(ω).
(ω + 3i)
25 π(ω + 3i)
|
{z
}
Singularity-free in Im(ω)≥0
(10.7)
10.1. The Wiener–Hopf method
277
0.8
0.6
0.4
0.2
-3
-2
-1
0
1
2
3
(a) κ(x) = e−|x|
(b) κ̂(ω) =
q 2
π
1
ω 2 +1
0.15
0.1
0.05
-3
-2
-1
0
(c) g+ (x) =
1
−2x
xe
0
2
3
,x≥0
,x<0
(d) ĝ + (ω) =
√ −1
2π(ω−2i)2
Figure 10.1. The functions κ(x) and g+ (x) in Example 10.3 and their Fourier transforms.
Since the functions on both sides are analytic and coincide on the real axis, they must be
two different representations of one and the same function, i.e., the analytic continuation
of one another, as is illustrated in Figure 10.2.
The LHSs and RHSs are singularity-free and decay to zero on complementary parts of
the complex plane, so their underlying function, E(ω), must be a bounded entire function
on C, i.e., by Liouville’s theorem a constant. The decay to zero tells us that this constant is
zero; E(ω) ≡ 0. Therefore,
fˆ+ (ω) =
(2iω − 11)(ω − i)
√
,
25 2π(ω − 2i)2 (ω − 3i)
which, as required, is analytic for Im(ω) ≤ 0.
Step 3: Inverse Fourier transform
The Fourier transform inversion formula yields
ˆ ∞
1
(2iω − 11)(ω − i)
√
f+ (x) = √
eixω dω.
2π −∞ 25 2π(ω − 2i)2 (ω − 3i)
(10.8)
278
Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
fˆ− (ω) anal y t ic
Im ω
over l ap
Re ω
fˆ+ (ω) anal y t ic
Figure 10.2. Analyticity regions for Example 10.3.
Since the integrand vanishes away from the origin in the upper half of the complex plane,
one can close the contour with a semicircle centered at the origin with infinite radius and
use residue calculus to compute the solution:
√
(2iω − 11)(ω − i)
√
f+ (x) = 2πi Res
eixω , ω = 2i
25 2π(ω − 2i)2 (ω − 3i)
(2iω − 11)(ω − i)
√
+ Res
eixω , ω = 3i
25 2π(ω − 2i)2 (ω − 3i)
1 −3x
e
(−34 + ex (32 − 15x)).
=
25
Although this expression
for f+ (x) solves the posed problem, we can similarly from (10.7)
√
i 2
2 x
−
ˆ
√
obtain f (ω) = − 25 π(ω+i) and f− (x) = − 25
e . As an algebra check, this can be veri´ ∞ −|x−t|
´ ∞ 1 −3t
fied by evaluating (for x < 0): f− (x) = −4 0 e
f+ (t)dt = −4ex 0 25
e (−34+
2 x
+
−
t
ˆ
ˆ
e (32 − 15t))dt = − 25 e . Figure 10.3 shows f (ω), f+ (x), f (ω) and f− (x).
Example 10.4. Consider the following ODE:
00
f (x) − f (x) = 0,
f 00 (x) − 4f (x) = 0,
x > 0,
x < 0.
(10.9)
Let us assume that f (0+ ) − f (0− ) = 0 and f 0 (0+ ) − f 0 (0− ) = 1, and that f (x) vanishes
as x → ±∞. Solve for f (x).84
Step 1: Wiener–Hopf equation
Let’s assume once again that the solution f (x) vanishes as x → ±∞, and let
f (x) , x ≥ 0,
f+ (x) =
0
, x < 0,
and
f− (x) =
0
, x ≥ 0,
f (x) , x < 0.
84 This problem is trivially solved by noting from (10.9) that f (x) = c ex + c e−x , f (x) = c e2x +
+
−
1
2
3
c4 e−2x . The four additional conditions give then c1 = c4 = 0 and c2 = c3 = − 31 . The Wiener–Hopf solution
is given here as a technique illustration.
10.1. The Wiener–Hopf method
279
0.05
-2
-1
1
2
-0.05
(a) fˆ+ (ω) =
(2iω−11)(ω−i)
√
.
25 2π(ω−2i)2 (ω−3i)
(b) f+ (x) =
-3
-2
1 −3x
(−34
25 e
-1
0
+ ex (32 − 15x)).
1
2
3
-0.02
-0.04
-0.06
√
2
(c) fˆ− (ω) = − 25√iπ(ω+i)
.
2 x
(d) f− (x) = − 25
e .
Figure 10.3. The functions fˆ+ (ω) and fˆ− (ω) in Example 10.3 and their half-line counterparts.
´∞
´0
Then fˆ+ (ω) = √12π 0 f+ (x)e−ixω dx and fˆ− (ω) = √12π −∞ f− (x)e−ixω dx are
singularity-free on Im(ω) ≤ 0 and Im(ω) ≥ 0, respectively. Applying the standard
Fourier transform on both √
differential equations and using integration by parts twice
√ gives
0
0
−f+
(0+ ) − iωf+ (0+ ) − 2π(1 + ω 2 )fˆ+ (ω) = 0 and f−
(0− ) + iωf− (0− ) − 2π(4 +
ω 2 )fˆ− (ω) = 0 if we also assume that
0
0
lim f+
(x) + iωf+ (x) e−iωx = lim f−
(x) + iωf− (x) e−iωx = 0.
x→∞
x→−∞
Adding both equations and applying the boundary conditions results in the Wiener–Hopf
problem
√
√
1 + 2π ω 2 + 1 fˆ+ (ω) + 2π ω 2 + 4 fˆ− (ω) = 0.
Step 2: Decomposition and analytic continuation
We know that fˆ+ (ω) is free of singularity for Im(ω) ≤ 0 and, similarly, that fˆ− (ω) is
singularity-free for Im(ω) ≥ 0. Once again with the intention of grouping terms that are
singularity-free on either domain,
wedivide both
√
√ sides of the
−equation by (ω + i)(ω − 2i)
1
ω−i ˆ+
ˆ
and obtain (ω+i)(ω−2i)
+ 2π ω−2i
f (ω) + 2π ω+2i
ω+i f (ω) = 0. A partial fractions
280
Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
1
1
1
decomposition of the first term gives (ω+i)(ω−2i)
= 3i
ω−2i −
equation becomes
√
ω − i ˆ+
1
1
+ 2π
f (ω)
3i
ω − 2i
ω − 2i
{z
}
|
1
ω+i
. The reordered
Singularity-free in Im(ω)≤0
√
= − 2π
|
ω + 2i
ω+i
fˆ− (ω) +
{z
1
3i
1
ω+i
= E(ω).
}
Singularity-free in Im(ω)≥0
Since we, in the split, also ensured that both parts go to zero far out, analytic continuation
and Liouville’s theorem now give E(ω) ≡ 0. This allows us to solve for fˆ+ (ω) and fˆ− (ω).
Step 3: Inverse Fourier transform
1
The last step is to apply the inverse Fourier transform to fˆ+ (ω) = − √2π(3i)(ω−i)
and
´
∞
1
1
1
iωx
−
ˆ
f (ω) = √2π(3i)(ω+2i) . We obtain f+ (x) = 2π −∞ − (3i)(ω−i) e dω and f− (x) =
´∞
1
1
iωx
dω. From this follows
2π −∞ (3i)(ω+2i) e

1

 − e−x
3
f (x) =

 − 1 e2x
3
, x ≥ 0,
, x < 0.
Example 10.5. Solve the following PDE boundary value problem
∆u = u,
−∞ < x < ∞, 0 < y < ∞
such that uy (x, 0) = 0 on y = 0, x < 0 and u(x, 0) = e−x on y = 0, x ≥ 0 (with ∆
∂2
∂2
denoting the Laplacian operator ∆ = ∂x
2 + ∂y 2 ).
Step 1: Wiener–Hopf equation
Applying the Fourier transform in x, we readily obtain
∂2
û(ω, y) = ω 2 + 1 û(ω, y),
2
∂y
which can be solved analytically as û(ω, y) = A(ω)e−y
impose that û(ω, y) be bounded, B(ω) = 0 and
û(ω, y) = A(ω)e−y
√
√
ω 2 +1
ω 2 +1
+ B(ω)ey
√
ω 2 +1
.
. Since we
(10.10)
To incorporate the boundary conditions, let
−x
e
, x ≥ 0,
0
, x ≥ 0,
µ+ (x) =
µ− (x) =
0
, x < 0,
u(x, 0) , x < 0,
−i
with half-range Fourier transforms µ̂+ (ω) = √2π(ω−i)
and µ̂− (ω), singularity-free in
Im(ω) ≤ 0 and in Im(ω) ≥ 0, respectively. The extended equation becomes û(ω, 0) =
10.1. The Wiener–Hopf method
√ −i
2π(ω−i)
281
+ µ̂− (ω). Further, let ν+ (x) =
n
uy (x, 0)
0
+
, x ≥ 0,
, x < 0,
−
ν− (x) = 0, and de-
note their respective Fourier transforms by ν̂ (ω) and ν̂ (ω) = 0. The corresponding
∂
û(ω, 0) = ν̂ + (ω). From (10.10), û(ω, 0) = A(ω) =
extended equation becomes ∂y
√
∂
√ −i
+ µ̂− (ω) and ∂y
û(ω, 0) = − ω 2 + 1 A(ω) = ν̂ + (ω). Combining the two
2π(ω−i)
equations once again provides a Wiener–Hopf problem
−
p
ω2
+1
−i
−
√
+ µ̂ (ω) = ν̂ + (ω),
2π(ω − i)
from which we need to extract both unknown functions µ̂− (ω) and ν̂ + (ω).
Step 2: Decomposition and analytic continuation
The equation above has branch points at ω = ±i and a pole at ω = i. In order to group
the terms that are singularity-free in the upper part of the complex plane (including the
real axis) and the functions that are analytic in the
√ lower part of the complex plane (also
including the real axis), we divide both sides by ω − i, giving
√
√
i ω+i
ν̂ + (ω)
√
− ω + i µ̂− (ω) = √
.
ω−i
2π(ω − i)
The only problematic
term is the first one on the LHS, for which we need to do a sum
√
ω+i
= h− (ω) + h+ (ω). This function h(ω) has a pole at ω = i and
split: h(ω) = (ω−i)
√
√
a branch point at ω = √−i. At the
pole ω = i, ω + i = 2i = 1 + i, suggesting the
√
(1+i)
ω+i
reformulation h(ω) = (ω−i)
= ω+i−(1+i)
+ (ω−i)
= h− (ω) + h+ (ω). This works, since
(ω−i)
−
ω = i has now become a removable singularity of h (ω). The reordered equation becomes
√
√
i−1
i ω+i
i−1
ν̂ + (ω)
√
−√
=√
−√
− ω + iµ̂− (ω) = E(ω).
ω−i
2π(ω − i)
2π(ω − i)
2π(ω − i)
|
{z
} |
{z
}
Singularity-free in Im(ω)≤0
Singularity-free in Im(ω)≥0
By Liouville’s theorem, E(ω) = 0. Now we can solve for µ̂− (ω). We have µ̂− (ω) =
i−1 √
i−1 √
√ i
− √2π(ω−i)
. Therefore, A(ω) = − √2π(ω−i)
.
2π(ω−i)
ω+i
ω+i
Step 3: Inverse Fourier transform
√
2
The inverse Fourier
transform of A(ω) e−y ω +1 gives the PDE solution as u(x, y) =
√
´
−y ω 2 +1
∞
(1−i) e
1
√
eixω dω. This function should be purely real-valued for x and y
2π −∞
(ω−i) ω+i
real, y > 0. We can indeed eliminate all complex quantities from this answer, to obtain
u(x, y) =
π
1
√
ˆ
2
∞
−∞
√
2
e−y ω +1
cos
(ω 2 + 1)3/4
1
arctan(ω) − ωx dω.
2
Figure 10.4 illustrates both this solution u(x, y) and its y-derivative
∂
∂y u(x, y).
282
Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
1
0
-0.5
0.5
-1
-1.5
2
0
-4
1
-2
0
2
4
2
-2
-4
1
-2
0
0
(a) PDE solution u(x, y).
2
(b) Its y-derivative
4
0
∂
∂y u(x, y).
Figure 10.4. The solution u(x, y) to Step 3: the PDE in Example 10.5 and its y-derivative
The solid red curves mark the two boundary conditions.
∂
u(x, y).
∂y
Generalization to case with divergent Fourier transforms
Up until now, we have required that all functions vanish as x → ±∞. From this it followed
that the real axis was included in both the upper and the lower analyticity domains. We
then knew that there was an overlap between those domains, and that is what allowed us
to use analytic continuation. Finally, it enabled us to have an easy half-line inverse Fourier
transform integral, along the real axis. This requirement is lifted in the following example,
allowing us to proceed even when the Fourier transforms have poles (or other singularities)
right on the real ω-axis. We introduce two theorems, with proofs given in Section 10.3.
Theorem 10.6. If |f (x)| < O (eαx ) for x → ∞ and |f (x)| < O eβx for x → −∞,
where the quantities α and β are called the exponential bound indices, then fˆ+ (ω) =
´∞
´0
√1
f (x)e−ixω dx and fˆ− (ω) = √12π −∞ f (x)e−ixω dx are singularity-free, re2π 0
spectively, in Im(ω) < −α and in Im(ω) > −β.
Furthermore, the inversion formulas are as follows.
ˆ
1
f− (x) = √
fˆ− (ω) eixω dω,
2π Γ
ˆ
1
f+ (x) = √
fˆ+ (ω) eixω dω,
2π Γ
where Γ lies within the common region of analyticity of fˆ+ (ω) and fˆ− (ω), i.e. (with a small
abuse of notation), −β < Im Γ < −α, as in Figure 10.5 (having assumed that α < β).
Theorem 10.7.
Example 10.8. Let f (x) = 1 + x for x > 0. Find the half-line Fourier transform fˆ+ (ω)
of f (x) along the positive real axis.
Let
f+ (x) =
1+x
0
, x ≥ 0,
, x < 0.
10.1. The Wiener–Hopf method
283
fˆ− (ω) anal y t ic
Im ω
Im ω
−iα
Γ
Γ
Re ω
Re ω
fˆ+ (ω) anal y t ic
(a) Integration path to recover f+ (x).
−iβ
(b) Integration path to recover f− (x).
Figure 10.5. Analyticity of half-line Fourier transforms.
´∞
The half-line Fourier transform of f (x) is fˆ+ (ω) = √12π 0 (1 + x) e−iωx dx. Integrating
´∞
−iωx x=∞
−ixω
by parts, 0 (1 + x) e−iωx dx = (1 + x) e −ω + e ω2 x=0 . Assuming that Im(ω) < 0,
−iω−1
the terms evaluated at x = ∞ vanish and fˆ+ (ω) = √
.
2πω 2
+
ˆ
Note that f (ω) can also be found from the generalized (half-line) Fourier transform
´∞
of f (x). Using the same technique as in Section 9.1.7, fˆ+ (ω) = limα→0+ √1
(1 +
x) e−αx e−ixω dx =
2π
−iω−1
√
.
2πω 2
0
−iω−1
Example 10.9. Given fˆ+ (ω) = √
, use the inversion formula to recover function
2πω 2
f+ (x), which is nonzero only on the positive real axis.
Since fˆ+ (ω) is analytic for Im(ω) < 0, then f+ (x) =
´
−iω−1
√1
√
2π Γ 2πω 2
ˆ+
eixω dω, where
Γ is a contour line that passes below all the singularities of f (ω), as drawn in Figure
10.5(a). If x > 0, then we can close the contour line Γ with a top semicircle as in Figure
10.6(a) and invoke Jordan’s lemma. Indeed, the contribution from the integral along the
semicircle vanishes, and by residue calculus, f+ (x) = 1 + x. However, if x < 0, one
can close the contour as in Figure 10.6(b), and again using Jordan’s lemma and residue
calculus, f+ (x) = 0. Therefore,
1 + x , x ≥ 0,
f+ (x) =
0
, x < 0.
Example 10.10. Solve the integral equation
ˆ ∞
f (x) = 4
e−|x−t| f (t) dt,
x ≥ 0.
(10.11)
0
This example differs from Example 10.3 in two key ways: (i) it has the opposite sign
for the integral, and (ii) the equation is homogeneous (no additional RHS function g(x)).
Step 1: Wiener–Hopf equation
´∞
As before, let 4 0 e−|x−t| f+ (t) dt = f+ (x) + f− (x), where
0
, x ≥ 0,
´∞
f− (x) =
4 0 e−|x−t| f (t) dt , x < 0.
284
Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
Im ω
Im ω
Re ω
Re ω
Γ
Γ
fˆ+ (ω) anal y t ic
(a) Integration path to recover f + (x)
for x < 0.
fˆ+ (ω) anal y t ic
(b) Integration path to recover f + (x)
for x > 0.
Figure 10.6. Inversion integration paths for Example 10.9.
After applying the Fourier transform, we
9.5 and (9.23) and (9.25))
q obtain (cf. Example
√
´ ∞ −|x−t|
2
1
+
ˆ
F {4 0 e
f (t) dt}(ω) = 4 2π
π ω 2 +1 f (ω) and F {f+ (x) + f− (x)}(ω) =
fˆ+ (ω) + fˆ− (ω), which simplifies to
ω2 − 7
ω2 + 1
fˆ+ (ω) = −fˆ− (ω).
(10.12)
Step 2: Decomposition and analytic continuation
2
−7
in (10.12) are on the real ω-axis, we cannot this time move
Since both zeros of ω1+ω
2
over to the RHS one factor from the numerator and one from the denominator, and get a
singularity-free band that includes the real ω-axis. We instead move just one factor (ω + i)
over, giving
2
ω −7
fˆ+ (ω) = −(ω + i)fˆ− (ω) = E(ω) ,
(10.13)
ω−i
where now
(i) we define f + and f − relative to a path Γ with −1 < Im Γ < 0, and
(ii) E(ω) is an entire function that can grow at most as |E(ω)| ∼ |ω|.
By the generalized Liouville theorem (Theorem 4.13), E(ω) = α + β ω. From (10.13)
it follows that |E(ω)|/|ω| → 0 up/down the complex ω-plane, implying β = 0. Thus
−α
−(ω+i)fˆ− (ω) = α (constant), giving fˆ− (ω) = (ω+i)
and similarly fˆ+ (ω) = −α ωω−i
2 −7 .
Step 3: Inverse Fourier transform
´
Applying the inverse Fourier transform to fˆ+ (ω) gives f+ (x) = √12π Γ fˆ+ (ω) eiωx dω,
where Γ is in the band of analyticity. For positive x, Jordan’s lemma can be used if
the contour
as in √
Figure 10.7(b). By residue calculus, f+ (x) =
√ is the upper√semicircle,
C cos x 7 + √17 sin x 7 (with C = iα 2π an arbitrary constant). For x < 0, Jordan’s
lemma can be used if the lower semicircle contour is chosen as in Figure 10.7(a). Since
this contour contains no singularity, f+ (x) = 0 (as expected).
10.1. The Wiener–Hopf method
285
fˆ− (ω) anal y t ic
Im ω
fˆ− (ω) anal y t ic
Im ω
Re ω
Re ω
Γ
Γ
fˆ+ (ω) anal y t ic
(a) Integration path to recover f+ (x)
for x < 0.
fˆ+ (ω) anal y t ic
(b) Integration path to recover f+ (x)
for x > 0.
Figure 10.7. Inversion integration paths.
In light of the solution above, let us revisit Example 10.3 in its homogeneous form:
Example 10.11. Solve the integral equation
ˆ ∞
f (x) = −4
e−|x−t| f (t) dt,
x ≥ 0.
(10.14)
0
By the same procedure that led to (10.12), we this time arrive at the Wiener–Hopf
formulation
ω + 3i
ω − 3i ˆ+
f (ω) = −fˆ− (ω),
ω+i
ω−i
which can be rearranged as
ω + i ˆ−
ω − 3i
+
ˆ
=−
f (ω)
f (ω) = E(ω),
ω−i
ω + 3i
{z
} |
{z
}
|
analytic for Im(ω)≤0
analytic for Im(ω)≥0
where E(ω) is an entire function. Also, since fˆ− (ω) = O( ω1 ), so is E(ω). By Liouville’s
theorem, E(ω) = 0, and it thereby follows that fˆ+ (ω) = 0 and that f+ (x) = 0 is the
unique solution.
The natures of the solutions in Examples 10.10 and 10.11 are fundamentally different,
although only a sign has been changed. We therefore consider the integral equation
ˆ ∞
f (x) = λ
e−|x−t| f (t) dt,
x ≥ 0.
(10.15)
0
The Wiener–Hopf formulation takes the form
2
ω + 1 − 2λ ˆ+
f (ω) = −fˆ− (ω).
ω2 + 1
286
Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
What made the difference between the previous cases of λ = −4 and λ = +4 was
whether the roots of ω 2 + 9 = 0 and ω 2 − 7 = 0, respectively, were on or off the real ωaxis. For general λ, the break point between these cases occurs at λ = 12 (for the solution
in this case, see Exercise 10.4.5). A more thorough discussion is presented in [22].
The differences in the structure of the solutions for varying λ can be formulated using
operator theory. Writing (10.15) as (I − λT ) f = 0, the key question becomes whether
(I − λT ) is singular or not. If written as T f = µf (with µ = 1/λ), the relation takes the
form of an eigenvalue problem for the integral operator T .
10.2 A brief primer on Riemann–Hilbert methods
10.2.1 Hilbert transform case
In this case, the real part of a function is given along the real axis, and the matching imaginary part is sought (or vice versa; the function is unique if we also know it to decay to zero
in the upper half-plane). By a conformal mapping, we can change the real axis to the unit
circle (or any other closed curve) C. For example,
w=−
z−i
z+i
brings the upper half-plane to the inside of the unit circle (with z = 0 → w = 1 and
z = ∞ → w = −1). Knowing Ref (z) along the real z-axis means now knowing it around
the unit circle in the w-plane. This perspective provides an additional opportunity to find
the matching imaginary part. Calling the transformed plane z (rather than w), andthe
function
f (z), we can Fourier expand around the periphery: Ref eiθ =
P∞in this z-planeP
∞
a0 + k=1 aP
k cos kθ +
k=1 bk sin kθ, and we note that this matches the Taylor expansion
∞
f (z) = a0 + k=1 (ak − ibk ) z k . From this follows not only the corresponding imaginary
part, but also f (z) throughout the interior of the unit circle.
10.2.2 Riemann’s case
We next separate f = u + iv, with u and v real, and denote the independent variable t to
remind ourselves that we will at first consider it only on the periphery of the unit circle.
One of many fundamental issues Riemann pioneered in his thesis was to find both u(t) and
v(t) when the linear relation
α(t) · u(t) + β(t) · v(t) = γ(t)
(10.16)
holds, with α(t), β(t), γ(t) given. The Hilbert transform case above corresponds to the
special case of α(t) ≡ 1 and β(t) ≡ 0, i.e.,
1 · u(t) + 0 · v(t) = γ(t).
(10.17)
It is customary at this point to call the desired function f (z) = Φ+ (z), with the “+”
signifying arguments (i.e., z-values) inside the unit circle C. The function
1
Φ− (z) = Φ+
z
10.2. A brief primer on Riemann–Hilbert methods
287
will then be defined outside C and is also analytic (cf. the last case in Section 2.1.2). When
z → t on C, it holds that Φ− (t) = Φ+ (t) = u(t) + iv(t). Equation (10.16) then implies
α(t) + iβ(t)
α(t) − iβ(t)
Φ+ (t) +
Φ− (t) = γ(t).
2
2
(10.18)
Riemann’s problem has now been reformulated as a connection between two analytic functions, Φ+ (t) and Φ− (t), singularity-free inside and outside C, respectively.
The next secn
tion outlines strategy to solve for Φ(z), which we define as Φ(z) =
analytic everywhere but on C, along which there is a discontinuity.
Φ+ (z)
Φ− (z)
inside C,
outside C,
10.2.3 More general Riemann–Hilbert problems
A slightly more general version of (10.18) is to find analytic functions Φ+ (t) and Φ− (t),
singularity-free in their respective domains, that along C (which we still take as the unit
circle) are related by
Φ+ (t) = g(t)Φ− (t) + f (t).
(10.19)
Here, g(t) and f (t) are given, complex-valued functions (required to be “Hölder continuous,” a much weaker condition than being differentiable). Finding solutions to (10.19)
becomes a two-step process, where we first consider the homogeneous case with f (t) ≡ 0.
In Riemann–Hilbert problems, the contour C need not be closed. The Wiener–Hopf
method (in Section 10.1) focuses on the issues that arise near the end points of an open
contour.
Homogeneous Riemann–Hilbert problem
Assuming g(t) to be nonzero, we look for a nonvanishing function L(z) that satisfies the
simplified problem
L+ (t) = g(t)L− (t).
(10.20)
log L+ (t) − log L− (t) = log g(t).
(10.21)
Taking the logarithm gives
From the discussion of principal value integrals in Section 5.1.4 illustrated in Figure 5.18,
we know how to construct such a function:
1
log L(z) =
2πi
ˆ
C
log g(t)
dt
t−z
(10.22)
(with L(z) indeed being nonvanishing, since log L(z) becomes finite). From this L(z), we
read off L+ (z) and L− (z).
One issue that requires some care is that g(t) might have encircled the origin when t
went around C, making log g(t) not return to its initial value. If log g(t) increased by 2πin
(n integer), we can instead of (10.20) solve N+ (t) = t−n g(t) N− (t) (which does not have
288
Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
this issue) and then obtain our intended L(z) as
N (z)
inside C,
L(z) =
z −n N (z) outside C.
Inhomogeneous Riemann–Hilbert problem
Wanting to find a solution to (10.19), we start by solving (10.20) with the same function
g(t). Eliminating g(t) between these two equations gives
Φ+ (t) Φ− (t)
f (t)
−
=
.
L+ (t)
L− (t)
L+ (t)
(10.23)
Writing M (z) = Φ(z)/L(z), this function jumps with the amount Lf+(t)
(t) across C (known,
since we know both f (t) and L+ (t)). The principal value approach provides again a solution:
ˆ
f (t)/L+ (t)
1
dt.
(10.24)
M (z) =
2πi C
t−z
With L(z) known, the solution Φ(z) to the inhomogeneous Riemann–Hilbert problem
(10.19) follows, as Φ(z) = M (z)L(z).
10.3 Supplementary materials
Proof of Theorem 10.6. |f+ (x)| < Ceαx as x → ∞. Therefore,
ˆ ∞
C
eαx e−ixω dx
|fˆ+ (ω)| < √
2π 0
ˆ ∞
C
√
<
eαx |e−ixω |dx
2π 0
ˆ ∞
C
eαx ex Im(ω) dx
=√
2π 0
−C
.
=√
2π(Im(ω) + α)
Thus the integral converges absolutely for Im(ω) < −α. The arguments for |fˆ− (ω)|
are similar.
Proof of Theorem 10.7. The positive half-line Fourier transform is given by fˆ+ (ω) =
´∞
√1
f+ (x)e−iωx dx. Let ω = µ + iη. fˆ+ (ω) is analytic for η < −α. Then
2π 0
ˆ ∞
1
fˆ+ (ω) = √
(eηx f+ (x)) e−iµx dx
2π 0
ˆ ∞
1
=√
(eηx f+ (x)) e−iµx dx.
(10.25)
2π −∞
One can now use the standard inversion formula and obtain
ˆ ∞
1
eηx f+ (x) = √
fˆ+ (ω)eiµx dµ.
2π −∞
10.4. Exercises
289
´∞
Multiplying through by e−ηx , f+ (x) = √12π −∞ fˆ+ (ω)ei(µ+iη)x dµ. A substitution back
to ω = µ + iη gives
ˆ ∞+iη
1
f+ (x) = √
fˆ+ (ω)eiωx dω,
2π −∞+iη
where η < −α. The proof for the inversion formula of f− (x) follows the same arguments.
10.4 Exercises
Exercise 10.4.1. Equation (9.34) provides a way to compute f + (z) and f − (z). Show that
these formulas, which make use of Fourier transforms, recover the same functions f + (z)
and f − (z), as will formulas (5.9) and (5.10).
Exercise 10.4.2. Use (9.34) to show that f + (z) and f − (z) decay and are singularity-free
in the lower and upper half complex planes, respectively.
Exercise 10.4.3. Prove Theorem 4.13.
Exercise 10.4.4. Using the Wiener–Hopf technique, solve for f (x) the ODE
00
f (x) + a2 f (x) = 0, x > 0,
f 00 (x) + b2 f (x) = 0, x < 0,
(10.26)
for Im a > 0 and Im b > 0, with the boundary and far-field conditions f (0+ )−f (0− ) = 0,
f 0 (0+ ) − f 0 (0− ) = 1, and |f (x)| → 0 as x → ±∞.
Exercise 10.4.5. Derive for (10.15) in the case of λ =
C (1 + x).
Exercise 10.4.6. Solve the following PDE:
2
∂2
∂
+ 2 u + λ2 u = 0,
∂x2
∂y
1
2
the general solution f+ (x) =
−∞ < x < ∞, 0 < y < ∞,
with Im λ = , for a small and positive real number, also such that uy (x, 0) = 0 on y =
0, x < 0, and u(x, 0) = cos x on y = 0, x ≥ 0, and finally with limy→∞ |u(x, y)| = 0.
Exercise 10.4.7. Solve the following PDE:
2
∂
∂2
+
u − u = 0,
∂x2
∂y 2
−∞ < x < ∞, 0 < y < ∞,
such that uy (x, 0) = 0 on y = 0, x < 0, and u(x, 0) = 1 on y = 0, x ≥ 0.
Exercise 10.4.8. Solve the following PDE:
2
∂
∂2
∂
+ 2 u−
u = 0,
∂x2
∂y
∂y
−∞ < x < ∞, 0 < y < ∞,
such that uy (x, 0) = 1 on y = 0, x < 0, and f (x, 0) = e−x on y = 0, x ≥ 0.
290
Chapter 10. Wiener–Hopf and Riemann–Hilbert Methods
Exercise 10.4.9. Given the
part of a function
Preal
P∞ that is around the periphery of a
∞
iθ
circle,
Ref
e
=
a
+
a
cos
kθ
+
0
k
k=1
k=1 bk sin kθ, show that f (z) = a0 +
P∞
k
(a
−
ib
)
z
.
k
k
k=1
Exercise 10.4.10. The Sokhotski–Plemelj theorem states that, for a given simple closed
¸ φ(t)
1
dt, then Φ+ (z) =
curve C and an analytic function φ(z) on C, if Φ(z) = 2πi
C t−z
ffl φ(t)
ffl φ(t)
1
1
1
1
2πi C t−z dt + 2 φ(z) and Φ− (z) = 2πi C t−z dt − 2 φ(z). Prove this theorem.
Exercise 10.4.11. Given a closed contour C, find an analytic function Φ(z) that satisfies
the relation Φ+ (t) − Φ− (t) = 1 along C. Is the solution unique?
Exercise 10.4.12. Given a closed contour C, find an analytic function L(z) that satisfies
the relation L+ (t) = ef (z) L− (t) along C. Is the solution unique?
Chapter 11
Special Functions
Defined by ODEs
One of the core resources of applied mathematics and theoretical physics is their collection
of standard and special functions. Many of these “special functions” can be introduced as
solutions to linear ordinary differential equations (ODEs) with variable coefficients, with
some examples given below in Sections 11.1–11.3. A typical form for a linear ODE in this
context is
u00 (z) + p(z) u0 (z) + q(z) u(z) = 0.
(11.1)
Already if p(z) and q(z) are linear functions in z, it is rare that solutions u(z) can be found
in closed form using only the standard functions from calculus. As in the previous chapters,
our focus below is on analytic function aspects of the topic, thus also including a discussion
of the nonlinear Painlevé equations in Section 11.5.
Special functions that arise from ODEs (either linear or nonlinear) are typically analytic,85 as follows from a very general theorem due to Cauchy.
Theorem 11.1. Let u(z) satisfy an ODE of the form
dn u
=F
dz n
u,
du
dn−1 u
, . . . , n−1 ; z ,
dz
dz
(11.2)
n−1
d
u
where F is analytic in each of its arguments. For any initial conditions on u, du
dz , . . . , dz n−1
at some point z0 , the solution to (11.2) then becomes an analytic function in some neighborhood of z0 .
It is also common that there is a factor multiplying the highest derivative term, such as
z 2 for Bessel’s equation (11.6) and z(1 − z) for the hypergeometric ODE (11.15). Zeros
of such factors create singularities at which Taylor expansions for solutions typically need
to be replaced by what are known as Frobenius expansions. If the ODE near such a point
z = z0 can be cast in the form
u00 (z) +
85 Recalling
p(z) 0
q(z)
u (z) +
u(z) = 0,
z − z0
(z − z0 )2
that our definition of analytic does not exclude singularities or restrictions for domains of validity.
291
292
Chapter 11. Special Functions Defined by ODEs
with p(z) and q(z) locally analytic, the two solutions will be of the form
Taylor series
α
u(z) = (z − z0 ) ·
,
around z = z0
where α is a root of α2 + (p(z0 ) − 1) α + q(z0 ) = 0 (with a slightly modified rule for the
second solution in case of double roots). The method of undetermined coefficients (Section
2.3.2) will give the Taylor expansions.
Second order linear ODEs with one or several singular points also show up in applications far more than one might at first have expected. The analytical theory surrounding
their solutions is very extensive, and we will make no attempt at all here to survey this
topic, especially since there are several excellent books that provide comprehensive treatments, e.g., [4, 25, 27, 30]. We will here limit ourselves to give some illustrative examples
of how insights about analytic functions can be brought to bear on the subject of ODEs.
11.1 Airy’s equation
11.1.1 Real-valued independent variable
The constant coefficient ODE y 00 √− αy = 0 has
√ for α > 0 exponentially growing or
+ αx
− αx
decaying solutions
y(x)
=
C
e
+
C
e
and for α < 0 oscillatory solutions
1
2
√
√
y(x) = C1 cos( −α x) + C2 sin( −α x). If we replace the constant α with the independent variable x, we arrive at Airy’s equation
y 00 (x) − x y(x) = 0.
(11.3)
Unsurprisingly, its solutions are of exponential-like character for x positive and oscillatory
for x negative. This equation is of great importance in many applications, since it forms the
simplest description of (and leading order approximation to) many phenomena that change
character between being oscillatory or not. Since it is a second order linear homogeneous
ODE, its general solution can be written as a linear combination of two independent ones,
traditionally called Ai(x) and Bi(x), respectively, looking as shown in Figure 11.1. In
order to understand this equation better, questions that need to be answered include the
following:
1. What is the unique ratio of initial conditions u0 (0)/u(0) that gives solutions that
decay for x → +∞?
1
Ai(x)
Bi(x)
0.5
0
-0.5
-10
-5
0
x
Figure 11.1. The Ai and Bi functions along the real axis.
5
11.2. Bessel functions
293
2. Just how fast does the Ai function decay and the Bi function grow for x → +∞
(presumably at some “superexponential” rates)?
3. Can we approximate the envelope and oscillation rates (including the phase angle)
for the oscillations as x → −∞?
An important first step is to express the Ai(x) and Bi(x) functions as integrals. For the
bounded Ai function, this can be done by a Fourier transform. Recalling Theorem 9.3, the
dŷ
= 0, which can be solved in closed form:
transform of (11.3) becomes (ik)2 ŷ(k) − i dk
´∞
3
3
i k /3
ŷ(k) = C e
. Inversion back to physical space gives y(x) = C −∞ ei k /3 ei kx dk,
which can be rearranged into
3
ˆ ∞
k
y(x) = 2C
cos
+ kx dk.
(11.4)
3
0
However, this integral is quite impractical as it stands, being rapidly oscillatory and not absolutely convergent. It also only represents the bounded Ai function and not the unbounded
Bi function. As in so many other cases, the simplest path towards progress turns out to go
via the complex plane.
11.1.2 Complex-valued independent variable
The complex counterpart to (11.3) becomes
u00 (z) − z u(z) = 0.
(11.5)
If we are given initial conditions for u(0) and u0 (0), the method of unknown coefficients
(Section 2.3.2) will recursively provide any number of coefficients in the everywhere convergent Taylor expansion of u(z). However, Taylor expansions can rarely give any insights
about functional behaviors at large distances. Figures 11.2–11.4 show that the two independent solutions Ai(z) and Bi(z) then take on very distinctive features. Techniques
to understand these will be described in Section 11.4 and Chapter 12. The first of these
sections describes a more general approach to obtain integral relations than the Fourier
transform based one that we used above, and Chapter 12 discusses asymptotic analysis for
exploring far-field behaviors. For both of these tasks, the extension to complex variables
is essential. Following these discussions, we will then return to the Airy functions as an
example of applying these methods.
11.2 Bessel functions
Bessel’s equation
d2 y
dy
+ (t2 − n2 ) y = 0
(11.6)
+t
dt2
dt
is another example of a variable coefficient linear second order ODE that arises in many
contexts. One of these is the separation of variables when solving Laplace’s equation in
cylindrical coordinates. In that case, the values for n will be integers, and we focus here
first on that case. Being a linear second order ODE, its general solution can be written
as a combination of two independent ones, typically denoted by Jn (t) and Yn (t). These
functions were first systematically explored around 1817 by the German astronomer Bessel
t2
294
Chapter 11. Special Functions Defined by ODEs
(a) Real part of Ai(z).
(b) Imaginary part of Ai(z).
Figure 11.2. Real and imaginary parts of Ai(z).
(in the context of investigating Kepler’s equations of planetary motion). Since they arise in
many situations, there are unsurprisingly also many different ways to introduce them. We
will proceed by considering the analytic function
1
f (z, t) = et (z− z )/2 ,
(11.7)
where t is some parameter.
11.2.1 Differential equation and series expansion
The function f (z, t) has an essential singularity at z = 0. Its Laurent expansion will
have coefficients that depend on the parameter t according to the generating function
11.2. Bessel functions
295
(a) Real part of Bi(z).
(b) Imaginary part of Bi(z).
Figure 11.3. Real and imaginary parts of Bi(z).
(z-transform)
1
f (z, t) = et (z− z )/2 =
∞
X
Jn (t) z n .
(11.8)
n=−∞
This notation is justified, as we will see next, by the fact that these functions Jn (t) indeed
satisfy (11.6). Direct differentiation of (11.7) shows that f (z, t) satisfies
t2
∂2f
∂f
∂
+t
+ t2 f = z
∂t2
∂t
∂z
∂f
z
.
∂z
(11.9)
296
Chapter 11. Special Functions Defined by ODEs
(a) Abs (Ai(z)).
(b) Abs (Bi(z)).
Figure 11.4. The magnitude and phase of the Ai(z) and Bi(z) functions.
If we here replace f by the RHS of (11.8) and collect together all terms according to their
powers of z, we get
∞ X
d Jn (t)
d2 Jn (t)
2
2
+
(t
−
n
)J
(t)
z n = 0.
+
t
t2
n
2
dt
dt
n=−∞
For this to hold identically in z (recalling that a Laurent expansion has unique coefficients),
all the coefficients must vanish, and it is therefore established that the functions Jn (t), as
defined by (11.8), indeed satisfy (11.6). We can also see directly from (11.7) that
J−n (t) = (−1)n Jn (t).
(11.10)
Either as outlined in Exercise 11.6.3 or by the same procedure as used in Example 4.25
applied to (11.8), we obtain the Laurent coefficients (i.e., the Bessel functions Jn (t)) in the
form of a Taylor expansions in t:
2k+n
∞
X
(−1)k
t
Jn (t) =
.
(11.11)
k! (k + n)! 2
k=0
11.2. Bessel functions
297
Figure 11.5. Front to back, the Jn (t) Bessel functions for n = 0, 1, 2, . . . , 10 and t real.
This shows that Jn (t) are analytic and entire functions in t. Figure 11.5 shows what they
look like, along the real t-axis, for different values of n, and Figure 11.6 displays real
parts and magnitudes for J0 (z) and J10 (z). The transitions seen in Figure 11.5 between
oscillatory and nonoscillatory character around t = ±n are due to the sign changes there
for the factor (t2 − n2 ) in (11.6). Another reason for nonoscillatory behavior around the
origin as n increases is that the leading Taylor term in (11.11) is O (tn ) . The color pattern
around the origin in Figure 11.6(d) confirms the origin for J10 (z) to be a zero of order 10.
11.2.2 Integral representations of Jn (t)
With (11.8) as the starting point, we can derive numerous formulas for the Jn (t) Bessel
functions. For example, (4.13) tells us immediately that
1
Jn (t) =
2πi
ˆ
C
1
et (z− z )/2
dz,
z n+1
(11.12)
where C is any contour that goes around the origin once, in the positive (counterclockwise)
direction. Choosing C as the unit circle gives further integral representations, such as
Jn (t) =
1
2π
ˆ
π
e−i (nθ−t sin θ) dθ =
−π
1
π
ˆ
π
cos(nθ − t sin θ)dθ.
(11.13)
0
11.2.3 Other types of Bessel functions
Since the ODE (11.6) is linear and of second order, it should have two independent solutions. We found one of these in the form of the Jn (t) Bessel functions (also known as
Bessel functions of the first kind). For example, with variation of parameters, one can find
the second solution family, similarly denoted as Yn (t).86 These are quite different in character: no longer entire functions, but with a branch point at the origin. Figure 11.7 shows
the first few along the positive real axis. Many other generalizations are described in the
literature (but go beyond our present goals).
86 Known
as Bessel functions of the second kind or as Neumann functions.
298
Chapter 11. Special Functions Defined by ODEs
5
5
0
-20
0
-10
-5
-20
0
-10
10
0
10
10
5
10
5
x
0
0
-5
x
20
20
y
-10
(a) Re J0 (z)
-5
y
-10
(b) Abs J0 (z).
5
5
0
-20
0
-10
-5
-20
0
-10
10
0
10
10
5
10
5
x
0
0
-5
x
20
-10
20
y
(c) Re J10 (z)
-5
-10
y
(d) Abs J10 (z).
Figure 11.6. Real parts and magnitudes/phase angles for J0 (z) and J10 (z).
If one lets n in (11.6) take noninteger values, (11.11) generalizes to
2k+n
∞
X
t
(−1)k
.
Jn (t) =
k! Γ(k + n + 1) 2
(11.14)
k=0
In this case, the linear dependence between Jn (t) and J−n (t) expressed in (11.10) gets broken, and these two (now multivalued) functions represent independent solutions to (11.6).
Figure 11.7. Front to back, the Yn (t) Bessel functions for n = 0, 1, 2, . . . , 10 and t real.
11.3. Hypergeometric functions
299
The case of an n half-integer leads to closed-form expressions, for example
2 1/2
cos t,
J−1/2 (t) = πt
2 1/2
J+1/2 (t) = πt
sin t,
2 1/2
J+3/2 (t) = − πt
cos t − 1t sin t ,
2 1/2
Y−1/2 (t) = πt
sin t,
2 1/2
Y+1/2 (t) = − πt
cos t,
2 1/2
Y+3/2 (t) = − πt
sin + 1t cos t .
Another variation that arises quite frequently in applications is a sign change in front
of the n2 term in (11.6), then giving as solutions what are known as the modified Bessel
functions Kn (t) and In (t).
11.3 Hypergeometric functions
This is a wide-ranging family of analytic functions, which not only satisfy certain linear
variable coefficient ODEs (this is how we at first will introduce them), but which have a
tendency to “pop up” in lots of unexpected contexts as well (with Example 11.3 just one
such illustration). There exist vast amounts of identities and formulas associated with these
functions, which we here will make no attempt to summarize. Symbolic algebra systems
quite often formulate their answers in terms of various hypergeometric functions, and these
systems are thus often the most practical means for further manipulating such expressions,
maybe allowing transformations into other types of regular or special functions.
11.3.1 Introduction of the 2 F1 function via a second order linear
ODE
Consider the ODE
z (1 − z)
d2 w
dw
+ {c − (a + b + 1) z}
− a b w = 0,
dz 2
dz
(11.15)
where a, b, c are scalar parameters, and w = w(z) is the function for which to solve. A natural first attempt to approximate solutions to (11.15) would be to substitute a Taylor series
for w(z) into (11.15) and then equate coefficients. P
However, it turns out to be somewhat
∞
better in this case to instead start with w(z) = z α n=0 an z n (with a0 6= 0), leading to
the indicial equation
α (α + c − 1) = 0,
giving two possible values for α, namely α = 0 and α = 1−c, together with the coefficient
recursion (α + n)(α + c − 1 + n)an = (α + a + n − 1)(α + b + n − 1)an−1 . The choice
α = 0 leads to the Taylor expansion
2 F1 (a, b; c; z)
a·b
a(a + 1) · b(b + 1) 2
z+
z
c · 1!
c(c + 1) · 2!
a(a + 1)(a + 2) · b(b + 1)(b + 2) 3
+
z + ··· ,
c(c + 1)(c + 2) · 3!
=1+
and the two independent solutions to (11.15) can now be given as
2 F1 (a, b; c; z)
z 1−c 2 F1 (1 + a − c, 1 + b − c; 2 − c; z).
(11.16)
300
Chapter 11. Special Functions Defined by ODEs
Solutions to many higher order ODEs can be expressed similarly if one generalizes
(11.16) to have p and q parameters, respectively, in the numerator and denominator of the
expansion:
a1 · a2 · · · · · ap
z
b1 · b2 · · · · · bq · 1!
a1 (a1 − 1) · a2 (a2 − 1) · · · · · ap (ap − 1) 2
+
z
(11.17)
b1 (b1 − 1) · b2 (b2 − 1) · · · · · bq (bq − 1) · 2!
a1 (a1 − 1)(a1 − 2) · a2 (a2 − 1)(a2 − 2) · · · · · ap (ap − 1)(ap − 2)
+
z3 + · · · .
b1 (b1 − 1) · (b1 − 2) · b2 (b2 − 1) · (b2 − 2) · · · · · bq (bq − 1)(bq − 2) · 2!
p Fq (a1 , . . . , ap ; b1 , . . . , bq ; z)
=1+
11.3.2 Examples of elementary functions expressed as
hypergeometric functions
A large number of elementary analytic functions have Taylor expansions that can be expressed in the form (11.17) with suitable fixed-parameter choices. Just a few examples
suffice to illustrate this:
log(1 + z) = z 2 F1 (1, 1; 2; −z),
3
1
, 1; ; −z 2 ,
arctan z = z 2 F1
2
2
1 1 3 2
, ; ;z ,
arcsin z = z 2 F1
2 2 2
(1 − z)α = 1 F0 (α; ; z),
3 z2
sin z = z 0 F1 ; ; −
,
2
4
1 z2
cos z = 0 F1 ; ; −
,
2
4
ez = 0 F0 (; ; z).
The list extends to Bessel and Airy functions, etc. One important aspect of the hypergeometric formalism is that its wide range of identities unifies features of many otherwise
seemingly diverse functions. Another is that it encompasses far more functions than have
been assigned commonly used acronyms.
11.3.3 Analytic continuation of hypergeometric functions
Omitting special cases when the series expansions truncate to a finite number of terms (for
example, if either a or b is zero or negative integers in (11.16)), the expansion (11.17)
shows the radius of convergence to be R = ∞ if p ≤ q, R = 1 if p = q + 1, and R = 0 for
p ≥ q + 2. In applications, the case of p = q + 1 tends to arise particularly frequently, and
especially p = 2, q = 1.
Theorem 11.2. The 2 F1 hypergeometric function satisfies
ˆ 1
Γ(c)
F
(a,
b;
c;
z)
=
tb−1 (1 − t)c−b−1 (1 − zt)−a dt.
2 1
Γ(b)Γ(c − b) 0
(11.18)
11.3. Hypergeometric functions
301
A proof is outlined in Exercise 11.6.6.
This integral representation (11.18) shows that, in general, z = 1 is a branch point for
2 F1 , and that it is its only singularity in the complex plane. Since the functional behavior
around the origin often is of interest, it is conventional to place a branch cut from +1 to
+∞ (following the positive real axis). A small remaining “snag” is that the integral, as it
stands, converges only when Re c > Re b > 0. Changing the integration contour to one of
Pochhammer type (cf. Example 5.33) overcomes this restriction.
Numerous important integral formulas follow from (11.18). As an example, we can
generalize the beta function
ˆ
1
tp−1 (1 − t)q−1 dt =
B(p, q) =
0
Γ(p)Γ(q)
,
Γ(p + q)
introduced in Section 6.1.2, to the incomplete beta function, for which the upper integration
limit is z rather than 1:
ˆ z
zp
(11.19)
B(p, q; z) =
tp−1 (1 − t)q−1 dt =
2 F1 (p, 1 − q; p + 1; z).
p
0
Example 11.3. Conformally map the inside of the unit circle to the inside of a regular
polygon with n corners.
We noted at the start of Section 8.4 that the Schwarz–Christoffel formula, when mapping from the unit circle, takes an almost identical form to when mapping from the upper
half-plane. For the present mapping case, the Schwarz–Christoffel mapping function turns
out to become
ˆ z
1 n
1
1
2 n
dt
1 2
wn (z) =
=
z
F
,
;
1
+
;
z
=
B
,
1
−
;
z
.
2 1
n 2/n
n n
n
n
n
n
0 (1 − t )
(11.20)
Figure 11.8(a) shows a polar grid in the z-plane, and parts (b)–(d) the resulting curve sets
in the w-plane for the cases of n = 3, 4, 6, respectively.
Another family of integral representations which also provides analytic continuations
of hypergeometric functions was devised by Barnes around 1900. The following example
can also be seen as an inverse Mellin transform.
Theorem 11.4. The 2 F1 hypergeometric function satisfies
2 F1 (a, b; c; z) =
1
2πi
ˆ
+i∞
−i∞
Γ(a + s)Γ(b + s)Γ(−s)
(−z)s ds ,
Γ(c − s)
(11.21)
where the contour from −i∞ to +i∞ goes around the three infinite rows of poles in the
way illustrated in Figure 11.9.
Because the gamma function is never zero, the denominator Γ(c − s) in (11.21) will
not cause any singularities.
302
Chapter 11. Special Functions Defined by ODEs
1
1.5
0.8
1
0.6
0.4
0.5
0.2
0
0
-0.2
-0.5
-0.4
-0.6
-1
-0.8
-1.5
-1
-1
-0.5
0
0.5
1
-1
(a) Initial domain in the z-plane.
-0.5
0
0.5
1
1.5
2
(b) Mapping given by w = w3 (z).
0.8
1
0.6
0.4
0.5
0.2
0
0
-0.2
-0.5
-0.4
-0.6
-1
-0.8
-1.5
-1
-0.5
0
0.5
1
(c) Mapping given by w = w4 (z).
1.5
-1
-0.5
0
0.5
1
(d) Mapping given by w = w6 (z).
Figure 11.8. The unit circle and the maps provided by the function wn (z) given in (11.20).
From the integral formulations of the 2 F1 function, one can derive a large number of
functional equations, such as
z
−a
2 F1 (a, b; c; z) = (1 − z)
2 F1 a, c − b; c;
z−1
z
−b
= (1 − z) 2 F1 c − a, b; c;
z−1
= (1 − z)c−a−b 2 F1 (c − a, c − b; c; z)
and
(−z)−a
1
sin(π(b − a))
2 F1 (a, b; c; z) =
2 F1 a, a − c + 1, a − b + 1;
π
Γ(b)Γ(c − a)
z
−b
(−z)
1
−
.
2 F1 b, b − c + 1, b − a + 1;
Γ(a)Γ(c − b)
z
When there are particular relations between a, b, c the list of functional relations becomes
nearly (or even literally) endless. Many of these relations provide analytic continuations,
11.4. Converting linear ODEs to integrals
303
6 ↑
←
4
poles of Γ(a+s)
↑
2
→
poles of Γ(-s)
-6
-4
-2
2
poles of Γ(b+s)
4
6
←
-2
→
-4
↑
-6
Figure 11.9. Schematic illustration of integration path for Barnes’s integral (11.21).
allowing values outside the unit circle to be obtained from values inside the unit circle,
where the Taylor series (11.16) converges.
11.3.4 An illustration of the 2 F1 hypergeometric function
Since this function has three scalar parameters a, b, c apart from the independent variable
z, we choose here just an arbitrary single case for illustration, a = 1, b = 2, c = 3, which
is typical in its general character. Figure 11.10 shows for this case a real part that peaks to
infinity at the branch point z = 1, while the imaginary part there remains finite.
11.4 Converting linear ODEs to integrals
In each of the Sections 11.1–11.3, we somehow converted the original ODEs to integrals.
This section describes a couple of systematic ways to do this.
11.4.1 Fourier–Laplace method
Given a variable coefficient linear ODE, the general idea in this approach is to look for
solutions of the form
ˆ
u(z) =
F (t) ez t dt,
(11.22)
C
304
Chapter 11. Special Functions Defined by ODEs
10
5
4
2
0
0
-5
-4
-2
-2
0
2
4
x
y
-4
(a) Re 2 F1 (1, 2, 3; z).
6
4
2
0
-2
4
-4
-6
-4
2
0
-2
-2
0
2
x
4
-4
y
(b) Im 2 F1 (1, 2, 3; z).
Figure 11.10. The real and the imaginary parts of 2 F1 (1, 2, 3; z).
where both the integration contour C and the function F (t) remain to be determined.87 The
following couple of examples explain the typical steps in the procedure.
Example 11.5. Determine an integral representation for the solutions to the Airy equation
(11.5).
´
Substituting (11.22) into the Airy equation (11.5) gives C (t2 − z)F (t) ez t dt = 0. We
next want to manipulate this expression so that z will appear in the exponent only. Noting
87 Since C can be a general contour in the complex plane, (11.22) generalizes both the Fourier and the Laplace
transforms. In some literature, the exponential is written as eizt rather than as ezt here (making no conceptual
difference in the method).
11.4. Converting linear ODEs to integrals
that z ez t =
d
zt
dt (e )
305
and then integrating by parts gives
ˆ dF
2
+ t F ez t dt − F (t) ez t C = 0.
dt
C
2
zt
This will be true for all z if it holds that (i) dF
takes the
dt + t F = 0 and (ii) F (t) e
same value at both ends of the contour C. Starting with item (i), the general solution of
dF
2
−t3 /3
(where c is some constant which can be omitted).
dt + t F = 0 is F (t) = c e
3
Regarding item (ii) we thus need to study the expression G(t) = F (t) ez t = e−t /3+z t
in a complex t-plane (temporarily viewing z as a complex constant). G(t) is clearly an
entire function of t, which far out in the t-plane rapidly decreases to zero within the three
7π
3π
sectors (valleys) (V1 ) − π6 < arg t < π6 , (V2 ) π2 < arg t < 5π
6 , (V3 ) 6 < arg t < 2 ,
and with high “mountain ridges” in between. Letting C form a closed loop will make
[F (t) ez t ]C vanish, but this is useless, since u(z) (as defined in (11.22)) will then also
vanish, resulting only in the trivial u(z) = 0 solution. A much better idea in this case is
to run the contour C from one valley to another. Figure 11.11 illustrates the magnitude of
3
the integrand, |e−t /3+z t |, in the case of z = 1 (with this specific z-value only causing
a slight deformation of the surface near the origin in the complex t-plane). Denoting by
C1 , C2 , C3 contours from valley V2 to V3 , V2 to V1 , and V1 to V3 , respectively, we obtain
the two standard Airy functions as
ˆ
ˆ
ˆ 3
1
1
−t3 /3+z t
Ai(z) =
e
dt, Bi(z) = −
−
e−t /3+z t dt. (11.23)
2πi C1
2π
C2
C3
5
0
V2
5
4
V1
3
2
1
0
V3
-1
-2
y
C
-3
1
-4
-5
-5
-4
-3
3
-2
-1
0
1
2
3
4
5
x
Figure 11.11. The function e−t /3+z t , displayed in a complex t-plane in the case of
z = 1. As usual, the real axis is marked by a thick red line. The green curve shows the lowest
possible choice for the contour C1 . While any contour from V2 to V3 would give the correct value for
Ai(z), this specific type of contour will play a key role in the analysis in Section 12.4.
306
Chapter 11. Special Functions Defined by ODEs
The details in these definitions are motivated by (i) ensuring
that the two solutions are in´∞
dependent and real along the real axis, (ii) making −∞ Ai(x)dx = 1, and (iii) making
the two functions share the same envelope for their oscillations as x → −∞ (further analyzed in Section 12.4.3). We can also note that deforming the contour C1 to run along the
imaginary axis (making its convergence properties much worse) produces (11.4).
Example 11.6. Find an integral representation of a solution to Bessel’s equation (11.6) in
the case of n = 0, i.e., satisfying
d2 u du
(11.24)
+
+zu = 0 .
dz 2
dz
´
Substituting (11.22) into (11.24) gives C F (t) zt2 + t + z ezt dt = 0. Integrating by
´
d
parts (using again z ez t = dt
(ez t ) ) gives C F 0 (t) (1 + t2 ) + F (t) t ezt dt = 0, with
1
.88
the ODE F 0 (t) (1 + t2 ) + F (t) t = 0 that is solved by (any multiple of) F (t) = √1+t
2
Hence,
ˆ
ezt
√
u(z) =
dt
(11.25)
1 + t2
C
z
solves (11.24). A suitable contour C will be one that encircles a branch cut placed along
the imaginary axis between t = ±i.
If we want toreconcile (11.25) with (11.12), it is natural to perform
the change vari√
able t = τ − τ1 21 (to match the form of the exponent). Then 1 + t2 = τ + τ1 12 ,
´
z(τ −1/τ )/2
dt = τ1 τ + τ1 12 dτ , and (11.25) becomes u(z) = C 0 e τ
dτ , with a path C 0 that
encircles the origin. This matches the n = 0 case of (11.12).
11.4.2 Euler’s method
Several options are available for choosing the integrand in (11.22). In Euler’s version, we
start instead with
ˆ
F (t) (t − z)µ dt
u(z) =
(11.26)
C
and proceed then in the same way: substitute into the ODE, integrate by parts, choose F (t),
path C, and now also the parameter µ so that the ODE becomes satisfied for all values of
z. The algebra becomes particularly nice if we can let C be a small circle around z, and
use negative integer values for µ, since (11.26) then matches (4.7).
Example 11.7. Find for each n = 0, 1, 2, . . . a solution un (z) to
d2 un (z)
dun (z)
+ 2(n + 1)un (z) = 0
+ 2z
dz 2
dz
(recalling that, if we have one solution, a linearly independent second one can be obtained
by variation of parameters).
88 The
ODE is separable. Writing it as
grating gives the result.
F 0 (t)
F (t)
t
= − 1+t
2 , i.e.,
d
dt
d
log F (t) = − 21 dt
log(1 + t2 ), and inte-
11.5. The Painlevé equations
307
2
d
d
d
µ
We substitute (11.26) into the ODE, note that dz
(t−z)µ = − dt
(t−z)µ , dz
=
2 (t−z)
2
d
µ
dt2 (t − z) , and then integrate by parts to obtain derivatives of F (t) instead of derivatives
of (t − z)µ (intending to use a closed loop and integer values for µ, what “comes out” when
integrating by parts cancels). This leads to
ˆ
(F 00 (t) + 2zF 0 (t) + 2(n + 1)F (t)) (t − z)µ dt = 0.
C
The z-factor in front of F 0 (t) needs to be removed, and we can do that by writing z =
t − (t − z), and do one more integration by parts, to obtain
ˆ
(F 00 (t) + 2tF 0 (t) + 2(n + µ + 2)F (t)) (t − z)µ dt = 0.
C
Separately for each n, we now choose µ = −n − 1, making the ODE for F (t) in all cases
become
F 00 (t) + 2tF 0 (t) + 2F (t) = 0,
2
´
2
e−t
which is solved by F (t) = e−t .89 Therefore un (t) = C (t−z)
n+1 dt, which by (4.7) (and
ignoring an irrelevant multiplicative factor) can be written as
un (z) =
dn −z2
e , n = 0, 1, 2, 3, . . . .
dz n
11.5 The Painlevé equations
With often centuries of history behind them and, more recently, with the immense power
provided by modern computers, most of the special functions of applied mathematics have
by now become thoroughly explored and understood. The Painlevé equations and their
solutions, known as the Painlevé transcendents, form an exception in this regard. They
were first considered just over 100 years ago, initially from a very theoretical perspective.
In spite of proving to be particularly difficult to analyze theoretically and also unusually
challenging to compute numerically (in both cases partly due to their commonly occurring
vast pole fields in the complex plane), their range of applications kept expanding, to the
point that they now are considered among the most important types of special functions.
Recent numerical advances have finally made them much more accessible than previously.
Based on a level of intuition that is hard to imagine, Paul Painlevé (later Prime Minister
2
of France) and some collaborators of his decided to consider all ODEs of the form ddzu2 =
F z, u, du
dz , where F is an arbitrary rational function of its three arguments z, u, and
du
(algebraic
function such that both the numerator and the denominator are polynomials).
dz
Typically, solutions to nonlinear ODEs of such a general form would be expected to feature
movable branch points (i.e., branch points at locations that depend on the ODE’s two initial
conditions). Remarkably, only 50 genuinely different equations of this form are possible
if one imposes the Painlevé property: The only movable singularities in the complex plane
are either poles or essential singularities (i.e., no movable singularities are allowed that
give rise to multivaluedness). It further transpires that essential singularities cannot arise in
this context, so all movable singularities to these equations are in fact poles. Out of these
50 equations, it has also been discovered that the solution to 44 of them could be expressed
89 This
d
follows quickly if one happens to note that 2tF 0 (t) + 2F (t) = 2 dt
(t F (t)).
308
Chapter 11. Special Functions Defined by ODEs
in terms of elementary or traditional special functions. The remaining six equations are
now known as the Painlevé equations:
PI
d2 u
dz 2
= 6u2 + z,
PII
d2 u
dz 2
PIII
d2 u
dz 2
PIV
d2 u
dz 2
PV
d2 u
dz 2
PV I
d2 u
dz 2
= 2u3 + zu + α,
2 1 du αu2 +β
= u1 du
− z dz + z + γu3 + uδ ,
dz
1
du 2
= 2u
+ 32 u3 + 4zu2 + 2(z 2 − α)u + βu ,
dz
(u−1)2
γu
β
1
1
du 2
1 du
= 2u
+ u−1
−
+
αu
+
dz
z dz
z2
u + z +
1
1
1
1
1
du 2
du
−
= 12 u1 + u−1
+ u−z
+
+
dz
z
z−1
u−z
dz
γ(z−1)
δz(z−1)
u(u−1)(u−z)
βz
α + u + (u−1)2 + (u−z)2 .
+ z2 (z−1)2
δu(u+1)
u−1 ,
Apart from two initial conditions (ICs), the equations feature different numbers of additional free parameters, ranging from none for PI to four for PIII , PIV , and PV I . While
most of the analytic functions we have considered so far have not had any free parameters
(beyond the independent variable z, for example sin z or Γ(z)), some have involved one or
more parameters (for example the two periods ω1 and ω2 for the Weierstrass ℘(z)-function
and the three parameters for the 2 F1 hypergeometric function). Here (including the two
ICs), we have from two to six parameters. This obviously adds greatly to the challenge of
exploring their full solution spaces.
We will below limit ourselves to showing a few highlights of the analytic functions that
arise as solutions to the PI and the PII equations.
11.5.1 The PI equation
We noted in Chapter 7 (see (7.8)) that the Weierstrass ℘-function satisfies the nonlinear
ODE
1
(11.27)
℘00 (z) = 6℘(z)2 − g2 ,
2
where the parameter g2 could be chosen arbitrarily. For all ICs, solutions to (11.27)
are therefore doubly periodic functions, with double poles repeating completely regularly
across the full complex z-plane. The difference to the PI equation
u00 (z) = 6u(z)2 + z
(11.28)
is that, in the last term, the constant − 21 g2 has been replaced by the independent variable
z. By local Laurent expansions, one sees that all singularities to (11.28) are again double
poles of strength 1 and residue zero (cf. Exercise 11.6.8). Far away from the origin, z
becomes locally like a large constant, and the pole fields become locally similar to those
of (11.27). However, as z grows, they generally get denser further out. A main difference,
however, is caused by the fact that the variable changes u → ζ 3 u, z → ζz with ζ 5 = 1
leaves (11.28) again satisfied, causing pole fields far out to differ between five sectors with
boundaries between them following the directions arg z = 1π/5, 3π/5, 5π/5, 7π/5, 9π/5.
These sectors are visible in the PI solution pictures in Figure 11.12, which show the pole
locations in four illustrative cases.
11.5. The Painlevé equations
309
Figure 11.12. Four pole field plots for the PI equation, displayed over the region
[−50, 50] × [−50, 50] in the complex plane. (a) Case with the ICs u(0) = u0 (0) = 0; the sector boundaries are here narrow pole-free strips. (b) ICs u(0) = u0 (0) = 2; the pole field structures are far out similar to the previous case, but the transitions between the sectors have become
gradual instead of sharply defined. (c) The tritronquée solution; the unique case for which four
of the five sectors have become completely pole-free; its ICs are u(0) ≈ −0.1875543083404949,
u0 (0) ≈ 0.3049055602612289. (d) A solution with ICs very close to the tritronquée case (here
u(0) = −0.1875, u0 (0) = 0.3049). Little has changed in the right pole field, but a pair of additional pole fields have very rapidly moved in from the top left and bottom left, respectively. Reprinted
with permission of Elsevier from [19].
The NIST Handbook of Mathematical Functions (2010) [35] was written just before
Painlevé solutions could readily be computed across the complex plane [19, 14]. It contains in its Chapter 32, on the Painlevé equations, an illustration similar to Figure 11.13,
showing along the real axis how the solution changes when u(0) = 0 and u0 (0) is varied in
close vicinity of a critical value for which the solution is smooth and nonoscillatory along
the negative real axis. Corresponding illustrations over the complex plane are far more
310
Chapter 11. Special Functions Defined by ODEs
Figure 11.13. Illustration along the real axis of the PI solutions with u(0) = 0 and with
u0 (0) varied as shown in the figure. Reprinted with permission of Elsevier from [19].
revealing; cf. Figures 11.14(a)–(b) (with the matching real axis intervals marked by thick
black curves). For u0 (0) = 1.8518, the oscillations for x negative are seen to be caused
by a pole field that symmetrically surrounds the negative real axis, but without any poles
on the axis itself. As u0 (0) increases slightly, this left pole field recedes rapidly out to minus infinity, and then moves back in again, however shifted vertically so there now have
become an infinity of poles on the negative real axis. The right pole field has remained
essentially stationary.
Theoretical work on the PI equation has to a large extent been limited to asymptotic
analysis of solutions far out in the different sectors, and on connection formulas linking
these results between sectors. For solution features at finite distances, computational methods are nowadays the dominant source of information.
11.5.2 The PII equation
In contrast to the PI equation, the remaining Painlevé equations have a few instances of
closed-form solutions (still like drops in a bucket only, compared to their full solution
spaces). In the case of the PII equation, there are two types of closed-form solutions:
(i) For α integer: There is in this case one rational solution for each α-value:
α=0 ,
α=1 ,
α=2 ,
α=3 ,
u(z) = 0,
1
u(z) = − ,
z
4 − 2z 3
,
u(z) =
4z + z 4
u(z) =
..
.
3z 2 (160 + 8z 3 + z 6 )
,
320 − 24z 6 − z 9
11.5. The Painlevé equations
311
Figure 11.14. The magnitude of same solutions as the bottom and top cases in Figure
11.13, here displayed (without phase angle coloring) over the complex plane rather than only along
the real axis: (a) u(0) = 0, u0 (0) = 1.8518, (b) u(0) = 0, u0 (0) = 1.8519. The crossing of the
thick black lines marks the origin z = 0. Reprinted with permission of Elsevier from [19].
(ii) For α “half-integer”: After defining
−z
−z
θ
θ
Ai
Bi
φ(z) = cos
+
sin
2
2
21/3
21/3
and
Φ(z) = φ0 (z)/φ(z),
one can find90 the following solutions:
α=
1
2
,
u(z) = −Φ,
α=
3
2
,
u(z) =
2Φ3 + zΦ − 1
,
2Φ2 + z
α=
5
2
,
u(z) =
4zΦ4 + 6Φ3 + 4z 2 Φ2 + 3zΦ + z 3 − 1
,
(4Φ3 + 2zΦ − 1)(2Φ2 + z)
..
.
..
.
Letting the free parameter θ move through the period [0, 2π] of Φ(z) causes poles to move
as seen for the case of α = 25 in Figure 11.15. When θ = 0, all poles are located in a band
90 Both
sequences depend on finding a single closed-form solution uα (z) for some α-value (such as here
α = 0 and α = 12 ), and then repeatedly using the Bäcklund transformation uα+1 (z) = −uα (z) −
2α+1
, giving a solution for α + 1.
2u0 (z)+2u (z)2 +z
α
α
312
Chapter 11. Special Functions Defined by ODEs
Figure 11.15. The Airy family of PII solutions in the case of α = 52 . All poles to PII are
simple, with residues +1 or −1. In this and the next two figures, these are colored blue and yellow,
respectively. Reprinted with permission of Springer Nature from [20].
surrounding the positive real axis. As θ increases, a curved band of poles enters from the
left, becomes part of two different entirely symmetric solution configurations at θ = π3 and
θ = 5π
3 , and finally moves back out left towards minus infinity as θ approaches 2π, leaving
for θ = 2π the same pole distribution as for θ = 0. However, if one looks carefully at
the pole field along the positive real axis, one can see that this has moved steadily to the
left during this process. The band of poles that came in from the left picked up, during
the interactions, five poles from the field along the positive real axis, included then in the
curved band, and then brought theses out to minus infinity.
No closed-form solutions are known to the PII equation other than the two classes just
noted above, for α integer and for α half-integer, respectively.
We conclude our PII equation summary by showing two other types of solutions that
are particularly important in applications, much due to being pole-free along the entire
real axis (however, with no closed-form expressions available). Figure 11.16 shows the
Hastings–McLeod solutions. There are two types of these for each α-value (real) in the
range |α| < 12 ; otherwise there is a unique one. There also exists for |α| < 1 a oneparameter family of Ablowitz–Segur solutions. These are smooth and bounded along the
entire real axis, oscillatory to the left and decaying to the right. Figure 11.17 shows a
typical such solution in the case of α = 0. In both of these cases, we see in the figures
again how extensions to the complex plane reveal features that are not apparent from the
properties along the real axis.
11.5. The Painlevé equations
313
Figure 11.16. Examples of Hastings–McLeod solutions, shown along the real axis, and
their associated pole fields. Reprinted with permission of Springer Nature from [20].
Figure 11.17. Example of an Ablowitz–Segur solution in the case of α = 0. Reprinted with
permission of Springer Nature from [20].
314
Chapter 11. Special Functions Defined by ODEs
11.5.3 The remaining Painlevé equations
In contrast to the PI , PII , and PIV equations, solutions to the PIII , PV , and PV I equations
need not be single-valued. For example, in the PIII case, we can note the term − z1 du
dz in
its RHS. Had this been the only term in the RHS, u(z) = log z would have been a solution.
This multivaluedness does not contradict the Painlevé property, since the location of the
branch point is fixed (at z = 0) and is not “movable” with the choice of ICs. Analysis will
reveal that solutions to PIII for different ICs and parameter values (for α, β, γ, δ) can have
any number of Riemann sheets, ranging from one to infinity. With the larger number of free
parameters in the higher-numbered Painlevé equations, it becomes increasingly difficult to
survey their complete solution spaces (although there has been significant recent progress
in this).
We have noted several times earlier that what can be seen about an analytic function
along the real axis contains only a tiny fraction of the information that can be gained by
inspecting it (and utilizing its properties) over the complex plane. In the case of the multivalued PIII , PV , and PV I functions, intriguing phenomena can occur on different Riemann
sheets even when little that is noteworthy is apparent on the primary sheet.
11.6 Exercises
Exercise 11.6.1. Many special functions are defined by second order linear ODEs. If one
somehow can find one particular solution y1 (x) to
a(x)y 00 (x) + b(x)y 0 (x) + c(x)y(x) = 0,
(11.29)
one can then find the general solution to the inhomogeneous ODE
a(x)y 00 (x) + b(x)y 0 (x) + c(x)y(x) = d(x)
(11.30)
by reduction of order.91 Substituting y(x) = y1 (x)u(x) into (11.29) gives a linear ODE in
u0 (x) from which follows y2 (x), and the general solution to (11.29) as y(x) = c1 y1 (x) +
c2 y2 (x). To this we need to add any one particular solution of (11.30), which can be
obtained by substituting into it either y(x) = y1 (x)v(x) or y(x) = y2 (x)v(x).
Determine thus the general solution to
x2 y 00 (x) − 4x y 0 (x) + 6y(x) = x5 ex ,
having noted that the homogeneous part (RHS = 0) is solved by y(x) = x2 .
Exercise 11.6.2. Starting from the integral formulation of the Airy function Ai (z) in
(11.23),´ show that
∞
(a) −∞ Ai (x)dx = 1
´∞
(b) 0 Ai (x)dx = 13 .
´∞
Hints: Part (a): Change the contour to the imaginary axis, and then use −∞ eixt dt =
´∞
´ e−t3 /3
1
2π δ(x); part (b): Show that 0 Ai(z)dz = − 2πi
dt; change the contour to a
t
C
straight line from e−2πi/3 ∞ towards the origin, an ε-radius circle segment, and a straight
line exit towards e+2πi/3 ∞. Show that the two straight line parts cancel.
91 A
special case of variation of parameters.
11.6. Exercises
315
Exercise 11.6.3. An alternative way to obtain the Taylor expansion for Jn (t) (see (11.11))
is to start from (11.12), change variable z = 2ξ/t, giving
Jn (t) =
1
2πi
n ˆ ξ −t2 /(4ξ)
t
e e
dξ,
2
ξ n+1
C
and then apply residue calculus. Carry this out.
Exercise 11.6.4. For example, using (11.11), show that
d
Jn (t) = 12 (Jn−1 (t) − Jn+1 (t)).
generally, that dt
d
dt J0 (t)
= −J1 (t) and, more
Exercise 11.6.5. With the use of computer software, explore some of the Bessel Yn (t) functions in the complex t-plane (plot their real and imaginary parts, and magnitude/phase).
´ 1 b−1
Γ(c)
Exercise 11.6.6. Prove the relation (11.18): 2 F1 (a, b; c; z) = Γ(b)Γ(c−b)
t (1 −
0
t)c−b−1 (1 − zt)−a dt.
Hint: Assume |z| < 1, and Taylor expand (1 − zt)−a around t = 0. Next, swap the order
between the sum and the integral, and use (6.5) to arrive at the Taylor series (11.16) for
the LHS of (11.18).
Exercise 11.6.7. Apply the Fourier–Laplace method to find an integral representation for
a solution to the ODE
z u00 (z) + (2n + 1)u0 (z) + z u(z) = 0,
n integer.
Hint: You should arrive at (t2 + 1)F 0 (t) − (2n − 1) t F (t) = 0, and u(z) =
1)n−1/2 ez t dt. Explain your choice for C.
´
C
(t2 +
Exercise 11.6.8. Show that each pole of the PI equation is double, with strength 1 and
residue 0.
Chapter 12
Steepest Descent for
Approximating Integrals
Contour integration allows many definite integrals to be evaluated in closed form. There
are also many cases where closed forms cannot be found, but for which the contour integration idea instead can be used to provide excellent approximations. This happens especially when there is some free parameter present, and we are interested in cases where,
as this parameter increases, the integrand either becomes locally spiked or increasingly
oscillatory.
As a preliminary to the topic of steepest descent, we first make some brief comments
about asymptotic analysis. This is a far-reaching subject that applies to many areas, such
as approximating solutions to ODEs, PDEs (for example, boundary layers in fluids), integrals, etc. Even with computational approaches nowadays becoming increasingly capable,
asymptotic analysis remains as important as ever. This approach usually gains rapidly
in accuracy just in those situations (approaching singular cases) when typical numerical
methods encounter their largest difficulties. Many good books on asymptotic methods are
available, including [4, 5, 9, 26, 32]. We start our discussion by some brief comments on
what is meant by an asymptotic (in contrast to a convergent) expansion, and we illustrate
the concept by the error function erf(x) and by the Euler–Maclaurin formula. After that,
we turn to approximating integrals. Section 12.3 focuses first on a purely real-valued case
in which an integrand becomes increasingly spiked. Utilizing the freedom of choosing integration paths in the complex plane for analytic functions, the method of steepest descent
(Section 12.4) turns the real-valued result into a tool of far greater generality.
12.1 Asymptotic vs. convergent expansions
P∞
PN
An infinite sum S = k=0 ak converges if S − k=0 ak → 0 as N → ∞. This readily
generalizes to the case when the terms are functions of some variable x with ak as their
coefficients. The sum
∞
X
S(x) =
ak ϕk (x)
(12.1)
k=0
PN
converges for a fixed x if S(x) − k=0 ak ϕk (x) → 0 as N → ∞. Given a value for x,
we can reach any accuracy we want by just letting N → ∞. For an asymptotic series, the
317
318
Chapter 12. Steepest Descent for Approximating Integrals
roles of x and N can be seen as reversed. In place of (12.1), we write
S(x) ∼
∞
X
ak ϕk (x)
(12.2)
ak ϕk (x) = o(ϕN (x))
(12.3)
k=0
if it holds, for any fixed N , that
S(x) −
N
X
k=0
when x goes to some limit (such as x → 0 or x → ∞).92 Both (12.1) and (12.2) are very
useful for analysis as well as for computation. The utility of (12.1) is obvious, especially if
the functions ϕk (x) are easy to compute, and S(x) is a lot more complicated (with Taylor
expansions a typical example). The utility of (12.2) is somewhat more subtle. The example
below is typical for this case.
Example 12.1. Compare along the real axis the´accuracies of the Taylor and the asymptotic
2
x
expansions for the error function erf(x) = √2π 0 e−t dt.
Along the real axis, the error function transitions smoothly from −1 to +1, as seen in
Figure 12.1. Since this is an entire function, its Taylor expansion
2
x3
x5
x7
erf(x) = √
x−
+
−
+ −···
(12.4)
3 · 1! 5 · 2! 7 · 3!
π
is everywhere convergent. Repeated integration by parts of the defining integral produces
an asymptotic expansion
2
e−x
erf(x) = 1 − √
x π
1−
1·3
1·3·5
1
+
−
+ −···
2
1
2
2
(2x )
(2x )
(2x2 )3
,
(12.5)
which, in contrast, never converges (since the factorial-type growth in the numerators always “wins out” over the geometric rate in the denominators). Nevertheless, this expansion
provides, for similar number of terms, superior accuracy as soon as x & 5, as illustrated
in Figure 12.2(b). For x-values increasing further still, the figure shows that the accuracy
increases extremely rapidly, even if the number of terms N is kept quite low. The fact that
(12.5) always diverges if we keep x fixed and let N → ∞ is irrelevant if our goal is to
estimate erf(x) for x large.
Example 12.2. In order to evaluate erf(4), estimate the number of terms N to use in (12.5)
and the resulting error.
The terms in the expansion (12.5) first decrease and then increase in magnitude. The
error is smallest when two successive terms have roughly equal magnitude. Hence, looking
at their ratio, N should be chosen such that
1≈
92 The
2x2
(2N − 3)!! (2N − 1)!!
/
=
,
2
N
−1
2
N
(2x )
(2x )
2N − 1
“little o” notation, f (x) = o(g(x)), means that f (x)/g(x) → 0 as x approaches some specified limit.
12.2. Euler–Maclaurin formula
319
1.5
1
erf(x)
0.5
0
-0.5
-1
-1.5
-5
-4
-3
-2
-1
0
1
2
3
4
5
x
Figure 12.1. The function erf(x) displayed along the real x-axis (previously displayed in
the complex plane in parts (c) and (d) of Figure 9.12).
suggesting for x = 4 the choice of N = 16, with a resulting error of
2
e−4 (2 · 16 − 3)!!
√
≈ 2.6 · 10−15 ,
4 π (2 · 42 )15
in excellent agreement with the values that can be read off from Figure 12.2(b).
While the region of convergence for a Taylor expansion always is the largest singularityfree circle surrounding the expansion point, applicability of asymptotic expansions (of the
form of inverse powers, such as (12.5)) occurs in sectors of the complex plane. Generalizing x to be a complex variable z, (12.5) is valid within the sector shaded gray in Figure
12.3(a), whereas a version with the leading “1” swapped for “−1”
2
e−z
erf(z) = −1 − √
z π
1−
1·3
1·3·5
1
+
−
+ −···
2
1
2
2
(2z )
(2z )
(2z 2 )3
(12.6)
is valid in the sector shown in part (b) of the figure. The expression Stokes phenomenon
refers both to sector structures of this type and to the fact that sectors of validity can overlap.
2
In the overlapping regions, the factor e−z grows massively large as z moves away from the
origin, making the leading constant irrelevant. A more extensive example of Stokes sectors
will be given in Section 12.4.3, when we use the method of steepest descent to analyze the
far-field behavior of solutions to Airy’s equation (illustrated earlier in Figures 11.2–11.4).
12.2 Euler–Maclaurin formula
Together with Stirling’s formula (Section 12.3.3), this may well be the best known asymptotic formula in the literature. It can be written in different ways, including
∞
X
k=n
ˆ
∞
f (k) =
n
1
1
1 000
1
f (x)dx + f (n) − f 0 (n) +
f (n) −
f (5) (n) + − . . . ,
2
12
720
30240
(12.7)
320
Chapter 12. Steepest Descent for Approximating Integrals
45
-5
0
-30
35
-20
5
30
N
20
40
10
-10 15
50
25
5
10
5
0
-10 -15
10
20
10
15
-5
20
5
1
2
3
4
5
6
7
8
9
10
x
(a) Taylor expansion.
-30
20
-60
45
10
5
0
-5
-10
-15
-20
50
40
35
25
-10
-30
-20
0
15
-15
-5
20
-60
10
5
N
30
10
5
1
2
3
4
5
6
7
8
9
10
x
(b) Asymptotic expansion.
Figure 12.2. log10 of the errors (shown numerically along the contour lines) when erf(x)
is approximated at location x using N terms in the Taylor expansion (12.4) and the asymptotic
expansion (12.5), respectively. The areas shaded gray indicate where the accuracy is better than
10−15 . In the asymptotic expansion case, there is typically, for a given value of x, a unique finite
value for N that gives the highest accuracy, as shown by the dashed red curve in part (b). This curve
marks where the contour lines for accuracy have a vertical tangent.
and, if both limits are finite,
m
X
k=n
ˆ
m
f (k) =
n
+
1
1
f (x)dx + [f (n) + f (m)] − [f 0 (n) − f 0 (m)]
2
12
1 000
1
[f (n) − f 000 (m)] −
[f (5) (n) − f (5) (m)] + − · · · .
720
30240
(12.8)
12.2. Euler–Maclaurin formula
321
+3 /4
+ /4
-3 /4
- /4
(a) Sector of validity for (12.5).
(b) Sector of validity for (12.6).
Figure 12.3. The sectors of validity (in the limit of |z| → ∞), shaded gray, for the two
expansions (12.5) and (12.6), respectively.
Two different derivations are given in Section 12.5.2. Both provide closed-form expressions for the coefficients in terms of Bernoulli numbers; cf. (5.12). For most functions
f (z), the expansion in the RHS will eventually diverge. However, like for most asymptotic
expansions that we will encounter, the error when truncating the expansion will typically
be of the same size as the first omitted term.
Example 12.3. Numerically approximate Euler’s constant
!
n
X
1
γ = lim
− log n .
n→∞
k
k=1
P∞
We can write the definition as γ = 1 + k=2 k1 + log(k
−
1)
−
log
k
. This sum
1
1
1
2
converges, since k + log(k − 1) − log k = k + log 1 − k = O(1/k ), but only very
slowly. We thus apply (12.7) to f (z) = z1 + log(z − 1) − log z. The convergence rate of
the new expansion improves if one sums some leading terms before applying the formula.
Choosing, for example, n = 10 gives
1+
ˆ
9
X
f (k) =
0.6317436767,
k=2
∞
f (x)dx = −0.0517553591,
10
1
f (10) = −0.0026802578,
2
1 0
f (10) = −0.0000925926,
12
1 000
f (10) = 0.0000001993,
720
1
−
f (5) (10) = −0.0000000015.
30240
−
322
Chapter 12. Steepest Descent for Approximating Integrals
Adding these gives a result that is only offPone unit in the last decimal digit; cf. (6.4). It
n
can be noted that direct summation of 1 + k=2 f (k) does not come this close to the limit
1
10
until n ≈ 2 10 .
Pn
Example 12.4. Evaluate k=1 k 3 .
´n
Pn
Choosing f (x) = x3 , equation (12.8) gives k=1 k 3 = 1 x3 dx + 21 (13 + n3 ) −
Pn
n(n+1) 2
1
1
2
2
. Recognizing n(n+1)
as k=1 k, we have thus
12 3(1 − n ) + 720 (6 − 6) =
2
2
arrived at
!2
n
n
X
X
3
k =
k .
k=1
k=1
Since f (z) in this example was a polynomial, the RHS of (12.8) vanished after a certain
number of terms, giving an exact result.93
12.3 Laplace integrals
We start by considering real-valued integrals along the real x-axis, of the Laplace integral
form
ˆ
b
f (x) es φ(x) dx,
I(s) =
(12.9)
a
in the limit of the parameter s → ∞. For Examples 12.6–12.8, considered later in this
section, the top row of subplots in Figure 12.4 illustrates the φ(x) functions, and the bottom row shows the integrands f (x) es (φ(x)−φ(c)) for the s-values s = 1, 3, 10, 30. For
improved visibility, we have for the integrands included a (x-independent) factor e−s φ(c) ,
where x = c is the location of the maximum for φ(x). In all cases, the asymptotic expansions for s → ∞ will come from the immediate vicinity of these x = c locations. In the
third example, we split the interval into two, [0, 1] and [1, ∞], and for the first one change
variable x → −x. Hence we can, without loss of generality, proceed by assuming that
the maximum value of φ(x) over the integration interval [a, b] occurs at its left boundary,
denoted by x = a. We next distinguish two main cases:
Case 1: φ0 (a) < 0.
Case 2: φ0 (a) = 0 and φ00 (a) < 0.
Case 2 can be generalized to having several leading derivatives zero at x = a. The next key
step in obtaining the asymptotic expansion for (12.9) will be to make a change of variables
in such a way that we then can apply Watson’s lemma.
12.3.1 Watson’s lemma
Theorem 12.5 (Watson’s lemma). If
ˆ b
I(s) =
f (ξ) e−s ξ dξ
(b > 0)
(12.10)
0
Pn
m+1
93 More generally (Faulhaber’s formula), for n, m
m
≥
1,
= nm+1 +
k=1 k
Pm−1 m Bk+1 m−k
n
(in the last sum, terms with k even vanish, since B3 , B5 , B7 , . . . are zero).
k=1
k
k+1
nm
+
2
12.3. Laplace integrals
323
Figure 12.4. The three columns of subplots correspond to Examples 12.6–12.8, respectively, that are discussed in Section 12.3.2. The top row of subplots shows the functions φ(x) that
are multiplied by s in the exponents, and the bottom row of subplots shows how the integrands
f (x) es φ(x) then vary across the respective intervals (scaled by the factor e−s φ(c) , in order to
display an s-independent value at the location x = c where φ(x) has its maximum). In all cases,
it is visually apparent that, as s → ∞, the integral will become increasingly well described by its
properties in the immediate vicinity of x = c.
and
f (ξ) = ξ α
∞
X
an ξ β n
(ξ & 0),
(12.11)
n=0
then
I(s) ∼
∞
X
an Γ(α + β n + 1)
sα+β n+1
n=0
(s → ∞).
(12.12)
Proof. (Sketch) The key observations behind demonstrating this result are as follows:
(i) As s → ∞, the behavior of f (ξ) beyond an immediate vicinity of ξ = 0 becomes
increasingly irrelevant to the value of I(s) (as defined by (12.10)); (ii) in I(s), we can
therefore substitute for f (ξ) the expansion (12.11); (iii) having done this, the actual value
of the upper integration limit b becomes also irrelevant, and can be replaced by b = ∞; and
(iv) all terms in the integral can at this point be evaluated in closed form by means of the
´∞
Γ-function relation 0 ξ z e−s ξ dξ = Γ(z+1)
(Re z > −1, Re s > 0; In the special case of
´ ∞ n −ssξz+1
n!
z = n integer, this becomes 0 ξ e
dξ = sn+1
). This gives the result (12.12).
324
Chapter 12. Steepest Descent for Approximating Integrals
While some other approaches (such as integration by parts) might work in certain cases
for approximating (12.10), Watson’s lemma is particularly powerful in that it works well
also when (12.11) involves fractional powers of ξ (as will happen in Case 2).
To find the appropriate change of variables so that Watson’s lemma can be applied
to (12.9), we first rewrite this integral (assumed already arranged so that maxx∈[a,b] φ(x)
occurs at x = a) as
ˆ b
s φ(a)
I(s) = e
f (x) es (φ(x)−φ(a)) dx,
(12.13)
a
and then make the change of variable
− ξ(x) = φ(x) − φ(a),
(12.14)
so that the exponent in the integral becomes −s ξ (as in (12.10)). The integral now takes
the form
ˆ 0+ f (x)
e−s ξ dξ,
(12.15)
I(s) = es φ(a)
− 0
φ
(x)
0
where the upper limit of the integral (denoted by 0+) will become irrelevant as s → ∞.
The key step that now remains is to express − φf0(x)
(x) as a power series expansion in ξ,
and Watson’s lemma will then give the complete asymptotic expansion.
12.3.2 Applications of Watson’s lemma
Example 12.6. Find the complete asymptotic expansion for I(s) =
s → +∞ (cf. the first column of subplots in Figure 12.4).
´ π/2
0
2
e−s sin
x
dx as
The integral is of the form (12.9) with φ(x) = − sin2 x and f (x) = 1, i.e., − φf0(x)
(x) =
π
1
2 sin x cos x . The function φ(x) has its maximum over [0, 2 ] at x = 0, so the variable change
(12.14) −ξ(x) = φ(x) − φ(0) gives ξ = sin2 x, and therefore sin x = ξ 1/2 , from which
f (x)
. The integral
= √1
it follows that − φf0(x)
(x) , expressed in ξ, becomes − φ0 (x)
2 ξ(1−ξ)
´
1
has become I(s) = 21 0 ξ −1/2 (1 − ξ)−1/2 e−s ξ dξ, which is in the perfect form for WatP∞ Γ(n+ 1 ) ξn
son’s lemma after using the binomial theorem expansion (1 − ξ)−1/2 = n=0 n! Γ 2 1 .
(2)
2
P∞
[Γ(n+ 12 )]
1
Therefore, I(s) ∼ 2 n=0 n! Γ 1 sn+1/2 for s → +∞.
(2)
In most cases, one cannot write down a closed-form expression for − φf0(x)
(x) in terms
of ξ(x). One can then instead proceed by using Taylor expansions, starting from (12.14)
and then expanding φ(x) around the critical point x = a:
− ξ(x) = φ(x) − φ(a) = d1 (x − a) + d2 (x − a)2 + · · ·
(explicitly as d1 =
Section 2.3.2).
φ0 (a)
1! ,
d2 =
φ00 (a)
2! ,
(12.16)
etc., or by the method of undetermined coefficients;
Case 1: φ0 (a) 6= 0. We can invert (12.16) to obtain a new set of coefficients {ai }, i =
1, 2, . . . , so that
(x − a) = a1 ξ + a2 ξ 2 + a3 ξ 3 + · · · , i.e., x = a + a1 ξ + a2 ξ 2 + a3 ξ 3 + · · · , (12.17)
12.3. Laplace integrals
325
as discussed in Sections 2.3.2 and 2.8.1 (i.e., by substituting (12.17) into (12.16) and equat
ing coefficients). Substituting into − φf0(x)
(x) gives a Taylor expansion in ξ around ξ = 0:
f (x)
= b0 + b1 ξ + b2 ξ 2 + · · · ,
(12.18)
− 0
φ (x)
and Watson’s lemma can be applied.
Case 2: φ0 (a) = 0. The first term in the RHS of (12.16) is now missing. The series can
again be inverted, but the form of (12.17) will have to be changed to also include halfinteger powers:
(x − a) = a1 ξ 1/2 + a2 ξ 2/2 + a3 ξ 3/2 + · · · .
(12.19)
The coefficients {ai } can again be obtained by substituting the expansion into (12.16) and
then equating coefficients. The counterpart to (12.18) takes now the form
f (x)
= b−1 ξ −1/2 + b0 + b1 ξ 1/2 + b2 ξ 2/2 + b3 ξ 3/2 + · · · ,
− 0
φ (x)
which again works just fine for Watson’s lemma. If this case arose from a situation with
a max point of φ(x) occurring at an interior location and the interval was split in two
subintervals, the signs for the even-numbered coefficients b0 , b2 , b4 , . . . get reversed when
the other subinterval (with the critical point at its right edge) is analyzed. Then, we simply ignore these coefficients and double the contributions from the odd-numbered ones
b−1 , b1 , b3 , . . . . We next give one example for each of Cases 1 and 2, respectively.
´ 2 1 −s arctan x
Example 12.7. Find some leading terms for I(s) = 1 1+x
e
dx (cf. the second
column of subplots in Figure 12.4).
1
This example features f (x) = 1+x
, and φ(x) = − arctan x. The variable change
π
(12.14) gives ξ = arctan x − 4 and
1
7
10
62
f (x)
1 + x2
=
= 1+ξ +2ξ 2 + ξ 3 + ξ 4 + ξ 5 −· · · .
− 0
=
φ (x)
1+x
cos ξ (sin ξ − cos ξ)
3
3
15
(12.20)
With φ(1) = − π4 giving a leading factor e−s π/4 , Watson’s lemma produces
1
1
4
14 80 496
I(s) = e−s π/4
+ 2 + 3 + 4 + 5 + 6 + ··· .
s s
s
s
s
s
f (x)
In most cases, we would not be able to express − φ0 (x) explicitly like this as a function
of ξ. We would then instead do Taylor series manipulations as follows (most conveniently
and reliably by using a symbolic algebra package such as Mathematica, which has each of
the steps as a built-in command). First expanding around x = a = 1 gives
π
1
1
1
1
= (x − 1) + (x − 1)2 − (x − 1)3 + 0 (x − 1)4 − (x − 1)5 + · · · ,
4
2
4
12
40
which can be inverted:
8
10
64
(x − 1) = 2ξ + 2ξ 2 + ξ 3 + ξ 4 + ξ 5 + · · · ,
3
3
15
8 3 10 4 64 5
2
i.e., x = 1 + 2ξ + 2ξ + ξ + ξ + ξ + · · · .
3
3
15
ξ = arctan x −
326
Chapter 12. Steepest Descent for Approximating Integrals
Substituting this expression for (x − 1) into − φf0(x)
(x) =
in ξ as obtained above in (12.20).
Example 12.8. Find some leading terms for I(s) =
column of subplots in Figure 12.4).
1+x2
1+x
´∞
0
leads to the same expansion
1 s (−x+log x)
dx
xe
(cf. the third
The function φ(x) = −x + log x has its maximum at x = 1, with φ(1) = −1. Hence,
we consider first the integral over [1, ∞], and make the change of variable −ξ(x) = φ(x)−
φ(1) = −x + 1 + log x, with Taylor expansion
ξ=
1
1
1
1
1
(x − 1)2 − (x − 1)2 + (x − 1)4 − (x − 1)5 + (x − 1)6
2
3
4
5
6
1
1
1
1
7
8
9
10
− (x − 1) + (x − 1) − (x − 1) + (x − 1) + · · · ,
7
8
9
10
giving when inverted94
√ 1/2 2ξ
ξ 3/2
2ξ 2
ξ 5/2
4ξ 3
139ξ 7/2
√ +
√
2ξ
+
+ √ −
+
−
3
9 2 135 540 2 8505 340200 2
2ξ 4
571ξ 9/2
√ + ···
+
−
25515 73483200 2
x=1+
(where we have included unusually many terms, since half of them soon will drop out).
−1
Substituting this expansion into − φf0(x)
(x) = 1−x gives
ξ −1/2
1
ξ 1/2
4ξ
ξ 3/2
4ξ 2
139ξ 5/2
f (x)
√ +
√
= √ − + √ −
+
−
− 0
φ (x)
3 6 2 135 216 2 2835 97200 2
2
8ξ 3
571ξ 7/9
√ + ··· .
+
−
25515 16329600 2
As described under Case 2 above, the expansion for the subinterval x ∈ [0, 1] will become
identical apart from a switch of sign for all terms with integer powers of ξ. When adding
the results from the two subintervals, all these terms vanish, and we double the coefficients
for the remaining ones (with half-integer powers of ξ). Recalling the leading factor esφ(a)
in (12.13), Watson’s lemma gives
r
I(s) ∼ e
−s
2π
s
1
1
139
571
1+
+
−
−
+ ··· .
12 s 288 s2
51840 s3
2488320 s4
(12.21)
12.3.3 Stirling’s formula
As a follow-up
to Example 12.8, we can note
´∞
´ ∞that the change of variable s x = u in
I(s) = 0 x1 es (−x+log x) dx gives I(s) = s1s 0 us−1 e−u du = s1s Γ(s). We thus obtain
the famous Stirling formula for Γ(z) by multiplying (12.21) by ss . Since s! can be defined
94 Compare
with the second case described in Section 2.8.1.
12.3. Laplace integrals
327
as Γ(s + 1) also for noninteger s-values, it follows (having renamed s to n) that
n n √
139
571
1
1
−
−
+
·
·
·
. (12.22)
n! ∼
2πn 1 +
+
e
12 n 288 n2
51840 n3
2488320 n4
This expansion (12.22) can be recast in different ways, with
1
1
1
1
1
+
−
+···
(log 2π + log n) +
−
3
5
2
12 n 360 n
1260 n
1680 n7
(12.23)
particularly attractive in that all even powers of n have vanished. Since n! = nΓ(n), this
can alternatively be formulated as
log n! ∼ n log n − n +
1
1
1
1
1
(log 2π − log z) +
−
+
−
+··· ,
2
12 z 360 z 3 1260 z 5 1680 z 7
(12.24)
with the only difference (beyond changing n to z) the sign in front of the log z term. Exercises 12.6.7 and 12.6.8 show two different ways to extend (12.23) and (12.24) explicitly to
all orders.
The sector of validity for both (12.22) and (12.24) can be shown to be the entire complex
plane with the exception of the negative real axis. Both approximations are remarkably
accurate already for quite modest values of n, as seen in Figures 12.5(a)–(b). We display
here the relative errors that the expansions produce for positive values of n, with −15
roughly corresponding to the regular computer machine rounding level. This is for (12.22)
reached already for n (= s) around 6, at which point the bottom right subplot in Figure
12.4 shows that the integrand is still far from being sharply peaked. Nevertheless, the local
expansion around x = 1 suffices for typical machine precision accuracy.
In conclusion to the present
discussion of Stirling’s formula, we note from (12.22) that
nn
1
√1
−
=
O
,
and
therefore the following sum will converge (although very
n
n!e
n3/2
2πn
slowly):
∞ X
nn
1
√
−
≈ −0.08406950873 .
S=
n!en
2πn
n=1
P∞
1
The sum n=1 √2πn
is by itself divergent, but had it converged, a plausible value would
1
1
√
have been 2π ζ( 2 ). The expression S became around 2001 a striking example of “experimental mathematics” in action, when numerical computations unexpectedly showed it to
match S = − 23 − √12π ζ( 12 ) to vast numbers of digits. A strict proof followed about six
years later [6].
log Γ(z) ∼ z log z − z +
12.3.4 Laplace’s method in case of movable maxima
In Example 12.8, we started with
ˆ
I(s) =
0
∞
1 s (−x+log x)
e
dx,
x
(12.25)
which, with φ(x) = −x + log x, was in perfect form for the standard variable change
followed by Watson’s lemma. As an afterthought, we then noted that the variable change
sx = u led to´ the standard integral for the gamma function. Suppose that we instead start
∞
with Γ(s) = 0 us−1 e−u du, the question now becoming how to find an appropriate way
to reverse this last variable change, to get the integral back to a form suited for asymptotics.
328
Chapter 12. Steepest Descent for Approximating Integrals
5
-15
-3
-30
-10
-4
0
-25
-20
45
15
10
5
0
-5
50
40
35
-30
30
-10
-5
25
-2
-15
N
0
-2
5
0
-25
20
-20
15
-15
-15
-10
10
-10
-5
5
-10
-5
-5
0
2
4
6
8
10
12
14
16
18
20
n
(a) Asymptotic expansion (12.22).
-40
-35
-30
-25
-20
-15
-10
-5
0
-50
-45
45
20
15
10
5
50
40
35
-20
-15
0
0
5
-10
-3
25
-4
-30
N
-45
-25
-5
30
20
-35
15
-30
-25
10
5
-1
0
-20
-15
-25
-20
-15
-15
-10
-5
-10
0
4
6
8
10
12
14
16
18
20
n
(b) Asymptotic expansion (12.23).
Figure 12.5. log10 of the relative errors (shown numerically along the contour lines) when
the expansions (12.22) and (12.23) are applied to different values of n using N terms in the respective
asymptotic expansions. The areas shaded gray indicate where the accuracy is better than 10−15 .
´ ∞ −u
We can write the gamma function integral as Γ(s) = 0 e u es log u du, but φ(u) =
log u lacks a local max point.
´ ∞ Another idea is to collect the exponential factors and write
the integral as Γ(s) = 0 u1 e−u+s log u du. Now, the exponent −u + s log u does have
a max point, at u = s, but it moves position with s. To bring it to a fixed location, an
appropriate variable change
would in this case be u = xs (bringing it to x = 1). We have
´∞
now obtained Γ(s) = ss 0 x1 es (−x+log x) dx, which is in the form that Example 12.8
started with.
12.4. Steepest descent
329
12.3.5 Leading order approximations
If we only want the leading term of the asymptotic expansion, application of Watson’s
lemma becomes simple enough that we can write down the end result explicitly. For the
Laplace integral (12.9) we have the following:
Case 1: φ0 (a) < 0:
I(s) ∼ −
1 f (a) s φ(a)
e
.
s φ0 (a)
Case 2: φ0 (c) = 0, φ00 (c) < 0, where the critical point c is located inside [a, b]:
√
2πf (c) s φ(c)
p
e
.
I(s) ∼
−s φ00 (c)
(12.26)
(12.27)
For one-sided cases (c at either edge of [a, b]), the result should be halved.
This last Case 2 is a special case of φ(k) (c) = 0, k = 1, 2, . . . , p − 1, and φ(p) (c) < 0,
with p even, for which (assuming c is located inside [a, b]; else the result is halved) we have
the following:
2Γ(1/p)(p!)1/p f (c) s φ(c)
e
.
(12.28)
I(s) ∼
p [−s φ(p) (c)]1/p
12.4 Steepest descent
12.4.1 Concept and background
In the description above of Laplace integrals, we made no use of two key opportunities
provided by the theory of analytic functions:
(i)
The integration path can be altered freely (as long as we account for residues
when it crosses singularities).
(ii)
The real and the imaginary parts of analytic functions are tightly coupled
through the Cauchy–Riemann (C-R) equations (2.1).
Focusing first on item (ii), we consider an analytic function φ(z) and decompose it in its
real and imaginary parts
φ(z) = u(x, y) + i v(x, y) ,
where z = x + i y. As we noted already in Section 2.1.1, the C-R equations ux = vy ,
vx = −uy imply the following:
1. Both u(x, y) and v(x, y) are harmonic functions (satisfying uxx + uyy = 0 and
vxx + vyy = 0, respectively), so they cannot have any local maxima or minima.
h
iT h
i
vx
2. The C-R equations also imply that uuxy
= 0, telling us that the gradient
vy
vectors (directions of steepest descent) for u(x, y) and v(x, y) are orthogonal to each
other.
3. For any smooth functions in x, y, level curves are orthogonal to gradient directions.
330
Chapter 12. Steepest Descent for Approximating Integrals
Combining the observations just above, we next conclude the following:
4. Along the steepest descent paths of u(x, y), the function v(x, y) is constant, and vice
versa.
5. A point z with φ0 (z) = 0 (and φ00 (z) 6= 0) will be a saddle point for both u(x, y)
and v(x, y). The function u(x, y) will at such a point have two orthogonal steepest
descent/ascent directions, with a local maximum along one and a local minimum
along the other, and with v(x, y) constant along both (and equivalent for v(x, y)).
Figures 12.6 and 12.7 illustrate the general features described above in the special case of
φ(z) = z − log z .
(12.29)
(a) Real part of φ(z) = z − log z.
(b) Imaginary part of φ(z) = z − log z.
Figure 12.6. Real and imaginary parts of φ(z) = z − log z, with the saddle point at z = 1
marked with a black dot. The steepest descent path through it for Re φ(z) is shown in green and the
steepest ascent path in red. These same paths both become level curves for Im φ(z).
12.4. Steepest descent
331
3
2
y
1
0
-1
-2
-3
-2
0
2
x
Figure 12.7. Steepest descent paths for Re φ(z) (equal to level curves for Im φ(z)) shown
as solid curves, and level curves for Re φ(z) (equal to steepest descent curves for Im φ(z)) shown as
dashed curves. The blue line segment (also shown in Figure 12.6) is a tangent at the saddle point to
the steepest descent curve (green).
This function features a branch cut along the negative real axis (which is irrelevant for our
present discussion). The black dot marks the only location at which φ0 (z) = 0, which
occurs at z = 1. As seen in Figure 12.6(a), this is a saddle point for Re φ(z)—a local
minimum if we follow the real axis (steepest ascent path; red curve) near z = 1, and a local
maximum if we follow the steepest descent path (green curve) through that point. These
same paths are both constant level paths for Im φ(z); cf. part (b) of the figure.
Figure 12.7 displays φ(z) somewhat differently over the same region in the complex
plane. Instead of highlighting just one steepest descent path for Re φ(z) (level path for
Im φ(z)), we illustrate several of them as solid black curves. The dashed curves are similarly the level paths for Re φ(z) and therefore also steepest descent paths for Im φ(z). The
two curve sets (solid and dashed) are everywhere orthogonal to each other, apart from at
the saddle point z = 1 and at the singularity z = 0.
In the next subsection, we will see that steepest descent path for Re φ(z) through the
saddle point (marked green in Figures 12.6 and 12.7) plays a key role in the steepest descent
method for approximating Laplace integrals. The curved shape of this path somewhat
complicates the algebra compared to the previous cases of integrating along the real axis.
However, this path also being a level curve for Im φ(z) has the great benefit that es i Im φ(z)
´b
can be factored out of the integral a f (z) es φ(z) dz, making the integrand locally peaked
and nonoscillatory as s increases. Choosing to instead integrate along the local tangent line
at the saddle point (marked blue in the figures) gives readily the leading expansion term (as
shown in Section 12.4.4), but may require more work if many terms are needed.
12.4.2 Some applications of steepest descent
Integrals that become increasingly peaked in the s → ∞ limit
We consider again the Laplace integral (12.9):
ˆ b
f (z) es φ(z) dz,
I(s) =
a
(12.30)
332
Chapter 12. Steepest Descent for Approximating Integrals
but we denote the integration variable by z instead of x, as we will now utilize that the path
need not be the straight line from a to b (nor do a and b need to be on the real axis).
Example 12.9. Determine an asymptotic expansion for
1
1
=
Γ(s)
2πi
1
Γ(s)
as s → ∞, based on (5.19):
ˆ
ez z −s dz ,
(12.31)
C
where C is a Hankel contour, as illustrated in Figure 5.32 (there denoted by H instead of
by C; it enters from infinity in the third quadrant, goes around the branch point at z = 0,
and exits to infinity in the second quadrant).
The first step is to convert (12.31) to the Laplace form (12.30), which can be done with
the change of variable z → s z. This yields
ˆ
1
1
=
es (z−log z) dz ,
(12.32)
Γ(s)
2πi ss−1 C
with f (z) = 1, φ(z) = z − log z and with the same type of Hankel contour C. This
function φ(z) has already been illustrated in Figures 12.6 and 12.7. Of all possible paths
from the third to the second quadrant, going around the branch point at the origin, we note
that the highlighted steepest descent curve (green) has some unique features. With Im φ(z)
constant along this path, es Im(φ(z)) is also a constant, which thus can be factored out of the
integral. In the present case this is just a factor of 1. We thus replace (12.32) by
ˆ
1
1
=
es Re (z−log z) dz .
(12.33)
Γ(s)
2πi ss−1 C
With a purely real-valued factor now multiplying s in the exponent, we can apply the insights from Sections 12.3.1–12.3.5. There are now two main options for generating the
asymptotic expansion:
1. Integrate along the steepest descent path and then apply Watson’s lemma.
2. Return to (12.32) and follow the straight line segment that is tangent to the steepest
descent curve at the saddle point (marked blue in Figures 12.6 and 12.7).
In either case, the complete asymptotic expansion can be deduced from an infinitesimal
vicinity of the saddle point. The second approach might appear to be simpler, but intermediate expansions may require quite a large number of terms. We compare the two
approaches below.
Method 1: Follow the steepest descent curve—Use its analytic form: With φ(z) =
z − log z = x + iy − 21 log(x2 + y 2 ) − i arctan xy , we obtain the steepest descent curve
through the saddle point (located at z = 1) by solving Im φ(z) = 0, giving x = y/ tan y.
y
sin y
dx
dx
Along this curve, Re φ(z) = tan
y + log y and dz = dx + i dy = dy + i dy. Since dy
is an odd function of y, it will not contribute to the integral, and we obtain
1
1
=
Γ(s)
2πi ss−1
ˆ
es Re (z−log z) dz ∼
C
1
2π ss−1
ˆ
0+
f (y) es φ(y) dy,
0−
(12.34)
12.4. Steepest descent
333
y
sin y
with f (y) = 1 and φ(y) = tan
y + log y . Apart from the naming of the integration
variable as y instead of x, the problem is now in the standard form for making the key variable change (12.14) followed by using Watson’s lemma as in Example 12.8. The remaining
steps become
sin y
y
+ log
−1
tan y
y
y2
y4
y6
y8
y 10
691y 12
=− −
−
−
−
−
− ··· ,
2
36 405 4200 42525 294698250
p
ξ
ξ2
139ξ 3
571ξ 4
163879ξ 5
+
+
−
−
+ ··· ,
y = 2ξ 1 −
18 1080 680400 146966400 67898476800
y
f (y)
=
− 0
φ (y)
1 − 2y cot y + y 2 csc2 y
1
ξ
ξ2
139ξ 3
571ξ 4
=√
1− +
+
−
+ ··· ,
6 216 97200 16329600
2ξ
−ξ = φ(y) − φ(0) =
and therefore (by Watson’s lemma)
ˆ ∞
e s r s 1
2 es
f (y)
1
1
139
−s ξ
∼
− 0
e
dξ ∼
1−
+
+
Γ(s)
2π ss−1 0
φ (y)
s
2π
12s 288s2
51840s3
571
−
+ ··· .
2488320s4
(12.35)
Method 2: Follow the steepest descent curve—Rely on series expansions only: The
derivation above utilized closed-form expressions for the steepest descent curve. Such
expressions are often not available (or are not practical). It transpires that the local Taylor
expansion of φ(z) around the saddle point contains all the needed information. To illustrate
this, we focus again on the integral
ˆ
es Re (z−log z) dz
(12.36)
C
in (12.33). The standard variable change −ξ = φ(z) − φ(a) (cf. (12.14)), here with a = 1,
gives
−ξ = φ(z) − φ(1) =
1
1
1
1
(z − 1)2 − (z − 1)3 + (z − 1)4 − (z − 1)5 + − · · · ,
2
3
4
5
which can be inverted (best using a symbolic algebra package):
2
2 2
4 3
2
z = 1− ξ−
ξ −
ξ +
ξ4 + · · ·
3
135
8505
25515
i
1
1
139 7/2
571
±√
2ξ 1/2 − ξ 3/2 +
ξ 5/2 +
ξ
−
ξ 9/2 + · · · . (12.37)
9
540
340200
73483200
2
With ξ > 0 (real), this equation amounts to a parameterization of the upper and lower
branches
´ of the steepest descent curve in the z-plane. The integral (12.36) is now in the
form C e−s ξ dz, and we next need to change the dz to a dξ, which we get from differentiating (12.37) with respect to ξ. Before doing this, we can simplify by noting that dz on
334
Chapter 12. Steepest Descent for Approximating Integrals
the upper branch matches (−dz) on the lower branch.95 Adding the two, the contribution
from the first bracket in (12.37) cancels, and the one from the second should be doubled.
Following this observation, we obtain from (12.37)
1 1/2
i
1 3/2
139 5/2
571
−1/2
7/2
√
2ξ
− ξ
ξ
ξ
ξ
dz =
+
+
−
+ · · · dξ.
3
108
48600
8164800
2
The integral in (12.36) has now become equivalent to
ˆ 0+
1 1/2
1 3/2
139 5/2
571
i
−1/2
7/2
√
2ξ
− ξ
+
ξ
+
ξ
−
ξ
+ · · · e−sξ dξ,
3
108
48600
8164800
2
0
which is in the perfect form for applying Watson’s lemma. Remembering the factor 2πi 1ss−1
from (12.33) and also the factor eφ(1)s = es , we have again obtained the expansion (12.35).
Method 3: Follow the tangent at saddle point: Although the tangent (being a straight
line) is a simpler curve than the true steepest descent path, Im φ(z) is not constant along it,
preventing us from factoring es Im φ(z) out of the integral. When s increases, the integrand
gets not only increasingly peaked, but also increasingly oscillatory. The approximation
procedure becomes as a result somewhat different.
In order to locally follow the tangent in the present case, we let z = 1 + it, and consider
t along an infinitesimally short stretch around t = 0, along a real t-axis. The integral in
(12.32) can then be replaced by
ˆ
ˆ 0+
1
−1
1
s Re (z−log z)
=
e
dz
∼
es (1+it−log(1+it)) dt . (12.38)
Γ(s)
2πi ss−1 C
2π ss−1 0−
The key idea in this approach is to Taylor expand the exponent, and separate off the constant
and the quadratic terms:
es (1+it−log(1+it)) = e
3
2
4
5
6
7
8
s 1− t2 +s i t3 + t4 −i t5 − t6 +i t7 + t8 +···
2
s − st2
=e e
= es e
2
− st2
3
4
5
6
7
8
s i t3 + t4 −i t5 − t6 +i t7 + t8 +···
e
(12.39)
ist3
st4
ist5
s(3 + s)t6
1+
+
−
−
3
4
5
18
7
is(12 + 7s)t
s(60 + 47s)t8
+
+
+ ··· .
84
480
If our aim is just to obtain the leading term, we don’t need anything beyond the factor
st2
es e− 2 , making this approach then very convenient. Pursuing more terms, we note that,
following the leading “1” in the last expansion above, the next three terms are of size O(s),
then three of size O(s2 ), etc. Knowing that only a vicinity of t = 0 matters, we next
swap the integration interval from [−ε, ε] to [−∞, ∞], after which we can do term-by-term
integration in closed form thanks to the relation (α > 0):
ˆ ∞
n+1
2
2
if n even,
Γ( n+1
2 )/α
e−α t tn dt =
(12.40)
0
if n odd
−∞
95 Given the symmetry of the path (green curve in Figure 12.7), at matching points z and z on the curve, it will
hold that dz
= −Re dz
+ Im dz
.
dξ
dξ
dξ
12.4. Steepest descent
335
´∞
n+1
2
2 , n > −1).
(which generalizes for one-sided cases to 0 e−α t tn dt = 12 Γ( n+1
2 )/α
Since all terms with odd powers of t vanish, we obtain from (12.39), using even powers up
through t12 , that
r 3 1
5 3+s
7 60 + 47s
−1
1
π
1
s
e
−
+
=
2 1/2 +
s−1
3/2
5/2
Γ(s)
2π s
2
2 s
3 s
16 s7/2
s
3 1260 + s(1377 + 175s)
−
20
s9/2
11 90720 + s(120564 + 7s(4437 + 80s))
+
·
·
·
.
+
576
s11/2
(12.41)
This needs to be resorted into inverse powers of s. The increasing powers of s in the
numerators slow up the process, but the powers of s in the denominators grow faster in the
long run. When including still one more term in (12.41), we arrive at
e s r 1 1
1
1
139
(12.42)
=
1−
+
+
+ ··· .
Γ(s)
s
2πs
12s 288s2
51840s3
The need for a quite large number of terms in the internal expansions is not unusual and
can significantly increase the amount of work that is required when one chooses to follow
the tangent instead of the steepest descent curve. However, following the tangent can be
very convenient if we only want the leading term. Also, in some cases, it can provide
closed-form expressions for complete expansions (as we will see for the Airy function in
Section 12.4.3).
Example 12.10. Find the leading term in the expansions for the Jn (t) Bessel functions in
the two limits:
(a) For t fixed, and n → ∞.
(b) For n fixed, and t → ∞.
We first recall the illustration of these Bessel functions in Figure 11.5.
(a) Writing (11.11) as
#
n "
2
4
1 t
1
t
1
t
Jn (t) =
1−
+
− +···
n! 2
1!(n + 1) 2
2!(n + 1)(n + 2) 2
gives immediately
1
Jn (t) ∼
n!
n
t
2
as n → ∞.
(12.43)
See also Exercise 12.6.11.
(b)
(11.12) is in the perfect form for steepest descent.
We write it as Jn (t) =
´ Equation
1
1
1
1
t φ(z)
f
(z)
e
dz,
with
f
(z)
=
and
φ(z)
=
z
−
n+1
2πi C
z
2
z . The saddle points are
where φ0 (z) = 0, i.e., at z = ±i, with the corresponding steepest descent paths shown in
Figure 12.8. Since we aim for only the leading order term, it suffices to integrate along
each of the straight line segments, and only keep the leading contribution.
We focus first on the saddle at z = i: Changing variable z = i + s(i − 1) gives
dz = (i − 1)ds and φ(z) = i − s2 + O(s3 ). Thus,
ˆ
ˆ +ε
ˆ
2
1
1 −(n+1) π i t i +ε −t s2
2 e
∼
f (i) et(i−s ) (i − 1)ds =
e
e
ds.
2πi −ε
2πi
saddle+i
−ε
336
Chapter 12. Steepest Descent for Approximating Integrals
(a) Real part of φ(z) =
1
2
z−
1
z
.
2
y
1
0
-1
-2
-2
-1
0
1
2
x
(b) Paths seen from above.
Figure 12.8. (a) Real part of φ(z) = 12 z − z1 . (b) Solid and dashed black curves show
contour lines of Im φ(z) and Re φ(z), respectively. In both parts (a) and (b), the green curve shows
the steepest descent path through the saddles (red dots) and its local tangents there (blue lines).
The last integral is asymptotically the same as
ˆ
∞
2
e−t s ds =
−∞
r
π
, so
t
Similarly, for the second saddle,
ˆ
∼
saddle+i
´
saddle−i
∼
r
1 +(t−(n+1) π ) i
π
2
e
(i − 1)
.
2πi
t
1 −(t−(n+1) π
2 ) i (i
2πi e
p
+ 1) πt . The sum of
12.4. Steepest descent
337
0.4
Leading term approx.
True function J6 (t)
0.3
0.2
0.1
0
-0.1
-0.2
-0.3
-0.4
0
10
20
30
40
50
60
70
80
90
100
t
Figure 12.9. Comparison between J6 (t) and the leading order asymptotic term (12.44) for
the case of n fixed (here n = 6) and t increasing.
these two contributions simplifies to
r
Jn (t) ∼
2
nπ π cos t −
−
πt
2
4
as t → ∞.
(12.44)
Figure 12.9 illustrates for the case of n = 6 how well already this leading order asymptotic
formula (12.44) approximates the true function J6 (t) as t increases, with regard to both
amplitude and phase angle. Higher order versions of (12.44) require additional terms for
both of these quantities (see, for example, [35, equations 10.18.17 and 10.18.18]).
Integrals that become increasingly oscillatory in the s → ∞ limit
The key observation here is that an analytic function that is highly oscillatory in one direction in the complex plane may become rapidly decaying in another direction. The steepest
descent approach may then again be effective in obtaining asymptotic expansions, as illustrated in the next example.
Example 12.11. Find the asymptotic expansion for I(s) =
´1
0
cos sx2 dx as s → ∞.
Figure 12.10(a) illustrates this integrand for s = 10 and s = 100. The cost of direct
numerical integration increases rapidly with s, since the whole interval will then need to
become increasingly finely resolved. The integrand cos sx2 grows rapidly in size when
analytically continued away from the real axis, so an immediate change of integration path
´1
2
is unlikely to help. However, we can rewrite I(s) as I(s) = Re 0 eisx dx, and the form
is now that of a Laplace integral. Figures 12.10(b) and 12.11 illustrate the corresponding function φ(z) = i z 2 in the complex plane. If we, instead of the original path [0, 1]
(marked blue), follow the green path starting at z = 0 and then return via the other green
path ending at z = 1, the value of the integral has not changed, but we now have two
steepest descent paths. Only immediate vicinities of z = 0 and z = 1 will thus matter for
the asymptotic expansion. Had the value of Re φ(0) been different from that of Re φ(1),
we could for the asymptotic expansion have ignored the path with the lower value, but
Re φ(0) = Re φ(1) = 0, so we need to consider both paths separately (and then add the
contributions from the two paths). We write I(s) = Re (J(s) + K(s)) corresponding to
the two paths.
338
Chapter 12. Steepest Descent for Approximating Integrals
1
0.5
0
-0.5
s = 10
s = 100
-1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
(a) The integrand cos sx2 .
1.5
1
0.5
0
-0.5
-1
-1.5
-1
-0.5
0
0.5
1
1.5
2
(b) Integration paths.
Figure 12.10. (a) The integrand cos sx2 displayed for s = 10 and s = 100. (b) The
steepest descent paths for Re φ(z) with φ(z) = i z 2 (black, solid) and level curves (black, dashed).
The original integration path is here shown in blue, and the two steepest descent paths are shown in
green.
Path originating at z = 0: The variable change z = ei π/4 t gives dz = ei π/4 dt and
´ 0+
´∞
pπ
i π/4 p
2
2
π
1
J(s) ∼ ei π/4 0 e−s t dt ∼ ei π/4 0 e−s t dt = e 2
s , i.e., Re J(s) = 2
2s .
Path returning to z = 1: As an alternative to first splitting φ(z) in real and imaginary
parts and determining the steepest descent path, we immediately apply (12.14): −ξ =
12.4. Steepest descent
339
(a) Real part of φ(z) = i z 2 .
(b) Imaginary part of φ(z) = i z 2 .
Figure 12.11. Real and imaginary parts of φ(z) = i z 2 , with the two ends of the original
integration interval marked with red dots and the steepest descent paths for Re φ(z) shown by green
curves. The imaginary part is shown from a different perspective to better illustrate that these same
curves are level curves for Im φ(z).
1/2
−1/2
φ(z) − φ(1) = i z 2 − i and obtain z = (1 + iξ) , dz = 2i (1 + iξ)
dξ, and
´
P∞ (−i ξ)n Γ(n+ 12 )
−1/2 −ξ s
−1/2
i is ∞
K(s) ∼ 2 e 0 (1 + iξ)
e
dξ. With (1 + iξ)
= n=0
, Watn! Γ( 12 )
n
1
P∞ (−i) Γ(n+ )
son’s lemma gives K(s) ∼ − 2i ei s n=0 Γ 1 sn+12 .
(2)
340
Chapter 12. Steepest Descent for Approximating Integrals
20
-10
-5
-25
-30
-15
N
30
-35
-25
-20
0
40
3 40
10 20 0
50
10
10
20
30
-15
-10
-5
0
-20
-15
-10
40
50
60
-10
70
80
90
100
s
Figure 12.12. log10 of the relative errors (shown numerically along the contour lines)
when (12.45) is applied with different values of s using N terms in the asymptotic expansion. The
areas shaded gray indicate where the accuracy is better than 10−15 .
Accuracy of the expansion: In the same style as Figures 12.2 and 12.5, Figure 12.12
shows the error in the asymptotic expansion
"
#
r
ˆ 1
∞
i i s X (−i)n Γ n + 12
1 π
2
I(s) =
− Re e
cos sx dx ∼
(12.45)
2 2s
2
Γ 12 sn+1
0
n=0
as a function of s and when using terms up through n = N in the expansion. The right
edge (s = 100) corresponds to the red curve in Figure 12.10(a).
12.4.3 Steepest descent analysis of Ai(z)
The Airy function Ai(z) was introduced (and illustrated)´ in Section 11.1 as the unique
∞
bounded solution to u00 (z) − z u(z) = 0, normalized by −∞ Ai(z)dz = 1. Based on its
integral representation (11.23)
ˆ
3
1
e−t /3+z t dt
(12.46)
Ai(z) =
2πi C
we will next use steepest descent to find its asymptotic expansions for z → +∞ and
z → −∞. The integration path C in (12.46) connects “valley V2 ” to “valley V3 ,” as
illustrated in Figure 11.11. The first step is to change variables to bring (12.46) to standard
Laplace form with the exponential factor es φ(z) , where s is the parameter that goes to plus
infinity and z is the integration variable.
Ai(z) for z → +∞
Method 1: Follow steepest descent curve: In order for (12.46) to take the form of
a Laplace integral, z should be a factor in the exponent −t3 /3 + z t in (12.46), suggesting the change of variable t = z 1/2 τ with dt = z 1/2 dτ , which yields Ai(z) =
´ 1/2 +z3/2 (−τ 3 /3+τ )
1
e
dτ . We next arrive at the standard form by further renaming
2πi C z
z 3/2 as s and then τ as z:
ˆ
3
s1/3
I(s) = Ai(s2/3 ) =
es(−z /3+z) dz .
(12.47)
2πi C
12.4. Steepest descent
341
With φ(z) = −z 3 /3 + z, solving dφ
dz = 0 shows that the only two saddle points are
z = ±1. The lowest path between the valleys V2 to V3 will thus go through the saddle
at z = −1;pcf. Figures 12.13(a)–(b). The steepest descent path
p becomes Im φ(z) = 0,
i.e., x = − 1 + y 2 /3 and, along it, Re φ(z) = − 92 (3 + 4y 2 ) 1 + y 2 /3. The standard
Watson’s lemma variable change gives
p
2
2
−ξ = φ(y) − φ(0) = − (3 + 4y 2 ) 1 + y 2 /3 +
9
3
4
6
8
10
5y
7y
y
11y
91y 12
= y2 +
−
+
−
+
+ ··· ,
36
648 576 31104 1119744
which becomes, when inverted,
p
5ξ 77ξ 2 2431ξ 3 1062347ξ 4 14003665ξ 5 1168767425ξ 6
−
+
−
+
+· · · .
y = ξ 1− +
72 3456 248832 214990848 515978352 743008370688
dx
As in Example 12.9, Method 1, dz = dx + i dy = dx
dy + i dy, where dy can be omitted,
since it is an odd function that will vanish when√integrating. With f (y) = 1 and φ(y) =
p
3 9+3y 2
− 29 (3 + 4y 2 ) 1 + y 2 /3, we get − φf0(y)
(y) = 18y+8y 3 , i.e.,
5ξ 1/2
385ξ 3/2
17017ξ 5/2
1062347ξ 7/2
ξ −1/2
f (y)
−
+
−
+
− 0
=
φ (y)
2
48
6219
497664
47775744
−
154040315ξ 9/2
15193976525ξ 11/2
+
+ ··· ,
10319560704
1486016741376
and then, by Watson’s lemma,
e−2s/3
385
85085
37182145
5
I(s) = √ 1/6 1 −
+
−
+
2
3
48s 4608s
663552s
127401984s4
2 πs
5391411025
5849680962125
−
+
+ ··· .
6115295232s5
1761205026816s6
The expansion for Ai(z), z → +∞ follows now by substituting s → z 3/2 .
Method 2: Follow tangent at saddle point: With z = −1 + i t, we obtain φ(t) =
3
3
− z3 + z = − 23 − t2 − i 3t , and therefore
ˆ
ˆ
3
2 i s t3
s − 23 −t2 − i 3t
− 2s
3
e
dt ∼ e
e−s t e 3 dt {substitute t = s−1/2 u, dt = s−1/2 du}
C
∼e
− 2s
3
C
0+
ˆ
e
−u2
e
i√
u3
3 s
i s−1/2 du
0−
2s
∼i
e− 3
s1/2
∼i
e− 3
s1/2
2s
ˆ
n
∞
X
1 i u3
√
du {note that only terms
n! 3 s
−∞
n=0
n = 0, 2, 4, . . .
contribute}
∞
m
X Γ(3m + 1 )
1
2
−
.
(12.48)
(2m)!
9s
m=0
∞
2
e−u
342
Chapter 12. Steepest Descent for Approximating Integrals
30
20
10
0
-10
-20
-30
3
2
1
0
-1
-2
-3
0
-1
-2
-3
1
2
3
3
(a) Real part of φ(z) = − z3 + z.
3
2
1
0
-1
-2
-3
-3
-2
-1
0
1
2
3
(b) Steepest descent paths.
3
Figure 12.13. Displays of the function φ(z) = − z3 + z, with its saddle points at z = ±1
marked with red dots and the steepest descent path through the one at z = −1 shown in green.
The same path is a level curve for Im φ(z) and is therefore the appropriate integration path for the
present problem. (a) Surface plot of Re φ(z) and (b) steepest descent paths of Re φ(z) (solid) and
level curves of Re φ(z) (dashed).
12.4. Steepest descent
343
The idea behind the substitution in the first line of (12.48) becomes apparent in its lines
3 and 4. It is rare that the approach of following the tangent simplifies this much. For
z → +∞, we now obtain the complete expansion:
Ai(z) ∼
m
∞
1
z −1/4 − 2 z3/2 X Γ(3m + 12 )
√ e 3
.
− 3/2
(2m)!
2 π
9z
m=0
(12.49)
Ai(z) for z → −∞
Instead of (12.47), we need now to consider the s → +∞ limit of
ˆ
3
s1/3
2/3
I(s) = Ai(−s ) =
es(−z /3−z) dz ,
2πi C
(12.50)
for which φ(z) = −z 3 /3 − z. The change of sign of the last term of φ(z) will have
no bearing on the character of the valleys V2 and V3 , but it has caused the saddle points
to move from z = ±1 to z = ±i. The lowest path between the two valleys will now
be a composite of two steepest descent sections, with equally high saddles, as shown in
Figure 12.14(a)–(b). Methods 1 and 2 will work once we sum the contributions from the
two saddles. Omitting the algebra, it will transpire that Method 2 again will produce a
closed-form result:
"
∞
2m
2
(−z)−1/4
π X (−1)m Γ(6m + 21 )
1
√
cos
Ai(z) ∼
(−z)−3/2 −
−
3
4 m=0 (4m)!
π
9(−z)3/2
Γ( 12 )
+ sin
∞
π X
2
(−z)−3/2 −
3
4
(−1)m Γ(6m +
(4m + 2)!
Γ( 12 )
m=0
7
2)
(12.51)
2m+1 #
1
.
−
9(−z)3/2
Figure 12.15 shows that both this expansion (12.51) for z negative and (12.49) for z positive
can give excellent accuracies for |z| greater than about 10 or 15. In contrast, it can easily
be verified that the Taylor expansion of Ai(z) around the origin is barely useful beyond |z|
around 2.
The form of (12.51) does not convey very clearly that the Ai(z) function for z → −∞,
while decaying in magnitude, oscillates with a slowly changing frequency (as was seen in
Figure 11.1). Hence, it is preferable to re-express (12.51) in terms of magnitude and phase
angle. Denoting the two sums by Sp
1 and S2 , respectively, we thusapply the trigonometric
identity (cos α) S1 + (sin α) S2 = S12 + S22 sin α + arctan SS12 to obtain
Ai(z) = M (z) sin θ(z),
where, for z → −∞,
∞
X
1
1 · 3 · 5 · · · · · (6k − 1) 1
M (z) ∼
k!(96)k
z3
π(−z)1/2 k=0
2
and
θ(z) ∼
π 2
5
1105
82825
1282031525
+ (−z)3/2 1 +
+
+
+
+
·
·
·
.
4
3
32 z 3
6144 z 6
65536 z 9
58720256 z 12
344
Chapter 12. Steepest Descent for Approximating Integrals
20
15
10
5
0
-5
-10
-15
-20
3
2
1
0
-1
-2
-3
-1
-2
-3
1
0
2
3
3
(a) Real part of φ(z) = − z3 − z.
3
2
1
0
-1
-2
-3
-3
-2
-1
0
1
2
3
(b) Steepest descent paths.
3
Figure 12.14. Displays of the function φ(z) = − z3 − z, with its saddle points at z = ±i
marked with red dots. The appropriate lowermost path passes through both saddles and is shown in
green. (a) Surface plot of Re φ(z) and (b) steepest descent paths of Re φ(z) (solid) and level curves
of Re φ(z) (dashed).
12.4. Steepest descent
345
20
-10
-15
-5
20
-25
5
5
-3
-40
16
10
5
-4
18
-20
0
-3
15
0
14
-35
-10
-30
10
10
N
12
-25
-25
-20
-5
6
-15
5
8
0
-20
-15
4
-10
-15
-10
-10
2
-5
-5
0
-30
-25
-20
-15
-10
-5
x
(a) Asymptotic expansion z ∈ [−30, −1].
-35
0
-10
-25
-20
-5
-3
-15
18
15
10
5
0
20
16
-30
14
-25
-5
0
N
-10
10
-2
0
5
-1
5
12
-20
8
-15
6
2
-15
-10
4
-10
-5
-5
0
5
10
15
20
25
30
x
(b) Asymptotic expansion z ∈ [1, 30].
Figure 12.15. log10 of the relative errors (shown numerically along the contour lines)
when Ai(x) is approximated at locations z along the real axis when using N terms in each of the
expansions in (12.51) and N terms in the expansion (12.49), respectively. The areas shaded gray
indicate where the accuracy is better than 10−15 .
Although no practical closed-form expressions are available for θ(z), that is of little practical consequence, since this reformulation into magnitude and phase angle was mainly used
to provide qualitative insight. In general, closed-form expressions for asymptotic expansions are rarities in all but the simplest cases. What is important is that we have systematic methods available to generate some moderate number of leading terms in asymptotic
346
Chapter 12. Steepest Descent for Approximating Integrals
expansions. Although steepest descent provides a very powerful mathematical approach,
symbolic algebra packages are essential for utilizing this most effectively.
12.4.4 Leading order saddle point approximation
Just as for Laplace integrals (cf. Section 12.3.5), the saddle point method can be reduced to
a simple formula in case we only want the leading term in the approximation.
Theorem 12.12. With a saddle at z = z0 at which φ0 (z0 ) = 0, φ00 (z0 ) 6= 0,
ˆ
s
f (z) e
I(s) =
s φ(z)
dz ∼ ±e
s φ(z0 )
f (z0 )
C
2π
.
−s φ00 (z0 )
The ± sign choice depends on the direction in which C passes through the saddle point. If
− π2 ≤ arg(direction) < π2 , use +, else −.
Proof. Near the saddle φ(z) = φ(z0 ) + 12 (z − z0 )2 φ00 (z0 ) + · · · . We make the change of
p
τ , choosing for the
independent variable τ = (z − z0 ) −φ00 (z0 ), i.e., z = z0 + √ 100
square root the principal branch. Then, s φ(z) =
and
ˆ
ˆ
f (z0 )esφ(z0 )
p
−φ00 (z0 )
f (z) es φ(z) dz ∼
I(s) =
C
−φ (z0 )
1
2
s φ(z0 )− 2 sτ +· · · , dz
C1
!
1
e−s 2 τ
2
1
dτ ,
−φ00 (z0 )
=√
dτ,
where the new integration path goes from 0− to 0+ , which, without affecting the asymptotics, can be changed toq−∞ to ∞. The first bracket inside the integral factors out, and
the second integrates to
2π
s .
12.5 Supplementary materials
12.5.1 The Riemann–Siegel formula
One might have expected that the first known use of the saddle point method would have
been on some simple demonstration example, but it was instead on a case of great mathematical beauty, found by Siegel around 1932 in unpublished notes by Riemann, dating
from the 1850s.96 Numerical hand calculations by Riemann, utilizing what has now become known as the Riemann–Siegel formula, most likely contributed to him proposing the
Riemann hypothesis.
In order to understand the ζ(s) function (with s = σ + it) for 0 ≤ σ ≤ 1 and t large,
direct use of (6.19) is impractical, since no choice of path makes the integrand sharply
peaked. However, writing (5.17) as
ζ(s) = −
96 Laplace
Γ(1 − s)
2πi
ˆ
Γ
(−z)s−1 1
dz
ez − 1 z
(12.52)
used earlier a similar method in the special case of integrals along the real axis. Others, before Carl
L. Siegel, attempted to understand Riemann’s difficult-to-interpret notes, but he was the first one to fully succeed.
12.5. Supplementary materials
347
(with the contour Γ opening to the right) and then rewriting the factor
1
ez −1
as
N
X
1
e−N z
=
e−nz + z
z
e − 1 n=1
e −1
brings (12.52) to
ζ(s) =
ˆ
N
X
Γ(1 − s)
e−N z (−z)s−1 1
1
−
dz.
s
n
2πi
ez − 1
z
H
n=1
−N z
(12.53)
s−1
1
Figure 12.16 shows this integrand e ez(−z)
−1
z in the case of N = 3 and s = σ + it
1
with σ = 2 and
t = 100. There has now appeared a small choice of saddles, one near
z = − 21 + it /N . Going through this saddle (along a straight line path, angled 45o to
the axes) also encloses a number M of poles caused by the factor ez1−1 , with residues that
need to be compensated for. Accounting for these, (12.53) takes the form
ζ(s) =
ˆ
N
M
1−s X
X
1
1
Γ(1 − s)
e−N z (−z)s−1 1
s− 12 Γ( 2 )
+π
−
dz. (12.54)
s
s
1−s
n
Γ( 2 ) n=1 n
2πi
ez − 1
z
H
n=1
It is very practical to choose N =
hq
t
2π
i
(as used in this example; [·] denotes the integer
part), since this turns out to give M = N . Furthermore, in the case of σ = 21 (i.e.,
evaluating ζ(s) on the critical line), the two sums are complex conjugates of each other,
meaning that just one needs to be evaluated. By then making a saddle point approximation
Figure 12.16. The integrand
100i and N = 3.
e−N z (−z)s−1 1
ez −1
z
in (12.53) displayed in the case of s =
1
2
+
348
Chapter 12. Steepest Descent for Approximating Integrals
(for which we here omit the details), ζ(s) has become expressed in a form that is very well
suited for studying the zeta function for large distances up/down the critical strip. When t
is increased, the number of terms in the sum(s) increases only as O(t1/2 ). The gap between
the two regions of very large function values slides up in the complex z-plane, but remains
“clear” for all t.
12.5.2 Select proofs
Euler–Maclaurin formula—derivation via exponential basis functions
´∞
P∞
To arrive at (12.7), we look for an expansion of k=n f (k) − n f (x)dx that is linear
in f , meaning that if it holds for two functions f1 and f2 , it will also hold for any linear
combination αf1 + βf2 . We then recall from the Fourier transform that any function over
[−∞, ∞] can be thought of as a linear superposition of Fourier modes eiwx and, from the
Laplace transform, that functions over [0, ∞] (or equivalently over [n, ∞])P
can as well if
∞
we
allow
w
to
also
have
an
imaginary
part.
It
is
thus
sufficient
to
consider
k=n f (k) −
´∞
wx
f
(x)dx
for
functions
of
the
type
f
(x)
=
e
with
w
complex
(and
Re
w
≤ 0). For
n
dk
k wx
these functions, it holds that dxk f (x) = w e , and we can evaluate the sum and the
integral analytically. Utilizing (5.12) and the fact that odd numbered Bernoulli numbers
beyond B1 vanish, we obtain
ˆ ∞
∞
X
1
1
+
f (k) −
f (x)dx = enw − w
e −1 w
n
k=n
= −enw
∞
X
Bk
k=1
k!
∞
wk−1 =
X B2k
1
f (n) −
f (2k−1) (n).
2
(2k)!
k=1
(12.55)
The fact that the power series expansion converges only for |w| < 2π puts a restriction on
the validity of (12.55), making it only asymptotically correct.
Euler–Maclaurin formula—derivation via contour integration
P∞
The “routine” contour integration approach for evaluating a sum k=n f (k) is to integrate
f (z)π cot πz around a path that includes the points z = n, n + 1, n + 2, . . . and divide
by 2πi (cf. Section 5.2). Hence, consider the closed path C1 , C0 , C2 , C3 shown in black
in Figure 12.17, and assume (i) that f (z) is singularity-free inside it, and (ii) that f (z)
decays fast enough that the contribution from C3 vanishes when chosen far enough out.
With the radius ε of the half-circle C0 small, the contribution from it will be πi f (n).97 It
only remains to find contributions from C1 and C2 .
We focus first on C1 : Along this path z = n + iy, and therefore
2
π cot πz = −iπ 1 + 2πy
e
−1
(cf. Figure 5.25(b)). Thus
ˆ
ˆ
f (z)π cot πzdz = −π
C1
97 Equivalently,
value type.
ε
ˆ
∞
f (n + iy)dy − 2π
ε
∞
f (n + iy)
dy.
e2πy − 1
we can let the path go straight through the pole at z = n and let the integral be of principal
12.5. Supplementary materials
349
C3
C1
C0
0
C3
n
...
n+1
C2
C3
Figure 12.17. The integration paths considered in Section 12.5.2.
´
Regarding the first of the two integrals in the RHS, consider f (z)dz around the dashed
green path in Figure 12.17. With no singularities inside
´ ∞ the path, and with
´ ∞the integral
vanishing along the top and right sides, it holds that 0 f (n + iy)dy = −i n f (x)dx.
Combining this with a similar argument for C2 gives
ˆ
ˆ
C1 +C2
ˆ
∞
f (z)π cot πzdz = 2πi
f (x)dx+2πi
n
0
∞
dy
f (n + iy) − f (n − iy)
.
2πy
i
e
−1
By Taylor expansion,
∞
X
y 2k+1 (2k+1)
f (n + iy) − f (n − iy)
=2
(−1)k
f
(n) ,
i
(2k + 1)!
k=0
and therefore
ˆ
∞
0
∞
X 2(−1)k
f (n + iy) − f (n − iy)
dy
=
f (2k+1) (n)
2πy
i
e
−1
(2k + 1)!
k=0
ˆ
0
∞
y 2k+1 dy
.
e2πy − 1
Remembering from Exercise 6.5.11 that this integral can be expressed in terms of an even
order Bernoulli number, we can collect all pieces together and obtain the same result as in
(12.55):98
ˆ ∞
∞
∞
X
X
1
B2k (2k−1)
f (k) =
f (x)dx + f (n) −
f
(n) .
(12.56)
2
(2k)!
n
k=n
98 This
k=1
contour integration derivation of the Euler–Maclaurin formula was given by G. Plana in 1820 and inde´ ∞ f (iy)−f (−iy)
P
1
k
pendently by N.H. Abel in 1823. Abel also gave ∞
dy (under
k=0 (−1) f (k) = 2 f (0) + i 0
2 sinh πy
appropriate convergence assumptions).
350
Chapter 12. Steepest Descent for Approximating Integrals
12.6 Exercises
Exercise 12.6.1. The Bernoulli polynomials Bk (x) are defined by their generating function
as
∞
X
t etx
tk
=
.
(12.57)
Bk (x)
t
e −1
k!
k=0
(a) Verify that Bk (0) = Bk , where Bk are the Bernoulli numbers, as defined in (5.12).
dBk (x)
(b) Show that
= k Bk−1 (x), k = 1, 2, 3, . . . .
dx
(c) Given the first Bernoulli numbers B0 = 1, B1 = − 12 , B2 = 16 , B3 = 0, B4 =
1
− 30 , . . ., calculate the matching Bernoulli polynomials.
(d) Show that Bk+1 (x + 1) − Bk+1 (x) = (k + 1) xk , and from this deduce that
n
X
km =
k=1
1
[Bm+1 (n + 1) − Bm+1 (1)]] , m = 0, 1, 2, 3, . . . .
m+1
(e) Use the results above in this exercise to confirm the result of Example 12.4.
Exercise 12.6.2. On an equispaced grid with xk = x0 + kh, k = 0, 1, . . . , N , the Euler–
Maclaurin formula (12.8) provides the error expansion for the trapezoidal rule:
ˆ
xN
x0
!
h
f (xk ) − [f (x0 ) + f (xN )]
f (x)dx − h
2
k=0
i
i
2 h
h
h4 h (3)
≈+
f (1) (x0 ) − f (1) (xN ) −
f (x0 ) − f (3) (xN )
(12.58)
12
720
i
h
i
h6 h (5)
h8
+
f (x0 ) − f (5) (xN ) −
f (7) (x0 ) − f (7) (xN ) + − · · · ,
30240
1209600
N
X
with coefficients obtained from the generating function
1
z 1
1
1 3
1
1
1
1 1
z−
z +
z5 −
z7 + − · · · .
− − = coth − =
−z
1−e
z 2
2
2 z
12
720
30240
1209600
(12.59)
When instead using the midpoint rule over the same interval
[x
,
x
],
the
nodes
are
placed
0
N
at the equispaced interior locations ξk = x0 + k + 12 h, k = 0, 1, . . . , N − 1, and the
error expansion becomes
ˆ
xN
f (x)dx −
h
x0
N
−1
X
k=0
2 h
!
f (ξk )
i 7 h4 h
i
h
f (1) (x0 ) − f (1) (xN ) +
f (3) (x0 ) − f (3) (xN ) (12.60)
24
5760
i
31 h6 h (5)
(5)
+−
f (x0 ) − f (xN )
967680
i
127 h8 h (7)
f (x0 ) − f (7) (xN ) + − · · · .
+
154828800
≈−
12.6. Exercises
351
Determine from what generating function the error expansion coefficients arise in this
case.99
Exercise 12.6.3. Using the midpoint rule over [x0 , ∞], with nodes at ξk = x0 + k + 12 h,
k = 0, 1, 2, . . ., and assuming that f (z) decays to zero for Re z > x0 , show the exact error
formula
!
ˆ ∞
ˆ ∞
∞
X
f (x0 + iy) − f (x0 − iy)
f (x)dx − h
f (ξk ) = i
dy.
1 + e2πy/h
0
x0
k=0
Exercise 12.6.4. The counterpart to the Euler–Maclaurin formula (12.7) for an alternating
series is100
∞
X
(−1)n
1
1
1 (5)
k
(−1) f (k) =
f (n) − f 0 (n) + f 000 (n) −
f (n)
2
2
24
240
k=n
(12.61)
17 (7)
f (n) − + · · · .
+
40370
Note that this expression contains no integral term.
(a) Relate the coefficients in (12.61) to the Taylor coefficients of an elementary function.
(b) Find a closed form expression for the expansion coefficients.
Hint: The identity ez1+1 = ez1−1 − e2z2−1 will be helpful.
Exercise 12.6.5. Show that
Pn
k 2
k=1 (−1) k
Pn
/ ( k=1 k) = (−1)n .
Exercise 12.6.6. Combine the method in Section 12.5.2 with the result of Example 5.14 to
justify Ramanujan’s formula (3.11).
ExerciseP12.6.7. Derive Stirling’s formula (12.23) by applying the Euler–Maclaurin forn
mula to k=2 log k. This gives all terms, apart from the constant term 12 log 2π, which can
then be deduced, for example, by using (6.7).
Exercise 12.6.8. Show the following
complete expansion of Stirling’s formula: log Γ(z) ∼
P∞
B2k
(z− 21 ) log z−z+ 12 log 2π+ k=1 2k(2k−1)
, where B2k denotes the (2k)th Bernoulli
z 2k−1
number by following the hint below.
Hint: With some effort (you do not need to carry out this step), (6.2) can be rearranged into
´∞
0
(z)
d
ψ(z) = dz
log Γ(z) = ΓΓ(z)
= log z + 0 1t − 1−e1 −t e−zt dt. With the definition of the
d
Bernoulli numbers given in (5.12), the full expansion for dz
log Γ(z) follows from Watson’s
lemma. Term-by-term integration gives the desired result, apart from the constant, which
can then be deduced from (6.7) (see also Exercise 12.6.9).
99 Equation (12.60) is commonly known as “the second Euler–Maclaurin formula.”
The closed
form for the error expansion is similar to the one for the regular Euler–Maclaurin expansion:
P
B2k h2k
− ∞
1 − 21−2k f (2k−1) (x0 ) − f (2k−1) (xN ) . The two versions are equivalent, as follows
k=1 (2k)!
1
from a slight generalization of the trivial identity 2 f (0) + f (1) + f (2) + f (3) + f (4) + · · · − 21 f (0) +
f (2) + f (4) + · · · = f (1) + f (3) + f (5) + · · · .
100 Also known as the Euler–Boole method.
352
Chapter 12. Steepest Descent for Approximating Integrals
Exercise 12.6.9. The duplication formula for the gamma function
Γ(2z) = π
1
Γ(z)Γ z +
2
−1/2 2z−1
2
π 1/2 Γ(2z)
(z+ 21 )
(see (6.7)) is equivalent to the ratio f (z) =
22z−1 Γ(z)Γ
being identically equal to 1.
Prove this result by the following steps: (i) Show that f (z) is an entire function that never
takes the value zero, (ii) consider log f (z) for |z| large, and show from the leading terms of
(12.24) that this quantity is bounded, (iii) apply Liouville’s theorem (Theorem 4.12), and
(iv) verify that f (z) = 1 for some value of z.
Exercise 12.6.10. Show that keeping just the leading term in the general approach in
Section 12.3 indeed produces the formulas (12.26) and (12.27).
Exercise 12.6.11. Derive (12.43) by instead starting from
(t/2)n
Jn (t) = √
πΓ(n + 12 )
ˆ
π
cos(t cos θ) sin2n θ dθ,
0
valid for n > − 21 .
√
´∞
Exercise 12.6.12. Show that, to leading order as s → ∞, I(s) = 0 e−x−s/ x dx ∼
p π 2/3 1/3 −3(s/2)2/3
s e
.
3 2
Hint: As the integral stands, it is a case of a movable maximum, as discussed in Section
12.3.4.
Exercise 12.6.13. Show that for x → ∞
ˆ
π/2
I(x) =
√
−x (sin t)4
sin t e
0
Γ 78
3Γ 11
Γ 38
8
+ ··· .
dt ∼ 3/8 + 7/8 +
4x
8x
32x11/8
Exercise 12.6.14. Show that for x → ∞
ˆ
π/2
I(x) =
e
i x cos t
0
r
dt ∼
π i (x− π )
4
e
+ ··· .
2x
Exercise 12.6.15. Show that for x → ∞
ˆ
∞
I(x) =
e
−∞
i t3 /3
√
i π
2 3/2 π
sin xt dt ∼ 1/4 cos
x −
+ ··· .
3
4
x
´1
Exercise 12.6.16. Show that the large-s expansion for I(s) = 0 eisx log x dx starts
I(s) ∼ − i log s+iγ+π/2
+ O s12 .
s´
∞ −x
Hint: The result 0 e log x dx = −γ may become useful (cf. Exercise 6.5.4).
12.6. Exercises
353
Figure 12.18. The function Re φ(z) given in the hint of Exercise 12.6.20.
Exercise 12.6.17. Find the leading term in the large-s expansion for the following functions:
´ π/4
(a)
cos(s t2 ) tan2 t dt,
´0π is cos t
(b)
e
dt.
0
P∞
n
Exercise 12.6.18. Let f (z) =
n=0 an z be an entire function. By (4.10), an =
f
(ξ)
1
2πi Γ ξ n+1 dξ, where Γ goes around the origin once in the positive direction. Consider
f (z) = ez and use the saddle point method to obtain the leading approximation to an for
n large. The result should match the leading term in Stirling’s formula.
Exercise 12.6.19. Verify the first two terms in the asymptotic expansion
√
ˆ 1√
i
π
t eist dt ∼ − eis −
I(s) =
(1 − i)
s
(2s)3/2
0
(for s → ∞).
´1
2
Exercise 12.6.20. Find the leading order approximation to I(s) = 0 e−4sz cos(s(5z −
3
z ))dz as s → ∞.
´1
Hint: First show that I(s) can be rewritten as I(s) = 21 e−2s −1 es φ(z) dz with φ(z) =
(1 + iz)2 (2 + iz). Figure 12.18 shows Re φ(z) with its two saddles and, for each, the
steepest descent/ascent curves.
Exercise 12.6.21. Theorem 9.1 was followed by a brief discussion about the decay rates
of Fourier coefficients ck for 2π-periodic functions. In case f (x) furthermore is an entire
function, the rates become particularly fast. Show that ck 2k k! → 1 as k → ∞ in the case
of f (x) = ecos x .
Hint: Combine the results of (11.13) and (12.43).
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Index
Abel transform, see transform,
Abel
Abel, Niels Henrik, 13, 349
Abel–Ruffini theorem, 13
Ablowitz–Segur solution, 312
Airy function, 311, 340
Airy’s equation, 292, 304
analytic continuation, 22, 71,
279, 300
Borel summation, 83, 91
circle-chain, 75
contour integration, 152–160
functional equation, 78
Padé approximations, 85
partitioning of interval, 81
replace Taylor coefficients, 81
Schwarz reflection principle,
77
subtraction, 82
analytic function, 2, 19–22
analytic geometry, 8
Argand plane, 3
argument, 5
argument principle, see
Cauchy’s argument
principle
arithmetic-geometric mean, 200
asymptotic analysis, 317–349
B-spline, 237, 261
Bäcklund transformations, 311
Barnes’s integral, 301
Basel problem, 81, 150
Bedrosian’s theorem, 271
Bernoulli
Jacob, 65
Johann, 23
number, 65, 151, 176, 188,
350
polynomial, 350
Bessel function, 265, 293–299,
335
Bessel’s equation, 293, 306
beta function, 159, 175–176
incomplete, 175, 301
binomial coefficient, 151
Blaschke factor, 213
Borel summation, see analytic
continuation
branch cut, 41, 43, 44, 132
branch point, see singularity,
branch point
Cardano, Girolamo, 2, 13
cardinal data, 242
Caserati–Weierstrass theorem,
112, 114, 204
Cauchy’s
argument principle, 100, 144
integral formula, 101, 104
theorem, 96
theorem for ODEs, 291
Cauchy’s estimate, 115
Cauchy–Hadamard theorem, 59
Cauchy–Riemann equations,
20–22, 61, 329
circle-chain method, see
analytic continuation
code for plots
Mathematica, 60
MATLAB, 59
color wheel, 17
complex
conjugate, 5
integration, 93
numbers, 1–11
plane, 3
complex derivative, 19
359
conformal mapping, 16, 209
applications, 219
bilinear function, 212
regular polygon, 215, 301
stereographic projection, see
stereographic projection
continued fractions, 85
contour
Bromwich, 247
Hankel, 156, 186, 332
integration, 97
keyhole, 131, 138, 139, 249
Pochhammer, 159
convergence
infinite products, 165
pointwise, 53
uniform, 53
convolution theorem
Fourier transform, 239
Laplace transform, 254
Cotes, Roger, 23
cubic equation, 2
general solution, 13
Dawson’s function, 169, 262
de Moivre’s formula, 6, 28
delta function, 247, 261
derivative, 19
Dirichlet series, 80
domain
doubly connected, 215
simply connected, 94, 214
elliptic function, 191
Jacobi, 197–198, 222
order, 193
Weierstrass, 194
entire function, 20, 203, 204,
236
360
equilateral triangle, 13, 14
error function, 144, 245, 262,
318, 320
Euler numbers, 65
Euler’s
constant, 174, 321
identity, 23
line, 63
method, 306
relations, 23
Euler, Leonhard, 63, 65, 81
Euler–Boole method, 351
Euler–Maclaurin formula, 319,
348–349, 351
second form, 351
expansion
asymptotic, 318
convergent, 317
exponential function, 23
exponential order, 245
Faulhaber’s formula, 322
Fermat prime, 11
Feynman’s trick, 128, 270
Fourier
series, 231
transform, see transform,
Fourier
Fourier–Laplace method, 303
Fubini’s theorem, 58
function
generalized, 247
kernel, 231
functional equation
2 F1 , 302
gamma function, 78
zeta function, 78, 185
fundamental theorem of algebra,
103, 115, 147
gamma function, 55, 78, 81,
140, 173–175, 326
duplication formula, 175, 352
product form, 174
Gamow, George, 8
Gauss constant, 201, 208
Gauss, Carl Friedrich, 11, 200
Gauss–Lucas theorem, 169
Gaussian, 169, 233, 234, 245,
262, 263
generating function, 264, 294,
350
Gibbs’ phenomenon, 53, 269
Index
Green’s theorem, 96, 114
Hadamard, quote, x
half-line splitting, 244, 274
half-plane splitting
product/quotient, 274
sum, 143, 242, 274
Hankel transform, see
transform, Hankel
Hardy, Godfrey H., 92
harmonic
conjugate, 21
functions, 21, 329
sum, 257
Hastings–McLeod solution, 312
Hilbert transform, see
transform, Hilbert
holomorphic, 20
hypergeometric function,
299–303
2 F1 , 299, 300, 303
p Fq , 300
indicial equation, 299
integral equation
Mellin transform, 258
inverse function, 28, 324
Jacobi
elliptic functions, 197, 222
theta function, 189
Jensen’s formula, 118, 166
Jordan’s lemma, 125, 142
l’Hôpital’s rule, 121
Lagrange’s inversion formula,
56
Lambert W-function, 181–184
Laplace
integral, 322, 331
transform, see transform,
Laplace
Laurent expansion, 38, 106–112,
120, 135, 195, 294
coefficients, 108
level curves, 21
linear fractional transform, see
transform, linear
fractional
Liouville’s theorem, 103, 162,
164, 165, 193, 203, 276
generalized, 103, 284, 289
logarithm function, 28
Maclaurin expansion, 22
magnitude, 5
max/min theorems, 104
mean value theorem, 104
Meissel–Mertens constant, 188
Mellin transform, see transform,
Mellin
meromorphic function, 40, 101,
192, 204, 258
midpoint rule, 350, 351
Mittag–Leffler expansion, 160,
162–165
Möbius transform, see
transform, Möbius
modular function, 203
moments, 270
Morera’s theorem, 97, 204
natural boundary, see
singularity, natural
boundary
number systems, 1
ODE, 291
linear, 37, 254, 291, 299, 303
nonlinear, 37, 195, 198, 307
open region, 20
Ostrowski–Hadamard gap
theorem, 90
Padé approximation, see
analytic continuation
Padé table, 86
Painlevé
equations, 307–314
property, 307
paradox, 61
Parseval’s relations, 238
partial fractions, 123, 252, 254
PDE, 236, 289
period
box, 192
strip, 191
period strip, 163
periodic function
doubly periodic, 191
simply periodic, 191
phase angle, 17
phase portrait, 17
Phragmén–Lindelöf theorem,
105
Picard’s theorem, 112, 203
Poisson’s
Index
integral formula, 105, 112
summation formula, 241, 266
polar
form, 5
rules, 5
pole, 10, 162
simple, 40, 123
polygon, regular, 11
prime number, 11, 12, 178
theorem, 179
principal value integral, 141,
143
product
infinite, 160, 174, 179
product rule, 5
Ptolemy’s theorem, 66
quadrant, 8
quaternions, 13
Radon transform, see transform,
Radon
Ramanujan
formula, 84, 140, 351
Srinivasa, 92
Ramanujan’s master theorem,
85, 256, 270
reduction of order, 314
residue, 111
calculus, 119
shortcuts, 120
resonance, 254
Riemann
Bernhard, 173
sheet, 42–44, 48, 314
Riemann’s
hypothesis, 178, 258
mapping theorem, 214, 223
Riemann–Hilbert, 273
Riemann–Siegel
formula, 346
theta function, 180
Riesz function, 258
361
roots of unity, 6, 11
Rouche’s theorem, 144
ruler and compass, 11
saddle point, 330, 336, 341
Schwarz lemma, 214
Schwarz reflection principle, 5,
77, 203
Schwarz–Christoffel, 217, 222,
301
sequence of functions, 51
singularities, 10
singularity, 38–42, 110–112
branch point, 41, 111, 307
cluster, 42
essential, 40, 111
isolated, 38
natural boundary, 42, 73, 189
pole, 40, 110
removable, 38, 110
Sokhotski–Plemelj theorem, 290
steepest descent, 317, 329–346
stereographic projection, 10, 14,
16, 42, 45, 67, 69, 205,
212, 214, 215
Stieltjes function, 87
Stirling’s formula, 256,
326–327, 351
Stokes phenomenon, 319
sums
infinite, 148
Taylor series, 21, 22, 105
coefficients, 108
integral test, 59
inversion, 56
limit comparison test, 59
radius of convergence, 21, 33,
105
ratio test, 59
root test, 59
torus, 205
transform
Abel, 265
Fourier, 231–245, 275
generalized, 241
Hankel, 265
Hilbert, 259–262
Laplace, 245–255
inverse, 247–253
linear fractional, 212
Mellin, 255–259
Möbius, 212
Radon, 265
z, 264, 295
transforms, 231
trapezoidal rule, 350
treasure hunt problem, 8
trigonometric
function, 23
integral, 126
undetermined coefficients, 35,
56, 115
Vitali’s theorem, 53
Wallis product, 162
Watson’s lemma, 322–327
Weierstrass
M-test, 59
product, 160–162
℘-function, 194–197, 308
Wessel, Caspar, 3
Wiener–Hopf, 273
winding number, 129, 144
z-transform, see transform, z
zeta function, 78, 176–181
continuation, 80, 158
critical line, 180
critical strip, 177
functional equation, 185
zeros, 176, 177
At almost all academic institutions worldwide, complex variables and analytic
functions are utilized in courses on applied mathematics, physics, engineering, and
other related subjects. For most students, formulas alone do not provide a sufficient
introduction to this widely taught material, yet illustrations of functions are sparse in
current books on the topic. This is the first primary introductory textbook on complex
variables and analytic functions to make extensive use of functional illustrations.
Aiming to reach undergraduate students entering the world of complex variables and
analytic functions, this book
• utilizes graphics to visually build on familiar cases and illustrate how these same
functions extend beyond the real axis;
• covers several important topics that are omitted in nearly all recent texts, including
techniques for analytic continuation and discussions of elliptic functions and of
Wiener–Hopf methods; and
• presents current advances in research, highlighting the subject’s active and
fascinating frontier.
The primary audience for this textbook is undergraduate students taking an
introductory course on complex variables and analytic functions. It is also geared
towards graduate students taking a second semester course on these topics, engineers
and physicists who use complex variables in their work, and students and researchers
at any level who want a reference book on the subject.
Bengt Fornberg has been a professor of applied mathematics at the
University of Colorado, Boulder since 1995. His primary research
interests focus on computational methods for solving PDEs and
numerical methodologies related to analytic functions. He has
held positions at the European Organization for Nuclear Research
(CERN), California Institute of Technology, and Exxon Corporate
Research. In 2014, he was elected a SIAM Fellow.
Cécile Piret is a faculty member in the department of mathematical
sciences at Michigan Technological University. A former postdoctoral
fellow at Catholic University of Louvain and the National Center for
Atmospheric Research (NCAR), her research interests revolve around
the development of high order methods for numerically solving
differential equations.
For more information about SIAM books, journals,
conferences, memberships, or activities, contact:
Society for Industrial and Applied Mathematics
3600 Market Street, 6th Floor
Philadelphia, PA 19104-2688 USA
+1-215-382-9800 • Fax +1-215-386-7999
siam@siam.org • www.siam.org
OT165
ISBN: 978-1-611975-97-0
90000
9781611975970