Problem Books in Mathematics
Tomasz Radożycki
Solving
Problems in
Mathematical
Analysis, Part III
Curves and Surfaces, Conditional
Extremes, Curvilinear Integrals, Complex
Functions, Singularities and
Fourier Series
Problem Books in Mathematics
Series Editor:
Peter Winkler
Department of Mathematics
Dartmouth College
Hanover, NH 03755
USA
More information about this series at http://www.springer.com/series/714
Tomasz Radożycki
Solving Problems
in Mathematical Analysis,
Part III
Curves and Surfaces, Conditional Extremes,
Curvilinear Integrals, Complex Functions,
Singularities and Fourier Series
Tomasz Radożycki
Faculty of Mathematics and Natural
Sciences, College of Sciences
Cardinal Stefan Wyszyński University
Warsaw, Poland
Scientific review for the Polish edition: Jerzy Jacek Wojtkiewicz
Based on a translation from the Polish language edition: “Rozwiazujemy
˛
zadania z analizy
matematycznej” cz˛eść 3 by Tomasz Radożycki Copyright ©WYDAWNICTWO OŚWIATOWE “FOSZE” 2015 All Rights Reserved.
ISSN 0941-3502
ISSN 2197-8506 (electronic)
Problem Books in Mathematics
ISBN 978-3-030-38595-8
ISBN 978-3-030-38596-5 (eBook)
https://doi.org/10.1007/978-3-030-38596-5
Mathematics Subject Classification: 00-01, 00A07, 40-XX, 30-XX, 41-XX
© Springer Nature Switzerland AG 2020
This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of
the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation,
broadcasting, reproduction on microfilms or in any other physical way, and transmission or information
storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology
now known or hereafter developed.
The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication
does not imply, even in the absence of a specific statement, that such names are exempt from the relevant
protective laws and regulations and therefore free for general use.
The publisher, the authors, and the editors are safe to assume that the advice and information in this book
are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or
the editors give a warranty, expressed or implied, with respect to the material contained herein or for any
errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional
claims in published maps and institutional affiliations.
This Springer imprint is published by the registered company Springer Nature Switzerland AG.
The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland
Preface
This book is the third and last part of the collection of problems for students in
their first 2 years of undergraduate mathematical analysis. It is intended to include
all the material required from the students in the three–four semesters, although
more ambitious readers will probably discover some gaps (e.g., Lebesgue’s integral,
Fourier’s transform, distributions). Due to the limited volume of the book, however,
I had to make a choice.
This last part is intended for second-year students who have already gained
significant experience in operating various formulas. Very detailed transformations,
which appeared in the previous two parts, would lead to excessive growth of the
volume. Therefore, when possible, I omit some computational details, referring to
the results obtained in the previous volumes. I also assume that the reader is already
familiar with some known integrals (e.g., Gaussian) and that they do not require any
separate derivation.
The chapters devoted to differential forms and oriented integrals are presented
in a “parallel” way, so that the reader can easily see the relation between the vector
description and that using forms.
Warsaw, Poland
Tomasz Radożycki
v
Contents
1
Examining Curves and Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1
Finding Curvature and Torsion of Curves . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2
Examining k-Surfaces in N Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3
Examining Ruled Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4
Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
2
18
31
37
2
Investigating Conditional Extremes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1
Using the Method of the Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . .
2.2
Looking for Global Extremes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3
Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
42
56
65
3
Investigating Integrals with Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
3.1
Examining Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
3.2
Differentiating with Respect to Parameters . . . . . . . . . . . . . . . . . . . . . . . . . 76
3.3
Integrating over Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
3.4
Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
4
Examining Unoriented Curvilinear Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1
Finding Area of Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2
Calculating Various Curvilinear Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3
Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
103
104
114
121
5
Examining Differential Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1
Studying the Exterior Forms Operating on Vectors. . . . . . . . . . . . . . . . .
5.2
Performing Various Operations on Differential Forms . . . . . . . . . . . . .
5.3
Calculating Exterior Derivatives. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4
Looking for Primitive Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5
Finding Potentials in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.6
Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
123
125
133
145
149
164
176
vii
viii
Contents
6
Examining Oriented Curvilinear Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1
Calculating Integrals over Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2
Calculating Integrals over Surfaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3
Using Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4
Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
179
181
189
198
217
7
Studying Functions of Complex Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1
Examining the Holomorphicity of Functions . . . . . . . . . . . . . . . . . . . . . . .
7.2
Finding Domains of Convergence of Complex Series . . . . . . . . . . . . . .
7.3
Calculating Contour Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.4
Using Cauchy’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.5
Looking for Images of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.6
Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
219
220
233
237
245
258
264
8
Investigating Singularities of Complex Functions . . . . . . . . . . . . . . . . . . . . . .
8.1
Identifying the Types of Singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2
Expanding Functions into Laurent Series . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3
Using the Residue Theorem to Calculate Definite Integrals . . . . . . .
8.4
Using Residue Theorem to Find Sums of Series. . . . . . . . . . . . . . . . . . . .
8.5
Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
267
268
278
289
304
312
9
Dealing with Multivalued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.1
Analytically Continuing Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2
Calculating Integrals Involving Functions with Branch Points . . . .
9.3
Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
315
316
327
357
10
Studying Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.1 Examining Expandability of Functions into Fourier Series . . . . . . . .
10.2 Finding Fourier Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3 Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
359
360
363
373
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375
Definitions and Notation
In this book series the notation and conventions adopted in the former two parts are
used. Several additional indications are given below.
• Partial derivatives are often denoted in the following brief way, e.g., ∂/∂x := ∂x ,
∂/∂ϕ := ∂ϕ , etc.
• The versors of the coordinate axes are denoted as ex , ey , ez . Similarly for the
spherical variables the following notation is used: er , eθ , eϕ defined with the
formulas (5.1.22), (5.1.23), and (5.1.24).
• The complex plane is denoted with C and the compactified plane with C̄.
• In the chapters dealing with the functions of the complex variable (i.e., 7, 8, and
9), in contrast to the other chapters of the book and to the previous volumes, the
symbol “ln” denotes the natural logarithm, in order to avoid collision with the
symbol “log” reserved for the complex logarithm.
ix
Chapter 1
Examining Curves and Surfaces
The present chapter is concerned with basic properties of curves and surfaces. By a
curve in RN , we understand the continuous function γ : [a, b] → RN (in R3 we
will write γ (s) or r(s)). Thus, one can say that the curve is the image of the interval
[a, b]. If the function γ is differentiable, the curve is also called differentiable.
If a given curve has cusps, it can be piecewise differentiable. If the function γ is
of the class C n , the curve is also called of the class Cn . If a curve is of sufficiently
high class for a given problem, it is called a smooth curve. Naturally it may also be
piecewise smooth. A curve in R3 is often defined as an intersection of two surfaces.
Unless stated otherwise, it will be assumed below
that N = 3.
The curvature of a curve is defined as κ = d T /ds = T (s), where s ∈ [a, b]
denotes the parameter and T (s) is the vector tangent to the curve at the point labeled
by s, normalized to unity. With the appropriate choice of s, one has T (s) = γ (s).
The vector N(s)
= κ −1 T (s) is the normal vector again normalized to unity.
The so-called binormal vector is defined as B = T × N and is the unit vector too.
constitutes
The orthonormal system composed of these three vectors (T , N , and B)
the Frenet frame. In turn, the planes designated by the Frenet frame, respectively
called the normal plane (that perpendicular to T ), the straightening plane (that
and the strictly tangent plane (that perpendicular to B),
create
perpendicular to N),
the Frenet trihedron. In some textbooks, however, the Frenet trihedron refers to
vectors T , N , and B and not planes.
The following system of Frenet’s equations can be derived (see Problem 2 for
the details):
T (s) = κ N(s),
(1.0.1a)
N (s) = −κ T (s) + τ B(s),
(1.0.1b)
B (s) = −τ N (s).
(1.0.1c)
Naturally κ and τ are in general s-dependent.
© Springer Nature Switzerland AG 2020
T. Radożycki, Solving Problems in Mathematical Analysis, Part III, Problem Books
in Mathematics, https://doi.org/10.1007/978-3-030-38596-5_1
1
2
1 Examining Curves and Surfaces
The torsion of the curve (denoted with τ ) is defined by the formula: τ = −N · B or B = −τ N . The notions of curvature and torsion are addressed in more detail in
the following section, when solving concrete problems.
A k-surface in RN is such subset V ⊂ RN that locally (i.e., at least on some
neighborhood of each point cut to V ) constitutes the graph of a certain function. If
these functions are of the class C 1 , this surface will be called smooth.
A k-surface can also be described in the parametric way, via the relations:
⎧
x1 = g1 (τ1 , τ2 , . . . , τk ),
⎪
⎪
⎨
x2 = g2 (τ1 , τ2 , . . . , τk ),
⎪
...
⎪
⎩
xN = gn (τ1 , τ2 , . . . , τk ),
(1.0.2)
which can be collectively written in the form:
x = G(τ ),
(1.0.3)
for τ ∈ D ⊂ Rk . This representation should be treated in the local sense too. The
condition for (1.0.3) to define a certain smooth k-surface is rank(G ) = k, where G
is the Jacobian matrix.
A ruled surface in R3 is a surface that has the parametrization in the form:
r(u, v) = α (u) + v β(u),
(1.0.4)
where α (u) is called the base curve or directrix and β(u)
the director curve. If a
given surface has two distinct (inequivalent) parametrizations like (1.0.4), then it is
called doubly ruled.
1.1 Finding Curvature and Torsion of Curves
Problem 1
Given the curve on the plane in the parametric form: x(ς ) = 3ς , y(ς ) = 2 ς 3 ,
where ς ∈ [0, ∞[. Its curvature at the point of coordinates (3, 2) will be found.
Solution
In order to imagine what quantity we mean by talking about “curvature” let us
refer to the elementary knowledge of physics. We know that a body performing
the circular motion has the so-called centripetal acceleration equal to
1.1 Finding Curvature and Torsion of Curves
ad =
3
v2
,
R
(1.1.1)
where v is the constant speed of the body and R is the circle’s radius. It is R that
interests us with the difference that, in general, it will change from point to point.
It can be calculated from (1.1.1):
1
ad
1 d v (1.1.2)
= 2 = 2 .
R
v
v dt
Since the speed is constant, this expression can be saved in the form
1 d v 1 d(
1
v /v) =
.
=
·
·
R v 2 dt v
dt (1.1.3)
The velocity vector v(t) is, of course, tangent to the trajectory at the temporary
position of the body. In turn, the vector v/v is a unit tangent vector (surely, it is
assumed here, that v = 0). It will be denoted with the symbol T in order to avoid
collision with time for which the lowercase letter t is reserved. Therefore, one has
1 d T 1
(1.1.4)
= ·
.
R v dt The distance s overpassed by the body on the circle or along any curve is related
to the speed v in the obvious way:
v=
ds
.
dt
(1.1.5)
Hence, instead of calculating the derivatives with respect to time t, one can equally
well calculate the derivatives over s, using the formula for the differentiation of a
composite function. First, denoting the temporary position of the body with r(t),
one can write that
v
d r/dt
d r
T = =
=
,
v
ds/dt
ds
(1.1.6)
and then
1 d T ds 1 d T
1
= ·
· = ·
R v ds dt v ds
d T v = .
ds (1.1.7)
The vector d T /ds which appears here is nothing other than a vector perpendicular to the curve at the same point. In fact, it is proportional to the change of the
vector T , which by definition has the fixed length equal to 1. Thereby, d T may not
4
1 Examining Curves and Surfaces
Fig. 1.1 The translation of
the tangential vector along
the curve
have any component tangent to the trajectory but only perpendicular ones. It is easy
to show this in a formal way by writing
T · T = 1
⇒
T · T
= 2T · T = 0.
(1.1.8)
The considered vector will appear repeatedly in our deliberations, so let us introduce
the symbol:
d T
N = R
,
ds
(1.1.9)
where the presence of R ensures its normalization: N · N = 1.
The result (1.1.7) becomes natural if one looks at the situation sketched in
Fig. 1.1, where the trajectory of a moving body is shown. At the two very close
moments of time, the body is located successively at points denoted as A and B.
The unit vectors tangent to this trajectory at these two points are TA and TB . The
normal vectors (the so-called “principal” ones that are lying in the plane spanned
by TA and TB ) are depicted too. The angle ϕ between two tangential vectors and
two normal vectors is the same because they are orthogonal to each other in pairs.
Imagine now that from the points A and B the two straight lines along perpendicular
vectors NA and NB are drawn. If only the curve is actually “curved” then these two
lines intersect at a certain point (marked in the figure with O). The distance |OA|
(and in fact also |OB|, since we consider the limit B → A) is the radius R of
curvature measured at the point A.
Let us denote with s the length of the arc contained between the points A and B.
Then one has
φ=
s
.
R
(1.1.10)
On the other hand, the same angle can be obtained from the vectors TA and TB too,
the second one having been shifted to A in parallel (the shifted vector is indicated
with the dashed line). Since φ → 0, so
1.1 Finding Curvature and Torsion of Curves
φ=
5
| T |
|TB − TA |
.
=
1
|TA |
(1.1.11)
Comparing these two formulas, it can be seen that
T s
1
= | T |, or
=
= |T |,
R
R
s s→0
(1.1.12)
i.e., again the formula (1.1.7) is obtained. The quantity 1/R is henceforth denoted
with the symbol κ and called the “curvature.” The milder the curve is bent, the
smaller the value of the curvature κ has, and the larger the curvature radius R is,
and vice versa: for a tightly bent curve, the curvature is large and the radius of
curvature is small. In particular for a straight line one obtains κ = 0 and R = ∞
because, regardless of the distance s at which one will slide along it, the angle φ,
by which tangent vector is rotated, will be equal to zero. Incidentally, this is just the
sense of the (local) curvature radius: the shift along the curve at the infinitesimal
distance δs generates the rotation of the tangential vector by the angle δφ = δs/R.
This vector would be rotated by the same angle if it was shifted (at the same
infinitesimal distance) along the arc of the circle of radius R instead of the curve
in question. This situation is schematically presented in Fig. 1.2.
After these general considerations, we can now proceed to solve the specific
problem. The above results indicate that one needs to find the tangential vector T
and then calculate its derivative with respect to the parameter s, introduced below.
As we know, s has the meaning of the distance traveled along the curve. Such
quantities were dealt with in Sect. 3.1 of the second part of this book series. Using
the formula (3.1.7) given at that time and denoting the integration variable with σ ,
one can write
y
Fig. 1.2 The circle tangent to
the curve at some point
(x(ς0 ), y(ς0 )) of the same
curvature radius R
R
circle
curve
y 0
x 0
x
6
1 Examining Curves and Surfaces
ς
ς
x (σ )2 + y (σ )2 dσ =
s(ς ) =
0
√
9 + 9σ dσ
0
ς
=3
√
ς
1 + σ dσ = 2(1 + σ )3/2 = 2(1 + ς )3/2 − 2,
0
(1.1.13)
0
primes in the first line denoting derivatives with respect to σ . This relation can be
very easily reversed, yielding
s
2
ς (s) = 1 +
2/3
− 1.
(1.1.14)
As a consequence, one finds
x(s) = 3ς (s) = 3 1 +
s
2
2/3
− 3,
y(s) = = 2(ς (s))3/2 = 2
1+
s
2
2/3
3/2
−1
(1.1.15)
and after having differentiated:
d r
= (1 + s/2)−1/3 1, (1 + s/2)2/3 − 1 .
T (s) =
ds
(1.1.16)
When calculating the length of this vector, the reader can easily ascertain that it
is in fact a unit vector (T · T = 1). Now one has to further differentiate over s in
accordance with the formula (1.1.7):
d T
1
= (1 + s/2)−4/3 −1, 1/ (1 + s/2)2/3 − 1 ,
(1.1.17)
ds
6
and finally it is obtained that
d T 1
=
κ=
=
R ds 1
.
(1.1.18)
6(1 + s/2) (1 + s/2)2/3 − 1
Something more than what the text of the problem had required has been found:
the curvature can be known at any point of the curve. In particular the point of
coordinates (3, 2) corresponds to the value of the parameter
ς0 = 1, or alternatively
√
using (1.1.13), the point corresponds to s0 = 2(2 2 − 1). By inserting this value
into the formula (1.1.18), one gets
κ=
1
√ ,
12 2
or
√
R = 12 2.
(1.1.19)
1.1 Finding Curvature and Torsion of Curves
7
Problem 2
The curvature and torsion of the curve given in the parametric form:
x(ς ) = sin ς,
y(ς ) = cos ς,
z(ς ) = cosh ς,
(1.1.20)
where ς ∈ [0, ∞[, will be found.
Solution
If a curve does not lie in a plane as it did in the previous exercise but is a truly
spatial curve, in addition to the curvature, it can still undergo some twisting. The
fact that it is described with the use of three coordinates, as in (1.1.20) and not two,
does not determine the spatial character of the curve yet. It can still happen that the
curve is flat while the plane in which it is contained gets inclined. For example, if
one relocates a circle—which undoubtedly is a flat curve—so that it will be tilted
with respect to all three coordinate axes, it will not become the spatial curve! A truly
spatial curve is such that it cannot be contained in any plane, which means that it is
subject to a torsion. The convenient parameter τ , which measures this quantity, was
defined in the theoretical summary. Below we will be concerned with this notion in
detail, starting with some general considerations.
In the previous problem, two important vectors were introduced: the tangent one
T and the normal one N . To complete the set, let us now append the third vector
(called binormal): B = T × N . By definition it is orthogonal to the curve (as it is
orthogonal to T ) and to N and, therefore, also to the plane in which (locally!) the
curve is lying. When the curve is being twisted, this plane is rotating together with
the binormal vector. This situation is depicted in Fig. 1.3.
Note that thanks to the prior normalization of the vectors T and N as well as
their orthogonality one has
B · B = (T × N ) · (T × N ) = |T |2 |N |2 − (T · N )2 = 1.
=1
=1
(1.1.21)
=0
The torsion of the curve should be connected with the rotation of the binormal
vector, because this means the identical rotation of the plane, in which the curve
is lying locally, i.e., the plane defined by the vectors T and N . In view of the
normalization condition (1.1.21), any change of the vector B indicates its rotation
(the same was pointed out in the preceding problem relative to the vector T ). Let us
denote
d B (1.1.22)
|τ | = ,
ds 8
1 Examining Curves and Surfaces
Fig. 1.3 Modification of the
tangential, normal, and
binormal vectors when
moving along the curve
T
B
N
B
N
T
the parameter s (the distance traveled along the curve) having been introduced in
the previous exercise.
What is the interpretation of the parameter τ ? Well, one knows that if a given
vector u is rotating by an infinitesimal angle d α (the vector symbol refers here to
the direction of the rotation axis in accordance with the right-hand screw rule), then
the change in the vector u is given by
d u = d α × u.
(1.1.23)
In our case the vector B is rotating around the perpendicular axis, designated by T .
Hence it can be inferred from (1.1.23) that
d u = |
u|.
(1.1.24)
dα We are interested in the twist of the vector u = B along the curve. So we have
d B =1
= |B|
dα d B ds ⇒ ·
= 1,
ds dα (1.1.25)
i.e.,
dα |τ | = .
ds
(1.1.26)
1.1 Finding Curvature and Torsion of Curves
9
Thus, the needed interpretation has been found: the parameter τ tells us by what
angle the vector B is rotated if one moves along the curve by the distance ds. This
parameter is called the torsion. One can additionally fix the sign of τ . The vector
B(s)
is constantly orthogonal to T (s), so B (s) is as well. In turn, the constant
imposes the orthogonality condition between B (s) and B(s).
normalization of B(s)
it
Consequently, B (s) must be proportional to N (s), since within the trio T , N , B,
is the only remaining linearly independent vector. Usually, one assumes
B (s) = −τ N (s),
(1.1.27)
which stays in harmony with (1.1.22), but also fixes the sign of τ .
As we already know, the vectors T (s), N (s), and B(s)
for each value of s
constitute the complete orthonormal system. Any other vector must be their linear
combination, which has been exploited above to write down B (s). Naturally the
same applies also to the derivatives T (s) and N (s). From the Eq. (1.1.9), one
already has
T (s) = κ N(s).
(1.1.28)
Additionally N · N = 0 due to normalization, so
N (s) = a T (s) + bB(s).
(1.1.29)
The constants a and b are easily determined from the conditions of orthogonality.
For one has
N · T = 0 ⇒ [N · T ] = 0
⇒ N · T = −N · T = −κ|N |2 = −κ.
(1.1.30)
Similarly,
=0
N · B = 0 ⇒ [N · B]
⇒ N · B = −N · B = τ |N |2 = τ
(1.1.31)
and finally
N (s) = −κ T (s) + τ B(s).
(1.1.32)
Equations (1.1.27), (1.1.28), and (1.1.32) are called Frenet’s equations and were
given in the theoretical summary at the beginning of this chapter.
Now let us have a look at the applications of the above in the current problem.
One has to start by tying the parameter s with ς using the following formula:
10
1 Examining Curves and Surfaces
ς
ς
x (σ )2 + y (σ )2 + z (σ )2 dσ =
s(ς ) =
0
cos2 σ + sin2 σ + sinh2 σ dσ
0
ς
=
ς
1 + sinh σ dσ =
2
0
ς
= sinh σ = sinh ς,
ς
cosh2 σ dσ =
0
cosh σ dσ
0
(1.1.33)
0
which is the generalization of (1.1.13) for a curve in three dimensions. Reversing
this relation, one obtains
ς (s) = log(s +
s 2 + 1),
(1.1.34)
where the function on the right-hand side is nothing other than arsinh s, i.e., the
inverse of the hyperbolic sine (see formula (14.1.54) in Part I).
The curve parametrized with this new parameter will now be defined in the
following way:
r(s) = sin ς (s), cos ς (s), s 2 + 1 ,
(1.1.35)
where for the first two coordinates the explicit expression ς (s) is not substituted
so as not to overcomplicate the formula, and the last coordinate derives from the
hyperbolic version of the Pythagorean trigonometric identity.
By calculating the derivative of (1.1.35) with respect to s, the normalized tangent
vector can be found:
d r = ς (s) cos ς (s), −ς (s) sin ς (s), s/ s 2 + 1
T (s) =
ds
1
= √
[cos ς (s), − sin ς (s), s],
(1.1.36)
2
s +1
√
where we have used the derivative: ς (s) = 1/ s 2 + 1. It is easy to find out that
this vector is in fact normalized to unity. For one has
T (s)·T (s) =
s2
1
1
(cos2 ς (s)+sin2 ς (s)+s 2 ) = 2
(1+s 2 ) = 1.
+1
s +1
(1.1.37)
As it has been argued, the curvature is given by
d T 1
κ=
=
,
R
ds (1.1.38)
1.1 Finding Curvature and Torsion of Curves
11
so the derivative of the vector (1.1.36) has to be found. We obtain
d T
1
−
s
cos
ς
(s)
−
s 2 + 1 sin ς (s),
=
ds
(s 2 + 1)3/2
s sin ς (s) − s 2 + 1 cos ς (s), 1 ,
(1.1.39)
and then
d T d T
2
.
·
= 2
ds ds
(s + 1)2
(1.1.40)
Thereby, the curvature κ is found to be:
√
2
1
.
κ(s) =
= 2
R(s)
s +1
(1.1.41)
It naturally depends on the parameter s and, therefore, on the point of the curve. It
becomes clearer if one looks at the curve presented in Fig. 1.4. It is also obvious that
this is in fact a truly spatial curve and, therefore, one should get the nonzero value
of the torsion (but still dependent on the parameter s).
The torsion τ will be found by calculating the derivative of the binormal vector
so first the vector itself should be found from the formula:
B,
B(s)
= T (s) × N (s),
Fig. 1.4 The curve defined
with the parametric
equations (1.1.20). Obviously
it is not a flat curve
(1.1.42)
z
y
x
12
1 Examining Curves and Surfaces
where N is the vector (1.1.39) normalized to unity, i.e.,
1
− s cos ς (s) − s 2 + 1 sin ς (s),
N (s) = √ √
2 s2 + 1
s sin ς (s) − s 2 + 1 cos ς (s), 1 .
(1.1.43)
Calculating the cross product does not present major difficulties and the vector B is
obtained in the form:
1
s cos ς (s) − s 2 + 1 sin ς (s),
B(s)
= √ √
2 s2 + 1
−s sin ς (s) − s 2 + 1 cos ς (s), −1 .
(1.1.44)
As it was previously indicated, this vector is found in the normalized form, and
the reader is encouraged to check it by direct calculation. The last step is to calculate
the derivative of the above expression with respect to s:
s
d B
= √
− s cos ς (s) − s 2 + 1 sin ς (s),
ds
2(s 2 + 1)3/2
s
s sin ς (s) − s 2 + 1 cos ς (s), 1 = 2
N (s),
s +1
(1.1.45)
from which it occurs that the torsion is given by the formula:
τ (s) = −
s
.
s2 + 1
(1.1.46)
In particular at the very start of the curve, i.e., for s = 0, one gets
κ(0) =
√
2,
or
1
R(0) = √ ,
2
τ (0) = 0.
(1.1.47)
At this point, the curvature reaches the largest value (κ(s) is a decreasing function,
as can be seen from (1.1.41)), but the graph is not twisted there. In turn the torsion
of the curve is maximal for s = 1 (this is the maximum of the function τ (s)) and is
equal to τ (1) = −1/2. Then also
1
κ(1) = √ ,
2
or
R(1) =
√
2.
(1.1.48)
It is still worth seeing at the end how the Frenet frame and trihedron look at
some point. By substituting, for example, s = 0 into the formulas (1.1.36), (1.1.43),
and (1.1.44), we obtain
1.1 Finding Curvature and Torsion of Curves
13
T (0) = [1, 0, 0],
√
√
N (0) = [0, −1/ 2, 1/ 2],
√
√
B(0)
= [0, −1/ 2, −1/ 2].
(1.1.49)
All these vectors are clearly orthogonal.
The point on the curve, for which s = 0, has the coordinates (0, 1, 1). In Fig. 1.4,
this point is seemingly located on the y-axis, but this is only an illusion resulting
from the logarithmic scale adopted for the z-axis, which must reflect the fact that
the variable z grows exponentially (strictly speaking, as the hyperbolic cosine). In
fact, the curve starts at a point located above the y-axis, i.e., for z = 1.
As the reader surely knows, the equation of the plane passing through a point
(x0 , y0 , z0 ) and perpendicular to a given vector v can be written in the form:
v · [x − x0 , y − y0 , z − z0 ] = 0.
(1.1.50)
Thus, for the normal plane one gets the equation:
T · [x − 0, y − 1, z − 1] = 0
⇒ x = 0.
(1.1.51)
⇒ y − z = 0,
(1.1.52)
⇒ y + z = 2.
(1.1.53)
Similarly for the straightening plane one finds
N · [x − 0, y − 1, z − 1] = 0
and for a strictly tangent plane:
B · [x − 0, y − 1, z − 1] = 0
Problem 3
The curvature and torsion of the curve representing the intersection of the sphere
x 2 + y 2 + z2 = 9 with the surface of the cylinder (x − 3/2)2 + y 2 = 9/4 (the
so-called Viviani curve) will be found.
Solution
In the first place, the convenient parametrization for our curve should be chosen. We
are going to refer below to the well-known physical notions involving the motion of
a point-like mass along a curve, so unlike previous exercises, the parameter will be
14
1 Examining Curves and Surfaces
denoted with the symbol t and called “time.” It seems to be most suitable if, for x
and y, the parametrization related to the cylindrical surface is adopted, i.e.,
x(t) =
3
(1 + cos t),
2
y(t) =
3
sin t,
2
(1.1.54)
and the variable z is obtained from the equation of the sphere:
9
9
(1 + cos t)2 + sin2 t + z2 = 9.
4
4
(1.1.55)
This entails that
z2 =
9
(1 − cos t) = 9 sin2 (t/2),
2
(1.1.56)
where the known formula 1 − cos α = 2 sin2 (α/2) has been used. As a result, one
gets
z(t) = 3 sin(t/2),
(1.1.57)
where the sign has been chosen arbitrarily since it does not affect the solution of
the problem—when flipping the sign, the two “loops” of the Viviani curve shown in
Fig. 1.5 simply swap their roles. Due to the presence of 1/2 under the sine, in order
to obtain the entire curve, one must allow for t ∈ [0, 4π ].
z
x
Fig. 1.5 The Viviani curve defined in the text of the exercise
y
1.1 Finding Curvature and Torsion of Curves
15
At this point in the previous problems, we were changing the parametrization.
Instead of the original parameter, the “traveled” distance measured along the curve
was introduced. Such a parametrization is called the arc length parametrization. In
this exercise, however, we will proceed in a different manner: the “time” parameter
t still will be used.
Let us start with the tangent vector, i.e., the velocity vector:
v(t) = r˙ (t) =
3
[− sin t, cos t, cos(t/2)].
2
(1.1.58)
The vector of acceleration
3
a (t) = v˙ (t) = − [2 cos t, 2 sin t, sin(t/2)]
4
(1.1.59)
does not need to be orthogonal to the curve (as opposed to T (s)) because v is not
normalized to any particular value. During the motion, the velocity can be changed,
as refers to both its direction and its value. So, in general, the vector of acceleration
can have both components: perpendicular and tangent to the trajectory.
Let us now find the derivative of the acceleration:
3
a˙ (t) = v¨ (t) = [4 sin t, −4 cos t, − cos(t/2)].
8
(1.1.60)
Note now that the determinant of the matrix created of the vectors v, a , and a˙ does
not identically vanish. For one has
⎡
⎤
⎡
⎤
v(t)
− sin t cos t
cos(t/2)
27
81
det ⎣ a (t) ⎦ = −
det ⎣ 2 cos t 2 sin t sin(t/2) ⎦ =
cos(t/2),
64
32
a˙ (t)
4 sin t −4 cos t − cos(t/2)
(1.1.61)
which means that these three vectors are in general linearly independent, i.e., the
curve has actually the spatial character. For a flat curve, the vector a˙ would have
to lie in the plane spanned by v and a and, therefore, would constitute their linear
combination. The other way of expressing the spatial character of the curve is
a˙ · (
v × a ) = 0.
(1.1.62)
Now the question arises, how to construct the curvature κ out of the vectors
introduced above. The answer is provided by the Eq. (1.1.2), where in place of the
centripetal acceleration ad , the component of the acceleration that is perpendicular
to the velocity (i.e., a⊥ ) should be substituted. Denoting the parallel component with
the symbol a , one gets from the properties of the cross product
v
v
v
v
× a = × (
a + a⊥ ) = × a + × a⊥ ,
v
v
v
v
=0
(1.1.63)
16
1 Examining Curves and Surfaces
which implies that
a⊥ := |
a⊥ | =
|
v × a⊥ |
|
v × a |
=
.
v
v
(1.1.64)
Now making use of the formula (1.1.2), one can write
κ=
|
v × a |
1
a⊥
.
= 2 =
R
v
v3
(1.1.65)
The vectors v and a are given by the formulas (1.1.58) and (1.1.59), so in order
to find the curvature, first one has to calculate the product:
v × a =
9
[2 sin t cos(t/2) − cos t sin(t/2), −2 cos t cos(t/2) − sin t sin(t/2), 2],
8
(1.1.66)
and then
3 √
|
v | = √ cos t + 3,
2 2
9
4 cos2 (t/2) + sin2 (t/2) + 4
|
v × a | =
8
9 √
= √
3 cos t + 13.
8 2
(1.1.67)
Now the formula (1.1.65) may be used, which entails
2
κ= ·
3
3 cos t + 13
.
(cos t + 3)3
(1.1.68)
Note that this expression is always positive and, therefore, the curvature does not
vanish anywhere (the curve nowhere “straightens out”). In particular, let us consider
the point of coordinates x = 3 and y = z = 0. This is the point on the Viviani curve
obtained for t = 0, t = 2π , or t = 4π . As it seems from the figure, at these points,
the curvature should be equal to the curvature of the sphere, i.e., R = 3 (κ = 1/3).
Substituting into the formula (1.1.68) the values 0, 2π , or 4π for t, one can easily
verify that actually κ = 1/3.
Now let us pass to the torsion of the curve. The formula for τ analogous
to (1.1.65) has the form
τ=
(
v × a ) · a˙
.
|
v × a |2
(1.1.69)
Note that the numerator of this expression is in fact the determinant (1.1.61) dealt
with earlier, so it is not a surprise that (1.1.69) constitutes a measure of the spatial
1.1 Finding Curvature and Torsion of Curves
17
nature of the curve. In order to strictly justify it, one should first note that the
acceleration of a body (as a vector) has components tangent to the trajectory and
perpendicular to it, which can be written in the form
v2 a = v̇ T +
N.
R
(1.1.70)
The velocity instead is fully tangent, i.e., v = v T . It is easy to calculate the cross
product; one needs
v3 v2 N =
B,
v × a = v T × v̇ T +
R
R
(1.1.71)
It entails also that |
since T × T = 0, and T × N = B.
v × a | = v 3 /R.
When differentiating the expression (1.1.70) with respect to time in order to
obtain a˙ , one can see that several terms emerge. However, only one of them is
needed: exclusively that whose scalar product with B does not vanish. The others
do not contribute to the numerator of (1.1.69). The term in question is that in which
the derivative is acting on the vector N at the very end of the formula (1.1.70).
As a result, it is obtained
Other vectors, i.e., T , T˙ , and N , are orthogonal to B.
from (1.1.69):
τ=
v 3 /R B · (v 2 /R N˙ )
1
1
d B · N,
= B · N˙ = − B˙ · N = −
3
2
v
v
ds
(v /R)
(1.1.72)
i.e., the expression defined in the previous problem by the formula (1.1.27). When
the fact of
transferring the above derivative over time from the vector N onto B,
their orthogonality has been used:
0=
d ˙
[B · N ] = B˙ · N + B · N,
dt
(1.1.73)
and when replacing the derivative over t with that over s, the relation
ds d
d
d
=
·
=v .
dt
dt ds
ds
(1.1.74)
After making use of (1.1.58), (1.1.59), and (1.1.60), one finally finds
τ=
4 cos(t/2)
.
3 cos t + 13
(1.1.75)
It should be stressed at the end that the formulas (1.1.65) and (1.1.69) obtained
when solving this problem allow us to find the curvature and torsion without
introducing the “arc length” parameter.
18
1 Examining Curves and Surfaces
1.2 Examining k-Surfaces in N Dimensions
Problem 1
It will be examined whether the elliptic hyperboloid defined by the equation
x 2 /4 + y 2 /9 − z2 = 1
(1.2.1)
is a surface in R3 .
Solution
The set defined by the Eq. (1.2.1) cut with any “vertical” plane, i.e., that parallel to
the z-axis, becomes a hyperbola, and with any “horizontal” plane (i.e., z = const)—
an ellipse. Hence the appropriate name is used: elliptic hyperboloid. In order to
examine whether this shell is a surface, one must first recall what is understood
by k-dimensional surface in RN . As it was told in the theoretical introduction, this
notion means a certain subset (denoted below with V ) of the space RN such that
in some neighborhood of each of its points (strictly speaking the intersection of the
neighborhood in RN with V ) is a graph of some mapping. For example, in R3 as
a 2-surface—and this is just the case of this exercise—one classifies such set, that
everywhere constitutes a graph of one of the functions: z(x, y), y(x, z), or x(y, z).
What requires the emphasis is that such function does not need to be defined on the
entire set V , but rather one expects the existence of at least one of them for each open
and small “patch” defined for any point of the set V . In addition, the smoothness of
these functions will be demanded, i.e., the existence and continuity of all partial
derivatives. If this condition is fulfilled, the surface will be called smooth. In our
further considerations, we will only deal with such smooth surfaces (later there will
also occur piecewise smooth surfaces), hence it is assumed by default, even if it is
not explicitly articulated.
So, when can one be sure that the equation F (x, y, z) = 0 defines locally at
least one of the functions: z(x, y), y(x, z), or x(y, z)? In order to answer this
question, it is best to refer to the implicit function theorem, known to the reader
from Chap. 8 of the second part of this book series. There, the independent variables
were denoted with the symbol X (for example, this can be the variables x and y),
and the dependent variables with the symbol Y (e.g., z). The theorem stated that if
F is of the class C 1 , then for the existence of the function Y (X) (i.e., z(x, y) in our
example), it is necessary and sufficient that
∂F = 0.
∂Y (X0 ,Y0 )
(1.2.2)
1.2 Examining k-Surfaces in N Dimensions
19
The symbol (X0 , Y0 ) means here simply (x0 , y0 , z0 ), i.e., a point belonging to the
set V , the neighborhood of which is of interest to us. Since it has not been decided
which variable will be treated as the dependent one (and this is in fact irrelevant for
our findings), it is sufficient simply that one of the following conditions be met:
∂F = 0,
∂x (x0 ,y0 ,z0 )
∂F = 0,
∂y (x0 ,y0 ,z0 )
∂F = 0,
∂z (x0 ,y0 ,z0 )
(1.2.3)
which is equivalent to the requirement that at each point (x0 , y0 , z0 ) belonging to
V , the Jacobian matrix
∂F ∂F ∂F
,
,
(1.2.4)
F =
∂x ∂y ∂z
have the rank equal to 1. In our case, after plugging in the expression (1.2.1) one
finds
1 2
F =
x, y, −2z ,
(1.2.5)
2 9
and it is obvious that the rank equals zero only when x = y = z = 0. This point
of the space R3 does not, however, belong to V , since the Eq. (1.2.1) is not satisfied
there. Thus, it has been shown that the rank of the matrix F on the set V is equal to
1, which entails that V is the two-dimensional surface in R3 (Fig. 1.6).
z
Fig. 1.6 The surface
described by the Eq. (1.2.1)
x
y
20
1 Examining Curves and Surfaces
One could terminate the solution of this problem here, but we will try below
to look at this issue from the other side. Not always does one have to deal with
the situation in which the subset is defined by an equation F (x1 , x2 , . . . , xn ) = 0
(or a system of equations of this type). Relatively common is to face the parametric
description of the set V by the following system of relations as mentioned at the
beginning of this chapter:
⎧
x1 = g1 (τ1 , τ2 , . . . , τk ),
⎪
⎪
⎨
x2 = g2 (τ1 , τ2 , . . . , τk ),
⎪
...
⎪
⎩
xN = gN (τ1 , τ2 , . . . , τk ),
(1.2.6)
collectively written in the form:
x = G(τ ),
(1.2.7)
where τ belongs to a certain set D ⊂ Rk , and x ∈ RN . It is again assumed that the
function G is smooth. The number of parameters (i.e., k), if a subset V turns out to
be a surface, will determine its dimension. Naturally in this case, it is subject to the
condition k ≤ N .
Equations (1.2.6) should be treated locally, which means that for each “patch”
of the set V , the function G can be distinct. Not always can any universal
parametrization for an entire set be found. In such a case, the procedure, that we
deal with below, has to be carried out for each of the functions G and for each
“patch” separately.
In order to determine whether the set is a surface, one can now—rather than
examining the function F —analyze the function G. The appropriate theorem—
probably known to the reader from the lecture—says that if the rank of the Jacobian
matrix G is maximal at a given point, then on its appropriate neighborhood the
set V is the graph of a certain mapping and constitutes, therefore, a surface. The
Jacobian matrix G has N rows and k columns, so (since k ≤ N ) the maximal rank
equals k.
What is the geometrical interpretation of this condition? Well, if we change
only one of the parameters (e.g., τi ), and fix the remaining k − 1 ones, then the
equations (1.2.6) specify the curve in RN . The quantity ti = ∂x/∂τi is nothing
other than a vector tangent to the curve. The matrix G has the form:
G = [t1 , t2 , . . . , tk ] ,
(1.2.8)
and, therefore, its columns constitute the tangential vectors. The requirement of the
maximal rank is in fact equivalent to the requirement that all tangent vectors are
linearly independent. If this is not the case, one “loses” some dimension and the shell
gets degenerated (“cusps” or “self-intersections” may appear as in Fig. 1.8).
1.2 Examining k-Surfaces in N Dimensions
21
In order to examine the set V in this manner, one must first propose some
parametrization. The structure of the expression x 2 /4 + y 2 /9 suggests as the first
choice:
x = 2r cos ϕ,
y = 3r sin ϕ,
(1.2.9)
so that the equation F = 0 takes the form:
r 2 − z2 = 1.
(1.2.10)
Then, if we recall the hyperbolic version of the Pythagorean trigonometric identity,
we can write
r = cosh w,
z = sinh w
(1.2.11)
and get the following result:
⎡
⎤ ⎡
⎤ ⎡
⎤
x(w, ϕ)
g1 (w, ϕ)
2 cosh w cos ϕ
⎣ y(w, ϕ) ⎦ = ⎣ g2 (w, ϕ) ⎦ = ⎣ 3 cosh w sin ϕ ⎦ .
z(w, ϕ)
sinh w
g3 (w, ϕ)
(1.2.12)
As it is easy to find out, this is the universal parametrization for the entire set V :
for each point in V , one finds the values of the parameters w and ϕ in a unique way.
What remains is to examine the rank of the matrix G . Calculating the derivatives
with respect to parameters, one obtains
⎡
⎤
2 sinh w cos ϕ −2 cosh w sin ϕ
G = ⎣ 3 sinh w sin ϕ 3 cosh w cos ϕ ⎦ .
cosh w
0
(1.2.13)
In order to be convinced that the rank of this matrix is maximal, i.e., equal to 2,
one needs to subsequently calculate the two-dimensional minors. Out of G , three
square 2 × 2 submatrices can be created if the first, the second, or the third row is
deleted. The following determinants are then obtained (indices denote columns that
remained after the deletion):
d12 = 6 sinh w cosh w cos2 ϕ + 6 sinh w cosh w sin2 ϕ = 6 cosh w sinh w,
d13 = 2 cosh2 w sin ϕ,
d23 = −3 cosh2 w cos ϕ.
(1.2.14)
It is sufficient that locally at least one of them does not vanish, and the rank of matrix
G will be maximal. Note that
(d23 /3)2 + (d13 /2)2 = cosh4 w > 0.
(1.2.15)
22
1 Examining Curves and Surfaces
These two minors cannot, therefore, simultaneously vanish (the function cosh w
never equals zero) and consequently the rank is equal to 2. Again, we conclude that
the subset V is a two-dimensional surface.
It should be noted at the end that it can sometimes happen that at a certain point
the matrix G will not have the maximal rank, but it will merely be a consequence
of choosing the wrong parametrization (at a given point). Then, one should try
another parametrization on the neighborhood of this point. As an example, let us
look at the standard parametrization of a sphere: x(θ, ϕ) = R sin θ cos ϕ, y(θ, ϕ) =
R sin θ sin ϕ, z(θ, ϕ) = R cos θ and the point which can conventionally be called
the “north pole” (for θ = 0). Obviously at this point nothing unusual happens on
the sphere—it does not differ from other points. It is the parametrization which
gets “degenerated” there because the value of the parameter ϕ is not unequivocally
defined. Another example of this kind appears in Problem 4c at the end of this
chapter.
Problem 2
It will be examined if the intersection of two spheres:
(x − 1)2 + y 2 + z2 = 4,
(x + 1)2 + y 2 + z2 = 4
(1.2.16)
3 . If so, vectors that span the tangent planes at the points A(0,
is a surface
√in R √
and B(0, 3/2, 3/2) will be found.
√
3, 0)
Solution
If one defines the functions f1 and f2 with the formulas:
f1 (x, y, z) = (x − 1)2 + y 2 + z2 − 4,
f2 (x, y, z) = (x + 1)2 + y 2 + z2 − 4,
(1.2.17)
it becomes clear that the function F from the previous exercise has now the form:
f1 (x, y, z)
,
F (x, y, z) =
f2 (x, y, z)
(1.2.18)
and the equation F (x, y, z) = 0 is a brief recording of the system of equations (1.2.16). One can expect, at least from the formal point of view, that these
equations will allow us to determine two of the variables x, y, z as the functions of
1.2 Examining k-Surfaces in N Dimensions
23
the third one. Thereby, there are two type-Y variables and one type-X variable. If
each point of the set defined by the equation F (x, y, z) = 0 (marked below as V )
has a neighborhood on which it is possible, it is a graph of some mapping Y (X)
and one can call it a surface. From the implicit function theorem, we know that this
takes place if at each of the points of the set V , the condition
det
∂F
= 0
∂Y
(1.2.19)
is satisfied.
As we remember, the function F has two components and the symbol Y also
refers to two variables, so the matrix symbolically denoted as ∂F /∂Y is a square
2 × 2 matrix. There is no obstacle in the calculation of its determinant. The
condition (1.2.19) simply means that the matrix is nonsingular. It should be stressed,
however, that this condition is local in nature and in the neighborhoods of different
points other variables can be considered as dependent ones (i.e., as being of the
type Y ). To take account for this possibility, let us first calculate the entire Jacobian
matrix:
⎡ ∂f ∂f
1
1
⎢ ∂x ∂y
F = ⎣ ∂f ∂f
2
2
∂x ∂y
∂f1 ⎤ ∂z ⎥ = 2(x − 1) 2y 2z .
∂f2 ⎦
2(x + 1) 2y 2z
∂z
(1.2.20)
When choosing two variables, one actually picks out two columns of this matrix
and such a reduced array—i.e., after the deletion of the unnecessary column—must
be nonsingular. This corresponds to calculating two-dimensional minors, of which
at least one should be (locally) nonzero. Using the language of algebra, one could
simply say that (1.2.20) must have the rank equal to 2.
Calculating the subsequent minors (indices refer to the labels of columns
remaining after deleting one), one finds
d12 = 4(x − 1)y − 4(x + 1)y = −8y,
d13 = 4(x − 1)z − 4(x + 1)z = −8z,
d23 = 4yz − 4yz = 0.
(1.2.21)
For all three being simultaneously equal to zero, the condition y = z = 0
would have to be met. After plugging these values into the equations (1.2.16), one,
however, comes to the contradiction:
(x − 1)2 = 4 ∧ (x + 1)2 = 4.
(1.2.22)
Upon adding these equations to each other, one obtains x 2 = 3, and upon
subtracting: x = 0. Thereby, in the set V , no point where all three minors (1.2.21)
24
1 Examining Curves and Surfaces
z
Fig. 1.7 The
one-dimensional surface (i.e.,
the curve) which is the
intersection of the two
spheres (1.2.16)
x
y
would vanish can be found. This means that the set V is a surface. It is depicted in
Fig. 1.7 with the thick line.
One still has to find the tangent space by which we understand the set of tangent
vectors (i.e., the vector space) at a given point. Let us start with the point A. First,
one needs to make sure that√the given point is located on the surface. Substituting
its coordinates x = 0, y = 3, z = 0 into the equations (1.2.16), it is easily seen
that both are satisfied.
The tangent vectors that are looked for span the kernel of the matrix F :
(1.2.23)
F A · v = 0.
The rows of this matrix are in fact gradients of functions f1 and f2 (see formula (2.0.1)), and hence the vectors are normal to the surfaces defined by the
may not have any component tangent
equations f1 = const and f2 = const (∇f
to the surface f = const, since when moving along it, the value of the function f
does not change by definition). The condition (1.2.23) simply means that the wanted
vector v must be orthogonal to both of these normal vectors.
When plugging the coordinates of the point A into (1.2.20) and writing the vector
v in the form v = [a, b, c], one obtains the equation:
⎡ ⎤
√ a
−2 2√3 0 ⎣ ⎦
0
,
(1.2.24)
b =
2 2 30
0
c
from which it immediately follows that a = b = 0. The third coordinate, c, remains
undetermined, but one can set it as equal to 1 since the normalization of this vector
is inessential (all normalizations are equally good). As a result one can see that the
vector tangent to the surface of V at the point a is
1.2 Examining k-Surfaces in N Dimensions
25
⎡ ⎤
0
⎣
vA = 0 ⎦ .
1
(1.2.25)
It spans the one-dimensional vector space which can be denoted as TA (V ).
The point B belongs to V too, which can be easily checked. The following
equation is required to be fulfilled at this point:
F B · v = 0,
(1.2.26)
i.e.,
⎡ ⎤
√ √ a
−2 √6 √6 ⎣ ⎦
0
.
b =
6 6
2
0
c
(1.2.27)
This time, one gets a = 0 and b = −c. Setting c = 1, we have
⎡
⎤
0
vB = ⎣ −1 ⎦ .
1
(1.2.28)
This vector spans the space TB (V ).
The reader could note at this point that instead of looking for the kernel of
the Jacobian matrix, the knowledge acquired when investigating curves might be
used. Then, the tangent vectors were simply derivatives of r(ς ) with respect to the
parameter ς . Below we will see that the tangent space can actually be found with this
method too. First, however, one needs to dispose of a parametric description of V . It
can be obtained relatively simply. If the two equations (1.2.16) are subtracted from
each other, one immediately gets x = 0. After inserting this result into any of the
equations one obtains in turn y 2 + z2 = 3. Now the parametric relations can be
easily proposed:
x(ς ) = 0,
y(ς ) =
√
3 cos ς,
z(ς ) =
√
3 sin ς,
(1.2.29)
and then the tangent vector is found:
⎡
⎤
√0
v(ς ) = ⎣ − 3 sin ς ⎦ .
√
3 cos ς
(1.2.30)
Since the point A corresponds to the value ς = 0 and the point B to the value
ς = π/4, after inserting them into (1.2.30), the tangent vectors are immediately
obtained in the form:
26
1 Examining Curves and Surfaces
⎡ ⎤
0
√
⎣
vA = 3 0 ⎦ ,
1
vB =
⎡
⎤
0
3⎣
−1 ⎦ .
2
1
(1.2.31)
They differ from the previously found ones only by normalization factors.
Problem 3
It will be examined for which positive values of parameters r0 and R the torus
defined by
(ρ − R)2 + z2 = r02 ,
(1.2.32)
where ρ = x 2 + y 2 , is the surface in R3 . Then the vectors spanning the tangent
space at the point of coordinates (R − r0 , 0, 0) will be found.
Solution
This time the function F dealt with in the previous exercises has the form:
F (x, y, z) = (ρ − R)2 + z2 − r02 .
Let us find the Jacobian matrix:
R
R
2y 1 −
2z .
F = 2x 1 −
ρ
ρ
(1.2.33)
(1.2.34)
It is obvious that all partial derivatives exist, apart from ρ = 0, which corresponds
to the straight line x = y = 0. In order to check whether points of these coordinates
are located on the torus, let us substitute these values into the Eq. (1.2.32) and see
whether it is possible to be satisfied. In this way we get the condition:
z2 = r02 − R 2 ,
(1.2.35)
and if the right-hand side is nonnegative, i.e., for r0 ≥ R, the solutions for z do
exist. The condition r0 ≥ R means, however, that the torus has self-intersections. In
order not to appeal solely to the reader’s imagination, in Fig. 1.8 the cross sections
of the torus for r0 ≥ R and r0 < R are depicted. The points where the surface is
intersected with itself do not have neighborhoods on which the shell would be the
graph of a function. Thus, it is seen that in this case the set of points described with
1.2 Examining k-Surfaces in N Dimensions
27
Fig. 1.8 The cross sections of a torus in the case of self-intersections (on the left) and without
self-intersections (on the right)
the Eq. (1.2.32) is not a surface in R3 . On the other hand for r0 < R, there are no
intersections, and the rank of the Jacobian matrix (1.2.34) always equals 1, which
means that one is dealing with a surface.
The same conclusion emerges by means of the parametric description. The
parametrization of a torus was introduced in Part II (see formula (13.2.21)):
x = (R + r0 cos θ ) cos ϕ,
y = (R + r0 cos θ ) sin ϕ,
z = r0 sin θ.
(1.2.36)
These relations can be collectively written, when defining the function G:
⎡
⎤
(R + r0 cos θ ) cos ϕ
G(θ, ϕ) = ⎣ (R + r0 cos θ ) sin ϕ ⎦ .
r0 sin θ
(1.2.37)
If the set is a surface, then, as we know, the partial derivatives with respect to
parameters are tangent vectors. They are columns of the Jacobian matrix:
⎡
∂x
⎢ ∂θ
⎢ ∂y
⎢
G =⎢
⎢ ∂θ
⎣ ∂z
∂θ
∂x
∂ϕ
∂y
∂ϕ
∂z
∂ϕ
⎤
⎥ ⎡ −r sin θ cos ϕ −(R + r cos θ ) sin ϕ ⎤
⎥
0
0
⎥ ⎣
⎥ = −r0 sin θ sin ϕ (R + r0 cos θ ) cos ϕ ⎦ .
⎥
⎦
0
r0 cos θ
(1.2.38)
When calculating the subsequent minors, one sees that the rank of this matrix is
maximal (i.e., equal to 2):
d12 = −(R + r0 cos θ )r0 sin θ,
d13 = (R + r0 cos θ )r0 cos θ sin ϕ,
d23 = −(R + r0 cos θ )r0 cos θ cos ϕ,
(1.2.39)
28
1 Examining Curves and Surfaces
since it is easy to check that
2
2
2
+ d13
+ d23
= (R + r0 cos θ )2 r02 ,
d12
(1.2.40)
and if r0 < R, all three minors cannot simultaneously vanish. On the other hand for
r0 ≥ R, the matrix G gets degenerated in a visible way for cos θ = −R/r0 , i.e.,
at the points of self-intersections. One can be certain of it by substituting this value
into the formulas (1.2.36) and obtaining
x = y = 0,
z = ± r02 − R 2 .
(1.2.41)
These are the same points that have already been found in the first part of our
solution.
Assuming below that r0 < R, we are going to find the tangent space at the point
of coordinates (R − r0 , 0, 0). As written above, the tangent vectors can be read off
from the columns of the matrix (1.2.38) and the simultaneous substitution of the
appropriate values of the parameters θ and ϕ. For the point in question, one has
θ = π,
ϕ = 0,
(1.2.42)
which leads to the following tangent vectors:
⎡ ⎤
0
v1 = ⎣ 0 ⎦ ,
1
⎡ ⎤
0
v2 = ⎣ 1 ⎦ ,
0
(1.2.43)
the inessential normalization constants having been omitted. The reader is encouraged to check that these vectors actually span the kernel of the matrix (1.2.34), if in
place of coordinates (x, y, z) one substitutes (R − r0 , 0, 0).
Problem 4
It will be examined whether the intersection of the hyperboloid x 2 +y 2 +z2 −w 2 = 1
and the hyperplane x − y − z = 1 is√
a surface in R4 . If so, the vectors spanning the
tangent space at the point (1, 1, −1, 2) will be found.
Solution
This time the issue takes place in four-dimensional space. However, there are
two conditions provided in the text of the problem which restrict the number of
1.2 Examining k-Surfaces in N Dimensions
29
independent variables to two (4 − 2 = 2). Therefore, we are going to examine
whether the set V defined through the above equations is a 2-surface in R4 . However,
before starting, one should ensure that this set is not empty. To this end, let us
examine whether the system of equations
x 2 + y 2 + z2 − w 2 = 1,
x−y−z=1
(1.2.44)
has any solutions. Solving the second equation for z and plugging it into the first
one, one gets
x 2 + y 2 + (x − y − 1)2 − w 2 = 1,
(1.2.45)
w2 = x 2 + y 2 + (x − y − 1)2 − 1.
(1.2.46)
i.e.,
It is clear that for arbitrarily large values of the variables x and y the right-hand
side is positive, and the equation has the solutions for w. It is also obvious that
the solutions of the equation x − y − z = 1 for z always exist. This leads to the
conclusion that the equations that define V are consistent.
The function F introduced in the previous problems this time has the form:
x 2 + y 2 + z2 − w 2 − 1
F (x, y, z, w) =
,
x−y−z−1
(1.2.47)
and its derivative:
F =
2x 2y 2z −2w
.
1 −1 −1 0
(1.2.48)
In order to determine the rank of the Jacobian matrix, one should now calculate
successively the determinants of submatrices 2 × 2 obtained by the deletion of two
columns. Even though there are as many as six matrices, writing out the respective
minors does not pose any problem:
d12 = −2(x + y),
d13 = −2(x + z),
d23 = −2(y − z),
d14 = −d24 = −d34 = 2w.
(1.2.49)
For the rank of the matrix (1.2.48) to be smaller than 2, all of them should
simultaneously vanish. Let us consider if in the set V there exists a point with these
30
1 Examining Curves and Surfaces
properties. First of all it should be required that w = 0. Then the equations resulting
from the condition of the vanishing of d12 , d13 , and d23 would lead to y = z = −x.
Upon inserting these values into the equations defining the set V , one gets the system
3x 2 = 1,
3x = 1,
(1.2.50)
which is manifestly inconsistent. The conclusion which follows may be stated in the
form: at each of the points of the set V , the rank of the matrix F equals 2, which
entails that on some neighborhood of each point, the set V constitutes the graph of
a certain mapping, although in neighborhoods of distinct points generally different
mappings can be defined. If so, V is the 2-surface in R4 .
Leaving to the reader to propose the appropriate parametrization and to solve the
problem using the second approach, we are going to find below the tangent vectors
as those belonging to the kernel of the Jacobian matrix. Let us √
denote such general
vector with the symbol v = [a, b, c, d]. At the point (1, 1, −1, 2) one has
F
√ 2 2 −2 −2 2
=
,
1 −1 −1 0
√
(1,1,−1, 2)
(1.2.51)
and the requirement that
F (1,1,−1,√2) · v = 0
(1.2.52)
leads to the two equations for the coordinates a, b, c, d:
√
2a + 2b − 2c − 2 2 d = 0,
a − b − c = 0.
(1.2.53)
Now the values of c and d can be found in an easy way:
c = a − b,
d=
√
2 b,
(1.2.54)
and finally we are able to write down a vector v as the linear combination of two
tangent base vectors:
⎡
⎤
⎡ ⎤
⎤
⎡
a
1
0
⎢ b ⎥
⎢0⎥
⎢ 1 ⎥
⎥
⎢ ⎥
⎥
⎢
v=⎢
⎣ a − b ⎦ = a ⎣ 1 ⎦ + b ⎣ −1 ⎦ .
√
√
2b
2
0
(1.2.55)
1.3 Examining Ruled Surfaces
31
1.3 Examining Ruled Surfaces
Problem 1
It will be examined whether the cone described by the equation z2 = x 2 + y 2 is a
ruled surface.
Solution
We will now be limited to two-dimensional surfaces in R3 . As we know, they can be
described parametrically through the relation r(u, v), where symbols u and v denote
the parameters. Of course there exist various parametrizations, but this section is
devoted to those that can be given the following form:
r(u, v) = α (u) + v β(u),
(1.3.1)
spoken of in the theoretical summary. For any fixed value of the parameter u
(
α and β will then become some constant vectors) this formula describes the
ordinary straight line in R3 . Now, if one allows the parameter u to evolve too,
then the formula (1.3.1) will mean that the whole surface has to be constructed
of straight lines only. They come out of the curve α (u) in the direction defined by
the vector β(u).
Such a surface is called a ruled surface. This kind of surface can
play an important role in architecture (e.g., describe different types of vaults) or in
the construction of boats, when it becomes necessary because of the material used
for the hull. Working to find the solution of this exercise, one sees that the shell (i.e.,
the cone) is axially symmetric. Because of it, let us write first
x(ρ, ϕ) = ρ cos ϕ,
y(ρ, ϕ) = ρ sin ϕ,
(1.3.2)
and after plugging that into the cone equation, we find
z2 = ρ 2 cos2 ϕ + ρ 2 sin2 ϕ = ρ 2 ,
(1.3.3)
which leads to z = ±ρ. If “+” were chosen, only the upper half of the cone sketched
in Fig. 1.9 would be described, and for “−” only the lower half. In order to describe
the entire cone, let us set z = ρ, but unlike the regular polar parametrization, both
positive and negative values of ρ will be allowed. The range of variability of the
parameters is, therefore, as follows:
ρ ∈ R,
ϕ ∈ [0, 2π [.
(1.3.4)
32
1 Examining Curves and Surfaces
Fig. 1.9 The cone as a ruled
surface
This parametrization can be now easily given the desirable form (1.3.1):
⎡
⎤ ⎡ ⎤
⎡
⎤
x(ρ, ϕ)
0
cos ϕ
r(ϕ, ρ) = ⎣ y(ρ, ϕ) ⎦ = ⎣ 0 ⎦ +ρ ⎣ sin ϕ ⎦ .
z(ρ, ϕ)
0
1
α
(1.3.5)
β
One can say that by changing the parameter ϕ, various straight lines are chosen,
and by modifying ρ, one moves along a given line. The cone referred to in the
text of this exercise is shown in Fig. 1.9 together with several exemplary straight
lines (1.3.1). As one can see, the base curve (see theoretical introduction) is
degenerated here to a single point.
If the parameter ρ was shifted by a certain number, for example by 1 (ρ → ρ+1),
then instead of (1.3.5) the following parametrization would be obtained:
⎡
⎤ ⎡
⎤
⎡
⎤
x(ρ, ϕ)
cos ϕ
cos ϕ
r(ϕ, ρ) = ⎣ y(ρ, ϕ) ⎦ = ⎣ sin ϕ ⎦ +ρ ⎣ sin ϕ ⎦ .
z(ρ, ϕ)
1
1
α
(1.3.6)
β
This parametrization is equally good leading to the same straight lines (that is, it has
nothing to do with the doubly ruled surfaces dealt with in the next problem), and the
base curve becomes now the circle x(ϕ) = cos ϕ, y(ϕ) = sin ϕ lying in the plane
z = 1.
1.3 Examining Ruled Surfaces
33
Problem 2
It will be examined whether the one-sheeted hyperboloid x 2 + y 2 − z2 = 1 is a ruled
and doubly ruled surface.
Solution
For the hyperboloid described with the formula given in the text of this problem it
is natural to use the following parametrization:
x(u, ϕ) = cosh u cos ϕ,
y(u, ϕ) = cosh u sin ϕ,
z(u, ϕ) = sinh u,
(1.3.7)
where
u ∈ R,
ϕ ∈ [0, 2π [.
(1.3.8)
It stems from the observation that the expression in question depends on x and
y only through the combination x 2 + y 2 =: r 2 , which suggests the use of the
Euclidean trigonometric identity, and then it depends on r and z through r 2 − z2 ,
which in turn hints at its hyperbolic counterpart. The reader can easily be convinced
by inserting (1.3.7) into the equation of the hyperboloid, and it will be automatically
satisfied. Unfortunately, the chosen parametrization, although getting imposed, does
not have the required form (1.0.4). This is not to say, of course, that the considered
hyperboloid is not a ruled surface since there can exist different parametrizations.
At this point, it might help to look at a figure. If a parametrization in terms of
straight lines (1.0.4) is to exist, these lines should look as it is illustrated in Fig. 1.10.
Surely, having found this type of straight lines, we know that a second family of
them, inclined in the opposite direction, must exist. Obviously it stems from the
symmetry of the surface and would simultaneously mean that it is doubly ruled.
The figure suggests that the lines one is looking for have to go out of the circle
x 2 + y 2 = 1 located in the plane z = 0. It is the circle of the strongest contraction
of the hyperboloid. If so, that is to say, the base curve of the surface has been found.
Using the notation of the Eq. (1.0.4), one can write
⎡
⎤
cos θ
α (θ ) = ⎣ sin θ ⎦ .
(1.3.9)
0
Let us then denote β = [a, b, c]. Then one must
One still has to find the vector β.
have
⎡
⎤ ⎡
⎤
⎡ ⎤
x(t, θ )
cos θ
a
⎣
⎦
⎣
⎦
⎣
r(θ, t) = y(t, θ ) = sin θ + t b ⎦ .
(1.3.10)
z(t, θ )
0
c
α
β
34
1 Examining Curves and Surfaces
Fig. 1.10 The hyperboloid as
a ruled surface
Let us now check if it is possible to adjust the numbers a, b, c so that (1.3.10)
represents the parametrization of the hyperboloid
The straight line, we are looking for (and, therefore, the vector β as well)
at the point at which it intersects the base curve, i.e., at (cos θ, sin θ, 0), must
be perpendicular to the vector of coordinates [cos θ, sin θ, 0]. This is due to the
reflection symmetry z → −z of the solid. Thus the following condition must hold:
β · [cos θ, sin θ, 0] = 0
⇒ a cos θ + b sin θ = 0.
(1.3.11)
One can choose a = − sin θ , b = cos θ , and c remains undetermined for a while.
Any other choice of the parameters a and b leads only to the rescaling of t in
the Eq. (1.3.10). One should now use the fact that the coordinates x, y, z satisfy
the hyperbola equation. This requirement allows us to determine the value of c.
Therefore, the following equation has to be satisfied:
(cos θ − t sin θ )2 + (sin θ + t cos θ )2 − t 2 c2 = 1,
(1.3.12)
which, after some simple transformations, implies that c2 = 1, i.e., c = ±1. Thus,
we were able to pursue our plan. The two different parametrizations that were looked
for are:
⎡
⎤ ⎡
⎤
⎡
⎤
x(t, θ )
cos θ
− sin θ
r(θ, t) = ⎣ y(t, θ ) ⎦ = ⎣ sin θ ⎦ + t ⎣ cos θ ⎦,
z(t, θ )
0
±1
α
(1.3.13)
β
the former being demonstrated in the figure. Thereby the hyperboloid has proved to
be the doubly ruled surface.
1.3 Examining Ruled Surfaces
35
Problem 3
It will be examined whether the hyperbolic paraboloid defined through the relation
z = 2xy is a ruled surface.
Solution
One can start the search of the parametrization in the form of straight lines with the
observation that the equation z = 2xy is linear, separately, in y and in z. Therefore,
if the value of x was fixed, for example by putting x = u, an equation which defines
a plane perpendicular to the x-axis, the obtained condition z = 2uy describes a
plane as well. The intersection of the two planes is a straight line. Now, changing
the value of u, one will get various lines contained in the paraboloid, i.e., the needed
parametrization.
In order to find the base curve, let us consider the intersection of the family of
straight lines:
x = u,
(1.3.14)
z = 2uy
with the plane z = 0. Obviously the result is x = u, y = z = 0, and thus the base
curve is α (u) = [u, 0, 0].
For each fixed u, it must be parallel to the
Now, one has to find the vector β.
straight line (1.3.14). In order to determine the vector, it is sufficient to select the
two points on this line, for example, (u, 0, 0) and (u, 1, 2u). By subtracting their
coordinates, one gets
⎡ ⎤
0
β(u)
= ⎣ 1 ⎦.
(1.3.15)
2u
This allows us to write the needed parametrization in the form:
⎡
⎤ ⎡ ⎤
⎡ ⎤
x(u, v)
u
0
r(u, v) = ⎣ y(u, v) ⎦ = ⎣ 0 ⎦ + v ⎣ 1 ⎦,
z(u, v)
0
2u
α
(1.3.16)
β
and thus conclude that the surface in question is a ruled one. By inserting the above
expressions for x(u, v), y(u, v), and z(u, v) into the equation of the paraboloid, one
can easily verify that it is automatically satisfied.
In Fig. 1.11, the examined surface is depicted along with a few straight lines of
the type (1.3.16).
36
1 Examining Curves and Surfaces
Fig. 1.11 Hyperbolic
paraboloid as a ruled surface
Problem 4
Given the point P of coordinates (2, 1, 1) on the one-sheeted hyperboloid defined
by the equation x 2 + y 2 − 4z2 = 1. All generating lines passing through this point
will be found.
Solution
This problem can be solved with the use of the parametrization similar to (1.3.13).
The considerations leading to it will not be repeated, because the modification of
the formulas is obvious: due to the presence of “4” in the hyperboloid equation, the
coordinate z is now simply multiplied by 1/2:
⎡
⎤ ⎡
⎤
⎡
⎤
x(t, θ )
cos θ
− sin θ
r(θ, t) = ⎣ y(t, θ ) ⎦ = ⎣ sin θ ⎦ + t ⎣ cos θ ⎦ .
z(t, θ )
0
±1/2
(1.3.17)
Let us first consider the parametrization obtained by choosing “+.” In the first
place, one needs to determine what the values of the parameters t and θ at P are. To
this end, the following set of equations is to be solved:
⎧
⎨ cos θ − t sin θ = 2,
sin θ + t cos θ = 1,
⎩
t/2 = 1.
(1.3.18)
We immediately get t = 2, and plugging this value into the first two equations we
find: sin θ = −3/5, cos θ = 4/5. The value of the angle θ does not need to be
known and, moreover, it is not any “nice” number. It is sufficient to use the obtained
values of the trigonometric functions.
By inserting these numbers into (1.3.17), one gets the equation for the first line:
1.4 Exercises for Independent Work
37
Fig. 1.12 The straight lines
contained in the hyperboloid
x 2 + y 2 − 4z2 = 1 passing
through the point P (2, 1, 1)
P
⎡
⎤ ⎡
⎤
⎡
⎤
x(t)
4/5
3/5
r(t) = ⎣ y(t) ⎦ = ⎣ −3/5 ⎦ + t ⎣ 4/5 ⎦ .
z(t)
0
1/2
(1.3.19)
In order to find the second one, we again use the parametrization (1.3.17), but
now choose −1/2. The requirement for the straight line to pass through the point P
leads to the system:
⎧
⎨ cos θ − t sin θ = 2,
sin θ + t cos θ = 1,
⎩
−t/2 = 1.
(1.3.20)
This time one has t = −2, sin θ = 1, cos θ = 0 and the needed equation becomes
⎡
⎤
⎤ ⎡ ⎤
0
−1
x(t)
r(t) = ⎣ y(t) ⎦ = ⎣ 1 ⎦ + t ⎣ 0 ⎦ .
0
−1/2
z(t)
⎡
(1.3.21)
Both lines, along with the discussed surface, are shown in Fig. 1.12.
1.4 Exercises for Independent Work
Exercise 1 Assuming that a curve is given parametrically by r(ς ), demonstrate the
following equations for the curvature and torsion:
κ=
|r × r |
,
|r |3
τ=
(r × r ) · r .
|r × r |2
38
1 Examining Curves and Surfaces
Exercise 2 Given a curve in the parametric form: x(ς ) = cos ς , y(ς ) = sin ς ,
z(ς ) = ς 2 , where ς ∈ R. Find the curvature and the torsion of the curve at the
points, for which ς = π/2 and ς = 1.
Answers
κ(π/2) = (5√+ π 2 )1/2 /(1 + π 2 )3/2 , τ (π/2) = π/(5 + π 2 );
κ(1) = 3/(5 5), τ (1) = 2/9.
√
Exercise 3 Given a curve in the parametric form: x(ς ) = ς , y(ς ) = log ς ,
z(ς ) = ς , where ς ∈]0, ∞[. Find the curvature and the torsion at the points on the
curve, for which ς = 1 and ς = 1/4.
Answers
√
√
κ(1) = 2 2/9, τ (1) = −1/9; κ(1/4) = 1/(3 2), τ (1/4) = −16/81.
Exercise 4 Verify whether
(a) The set defined with the equation x 3 = z2 − y 2 − 2y,
(b) The cone x 2 + y 2 = z2 ,
(c) The set defined parametrically: x(u, v) = 1/2 sinh u cos v,
y(u, v) = 1/3 sinh u sin v, z(u, v) = cosh u, where u ∈ [0, ∞[ and v ∈
[0, 2π [,
is a surface in R3 .
Answers
(a) Yes.
(b) No (because of the point (0, 0, 0)).
(c) Yes.
Exercise 5 Find the vectors spanning the space tangent to the surface:
√
(a) x 2 + 2y 2 + 3z2 = 6 in R3 at the point (0, 3/2, 1),
2
2
2
2
4
(b) x + y − z − w = 1 in R at the point (1, 1, 0, 1).
1.4 Exercises for Independent Work
39
Answers
√
(a) v1 = [1, 0, 0], v2 = [0, 3/2, −1].
(b) v1 = [1, 0, 0, 1], v2 = [0, 1, 0, 1], v3 = [0, 0, 1, 0].
Exercise 6 Check whether the given surfaces are ruled ones and if so, find the
appropriate parametrizations:
(a) The cylinder x 2 + y 2 = 4,
(b) The set defined with the equation 4z = x 2 − y 2 .
Answers
(a) Ruled surface, parametrization:
r(θ, t) = [2 cos θ, 2 sin θ, 0] + t[0, 0, 1].
(b) Ruled surface,
parametrization: r(u, t) = [u, u, 0] + t[1, −1, u].
Chapter 2
Investigating Conditional Extremes
The second chapter is concerned with the so-called conditional extremes, i.e.,
those appearing due to the presence of some additional conditions to be fulfilled.
This concept is explained in full detail when solving the first problem. Below some
principal notions are collected.
Given a differentiable scalar function f : RN ⊂ D → R, where D is an open set
and (x1 , x2 , . . . , xN ) ∈ RN denote the Cartesian coordinates. The gradient of this
function is the vector field composed of all partial derivatives, i.e., in the form:
:=
gradf := ∇f
∂f ∂f
∂f
.
,
,...,
∂x1 ∂x2
∂xn
(2.0.1)
This vector at any point of D indicates the direction of the fastest increase of the
value of the function f (x1 , x2 , . . . , xN ).
The Lagrange multipliers method of finding the conditional extremes of the
function f (x1 , x2 , . . . , xN ) subject to the set of additional conditions
gi (x1 , x2 , . . . , xN ) = 0, i = 1, . . . , r,
(2.0.2)
which describe r smooth surfaces in RN , consists of the following steps.
1. A certain new function F (x1 , x2 , . . . , xN ) called the Lagrange function is
defined according to the relation:
F (x1 , x2 , . . . , xN ) = f (x1 , x2 , . . . , xN ) −
r
!
λi gi (x1 , x2 , . . . , xN ),
(2.0.3)
i=1
where λi for i = 1, . . . , r are certain unknown real numbers.
© Springer Nature Switzerland AG 2020
T. Radożycki, Solving Problems in Mathematical Analysis, Part III, Problem Books
in Mathematics, https://doi.org/10.1007/978-3-030-38596-5_2
41
42
2 Investigating Conditional Extremes
2. All stationary points (see Chap. 7 of Part II) are found by solving the set of N
equations written in short as
∇F (x1 , x2 , . . . , xN ) = 0
(2.0.4)
supplemented by r conditions (2.0.2). One has in common N + r equations and
N + r unknowns (x1 , . . . , xN , λ1 , . . . , λr ).
3. The behavior of the function F at a given stationary
point P is examined by
analyzing the sign of the quadratic form hT · F P · h. This method is known
from the second part of the book series with the slight modification that now the
allowed vectors h have to be tangent to all surfaces (2.0.2) at P , i.e., the following
conditions are to be satisfied:
gi P · h = 0, i = 1, . . . , r.
(2.0.5)
The justification and application of this method are discussed comprehensively
below.
2.1 Using the Method of the Lagrange Multipliers
Problem 1
The local extremes of the function f (x, y) = x + y on the circle x 2 + y 2 = 4 will
be found.
Solution
In Part II in Chap. 7, we were looking for extremes of multivariable functions. It was
then found that at an extremal point all partial derivatives must be equal to zero, i.e.,
= 0.
the gradient (2.0.1) of the function vanishes: ∇f
However, it can happen—and this is just the case we are concerned with now—
that one is interested not in an extreme of a given function as such, but only as
subject to a certain condition. Let us imagine for example that we participate in
a mountain excursion along a tourist trail. The path on which we are moving can
raise and lower, which we perceive as achieving certain local maxima or minima,
although in fact, we reach neither the top of any hill nor the lowest point of any
valley. The path is merely leading us this way, and one cannot get off the trail and
move perpendicularly to it. Let us look at the exemplary in Fig. 2.1. It shows a
fragment of some map with a few terrain contours sequentially labeled (with the
increasing height): h1 , h2 , up to h5 . In the considered area, there is no mountaintop.
2.1 Using the Method of the Lagrange Multipliers
43
Fig. 2.1 The exemplary trail on a mountain side marked with the dashed line. The symbols h1 , h2 ,
etc. refer to the increasingly “higher” contours. The point P constitutes the conditional maximum
However, when walking along the trail marked with a dashed line, we achieve the
maximum at the point P where our path reaches the highest contour.
The “maxima” or “minima” one is overcoming are just the so-called conditional
extremes mentioned in the theoretical introduction at the beginning of this chapter
(they are met under the condition that one is going along the trail). When solving
this problem, we will learn how to find them. It should be stressed, however, that we
will be primarily interested only in cases in which the condition (conditions) cannot
be explicitly solved and, thereby, the problem cannot be boiled down to that of the
unconditional extremes of a function of fewer variables. The examples dealt with in
this section will be treated as such that the elimination of any variable is impossible
or unviable.
It is easy to see that the function given in the text of this problem, i.e., f (x, y) =
x + y, has the graph which constitutes an inclined plane and, therefore, it does not
have extremes anywhere. Both partial derivatives are constantly equal to unity and it
is clear that they cannot vanish. If we were looking for the unconditional extremes,
then our conclusion would be: the function f has no local extremes. On the other
hand, however, if one is allowed to move only along the circle x 2 + y 2 = 4 and
at the end we return to the starting point, as in this exercise, then certainly one
has to meet somewhere the maximal and minimal values of the function (they even
can be met several times). The question arises, how to find these points and what
should be modified within the procedure used in the previous part of the book. The
answer seems to be quite obvious. Now it is not the gradient of the function that
must vanish, but only its component tangent to the path! For, we are not interested
if and how the value of the function varies in a direction perpendicular to the trail,
because one cannot get off it. It is still important to stress that by the “path” or “trail”
the curve lying in the plane xy is understood (i.e., contained in the domain of the
function), just as it would be plotted on the flat map. Any rising and lowering when
moving is included in the values read off the contours of the function f during the
excursion.
44
2 Investigating Conditional Extremes
Fig. 2.2 The decomposition
of the gradient of the function
f onto components tangent
and perpendicular to the path.
In the general case, this path
obviously does not have to be
closed
Let us suppose that the two-dimensional gradient of the function f looks on
the plane xy, as it is drawn in Fig. 2.2. It can be decomposed onto the component
tangent to the trajectory and perpendicular to it. We are interested in the former one.
It can be expressed in a natural way as
∇f
− ∇f
= ∇f
⊥
.
(2.1.1)
on the right-hand side can readily be calculated, since the
The vector ∇f
formula for the function f is known. However, how can one find its perpendicular
component? Well, one should use the equation that defines the path which has the
one knows that the vector perpendicular to
form g(x, y) = 0. When calculating ∇g,
cannot
the curve g is obtained. This is because of the obvious observation that ∇g
have any tangential component, since the function g does not change its value (i.e.,
is constantly equal to zero), when moving along the curve. It was already mentioned
in Sect. 1.2. Consequently the vector ∇f
must be proportional to the gradient
⊥
of g (these vectors may differ only by a fixed ratio). One can, therefore, write
∇f
⊥
= λ∇g,
(2.1.2)
with λ denoting a constant. This means that at extremes, we are looking for, the
following condition must be satisfied:
∇f
=0
− λ∇g
= ∇(f
− λg) = 0.
⇒ ∇f
(2.1.3)
This latter form suggests that maybe instead of dealing with the function f (x, y),
from the very beginning one should introduce the auxiliary function with the
formula:
F (x, y; λ) := f (x, y) − λg(x, y),
(2.1.4)
2.1 Using the Method of the Lagrange Multipliers
45
= 0), and
examining its extremes as “unconditional” ones (with the requirement ∇F
determining the additional unknown λ using the condition g(x, y) = 0. This method
is called the Lagrange multipliers method as it was mentioned in the theoretical
introduction. In the present case, the multiplier is only one—it is just the number
λ—because the surface (curve), to which the domain of the function was reduced,
is defined by one condition: g(x, y) = 0. However, one can easily generalize
this formula—and the reader has certainly encountered the appropriate expression
during lectures of analysis—for the case when things take place in RN , and the
surface is an intersection of r sets described with the equations:
g1 (x1 , x2 , . . . , xN ) = 0, g2 (x1 , x2 , . . . , xN ) = 0, . . . , gr (x1 , x2 , . . . , xN ) = 0.
(2.1.5)
Then one will have r Lagrange multipliers, and the function F is defined as
(see (2.0.3))
F (x1 , x2 , . . . , xN ; λ1 , λ2 , . . . , λr ) := f (x1 , x2 , . . . , xN ) − λ1 g1 (x1 , x2 , . . . , xn )
−λ2 g2 (x1 , x2 , . . . , xN ) − · · · − λr gr (x1 , x2 , . . . , xN ).
(2.1.6)
Thereby, in our problem r additional unknowns appear, but fortunately we have at
our disposal the very same number of additional equations (2.1.5).
Applying the method described above to the current exercise, we first define the
function F :
F (x, y; λ) = x + y − λ(x 2 + y 2 − 4),
(2.1.7)
and then require that all its derivatives vanish. Including the additional condition the
following set of equations is obtained:
⎧
⎨ 1 − 2λx = 0,
1 − 2λy = 0,
⎩ 2
x + y 2 = 4.
(2.1.8)
Subtracting the two former equations from each other, one gets
2λ(x − y) = 0.
(2.1.9)
The quantity λ cannot be equal to zero, as this would be in conflict with the first and
second equations,√so the only possibility remains x = y. The last equation (2.1.8)
then gives x = ± 2. Thereby, two points A and B suspected of being the extremes
are found:
√ √
A( 2, 2),
√
√
B(− 2, − 2).
(2.1.10)
46
2 Investigating Conditional Extremes
√
At these points the
√values of the Lagrange multipliers can be found: λA = 1/(2 2)
and λB = −1/(2 2).
One can now proceed to examine whether the points A and B are actually local
extremes. The method, well known to us (see the second part of the book), based on
the investigation of the matrix F can be used. This matrix has the form:
⎡
∂ 2F
⎢ ∂x 2
F = ⎢
⎣ ∂ 2F
∂x∂y
⎤
∂ 2F
∂x∂y ⎥
⎥ = −2λ 0
.
∂ 2F ⎦
0 −2λ
∂y 2
(2.1.11)
It now becomes clear why we needed the values of the Lagrange multiplier. Even
after having found the points (2.1.10) but without knowing λ, one would be unable
to assess whether or not they are extremes and of which
√ kind.
Considering the first point, after setting λ = 1/(2 2), one easily sees that the
matrix is negative-definite:
⎤
⎡
1
⎢ − √2 0 ⎥
⎥
F = ⎢
⎣
1 ⎦.
A
0 −√
2
(2.1.12)
It is obvious since both its eigenvalues are negative. Such a situation corresponds to
a maximum of the function. Remember, however, that it would be sufficient that the
matrix be negative-definite only in the subspace of vectors that are tangent to the
path. In this example, one was not obliged to make this further reduction because
the sign of the matrix F is determined for any vector. However, one should not
always count on such a comfortable result, so we are going to show below how to
perform the appropriate reduction.
As we know from Sect. 7.2 of Part II, examining the sign on the matrix F corresponds to examining that of the quadratic form:
h · F · h,
T
A
hx
.
h=
hy
where
(2.1.13)
This time, however, only tangent vectors h come into play, i.e., those fulfilling the
condition:
√
√
∇g · h = 0, i.e., 2 2hx + 2 2hy = 0,
(2.1.14)
A
which implies that hy = −hx. By inserting this result into the quadratic
form (2.1.13), one gets
2.1 Using the Method of the Lagrange Multipliers
⎡
⎤
1
⎢ − √2 0 ⎥ hx
T
⎢
⎥
h · F · h = hx , hy ⎣
1 ⎦ hy
A
0 −√
2
√
1 2
= − √ (hx + h2y ) = − 2 h2x .
2
47
(2.1.15)
Since hx may not equal zero (otherwise hy = 0 and in consequence h would
become a null vector), the above expression is always negative,
√ which means that
the function f has at A the conditional maximum (fmax = 2 2). As to the point B,
we proceed in an analogous way. The matrix of the second derivatives this time has
the form:
⎤
⎡
1
0
√
⎥
⎢ 2
⎥
F = ⎢
(2.1.16)
⎣
1 ⎦
B
0 √
2
and is positive-definite, which can be seen already at this stage. Nonetheless,
reducing it to the subspace of the tangent vectors, we first require
∇g · h = 0,
i.e.
√
√
− 2 2hx − 2 2hy = 0,
(2.1.17)
B
which entails that hy = −hx . Then for the (reduced) quadratic form one finds
⎡
⎤
1
0
√
⎢ 2
⎥ hx
T
⎢
⎥
h · F · h = hx , hy ⎣
1 ⎦ hy
B
0 √
2
√
1
= √ (h2x + h2y ) = 2 h2x > 0.
2
(2.1.18)
It is then obvious
that the function has at this point the conditional minimum
√
(fmin = −2 2).
At the end, it is worth adding that if the equation g = 0 describes a compact set
(as in the present case), examining of the second derivatives may be abandoned. In
order to determine whether a given function has extremes, it is sufficient to find and
compare its values achieved at all “suspected” points (see Sect. 7.1 in Part II).
48
2 Investigating Conditional Extremes
z
x
extremes
y
Fig. 2.3 The conditional extremes of function f (x, y). The graph is shown “from underneath”
Problem 2
The critical points of the function f (x, y, z) = x 2 + 2y 2 + z on the surface
g(x, y, z) = x 2 + y 2 − z2 + 2z = 0 will be found and explored.
Solution
This time we are dealing with a function of three variables, so it would be difficult
to draw its graph. However, the method of the Lagrange multipliers can still be
used. To do this, one should define the auxiliary function F (x, y, z; λ), according to
formulas (2.0.3) or (2.1.6):
F (x, y, z; λ) = x 2 + 2y 2 + z − λ(x 2 + y 2 − z2 + 2z).
(2.1.19)
= 0 together with the condition g(x, y, z) = 0 leads to
The requirement that ∇F
the system of equations:
⎧
2x − 2λx = 2(1 − λ)x = 0,
⎪
⎪
⎨
4y − 2λy = 2(2 − λ)y = 0,
⎪
1 + 2λz − 2λ = 0,
⎪
⎩ 2
x + y 2 − z2 + 2z = 0.
(2.1.20)
The first equation yields two possibilities: λ = 1 or x = 0. Let us consider them in
turn.
1. λ = 1.
Upon inserting this value into the second and third equations, one immediately
obtains y = 0 and z = 1/2. However, with these values, the last equation
2.1 Using the Method of the Lagrange Multipliers
49
cannot be satisfied, since one obtains a contradiction: x 2 + 3/4 = 0. Thereby, the
possibility λ = 1 must be rejected.
2. x = 0.
The second equation offers now two options: λ = 2 or y = 0. Assuming the
first possibility, from the third equation one obtains z = 3/4, but the fourth one
again leads to a contradiction: y 2 + 15/16 = 0. We are then left with y = 0.
After having inserted the values x = y = 0, the fourth equation takes the form
z(2 − z) = 0, which implies that z = 0 or z = 2.
Summarizing, the two “suspected” points have been found:
A(0, 0, 0), λA =
1
2
and
1
B(0, 0, 2), λB = − ,
2
(2.1.21)
the values of the Lagrange multiplier having been established with the use of the
third equation (2.1.20).
Now one needs to analyze the matrix of second derivatives. Simple calculations
give
⎡
⎤
2(1 − λ)
0
0
F = ⎣
(2.1.22)
0
2(2 − λ) 0 ⎦ .
0
0
2λ
When examining the first point, the value 1/2 should be plugged in for λ. Then
one gets the matrix
⎡
⎤
100
F = ⎣ 0 3 0 ⎦ ,
(2.1.23)
A
001
which obviously is positive-definite. Even without any reduction to the subspace
of the vectors tangent to g, it can be seen that at the point A the function f has a
minimum. Nevertheless, let us find the tangent vectors. If such vector is denoted as
h = [hx , hy , hz ], the following condition must be satisfied:
⎡ ⎤
hx
∇g · h = 0, i.e. [2x, 2y, −2z + 2] · ⎣ hy ⎦
A
A
hz
⎡ ⎤
hx
= [0, 0, 2] · ⎣ hy ⎦ = 2hz = 0,
hz
(2.1.24)
which implies that hz = 0, and hx , hy are arbitrary. This means that there are two
linearly independent tangent vectors, e.g., [1, 0, 0] and [0, 1, 0]. This should not be
surprising because in the present problem the condition g = 0 describes a twodimensional surface, so at each point there is a two-dimensional tangent plane.
50
2 Investigating Conditional Extremes
The reduced quadratic form:
⎡
⎤⎡ ⎤
100
hx
hT · F · h = hx , hy , 0 ⎣ 0 3 0 ⎦ ⎣ hy ⎦
A
001
0
= h2x + 3h2y
(2.1.25)
is positive-definite and, therefore, it corresponds to the minimum of the function f .
In the case of the second “suspected” point, the value −1/2 should be substituted
for λ into (2.1.22), giving
⎡
⎤
30 0
F = ⎣ 0 5 0 ⎦ .
B
0 0 −1
(2.1.26)
This matrix is neither positive nor negative, since it has eigenvalues of different
signs. Therefore, the reduction to the two-dimensional subspace is necessary. This
time any tangent vector must satisfy the requirement:
∇g · h = 0,
(2.1.27)
B
or
[2x, 2y, −2z + 2]
B
⎤
⎡ ⎤
hx
hx
· ⎣ hy ⎦ = [0, 0, −2] · ⎣ hy ⎦ = −2hz = 0,
hz
hz
⎡
(2.1.28)
leading again to the conclusion that hz = 0, and hx , hy are arbitrary. The reduced
quadratic form is
⎡
⎤⎡ ⎤
30 0
hx
hT · F · h = hx , hy , 0 ⎣ 0 5 0 ⎦ ⎣ hy ⎦
B
0 0 −1
0
= 3h2x + 5h2y .
(2.1.29)
As can be seen, after the reduction it has become positive-definite, and thus one is
again dealing with a minimum of the function f .
It may seem puzzling to the reader how it is possible that the function has only
two minima without any maxima or even saddle points in between. It seems to
be counterintuitive. However, a careful look at the set defined with the equation
g(x, y, z) = 0 explains this apparent paradox. This area consists of two separate
sheets (it is simply a two-sheeted hyperboloid) and on each of them the function f
has one minimum. A similar situation will be also encountered in the next problem.
2.1 Using the Method of the Lagrange Multipliers
51
Problem 3
The critical points of the function f (x, y, z) = 2/x 2 +y +4z2 on the surface defined
with the system of equations:
x 2 − y 2 + z2 = 0,
x 2 + y 2 + z2 = 2
(2.1.30)
will be found and examined.
Solution
Let us introduce the notation:
g1 (x, y, z) = x 2 − y 2 + z2
and
g2 (x, y, z) = x 2 + y 2 + z2 − 2.
(2.1.31)
Both these functions are defined on R3 so each of the conditions gi (x, y, z) = 0
defines a two-dimensional surface, and both of them together—a curve (or a sum
of curves) being their intersection. Thereby, we will be looking for the extremes of
the function f (x, y, z) on a certain curve. It is easy to see that due to the condition
x = 0 resulting from the domain of the function, (2.1.30) does not define closed
curves.
Due to the presence of two conditions, the two Lagrange multipliers will be
introduced:
F (x, y, z; λ1 , λ2 ) = f (x, y, z) − λ1 g1 (x, y, z) − λ2 g2 (x, y, z)
=
2
+ y + 4z2 − λ1 (x 2 − y 2 + z2 ) − λ2 (x 2 + y 2 + z2 − 2).
x2
(2.1.32)
The requirement of the vanishing of the gradient of the function F leads, together
with the conditions (2.1.30), to the set of equations:
⎧
⎪
−4/x 3 − 2λ1 x − 2λ2 x = 0,
⎪
⎪
⎪
⎪
⎨ 1 + 2y(λ1 − λ2 ) = 0,
8z − 2λ1 z − 2λ2 z = 2z(4 − λ1 − λ2 ) = 0,
⎪
⎪
⎪
x 2 − y 2 + z2 = 0,
⎪
⎪
⎩ x 2 + y 2 + z2 = 2.
(2.1.33)
Solving this type of equations can be tedious and cumbersome, so a good idea is
to start with the simplest one, i.e., the third one, thanks to its product form. It entails
that z = 0 or λ1 + λ2 = 4. Let us examine these possibilities.
52
2 Investigating Conditional Extremes
1. z = 0.
Plugging this value into the last two equations leads to a subsystem which is easy
to solve:
2
2
x − y 2 = 0,
x = 1,
⇒
(2.1.34)
x 2 + y 2 = 2,
y 2 = 1,
the equations on the right having been obtained by adding and subtracting those
on the left. Their solutions can be written out: x = ±1 and y = ±1. One still has
to check if for these values of x, y, and z the first two of the equations (2.1.33)
can be satisfied as well, and to find the corresponding values of the parameters λ1
and λ2 . This task is very easy, since the equations obtained by using the Lagrange
multipliers method are always linear in these quantities. In this way the following
set of “suspected” points is obtained:
5
3
A(1, 1, 0), λ1A = − , λ2A = − ,
4
4
5
3
B(−1, 1, 0), λ1B = − , λ2B = − ,
4
4
3
5
C(1, −1, 0), λ1C = − , λ2C = − ,
4
4
3
5
D(−1, −1, 0), λ1D = − , λ2D = − .
4
4
(2.1.35)
2. λ1 + λ2 = 4.
Inserting this value into the first of the equations (2.1.33) leads to the condition
−4/x 3 − 8x = 0, which cannot be fulfilled (both terms on the left-hand side are
positive—for negative x—or negative—for positive x).
The points A, B, C, D are then the only “suspected” ones. Let us now examine
them in detail. First one has to calculate the matrix of the second derivatives:
⎤
⎡
12
−
2λ
−
2λ
0
0
1
2
⎥
⎢ x4
⎥,
F = ⎢
(2.1.36)
⎦
⎣
0
2λ − 2λ
0
1
0
2
0
8 − 2λ1 − 2λ2
and then all the points are dealt with in turn.
1. The point A(1, 1, 0), λ1A = −5/4, λ2A = −3/4.
In this case, one has
⎡
⎤
16 0 0
F = ⎣ 0 −1 0 ⎦ .
A
0 0 12
(2.1.37)
2.1 Using the Method of the Lagrange Multipliers
53
This matrix is neither positive- nor negative-definite (it has eigenvalues of
different signs), but remember that we are interested in its sign only after the
reduction to the subspace of tangent vectors. This subspace is naturally one
dimensional, as the conditions (2.1.30) describe a curve (i.e., a one-dimensional
“surface”) in R3 . So one has to find one tangential vector h = [hx , hy , hz ]
satisfying the set of equations:
⎧ ⎪
⎪
⎨ g1 · h = 0,
A
⎪
⎪
⎩ g2 · h = 0,
(2.1.38)
A
i.e.,
2(hx − hy ) = 0,
2(hx + hy ) = 0.
(2.1.39)
This system immediately implies that hx = hy = 0 and hz is arbitrary (but
nonzero). The reduced quadratic form is, therefore,
⎡
⎤⎡ ⎤
16 0 0
0
hT · F · h = 0, 0, hz ⎣ 0 −1 0 ⎦ ⎣ 0 ⎦
B
0 0 12
hz
= 12 · h2z > 0.
(2.1.40)
The conclusion can be drawn that at the point A the function f (x, y, z) has a
conditional minimum.
2. The point B(−1, 1, 0), λ1B = −5/4, λ2B = −3/4.
The matrix of the second derivatives is identical as that found for A. One only
needs to perform the appropriate reduction. As usual it is required that
⎧ ⎪
⎪
⎨ g1 · h = 0,
B
⎪
⎪
⎩ g2 · h = 0,
(2.1.41)
B
i.e.,
−2(hx + hy ) = 0,
−2(hx − hy ) = 0.
(2.1.42)
The same conclusion is then obtained: hx = hy = 0 and hz is arbitrary. The
reduced quadratic form is again given by (2.1.40), which implies that at the point
B the function reaches a conditional minimum.
54
2 Investigating Conditional Extremes
3. The point C(1, −1, 0), λ1C = −3/4, λ2C = −5/4.
The conditions
⎧ ⎪
⎪
⎨ g1 · h = 0,
C
⎪
⎪
⎩ g2 · h = 0,
(2.1.43)
C
in the same way as before lead to hx = hy = 0 and the reduced quadratic form is
⎡
⎤⎡ ⎤
16 0 0
0
hT · F · h = 0, 0, hz ⎣ 0 1 0 ⎦ ⎣ 0 ⎦
B
0 0 12
hz
= 12 · h2z > 0.
(2.1.44)
Thereby, a new conditional minimum has been found.
4. The point D(−1, −1, 0), λ1D = −3/4, λ2D = −5/4.
The results are the same as for C.
In conclusion, one can see that four minima and no maxima have been found.
It might seem that this stays in contradiction with the observation made in the
first exercise. Indeed, if we go for an excursion along a certain path and at the
end we come back to the starting point, it is inevitable to reach both the smallest
and the largest height. So why this time only minima have been encountered? The
answer is similar to the one at the end of the previous problem: in the case of
the above path we do not come back to the starting point! Having a closer look
at the conditions (2.1.30), it can be concluded that they do not represent a closed
curve but four open fragments (semicircles without ends). It was mentioned at the
very beginning. This is a consequence of the fact that the plane yz is excluded from
the domain of the function. As a result of the curve, which is the intersection of
the sphere with the cone surface and is defined with the equations (2.1.30)—this is
simply two circles—four points are removed: (0, 1, 1), (0, 1, −1), (0, −1, 1), and
(0, −1, −1).
Problem 4
The extremes of the function given by
f (x1 , x2 , . . . , xN ) = −
N
!
xi log xi ,
(2.1.45)
i=1
defined on the set ]0, 1[N , will be found, subject to the condition x1 + x2 + . . . + xN
= 1.
2.1 Using the Method of the Lagrange Multipliers
55
Solution
The function (2.1.45) is called the information entropy and the variable xi is
the probability of the occurrence of an ith event from among the set of N possible
ones. The probability of whatever event to happen is, naturally, equal to 1 and hence
the side condition. The problem encountered in this exercise can be then stated as
follows: at what distribution of probabilities will the entropy take its extreme values?
If necessary, the domain of the function f can be easily extended to the set
[0, 1]N , by the appropriate limits. Since
lim x log x = 0,
(2.1.46)
x→0+
if either of the following numbers x1 , x2 , . . . , xN equals zero, the corresponding
term will simply disappear from the sum (2.1.45).
Now let us define the function F , introducing the Lagrange multiplier λ:
F (x1 , x2 , . . . , xN ; λ) = −
N
!
N
!
xi log xi − λ(
xi − 1).
i=1
(2.1.47)
i=1
After differentiating over consecutive variables, one gets the set of equations:
⎧
⎪
⎪ log x1 + 1 + λ = 0,
⎪
⎪
⎪
⎨ log x2 + 1 + λ = 0,
···
⎪
⎪
⎪ log xN + 1 + λ = 0,
⎪
⎪
⎩ x + x + . . . + x = 1.
1
2
N
(2.1.48)
These equations ensure that at the critical point the values of all variables are
identical. Taking into account the last equation, one obtains
x1 = x2 = . . . = xN =
1
.
N
(2.1.49)
One still has to determine the nature of this point. It is easy to calculate the matrix
of the second derivatives:
⎤
⎡
1
−
0 ··· 0
⎥ ⎡ −N 0 · · · 0 ⎤
⎢ x1
⎥
⎢
1
⎥
⎢
0 −N · · · 0 ⎥
⎢ 0 − ··· 0 ⎥ ⎢
⎥.
(2.1.50)
F =⎢
⎥=⎢
x2
⎣
⎦
⎥
⎢
·
·
·
·
·
·
·
·
·
·
·
·
⎢ ··· ··· ··· ··· ⎥
⎣
0 0 · · · −N
1 ⎦
0
0 ··· −
xN
56
2 Investigating Conditional Extremes
It is negative-definite in an obvious way, even without any reduction to the subspace
of tangent vectors. This means that at the critical point the function has a maximum
and f (1/N, 1/N, . . . , 1/N) = log N.
This result indicates that the entropy achieves the largest value if the probabilities
are uniformly distributed. This simultaneously means that our knowledge about
the system is the smallest. Therefore, one can say that the entropy is a measure
of our ignorance. Any inequality between the values of probabilities x1 , x2 , . . . , xN
means that we have gained some additional knowledge (as compared to the complete
ignorance) and the value of the entropy gets smaller. It can easily be seen that in the
extreme case, when our knowledge about the system becomes full—that is, if one of
the xi ’s equals 1 (then the others must vanish) and one can be sure that a particular
event occurs—the entropy vanishes. This zero is the minimal value of the entropy
since, because of the condition xi ≤ 1, the expression (2.1.45) is nonnegative.
2.2 Looking for Global Extremes
Problem 1
The global extremes of the function given with the formula
f (x, y) = x 2 + (x − y)2
(2.2.1)
A = {(x, y) ∈ R2 | x 2 ≥ y 2 ∧ |x| ≤ 4}.
(2.2.2)
will be found on the set
Solution
In Sect. 7.1 of Part II, we were concerned with maxima and minima of functions
on compact sets. Now we come back to this topic, using the Lagrange multipliers
method introduced in this chapter. As we remember, searching for global extremes
of functions on compact sets consists of two steps. First one had to identify all
critical points located in the interior of a given set, and second one needed to
examine the behavior of the function on the boundary. And this is just the place
where conditional extremes enter into play.
We need to start by drawing the set A. The set of equations:
x2 ≥ y2,
|x| ≤ 4
(2.2.3)
2.2 Looking for Global Extremes
57
Fig. 2.4 Graphical representation of the set A
can be rewritten in the form of an alternative:
(y ≤ x ∧ y ≥ −x ∧ 0 ≤ x ≤ 4) ∨ (y ≥ x ∧ y ≤ −x ∧ −4 ≤ x ≤ 0) .
(2.2.4)
It is now easy to demonstrate graphically the obtained predicate, as well as the set A.
It is made up of two triangles, as shown in Fig. 2.4. Let us then proceed to investigate
the function f .
1. The interior of A.
In this case we deal with unconditional extremes. By following exactly the same
way as in Sect. 7.1 of the previous part of this book series, where this procedure
is described in detail, one finds the critical points, and, without checking if
they correspond to extremes or saddle points, the appropriate values of the
function are calculated. Later this list will be compared to the values achieved
at “suspected” points located on the boundary. Thus one has the set of equations:
⎧
∂f
⎪
⎪
= 4x − 2y = 0,
⎪
⎨ ∂x
⎪
⎪
∂f
⎪
⎩
= 2y − 2x = 0,
∂y
(2.2.5)
58
2 Investigating Conditional Extremes
whose only solution is x = y = 0. This point, however, is not situated in the
interior of A. It means that “suspected” points do not appear in the interior and
the global extremes are located on the border.
2. The border of A.
As shown in the figure, the border is composed of two segments contained in
the vertical straight lines x = ±4 and the two contained in the inclined lines,
described jointly with the equation x 2 − y 2 = 0. In the first case, one simply
plugs x = ±4 into the formula for the function f and obtains a one variable
function to examine. And this is what is done below. In the second case, one could
do the same. The side condition is simple enough that it could be accomplished
without difficulties. Remember, however, that not always will it be possible and
one needs to learn how to solve more complex cases as well. For this reason we
denote g(x, y) = x 2 − y 2 and treat this part of the problem as an exercise for
finding extremes subject to the condition g(x, y) = 0 and, therefore, for using
the Lagrange multipliers. We start, however, with the easier part.
With x = 4, one gets the quadratic trinomial in the variable y in the form
f (4, y) = 16 + (4 − y)2 ,
(2.2.6)
from which it is visible that the minimum (vertex) corresponds to y = 4. The
point of coordinates (4, 4) belongs to the set A. The value of f (4, 4) = 16 is the
first one that should be noted as the potential extremal value. In turn for x = −4,
one gets
f (−4, y) = 16 + (4 + y)2 ,
(2.2.7)
and the minimum (again equal to 16) is achieved at the point (−4, −4). One has
yet to record the values obtained at two other “corners” of the set A:
f (4, −4) = f (−4, 4) = 80.
As we know from Part II, these values could not be found with the use of the
differential approach.
Now we are going to use the method of the Lagrange multipliers for two other
edges of the set A. Let us define the function
F (x, y; λ) = f (x, y) − λg(x, y) = x 2 + (x − y)2 − λ(x 2 − y 2 )
(2.2.8)
and require its gradient to vanish. Together with the side condition, the following
system of equations is obtained:
⎧
⎨ (2 − λ)x − y = 0,
x − (1 + λ)y = 0,
⎩ 2
x − y 2 = 0.
(2.2.9)
2.2 Looking for Global Extremes
59
One obvious solution of this system is x = y = 0, so the new value should be
appended to our list: f (0, 0) = 0. Since the function defined in the content of the
problem is nonnegative, this value certainly corresponds to the global minimum.
Moreover, this minimum is unique. To find it out, it is sufficient to insert y = x
and y = −x into the first two equations. In the first case, one gets
(1 − λ)x = 0,
⇒ x = 0,
(2.2.10)
−λx = 0,
and in the second
(3 − λ)x = 0,
(2 + λ)x = 0,
⇒ x = 0.
(2.2.11)
If one does not wish to resolve the condition x 2 − y 2 = 0, which may not
be possible in some more complicated case, the same result can be obtained
upon multiplying the first two equations of (2.2.9) by each other (after having
separated the terms with x and y on opposite sides) and then using the fact that
x 2 = y 2 . This leads to the condition (1−2λ)x 2 = 0, and hence either λ = 1/2 or
x = 0 (and consequently y = 0). This former case, however, again leads to the
result x = y = 0, since after substituting into the first two equations one obtains
5
4 x = 0.
In conclusion, the largest value assumed by the function f on the set A equals
80 (at the points (4, −4) and (−4, 4)), and the global minimum corresponds to the
value 0 (at the point (0, 0)).
Problem 2
The global extremes of the function f (x, y, z) = x + y − z on the set
A = {(x, y, z) ∈ R3 | x 2 + y 2 + z2 ≤ 9}
(2.2.12)
will be found.
Solution
This time the set A is a closed ball. To establish that no critical points of the function
f appear in its interior is not a problem. When calculating the partial derivatives,
one finds that all of them are nonzero
∂f
∂f
∂f
=
=−
= 1.
∂x
∂y
∂z
(2.2.13)
60
2 Investigating Conditional Extremes
Thereby, only the border of the set A has to be examined, which leads to the standard
issue of looking for conditional extremes with the side condition in the form:
g(x, y, z) := x 2 + y 2 + z2 − 9 = 0.
(2.2.14)
As usual the function F is defined as
F (x, y, z; λ) = f (x, y, z)−λg(x, y, z) = x+y−z−λ(x 2 +y 2 +z2 −9)
(2.2.15)
= 0, we get the set of equations:
and after having imposed the requirement ∇F
⎧
2λx = 1,
⎪
⎪
⎨
2λy = 1,
⎪
2λz = −1,
⎪
⎩ 2
x + y 2 + z2 = 9,
(2.2.16)
the side condition having been appended. The values of x, y, and z easily obtained
from the first three equations can be plugged into the last one:
1
2λ
2
+
1
2λ
2
+
−1
2λ
2
= 9.
(2.2.17)
Solving it for λ, one finds
√
√
3
3
∨ λ=−
.
⇒ λ=
6
6
3
=9
4λ2
(2.2.18)
Having found these values of the Lagrange multiplier, one can now get from the first
three equations
“suspected” points. There are two of
√ √(2.2.16)
√ the coordinates
√ of√the √
them: P ( 3, 3, − 3) and Q(− 3, − 3, 3). All that remains is to calculate
and compare the values of the function f at these points:
√
√
√
√
√
√
√ √
(2.2.19)
f ( 3, 3, − 3)) = 3 3, f (− 3, − 3, + 3)) = −3 3.
We conclude that at the point P the function f has the global maximum and at Q
the global minimum.
Problem 3
The global extremes of the function
2
z2
2
2
− 2xy
f (x, y, z) = 6(x + y ) + 3z − 4xy − 3x + 3y +
2
2
2
2
(2.2.20)
2.2 Looking for Global Extremes
61
will be found on the set
A = {(x, y, z) ∈ R3 | x 2 + y 2 + z2 /4 ≤ 1}.
(2.2.21)
Solution
The former examples were relatively easy to solve due to the simplicity of the
functions. At the end of this section we will, therefore, be concerned with a
somewhat more intricate case. The procedure to be followed, however, remains the
same: to find critical points in the interior of the set A (the closed ellipsoid) and
then to examine the behavior of the function at the border using the method of the
Lagrange multipliers.
1. The interior of the set A.
At the critical points, all three partial derivatives must be equal to zero. One,
therefore, obtains the set of equations:
⎧
∂F
⎪
⎪
= 4(3x − y)(1 − 3x 2 − 3y 2 − z2 /2 + 2xy) = 0,
⎪
⎪
∂x
⎪
⎪
⎪
⎪
⎪
⎨ ∂F
= 4(3y − x)(1 − 3x 2 − 3y 2 − z2 /2 + 2xy) = 0,
(2.2.22)
⎪
∂y
⎪
⎪
⎪
⎪
⎪
⎪
⎪
∂F
⎪
⎩
= 2z(3 − 3x 2 − 3y 2 − z2 /2 + 2xy) = 0.
∂z
The fulfillment of these equations requires the vanishing of the relevant factors
on their left-hand sides. One has in total 2 × 2 × 2 = 8 possibilities, which will
be examined one by one, starting with the simplest one.
(a) y = 3x, x = 3y, z = 0.
These conditions mean that x = y = z = 0. The critical point P1 (0, 0, 0)
belongs to the set A, so it must be included in the list of “suspected” points.
We have f (0, 0, 0) = 0 and this is the first one from among the values that
should be noted.
(b) 3x 2 + 3y 2 + z2 /2 − 2xy = 1, x = 3y z = 0.
Substituting x = 3y and z = 0 into√ the first condition, we obtain
24y 2 = 1, which implies that y = ± 6/12. In √
this way
√ two critical
points√are found,
both
lying
in
the
interior
of
A:
P
(
6/4,
6/12, 0) and
2
√
P3 (− 6/4, − 6/12, 0). When calculating the value of the function f for
both of them, one finds
√
√
√
√
f ( 6/4, 6/12, 0) = f (− 6/4, − 6/12, 0) = 1.
These values should be noted too.
(2.2.23)
62
2 Investigating Conditional Extremes
(c) y = 3x, 3x 2 + 3y 2 + z2 /2 − 2xy = 1, z = 0.
It is visible that, as compared to the previous case, x and
√ y simply
√ swap their
roles.√Then there
are
two
further
critical
points:
P
(
6/12,
6/4, 0) and
4
√
P5 (− 6/12, − 6/4, 0). Because of the symmetry of the formula (2.2.20)
when replacing x ↔ y, we know that at both of them the value of the
function equals 1.
(d) 3x 2 + 3y 2 + z2 /2 − 2xy = 1, 3x 2 + 3y 2 + z2 /2 − 2xy = 1, z = 0.
Two of the above conditions overlap, so the solutions will constitute the
entire curve and not discrete points only. After eliminating z, one gets the
condition, which can be given the form
3x 2 + 3y 2 − 2xy = 1.
(2.2.24)
If one now introduced some new variables defined with the relations ξ =
x + y and η = x − y in place of x and y, the Eq. (2.2.24) would become an
ellipse equation:
ξ 2 + 2η2 = 1.
(2.2.25)
Because of the linear nature of relations between variables x, y and ξ, η,
the axes of the new system are only inclined respective to the old ones. In
particular, the ξ axis, which corresponds to the condition η = 0, is just the
straight line y = x, and η, for which ξ = 0, is the line y = −x. The axes
of both systems, together with the curve (2.2.25) lying in the plane z = 0,
are depicted in Fig. 2.5. The intersection of the set A with the plane z = 0 is
also drawn. The figure shows that the ellipse (2.2.25) is entirely contained in
the set A. That this is really the case, one can also easily be convinced in an
analytical way, rewriting the Eq. (2.2.24) in the form
x2 + y2 +
1
1
(x − y)2 = ,
2
2
(2.2.26)
which obviously entails that x 2 +y 2 < 1, i.e., the ellipse lies inside the set A.
In order to obtain the value of the function f at any point of this curve, it
is sufficient to arrange the terms in (2.2.20) in the following way:
f (x, y, z) = 2[3(x 2 + y 2 ) − 2xy ] + 3 z2
=1
=0
1 2 2
2
2
z
− 3x + 3y − 2xy + = 1.
2
=1
(2.2.27)
=0
Thereby, the value of 1 is obtained as that of the possible global extreme.
2.2 Looking for Global Extremes
63
Fig. 2.5 The curve (2.2.24)
together with the intersection
of the set A with the plane
z=0
(e) y = 3x, x = 3y, 3x 2 + 3y 2 + z2 /2 − 2xy = 3.
As in the
√ item (a), one immediately gets x = y = 0, which implies that
z = ± 6. The points of these coordinates are located, however, away from
the set A.
(f) All other cases lead to contradictions, since they require that the following
conditions are simultaneously satisfied: 3x 2 + 3y 2 + z2 /2 − 2xy = 1 and
3x 2 + 3y 2 + z2 /2 − 2xy = 3.
2. The border of the set A.
Let us now denote g(x, y, z) = x 2 + y 2 + z2 /4 − 1 and define the function:
F (x, y, z; λ) = f (x, y, z) − λg(x, y, z).
(2.2.28)
We are not going to explicitly write out its form, but rather to provide the
equations obtained after the differentiation:
⎧ ∂F
⎪
= 4(3x − y)(1 − 3x 2 − 3y 2 − z2 /2 + 2xy) − 2λx = 0,
⎪
⎪
⎪
∂x
⎪
⎪
⎪
⎪
⎪
⎪
∂F
⎪
⎪
⎪
= 4(3y − x)(1 − 3x 2 − 3y 2 − z2 /2 + 2xy) − 2λy = 0,
⎨
∂y
⎪
⎪
⎪
∂F
⎪
⎪
⎪
= 2z(3 − 3x 2 − 3y 2 − z2 /2 + 2xy) − λz/2 = 0,
⎪
⎪
∂z
⎪
⎪
⎪
⎪
⎪
⎩ 2
x + y 2 + z2 /4 = 1.
(2.2.29)
64
2 Investigating Conditional Extremes
This system is quite intricate, but one has to be able to deal with such a system
as well. To start with, let us try to simplify it a bit. First of all, the quantity z2
can be removed from the brackets in the first three equations thanks to the side
condition. This yields after some minor transformations
⎧
2(3x − y)[(x − y)2 + 1] + λx = 0,
⎪
⎪
⎨
2(3y − x)[(x − y)2 + 1] + λy = 0,
⎪
z[(x − y)2 − 1 + λ/4] = 0,
⎪
⎩ 2
x + y 2 + z2 /4 = 1.
(2.2.30)
Now the first two equations can be replaced by taking their sum and difference.
It should be pointed out that the operations which are performed on the equations
do not constitute any universal procedure; they are merely the result of observing
that a simpler system is obtained. So if the reader ever meets the necessity of
solving this kind of a complex system, the first step should be to carefully look
at the structure of the equations and use one’s imagination to think of a way to
simplify it. Then, we obtain
⎧
(x + y)[4(x − y)2 + 4 + λ] = 0,
⎪
⎪
⎨
(x − y)[8(x − y)2 + 8 + λ] = 0,
⎪ z[4(x − y)2 − 4 + λ] = 0,
⎪
⎩ 2
x + y 2 + z2 /4 = 1.
(2.2.31)
Now let us observe that the expressions in the brackets in the first two equations
cannot simultaneously vanish. Therefore, the following possibilities come into
play
(a) x + y = 0 and x − y = 0.
These conditions mean that x = y = 0 and the last two equations give
z = ±2, λ = 4. Thus, one gets two successive “suspected” points:
P6 (0, 0, 2)
and
P7 (0, 0, −2).
(2.2.32)
Let us still note the values of the function f at these points:
f (0, 0, 2) = f (0, 0, −2) = 8.
(2.2.33)
(b) x − y = 0 and 4(x − y)2 + 4 + λ = 0.
We immediately get λ = −4, and z = 0 from the third of the equations (2.2.31). The side condition gets simplified to 2x 2 = 1 and, in consequence, the new “suspected” points are found:
√
√
P8 (1/ 2, 1/ 2, 0)
and
√
√
P9 (−1/ 2, −1/ 2, 0).
(2.2.34)
2.3 Exercises for Independent Work
65
The values of the function at these points are as follows:
√
√
√
√
f (1/ 2, 1/ 2, 0) = f (−1/ 2, −1/ 2, 0) = 0.
(2.2.35)
(c) x + y = 0 and 8(x − y)2 + 8 + λ = 0.
This implies that λ = −32x 2 − 8 and the third equation of (2.2.31) can be
given the form:
− z(16x 2 + 12) = 0.
(2.2.36)
The only solution is now obviously z = 0, and consequently the side
condition requires 2x 2 = 1. This means that two more interesting points
have been found:
√
√
P10 (1/ 2, −1/ 2, 0)
and
√
√
P11 (−1/ 2, 1/ 2, 0)
(2.2.37)
with the function values:
√
√
√
√
f (1/ 2, −1/ 2, 0) = f (−1/ 2, 1/ 2, 0) = −8.
(2.2.38)
At the end, the outcome needs to be reviewed. It shows that the largest value
(equal to 8) is adopted by the function
√ f at the
√ points (0, 0, ±2),
√ and√the smallest
value (equal to −8) at the points (1/ 2, −1/ 2, 0) and (−1/ 2, 1/ 2, 0).
2.3 Exercises for Independent Work
Exercise 1 Find conditional extremes of the functions:
(a) f (x, y) = x 2 + y 2 on the set A = {(x, y) ∈ R2 | x 4 − y 4 = 1},
(b) f (x, y, z) = x + y + z on the set A = {(x, y, z) ∈ R3 | x 2 + y 2 + z2 = 9},
(c) f (x, y) = x(y + z) on the set A = {(x, y, z) ∈ R3 | yz = 1 ∧ x + y + z = 0}.
Answers
(a) Local minima at the points (±1,
√ 0).√ √
(b) Local
maximum
at
the
point
(
3, 3, 3), local minimum at the point
√
√
√
(− 3, − 3, − 3).
(c) Local maxima at the points (−2, 1, 1), (2, −1, −1).
66
2 Investigating Conditional Extremes
Exercise 2 Find global extremes of the functions:
(a) f (x, y, z) = x 2 − y 2 , on the set A = {(x, y, z) ∈ R3 | x 2 /4 + y 2 + z2 ≤ 1},
(b) f (x, y, z) = x + y + z2 , on the set A = {(x, y, z) ∈ R3 | x 2 + y 2 ≤ 1 ∧ 0 ≤
z ≤ 1}.
Answers
(a) Global minima at the points (0, ±1, 0), global maxima at the points
(±2, 0, 0).
√
√
(b) Global minimum
at the point (− 2/2, − 2/2, 0), global maximum at
√
√
the point ( 2/2, 2/2, 1).
Chapter 3
Investigating Integrals with Parameters
In this chapter we are interested in functions defined as integrals
b
F (t) =
f (x, t) dx,
(3.0.1)
a
where a and b are some numbers but may be infinite too. Our main concern constitutes the limits of such functions, their continuity, differentiability, or integrability.
These properties are also applied to find the values of certain integrals.
The following continuity theorem holds:
Given a function f (x, t) continuous as a function of two variables on the rectangle
[a, b] × [A, B]. Then the function F (t) defined with the formula (3.0.1) is
continuous on [A, B], which also means that
b
lim F (t) =
b
lim f (x, t) dx =
t→t0
f (x, t0 ) dx = F (t0 ),
t→t0
a
(3.0.2)
a
for any t0 ∈ [A, B].
In the case when b = ∞, the above theorem takes the following form:
Given a function f (x, t) continuous as a function of two variables on the set
[a, ∞] × [A, B] and the integral
∞
F (t) =
f (x, t) dx,
(3.0.3)
a
uniformly convergent with respect to t ∈ [A, B]. Then the function F (t) is
continuous on [A, B].
© Springer Nature Switzerland AG 2020
T. Radożycki, Solving Problems in Mathematical Analysis, Part III, Problem Books
in Mathematics, https://doi.org/10.1007/978-3-030-38596-5_3
67
68
3 Investigating Integrals with Parameters
The notion of the uniform convergence for sequences and series is known
to the reader from Chap. 15 in Part I. Here, in turn, it refers to the continuous
variable b. The integral (3.0.2) is called uniformly convergent if the function
"b
G(b, t) := a f (x, t)dx is uniformly convergent to the function F (t) given with
the formula (3.0.3) for b → ∞, i.e., if
∀>0 ∃K>0 ∀b>K |G(b, t) − F (t)| < .
(3.0.4)
As regards the differentiability of the function (3.0.1), the so-called Leibniz rule
holds:
Given a function f (x, t) continuous in x on [a, b] for any fixed t ∈ [A, B]. Then,
if the partial derivative ∂f/∂t exists and is continuous as a function of two variables
on the rectangle [a, b] × [A, B], the function F (t) defined with the formula (3.0.1)
is differentiable for any t ∈]A, B[, and
b
d
F (t) =
dt
∂
f (x, t) dx.
∂t
(3.0.5)
a
This theorem is sometimes reformulated so that the continuity of f (x, t) as a
function of two variables is assumed. It can be shown that this formulation is
equivalent.
For the case of the integral (3.0.3), the appropriate theorem called again the
Leibniz rule has the form:
If the integrand function f (x, t) and partial derivative ∂f (x, t)/∂t are continuous
on the set [a, ∞[×[A, B], the improper integral (3.0.3) is convergent for any t ∈
[A, B] and the improper integral
∞
∂
f (x, t) dx
∂t
(3.0.6)
a
is uniformly convergent on [A, B], then the function F (t) is differentiable for t ∈
]A, B[ and
d
F (t) =
dt
∞
∂
f (x, t) dx.
∂t
(3.0.7)
a
As to the integrability, this issue was considered in Chap. 12 of the second part.
For our purposes, the following Fubini theorem is sufficient: for a continuous (i.e.,
integrable) function f (x, t) both iterated integrals exist and are equal, so one can
write
3.1 Examining Limits and Continuity
b
69
B
dt f (x, t) =
dx
a
B
A
b
dt
dx f (x, t).
(3.0.8)
a
A
In the case of the improper integral over x, the following formula still can be used:
∞
B
dt f (x, t) =
dx
a
∞
B
A
dt
A
dx f (x, t),
(3.0.9)
a
provided, apart from the above assumptions, the integral
∞
f (x, t) dx
(3.0.10)
a
is uniformly convergent.
3.1 Examining Limits and Continuity
Problem 1
The limit lim F (t), where the function F (t) is given by the formula:
t→1
1
F (t) =
0
xt
dx,
1+x
(3.1.1)
will be found.
Solution
As already mentioned above, in this chapter we are concerned with integrals
dependent on a certain additional parameter, and in particular we are interested in
such properties of emerging functions like continuity or differentiability with respect
to this variable. There are many cases in which, when dealing with a two-variable
function, one is able to explicitly carry out the integration and then examining the
function F (t) does not pose any problems. This kind of example remains beyond our
interest below. We will be focused only on those for which are needed to determine
the properties of the function F (t) without executing the integral.
70
3 Investigating Integrals with Parameters
The first and simplest task standing in front of us is how to perform the limit t →
1 in the expression (3.1.1). To this end, one would be willing to simply put t = 1
"1
on the right-hand side and to perform the emerging easy integral 0 x/(1 + x) dx.
In this way, however, one would not calculate limt→1 F (t) but rather F (1). As the
reader might remember, both of these values must be actually equal, provided the
function F (t) is continuous. Hence, if one wants to perform the limit in this way,
one must first make sure of the continuity of F (t) at least on a certain neighborhood
of the point t = 1. The following statement, already formulated in the theoretical
introduction above, comes to our aid: if a given function f (x, t) is continuous as a
function of two variables on the set [a, b] × [A, B], then the function
b
F (t) =
f (x, t) dx
(3.1.2)
a
is continuous (naturally as a function of one variable t) on the interval [A, B].
When we write “continuous as a function of two variables,” we have in mind not
the continuity in each variable separately, but the genuine continuity on the plane
dealt with in Chap. 4 of the second part of this book series.
In our example, one has [a, b] = [0, 1], and as [A, B] any closed interval may be
chosen, provided it contains in its interior the essential point t = 1. Of course there
is no need to consider an interval that would be too broad, for example, containing
the negative values of t, since the easy and nice integral (3.1.1) would be converted
into a much more troublesome improper integral. Therefore, our simple choice is
[A, B] = [1/2, 3/2].
According to the above-mentioned theorem, one has to verify the twodimensional continuity of the function f (x, t) on the set D := [0, 1] × [1/2, 3/2].
Using the method from Part II, one can estimate that for any (x1 , t1 ) ∈ D and
(x2 , t2 ) ∈ D:
x t2
x1t1 (1 + x1 )x2t2 − (1 + x2 )x1t1 2
|f (x2 , t2 ) − f (x1 , t1 )| = −
=
1 + x2
1 + x1 (1 + x2 )(1 + x1 )
≤ (1 + x1 )x2t2 − (1 + x2 )x1t1 = (1 + x1 )(x2t2 − x2t1 + x2t1 ) − (1 + x2 )(x1t1 − x2t1 + x2t1 )
= (1 + x1 )(x2t2 − x2t1 ) − (1 + x2 )(x1t1 − x2t1 ) + x2t1 (x1 − x2 )
≤ |1 + x1 | x2t2 − x2t1 + |1 + x2 | x1t1 − x2t1 + x2t1 |x1 − x2 | . (3.1.3)
Now, if (x1 , t1 ) → (x2 , t2 ), all three above terms converge to zero. The first one due
to the (one-dimensional) continuity of the exponential function of the base x2 , the
second one because of the (again one-dimensional) continuity of the power function
with positive (!) exponent t1 , and the last one thanks to the presence of |x1 − x2 |.
3.1 Examining Limits and Continuity
71
As one can see, the assumptions of the theorem are met and, thereby, the function
F (t) is continuous in the neighborhood of the unity. Then, the limit (3.1.1) can be
calculated as follows:
1
lim F (t) = F (1) =
t→1
0
1
x
dx =
1+x
0
1
1−
1+x
1
= (x − log(1 + x)) = 1 − log 2.
dx
(3.1.4)
0
Problem 2
The continuity of the function F (y) defined with the formula:
π
F (y) =
sin2 (x 2 y) dx
(3.1.5)
0
will be examined, for any y ∈ R.
Solution
This exercise is very similar to the previous one, with the difference that now we
are interested in the continuity of the function F (y) not at a given point but for any
real value of y. If one wished to apply directly the theorem used in the previous
example, one would need to reduce the problem to testing the continuity on a closed
interval. It can be done very easily. Let us first choose any y, then fix the values
of two constants A and B, so that A < y < B, and finally examine the continuity
of the integrand function in both variables on the rectangle [0, π ] × [A, B]. If this
continuity was demonstrated, the continuity of the function F (y) on the interval
[A, B] would be proved as well. As long as no specific values for y, A, and B have
been chosen, the same reasoning remains legitimate for any y ∈ R. Thereby, it
would mean that the continuity of the function F for each real argument is proved.
It is assumed, therefore, that y ∈ [A, B], A and B are not subject to any
restriction except that A < B. Let us now examine the (two-dimensional) continuity
of the function
f (x, y) = sin2 (x 2 y)
(3.1.6)
72
3 Investigating Integrals with Parameters
on the rectangle [0, π ] × [A, B]. To this end, let us transform the expression as
follows:
|f (x2 , y2 ) − f (x1 , y1 )| = sin2 (x22 y2 ) − sin2 (x12 y1 )
=
1 1 1 − cos(2x22 y2 ) − 1 + cos(2x12 y1 ) = cos(2x12 y1 ) − cos(2x22 y2 ) ,
2
2
(3.1.7)
where the well-known formula:
sin2 α =
1 − cos(2α)
2
(3.1.8)
has been used. Now, if one applies the formula
cos α − cos β = −2 sin
α+β
α−β
sin
,
2
2
(3.1.9)
the following estimate can be done:
|f (x2 , y2 ) − f (x1 , y1 )| = sin(x12 y1 + x22 y2 ) sin(x12 y1 − x22 y2 )
≤1
≤ sin(x12 y1 − x22 y1 + x22 y1 − x22 y2 )
= sin(x12 y1 − x22 y1 ) cos(x22 y1 − x22 y2 ) + cos(x12 y1 − x22 y1 ) sin(x22 y1 − x22 y2 )
≤ sin[(x12 − x22 )y1 ] cos[x22 (y1 − y2 )] + cos[(x12 − x22 )y1 ] sin[x22 (y1 − y2 )]
≤1
≤1
≤ sin[(x12 − x22 )y1 ] + sin[x22 (y1 − y2 )] ≤ x12 − x22 |y1 | + x22 |y1 − y2 |
−→
0.
(3.1.10)
≤ max{|A|, |B|} x12 − x22 + π 2 |y1 − y2 |
(x2 ,y2 )→(x1 ,y1 )
The continuity of the function f (x, y) on the rectangle [0, π ] × [A, B] is then
demonstrated, and hence the continuity of the function F (y) on the interval [A, B]
as well. As to A and B nothing special was assumed, so in line with the comments
made at the beginning of this exercise, this implies the continuity of the function
defined with the integral (3.1.5) on the entire real axis.
3.1 Examining Limits and Continuity
73
Problem 3
The limit lim (t) will be found, where (t) is given by the formula:
t→n
∞
x t−1 e−x dx,
(t) =
(3.1.11)
0
with t > 0, n being a natural number.
Solution
The function (t) defined above bears the name of the Euler’s Gamma function
and is one of the most important special functions applicable in many branches of
mathematics and physics. As it will be seen in a moment, it constitutes the extension
of the operation called the factorial beyond natural numbers.
When looking at the formula (3.1.11), one notices the difficulty, which was not
encountered in the previous problems: the integral on the right-hand side is now
an improper integral for two reasons. First, the integration interval is infinite, and
second the function can diverge in the lower limit of the integration (and it is really
the case for t < 1). Therefore, one cannot directly use the theorem applied in the
preceding problems and instead one has to rely on another one formulated in the
theoretical introduction.
Before recalling the needed theorem, it is a good idea to slightly prepare the
expression (3.1.11), in order not to have to simultaneously deal with both the above
difficulties. Let us then rewrite the function (t) in the form:
∞
1
(t) =
x
t−1 −x
dx +
e
0
x t−1 e−x dx ,
1
(3.1.12)
2 (t)
1 (t)
thanks to which in the expression for 1 (t) only the lower limit of integration, and
in that for 2 (t) only the upper limit will be of interest for us.
1. 1 [t].
In order to get rid of the trouble, let us integrate by parts the expression for 1 (t):
1
1 (t) =
1
x
t−1 −x
e
1
dx =
t
0
1 t −x 1 1
= xe +
t
t
0
t −x
x e dx
0
1
1
t −x
xe
0
1
1
dx =
+
te
t
x t e−x dx.
0
(3.1.13)
74
3 Investigating Integrals with Parameters
The reader might ask what has been gained through this operation. The answer is
the following: in the integrand expression, the unity in the exponent disappeared
(one has now x t instead of x t−1 ) and the function is no longer divergent at x →
0+ even for small values of t. For, we remember that t is positive.
The separated term 1/(te) is continuous for t > 0 in a visible way. The
continuity of the function 1 (t) is, therefore, conditioned by the continuity of
"1
the integral 0 x t e−x dx, to which the theorem used in the previous problems
can be applied. For this purpose, as we remember, it is sufficient to check the
continuity of the expression x t e−x as the function of two variables on a certain
rectangle [a, b] × [A, B]. According to the text of the exercise, we are interested
in the limit t → n, where n ∈ N, so for convenience the rectangle may be chosen
in the form [0, 1] × [n − 1/2, n + 1/2]. Now let (x1 , t1 ) and (x2 , t2 ) be two points
of this rectangle. The following estimate can be done:
t2 −x
x e 2 − x t1 e−x1 2
1
= x2t2 e−x2 − x2t2 e−x1 + x2t2 e−x1 − x2t1 e−x1 + x2t1 e−x1 − x1t1 e−x1 = x2t2 (e−x2 − e−x1 ) + (x2t2 − x2t1 )e−x1 + (x2t1 − x1t1 )e−x1 ≤ x2t2 (e−x2 − e−x1 ) + (x2t2 − x2t1 )e−x1 + (x2t1 − x1t1 )e−x1 −→
(x2 ,t2 )→(x1 ,t1 )
0.
(3.1.14)
This last result has been obtained, as:
• e−x2 − e−x1 −→ 0, pursuant to the continuity of the exponential function,
x2 →x1
• x2t2 − x2t1 −→ 0, pursuant to the continuity of the exponential function, and
t2 →t1
• x2t1 − x1t1 −→ 0, pursuant to the continuity of the power function.
x2 →x1
The integrand function is, therefore, continuous as the function of two variables
on the rectangle concerned, which entails the conclusion that 1 (t) is continuous
on the interval [n − 1/2, n + 1/2] and the limit t → n may be found by simply
plugging t = n into the integral.
2. 2 [t].
In order to verify the continuity of the second integral in (3.1.12), let us invoke
the following theorem formulated at the beginning of this chapter: if the function
f (x, t) is continuous as the function of two variables on the set [a, ∞[×[A, B]
and if the integral
∞
F (t) =
f (x, t) dx
a
(3.1.15)
3.1 Examining Limits and Continuity
75
is uniformly convergent with respect to the parameter t ∈ [A, B], then the
function F (t) is continuous on this interval. The first condition can be easily
checked in the same way as above, so we will be focused on the second one.
The notion of the uniform convergence for discrete variable n appeared in
the last chapter of Part I. For a continuous parameter t, it was formulated in
the theoretical introduction. Let us, however, recall it here. The integral (3.1.15)
"b
is said to be uniformly convergent if the function G(b, t) = a f (x, t)dx is
uniformly convergent to the function F (t) given by (3.1.15) for b → ∞, i.e., if
∀>0 ∃K>0 ∀b>K |G(b, t) − F (t)| < .
(3.1.16)
The reader might object, pointing to the fact that to demonstrate (3.1.16)
may not be easy either. Fortunately, the following Weierstrass test (or, in other
words, the theorem) comes to the aid: if there exists an integrable majorant, i.e.,
a function ϕ(x) satisfying
|f (x, t)| ≤ ϕ(x)
(3.1.17)
for all x ≥ a and arbitrary t ∈ [A, B], then the integral (3.1.15) is uniformly
convergent.
So let us look for a majorant for the function f (x, t) = x t−1 e−x for x ≥ 1
and t ∈ [n − 1/2, n + 1/2]. Note that if x ≥ 1 the following inequality holds:
x t−1 ≤ x n+1/2−1 = x n−1/2 .
(3.1.18)
This last exponent is positive, as n ≥ 1, so one has
f (x, t) ≤ x n−1/2 e−x .
(3.1.19)
A candidate for a majorant is, therefore,
ϕ(x) = x n−1/2 e−x ,
(3.1.20)
but one needs to make sure that it is integrable. The “know-how” gained on this
subject when solving the problems of Chap. 2 in the second part of the book
series allows us to immediately draw a conclusion in this case. In the lower
limit of the integration interval, no problems emerge, since ϕ(x) is bounded. In
turn for x → ∞, the function exponentially decreases, i.e., sufficiently quickly.
The continuity of (3.1.20) for x ≥ 1 does not raise any doubts either, since we
are dealing with the product
" ∞ of the power and exponential functions. Thus we
come to the conclusion: 1 ϕ(x)dx exists. Consequently the Weierstrass test
ensures the uniform convergence of the integral (3.1.15) and the continuity of
the function 2 (t) as well.
76
3 Investigating Integrals with Parameters
Since the continuity of the functions 1 (t) and 2 (t) on the interval [n−1/2, n+
1/2] has been proved, so the function (t) is continuous too. One, therefore, finds
∞
x n−1 e−x dx,
lim (t) = (n) =
t→n
(3.1.21)
0
and after n − 1 integrations according to the pattern below, one comes to
∞
(n) =
∞
x
n−1 −x
e
0
= −x
x n−1 [e−x ] dx
dx = −
0
∞
+ (n − 1)
∞
n−1 −x e
0
∞
x
n−2 −x
e
x n−2 [e−x ] dx
dx = −(n − 1)
0
0
∞
e−x dx = (n − 1)!
= . . . = (n − 1)(n − 2) · . . . · 1 ·
0
=1
(3.1.22)
This result justifies the treatment of the Euler’s Gamma function as a generalization
of the concept of the factorial beyond natural numbers as mentioned previously.
3.2 Differentiating with Respect to Parameters
Problem 1
The integral:
π/2
arctan(ξ tan x)
dx,
tan x
F (ξ ) =
(3.2.1)
0
will be calculated for ξ > 0.
Solution
In this problem we will make use of the so-called Leibniz rule, formulated at the
beginning of this chapter. Let us recall its form below. Assume that a given function
f (x, t) is continuous in the variable x on the interval [a, b] for any t ∈ [A, B].
3.2 Differentiating with Respect to Parameters
77
Besides, it is assumed that the partial derivative ∂f/∂t is continuous as the function
of two variables on the set [a, b] × [A, B]. Then in the interval ]A, B[, there exists
the derivative dF /dt, and it can be calculated according to the formula:
b
d
F (t) =
dt
∂
f (x, t) dx.
∂t
(3.2.2)
a
Now one needs to check whether the above assumptions are met for
f (x, ξ ) =
arctan(ξ tan x)
.
tan x
(3.2.3)
Inside the interval [0, π/2] the above function is, naturally, continuous in the
variable x pursuant to the continuity of the functions tangent and arcus tangent.
The problem emerges only at the ends, where it is not defined at all. In order to take
account of these points, let us modify the definition of the function f :
⎧
⎨ arctan(ξ tan x)/tan x for x ∈]0, π/2[,
f (x, ξ ) = ξ
for x = 0,
⎩
0
for x = π/2.
(3.2.4)
The inclusion of these end points has no consequences for the value of the
integral (3.2.1), but at the same time ensures that the integrand function is
continuous in the variable x (for any ξ ) on the entire interval [0, π/2]. One can
be easily convinced about that, when using the methods of Part I (see Sect. 8.2).
Now, as required by the Leibniz rule given above, we must address the partial
derivative ∂f/∂ξ :
∂f (x, ξ )
=
g(x, ξ ) :=
∂ξ
1/(ξ 2 tan2 x + 1) for x ∈ [0, π/2[,
0
for x = π/2,
(3.2.5)
the case x = 0 having been covered by the upper formula. One needs to make sure
that such a function is continuous as the function of the variables x, ξ on the set
[0, π/2] × [A, B], with no restrictions on A and B except that 0 < A < B < ∞.
This task is not difficult. For x = π/2, it is enough to choose any sequence of points
(xn , ξn ) convergent to (x, ξ ) and write
1
1
|g(xn , ξn ) − g(x, ξ )| = 2 2
− 2 2
ξn tan xn + 1 ξ tan x + 1 2 2
ξ 2 tan2 x − ξn2 tan2 xn
≤ ξ tan x − ξ 2 tan2 xn −→ 0,
= 2 2
n
n→∞
(ξn tan xn + 1)(ξ 2 tan2 x + 1) (3.2.6)
78
3 Investigating Integrals with Parameters
thanks to the continuity of the power and tangent functions. At the point (π/2, ξ )
the function g is continuous too, since one has
1
1
≤
−
0
−→ 0,
2
2
2
ξn tan xn + 1
(ξ/2) tan2 xn + 1 n→∞
(3.2.7)
where the fact that (for respectively large n) one has ξn > ξ/2 (once ξn −→ ξ ))
n→∞
has been made use of.
The partial derivative has turned out to be continuous as the function of two
variables on the set [0, π/2] × [A, B] and, therefore, the formula (3.2.2) may be
used, bearing in mind that here t = ξ :
|g(xn , ξn ) − g(π/2, ξ )| = ∞
π/2
F (ξ ) =
0
=
1
dx
2
2
ξ tan x + 1
1
1 − ξ2
∞
0
=
(u2
u=tan x
0
1
ξ2
− 2 2
2
u +1 ξ u +1
1
du
+ 1)(ξ 2 u2 + 1)
=
∞
1
u
−
ξ
arctan(u
ξ
))
(arctan
2
1−ξ
0
π
=
.
2(1 + ξ )
(3.2.8)
The value of the derivative having been found, one now obtains by simple
integration:
F (ξ ) =
π
log(1 + ξ ) + C.
2
(3.2.9)
It remains only to determine the value of the constant C. The above formula easily
shows that C = F (0). So, if one could independently find the value F (0), our
problem would be solved. At this place, one might like to make use of (3.2.1),
writing
π/2
π/2
arctan(ξ tan x)
dx =
tan x
C = F (0) = lim F (ξ ) = lim
ξ →0
ξ →0
0
0 dx = 0,
(3.2.10)
0
however, it is not obvious whether the limit may be inserted under the symbol of the
integration, that is, whether the function F is continuous at zero. Its differentiability
(which also entails the continuity) has been verified only for ξ > 0. We know,
however, from the previous section that it would be sufficient to examine the twodimensional continuity of the function f (x, ξ ) on the set [0, π/2] × [A, B], where
this time we can accept 0 ≤ A < B < ∞.
Again the sequence of points (xn , ξn ) convergent to (x, ξ ) is chosen (all these
points lying in the set [0, π/2] × [A, B]) and for x = 0, π/2 the following estimate
can be done:
3.2 Differentiating with Respect to Parameters
79
arctan(ξn tan xn ) arctan(ξ tan x) |f (xn , ξn ) − f (x, ξ )| = −
tan xn
tan x
tan x arctan(ξn tan xn ) − tan xn arctan(ξ tan x) = tan xn tan x
tan x[arctan(ξn tan xn )−arctan(ξ tan x)]+[tan x −tan xn ]arctan(ξ tan x) = tan xn tan x
≤
|tan x| · |arctan(ξn tan xn ) − arctan(ξ tan x)|
|tan xn | · |tan x|
+
|tan x − tan xn | · |arctan(ξ tan x)|
−→ 0.
n→∞
|tan xn | · |tan x|
(3.2.11)
As it can be easily checked, both of the terms tend to zero due to the continuity of
the tangent and arcus tangent functions.
For x = 0 and x = π/2 to show the two-dimensional continuity is not a problem
either. In the first case, one has
arctan(ξn tan xn )
arctan(ξn tan xn )−ξ tan xn |f (xn , ξn )−f (0, ξ )| = −ξ = tan xn
tan xn
arctan(ξn tan xn ) − ξn tan xn + (ξn − ξ )tan xn = tan xn
arctan(ξn tan xn ) − ξn tan xn tan xn −→ 0.
≤
(3.2.12)
+ |ξn − ξ | · tan x n→∞
tan xn
n
The reader should not have any difficulties with the justification of this result. The
last term goes to zero in an obvious way. The same is true for the first one, which
can be easily established by using the Taylor expansion of the arcus tangent function
(this method of finding limits was discussed in Part I in Sect. 11.3) in accordance
with the formula:
arctan y = y −
y3
y5
+
+ ... .
3
5
(3.2.13)
One can see that the first term in the numerator in (3.2.12) cancels, and only the
second, i.e., (ξn tan xn )3 “survives.”
In turn, for x = π/2, our job is even easier (this time we, obviously, have
(xn , ξn ) −→ (π/2, ξ )):
n→∞
arctan(ξn tan xn )
π/2 −→ 0,
|f (xn , ξn ) − f (π/2, ξ )| = − 0 ≤ tan xn
tan xn n→∞
(3.2.14)
since −π/2 < arctan y < π/2, and the tangent in the denominator diverges to
infinity.
80
3 Investigating Integrals with Parameters
Thereby, the two-dimensional continuity of the function f (x, ξ ) on the set
[0, π/2] × [A, B] has been demonstrated, which implies the continuity of the
function F (ξ ) at zero. Pursuant to (3.2.10), the constant C = 0 and we find
π/2
arctan(ξ tan x)
π
dx =
log(1 + ξ ).
tan x
2
(3.2.15)
0
Problem 2
The integral:
∞
e−x
I=
2 −1/x 2
dx
(3.2.16)
0
will be calculated.
Solution
In this problem some new elements are encountered. First, one has to deal with the
improper integral, so the Leibniz rule as it stands in the previous examples is not
adequate. Second, (3.2.16) is not an integral with a parameter, but simply a number.
However, since this exercise appears in this chapter, the reader probably can guess
that this parameter shall appear in a moment. Indeed, rather than directly calculating
the integral (3.2.16), one can try to explicitly find the function:
∞
e−x
F (ξ ) =
2 −ξ/x 2
dx.
(3.2.17)
0
It might seem to be illogical. In general it should be easier to find a single number
than the entire function. Why to complicate our task, then? Well, the problem lies
in the fact that one does not know how to directly calculate the integral I in the
given form. Introducing the parameter ξ will enable us to perform some additional
operations like differentiation over ξ (if only the assumptions of the appropriate
theorem are met), which can help to transform the integrand expression.
Now let us restate the theorem formulated in the theoretical introduction, which
will be helpful. It specifies when one is allowed to differentiate the function
∞
F (t) =
f (x, t) dx
a
(3.2.18)
3.2 Differentiating with Respect to Parameters
81
by applying the formula
d
F (t) =
dt
∞
∂
f (x, t) dx,
∂t
(3.2.19)
a
i.e., by differentiating the integrand expression. Suppose that one is interested in the
dependence F (t) for t lying within a certain interval [A, B]. The conditions to be
met are the following:
1. The integrand function f (x, t) and its partial derivative ∂f (x, t)/∂t are continuous on the set [a, ∞[×[A, B].
2. The improper integral (3.2.18) is convergent for each t ∈ [A, B].
3. The improper integral (3.2.19) is uniformly convergent on [A, B].
Let us examine whether these conditions are satisfied by our function
#
f (x, ξ ) =
e−x
0
2 −ξ/x 2
for x > 0,
for x = 0,
(3.2.20)
whose definition, as in the previous problem, has been extended to the case of x = 0
as well. Let us note at the beginning that finally we are interested in the value ξ = 1.
Therefore, our discussion can be limited to the interval [A, B] = [1/2, 3/2]. The
choice of these specific numbers is of no particular significance. It allows, however,
to leave outside the interval the point ξ = 0, which could cause some unnecessary
problems.
Now we are going to examine the continuity of the function f (x, ξ ). We start
with considering the case x > 0. Let us choose a point (x, ξ ) and a certain arbitrary
sequence (xn , ξn ) convergent to it. Without loss of generality one can assume here
and in further considerations that ξn ∈ [1/4, 7/4] ⊃ [1/2, 3/2]. Let us perform the
following transformations:
2
2
2
2
|f (xn , ξn ) − f (x, ξ )| = e−xn −ξn /xn − e−x −ξ/x 2
2 2
2
2
2
2
= e−x −ξ/x ex −xn e(ξ −ξn )/xn eξ(1/x −1/xn ) − 1
2 2
2
2
2
(3.2.21)
≤ ex −xn e(ξ −ξn )/xn eξ(1/x −1/xn ) − 1 .
Note that if n → ∞, each of the exponential factors goes to 1. Since x = 0,
and consequently xn = 0 for sufficiently large n, no problem with vanishing
denominators arise. In the consequence the entire expression under the absolute
value converges to 0. This ensures the continuity of the function f (x, ξ ) on the set
]0, ∞[×[1/2, 3/2].
82
3 Investigating Integrals with Parameters
In the case when x = 0, one again has
2
2
|f (xn , ξn ) − f (0, ξ )| = e−xn −ξn /xn − 0
= e−xn e−ξn /xn −→ 0,
2
2
n→∞
(3.2.22)
because the former exponential factor tends to 1 and the latter to 0 (remember that
ξn > 1/4).
Now let us turn to the derivative with respect to the parameter:
∂f (x, ξ )
=
g(x, ξ ) :=
∂ξ
#
− x12 e−x
0
2 −ξ/x 2
for x > 0,
for x = 0.
(3.2.23)
The presence of the additional factor 1/x 2 does not make our estimate more
complicated. One has (again for x > 0)
1
1
2
2
2
2
|g(xn , ξn ) − g(x, ξ )| = 2 e−xn −ξn /xn − 2 e−x −ξ/x xn
x
2
1
2
2 x
2
2
2
2
2
= 2 e−x −ξ/x 2 ex −xn e(ξ −ξn )/xn eξ(1/x −1/xn ) − 1 −→ 0,
n→∞
x
xn
(3.2.24)
similarly as before, since x 2 /xn2 −→ 1.
n→∞
Now the case x = 0 should be addressed. One has
1 −x 2 −ξ /x 2
n
n
n
|g(xn , ξn ) − g(0, ξ )| = 2 e
− 0
xn
=
1 −xn2 −ξn /xn2
1
2
e e
≤ 2 e−(1/4)/xn ,
xn2
xn
(3.2.25)
since e−xn ≤ 1, and ξn ≥ 1/4. The obtained expression leads to the limit of
the type ∞ · 0. It is, however, a product of a power factor and an exponential
factor, and as we know from Part I, in such a case the latter determines the result.
Consequently the limit equals zero, as is needed. In conclusion, one can say that
both the function f (x, ξ ) and its derivative with respect to ξ are continuous on the
set [0, ∞[×[1/2, 3/2]. The first step of our proof is then completed.
The second step requires examining the convergence of the improper integral (3.2.17) for an arbitrary value of ξ (but belonging to the given interval).
Since the integrand function is continuous in the variable x, so the eventual doubts
may concern only the upper limit of the integration. They can be immediately
dispelled thanks to the experience gained from the previous volume of the book (see
2
Chap. 2). For, the function f (x, ξ ) is the product of a bounded factor (e−ξ/x ) and a
2
3.2 Differentiating with Respect to Parameters
83
factor decreasing exponentially—and thus sufficiently quickly—to zero (e−x ). The
convergence of the integral in this case is evident.
In the last point, one has to investigate the uniform convergence of the integral:
2
∞
1 −x 2 −ξ/x 2
e
dx.
x2
(3.2.26)
0
As we know from the previous section, for this purpose one has to find a majorant
(see formula (3.1.17)) of the function:
g(x, ξ ) =
1 −x 2 −ξ/x 2
e
.
x2
(3.2.27)
Since it is decreasing in the variable ξ and 1/2 ≤ ξ ≤ 3/2, one has:
g(x, ξ ) ≤
1 −x 2 −(1/2)/x 2
e
=: ϕ(x).
x2
(3.2.28)
The function on the right-hand side does not depend on ξ any longer and is
integrable on the interval [0, ∞[. This can be clearly seen if this interval is broken
down for example onto [0, 1] and ]1, ∞[. On the former of these intervals, the
integral exists, since the continuous function is being integrated on the compact
2
set (the majorant ϕ(x) at zero is given the value 0), and on the latter |ϕ(x)| ≤ e−x
occurs, so the integrability is obvious. Thereby, the integral (3.2.26) is uniformly
convergent.
In this way all the necessary conditions have been verified, so one can proceed
with the implementation of our plan of calculating dF (ξ )/dξ rather than I . At the
same time, it becomes clear why the parameter ξ was introduced into the integral.
We have
∞
F (ξ ) = −
1 −x 2 −ξ/x 2
e
dx
x2
=
√
u= ξ /x
0
1
−√
ξ
∞
e−u
2 −ξ/u2
1
du = − √ F (ξ ).
ξ
0
(3.2.29)
Clearly we have succeeded in deriving a simple differential equation for the function
F (ξ ):
1
F (ξ ) = − √ F (ξ ),
ξ
(3.2.30)
which can be easily solved by separating the variables:
dξ
dF
= −√
F
ξ
⇒ F (ξ ) = Ce−2
√
ξ
.
(3.2.31)
84
3 Investigating Integrals with Parameters
The final step is to determine the integration constant C. One might be tempted
to simply put ξ = 0 on both sides of (3.2.17), but it is not known, however, if the
function F (ξ ) is continuous at zero (in our previous considerations it was assumed
for convenience that ξ ∈ [1/2, 3/2]). If one now wished to demonstrate this continuity by using the theorem formulated previously (see formulas (3.0.3), (3.1.15),
etc.) and by extending the interval for example to [0, 2], one would encounter a
serious difficulty: it is unclear how to demonstrate the continuity of the function
f (x, ξ ) at the point (0, 0). In the case when ξn −→ 0, our claim expressed in
n→∞
the formula (3.2.22) breaks down. What is more, this problem comes not from
the inadequacy of our proof, but simply from the fact that the function f (x, ξ ) is
actually not continuous at the origin. In order to somehow deal with this, let us split
up the integral in the following way:
∞
F (ξ ) =
e
−x 2 −ξ/x 2
dx +
0
e−x
2 −ξ/x 2
(3.2.32)
dx ,
F2 (ξ )
F1 (ξ )
where in a moment is going to converge to zero. The continuity of the function
F2 for ξ → 0 does not raise any suspicion. The integrand expression is continuous
on the set [, ∞[×[0, 2], and the integrable majorant can be easily indicated in the
2
form e−x . One can, therefore, write
C = lim F (ξ ) = lim (F1 (ξ ) + F2 (ξ )) = lim F1 (ξ ) + F2 (0),
ξ →0
ξ →0
ξ →0
(3.2.33)
which implies that
|C − F2 (0)| = lim
e
ξ →0
−x 2 −ξ/x 2
dx ≤ lim
1 dx = .
ξ →0
0
(3.2.34)
0
Letting go to zero, one gets the well-known Gaussian integral:
∞
C = lim F2 (0) = lim
→0
∞
e
→0
−x 2
e−x dx =
2
dx =
π
,
2
(3.2.35)
0
and the final result is
F (ξ ) =
π −2√ξ
e
2
⇒ I = F (1) =
π
.
2e2
(3.2.36)
3.2 Differentiating with Respect to Parameters
85
Problem 3
The integral:
∞
I (a, b) =
e−ax − cos bx
dx,
x2
2
(3.2.37)
0
will be calculated for a ∈ [a1 , a2 ], where 0 < a1 < a2 < ∞ and b ∈ [b1 , b2 ],
where 0 < b1 < b2 < ∞.
Solution
This time the integral depends on two parameters a and b, but one need not worry
about this. The behavior of the function can be examined, fixing one parameter, if
necessary. In the present problem, however, one can easily get rid of the second
parameter, by the appropriate rescaling of the integration variable. To this goal, let
us define the new variable y = bx, getting the result:
∞
I (a, b) = b
e−(a/b
2 )y 2
− cos y
y2
dy.
(3.2.38)
0
Introducing temporarily the symbol c = a/b2 , one obtains
∞
I (a, b) = bF (c),
where
F (c) =
e−cy − cos y
dy,
y2
2
(3.2.39)
0
which means that the issue has been transformed to a problem with one parameter c.
Now the procedure of the previous exercise can be applied. It can be seen that the
possibility of the differentiation of the integrand function with respect to c would
be very promising, since then the denominator y 2 would disappear, and the integral
would become easy to calculate. Let us denote
⎧ −cy 2
− cos y
⎨e
for y > 0,
f (y, c) =
y2
⎩
1/2 − c
for y = 0,
(3.2.40)
this lower value having been established with the use of the limit
e−cy − cos y
1
= − c.
2
y2
2
lim
y→0+
(3.2.41)
86
3 Investigating Integrals with Parameters
Let us now examine the continuity of (3.2.40) on the set [c1 , c2 ] × [0, ∞[, where
0 < c1 < c2 < ∞. First, suppose that y = 0 and let (yn , cn ) be a sequence of points
convergent to (y, c). At least for large values of n one has yn = 0 and it is clear that
2
e−cn yn2 − cos y
e−cy − cos y n
|f (yn , cn ) − f (y, c)| = −
(3.2.42)
−→ 0.
n→∞
yn2
y2
In turn for y = 0 one can make the following estimate:
e−cn yn2 − cos y
1
n
|f (yn , cn ) − f (0, c)| = +
c
−
2
yn2
e−cn yn2 − 1 + cy 2
cos yn − 1 − yn2 /2 n
=
−
yn2
yn2
e−cn yn2 − 1 + cy 2 cos y − 1 − y 2 /2 n
n
n
−→ 0,
≤
+ n→∞
yn2
yn2
(3.2.43)
since each of the numerators for yn → 0 behaves as yn4 (this results from the simple
Taylor expansion).
The next step requires the verification whether for arbitrary c ∈ [c1 , c2 ] the
improper integral (3.2.39) exists. As the integrand function is continuous (the point
y = 0 included) which has been shown above, only the upper limit of the
integration needs to be considered. Because of the presence of the denominator y 2 ,
the convergence of the integral is ensured (the function in the numerator is bounded).
As we remember from Part II, the power y 1+ would be sufficient.
In the last step, the uniform convergence of the following integral should be
checked:
∞
g(y, c) dy,
where
g(y, c) :=
∂f (y, c)
2
= −e−cy .
∂c
(3.2.44)
0
Due to the fact that c ∈ [c1 , c2 ] with c1 > 0, the natural majorant is
ϕ(y) = e−c1 y ,
2
(3.2.45)
clearly integrable over the variable y on the interval [0, ∞[.
All assumptions having been met, and one can now differentiate the integrand
function with respect to c, obtaining the Gaussian integral:
∞
F (c) =
0
∂f (y, c)
dy = −
∂c
∞
e−cy dy = −
2
0
1
2
π
.
c
(3.2.46)
3.2 Differentiating with Respect to Parameters
87
Now it is straightforward to find F (c):
F (c) = −
1
2
√
⇒ F (c) = − π c + D.
π
c
(3.2.47)
The constant D still remains to be determined. To this end one would like to
put c = 0, but so far the continuity of the function F (c) at zero has not been
demonstrated (formula (3.2.47) is by now valid only for c > 0). Therefore, one
is not allowed yet to write that
∞
F (0) =
$ √
%
1 − cos y
−
dy
=
lim
F
(c)
=
lim
π
c
+
D
= D.
↑ c→0
c→0
y2
(3.2.48)
0
The equality marked with an arrow still requires justification. We will try to do it in
the same way as in the previous exercise. First let us split the integral (3.2.39) into
the sum of two terms:
∞
e−cy − cos y
e−cy − cos y
F (c) =
dy
+
dy ,
y2
y2
0
2
2
(3.2.49)
F2 (c)
F1 (c)
and then both of them will be examined separately.
The function F2 (c) is continuous at zero, since first the function f (y, c) is
continuous on the set [, ∞[×[0, c2 ] (the result (3.2.42) still applies), and second
the integral
∞
e−cy − cos y
dy
y2
2
(3.2.50)
is uniformly convergent with respect to c because there exists an integrable
majorant, for example:
e−cy 2 − cos y 1 + 1
2
= 2.
≤
2
2
y
y
y
(3.2.51)
One can then write
D = lim F (c) = lim (F1 (c) + F2 (c)) = lim F1 (c) + F2 (0),
c→0
c→0
c→0
(3.2.52)
88
3 Investigating Integrals with Parameters
which entails
e−cy 2 − cos y e−cy 2 − 1 + 1 − cos y = lim |D − F2 (0)| = lim dy
dy
c→0 c→0 y2
y2
0
0
⎤
⎡ e−cy 2 − 1 1 − cos y dy ⎦ .
≤ lim ⎣ dy
+
(3.2.53)
c→0
y2
y2
0
0
If one uses now the two known inequalities (we are interested in the case x > 0):
1 − e−x ≤ x
1 − cos x ≤ x 2 ,
and
(3.2.54)
easily demonstrated by methods of Sect. 10.1 of the first part of this book series, one
can write:
⎤
⎡ 2
cy y2 |D − F2 (0)| ≤ lim ⎣ 2 dy + 2 dy ⎦ = lim (c + 1) = .
c→0
c→0
y
y
0
0
(3.2.55)
Letting go to zero, one obtains the result:
∞
D = lim F2 (0) = lim
→0
→0
⎡
= lim ⎣
→0
∞
1 − cos y
dy = lim
→0
y2
cos y − 1 ∞
+
y
∞
⎤
−1
(1 − cos y)
y
sin y ⎦ π
dy = ,
y
2
(3.2.56)
the value of the well-known integral
∞
π
sin x
dx = ,
x
2
(3.2.57)
0
having been used here. The reader has certainly encountered it during lectures of
analysis, and its value will be calculated in Sect. 7.4.
Going back to the formulas (3.2.39) and (3.2.47), it is finally found
√
a
π
πb
a
I (a, b) = bF 2 = b − π 2 +
= − πa +
.
2
2
b
b
(3.2.58)
3.2 Differentiating with Respect to Parameters
89
Problem 4
The integral:
∞
I=
arctan(αx) − arctan(βx)
dx,
x
(3.2.59)
0
will be calculated for positive values of the parameters α and β.
Solution
Integrals of the type (3.2.59) are called the Froullani’s integrals. It is a common term
for such expressions as
∞
K(α, β) =
f (αx) − f (βx)
dx.
x
(3.2.60)
0
Upon appropriate assumptions regarding the function f and the constants α, β, one
is able to show that
α
K(α, β) = f (0+ ) − f (+∞) log ,
β
(3.2.61)
the expression in square brackets being understood as the appropriate limits.
Let us consider the specific example given in the content of this problem. All of
the necessary theorems have already been formulated in the previous exercises, so
one can immediately start to study the integral. We assume that
α ∈ [αmin , αmax ],
β ∈ [βmin , βmax ],
(3.2.62)
where αmin , βmin > 0.
We would like to differentiate the integral (3.2.59) with respect to one of the
parameters (e.g., α), so we need to go through the procedure described on page 81.
This differentiation will make it easier to find the integral I , since it will transform
the “unpleasant” integrand function into a rational function, for which, as the
reader already knows, the indefinite integral can always be calculated, in so far as
one knows how to decompose the polynomial in the denominator into elementary
factors. Here the situation is even simpler:
90
3 Investigating Integrals with Parameters
dI
=
dα
∞
∂ arctan(αx) − arctan(βx)
dx =
∂α
x
0
∞
1
π
= arctan(αx) =
.
α
2α
0
∞
1
α2x 2
0
+1
dx
(3.2.63)
At this point one cannot yet be certain, however, if the above formula is true. For
that purpose all the required conditions must be verified.
1. We check the continuity of the integrand function and of its derivative over α on
the set [0, ∞[×[αmin , αmax ].
Denoting
⎧
⎨ arctan(αx) − arctan(βx)
for x > 0,
f (x, α) :=
x
⎩α − β
for x = 0
(3.2.64)
and selecting a sequence (xn , αn ) −→ (x, α), we will estimate the expression:
n→∞
|f (xn , αn ) − f (x, α)| .
(3.2.65)
We are not going to differentiate or take any limits over the parameter β, so
its value can be treated as fixed and, therefore, it is not explicitly written as an
argument of the function f . One can, of course, work vice versa: to fix α and
perform all operations on the parameter β.
First, let us assume that x > 0. Then, at least for large n, one must have xn > 0
as well. Now the expression (3.2.65) can be given the form:
f (xn , αn ) − f (x, α) = 1 x(arctan(αn xn ) − arctan(αx))
xxn
+ (x − xn )arctan(αx) − x(arctan(βxn ) − arctan(βx))
(3.2.66)
+ (xn − x)arctan(βx).
The transformations that lead to this form are very similar to those of the previous
problems, so their details have been omitted. Roughly speaking, they consist of
adding and subtracting identical terms, so as to obtain (3.2.66).
When n → ∞, all the terms under the sign of the modulus vanish, regardless
of the specific form of (αn , xn ), which entails the two-dimensional continuity of
the function f . The particular attention requires only the point x = 0, dealt with
below. Let us estimate:
f (xn , αn ) − f (0, α) = arctan(αn xn ) − arctan(βxn ) − α + β xn
3.2 Differentiating with Respect to Parameters
≤
arctan(αn xn ) − αxn xn
91
+
arctan(βxn ) − βxn xn
−→ 0.
n→∞
(3.2.67)
One can easily be convinced that the above result is correct, for example by
using the Taylor expansion of the function arctan (as a result, both numerators
will behave as xn3 ). Thus the continuity of the function f (x, α) on the set
[0, ∞[×[αmin , αmax ] has been established.
The continuity of the partial derivative
1
∂f
= 2 2
∂α
α x +1
(3.2.68)
is obvious, so the reader will not encounter any problems when checking it.
2. We check the convergence of the indefinite integral (3.2.59) for each α ∈
[αmin , αmax ].
In order to investigate the convergence of the integral, a certain known
trigonometric formula will be applied, made use of in the first two parts of the
book series (correct for u · v > −1):
arctan u − arctan v = arctan
u−v
.
1 + uv
(3.2.69)
The modulus of the numerator in (3.2.59) can now be given the form:
(α − β)x |α − β|x
|arctan(αx) − arctan(βx)| = arctan
= arctan
.
2
1 + αβx
1 + αβx 2
(3.2.70)
Since α ≥ αmin > 0 and β ≥ βmin > 0, it is clear that, for large values of x, the
behavior of the argument of the arcus tangent (and with it the function itself) is
not worse than const/x, i.e.,
|α − β|x
|α − β| 1
· ,
≤
2
αβ
x
1 + αβx
(3.2.71)
which ensures the convergence of the integral (3.2.59) in the upper limit. As to
the lower limit, i.e., zero, no problem emerges, since it was previously shown
that the function f (x, α) is continuous at this point.
3. We check the uniform convergence of the integral (3.2.63) on the interval
[αmin , αmax ].
Demonstrating the uniform convergence of the integral (3.2.63) is painless, as
it has the obvious integrable majorant (thanks to the fact that αmin = 0):
92
3 Investigating Integrals with Parameters
∂ arctan(αx) − arctan(βx) 1
1
=
α 2 x 2 + 1 ≤ α 2 x 2 + 1 =: ϕ(x).
∂α
x
min
(3.2.72)
All three conditions having been verified, and the formula (3.2.63) becomes true.
Integrating over α the equation
dI
π
=
,
dα
2α
(3.2.73)
one comes to
I (α) =
π
log α + C,
2
(3.2.74)
where the constant C, independent of α, may of course depend on the second
parameter, i.e., β. In order to find its value, let us set α = β. Because the
differentiability of the function I (α) has already been shown, it must also be
continuous and one is allowed to make the transition α → β. In this way one gets
π
log β + C = lim F (α) = F (β) = 0,
α→β
2
(3.2.75)
which implies that
C=−
π
log β,
2
(3.2.76)
and the final result takes the form:
I = π/2 log(α/β).
(3.2.77)
At the end, it should be noted that this expression is antisymmetric when
replacing α ↔ β, as it was expected from (3.2.59). This type of checking
whether the given result retains the symmetries of the initial expression, it is always
desirable, as it allows us to avoid some calculational mistakes.
Problem 5
The integral
∞
I=
e−2x − e−x
cos x dx
x
0
will be calculated, by introducing an appropriate parameter.
(3.2.78)
3.2 Differentiating with Respect to Parameters
93
Solution
When looking at the integral to be calculated, it is seen that the main factor that
bothers us is x in the denominator. The integrals of the functions such as eax cos(bx)
can be easily calculated by parts, whereas no primitive function for the integrands
like eax /x or cos(bx)/x can explicitly be found since they are expressed through
special functions.
How x can be removed from the denominator, then? The answer should be
imposed upon the reader: to temporarily introduce into one of the exponents
an additional parameter (this idea appeared already in Problem 2) and next to
differentiate with respect to it, in which case x will disappear from the denominator.
Therefore, instead of (3.2.78) let us find:
∞
F (α) =
e−2x − e−αx
cos x dx,
x
(3.2.79)
0
where the parameter α will be assumed to belong to the interval [1/2, 5/2]. The
only requirement for this interval is that it must contain the number 1, because this
value of the parameter will be of interest to us (since one has I = F (1)). At the end,
it will become clear that it is good to have the number 2 inside the interval as well.
Let us define the function
⎧
⎨ e−2x − e−αx
cos x for x > 0,
(3.2.80)
f (x, α) =
x
⎩
α−2
for x = 0
and verify all the conditions set out on page 81. The first two create no special
difficulties.
The examining of the continuity of the function f and its partial derivative
∂f/∂α on the set [0, ∞[×[1/2, 5/2] is similar as in the previous examples, so the
appropriate calculations are left to the reader. In turn, the integral
∞
f (x, α) dx
(3.2.81)
0
is clearly convergent in the upper limit, due to the presence of the exponential
factors, and in the lower limit because of the continuity of the integrand function (3.2.80). Therefore, for a moment we will stop only at the third point, which
requires to investigate the uniform convergence of the integral
94
3 Investigating Integrals with Parameters
∞
∂f (x, α)
dx.
∂α
(3.2.82)
0
As we remember, it is sufficient to indicate an integrable majorant for ∂f/∂α. It
does not pose any particular difficulty either. Thanks to the condition α ≥ 1/2 and
having in mind that the function e−αx is decreasing in α (with x > 0 fixed), one can
write
∂f (x, α) −αx
| cos x| ≤ e−αx ≤ e−x/2 ,
(3.2.83)
∂α = e
and this last function constitutes the majorant that is looked for.
These results justify the formula:
∞
F (α) =
∂f (x, α)
dx =
∂α
0
∞
e−αx cos x dx =
0
α
,
1 + α2
(3.2.84)
where the latter result can be easily obtained by (twice) integrating by parts (see
Sect. 14.1 in Part I).
Now one can integrate both sides of the Eq. (3.2.84), obtaining the formula for
the function F :
F (α) =
1
log(1 + α 2 ) + C.
2
(3.2.85)
In order to determine the value of the constant C, we put α = 2. The integrand
function in (3.2.79) then vanishes, so F (2) = 0. Because the function F is
continuous at this point (it is, after all, differentiable there), one can write
0 = F (2) = lim F (α) = lim
α→2
α→2
1
log 5
log(1 + α 2 ) + C =
+ C,
2
2
(3.2.86)
and
C=−
log 5
.
2
Thereby, the constant C is found. The last step requires letting α go to 1 in order
to obtain the value of I . In the same way as above, one gets
I = F (1) =
log 2 log 5
log(5/2)
−
=−
.
2
2
2
(3.2.87)
3.2 Differentiating with Respect to Parameters
95
Problem 6
It will be proved that the functions:
∞
t
√
F1 (t) = π
e
−x 2
dx
and
F2 (t) =
0
sin(tx) −x 2 /4
dx,
e
x
(3.2.88)
0
for t ∈ R, are identically equal.
Solution
This last problem in this section slightly differs from the former ones. First, our
goal is not to calculate both integrals, but only to check whether they are equal (in
fact, they cannot be calculated in an explicit way, i.e., by expressing them through
elementary functions). Second, in one of the functions, t does not appear as an
argument but at the upper limit of the integration.
The idea of the solution of this problem is the following: if the functions f1 (t)
and f2 (t) were to be identically equal (i.e., for any t ∈ R), by subtracting them from
each other one would get a constant (strictly speaking zero), which would entail the
condition
d
[F1 (t) − F2 (t)] = F1 (t) − F2 (t) = 0.
dt
(3.2.89)
And vice versa: if the above equation is valid, the functions may differ from each
other at most by a constant. Our aim is thus, primarily, the calculation of both
derivatives.
To calculate F1 (t) is not a problem: it is just the integrand function, where t is
put in place of the integration variable:
F1 (t) =
√
πe−t .
2
(3.2.90)
This is because we know that the indefinite integral is simply the primitive function
for the integrand expression.
When dealing with F2 (t), one needs to be more careful. In order to achieve our
goal it is necessary to differentiate under the integral, which requires to ensure that
it is allowed. Again the conditions formulated on page 81 should be verified. As a
range of variability of t, we accept the interval [A, B], where A < B, but no other
restrictions are imposed on A and B.
96
3 Investigating Integrals with Parameters
Let us define the function:
⎧
⎨ sin(tx) −x 2 /4
e
for x > 0,
f (x, t) =
x
⎩t
for x = 0.
(3.2.91)
Then one has
∞
F2 (t) =
(3.2.92)
f (x, t)dx.
0
Both the function f (x, t) and its derivative ∂f/∂t are continuous on the set
[0, ∞[×[A, B]. The verification of this fact proceeds in the same way as in the
previous problems so it is omitted here and left to the reader. The improper
integral (3.2.92) is convergent for each t ∈ [A, B], due to the presence of the
exponential factor in the function f (x, t). It remains, therefore, to find an integrable
majorant for ∂f/∂t. This task is not a problem either, for one has
∂f (x, t) 2
−x 2 /4
≤ e−x /4 =: ϕ(x).
(3.2.93)
∂t = |cos(tx)| e
All the necessary assumptions are met, so one can write
∞
F2 (t) =
∂f (x, t)
dx =
∂t
0
∞
cos(tx)e−x
2 /4
(3.2.94)
dx.
0
Now, in order to calculate the integral (3.2.94), we are going to solve an auxiliary
problem. In the same way as above, one can be convinced that (3.2.94) can be
differentiated under the integral as well:
d F (t) =
dt 2
∞
∂
2
cos(tx)e−x /4 dx = −
∂t
0
∞
sin(tx)xe−x
2 /4
dx.
(3.2.95)
0
The additional factor x is easily removed upon writing
xe−x
2 /4
= −2d(e−x
2 /4
)/dx,
thanks to which the integration by parts can be applied:
d F (t) = 2
dt 2
∞
0
d −x 2 /4
e
sin(tx)
dx = −2t
dx
∞
cos(tx)xe−x
2 /4
dx = −2tF2 (t).
0
(3.2.96)
3.3 Integrating over Parameters
97
A very simple differential equation for the function F2 (t) (it does not matter whether
the function is called y(x), f (z), F2 (t), or whatever) has been obtained, with the
obvious solution:
F2 (t) = C̃e−t .
2
(3.2.97)
In order to determine the constant C̃, one can plug t = 0 into the Eq. (3.2.94).
The integral in (3.2.94) then becomes purely Gaussian and one gets
∞
F2 (0)
e−x
= C̃ =
2 /4
dx =
√
1√
4π = π ,
2
(3.2.98)
0
1/2 originating from the integration over the half of the interval ] − ∞, ∞[. Thus,
we have obtained the derivative
F2 (t) =
√ −t 2
πe ,
(3.2.99)
i.e., the expression identical to (3.2.90). As a consequence, one has
F1 (t) − F2 (t) =
d
[F1 (t) − F2 (t)] = 0
dt
⇒ F1 (t) = F2 (t) + C.
(3.2.100)
The final step is to show that C = 0, i.e., that F1 (t) = F2 (t). In order to achieve it,
let us simply find the value of both functions at one selected point t. The easiest is
to do it for t = 0. It is clear that both F1 (0) = 0 and F2 (0) = 0. Hence the constant
C vanishes and the equality between the two functions F1 and F2 is established.
3.3 Integrating over Parameters
Problem 1
The integral:
1
I (α, β) =
0
will be calculated, where β > α > 0.
xα − xβ
dx,
log x
(3.3.1)
98
3 Investigating Integrals with Parameters
Solution
In the previous sections we were considering the issue, under what conditions one
is allowed to switch the order of performing a limit or differentiation with respect
to the parameter t with the integration over the variable x. Now we move on to the
integration over t and the question arises if and when one can use the formula:
b
B
dt f (x, t) =
dx
a
B
A
b
dt
dx f (x, t).
(3.3.2)
a
A
The answer should already be known after carefully studying Part II of this book
series, where we were dealing with concrete examples of the two-dimensional
integrability (see Sect. 12.1). There, we learned that if a given function is integrable
on the set [a, b] × [A, B], then both iterated integrals exist and are equal to each
other, which means that the above formula is true. In particular, this happens when
the integrand function is continuous on the set [a, b] × [A, B].
Our main concerns below are the applications of this formula to the calculation of
certain one-dimensional integrals such as (3.3.1). In order to convert it into a double
integral, let us use the known formula:
x t dt =
xt
log x
(3.3.3)
plus an irrelevant constant. Now one can easily observe that
β
x t dt =
α
x t β
xβ − xα
,
=
log x α
log x
(3.3.4)
thanks to which the expression (3.3.1) can be given the form:
β
1
I (α, β) = −
dt x t .
dx
0
(3.3.5)
α
Then one can try to use the theorem recalled above and at the beginning of the
chapter. Unfortunately the integrand expression
f (x, t) = x t
(3.3.6)
is not continuous on the set concerned, i.e., [0, 1] × [α, β], even if one supplement
its definition at the origin in some way. The problem is that going to the point (0, 0)
along both axes of the coordinate system, one gets two different limits: 1, along the
3.3 Integrating over Parameters
99
x-axis, and 0, along the t-axis. Thus the effort spared on proving the (nonexistent)
continuity of the function f would be wasted. The origin is the only point of the
set [0, 1] × [α, β] that causes difficulties. In all other points, the continuity of the
function f is clear and we are not going to deal with them.
In Problem 3 in Sect. 12.1 of the preceding part of this book series, we were
able to cope with this kind of troublesome point by separating it from the rest of
the set with a curve γ , which later was contracted to zero (see Fig. 12.1). It was
essential for the integral on this small area to be convergent to zero together with
the contraction of γ to a point. The same is true in the current problem as well
because the expression x t is bounded on our set. This consequently means that the
function is integrable on the rectangle [0, 1]×[α, β] and both iterated integrals exist
and are equal.
In (3.3.5), both integrations may now be “legally” swapped, which yields
β
I (α, β) = −
dx x = −
t
dt
α
α
α
0
β
=−
β
1
1
x t+1 dt
t + 1 0
β
1
α+1
dt = − log(t + 1) = log
.
t +1
β +1
α
(3.3.7)
The fraction under the logarithm is of course smaller than 1 (as α < β),
and therefore the integral I (α, β) < 0. This was expected because the integrand
expression in (3.3.1) is negative due to the fact that log x < 0 in the interval ]0, 1[.
As one can see, the integrals cumbersome to calculate in one order of integrations
have proved to be very simple in the other. This constitutes the clue of the method
presented in this section. It is also closely related to the method of the differentiation
over the parameter, as described in the previous section. The reader is encouraged
to find the result (3.3.7) by calculating ∂I /∂α.
Problem 2
Assuming that a, b, c > 0, the integral:
∞
I (a, b, c) =
0
will be calculated.
e−ax − e−bx
sin(cx) dx
x
(3.3.8)
100
3 Investigating Integrals with Parameters
Solution
This problem is a variation of the fifth exercise of the previous section, in which
we were concerned with the differentiation with respect to the parameter under the
integral. This time we would like to change the order of integrations. To achieve this
goal let us write
b
e−ax − e−bx
=
x
e−xt dt,
(3.3.9)
a
thanks to which the cumbersome denominator is removed. After inserting the righthand side into (3.3.8) one obtains
∞
I (a, b, c) =
b
dx sin(cx)
dt e
a
0
∞
b
−xt
=
dx sin(cx)e−xt ,
dt
a
(3.3.10)
0
but the “legitimacy” of the last equality still has to be established. The reader has
certainly noticed that the new element in relation to the previous problem constitutes
the improper character of the x integration. In this case, the appropriate theorem,
recalled in the theoretical introduction, requires that apart from the continuity of the
integrand function
f (x, t) = sin(cx)e−xt
(3.3.11)
on the set [0, ∞[×[a, b], the uniform convergence of the integral
∞
(3.3.12)
f (x, t) dx
0
must be proved. The continuity of the function f is obvious and it can be demonstrated in the same way as in the previous problems. In turn the uniform convergence
of (3.3.12) stems from the existence of the integrable majorant (remember that
t ≥ a):
|f (x, t)| = sin(cx)e−xt ≤ e−ax =: ϕ(x).
(3.3.13)
Thereby, the formula (3.3.10) has become “legal” and one is left with executing
the integrals in the given order. That with respect to x is already well known to the
reader: it can be calculated by integrating by parts twice, which gives
∞
sin(cx)e−xt dx =
0
c
.
t 2 + c2
(3.3.14)
3.4 Exercises for Independent Work
101
Finally, calculating the integral over t, one obtains the result:
b
I (a, b, c) =
a
t b
a
c
1
b
dt = c arctan = arctan − arctan .
2
2
c
c a
c
c
t +c
(3.3.15)
3.4 Exercises for Independent Work
Exercise 1 Demonstrate the continuity of the function
2
x 2 + ξ 2 dx
f (ξ ) =
−2
on the interval ] − 1, 1[ and of the function
2
g(ξ ) =
−1
x2
dx
+ ξ 2x4
1 + x2
on R.
Exercise 2 Differentiating with respect to the parameter, calculate the integrals:
π/2
log(sin2 x + ξ 2 cos2 x) dx, where ξ > 0,
(a)
0
π
log(1 − 2ξ cos x + ξ 2 ) dx, where ξ > 1.
(b)
0
Answers
(a) π log[(1 + ξ )/2].
(b) 2π log ξ .
Exercise 3 Integrating over the parameter, calculate the integrals:
1
(x α − x β )
(a)
0
sin(log x)
dz for α, β > 0,
log x
102
3 Investigating Integrals with Parameters
∞
e−a/x − e−b/x
2
(b)
2
dx for a, b > 0.
0
Answers
(a) arctan[(β
√ √ −√α)/(αβ + α + β + 2)].
(b) π( b − a).
Chapter 4
Examining Unoriented Curvilinear
Integrals
In this chapter, we are concerned with the calculations of unoriented integrals in
R3 such as areas, center-of-mass locations, or moments of inertia of the sheets of
surfaces (curved, in general). The following formulas will be of use.
Let the surface be defined with the equation z = f (x, y), where (x, y) ∈ D ⊂
R2 are Cartesian coordinates and f is a certain differentiable function. Then the
surface area of the curved sheet corresponding to the domain D may be calculated
from the formula:
2 2
∂f
∂f
S=
+
+ 1 dx dy.
(4.0.1)
∂x
∂y
D
This expression is more carefully analyzed when solving the first problem. If the
surface is defined in the parametric way via the relation r(u, v), where r = [x, y, z],
the appropriate formulas for our choice have the form:
tu × tv dudv =
S=
D
=
tu2 tv2 − (tu tv )2 dudv
D
∂ r
∂u
2 ∂ r
∂v
2
−
∂ r ∂ r
·
∂u ∂v
2
dudv,
(4.0.2)
D
where tu = ∂ r/∂u, tv = ∂ r/∂v are vectors tangent to the surface and associated
with the parameters u, v ∈ D.
© Springer Nature Switzerland AG 2020
T. Radożycki, Solving Problems in Mathematical Analysis, Part III, Problem Books
in Mathematics, https://doi.org/10.1007/978-3-030-38596-5_4
103
104
4 Examining Unoriented Curvilinear Integrals
Let us introduce the metric tensor ĝ:
⎤
⎡ 2
∂ r
∂ r ∂ r
·
⎢ ∂u
∂u ∂v ⎥
⎥
ĝ = ⎢
⎣ ∂ r ∂ r ∂ r 2 ⎦ ,
·
∂u ∂v ∂v
(4.0.3)
which defines the distance between adjacent points (calculated along the surface),
for which the parameters differ by du and dv, according to
du
ds 2 = [du, dv] · ĝ ·
.
(4.0.4)
dv
Then the formula (4.0.2) may be given the form:
det ĝ dudv.
S=
(4.0.5)
D
A certain function ρ(x, y) (for instance, the mass or charge density) can be
integrated over the surface following the formula
M=
ρ(x, y)
∂f
∂x
2
+
∂f
∂y
2
+ 1 dx dy
(4.0.6)
D
or in the parametric formulation
ρ(u, v) tu × tv dudv,
M=
(4.0.7)
D
where the symbol M refers here to the mass of the sheet, but may denote any other
quantity expressible in the form of a surface integral of a scalar function. Either
variant of (4.0.2) may be used as well.
4.1 Finding Area of Surfaces
Problem 1
The surface area of the sheet:
S = {(x, y, x) ∈ R3 | x 2 = zy},
cut out with the planes: z = 0, y = 0, and 2y + z = 2, will be found.
(4.1.1)
4.1 Finding Area of Surfaces
105
Fig. 4.1 The projection of a
curved surface onto the plane
Solution
If one wishes to calculate the area of a curved sheet constituting the graph of a
(smooth) function f (x, y), where (x, y) ∈ D, the formula (4.0.1) may be used.
It can be easily justified if one looks at Fig. 4.1. Let us start with mentally dividing
the curved surface into “patches” so small that each of them can be considered as
flat. In such a case, the area will be simply the sum of individual patches, and the
formula will be exact when they become infinitesimally small. This naturally means
that in place of the sum an integral appears.
One of such patches is marked in the figure in gray. The vector normal to it is
inclined to the z-axis by the angle θ (in general, different for various patches). The
area of the i-th patch projected onto the plane xy (be it σi proj ) is expressed by the
area of the patch itself (i.e., σi ) through the formula:
σi proj = σi cos θi .
(4.1.2)
Then, the area to be found can be given the form:
S=
!
i
σi =
! σi proj
i
cos θi
.
(4.1.3)
Note that with our assumptions cos θi never equals zero. Otherwise this would
mean that the normal to the patch is horizontal, and thus one of the tangents
would parallel to the z-axis. The function f with these properties would be non-
106
4 Examining Unoriented Curvilinear Integrals
differentiable, which was excluded at the beginning. Such cases will be covered
in the following exercises, and now we notice that presently the problem of the
vanishing denominator of (4.1.3) does not emerge.
By choosing the patches corresponding to the infinitesimal rectangles on the xyplane (i.e., with dimensions dx × dy), in place of (4.1.3) one gets the integral:
1
dx dy.
cos θ
S=
(4.1.4)
D
In order to determine cos θ , we are going to find the normal vector shown in
is orthogonal
Fig. 4.1. As one knows from the previous chapters, the vector ∇g
to the surface of g(x, y, z) = const. It is then sufficient to rewrite the equation
z = f (x, y) in the form
z − f (x, y) = 0,
(4.1.5)
∂f
∂f
n = ∇g(x, y, z) = − , − , 1 .
∂x
∂y
(4.1.6)
=:g(x,y,z)
to easily obtain
The cosine of the angle at which it is inclined with respect to the z-axis is given by
the scalar product:
cos θ =
n · ez
=
|
n|2
1
.
(4.1.7)
(∂f/∂x)2 + (∂f/∂y)2 + 1
After plugging this result into (4.1.4), one gets the needed formula (4.0.1).
Now one can move on to applying the obtained result to the present exercise. In
Fig. 4.2, the surface defined in the text of this problem is drawn. It is true that it does
not constitute the graph of the function x(y, z), but thanks to the symmetry, when
replacing x → −x, it will suffice to simply limit oneself to the case x > 0 (then
the graph shall represent a function), and multiply the obtained surface area by 2.
Surely, equally well one could (and the reader is encouraged to do it) in place of the
function x(y, z) consider z(x, y) or y(x, z).
The choice having been made, one has
√
x(y, z) = + yz
(4.1.8)
4.1 Finding Area of Surfaces
107
x
z
y
Fig. 4.2 The surface given by the formula (4.1.1)
which constitutes just the function f of the formula (4.0.1). Thus, the expression for
the area will take the form
2 2
∂x
∂x
+
+ 1 dydz.
(4.1.9)
S=2
∂y
∂z
D
Stemming from the content of the exercise, the following conditions must be
simultaneously satisfied:
y > 0,
2y + z < 2.
(4.1.10)
0 < z < 2(1 − y),
(4.1.11)
z > 0,
They entail that
0<y<1
and
thanks to which one can write
2(1−y) 1
2
∂x 2
∂x
S = 2 dy
dz
+
+1
∂y
∂z
0
0
2(1−y)
1
=2
dy
0
dz
0
1
√
2
z
y
2
+
1
√
2
y
z
2
+1
108
4 Examining Unoriented Curvilinear Integrals
2(1−y)
1
√
= 2
dy
0
=
0
0
0
dy
0
dz
z2 + y 2 + 2zy
√
zy
0
2(1−y)
1
y+z √
= 2
dz √
yz
dy
dy
0
2(1−y)
1
√
2
dz
2(1−y)
1
√
y
z
+ +2= 2
y
z
dz
√
√ y
z
.
√ +√
z
y
(4.1.12)
0
The integral with respect to z can be easily performed:
1
√
S= 2
&
√ '
2 z3 2(1−y)
√
dy
2 zy + √
3 y 0
0
1
=4
&
'
2 (1 − y)3
y(1 − y) +
dy.
√
3
y
(4.1.13)
0
Now let us note that in the second term the following integration by parts can be
carried out:
1
(1 − y)3
dy =
√
y
0
√ √ 1
3
y] dy = 2 (1 − y) y 1
(1 − y)3 [2
0
0
1
3
+2
2
1
(1 − y)y dy = 3
0
(1 − y)y dy
(4.1.14)
0
and the expression for the area can be written in the simple form:
1
S = 12
3
π
y(1 − y) dy = 12 = π.
8
2
(4.1.15)
0
The value of the last integral has been found using the methods discussed in the
first part of the book series in Sects. 14.5 and 14.6. The easiest way is to write
y(1 − y) =
1
1 2
− y−
,
4
2
(4.1.16)
then make the substitution
y−
1
1
= sin t,
2
2
(4.1.17)
4.1 Finding Area of Surfaces
109
and integrate over the parameter t:
1
1
y(1 − y) dy =
0
1 2
1
1
− y−
dy =
4
2
2
0
π/2
1
=
4
cos t
−π/2
1
(1 − sin2 t) dt
4
π/2
1
cos t dt =
8
[1 + cos(2t)] dt =
2
−π/2
π/2
−π/2
π
.
8
(4.1.18)
Problem 2
The area of the helical surface parametrized as:
x(u, v) = u cos v,
y(u, v) = u sin v,
z(u, v) = 4v,
(4.1.19)
cut out with the cylinder x 2 + y 2 = 9 and the planes z = 0 and z = 8π will be
found.
Solution
As we know, surfaces can be described in a parametric way, by specifying (in 3
dimensions) the dependences x(u, v), y(u, v), and z(u, v) for (u, v) belonging to a
certain domain D. In such a case, the area of a given sheet may be calculated as the
sum of areas of certain parallelograms, one of which, built on vectors tangent to the
surface, is shown in Fig. 4.3.
In what follows, we will be limited to smooth surfaces only, for which the
derivatives of the functions x, y, and z over the parameters u and v exist and
are continuous. As we know from the first chapter, the vector t tangent to a
given curve r(u) can be obtained by calculating the derivative with respect to the
parameter:
t =
d
r(u).
du
(4.1.20)
It is analogous to finding the velocity vector of a particle at a given point by
differentiating its position vector over time.
Various curves can be identified on the surface r(u, v) as well. For example, if the
value of the parameter v is fixed, the vector r will depend only on u, thus describing
a curve lying on the surface in question. Any vector tangent to this curve (denoted
as tu ) is tangent to the surface as well. Similarly, the value of u can be fixed and the
second independent tangent vector (i.e., tv ) appears. One has then
110
4 Examining Unoriented Curvilinear Integrals
Fig. 4.3 The method of calculating the area of a surface defined in the parametric way
tu =
∂ r
,
∂u
tv =
∂ r
.
∂v
(4.1.21)
Such vectors and the corresponding curves are shown in Fig. 4.3.
The infinitesimal vectors tu du and tv dv form a parallelogram on the surface; its
area being expressed by the vector product:
(4.1.22)
dS = tu × tv dudv.
The total area of a sheet of a given surface can now be found by calculating
the integral over the entire domain of parameters D (i.e., by adding the areas of
infinitesimal parallelograms):
tu × tv dudv.
S=
(4.1.23)
D
This justifies the first formula of (4.0.2) given in the theoretical introduction at the
beginning of this chapter. This expression, if necessary, can be given other forms
the following
contained in the quoted formula as well. As for any vectors a and b,
identity holds:
2
2,
a b)
(4.1.24)
a × b = a 2 b2 − (
then one also gets
tu2 tv2 − (tu tv )2 dudv =
S=
D
∂ r
∂u
2 ∂ r
∂v
2
−
∂ r ∂ r
·
∂u ∂v
2
dudv.
D
(4.1.25)
4.1 Finding Area of Surfaces
111
Moreover, let us note that the expression under the square root constitutes the
determinant of the matrix ĝ:
⎡
⎤
∂ r 2 ∂ r ∂ r
·
⎢ ∂u
∂u ∂v ⎥
⎥
ĝ = ⎢
⎣ ∂ r ∂ r ∂ r 2 ⎦ .
·
∂u ∂v ∂v
(4.1.26)
This matrix, playing a very important role in curved spaces, is called—as
mentioned—the metric tensor. Consequently, one can also write
det ĝ dudv.
S=
(4.1.27)
D
Thereby, if one knows the metric tensor, one can immediately start to calculate the
surface area.
Applying these formulas to the surface defined in the text of this exercise is no
longer a problem. Naturally we start by calculating both tangent vectors:
z
y
x
Fig. 4.4 The surface described parametrically with the equations (4.1.19)
112
4 Examining Unoriented Curvilinear Integrals
∂ r
= [cos v, sin v, 0],
∂u
∂ r
tv =
= [−u sin v, u cos v, 4].
∂v
tu =
(4.1.28)
The vector product tu × tv has the form:
tu × tv = [4 sin v, −4 cos v, u],
(4.1.29)
which allows us to make use of the formula (4.1.23). From the text of this problem,
one infers that
u ∈ [0, 3]
and
3
8π
v ∈ [0, 8π ],
(4.1.30)
so the area is given by
3
S=
8π
du
0
dv tu × tv =
0
du
0
dv
3
u2
+ 16 = 8π
0
u2 + 16 du.
0
(4.1.31)
This latter integral can be calculated in a number of ways, and we are going to
reacquaint ourselves with the Euler’s substitutions discussed in the first part of this
book series. In particular, the so-called first substitution (see (14.5.2) in the Part I)
can be applied by introducing a new variable t via the equation:
u2 + 16 = t − u.
(4.1.32)
Then it can easily be found that
8
t
u= −
2
t
1
16
⇒ du =
1 + 2 dt,
2
t
(4.1.33)
the integration over t running from 4 to 8. After inserting these expressions
into (4.1.31), one gets the rational function to integrate (after all this is just the
idea of Euler’s substitutions):
8
1
S = 8π
2
= 2π
4
16
t
8
1+ 2
t− +
dt = 2π
2
t
t
8
32 256
t+
+ 3 dt
t
t
4
t2
128 8
+ 32 log t − 2 = 4π(15 + 16 log 2).
2
t
4
(4.1.34)
4.1 Finding Area of Surfaces
113
At the end, it is worth figuring out how the metric tensor looks like on the
surface (4.1.19). As it is clear from the formula (4.1.26), it is constructed of the
products of tangent vectors, the calculation of which is very simple so we find:
ĝ =
tu · tu tu · tv
tv · tu tv · tv
=
1
0
.
0 u2 + 16
(4.1.35)
The vanishing of the off-diagonal terms reflects the fact that the curves on the surface
defined with the conditions u = const and v = const are perpendicular to each
other at anypoint of intersection. When looking at the tensor (4.1.35), the familiar
expression det ĝ in the integral (4.1.31) can be immediately recognized.
Problem 3
The area of the sheet of a sphere contained between two meridians and two parallels
will be found.
Solution
To solve this problem, the formula (4.1.27) will be used. The natural parametrization
of the sphere of radius R is given by the formulas:
x(θ, ϕ) = R sin θ cos ϕ,
y(θ, ϕ) = R sin θ sin ϕ,
z(θ, ϕ) = R cos θ,
(4.1.36)
where the azimuthal angle ϕ corresponds to the longitude, and the angle θ
corresponds to the latitude measured, however, from the north pole and not from
the equator. The sheet of the surface, we are interested in, is drawn. The integration
limits are clear: θ1 ≤ θ ≤ θ2 and ϕ1 ≤ ϕ ≤ ϕ2 .
Now, one needs to compute both tangent vectors. After the simple differentiations, one gets
∂ r
= [R cos θ cos ϕ, R cos θ sin ϕ, −R sin θ ],
∂θ
∂ r
= [−R sin θ sin ϕ, R sin θ cos ϕ, 0].
tϕ =
∂ϕ
tθ =
(4.1.37)
Calculating the appropriate scalar products, the metric tensor can explicitly be
found:
2
0
tθ · tθ tθ · tϕ
R
ĝ =
=
,
(4.1.38)
0 R 2 sin2 θ
tθ · tϕ tϕ · tϕ
and then det ĝ = R 4 cos2 θ .
114
4 Examining Unoriented Curvilinear Integrals
Fig. 4.5 The sheet of the
spherical surface between two
meridians and two parallels
Finally, the formula for the area looked for takes the form:
ϕ2
θ2
S=
dθ
dϕ
θ2
R 4 sin2 θ
= R (ϕ2 − ϕ1 )
2
ϕ1
θ1
sin θ dθ
θ1
= R (ϕ2 − ϕ1 )(cos θ1 − cos θ2 ).
2
(4.1.39)
4.2 Calculating Various Curvilinear Integrals
Problem 1
The integral:
(x + y)2 d 2 S
I=
(4.2.1)
S
will be calculated, where
(
)
S := (x, y, z) ∈ R3 | z = 2(x 2 + y 2 ) ∧ x 2 + y 2 ≤ 1 .
(4.2.2)
4.2 Calculating Various Curvilinear Integrals
115
Fig. 4.6 The sheet of surface
over which the integral (4.2.1)
is being calculated
Solution
In the case when instead of calculating the area of a given surface, our task consists
of integrating a function f (x, y, z) over the surface, only a slight modification of
the formulas applied in the previous section is required. If the surface is given in a
parametric way via the relations x(u, v), y(u, v), and z(u, v), then in place of the
integral (4.1.23) one needs now to calculate
f (u, v) tu × tv dudv.
(4.2.3)
D
Here the same symbol of the function f has been preserved to denote
f (u, v) = f (x(u, v), y(u, v), z(u, v)).
It is not quite accurate from the mathematical point of view, but often used in
practice (for example, if a physicist calculates the surface integral of the electric
field, he uses the same symbol E, regardless of on how many and which arguments
it depends).
Similarly, instead of (4.1.27), one can write
f (u, v) det ĝ dudv.
D
(4.2.4)
116
4 Examining Unoriented Curvilinear Integrals
In the case of the sheet of surface given in the content of this exercise the simplest
parametrization has the form:
x(ρ, ϕ) = ρ cos ϕ,
y(ρ, ϕ) = ρ sin ϕ,
z(ρ, ϕ) = 2ρ 2 .
(4.2.5)
With this choice, not only the formulas are relatively simple, but also the integration
limits do not depend on the varying parameters:
0 ≤ ρ ≤ 1,
0 ≤ ϕ < 2π.
(4.2.6)
The integrand function in our problem is f (x, y, z) = (x + y)2 , so
1
I=
2π
dρ
0
0
dϕ ρ 2 (cos ϕ + sin ϕ)2 tρ × tϕ .
(4.2.7)
f (ρ,ϕ)
Now both tangent vectors have to be determined together with their cross product.
Performing the appropriate differentiations, one easily gets
tρ =
∂ r
= [cos ϕ, sin ϕ, 4ρ],
∂ρ
tϕ =
∂ r
= [−ρ sin ϕ, ρ cos ϕ, 0],
∂ϕ
tρ × tϕ = [−4ρ 2 cos ϕ, 4ρ 2 sin ϕ, ρ]
(4.2.8)
and consequently
tρ × tϕ =
16ρ 4 + ρ 2 = ρ
16ρ 2 + 1.
(4.2.9)
This expression is also equal to the determinant of the metric tensor, which can be
easily checked by the reader with the use of the formula (4.1.35).
In order to find the value of I , the integrals with respect to the parameters ϕ and
ρ need to be subsequently calculated:
1
I =
2π
dϕ ρ 2 (cos ϕ + sin ϕ)2 ρ
dρ
0
0
1
=
2π
dϕ [1 + sin(2ϕ)]ρ 3 16ρ 2 + 1.
dρ
0
16ρ 2 + 1
(4.2.10)
0
The Euclidean trigonometric identity and the formula for the sine of the doubled
angle have been used here.
4.2 Calculating Various Curvilinear Integrals
117
The reader certainly notices now that the integral of the expression sin(2ϕ) over
the angle ϕ must vanish since the integration runs over the entire period of the sine
function (and even over two periods due to the presence of “2” in its argument). As
a result, the integration with respect to ϕ becomes trivial. In the remaining integral
over ρ, we are going to make the substitution ρ 2 = w, and next 16w + 1 = v,
obtaining finally
1
I = 2π
1
ρ
3
16ρ 2
+ 1 dρ = π
0
√
π
w 16w + 1 dw =
256
0
π
256
v 3/2 − v 1/2 dv =
√
(v − 1) v dv
1
17
=
17
π
256
1
2 5/2 2 3/2 17
v − v
5
3
1
√
π
391 17 + 1 .
=
960
(4.2.11)
Problem 2
The center of mass of the homogeneous hemisphere of mass M defined with the
equation x 2 + y 2 + z2 = R 2 for z ≥ 0 will be found.
Solution
In the case of a surface mass distribution, to find the center of mass the following
formula can be used:
rCM =
1
M
σ (r ) r d 2 S,
(4.2.12)
B
where the symbol σ stands for the surface mass density and B refers to the
considered sheet. If the surface is parametrized with u, v ∈ D, then one gets
rCM =
tu × tv σ (u, v)r (u, v) dudv
1
M
D
1
=
M
D
σ (u, v) standing for σ (r (u, v)).
ĝ σ (u, v)r (u, v) dudv,
(4.2.13)
118
4 Examining Unoriented Curvilinear Integrals
In our present problem, the parametrization that gets imposed is that defined
by (4.1.36), where now
0 ≤ ϕ < 2π,
0≤θ ≤
π
.
2
(4.2.14)
For this parametrization both tangent vectors have already been calculated
(see (4.1.37)), as well as the metric tensor (see (4.1.38)). We are going to make
use of the latter, writing
π/2
2π
rCM
1
=
M
dϕ
dθ
0
(4.2.15)
ĝ σ (θ, ϕ)r (θ, ϕ).
0
This is, in fact, three integrals, each for a different component of the vector rCM .
However, two of them do not need to be calculated, since the result of the integration
is immediately known. Thanks to the axial symmetry displayed by the body, one
must inevitably get
xCM = 0
and
yCM = 0.
(4.2.16)
Thereby, it remains only to calculate
π/2
2π
zCM
1
=
M
dϕ
dθ
0
(4.2.17)
ĝ σ (θ, ϕ)z(θ, ϕ).
0
The determinant of the metric tensor is equal to R 2 sin θ , and z(θ, ϕ) = R cos θ .
We still do not know the surface mass density σ (θ, ϕ), but since it is a constant,
it can easily be calculated as a quotient of the total mass M and the area equal to
2π R 2 (i.e., a half of the surface area of the sphere):
σ =
M
.
2π R 2
(4.2.18)
Plugging these expressions into the integral (4.2.17), one gets
π/2
2π
zCM
M
1
·
=
R3
M 2π R 2
dϕ
0
π/2
R
2π
dθ sin θ cos θ =
2π
0
0
1
R
sin(2θ ) dθ = ,
2
2
=1/2
(4.2.19)
and finally
R
.
= 0, 0,
2
rCM
(4.2.20)
4.2 Calculating Various Curvilinear Integrals
119
Problem 3
The moment of inertia (with respect to the z-axis) of the sheet of the elliptic
paraboloid defined with the equation:
z=
x2
y2
+ ,
4
2
(4.2.21)
cut out with the elliptical cylinder x 2 /4 + y 2 = 1 will be found.
Solution
Based on the definition of the elliptic cylinder, we choose the parametrization for x
and y in the form:
x(ρ, ϕ) = 2ρ cos ϕ,
y(ρ, ϕ) = ρ sin ϕ,
and then the paraboloid equation will be solved for z:
1
2
2
2
z(ρ, ϕ) = ρ cos ϕ + sin ϕ .
2
Obviously one has 0 ≤ ρ ≤ 1 and 0 ≤ ϕ < 2π .
Fig. 4.7 The sheet of
surface (4.2.21) cut out with
the elliptic cylinder
x 2 /4 + y 2 = 1
(4.2.22)
(4.2.23)
120
4 Examining Unoriented Curvilinear Integrals
The tangent vectors associated with the parameters ρ and ϕ have the following
forms:
∂ r
1 2
2
tρ =
= 2 cos ϕ, sin ϕ, 2ρ cos ϕ + sin ϕ ,
∂ρ
2
tϕ =
∂ r
= [−2ρ sin ϕ, ρ cos ϕ, −ρ 2 sin ϕ cos ϕ].
∂ϕ
(4.2.24)
Their cross product is equal to
tρ × tϕ = [−2ρ 2 cos ϕ, −2ρ 2 sin ϕ, 2ρ],
(4.2.25)
tρ × tϕ = 2ρ 1 + ρ 2 .
(4.2.26)
and hence
Let us calculate now the total mass of the solid (denoted with B) using the
formula:
M=
σ (r ) d 2 S,
(4.2.27)
B
the symbol σ standing for the surface mass density. In our case it is a constant, so
one can bring it out from under the integral. After having used the parametrization
given above and upon introducing a certain new variable with the formula u = ρ 2 ,
one gets
2π
M=σ
1
1
1 + ρ2
dρ 2ρ
= 4π σ ρ
0
0
0
|tρ ×tϕ |
1
4π σ
4π σ √
3/2 (1 + u) =
(2 2 − 1).
=
3
3
0
dϕ
1
1 + ρ 2 dρ
= 2π σ
√
1 + u du
0
(4.2.28)
When calculating the moment of inertia, the additional factor x 2 + y 2 (the square
of the distance of a given element of the solid from the rotation axis) should be
inserted under the integral of the type (4.2.27). Now, since
x 2 + y 2 = ρ 2 (3 cos2 ϕ + 1) =
1 2
ρ (3 cos(2ϕ) + 5),
2
(4.2.29)
4.3 Exercises for Independent Work
121
so it can be easily obtained that
2π
I =σ
1
dρ 2ρ 1 + ρ 2
dϕ
0
1 2
ρ (3 cos(2ϕ) + 5)
2
0
1
1
ρ 3 1 + ρ 2 dρ = 5π σ
= 10π σ
0
√
u 1 + u du,
(4.2.30)
0
where the fact that the integral of the cosine function over the entire (or multiple)
period equals zero has been used. Upon introducing a new integration variable v =
1 + u, one finally obtains
2
I = 5π σ
√
(v − 1) v dv = 5π σ
1
√
2+1
4π σ √
M,
=
( 2 + 1) = √
3
2 2−1
2 5/2 2 3/2 2
v − v
5
3
1
(4.2.31)
where the mass density has been eliminated due to (4.2.28).
The reader may wonder why the moment of inertia is expressed in our formulas
in the same units as the mass. The answer to this question is contained already in
the formula (4.2.21) and also in the elliptic cylinder formula. It simply stems from
the fact that the variables x, y, z, and therefore ρ as well, are here dimensionless
quantities. Otherwise, these equations would have to include some additional
parameters.
4.3 Exercises for Independent Work
Exercise 1 Using the form of the metric tensor on the sphere of radius R
(see (4.1.38)), derive the expression for its surface area.
Answer
S = 4π R 2 .
122
4 Examining Unoriented Curvilinear Integrals
Exercise 2 Find the area S of:
(a) the solid constituting the intersection of the two perpendicular cylinders: x 2 +
y 2 ≤ 1/4, x 2 + z2 ≤ 1/4,
(b) the sheet of the surface x 2 + y 2 + z2 = R 2 , cut out with the elliptic cylinder
4x 2 /R 2 + y 2 /R 2 ≤ 1,
(c) the sheet of the paraboloid z = x 2 + y 2 for z ≤ 2,
(d) the surface defined parametrically with the equations x(u, v) = v cos u,
y(u, v) = v sin u, z(u, v) = 2v, where 0 ≤ u < 2π , 0 ≤ v ≤ R.
Answers
(a)
(b)
(c)
(d)
S = 4.
S = 4π R 2 /3.
S
√= 13π/3.
5π R 2 .
Exercise 3 Find the center-of-mass location of:
(a) the sheet of the surface z = x + 2y, where x 2 + y 2 ≤ x, with the surface mass
density σ (x, y, z) = z2,
(b) the sheet of the surface y = x 2 cut out with the planes
x = 0, z = 0, x = 1,
√
z = 1, with the surface mass density σ (x, y, z) = 1 + 4y.
Answers
(a) xCM = 2/3, yCM = 1/3, zCM = 4/3π .
(b) xCM = 9/14, yCM = 17/35, zCM = 1/2.
Exercise 4 Find the moment of inertia with respect to the z-axis of:
(a) the
uniform sheet of the surface of mass M defined with the equation z =
4 − x 2 − y 2 , cut out with the planes: x = ±1, y = ±1,
(b) the uniform sheet of the surface of mass M defined with the equation (x 2 +
y 2 )3/2 + z = 1, where z ≥ 0.
Answers
(a) I = 2M/3.
√
√
√
(b) I = 2(10 10 − 1)/(9(3 10 + log(3 + 10))) M.
Chapter 5
Examining Differential Forms
This chapter is devoted to the differential forms. We will learn how to calculate
their values when acting on vectors, and how to find their exterior products, exterior
derivatives, etc. We will also become familiar with the notion and the methods of
determining the primitive forms.
Let V be a certain vector space over a field K (real or complex). A form of
degree 1 or a 1-form ω is simply a linear function ω : V → K. In general an
k
exterior form of degree k or a k-form ω is a linear function (in each argument)
k
defined on the Cartesian product V k = V × V × . . . × V (i.e., ω: V k → K) and
antisymmetric with respect to its arguments.
A differential k-form is defined on a certain smooth surface (manifold)—subset
of Rm —and acts on the vectors of the tangent space. Its most general form is
m
!
k
ω = ωi1 ,i2 ,...,ik (x1 , . . . , xm ) dxi1 ∧ . . . ∧ dxik ,
(5.0.1)
i1 ,i2 ,...,ik =1
i1 <i2 <...<ik
where ωi1 ,i2 ,...,ik (x1 , . . . , xm ) are certain smooth functions defined on the surface
and dxi are base 1-forms satisfying dxi (v) = vi , where vi are components of
the vector v in a certain fixed base. It is discussed in detail in Problem 1 below.
k
l
The symbol ∧ is explained as follows. Given two forms: ω1 and ω2 , where
k
k
ω1 : v1 , . . . , vk −→ ω1 (v1 , . . . , vk ),
(5.0.2)
and
l
l
ω2 : vk+1 , . . . , vk+l −→ ω2 (vk+1 , . . . , vk+l ).
© Springer Nature Switzerland AG 2020
T. Radożycki, Solving Problems in Mathematical Analysis, Part III, Problem Books
in Mathematics, https://doi.org/10.1007/978-3-030-38596-5_5
(5.0.3)
123
124
5 Examining Differential Forms
The exterior product (denoted with ∧) of these two forms is the new form of degree
k + l, i.e.,
k+l
k
l
ω =ω1 ∧ ω2 ,
defined as
k+l
k+l
ω : v1 , . . . , vk+l −→ ω (v1 , . . . , vk+l )
! k
l
= σP ω1 (vi1 , . . . , vik ) ω2 (vik+1 , . . . , vik+l ).
(5.0.4)
P
i1 <i2 <...<ik
ik+1 <ik+2 <...<ik+l
This sum runs over all permutations (hence the symbol P ) of the set of vectors
v1 , . . . , vk+l , with the restrictions imposed by the conditions i1 < i2 < . . . < ik and
ik+1 < ik+2 < . . . < ik+l . The symbol σP stands for the sign of the permutation P .
Given a smooth mapping
: Rn ⊂ U −→ V ⊂ Rm ,
(5.0.5)
where U and V are certain open sets. Let this mapping be defined as the set of m
relations:
x1 = ϕ1 (y1 , . . . , yn ),
x2 = ϕ2 (y1 , . . . , yn ),
··· ,
xm = ϕm (y1 , . . . , yn ).
(5.0.6)
k
The pullback of the form ω defined on V , to the set U , is denoted as ∗ω, and
defined with the formula:
!
∂(ϕi1 , . . . , ϕik )
k
dyj1 ∧ . . . ∧ dyjk .
∗ ω = ωi1 ,i2 ,...,ik ((y1 , . . . , yn ))
∂(yj1 , . . . , yjk )
(5.0.7)
1≤i1 <...<ik ≤m
1≤j1 <...<jk ≤n
The exterior derivative of a k-form in Rn given by (5.0.1) is the form of the
degree k + 1, defined in the following way:
k
dω =
n
!
∂ωi
j,i1 ,i2 ,...,ik =1
i1 <i2 <...<ik
(x1 , . . . , xn )
dxj ∧ dxi1 ∧ . . . ∧ dxik .
∂xj
1 ,i2 ,...,ik
(5.0.8)
5.1 Studying the Exterior Forms Operating on Vectors
125
Let D ⊂ Rn be a certain open and connected set. It is called contractible to a
point x0 ∈ D if there exists a continuous function f : [a, b]×D ∈(t, x) → f (t, x) ∈
D such that for all x ∈ D one has f (a, x) = x0 and f (b, x) = x.
k
k
Let ω be a smooth, closed form (i.e., such that d ω = 0), defined on a certain
open set D ⊂ Rn , contractible to a point. Then the Poincaré lemma states that this
k−1
form is exact, i.e., there exists such form ω (called the primitive form) that
k
k−1
ω= d ω .
(5.0.9)
5.1 Studying the Exterior Forms Operating on Vectors
Problem 1
The value of the 1-form in R3 :
1
ω= xy dx + yz dy + (z − x) dz,
(5.1.1)
will be found on the vectors vA = [0, 1, 2], vB = [1, 1, 2], and vC = [−1, 1, −1]
bound at the points A(0, 0, 1), B(2, 1, 0), and C(3, 2, 1), respectively.
Solution
The objects concerned with below, i.e., the so-called forms, are rather abstract ones.
Therefore, this concept will be discussed more loosely than it was done in the
theoretical introduction above.
A form is a certain function, which acts in a vector space V and attributes a
number (real or complex) to a given vector (or vectors). In particular the object
known as the form of degree k has k arguments and, therefore, is defined on the
Cartesian product V k . Among various forms, the important role is played by linear
ones, i.e., exhibiting the property:
ω(αv + βw) = αω(v) + βω(w).
(5.1.2)
The symbol ω denotes here the form, v and w stand for vectors of the vector space,
and α and β are scalars (i.e., numbers). If a linear form is to be of the degree k, then
the condition (5.1.2) should hold for each argument individually. An example of the
linear form of degree 1 can be given (denoted usually with ei ) which attributes to a
vector one (namely the ith) of its coordinates in the previously chosen basis, i.e.,
ei (v) = v i .
(5.1.3)
126
5 Examining Differential Forms
The vector coordinates are numbers, and one also has
ei (αv + βw) = αv i + βw i = αei (v) + βei (w),
(5.1.4)
so in fact, we are dealing with a linear form of the 1st degree.
Naturally other forms are possible as well. If, for example, the space of real
polynomials was treated as a vector space (the reader has, for sure, become
accustomed to this space within the lecture of algebra), then the mapping ω
defined as
1
ω(v) :=
v(x) dx,
(5.1.5)
−1
where v(x) is a polynomial, is again a linear form of the 1st degree. In turn the
mapping τ operating on pairs of vectors (polynomials) v and w as follows
1
τ (v, w) :=
v(x)w(x) dx
(5.1.6)
−1
constitutes the linear form of the degree 2. If, however, a form is to be “exterior,”
one would additionally require the antisymmetry when swapping its arguments:
τ (v, w) = −τ (w, v). For example, such a condition would be met by the form
1
τ (v, w) :=
[v(x)w(0) − w(x)v(0)] dx.
(5.1.7)
−1
In this and in the following problems of this chapter, we will be concerned with
the exterior differential forms of the degree k, in short called differential k-forms or
simply k-forms. They are defined on smooth surfaces (manifolds), and act on the
vectors of the tangent space.
To further clarify what we mean, let us imagine a smooth function f (x, y, z)
defined on the surface of a sphere. The differential of this function at some point
P (x0 , y0 , z0 ), given by
∂f ∂f ∂f df =
dx +
dy +
dz,
∂x P
∂y P
∂z P
(5.1.8)
specifies its increase if one moves from the point P by an infinitesimal vector
[dx, dy, dz]. However, this vector has some limitations. Since the function f is
defined only on the surface of a sphere, one cannot move perpendicularly to it. As
a consequence this vector must be tangential to the surface. If one moved by some
5.1 Studying the Exterior Forms Operating on Vectors
127
other tangent vector, e.g., v = [vx , vy , vz ], this increase would be expressed by the
formula
∂f ∂f ∂f dfv =
vx +
vy +
vz .
(5.1.9)
∂x P
∂y P
∂z P
Hence, it is convenient to agree that the symbols dx, dy, and dz (or more generally
dxi , i = 1, . . . , N in a N -dimensional space) are, in fact, forms of the 1st degree,
which operate on tangent vectors exactly as it was defined in the Eq. (5.1.3), i.e.,
dx(v) = vx ,
dy(v) = vy ,
dz(v) = vz .
(5.1.10)
Then the expression dfv simply denotes the result of the operation of the form df
defined by the formula (5.1.8) on a tangent vector v:
dfv = df (v).
(5.1.11)
Such form certainly depends on the point P on the sphere because the partial
derivatives
∂f ∂f ∂f ,
,
∂x P
∂y P
∂x P
are P dependent. The same reasoning can be carried out for other smooth surfaces.
After this introduction, we can address the current problem, which is very simple.
1
The 1-form ω is defined on a certain one-dimensional surface (i.e., on a curve). We
do not know which, but it is not needed in this exercise. This is because we already
know the tangent vector, e.g., vA (and successively vB and vC ), and the point of
attachment of the tangent space, i.e., the contact point A (and respectively B and
C). Therefore, one obtains
1
ω (vA ) = 0 · 0 · dx(vA ) + 0 · 1 · dy(vA ) +(
1 − 0 ) · dz(vA ) = 2.
xA
yA
=0
yA
zA
=1
zA
xA
=2
(5.1.12)
In an analogous way one gets
1
ω (vB ) = 2 · 1 · dx(vB ) +1 · 0 · dy(vB ) +(0 − 2) · dz(vB ) = −2
=1
=1
(5.1.13)
=2
and
1
ω (vc ) = 3 · 2 · dx(vC ) +2 · 1 · dy(vC ) +(1 − 3) · dz(vC ) = −2.
=−1
=1
=−1
(5.1.14)
128
5 Examining Differential Forms
Problem 2
The value of the 2-form in R3
2
ω= (x + y) dx ∧ dz + (y − z) dy ∧ dz ,
(5.1.15)
will be found for the pair of vectors u = [2, 1, 1], v = [0, 1, 1] bound at the point
A(1, 2, 1).
Solution
Before we start to solve this problem, let us recall the meaning of the symbol ∧,
which has already appeared in the theoretical introduction. Suppose that there are
k
l
two forms: ω1 and ω2 . The former, as a form of the degree k, operates on the set of
k vectors and the latter similarly of the l vectors, according to (5.0.2) and (5.0.3).
The symbol ∧ denotes the so-called exterior product, the result of which is a form
of the degree k + l:
k+l
k
l
ω =ω1 ∧ ω2 ,
which operates on the set of k + l vectors. Let us rewrite the appropriate
formula (5.0.4):
k+l
k+l
ω : v1 , . . . , vk+l −→ ω (v1 , . . . , vk+l )
! k
l
= σP ω1 (vi1 , . . . , vik ) ω2 (vik+1 , . . . , vik+l ).
(5.1.16)
P
i1 <i2 <...<ik
ik+1 <ik+2 <...<ik+l
Analyzing this expression, one can illustratively say that, out of the ordered set of
k + l vectors, a subset of k is chosen and moved to the beginning of the set, however,
k
without altering their internal ordering. These k vectors saturate the first form (ω1 ).
The remaining l vectors (again without changing their ordering) must saturate the
l
second form (ω2 ). This order is essential, since exterior forms change their signs
when any two vectors (i.e., arguments) switch their places. Any such shift of k
vectors at the beginning leads to a certain new permutation of the set v1 , . . . , vk+l .
The definition (5.1.16) requires that each term of the sum is multiplied by the sign
k+l
of the appropriate permutation (denoted with σP ). With this, the new form ω
preserves the property of the so-called skew symmetry for the full set of k +l vectors
5.1 Studying the Exterior Forms Operating on Vectors
129
(or simply antisymmetry). The summation is completed after all possible k-element
combinations of the k + l-element set have been exhausted. The exterior products
of forms will be dealt with in more detail in the next section.
Returning to our exercise, we must consider how the form dxi ∧ dxj operates on
a pair of vectors u, v. Of course, we know that
dxi (u) = ui ,
dxj (u) = uj
(5.1.17)
and similarly for vector v. Using the definition (5.1.16), one gets
dxi ∧ dxj (u, v) = dxi (u)dxj (v) − dxi (v)dxj (u) = ui vj − vi uj .
(5.1.18)
For the vectors u and v as given in the text of the exercise, one finds
dx ∧ dy(u, v) = ux vy − vx uy = 2 · 1 − 0 · 1 = 2,
dy ∧ dz(u, v) = uy vz − vy uz = 1 · 1 − 1 · 1 = 0.
(5.1.19)
In view of the fact that these vectors are bound at the point A(1, 2, 1) (i.e., they
are vectors of a space tangent to a hypothetical surface at A), one obtains
2
ω (u, v) = (1 + 2) · 2 + (2 − 1) · 0 = 6.
(5.1.20)
Problem 3
The values of the 1-forms dr, dϕ, and dθ in R3 on the versors er , eϕ , and eθ will be
found, where r, ϕ, and θ denote the spherical variables.
Solution
This problem will be solved by referring to the relation (5.1.17) formulated for
Cartesian coordinates. In R3 , it corresponds to the following formulas:
dx(ex ) = 1, dx(ey ) = 0, dx(ez ) = 0,
dy(ex ) = 0, dy(ey ) = 1, dy(ez ) = 0,
dz(ex ) = 0, dz(ey ) = 0, dz(ez ) = 1.
(5.1.21)
Incidentally, the above equations state that the differentials dx, dy, dz form a basis
in the space of 1-forms (which is a vector space as well), the so-called dual basis to
that composed of vectors ex , ey , ez in the vector space, in which these forms operate.
130
5 Examining Differential Forms
Fig. 5.1 The set of versors
er , eϕ , and eθ
Here, however, the properties of dual spaces and their bases are not the subject of
our interest. One has to perform two steps:
• to express the versors er , eϕ , and eθ through ex , ey , and ez ,
• to express 1-forms dr, dϕ, and dθ through dx, dy, and dz.
Let us start with the versors. The symbol er denotes the unit radial vector. It must,
therefore, have the following form, certainly known to the reader:
er = sin θ cos ϕ ex + sin θ sin ϕ ey + cos θ ez .
(5.1.22)
Making twice use of the Euclidean trigonometric identity, one can easily show that
in fact er · er = 1.
The vector eϕ is tangent to a given “parallel,” as it is shown in Fig. 5.1, so it
cannot have any component along ez . It is lying entirely in the plane parallel to that
determined by axes x and y and is given by
eϕ = − sin ϕ ex + cos ϕ ey .
(5.1.23)
In turn the vector eθ is tangent to a given “meridian.” If projected onto the plane
xy it points radially, but it also has a component of ez directed downward:
eθ = cos θ cos ϕ ex + cos θ sin ϕ ey − sin θ ez .
(5.1.24)
Naturally one has eϕ · eϕ = 1 and eθ · eθ = 1. All these versors are mutually
orthogonal.
5.1 Studying the Exterior Forms Operating on Vectors
131
Now the 1-forms dr, dϕ, and dθ must be addressed. For this purpose, the
formulas (5.1.8), (5.1.9), and (5.1.11) will be used. First let us treat
r(x, y, z) =
x 2 + y 2 + z2
(5.1.25)
as a function f (x, y, z) obtaining
dr =
∂r
∂r
x
y
z
∂r
dx +
dy +
dz = dx + dy + dz
∂x
∂y
∂z
r
r
r
= sin θ cos ϕ dx + sin θ sin ϕ dy + cos θ dz.
(5.1.26)
The formula (4.1.36) was used here, where, of course, r was put in place of R.
Now we are already in a position to calculate the result of the operation of the
form dr onto the versors (5.1.22–5.1.24). In accordance with (5.1.26), it is sufficient
to find how the forms of dx, dy, and dz act on them. Since these forms are linear,
one has
dx(er ) = dx(sin θ cos ϕ ex + sin θ sin ϕ ey + cos θ ez )
= sin θ cos ϕ dx(ex ) + sin θ sin ϕ dx(ey ) + cos θ dx(ez ) = sin θ cos ϕ,
dy(er ) = dy(sin θ cos ϕ ex + sin θ sin ϕ ey + cos θ ez )
= sin θ cos ϕ dy(ex ) + sin θ sin ϕ dy(ey ) + cos θ dy(ez ) = sin θ sin ϕ,
dz(er ) = dz(sin θ cos ϕ ex + sin θ sin ϕ ey + cos θ ez )
= sin θ cos ϕ dz(ex ) + sin θ sin ϕ dz(ey ) + cos θ dz(ez ) = cos θ,
(5.1.27)
where the relations (5.1.21) have been used. Thus one gets
dr(er ) = sin θ cos ϕ dx(er ) + sin θ sin ϕ dy(er ) + cos θ dz(er )
= sin2 θ cos2 ϕ + sin2 θ sin2 ϕ + cos2 θ = 1.
(5.1.28)
Similarly
dx(eϕ ) = dx(− sin ϕ ex + cos ϕ ey ) = − sin ϕ dx(ex ) + cos ϕ dx(ey ) = − sin ϕ,
dy(eϕ ) = dy(− sin ϕ ex + cos ϕ ey ) = − sin ϕ dy(ex ) + cos ϕ dy(ey ) = cos ϕ,
dz(eϕ ) = dz(− sin ϕ ex + cos ϕ ey ) = − sin ϕ dz(ex ) + cos ϕ dz(ey ) = 0
(5.1.29)
132
5 Examining Differential Forms
and consequently,
dr(eϕ ) = sin θ cos ϕ dx(eϕ ) + sin θ sin ϕ dy(eϕ ) + cos θ dz(eϕ )
= − sin θ cos ϕ sin ϕ + sin θ sin ϕ cos ϕ + cos θ · 0 = 0. (5.1.30)
In the identical way, the same result can be obtained for the vector eθ :
dr(eθ ) = 0.
(5.1.31)
In the case of the form dϕ as a function f (x, y, z), one can choose
y
ϕ(x, y, z) = arctan .
x
(5.1.32)
In the domain where x is supposed to be equal to zero, one can equally well use the
function
x
(5.1.33)
ϕ(x, y, z) = arcctan ,
y
which leads to the same results. One obtains
dϕ = −
sin ϕ
cos ϕ
dx +
dy.
r sin θ
r sin θ
(5.1.34)
Since how the forms dx, dy, dz operate on the vectors er , eϕ , and eθ is already clear,
one can immediately write down the final formulas:
sin ϕ
cos ϕ
dx(er ) +
dy(er ) = 0,
r sin θ
r sin θ
sin ϕ
cos ϕ
1
dx(eϕ ) +
dy(eϕ ) =
,
dϕ(eϕ ) = −
r sin θ
r sin θ
r sin θ
sin ϕ
cos ϕ
dx(eθ ) +
dy(eθ ) = 0.
dϕ(eθ ) = −
r sin θ
r sin θ
dϕ(er ) = −
(5.1.35)
One still has to consider the form dθ . In order to do it, let us use the function
x2 + y2
(5.1.36)
θ (x, y, z) = arctan
z
and proceed exactly as above. In this way, the reader will easily get the following
results:
dθ (er ) = 0,
dθ (eϕ ) = 0,
dθ (eθ ) =
1
.
r
(5.1.37)
5.2 Performing Various Operations on Differential Forms
133
5.2 Performing Various Operations on Differential Forms
Problem 1
The exterior product of two 1-forms in R3 :
1
1
ω1 = (x + y) dx + xz dy + z dz ,
ω2 = (x − y) dx + (x − z) dy + z dz
(5.2.1)
will be found.
Solution
The definition of the exterior product of forms, having already been recalled with the
formula (5.1.16), can be applied without obstacles in the present exercise. Among
other things, it entails that for any 1-form
1
1
1
1
ω ∧ τ= − τ ∧ ω .
(5.2.2)
For forms of higher degree, this property does not have to hold. In general one has
k
l
l
k
ω ∧ τ = (−1)kl τ ∧ ω .
(5.2.3)
The appearance of the factor (−1)kl is a consequence of the fact that if k ordered
vectors (i.e., arguments of the form) from among the set of k + l are to be moved
from the beginning to the end (without changing their internal ordering), one has to
perform exactly k ·l individual switchings. Each of the k vectors has to be “dragged”
by each of l vectors of the other group. And each such operation implies the reversal
of the sign of the permutation. In that way, after all switchings, the factor (−1)kl is
obtained. The reordering of groups of vectors discussed above is equivalent to the
swapping of forms in the exterior product but leaving vectors unchanged.
For k = l = 1, there appears the minus sign present on the right-hand side of the
formula (5.2.2) and it will always be so (and only then) when both forms are of odd
degree.
From the Eq. (5.2.2), we immediately get
dx ∧ dx = −dx ∧ dx
⇒ dx ∧ dx = 0
(5.2.4)
and similarly for the forms dy and dz. One also has
dy∧dx = −dx∧dy,
dz∧dx = −dx∧dz and dz∧dy = −dy∧dz.
(5.2.5)
134
5 Examining Differential Forms
Hence, wherever the exterior product of two identical differentials in the course
of our calculations appears, the appropriate expression vanishes. In addition, using
the formula (5.2.5), we will take care that the obtained result will always be written
in the canonical form, i.e., in such a way that the differentials dx, dy, and dz are
always in that particular order (dx standing before dy and before dz, and dy before
dz). The coefficient functions may freely be moved within the product. In view of
the above, one easily obtains
1
1
ω1 ∧ ω2 = [(x + y) dx + xz dy + z dz] ∧ [(x − y) dx + (x − z) dy + zdz]
= (x+y)(x−y) dx
∧ dx +(x + y)(x − z) dx ∧ dy + (x + y)z dx ∧ dz
=0
+ xz(x − y) dy ∧ dx +xz(x − z) dy ∧ dy +xz2 dy ∧ dz
=−dx∧dy
=0
+ z(x − y) dz ∧ dx +z(x − z) dz ∧ dy +z2 dz
∧ dz
=−dx∧dz
=−dy∧dz
=0
= (x 2 + xy − xz − yz − x 2 z + xyz) dx ∧ dy + 2yz dx ∧ dz
+ z(xz − x + z) dy ∧ dz.
(5.2.6)
As expected, it is a 2-form written in the canonical way.
Problem 2
The exterior product of the following 1-form and 2-form in R4 :
1
ω1 = y dx + z dy + x dz − y dw,
2
ω2 = yz dx ∧ dz + xz dy ∧ dw
(5.2.7)
will be found.
Solution
As we already know, the exterior product of a 1-form and a 2-form constitutes a
3-form. In order to find it in an explicit way, one has to use the properties stemming
from the “skew” symmetry of the product. In particular the following relations will
be needed:
5.2 Performing Various Operations on Differential Forms
135
dy ∧ dx ∧ dz = −dx ∧ dy ∧ dz,
dz ∧ dy ∧ dw = −dy ∧ dz ∧ dw,
dw ∧ dx ∧ dz = dx ∧ dz ∧ dw.
(5.2.8)
In the last case, no minus sign appears, since moving the differential dw to the
end involves the two-time change of the sign (after switching places with dx and
with dz). In addition, we know from the previous problem that if any of the
differentials appears twice in a product, the appropriate term vanishes. Omitting
such terms immediately, one gets
1
2
ω1 ∧ ω2 = (y dx + z dy + xdz − ydw) ∧ (yz dx ∧ dz + xz dy ∧ dw)
= xyz dx ∧ dy ∧ dw + yz2 dy ∧ dx ∧ dz + x 2 z dz ∧ dy ∧ dw
−y 2 z dw ∧ dx ∧ dz = xyz dx ∧ dy ∧ dw − yz2 dx ∧ dy ∧ dz
−x 2 z dy ∧ dz ∧ dw − y 2 z dx ∧ dz ∧ dw.
(5.2.9)
Note that if the forms were defined not in R4 but in R3 , only differentials dx, dy,
and dz would be available. The unique canonical product that might be constructed
in this case is dx ∧ dy ∧ dz. In any other, the differentials would have to be repeated
and the appropriate expression would vanish. This means that a 3-form in R3 can
have at most one term only. In the case of R4 , so many terms are allowed, as there are
ways to choose three symbols from among dx, dy, dz, and dw, in other words four.
Problem 3
The 1-form in R2 \ {(0, 0)}:
1
ω=
x
y
dx − 2
dy
x2 + y2
x + y2
(5.2.10)
will be rewritten in polar variables.
Solution
The polar variables r and ϕ are related to the Cartesian ones with the known
formulas:
x = r cos ϕ,
y = r sin ϕ,
(5.2.11)
136
5 Examining Differential Forms
which, for r = 0, can be easily inverted if necessary:
r=
x2 + y2,
y
ϕ = arctan .
x
(5.2.12)
Around x = 0, instead of the latter formula, one can use
x
ϕ = arcctan .
y
(5.2.13)
Acting just as in Problem 3 of the previous section (although using the inverted
formulas), one can write
dx =
∂x
∂x
dr +
dϕ = cos ϕ dr − r sin ϕ dϕ,
∂r
∂ϕ
dy =
∂y
∂y
dr +
dϕ = sin ϕ dr + r cos ϕ dϕ.
∂r
∂ϕ
(5.2.14)
The coefficient functions in (5.2.10) expressed in the variables r and ϕ have the
form:
x
cos ϕ
,
=
r
x2 + y2
y
sin ϕ
.
=
r
x2 + y2
(5.2.15)
It remains simply to plug the obtained results into (5.2.10) and rearrange the
terms. In this way, one gets
1
sin ϕ
cos ϕ
(cos ϕ dr − r sin ϕ dϕ) −
(sin ϕ dr + r cos ϕ dϕ)
r
r
cos(2ϕ)
dr − sin(2ϕ) dϕ.
(5.2.16)
=
r
ω=
Problem 4
The differential 2-form in R4 :
2
ω= (x + y) dx ∧ dz + (z + w) dy ∧ dz + (x − y) dx ∧ dw
(5.2.17)
will be rewritten in the variables: ξ = x − y, η = y − z, ζ = z − w, τ = x + w.
5.2 Performing Various Operations on Differential Forms
137
Solution
In the previous examples, the question of changing variables in differential forms
has already been encountered. In this exercise, one would like to discuss this
subject in a more formal way. Suppose that a certain general k-form in Rm is given
(see (5.0.1)):
!
k
ω = ωi1 ,i2 ,...,ik (x1 , . . . , xm ) dxi1 ∧ . . . ∧ dxik ,
(5.2.18)
1≤i1 ,i2 ,...,ik ≤m
i1 <i2 <...<ik
where the symbols ωi1 ,i2 ,...,ık (x1 , . . . , xm ) stand for the coefficient functions. For
example, in the case of the form (5.2.17) one has k = 2; m = 4; the variables
x1 , . . . , xm are simply called x, y, z, w; and the above-mentioned coefficient functions ω are respectively equal to x +y, z +w, and x −y (or 0, if a given combination
of differentials does not appear at all).
Consider in addition, a smooth mapping
: Rn ⊂ U −→ V ⊂ Rm ,
(5.2.19)
U and V denoting some open sets. Such a mapping is defined through m relations:
x1 = ϕ1 (y1 , . . . , yn ),
x2 = ϕ2 (y1 , . . . , yn ),
··· ,
xm = ϕm (y1 , . . . , yn ).
(5.2.20)
k
Now, our task is to “transfer” (or, as is often called, to “pull back”) the form ω
defined on the set V to U . For this “pulled-back” form, one usually uses the symbol
∗ω, and it can be found from formula (5.0.7) recalled below:
!
∂(ϕi1 , . . . , ϕik )
k
dyj1 ∧ . . . ∧ dyjk .
∗ ω = ωi1 ,i2 ,...,ik ((y1 , . . . , yn ))
∂(yj1 , . . . , yjk )
1≤i1 <...<ik ≤m
1≤j1 <...<jk ≤n
(5.2.21)
It looks rather complicated, so we are going to clarify it. As a matter of
fact, this procedure consists of two components. First, the coefficient functions
ωi1 ,i2 ,...,ık (x1 , . . . , xm ) have to be expressed through the variables y1 , . . . , yn using
the relations (5.2.20). This is just what the symbol ωi1 ,i2 ,...,ık ((y1 , . . . , yn )) stands
for. Second, one has to transform the exterior product of differentials, according to
the formula:
138
5 Examining Differential Forms
dxi1 ∧ . . . ∧ dxik =
!
1≤j1 <...<jk ≤n
∂(ϕi1 , . . . , ϕik )
dyj1 ∧ . . . ∧ dyjk .
∂(yj1 , . . . , yjk )
(5.2.22)
We recognize above the determinant of the Jacobian matrix of the mapping , well
known to us from the previous part of this book series and, the sum over
1 ≤ j1 < . . . < jk ≤ n
is just the sum over all possible ways of selecting k variables (after all a k-form is
being constructed) from among n variables. Note that one must have k ≤ n (and
similarly k ≤ m), since otherwise the number of variables would not be enough.
Then the differentials in the expression (5.2.22) would have to be repeated, and
thus the entire expression would vanish. The situation of repeated differentials was
excluded immediately, by writing 1 ≤ j1 < . . . < jk ≤ n, rather than j1 , . . . , jn =
1, . . . , n.
The above procedure will become clear when it is applied to the current exercise,
to which we proceed below. The variables x1 , x2 , x3 , x4 now bear the names
x, y, z, w. Similarly, in place of the variables y1 , y2 , y3 , y4 one has ξ, η, ζ, τ . Of
course, as already mentioned, there is m = n = 4 and k = 2.
In order to find the mapping , one has to reverse the relations between x, y, z, w
and ξ, η, ζ, τ provided in the text of the exercise. This is a very simple task and one
gets
1
(ξ + η + ζ + τ ) =: ϕ1 (ξ, η, ζ, τ ),
2
1
y = (−ξ + η + ζ + τ ) =: ϕ2 (ξ, η, ζ, τ ),
2
1
z = (−ξ − η + ζ + τ ) =: ϕ3 (ξ, η, ζ, τ ),
2
1
w = (−ξ − η − ζ + τ ) =: ϕ4 (ξ, η, ζ, τ ).
2
x=
(5.2.23)
The set U is here the entire space R4 , similarly as V .
Now we are going to find the “pulled-back” coefficient functions. The first one is
ωxz (x, y, z, w) = x + y,
(5.2.24)
the index xz coming from the fact that it stands in front of dx ∧ dz (in full
accordance with (5.2.18), one could equally well write 1 3 instead of xz). After
changing the variables, one gets
ωxz ((ξ, η, ζ, τ )) = ϕ1 (ξ, η, ζ, τ ) + ϕ2 (ξ, η, ζ, τ ) = η + ζ + τ,
(5.2.25)
5.2 Performing Various Operations on Differential Forms
139
and similarly
ωyz ((ξ, η, ζ, τ )) = ϕ3 (ξ, η, ζ, τ ) + ϕ4 (ξ, η, ζ, τ ) = −ξ − η + τ,
ωxw ((ξ, η, ζ, τ )) = ϕ1 (ξ, η, ζ, τ ) − ϕ2 (ξ, η, ζ, τ ) = ξ,
(5.2.26)
with all other functions vanishing.
Now we are going to find the Jacobian determinants. Unfortunately, there are
quite a lot, although very easy to calculate. A few selected ones will be computed
below, and for the others, only the results will be given. Rather than ϕ1 , ϕ2 , ϕ3 , and
ϕ4 in Jacobian matrices, the symbols x, y, z, and w are successively used. When
looking at the expression (5.2.17), one sees that we must first find the determinants:
∂(x, z)
∂x/∂ξ ∂x/∂η
1/2 1/2
= det
= det
= 0,
∂z/∂ξ ∂z/∂η
−1/2 −1/2
∂(ξ, η)
∂(x, z)
1
∂x/∂ξ ∂x/∂ζ
1/2 1/2
= det
= det
= ,
∂z/∂ξ ∂z/∂ζ
−1/2 1/2
∂(ξ, ζ )
2
∂(x, z)
1
∂x/∂ξ ∂x/∂τ
1/2 1/2
= det
= det
= ,
∂z/∂ξ ∂z/∂τ
−1/2 1/2
∂(ξ, τ )
2
∂(x, z)
1
∂x/∂η ∂x/∂ζ
1/2 1/2
= det
= det
= ,
∂z/∂η ∂z/∂ζ
−1/2 1/2
∂(η, ζ )
2
∂(x, z)
1
∂x/∂η ∂x/∂τ
1/2 1/2
= det
= det
= ,
∂z/∂η ∂z/∂τ
−1/2 1/2
∂(η, τ )
2
∂(x, z)
∂x/∂ζ ∂x/∂τ
1/2 1/2
= det
= det
= 0.
∂z/∂ζ ∂z/∂τ
1/2 1/2
∂(ζ, τ )
(5.2.27)
One still needs two similar sets of determinants for the pairs of variables y, z
and x, w. They can be calculated in the same way, so only their values are listed
below. For the first pair, one has
1
∂(y, z)
= ,
∂(ξ, η)
2
∂(y, z)
= 0,
∂(ξ, ζ )
∂(y, z)
= 0,
∂(ξ, τ )
∂(y, z)
1
= ,
∂(η, ζ )
2
∂(y, z)
1
= ,
∂(η, τ )
2
∂(y, z)
= 0,
∂(ζ, τ )
∂(x, w)
= 0,
∂(ξ, η)
∂(x, w)
= 0,
∂(ξ, ζ )
∂(x, w)
1
= ,
∂(ξ, τ )
2
∂(x, w)
= 0,
∂(η, ζ )
∂(x, w)
1
= ,
∂(η, τ )
2
∂(x, w)
1
= .
∂(ζ, τ )
2
(5.2.28)
and for the second one
(5.2.29)
140
5 Examining Differential Forms
We already dispose of all components, which must be inserted into the formula (5.2.21). Omitting the terms, for which the corresponding determinants vanish,
one finds
2
1
(η + ζ + τ ) (dξ ∧ dζ + dξ ∧ dτ + dη ∧ dζ + dη ∧ dτ )
2
1
+ (−ξ − η + τ ) (dξ ∧ dη + dη ∧ dζ + dη ∧ dτ )
2
1
+ ξ (dξ ∧ dτ + dη ∧ dτ + dζ ∧ dτ )
2
1
1
= (−ξ − η + τ ) dξ ∧ dη + (η + ζ + τ ) dξ ∧ dζ
2
2
1
1
+ (η + ζ + τ ) dξ ∧ dτ + (−ξ + ζ + 2τ ) dη ∧ dζ
2
2
1
1
(5.2.30)
+ (ζ + 2τ ) dη ∧ dτ + ξ dζ ∧ dτ.
2
2
∗ ω =
Applying the formula (5.2.21) is then quite simple, although it can be tedious. An
eventual alternative constitutes using of the formula analogous to (5.1.26), applied
to the variables x, y, z, w and ξ, η, ζ, τ in the following way:
dx =
∂x
∂x
∂x
∂x
1
1
1
1
dξ +
dη +
dζ +
dτ = dξ + dη + dζ + dτ,
∂ξ
∂η
∂ζ
∂τ
2
2
2
2
dy =
∂y
∂y
∂y
∂y
1
1
1
1
dξ +
dη +
dζ +
dτ = − dξ + dη + dζ + dτ,
∂ξ
∂η
∂ζ
∂τ
2
2
2
2
dz =
∂z
∂z
∂z
∂z
1
1
1
1
dξ +
dη +
dζ +
dτ = − dξ − dη + dζ + dτ,
∂ξ
∂η
∂ζ
∂τ
2
2
2
2
dw =
∂w
∂w
∂w
∂w
1
1
1
1
dξ +
dη +
dζ +
dτ = − dξ − dη − dζ + dτ.
∂ξ
∂η
∂ζ
∂τ
2
2
2
2
(5.2.31)
Now, instead of calculating the Jacobian determinants, the properties of the exterior
product known from the previous problems can be used. In particular
dx ∧ dz =
=
1
1
1
1
dξ + dη + dζ + dτ
2
2
2
2
∧
1
1
1
1
− dξ − dη + dζ + dτ
2
2
2
2
1
(dξ ∧ dζ + dξ ∧ dτ + dη ∧ dζ + dη ∧ dτ )
2
(5.2.32)
5.2 Performing Various Operations on Differential Forms
141
and
dy ∧ dz =
1
1
1
1
1
1
1
1
− dξ + dη + dζ + dτ ∧ − dξ − dη + dζ + dτ
2
2
2
2
2
2
2
2
1
(dξ ∧ dη + dη ∧ dζ + dη ∧ dτ ) ,
2
1
1
1
1
1
1
1
1
dξ + dη + dζ + dτ ∧ − dξ − dη − dζ + dτ
dx ∧ dw =
2
2
2
2
2
2
2
2
=
=
1
(dξ ∧ dτ + dη ∧ dτ + dζ ∧ dτ ) .
2
(5.2.33)
If one already has the “pulled-back” coefficient functions (formulas (5.2.25)
and (5.2.26)), it simply remains to plug the obtained expressions into (5.2.17) and
after some rearrangement of terms the result (5.2.30) is again derived. As the reader
surely notices, the advantage of this method consists of avoiding the calculation of
a large number of determinants.
Problem 5
The transformations of the electric and the magnetic fields to the coordinate system,
which moves along the x-axis with velocity v, will be found.
Solution
In Chap. 6, we will be concerned with the integrals of differential forms. Then the
reader will be accustomed to such concepts as “work form” or “flux form.” It will
become clear that the former is a 1-form, and the work is given by its integral over
a one-dimensional oriented surface (i.e., oriented curve). As to the flux form, it is a
2-form, and the flux of some physical quantity can be found by integrating it over a
sheet of a two-dimensional oriented surface. In general k-forms shall be integrated
over k-surfaces only. This integration is not the subject of our current interest. These
facts are mentioned here only because we would like to associate certain forms with
the electric and magnetic fields, and we need to decide how it should be done.
Imagine that a
First, let us have a look at the vector of the electric field E.
certain trial charge q is being shifted in this field. Apart from the coefficient, which
constitutes the value of this charge, the electric field is actually the same as the force.
In turn, the force having been integrated over the curve, along which the charge is
being shifted, corresponds to the work performed on the charge. Therefore, it is
natural to associate with the electric field a 1-form (a work form) in the following
way:
142
5 Examining Differential Forms
1
ωE = Ex dx + Ey dy + Ez dz.
(5.2.34)
On the other hand, a charged particle moving in a magnetic field B is subject to
the Lorentz force:
F = q v × B,
(5.2.35)
v standing for the particle velocity. This force is always perpendicular to v and
it cannot change the kinetic energy. Therefore, the magnetic field may not be
associated with that kind of form. The natural object here is rather the flux form,
i.e., the 2-form:
2
ωB = Bx dy ∧ dz + By dz ∧ dx + Bz dx ∧ dy.
(5.2.36)
The reader should pay attention to the “noncanonical” order of differentials
dz and dx in the second term. It will be justified in Sect. 5.5, and now the
formula (5.2.36) should be accepted as a rule, according to which one associates a 2form with a vector (in R3 ). The formula (5.2.36) can be justified also in another way,
namely by using a so-called inner product of a 2-form and a vector (here velocity
vector), but this is a topic not dealt with in this book.
Using the above forms and also others associated with the sources (charge
density, current density), one can rewrite in a compact way all equations of
electromagnetism. However, in the relativistic formulation, in which electric and
magnetic fields can swap their roles, a different association becomes more useful.
Both fields should be then treated on an equal footing and to both certain 2-forms
1
ought to be attributed. The 1-form ωE is very easily transformed into a 2-form by
taking the exterior product with one of the differentials dt, dx, dy, dz. Only the
first one comes into play, because other combinations would lead to those already
present in (5.2.36).
Let us then define the following, very useful 2-form, unifying the electric and
magnetic fields:
2
1
2
ωF := dt ∧ ωE + ωB .
(5.2.37)
When the reader is accustomed to exterior derivatives of forms (in the next section),
they will be easily convinced that the pair of the sourceless Maxwell’s equations can
be given the simple form:
2
d ωF = 0.
(5.2.38)
The second pair of equations can also be written in a similar manner, but this
requires the introduction of the so-called “Hodge-star” operation, not dealt with in
this book. The reader who feels unsatisfied with the justification of the form (5.2.37)
5.2 Performing Various Operations on Differential Forms
143
may treat it as “guessed” and legitimated a posteriori by the fact of its suitability for
the formulation of the equations of electromagnetism.
The relativistic transformations of the fields E and B to a system U moving
relative to an observer at rest in U can now be obtained by considering the “pullback” of the above form from U to U , the mapping being given as:
t + vx /c2
= γ (t + βx /c),
t= 2
2
1 − v /c
x + vt x= = γ (x + βt ),
1 − v 2 /c2
y = y,
z = z .
(5.2.39)
These formulas, relating the coordinates (t, x, y, z) in the system U with
(t , x , y , z ) in U , are the famous Lorentz transformations, well known to every
physicist, for the system U moving at a constant velocity v pointing along the
x-axis, where the axes of both systems remain parallel. The standard denotation has
been introduced here:
γ =
1
,
1 − v 2 /c2
β=
v
c
(5.2.40)
where c is the speed of light.
In order to find the electromagnetic fields in the system U , one can just read
2
them off from the “pulled back” form ∗ ωF . After this “pulling back” it has to
have the following form:
2
∗ ωF = Ex dt ∧ dx + Ey dt ∧ dy + Ez dt ∧ dz
−Bx dy ∧ dz − By dz ∧ dx − Bz dx ∧ dy .
(5.2.41)
On the other hand, this transformation can be explicitly performed, as described in
the previous problems, and the final result compared to (5.2.41). To this end, using
the relations (5.2.39), one can write
dt = γ (dt + β dx /c),
dx = γ (dx + βc dt ),
dy = dy ,
dz = dz ,
(5.2.42)
144
5 Examining Differential Forms
and then insert these differentials into (5.2.37). This is the alternative way of proceeding as compared to the calculation of a large number of Jacobian determinants,
already applied in the previous exercise. In this way one gets
2
∗ ωF = Ex γ (dt + β dx /c) ∧ γ (dx + βc dt ) + Ey γ (dt + β dx /c) ∧ dy +Ez γ (dt + β dx /c) ∧ dz − Bx dy ∧ dz
−By dz ∧ γ (dx + βc dt ) − Bz γ (dx + βc dt ) ∧ dy .
(5.2.43)
Now, one has to patiently reorganize all terms and write the result in the canonical
form. One should also remember that γ 2 (1 − β 2 ) = 1. The outcome is
2
∗ ωF = Ex (γ 2 − γ 2 β 2 ) dt ∧ dx + Ey γ (dt ∧ dy + β/c dx ∧ dy )
+Ez γ (dt ∧ dz + β/c dx ∧ dz ) − Bx dy ∧ dz
−By γ (dz ∧ dx − βc dt ∧ dz ) − Bz γ (dx ∧ dy + βc dt ∧ dy )
= Ex dt ∧ dx + γ (Ey − βcBz ) dt ∧ dy + γ (Ez + βcBy ) dt ∧ dz
Ex
Ey
Ez
− Bx dy ∧ dz − γ (By + β/c Ez ) dz ∧ dx Bx
By
− γ (Bz − β/c Ey ) dx ∧ dy .
(5.2.44)
Bz
Comparing this expression with (5.2.41), one finds the relations that are needed:
Ex = Ex ,
Ey = γ (Ey − βcBz ),
Bx = Bx ,
By = γ (By + β/c Ez ),
Ez = γ (Ez + βcBy ),
Bz = γ (Bz − β/c Ey ). (5.2.45)
These are the general formulas. In the case of specific fields that depend in a known
way on the positions and time in U , the unprimed arguments of fields standing on the
right-hand side ought to be replaced with the expressions (5.2.39)—in accordance
with the formula (5.2.21)
In the same way, one can transform the four-potential:
A = [A0 , A1 , A2 , A3 ],
(5.2.46)
by associating with it a 1-form:
1
ωA = A0 dt − A1 dx − A2 dy − A3 dz.
This task is left to the reader.
(5.2.47)
5.3 Calculating Exterior Derivatives
145
5.3 Calculating Exterior Derivatives
Problem 1
The exterior derivative of the 1-form in R2 :
1
ω= xy dx + y 2 dy
(5.3.1)
will be found.
Solution
In this section, we will become familiar with the exterior differentiation of forms. As
the reader knows from the theoretical introduction and probably from the lecture of
analysis, the exterior derivative of a k-form in Rn defined with the general formula,
n
!
k
ω = ωi1 ,i2 ,...,ik (x1 , . . . , xn ) dxi1 ∧ . . . ∧ dxik ,
(5.3.2)
i1 ,i2 ,...,ik =1
i1 <i2 <...<ik
is a form of degree k + 1 obtained through the differentiation of the coefficient
functions ωi1 ,i2 ,...,ik (x1 , . . . , xn ) according to the rule:
k
dω =
n
!
∂ωi
j,i1 ,i2 ,...,ik =1
(x1 , . . . , xn )
dxj ∧ dxi1 ∧ . . . ∧ dxik .
∂xj
1 ,i2 ,...,ik
(5.3.3)
i1 <i2 <...<ik
As one can see, the additional differential (dxj ) has appeared (and has been
placed at the beginning of all differentials) together with the sum over j = 1, . . . , n.
It does not mean, however, that the new form will contain n times more terms
k
than ω. It will often happen that this new differential is already present among
dxi1 , . . . , dxik , which implies that the corresponding term vanishes. This rule has
already appeared in the context of exterior products of differential forms.
If such a repetition does not occur, however, the given term will “survive” and
then it is worth writing it in the canonical form by shifting dxj to the appropriate
place. Surely one has to keep in mind the necessity of inverting the sign at each
“dragging” of differentials. These general principles are going to be followed in this
and in the following problems.
146
5 Examining Differential Forms
In the current example, we are dealing with a 1-form and, therefore, as a result
of the differentiation a 2-form will be obtained. This calculation is much easier than
the general formula (5.3.3) would suggest. In our case, n = 2 and the variables x1
and x2 bear the names of x and y, so one has
1
dω=
∂(xy)
∂(y 2 )
∂(xy)
∂(y 2 )
dx
∧
dx
+
dx
∧
dy
+
dy
∧
dx
+
dy ∧ dy
∂x
∂y
∂y ∂x
=0
=0
=y
=0
=x
= x dy ∧ dx = −x dx ∧ dy.
=2y
(5.3.4)
As one can see, a very simple result has been obtained due to the announced
repetitions of differentials in the exterior products, as well as the vanishing of
the derivatives of the coefficient functions. This kind of a result should have been
expected. After all, as an outcome of the integration, one gets a 2-form in the twodimensional space. In such a case, we have at our disposal only two differentials:
dx and dy. The only object that can be constructed out of them is simply
ω(x, y) dx ∧ dy
(5.3.5)
with certain coefficient function ω(x, y), equal to −x in our case.
Problem 2
The exterior derivative of the 2-form in R3 :
2
ω= (x + z) dx ∧ dy + exyz dz ∧ dx + x 2 dy ∧ dz
(5.3.6)
will be found.
Solution
In the present problem, we are to calculate the exterior derivative of a 2-form in R3 .
The result must be, as we know, a form of degree 3. As in R3 , the only differentials
at our disposal are dx, dy, and dz, and they are not allowed to be repeated, so one
certainly must have
2
d ω= ω(x, y, z) dx ∧ dy ∧ dz.
(5.3.7)
5.3 Calculating Exterior Derivatives
147
The problem is, therefore, reduced to that of determining the coefficient factor
ω(x, y, z).
Aimed as simplifying the calculations as much as possible, it should be noted
that the first term of the form (5.3.6) can be differentiated solely over z (since it
already contains the differentials of two other variables), the second one over y, and
the last one over x. Other expressions can be skipped, because they will surely prove
to vanish. Thereby, one gets
2
dω=
∂ (exyz )
∂ (x 2 )
∂ (x + z)
dz ∧ dx ∧ dy +
dy ∧ dz ∧ dx +
dx ∧ dy ∧ dz
∂z
∂y
∂x
= dz ∧ dx ∧ dy + xzexyz dy ∧ dz ∧ dx + 2x dx ∧ dy ∧ dz
$
%
= 1 + xzexyz + 2x dx ∧ dy ∧ dz.
(5.3.8)
Problem 3
The exterior derivative of the 3-form in R3 :
3
ω= xyz dx ∧ dy ∧ dz
(5.3.9)
will be found.
Solution
If the reader has attentively analyzed two recent examples, they should not have
any problems with solving this exercise even mentally. We are to differentiate a 3form and, therefore, a 4-form will be obtained. However, in R3 there are only three
differentials at our disposal: dx, dy, and dz. There is no way to construct a 4-form,
without repeating any of them. Thereby, the result of the differentiation must be
equal to zero:
3
d ω= 0.
(5.3.10)
One can easily confirm this conclusion with explicit calculations. For, we have
3
dω=
∂ (xyz)
∂ (xyz)
dx ∧ dx ∧ dy ∧ dz +
dy ∧ dx ∧ dy ∧ dz
∂x ∂y =0
+
∂ (xyz)
dz ∧ dx ∧ dy ∧ dz = 0.
∂z =0
=0
(5.3.11)
148
5 Examining Differential Forms
Problem 4
The exterior derivative of the k-form in RN :
2
ω= x1 x2 x3 dx2 ∧ dx3 + x2 x3 x4 dx3 ∧ dx4 + . . . + xN −2 xN −1 xN dxN −1 ∧ dxN
(5.3.12)
2
will be found, as well as the second derivative of this form, i.e., d 2 ω.
Solution
Like the previous one, this problem should not pose any difficulties. The form is the
sum of terms of the kind
xi−2 xi−1 xi dxi−1 ∧ dxi ,
where i = 3, . . . , N.
(5.3.13)
The coefficient functions depend only on three variables each, so the derivatives
with respect to the other ones are trivial. From among the three differentiations that
come into play, two lead to the repetitions of differentials (dxi−1 and dxi ). The only
nonzero term which emerges after the differentiation of (5.3.13) is, therefore,
d(xi−2 xi−1 xi dxi−1 ∧ dxi ) = xi−1 xi dxi−2 ∧ dxi−1 ∧ dxi ,
(5.3.14)
2
having already been written in the canonical form. The entire form ω is the sum of
similar terms but for i = 3, . . . , N , so one gets
2
d ω = x2 x3 dx1 ∧ dx2 ∧ dx3 + x3 x4 dx2 ∧ dx3 ∧ dx4 + . . .
+xN −1 xN dxN −2 ∧ dxN −1 ∧ dxN .
(5.3.15)
2
Now, if one wished to calculate the subsequent exterior derivative, i.e., dd ω
2
(or d 2 ω), it is easy to see that the outcome must vanish. The coefficient
functions in (5.3.15) depend only on variables that are already present among the
accompanying differentials. The differentiation must then lead to their repetitions in
each consecutive term, which implies:
2
d 2 ω= 0.
(5.3.16)
The result (5.3.16), while obtained by using a specific form, turns out to be
k
always legitimate, i.e., for any form ω. This is because even if single terms do not
vanish as in the example above, their cancellations are inevitable. Let us consider
5.4 Looking for Primitive Forms
149
any coefficient function ω(x1 , x2 , . . . , xN ). After having been differentiated twice
it generates the contribution in the form:
N
!
i,j =1
∂ 2ω
dxj ∧ dxj .
∂xi ∂xj
(5.3.17)
Now, since i and j both run from 1 to N , so the terms containing the same
differentials occur twice: once as dxn ∧ dxm , and for the second time as dxm ∧ dxn .
As we know, these terms differ only by the sign. In turn, we remember from Part II
that for a function which is differentiable twice, the sequence of derivatives is
inessential:
∂ 2ω
∂ 2ω
=
.
∂xn ∂xm
∂xm ∂xn
(5.3.18)
Consequently this always yields
∂ 2ω
∂ 2ω
dxn ∧ dxm +
dxm ∧ dxn = 0.
∂xn ∂xm
∂xm ∂xn
(5.3.19)
This fact will be used in the next section.
5.4 Looking for Primitive Forms
Problem 1
The primitive form for the following 1-form in R2 :
1
ω= (2x + y 2 ) dx + 2xy dy
(5.4.1)
will be found.
Solution
In this problem and in the subsequent ones, we will be concerned with, in a sense
k
opposite issue of the previous section. Suppose that one has a k-form ω. One is then
faced with two questions:
150
5 Examining Differential Forms
k−1
1. Is this form the exterior derivative of a k − 1-form, i.e., does a form σ exist
k
k−1
such that d σ =ω? And, if so,
k−1
2. How does one find σ in an explicit way?
k
If such a form does exist, it will be called the primitive form, and ω is told to
k−1
be exact. Surely there appears also an additional question, whether the form σ we
have found is unique.
The knowledge gained when solving the last exercise suggests a rather obvious
k
preliminary test. If the form ω is to be exact, its exterior derivative must vanish:
k
k−1
k−1
d ω= d(d σ ) = d 2 σ = 0,
(5.4.2)
as the second derivative of any form is zero. This implication, at least for the time
being, is effective only in one way: if the form is exact, then its derivative vanishes,
or, as one says, the form is closed. This does not mean yet that every closed form
must be exact, but the form, which is not closed, cannot have the primitive one.
Therefore, it is a good idea to start looking for a primitive form by verifying if
k
d ω= 0.
In the case of (5.4.1), omitting the terms, which would be found to vanish due to
the repetition of the identical differentials, one obtains
1
dω=
∂ (2xy)
∂ (2x + y 2 )
dy ∧ dx +
dx ∧ dy
∂y
∂x
= −2y dx ∧ dy + 2y dx ∧ dy = 0.
(5.4.3)
1
The form ω has proved to be closed, so it can also be exact. The question is:
are we sure it must? What comes to our aid is the so-called Poincaré lemma, which
may have various wordings, but it can be easily stated in the following manner: each
closed form in the area contractible to a point is exact, i.e., it does have a primitive
form. It should be recalled that this area is such that can be continuously deformed
to a point. For example, on a plane it may not have any “hole,” because a curve
encircling it could not be contracted to a point. If, however, the form is thought of
only locally and, thereby, one is free of the global topological properties, then any
closed form is exact. This expression of the Poincaré lemma can also be encountered
by the reader. In the following problems, if the area is not explicitly stated, we will
always have in mind the “local” exactness.
We already know that the form (5.4.1) has a primitive form. There still remains
1
an issue of how to find it. Well, since ω is the 1-form, the primitive form must be
a 0-form, which simply means a scalar function. Let us denote it with the symbol
(x, y). The 1-form in R2 can always be written as
5.4 Looking for Primitive Forms
151
1
ω= P (x, y) dx + Q(x, y) dy.
(5.4.4)
The coefficient functions P and Q can easily be read off from (5.4.1). Since
d =
∂
∂
1
dx +
dy = ω,
∂x
∂y
(5.4.5)
the following equations must hold:
∂ /∂x = P (x, y),
∂ /∂y = Q(x, y).
(5.4.6)
It is a system of two differential equations for an unknown function (x, y). Let us
examine these equations, substituting
P (x, y) = 2x + y 2 ,
Q(x, y) = 2xy.
(5.4.7)
At first the following equation is integrated with respect to x
∂
= 2x + y 2 ,
∂x
(5.4.8)
(x, y) = x 2 + xy 2 + C(y).
(5.4.9)
getting as a result
The symbol C stands here for an integration “constant” (from the point of view of
the integral over x), which in general can depend on the other variable (which could
be temporarily called a “parameter”), i.e., y. The function (x, y) as obtained above
should now be plugged into the second of the equations (5.4.6). This leads to
∂ 2
x + xy 2 + C(y) = 2xy,
∂y
(5.4.10)
i.e.,
2xy + C (y) = 2xy
⇒ C (y) = 0.
(5.4.11)
The quantity C has turned out to be actually a constant. Then, the primitive form
looked for is
(x, y) = x 2 + xy 2 + C
(5.4.12)
and is determined up to an additive constant. The reader is encouraged to check by
a direct calculation that it actually satisfies the Eq. (5.4.5).
152
5 Examining Differential Forms
Instead of solving the differential equations, the desired result (5.4.12) can also
be obtained from the following integral formula to be generalized to the greater
number of dimensions in the following examples. It has the following form:
y
x
(x, y) =
Q(x, y ) dy ,
P (x , y0 ) dx +
x0
(5.4.13)
y0
where (x0 , y0 ) = (x, y) is some point on the plane arbitrarily chosen. Upon
inserting the functions P and Q as given by (5.4.7) into the above formula, one
gets
y
x
(x, y) =
(2x
+ y02 ) dx x0
+
2xy dy = (x
y0
2
x
2 + x y0 )
x0
y
2 + xy y0
= x 2 + xy02 − x02 − x0 y02 + xy 2 − xy02 = x 2 + xy 2 −x02 − x0 y02 .
C
(5.4.14)
As one can see, the same result as before has been derived, and the role of the
“mysterious” point (x0 , y0 ) has boiled down to merely generate a constant. Note that
all terms that “mix” the end-point coordinates (x, y) and those of the initial one, i.e.,
(x0 , y0 ), have canceled out. It was not coincidental, since otherwise it would not be
possible to write the primitive form as the sum of a clearly defined function of x
and y and a constant. If this type of “mixed” term remained in the final formula, this
should provoke the reader to check his calculations for errors or to make sure that
the form (5.4.4) under examination has a primitive form at all.
The formula (5.4.13) which has been used is not the only possible one. If we
imagine a rectangle ABCD on the plane with vertices at the points A(x0 , y0 ),
B(x, y0 ), C(x, y), and D(x0 , y) (see Fig. 5.2 in the following exercise), then the
integrals (5.4.13) describe the path from A to C along two sides (AB and BC) of
this rectangle. Equally well, however, one could move along the sides AD and DC:
y
x
(x, y) =
P (x , y) dx ,
Q(x0 , y ) dy +
y0
(5.4.15)
x0
along the diagonal AC of the rectangle or even any other curve connecting the
vertices A and C. Each time the same result would be obtained, with one exception
mentioned earlier in the context of the Poincaré lemma. If the functions P and Q
were not defined in the contractible area, for example, if they were undefined at
some point O of the plane (and, therefore, the domain of the form would have a
“hole”), the curves of ends at A and C encircling the point O from different sides
(i.e., such that cannot be converted into each other) would lead to different values of
the integral (5.4.13). An example will be given in the next exercise.
5.4 Looking for Primitive Forms
153
It still remains at the end to clarify how one knows that the primitive form for a
1-form can always be obtained with the use of the formula (5.4.13). To justify it, let
us simply calculate d:
⎤
⎡ x
⎡ x
y
∂ ⎣
∂
⎣ P (x , y0 ) dx d =
P (x , y0 ) dx + Q(x, y ) dy ⎦ dx +
∂x
∂y
x0
y0
⎤
y
⎡
x0
y
Q(x, y ) dy ⎦ dy= ⎣P (x, y0 )+
+
y0
⎤
∂ Q(x, y ) ⎦
dy dx + Q(x, y) dy.
∂x
y0
(5.4.16)
The obvious fact that
x
d
dx
f (x )dx =
d
[F (x) − F (a)] = F (x) = f (x)
dx
a
has been used here, and F stands for the primitive function for f .
In order to further simplify the expression (5.4.16), it is sufficient to note that the
closeness condition for the form (5.4.4), without satisfying of which our problem
does not make sense at all, actually constitutes the requirement that for any x, y,
∂ P (x, y)
∂ Q(x, y)
=
.
∂y
∂x
(5.4.17)
If so, in (5.4.16) in place of ∂Q(x, y )/∂x one can insert ∂P (x, y )/∂y , which
yields
⎡
⎤
y
)
∂
P
(x,
y
d = ⎣P (x, y0 ) +
dy ⎦ dx + Q(x, y) dy
∂y y0
= [P (x, y0 ) + P (x, y) − P (x, y0 )] dx + Q(x, y) dy
1
= P (x, y) dx + Q(x, y) dy =ω .
(5.4.18)
Problem 2
The primitive form for the 1-form in R2 \ {(0, 0)}:
1
ω=
will be found.
y
x
dx − 2
dy
x2 + y2
x + y2
(5.4.19)
154
5 Examining Differential Forms
Solution
In accordance with the procedure laid down in the previous problem, we begin with
1
verifying whether the form ω is closed. For this purpose, one has to calculate the
exterior derivative (the vanishing terms are omitted):
1
dω=
∂
∂y
y
x2 + y2
∂
dy ∧ dx −
∂x
x
x2 + y2
dx ∧ dy
=
x 2 + y 2 − 2y 2
x 2 + y 2 − 2x 2
dy
∧
dx
−
dx ∧ dy
(x 2 + y 2 )2
(x 2 + y 2 )2
=
y2 − x2
y2 − x2
dx
∧
dy
−
dx ∧ dy = 0.
(x 2 + y 2 )2
(x 2 + y 2 )2
(5.4.20)
In order to find the primitive form , which is a 0-form, i.e., a function, the
formula (5.4.13) will be used, where
P (x, y) =
y
,
x2 + y2
Q(x, y) = −
x
.
x2 + y2
(5.4.21)
Let us now choose any point (x0 , y0 ), assuming, however, that x0 = 0 and y0 = 0.
The identical assumptions will be also adopted with respect to the point (x, y).
Otherwise, the sides of the rectangle, over which the integral is taken, could pass
through the origin, where the functions P and Q are not defined. These assumptions
are not essential for our future conclusions. They are merely ancillary and the whole
issue can be reformulated in a way avoiding them. However, the solution would then
become more complicated from the technical point of view.
Exploiting the formula (5.4.13) which corresponds to the path γ1 in Fig. 5.2, and
using the known value for the integral (for a = 0):
x2
x
1
1
dx = arctan ,
2
a
a
+a
(5.4.22)
one obtains
y
x
(x, y) =
x0
y0
dx +
x 2 + y02
y0
y y
−x
y0
x x
x
arctan
dy
=
arctan
−
y0
y0 x0 x
x y 0
x 2 + y 2
y0
x
x0
y
= arctan
− arctan
− arctan + arctan .
y0
y0
x
x
(5.4.23)
The first and the last terms of the above expression appear to apparently mix the
coordinates of both ends of the curve γ1 , but due to the identity:
5.4 Looking for Primitive Forms
155
Fig. 5.2 The example of the mutual position of points (x0 , y0 ), (x, y), and the origin on a plane
together with the curves γ1 , γ2 , and γ3
arctan u + arctan
π
1
=± ,
u
2
(5.4.24)
where the sign on the right-hand side is the same as that of u, they add to a constant.
If we were not attentive enough and did not notice that the domain of interest (i.e.,
the plane with the point (0, 0) removed), is not contractible to a point and, therefore,
does not meet the assumptions of the Poincaré lemma, we would be satisfied with
the result (5.4.23). However, we are going to continue and find the quantity (x, y)
once again, this time using the formula (5.4.15), i.e., integrating over the curve γ2 :
y
(x, y) =
y0
−x0
dy +
2
x0 + y 2
x
x0
x x
y
x0
y y y
arctan
dx
=
−
arctan
+
x0
x0 y0 y
y x0
x 2 + y 2
x0
y0
x
y
+ arctan
+ arctan − arctan .
= −arctan
x0
x0
y
y
(5.4.25)
The expressions obtained (i.e., (5.4.23) and (5.4.25)) appear to differ, which
should arouse our concern. To investigate it, let us calculate their difference denoted
with the symbol , by subtracting them from each other:
156
5 Examining Differential Forms
= arctan
y0
x
x0
y
− arctan
− arctan + arctan
y0
y0
x
x
x0
y
y0
x
− −arctan
+ arctan
+ arctan − arctan
x0
x0
y
y
y
x
y0
x
= − arctan + arctan
+ arctan
+ arctan
x
y
y0
x
x0
y
x0
y0
.
− arctan
+ arctan
+ arctan
+ arctan
x0
y
y0
x0
(5.4.26)
Since there are several possibilities of the mutual position of the points (x0 , y0 ),
(x, y), and (0, 0) on the plane, let us, for example, choose that which is shown in
Fig. 5.2. The reader will be certainly able to analyze all other variants. Consider now
the expressions in square brackets of (5.4.26).
1. Surely x/y > 0, so pursuant to (5.4.24)
arctan
y
x
π
+ arctan = .
x
y
2
2. One also has x/y0 < 0 and
arctan
x
π
y0
=− .
+ arctan
y0
x
2
3. Analogously y/x0 < 0, so
arctan
π
x0
y
=− .
+ arctan
x0
y
2
4. And finally x0 /y0 > 0, which gives
arctan
y0
π
x0
+ arctan
= .
y0
x0
2
Collecting all these results, we find
=−
π
π
π
π
− − − = −2π = 0.
2
2
2
2
(5.4.27)
Thus, the expressions obtained for differ by the constant −2π . It is true that
the primitive form is determined up to a constant, but once the starting point (x0 , y0 )
has been fixed, this freedom disappears and one should get a unique expression. The
1
reason for such abnormal behavior is known: it is the fact that the form ω was not
defined in a contractible domain (the “hole” at the origin). The primitive form exists only locally. In particular, the results obtained from the integrations (5.4.13)
5.4 Looking for Primitive Forms
157
and (5.4.15) over paths evading the “hole” from different sides cannot be considered
as identical. The curves γ1 and γ2 are not contained in a domain contractible to a
point, since each such set would have to inevitably include the “hole.”
However, it is easy to predict what would be obtained if the chosen curve where
denoted with γ3 (e.g., it could be the diagonal AC of the rectangle). Naturally the
result would be given by (5.4.25), as the curves γ2 and γ3 lie in the contractible set
(marked in the figure in gray). In this area, the primitive form exists not only locally
but even globally.
The reader might now raise the following question: since the 0-form can
be determined, even if it is just locally, ultimately which formula for it is correct: (5.4.23) or (5.4.25)? To resolve this, let us assume that the point A is very
close to C (as close as we will require). If the form is to be treated only locally,
this choice is justified. In that case, one can regard x/y, x0 /y, y0 /x, and x0 /y0
as positive and, as a result, all expressions in (5.4.26) inside the square brackets
are equal to π/2. Consequently = 0, which means that the expressions (5.4.23)
and (5.4.25) are (locally) identical.
When considering a very similar example, namely
1
σ=
x2
x
y
dx + 2
dy,
2
+y
x + y2
(5.4.28)
which is also defined on R2 \ {(0, 0)}, the reader can easily find out that integrals
evading the origin from other sides do not have to be different, even if the Poincaré
lemma cannot be applied. First, let us check that this form is closed:
∂
x
y
∂
1
dy ∧ dx +
dx ∧ dy
dσ =
∂y x 2 + y 2
∂x x 2 + y 2
=−
=
(x 2
2xy
2xy
dy ∧ dx − 2
dx ∧ dy
2
2
+y )
(x + y 2 )2
2xy
2xy
dx ∧ dy − 2
dx ∧ dy = 0.
(x 2 + y 2 )2
(x + y 2 )2
(5.4.29)
The primitive form can now be found with the use of the formula (5.4.13):
x
(x, y) =
x0
x
dx +
x 2 + y02
y
y0
y
dy x 2 + y 2
x
y
1
1
2
2 2
2 = log(x + y0 ) + log(x + y )
2
2
x0
y0
1
log(x 2 + y02 ) −
2
1
= log(x 2 + y 2 ) −
2
=
1
1
1
log(x02 + y02 ) + log(x 2 + y 2 ) − log(x 2 + y02 )
2
2
2
1
log(x02 + y02 ),
(5.4.30)
2
158
5 Examining Differential Forms
and then (5.4.15):
y
(x, y) =
y0
y
dy +
2
x0 + y 2
x
x 2
x0
x
dy + y2
y
x
1
1
2
2 2
2 = log(x0 + y ) + log(x + y )
2
2
y0
x0
1
log(x02 + y 2 ) −
2
1
= log(x 2 + y 2 ) −
2
=
1
1
1
log(x02 + y02 ) + log(x 2 + y 2 ) − log(x02 + y 2 )
2
2
2
1
log(x02 + y02 ).
(5.4.31)
2
Both results are then identical. They will also be the same for each path joining
the points A and C and not passing through the origin. Any other path is contained
in a certain contractible area either with the curve γ1 , or with γ2 (unless it makes
a “loop” around O, but in this case one can show that the contribution of the loop
vanishes). This result does not, however, contradict the Poincaré lemma. Just like
any other, this implication states what happens if the assumptions are met but does
not arbitrate the opposite.
Problem 3
The primitive form for the 2-form in R4 :
2
ω= −2 dx ∧ dz − 2 dy ∧ dw
(5.4.32)
will be found.
Solution
This time the primitive form for a 2-form is to be found, but the general rule
k
d 2 ω= 0 always applies. Therefore, as in the previous problems, we begin with the
calculation of the exterior derivative of (5.4.32). Since all its coefficient functions
are constants, any differentiation gives zero. The closeness condition
2
d ω= 0
(5.4.33)
1
is, therefore, satisfied and the primitive form σ exists, at least locally. In order to
find it, let us first try to apply the method of differential equations. The most general
1-form in R4 can be written as:
5.4 Looking for Primitive Forms
159
1
σ = A(x, y, z, w) dx + B(x, y, z, w) dy + C(x, y, z, w) dz + D(x, y, z, w) dw.
(5.4.34)
If one now requires that
1
2
d σ =ω,
(5.4.35)
the following equation is obtained:
∂B
∂A
∂C
∂A
∂D
∂A
−
dx ∧ dy +
−
dx ∧ dz +
−
dx ∧ dw
∂x
∂y
∂x
∂z
∂x
∂w
∂C
∂B
∂D
∂B
∂D
∂C
−
−
−
+
dy ∧ dz +
dy ∧ dw +
dz ∧ dw
∂y
∂z
∂y
∂w
∂z
∂w
= −2 dx ∧ dz − 2 dy ∧ dw,
(5.4.36)
leading to the system of differential equations for the unknown functions A, B,
C, D:
∂A
∂C
∂A
∂D
∂A
∂B
−
= 0,
−
= −2,
−
= 0,
∂x
∂y
∂x
∂z
∂x
∂w
∂C
∂B
∂D
∂B
∂D
∂C
−
= 0,
−
= −2,
−
= 0.
∂y
∂z
∂y
∂w
∂z
∂w
(5.4.37)
As the reader certainly notices at this point, the method of looking for the
primitive form by solving the differential equations becomes very complicated. It
is relatively simple and effective when one is looking for primitive form for a 1form, since there is only one unknown function (see Problem 1). However, when
one deals with forms of higher degrees, the number of coefficient functions to be
determined strongly increases and this method is not effective any more. In such
situations, it would be welcome to look for primitive forms with the use of a certain
integral formula as done in two previous exercises. Such a formula for any k-form
fortunately exists and the reader probably knows it from the lecture of analysis. We
will limit ourselves here to merely write it down, because its justification is beyond
the scope of this book, which is nothing but a set of problems. Well, let us assume
that some k-form in Rm is given:
m
!
k
ω = ωi1 ,i2 ,...,ik (x1 , . . . , xm ) dxi1 ∧ . . . ∧ dxik
(5.4.38)
i1 ,i2 ,...,ik =1
i1 <i2 <...<ik
satisfying the condition
k
d ω= 0.
(5.4.39)
160
5 Examining Differential Forms
Then its primitive form can be found from the formula:
k−1
σ =
m
!
⎛
k
!
(−1)l−1 ⎝
l=1
i1 ,i2 ,...,ik =1
⎞
1
ωi1 ,i2 ,...,ik (tx1 , . . . , txm )t k−1 dt ⎠ xil ×
0
i1 <i2 <...<ik
×dxi1 ∧ . . . ∧ dxil−1 ∧ dxil+1 ∧ . . . ∧ dxik .
(5.4.40)
The advice that can be given to the reader who is looking at this formula
reluctantly is not to try to encompass the entire expression, but consider what
k−1
subsequent (and simple!) steps make up the procedure of finding σ . First one has
to integrate with respect to the parameter t the set of the coefficient functions, i.e.,
to calculate for each of them the quantity:
1
ωi1 ,i2 ,...,ik (tx1 , . . . , txm )t k−1 dt,
(5.4.41)
0
where every argument is rescaled by the factor t. Second, in each term of the
k
form ω, one of the differentials should be removed (denoted here by dxil ), and
in its place appears xil (which can then be absorbed into some new coefficient
function). This removal goes as follows: first the differential dxil is moved to
the beginning (i.e., “dragged” through other differentials)—and hence the factor
(−1)l−1 emerges—and then replaced with xil . In this way the k-form becomes a
(k − 1)-form. And finally, this should be repeated for each term in the form and
for each differential it contains. The reader is encouraged to check with a direct,
although quite complex, computation that the new form obtained in this way actually
satisfies the Eq. (5.4.35).
Still pausing for a moment at the formula (5.4.40), one can ask whether it
restores the integral formulas used in the previous two problems, namely (5.4.13)
and (5.4.15). In the case when k = 1, it takes the form
0
σ=
m
!
1
ωi (tx1 , . . . , txm )xi dt.
(5.4.42)
i=1 0
The expression on the right-hand side is nothing more than the integral of the
coefficient functions along the straight line joining some starting point, which this
time is the coordinate system origin (for t = 0) with the point (x1 , . . . , xm ).
However, as we know, the choice of the specific path (as the sides of a rectangle
or a straight line) is irrelevant for the result (at least locally). Of course if one
wished to apply the formula (5.4.42) to the examples from the previous exercise,
the starting point should be shifted, because at the origin the coefficients were not
5.4 Looking for Primitive Forms
161
defined. However, this is a purely technical matter and we do not want to bother
the reader with that question. Let us note only that the expression (5.4.40) can be
very easily restated so as to avoid a troublesome point, if any. It can be, therefore,
concluded that the integral (5.4.42) is compatible with those used in previous tasks.
Now the formula can be applied to the form (5.4.32). In this case k = 2, and
m = 4, so there are only two indices i1 and i2 , for which the allowed values are:
1, 2; 1, 3; 1, 4; 2, 3; 2, 4; 3, 4. However, most of the coefficient functions vanish,
and the only nonzero are:
ω13 (x, y, z, w) = ωxz (x, y, z, w) = −2,
ω24 (x, y, z, w) = ωyw (x, y, z, w) = −2.
(5.4.43)
Thus, only two simple integrals remain to be found:
1
1
ωxy (tx, ty, tz, tw)t dt =
0
(−2)t dt = −1,
0
1
1
ωyw (tx, ty, tz, tw)t dt =
0
(−2)t dt = −1.
(5.4.44)
0
Upon inserting these results into the formula (5.4.40), one gets
1
σ = (−1)(x dz − z dx) + (−1)(y dw − w dy) = z dx + w dy − x dz − y dw.
(5.4.45)
1
It is easy to check, when calculating the exterior derivative of σ , that one actually
2
gets the form ω. As the reader should note, despite the apparent complexity of the
expression (5.4.40), the calculations have proved to be very simple.
1
A question can emerge, whether the obtained form σ is the only possible one.
In the case of the primitive form for a 1-form, it was found, when solving previous
problems, that it was determined only up to an additive constant. This time one has
2
1
more freedom: to σ one can add the exterior derivative of any 0-form , and ω
remains unchanged. For we have
1
1
2
2
d(σ +d) = d σ +d 2 =ω +0 =ω .
1
(5.4.46)
Therefore, each of the 1-forms σ + d is an equally good primitive form. These
issues, in the case of R3 which is the most interesting from the physical point of
view, will be dealt with in the next section. We will see then that, for example,
in electrodynamics, the freedom in determining primitive forms (contained in
162
5 Examining Differential Forms
the formula (5.4.46)) corresponds to the so-called gauge transformations of the
electromagnetic field, i.e., to such modification of the potentials, under which E
and B remain invariant.
Problem 4
The primitive form for the 3-form in R3 :
3
ω= y(x − 1) dx ∧ dy ∧ dz
(5.4.47)
will be found.
Solution
In this problem, since we are working with the 3-form in R3 , k = m = 3. In such a
situation, only one canonical combination of differentials is possible: dx ∧ dy ∧ dz,
and consequently there appears only one coefficient function:
ω123 (x, y, z) = ωxyz (x, y, z) = y(x − 1).
(5.4.48)
In accordance with the procedure outlined in the previous exercise, one first needs
to calculate the following integral over the parameter t:
1
1
ωxyz (tx, ty, tz)t dt =
2
0
1
xy 5 y 4 1
x
t − t =y
−
.
ty(tx − 1)t dt =
5
4
5 4
0
2
0
(5.4.49)
From among summations present in the formula (5.4.40), only that with respect to l
remains, because, as already mentioned, the first one reduces to a single term. Thus,
2
1
x
−
(x dy ∧ dz − y dx ∧ dz + z dx ∧ dy)
5 4
x
1
1
x
1
2 x
= xy
−
dy ∧ dz − y
−
dx ∧ dz + yz
−
dx ∧ dy.
5 4
5 4
5 4
σ =y
(5.4.50)
An easy calculation shows that the obtained 2-form meets the condition:
2
3
d σ =ω.
(5.4.51)
5.4 Looking for Primitive Forms
163
As the reader certainly figures out, this result is not unique. This is because if
2
1
2
1
instead of σ one takes σ +d τ , where τ is an arbitrary 1-form, the Eq. (5.4.51) will
still hold:
2
1
1
2
3
3
d(σ +d τ ) = d σ +d 2 τ =ω +0 =ω.
(5.4.52)
Problem 5
The primitive form for the 2-form in Rn :
1
ω= x1 dx1 ∧ dx2 + x2 dx2 ∧ dx3 + . . . + xN −1 dxN −1 ∧ dxN + xN dxN ∧ dx1
(5.4.53)
will be found.
Solution
The experience gamered when solving the preceding problems allows now for
the direct application of the formula (5.4.40) to the form given above. One can,
therefore, write
t
1
σ =
t
tx1 t dt (x1 dx2 − x2 dx1 ) +
0 ω12 (tx)
tx2 t dt (x2 dx3 − x3 dx2 ) + . . .
0 ω23 (tx)
t
+
txN −1 t dt (xN −1 dxN − xN dxN −1 )
0 ωN N−1 (tx)
t
txN t dt (xN dx1 − x1 dxN ),
+
(5.4.54)
0 ωN1 (tx)
where the symbol x stands collectively for all the arguments: x1 , . . . , xN . All the
"1
integrals with respect to the parameter t are then identical and equal to 0 t 2 dt =
1/3. Sorting out terms in (5.4.54), one obtains
1
σ =
1 2
(x dx2 − x1 x2 dx1 + x22 dx3 − x2 x3 dx2 + . . .
3 1
2
2
+xN
−1 dxN − xN −1 xN dxN −1 + xN dx1 − xN x1 dxN )
164
5 Examining Differential Forms
=
1 2
1
(x − x1 x2 ) dx1 + (x12 − x2 x3 ) dx2 + . . .
3 N
3
1 2
1 2
+ (xN
−2 − xN −1 xN ) dxN −1 + (xN −1 − xN x1 ) dxN .
3
3
1
(5.4.55)
2
There remains, as always, a simple task to check that in fact d σ =ω. To get this
done, it is enough to note that the coefficient in front of the product of differentials,
for example, dx1 ∧ dx2 , will come from two terms:
∂
∂x2
1 2
1
(xN − x1 x2 ) dx2 ∧ dx1 = x1 dx1 ∧ dx2
3
3
(5.4.56)
1 2
2
(x1 − x2 x3 ) dx1 ∧ dx2 = x1 dx1 ∧ dx2 .
3
3
(5.4.57)
and
∂
∂x1
1
As they are added, it can be seen that the first term of d σ takes the needed form:
x1 dx1 ∧ dx2 . Repeating the identical calculation for subsequent terms, one actually
gets the formula (5.4.53).
5.5 Finding Potentials in R3
Problem 1
The scalar potential for the vector field:
V = [z2 cos(x + y), z2 cos(x + y), 2z sin(x + y)]
(5.5.1)
will be found.
Solution
Aimed at physical applications, our special interest should be focused on the
differential forms in R3 . We have already encountered the formulation of the
equations of electromagnetism in this language, but the forms are also applicable
in other branches of physics, as for example thermodynamics or hydrodynamics.
The reader certainly has, in this context, such notions as “gradient,” “curl,” or
“divergence” in common (the former, after all, having appeared already in the
5.5 Finding Potentials in R3
165
previous chapters of this book). These concepts are closely related to the operations
of the exterior differentiation performed on the forms in R3 .
We begin with the short reminder of a couple of facts, hopefully known to reader.
First, let us consider a 0-form, i.e., a scalar function (x, y, z). As we know, its
exterior derivative is a 1-form obtained according to the formula (5.0.8):
d =
∂
∂
∂
dx +
dy +
dz.
∂x
∂y
∂z
(5.5.2)
If one looks at the coefficient functions which appeared as a result of the
differentiation, it is visible that they constitute the components of the gradient:
=
∇
∂ ∂ ∂
= F .
,
,
∂x
∂y ∂z
Fx
Fy
(5.5.3)
Fz
So one can say that the calculation of the exterior derivative of a 0-form corresponds
to calculating the gradient of the function. The function itself is called the scalar
potential of the vector field F .
Now consider a 1-form, whose coefficients, denoted as Ax (x, y, z), Ay (x, y, z),
and Az (x, y, z), are components of a certain vector A:
1
ω= Ax (x, y, z) dx + Ay (x, y, z) dy + Az (x, y, z) dz,
(5.5.4)
and calculate its exterior derivative, as it was done in Sect. 5.3. As it is known, in
this way one gets the following 2-form:
∂ Ay
∂ Ay
∂ Ax
∂ Ax
dy +
dz ∧ dx +
dx +
dz ∧ dy
dω=
∂y
∂z
∂x
∂z
∂ Az
∂ Az
dx +
dy ∧ dz
+
∂x
∂y
∂ Ay
∂ Az
∂ Az
∂ Ax
−
−
=
dy ∧ dz +
dz ∧ dx
∂y
∂z
∂z
∂x
1
+
Bx
∂ Ay
∂ Ax
−
dx ∧ dy.
∂x
∂y
By
(5.5.5)
Bz
Note the noncanonical order of differentials in the second term on the right-hand
side, already met in the formula (5.2.36).
166
5 Examining Differential Forms
The obtained coefficients Bx , By , and Bz constitute the components of a certain
well known in differential geometry as well as in various physical
vector B,
applications, called the curl (or the rotation) of the vector A:
∂ Ay ∂ Ax
∂ Az ∂ Ay
∂ Ax
∂ Az
−
,
−
,
−
.
B =∇ ×A=
∂y
∂z
∂z
∂x
∂x
∂y
(5.5.6)
Thus one sees that the calculation of the exterior derivative of a 1-form corresponds
to finding the curl of the vector function composed of its coefficients.
If in the formula (5.5.5) the canonical order of differentials was maintained,
it would not be possible to identify the emerging coefficient functions with the
components of the curl vector. Strictly speaking, they would not constitute a vector
at all, not properly (i.e., as vector components) transformed under rotations of the
coordinate system. The order dz ∧ dx, and with it also the sign of the corresponding
coefficient function, is then somewhat forced upon us.
The symbol × appears in the formula (5.5.6), since, using a certain mental
shortcut, one can say that curl is equal to the cross product of the vectors
=
∇
∂
∂
∂
,
,
∂x ∂y ∂z
(5.5.7)
In order to denote curl, the symbol rotA is used as well.
and A.
If the Eq. (5.5.6) is satisfied, one says that the vector A is a vector potential for
the field B.
At the end, let us consider a 2-form:
2
σ = Ex dy ∧ dz + Ey dz ∧ dx + Ez dx ∧ dy,
(5.5.8)
whose coefficient functions constitute a vector E = [Ex , Ey , Ez ]. Calculating the
exterior derivative, one finds
∂ Ey
∂ Ez
∂ Ex
2
σ
+
+
dx ∧ dy ∧ dz.
(5.5.9)
d =
∂x
∂y
∂z
The expression in brackets is called the divergence of the vector field E:
∂ Ey
∂ Ez
· E = ∂ Ex +
+
.
∇
∂x
∂y
∂z
(5.5.10)
The designation divE is used as well. Thus one can say that the exterior derivative
of a 2-form corresponds to the divergence of the vector field.
At the end let us note again that if in the expression (5.5.8) the differentials were
written in the order dx ∧ dz, then in the second term of (5.5.10) the minus sign
5.5 Finding Potentials in R3
167
would appear and the obtained quantity would not be a scalar (it would get changed
under rotations).
We can now move on to solve the present exercise, i.e., to find a scalar potential
for the given vector field. In other words, (5.5.1) is to be the gradient of a certain
function and this function is to be found. In the language of forms this means that
one is looking for the primitive form for a given 1-form. From the previous section,
we know that in the first instance the exterior derivative should be calculated, in
order to make sure that our job has a solution at all. In the vector language, the
appropriate requirement reads:
× V = 0.
∇
(5.5.11)
Let us then find the components of the curl (we denote ∂/∂x := ∂x and similarly for
the other variables):
× V
∇
x
× V
∇
y
× V
∇
z
= ∂y Vz − ∂z Vy = 2z cos(x + y) − 2z cos(x + y) = 0,
= ∂z Vx − ∂x Vz = 2z cos(x + y) − 2z cos(x + y) = 0,
= ∂x Vy − ∂y Vx = −z2 sin(x + y) + z2 sin(x + y) = 0.
(5.5.12)
As one can see, the condition (5.5.11) is satisfied, which means that the vector
field (5.5.1) has the scalar potential, at least locally.
Now let us find this potential. From the formula (5.5.4), one can see that, in the
case of a 1-form, the vector components and the appropriate coefficient functions
may be identified. A simple generalization of formula (5.4.13), made use of in the
previous section, for the case of 3 dimensions is then
y
x
(x, y, z) =
Vx (x , y0 , z0 ) dx +
x0
z
Vz (x, y, z ) dz .
Vy (x, y , z0 ) dy +
y0
z0
(5.5.13)
It means that the integrations are taken along the edges of some cuboid from the
vertex of coordinates (x0 , y0 , z0 ) to that of (x, y, z). Since we are interested in the
potential only in a local sense, in the neighborhood of the point (x, y, z), one can
assume that this former vertex is contained within it. The reader already knows that
a different path that connects these two points (e.g., a straight line) could be chosen.
This is what will be done in the last problem of this section.
In order to find the potential we need, it is enough to perform the above three
integrals plugging into them the components of the vector V read off from (5.5.1):
168
5 Examining Differential Forms
y
x
z02 cos(x (x, y, z) =
+ y0 ) dx +
x0
z
z02 cos(x
2z sin(x + y) dz
+ y ) dy +
y0
z0
x
y
z
2
2
2
= z0 sin(x + y0 ) + z0 sin(x + y ) + z sin(x + y)
x0
= z sin(x + y)
2
y0
−z02 sin(x0
+ y0 ) .
z0
(5.5.14)
C
All terms that “mixed” the coordinates of the start and end points have disappeared. We know already that it must have happened. The potential (5.5.14) is, of
course, determined up to an additive constant C, but this had been expected.
The reader is encouraged to calculate the gradient of the obtained function
(x, y, z) and ascertain that our calculations are correct and in fact one gets V .
Problem 2
The vector potential for the vector field:
B = [−2xz, 2yz, −2]
(5.5.15)
will be found.
Solution
Since we are looking for a vector potential, we already know from the previous
exercise that with the vector B a 2-form should be associated according to the
formula (5.5.8). Then it should be checked if this form is closed by calculating
its exterior derivative. This procedure corresponds to verifying the value of the
divergence of the vector B :
· B = ∂ (−2xz) + ∂ (2yz) + ∂ (−2) = −2z + 2z = 0.
∇
∂x
∂y
∂z
(5.5.16)
at least locally,
As it is seen, it turns out to vanish and, therefore, the vector B,
constitutes the rotation of a certain vector A (i.e., the wanted vector potential). The
result (5.5.16) is simply a consequence of the fact that the divergence of the curl of
any vector V vanishes:
· (∇
× V ) = 0
∇
1
or, which is equivalent, that d(d ω) = 0.
(5.5.17)
5.5 Finding Potentials in R3
169
the formula (5.4.40) should be used. Translated into the vector
In order to find A,
language, it is equivalent to (for k = 2, m = 3)
1
r) =
A(
r) × r dt.
t B(t
(5.5.18)
0
Note that in the case of a constant vector B (i.e., independent of r), it can be removed
from under the integral, which then gives 1/2, and one gets the formula well known
in electrodynamics:
r ) = 1 B × r.
A(
2
(5.5.19)
It simply corresponds to the value of the vector potential in the case of the constant
In the present case, however, B is not constant, so one has to
magnetic field B.
perform three integrations:
1
1
1
(−2xz)t 3 dt = − xz,
2
tBx (t r) dt = =
0
0
1
1
tBy (t r) dt = =
0
(2yz)t 3 dt =
0
1
1
tBz (t r) dt = =
0
1
yz,
2
(−2)t dt = −1.
0
(5.5.20)
Three components of a certain vector have been obtained, but in order to find the
vector potential in accordance with the formula (5.5.18), one still has to calculate the
cross product with r = [x, y, z]. This can be done in the standard manner, leading to
1 2
1 2
yz + y, xz − x, −xyz .
(5.5.21)
A(r ) =
2
2
Calculating the rotation of the above, the reader easily finds that
× A = B.
∇
(5.5.22)
It should be recalled here that the vector potential is defined only up to the
gradient of a certain scalar function (analogously the primitive form of a 2-form
was determined only up to the exterior derivative of a 0-form, cf. (5.4.46)). Since
the gradient of the rotation always vanishes, as one can see by a direct computation,
× (A + ∇)
× A + ∇
× ∇
= B.
∇
=∇
(5.5.23)
170
5 Examining Differential Forms
Problem 3
The vector potential for the vector field:
B = [zey − 1, 0, 1 − x cos y]
(5.5.24)
will be found.
Solution
We follow the scenario developed in the previous problems. First one must make
sure that the divergence of the vector B vanishes. So let us calculate:
· B = ∂ (zey − 1) + ∂ 0 + ∂ (1 − x cos y) = 0,
∇
∂x
∂y
∂z
(5.5.25)
Therefore, (5.5.18) can
which implies that locally B is the curl of a vector field A.
be applied, which requires us to calculate three integrals:
1
1
tBx (t r) dt =
0
1
(tze − 1)t dt = z
0
1
t ze dt −
2
ty
ty
0
2
2
2z 1
z
1 − + 2 ey − 3 − ,
=
y
y
2
y
y
1
0
1
tBy (t r) dt =
0
0 t dt = 0,
0
1
1
tBz (t r) dt =
0
t dt
1
(1 − tx cos(ty))t dt =
0
1
t dt − x
0
t 2 cos(ty) dt
0
2
2
1
x
− sin y − cos y + 2 sin y + .
=
y
y
2
y
(5.5.26)
The details of the elementary integration by parts, well known to the reader, have
been omitted.
Still, in accordance with the formula (5.5.18), one has to calculate the cross
product with the vector r = [x, y, z]. As a result, the vector potential A is obtained
with the components:
5.5 Finding Potentials in R3
171
2x
y
2x
cos y −
sin y − ,
y
y
2
z2
2
2
x2 2
2
2z2
x+z
Ay =
ey + 3 +
cos
y
−
sin
y
−
1
−
+
,
sin
y
−
2
2
y y
y
y
y
2
y
y
2
2
2z y
(5.5.27)
Az = z 1 − + 2 ey − 2 − .
y
2
y
y
Ax = x sin y +
An easy, but relatively long, calculation shows that our solution is correct because
× A = B.
∇
(5.5.28)
Problem 4
The scalar and vector potentials for the vector field:
V = [yz, 4y 3 − 12yz2 + xz, 4z3 − 12y 2 z + xy]
(5.5.29)
will be found.
Solution
The following question occurs in the present problem: could there exist a vector
field that simultaneously is the gradient of a scalar and the curl of a vector? At the
beginning, let us try to answer it in a more general way. If it is possible, then the two
conditions must be met:
V = ∇
and
× A,
V = ∇
(5.5.30)
which entails that
=∇
× A.
∇
(5.5.31)
However, the divergence of the right-hand side vanishes, as we know from the
previous examples. The same must, therefore, be true for the left side. Thus one
obtains the prerequisite:
2
2
2
· ∇
=∂ +∂ +∂ =
∇
∂x 2
∂y 2
∂z2
= 0.
(5.5.32)
172
5 Examining Differential Forms
The operator constructed of the second derivatives is called the Laplace operator
(or Laplacian) and the equation to be satisfied by —the Laplace equation. The
function, which satisfies it, is called the harmonic function. Our conclusion can be
summarized: the gradient of a harmonic function is simultaneously the curl of a
certain vector.
As an example, let us examine the scalar function in the form:
(x, y, z) = x 2 − 2y 2 + z2 .
(5.5.33)
It is a harmonic function, which can be checked with the simple calculation:
=
∂2
∂2
∂2
+
+
∂x 2
∂y 2
∂z2
(x 2 − 2y 2 + z2 ) = 2 − 4 + 2 = 0.
(5.5.34)
Let us find the gradient of :
= [2x, −4y, 2z].
∇
(5.5.35)
It is now straightforward to see that it is also the curl of the vector:
A = [−2yz, 0, 2xy].
(5.5.36)
Thus the issue formulated in the text of the exercise makes sense. It will be solved
in two steps. First, we will consider whether (5.5.29) has a scalar potential. Aimed
at this, let us calculate the curl of V :
∂ Vy ∂ V x
∂ Vz ∂ Vy
∂ Vx
× V = ∂ Vz −
,
−
,
−
∇
∂y
∂z
∂z
∂x
∂x
∂y
= [−24yz + x + 24yz − x, y − y, z − z] = [0, 0, 0].
(5.5.37)
The necessary condition (and sufficient one in the local case as well) is met, so one
can use the formula (5.5.13), which yields
y
x
(x, y, z) =
y0
y0 z0 dx +
Vz (x, y, z ) dz
Vy (x, y , z0 ) dy +
z0
y
x
x0
z
Vx (x , y0 , z0 ) dx +
x0
=
z
(4y 3 − 12y z02 + xz0 ) dy +
y0
(4z3 − 12y 2 z + xy) dz
z0
= y0 z0 (x − x0 ) + y 4 − y04 − 6(y 2 − y02 )z02 + xz0 (y − y0 )
+z4 − z04 − 6y 2 (z2 − z02 ) + xy(z − z0 )
= xyz − 6y 2 z2 + y 4 + z4 + −x0 y0 z0 + 6y02 z02 − y04 − z04 .
C
(5.5.38)
5.5 Finding Potentials in R3
173
= V is left for the reader.
As usual, checking that indeed ∇
One can now proceed to the second part of the job, i.e., to find the vector
potential. To be sure that it exists, let us calculate the divergence:
· V = +12y 2 − 12z2 + 12z2 − 12y 2 = 0.
∇
(5.5.39)
We conclude that the vector potential exists and it can be found with the use of the
We subsequently get
formula (5.5.18), where one has to put V in place of B.
1
1
tVx (t r) dt =
0
yzt 3 dt =
1
yz,
4
0
1
1
tVy (t r) dt =
0
(4t 3 y 3 − 12t 3 yz2 + t 2 xz)t dt =
4 3 12 2 1
y −
yz + xz,
5
5
4
(4t 3 z3 − 12t 3 y 2 z + t 2 xz)t dt =
4 3 12 2
1
z −
y z + xy.
5
5
4
0
1
1
tVz (t r) dt =
0
0
(5.5.40)
All that remains is to find the cross product of the above vector with r, which gives
A =
−
−
16 3 1 2 16 3
1
1
4
yz + xz +
y z − xy 2 , − yz2 + xz3
5
4
5
4
4
5
12 2
1
1
4
12
1
xy z + x 2 y, y 2 z − xy 3 +
xyz2 − x 2 z . (5.5.41)
5
4
4
5
5
4
× A = V .
As always, at the end it is worth verifying that ∇
Problem 5
The scalar potential for the vector field in R3 :
V (r ) = r sin r,
where r = |r |, will be found.
(5.5.42)
174
5 Examining Differential Forms
Solution
As it is known, one must check whether the curl of the field
V (x, y, z) = [x, y, z] sin2
x 2 + y 2 + z2
(5.5.43)
vanishes. This calculation is most easily carried out with the use of the antisymmetric symbol ij k , where i, j, k = 1, 2, 3. It is defined as follows:
ij k = 1,
(5.5.44)
when the system of indexes i j k is obtained from 1 2 3 by an even permutation, and
ij k = −1,
(5.5.45)
when this permutation is odd. In case of repeated indexes (e.g., when j = i, etc.),
its value amounts to 0. Of course, one has 123 = 1, as the permutation becomes
then an identity (and obviously is even). For other exemplary systems of indexes,
one has
213 = −1,
321 = −1,
312 = 1, etc.
(5.5.46)
In the first case, in order to restore the canonical sequence (i.e., 1 2 3), exactly one
transposition has to be done (1 ↔ 2), which means that the permutation is odd,
and hence the minus sign occurs. In the second case the three transpositions have
to be made. First, let us look at “1.” Since we want to move it to the beginning,
it must first be “dragged” through 2, and next through 3. But that is not the end,
as this would yield the sequence 1 3 2. The last transposition (3 ↔ 2) allows us to
obtain the required sequence. In total, one has three transpositions and, therefore, the
permutation is odd. In the case of the last example, two steps are needed: 3 ↔ 1 and
3 ↔ 2. The permutation is even and one gets +1. In the case of repeated indexes,
we have, for example
223 = 211 = 333 = 212 = . . . = 0.
(5.5.47)
Using the introduced symbol, which bears the name of the Levi–Civita’s symbol,
or strictly speaking, its i-th
it is very easy to write the cross product A × B,
component as
i = ij k Aj Bk .
(A × B)
(5.5.48)
It should be added here that physicists very often use the so-called “Einstein
summation convention,” according to which the presence of repeated indexes in
an expression entails the presumptive summation over all their permissible values.
5.5 Finding Potentials in R3
175
In particular, on the right-hand side of (5.5.48), there are two repeated indexes: j
and k. The index i is “free.” The range of variability of the indexes j and k is 1, 2, 3.
This means that the formula (5.5.48) should be read as
i=
(A × B)
3
3 !
!
(5.5.49)
ij k Aj Bk .
j =1 k=1
This convention is very useful because in the great majority of cases if repeated
indexes appear in an expression, then one is actually dealing with the summation. It
is then easier to write openly that there is no sum in those few exceptional situations,
than to place symbols of the sums in the vast majority. This convention will be used
henceforth.
The curl of the vector A can now be written in the form (a shorthand denotation
∂xj = ∂j is used):
i = ij k ∂j Ak ,
× A)
(∇
(5.5.50)
· A = ∂i Ai .
∇
(5.5.51)
and its divergence as
One can easily be convinced that the formula (5.5.50) gives the same result
as (5.5.6). For example, let i = 1, which corresponds the x-th component of the
curl. As none of the indexes in ij k may be repeated (otherwise one would get 0),
for j and k there are only two admissible values: 2 and 3. Thus
1 = 1j k ∂j Ak = 123 ∂2 A3 + 132 ∂3 A2 = ∂2 A3 − ∂3 A2 ,
× A)
(∇
(5.5.52)
or simply
x = ∂y Az − ∂z Ay ,
× A)
(∇
(5.5.53)
in full accordance with (5.5.6). We encourage the reader to verify in the same way
that the other two components of the curl are correctly obtained as well.
After this introduction, one can now proceed to solve the current exercise. Since
it is the scalar potential that has to be found, we must make sure that curl of the
vector V vanishes. Let us then calculate
V )i = ij k ∂j Vk = ij k ∂j (rk sin r) = ij k (δj k sin r+rj rk
(∇×
1
cos r).
r
(5.5.54)
The obvious relations having been used here:
∂j rk = δj k
and
∂j r =
rj
.
r
(5.5.55)
176
5 Examining Differential Forms
Let us notice now that the expression in brackets on the right-hand side of
the formula (5.5.54) is symmetrical upon swapping j and k. In turn ij k is
antisymmetric, which was spoken of above. This is because any transposition of
indexes leads to an odd permutation and consequently requires the change of the
sign. The right-hand side of (5.5.54) is then the contraction (i.e., the sum over both
indexes) of an antisymmetric object (let us temporarily denote it with Aj k ) with a
symmetric one (Sj k ). Such a contraction always vanishes, as
Aj k Sj k = −Akj Skj = −Aj k Sj k .
(5.5.56)
The former equality is due to the swapping of indexes and the latter is simply the
change of their names (and not any “dragging”!). Of course, it is inessential if the
index over which the summation runs is called j or k. On the other hand, the name
of a free index is indispensable, since it appears on both sides of the equation (see
e.g. (5.5.50)). As a result, the expression is equal to minus itself, i.e., it vanishes.
Thereby, it is seen that
× V )i = 0
(∇
(5.5.57)
and V has (locally) the scalar potential. In order to find it one can use the
formula (5.5.14), but instead of integrating over the edges of a cube, we shall
integrate along its diagonal, writing (the prime here represents the derivative with
respect to t)
1
1
V (t r) · r dt =
(r ) =
0
1
r sin(tr)t dt = r
2
0
2
−1
[cos(tr)] t dt
r
0
1
cos(tr) dt = −r cos r + sin r.
= −r cos r + r
(5.5.58)
0
Upon a simple calculation one can check that
r ) = V (r ).
∇(
(5.5.59)
5.6 Exercises for Independent Work
Exercise 1 Find the operation of the form
1
(a) ω= (2x + y) dx + y dy on the vector v1 = [1, 1] bound at the point (1, 0) and
on the vector v2 = [−1, 2] bound at the point (2, 1),
2
(b) ω= (x 2 + y 2 ) dx ∧ dy + xz dz ∧ dx + dy ∧ dz on the pair of vectors v1 =
[1, 1, 1], v2 = [1, 0, −1] bound at the point (1, 2, 3).
5.6 Exercises for Independent Work
177
Answers
1
1
(a) ω (v1 ) = 2, ω (v2 ) = 0.
2
(b) ω (v1 , v2 ) = −5.
2
1
2
Exercise 2 Find ω1 ∧ ω2 , where ω1 = x dx ∧ dy + y dz ∧ dx + y 2 dy ∧ dz and
1
ω2 = x dx + y dy + z dz.
Answer
2
1
ω1 ∧ ω2 = (xz + y 2 + xy 2 ) dx ∧ dy ∧ dz.
Exercise 3 Calculate the exterior derivative of the form
2
(a) ω= xy dx ∧ dz + xyz dy ∧ dz in R3 ,
1
(b) ω= (x1 + x2 ) dx1 + (x2 + x3 ) dx2 + . . . + (xn−1 + xn ) dxn−1 in Rn .
Answers
2
(a) d ω= (−x + yz) dx ∧ dy ∧ dz.
1
(b) d ω= −dx1 ∧ dx2 − dx2 ∧ dx3 − . . . − dxn−1 ∧ dxn .
Exercise 4 Using the form (5.2.47), find the transformation of the four-potential to
the coordinate system moving with velocity v along the x-axis.
Answer
A0 = γ (A0 − βc A1 ), A1 = γ (A1 − β/c A0 ), A2 = A2 , A3 = A3 .
Exercise 5 Using the formulas (5.2.37), (5.2.47), and the known relations between
the electric and magnetic fields and the four-potential (they should be found in a
textbook of electromagnetism), prove that
1
2
d ωA =ωF .
178
5 Examining Differential Forms
Exercise 6 Prove that the Eq. (5.2.38) corresponds to the pair of the sourceless
Maxwell equations.
Exercise 7 Considering the form
2
2
ωF ∧ ωF ,
prove that the quantity E · B is invariant under the Lorentz transformations.
where the components of the vector A are given
× A = B,
Exercise 8 Prove that ∇
by the formula (5.5.27), and the vector B by (5.5.24).
× A = V , where the components of the vector A are given
Exercise 9 Prove that ∇
by the formula (5.5.41), and the vector V with (5.5.29).
Exercise 10 Check whether the given vector fields have scalar potentials, and if so,
find them:
(a) V = [3x 2 + 2xy, x 2 ],
(b) V = [y 2 z − 1, 2xyz, xy 2 ].
Answers
(a) Yes, (x, y) = x 2 (x + y) + C.
(b) Yes, (x, y, z) = xy 2 z − x + C.
Exercise 11 Check whether the given vector fields have vector potentials, and if so,
find them:
(a) V = [−1, −1, −1],
(b) V = [z − x, 0, z − x],
(c) V = [z − y, 0, z − y].
Answers
y, z) = [y, z, x] + ∇(x,
(a) Yes, A(x,
y, z).
(b) Yes, A(x, y, z) = [xy, xz, yz] + ∇(x,
y, z).
(c) No.
Chapter 6
Examining Oriented Curvilinear
Integrals
In the present chapter, the question of the so-called “oriented integrals” over
surfaces is undertaken. The first notion to be introduced is that of the orientation.
It will be explained in detail when solving particular problems and the formal
definition is as follows. Given a linear k-dimensional space over R and its basis
e = (e1 , e2 , . . . , ek ). The elements of any other basis f = (f1 , f2 , . . . , fk ) are
linear combinations of ei , which can be written in the matrix form
f = Me.
The matrix M is quadratic and nonsingular, so either det M > 0 or det M < 0.
These two signs define two distinct orientations of the linear space. Consider now
a smooth k-dimensional surface Sk ⊂ Rn defined as
⎧
x1 = g1 (τ1 , τ2 , . . . , τk ),
⎪
⎪
⎨
x2 = g2 (τ1 , τ2 , . . . , τk ),
⎪
...
⎪
⎩
xn = gn (τ1 , τ2 , . . . , τk ),
(6.0.1)
x = G(τ ).
(6.0.2)
or in abbreviation:
It can happen that such universal mapping G does not exist. Then we assume that
the Eq. (6.0.2) is treated locally, and in fact, a set of distinct maps (an atlas) Gα
covering the entire
. surface Sk is given. Gα is, thereby, a map of a certain open set
Uα ⊂ Sk and α Uα = Sk .
© Springer Nature Switzerland AG 2020
T. Radożycki, Solving Problems in Mathematical Analysis, Part III, Problem Books
in Mathematics, https://doi.org/10.1007/978-3-030-38596-5_6
179
180
6 Examining Oriented Curvilinear Integrals
The orientation of the surface can be defined as the orientation of the tangent
spaces in the following way. Let us choose a certain point X0 ∈ Sk . The set of
tangent vectors at this point:
tαi :=
∂Gα ,
∂τi X0
i = 1, . . . , k,
(6.0.3)
can be treated as an ordered basis of the appropriate tangent space. The surface Sk is
told to be orientable and the orientation is fixed provided that for the entire surface
the maps Gα can be chosen such that the orientations of tangent spaces at any point
of Uα ∩ Uα agree. This means that det Mα α > 0, where eα = Mα α eα and eα (eα )
is the basis of the space of tangent vectors defined by Gα (Gα ).
The k-form of the type (5.0.1) (with the notation adjusted accordingly) can be
integrated over the k-dimensional oriented surface following the formula:
k
I=
ω=
n
!
∂(gi1 , . . . , gik )
dτ1 · . . . · dτk ,
ωi1 ,i2 ,...,ik (G(τ ))
∂(τ1 , . . . , τk )
(6.0.4)
D i1 ,i2 ,...,ik =1
Sk
i1 <i2 <...<ik
for (τ1 , τ2 , . . . , τk ) ∈ D. In the case when any universal map G(τ ) does not exist,
the corresponding integral should be found as the appropriate sum of integrals over
disjoint patches.
In the vector language, the formula (6.0.4) can be given the forms:
V (r ) d r =
S1
V (r (τ )) t dτ,
(6.0.5a)
D
V (r ) d 2 S =
S2
(in R2 or R3 ),
V (r (τ )) t1 × t2 dτ1 dτ2 ,
(in R3 ),
(6.0.5b)
D
where t, t1 and t2 stand for the tangent vectors according to (6.0.3).
Stokes’ theorem has the following form. Let Sk be a smooth, k-dimensional,
compact, orientable surface in Rn (hence k ≤ n), ∂Sk its k − 1-dimensional border,
k−1
and ω a differential form of the degree k − 1. Then
k−1
k−1
d ω=
Sk
ω,
(6.0.6)
∂Sk
provided the orientation of ∂Sk is inherited from Sk (in order not to broaden
too much the theoretical part, this issue will be explained when solving specific
examples).
6.1 Calculating Integrals over Curves
181
In the vector language, this theorem takes the following forms.
• The Green theorem (R2 ).
∂Q ∂P
−
∂x
∂y
/
d S=
2
S2
V d r,
(6.0.7)
∂S2
where V = [P , Q].
• Stokes’ theorem (R3 ).
/
× A d 2 S =
∇
S2
A d r.
(6.0.8)
∂S2
• The Gauss–Ostrogradsky theorem (R3 ).
B d 3 V =
∇
S3
/
B d 2 S.
(6.0.9)
∂S3
6.1 Calculating Integrals over Curves
Problem 1
The work done by the force F = x ex + y ey + z ez over the helix:
x(ϕ) = a cos ϕ,
y(ϕ) = a sin ϕ,
z(ϕ) = bϕ
(6.1.1)
will be found, where ϕ ∈ [0, 6π ], and a and b are positive constants.
Solution
In Sect. 3.1 of Part II, and also in Chap. 4 of the present part, we were calculating
various integrals over curves and surfaces. Their common feature was that they
were the so-called unoriented integrals. If one is interested in the length of a
certain fragment of a curve and the appropriate integral is found according to the
formula (3.1.2) or (3.1.7) of Part II, the direction along the curve one is moving is
irrelevant. Its length is, after all, always the same. Things look similar for the total
mass of such fragment (e.g., the mass of the wire) or the center-of-mass location.
182
6 Examining Oriented Curvilinear Integrals
However, sometimes it happens that the integral to be calculated depends on the
direction in which one “travels” along a curve, a standard example being the work
done by a force. Let us imagine that a body is moved between two points A and
B along a curve γ in the Earth’s gravitational field and as a result the body moves
away from it. It is clear that, in order to achieve this, one needs to supply a certain
energy or, in other words, the work performed by us is positive. However, if this
body was shifted along the same curve, but in the opposite direction (from B to A),
i.e., approaching the Earth, this work would be performed by the gravitational field,
which means that our work would now be negative. The integral representing this
work, upon reversing the orientation of the curve, would change its sign. For this
reason such an integral is called oriented.
Thereby, if one wants to calculate this kind of an integral, in addition to the
path itself its orientation has to be fixed. It is easy to guess that the orientation is
defined by a vector tangent to the curve, determining the direction of the motion.
If this vector is fixed at one point, the orientation is defined everywhere. It can be
displaced to any other point on the curve, with its tangential nature maintained and it
will not be changed into the opposite vector for sure. Each of these displaced vectors
could be selected as fixing the orientation of the curve equally well. This is shown
in Fig. 6.1.
Some complications may occur, if the considered curve is not smooth (e.g., it has
“corners”). As an example, one can recall here the edge of a square. Each of the
sides of this curve is smooth, but it is not known how to transfer the orientation (i.e.,
a tangential vector) continuously from one side to the other. In such a situation,
the curve should be divided into smooth pieces (the curve is, therefore, called
piecewise smooth), and then one ought to, so to say, “manually” conform their
orientations, guided, for example, by physical considerations (after all we know, in
which direction one has to move along the curve). Or one can temporarily smoothen
the curve by rounding the angles, only to correctly fit the orientations on each
fragment. The case of a piecewise smooth curve will be encountered in Problem 3.
Fig. 6.1 The displacement of
a tangential vector along the
curve γAB . The selected
vector, fixing the orientation,
is marked with the bold line.
Other vectors are displaced
vectors
6.1 Calculating Integrals over Curves
183
Instead of specifying a tangent vector, one can determine the orientation by
selecting the initial and end points of the curve: for example, the curve γAB leading
from A to B is contrarily oriented from γBA .
Let us assume that a curve is parametrized with the parameter t, i.e., the relations
r(t) are given for t ∈ [α, β], where r(α) are the coordinates of the starting
point (A) and r(β) of the end point (B). As the reader can see, having chosen
a specific parametrization, we simultaneously fixed the orientation. To find the
oriented integral over this curve, one can now use the well-known formula used
by physicists for calculating the work done by a force F :
β
F (r ) d r =
γAB
d r
dτ.
F (r (τ ))
dτ
(6.1.2)
α
As we remember from the first chapter, the expression d r/dτ is nothing more than
a vector tangent to the curve.
In the case specified in this exercise, the parameter is the angle ϕ and one has:
d r
= −a sin ϕ ex + a cos ϕ ey + b ez .
dϕ
(6.1.3)
Since ϕ ∈ [0, 6π ], the integral (6.1.2) is expressed as follows:
F (r ) d r
γAB
6π
[x(ϕ) ex + y(ϕ) ey + z(ϕ) ez ] · [−a sin ϕ ex + a cos ϕ ey + b ez ] dϕ
=
0
b2 2 6π
ϕ = 18π 2 b2 ,
(−a sin ϕ cos ϕ + a sin ϕ cos ϕ + b ϕ) dϕ =
2
0
6π
=
2
0
2
2
(6.1.4)
the relations (6.1.1) having been used.
As we know from the theoretical introduction, the oriented integrals can be
expressed in the language of differential forms as well, according to the formula (6.0.4). Deciphering this expression for needs of the current problem, one can
see that k = 1, m = 3, τ1 = ϕ, the function G is defined by the formulas (6.1.1),
and the subsequent coefficient functions of the 1-form associated with the vector F :
1
ωF = Fx dx + Fy dy + Fz dz
(6.1.5)
184
6 Examining Oriented Curvilinear Integrals
simply equal x, y, and z. The set D is nothing other than the interval [0, 6π ].
Therefore, the formula (6.0.4) boils down to
6π 1
ωF =
Sk
x(ϕ)
dx
dy
dz
+ y(ϕ)
+ z(ϕ)
dϕ.
dϕ
dϕ
dϕ
(6.1.6)
0
It is the expression identical to (6.1.4). As the reader sees, the initially “scarylooking” curvilinear oriented integrals translate in fact into definite integrals with
respect to the parameters.
Problem 2
The work done by the force F = (1 − y) ex + x 2 ey + (2xy + z) ez over the curve
representing the intersection of the plane x + y = 1 with the paraboloid z = x 2 + y 2
will be calculated from the point A(2, −1.5) to B(1, 0, 1).
Solution
This problem can be solved using the formula (6.1.2), if the curve γAB is
parametrized by the relation r(τ ). This curve is drawn in Fig. 6.2.
The simplest parametrization resulting from the equations in the text of the
exercise appears to be:
⎧
⎨ x(τ ) = τ,
(6.1.7)
y(τ ) = 1 − τ,
⎩
2
2
2
z(τ ) = τ + (1 − τ ) = 2τ − 2τ + 1,
in which case the tangent vector d r/dτ will have the form:
d r
= ex − ey + (4τ − 2) ez .
dτ
(6.1.8)
Let us assume in addition, that α = 2 and β = 1, since this choice ensures the
correct orientation of the curve: from A to B. The expression (6.1.2) will now have
the form:
1
F (r ) d r =
γAB
2
[τ ex + τ 2 ey + ez ] ·[ex − ey + (4τ − 2) ez ] dτ
F (r (τ ))
2
=−
d r/dτ
2
1 3 5 2
19
(−τ + 5τ − 2) dτ = − − τ + τ − 2τ = − .
3
2
6
1
2
1
(6.1.9)
6.1 Calculating Integrals over Curves
185
z
Fig. 6.2 The fragment of the
curve resulting as the
intersection of the plane
x + y = 1 and the paraboloid
z = x 2 + y 2 , between the
points A and B
A
B
y
x
Naturally, the same result is obtained by introducing the work form as
1
ωF = (1 − y) dx + x 2 dy + (2xy + z) dz,
(6.1.10)
"
1
and then calculating the integral γAB ωF with the use of the formula (6.0.4), upon
simplifying it similarly as in the previous problem.
The reader might now pose the following question: if some other curve connecting the points A and B was chosen, would the calculated work be identical?
After all, it is known that it is so in the gravitational or electrostatic fields. The
value of the work depends only on the initial and final points, and not on the path,
on which it is calculated. Moreover, this fact allows us to introduce the concept
of the potential energy. It is possible because these two fields are potential (i.e.,
irrotational) ones, which means that a potential can be defined for them. It was
spoken of in the previous chapter. Therefore, if there exists such a function (r )
r ) = F (r ), the integral over each curve (located in a domain contractible
that ∇(
to a point) joining the points A and B gives the same result.
186
6 Examining Oriented Curvilinear Integrals
In order to verify whether the vector field F (r ) is the gradient of a scalar field,
the curl of F has to be calculated. It is easy to check that the curl does not vanish,
so the considered field is not a potential one. Consequently, the integral calculated
along different paths connecting the same pair of points can give different results.
The force given in the text of the exercise is neither gravitational nor electrostatic for
sure. In turn, the field F from the previous job has vanishing curl. It is then possible
to define a potential, for instance in the form (x, y, z) = (x 2 + y 2 + z2 )/2. This
means that, instead of the helix, one could calculate the integral over any other curve
of identical ends, getting the same result.
Problem 3
The integral
"
V · d l will be calculated, where V = 2xy ex + x 2 ey , if the curve γ ,
γ
lying in the plane R2 , is:
(a) the segment of a straight line between the points A(0, 0) i B(1, 2),
(b) the polyline A(0, 0) → C(1, 0) → B(1, 2),
(c) the fragment of the parabola y = 2x 2 between the points A and B.
Solution
The vector V is associated with the 1-form:
1
ωV = Vx dx + Vy dy = 2xy dx + x 2 dy,
(6.1.11)
which can easily be seen as closed:
1
d ωV = (∂x Vy − ∂y Vx ) dx ∧ dy = (2x − 2x) dx ∧ dy = 0.
(6.1.12)
The expression in brackets could be called the two-dimensional curl of V .
The plane R2 , on which this form is operating, has no “holes” and is contractible
1
to a point. Thus the form ωV is locally and globally compact. For such a form the
value of the integral does not depend on the path, but only on its ends. Therefore,
we expect that all three curves given in the text above will lead to identical results.
These curves are drawn in Fig. 6.3. The appropriate integrals are calculated below
one by one.
6.1 Calculating Integrals over Curves
187
Fig. 6.3 The curves defined
in the text of the exercise
1. The curve γ1 .
The parametrization that gets imposed for the curve γ1 , which is a segment of an
oriented straight line from A to B, is
x(τ ) = τ,
y(τ ) = 2τ,
(6.1.13)
with τ ∈ [0, 1]. Thus one has
d r
= ex + 2 ey
dτ
(6.1.14)
and one can calculate
1
1
I1 =
ωA =
γ1
dx
dy
2x(τ )y(τ )
+ x(τ )2
dτ =
dτ
dτ
0
1
6τ 2 dτ = 2.
0
(6.1.15)
2. The curve γ2 .
The curve γ2 is only piecewise smooth, so one should break it down into
two smooth pieces: let γ2a be the segment AC and γ2b the segment CB. The
orientation for them both ought to be chosen in such a way that with the
increasing values of parameters one moves in the sequence: A → C → B.
For the curve γ2a , we then have
x(τ ) = τ,
y(τ ) = 0
(6.1.16)
188
6 Examining Oriented Curvilinear Integrals
and τ ∈ [0, 1], and then:
d r
= ex + 0 ey .
dτ
(6.1.17)
In turn for the curve γ2b , it is best to take
x(τ ) = 1,
y(τ ) = τ,
(6.1.18)
with τ ∈ [0, 2] which implies
d r
= 0 ex + ey .
dτ
(6.1.19)
In this way, the following value of the integral is obtained:
1
1
I2 =
ωA +
γ2a
1
ωA =
γ2b
2
(0·1+τ ·0) dτ +
(2τ ·0+1·1) dτ = 2.
2
0
(6.1.20)
0
As anticipated, the same result as before has been obtained.
3. The curve γ3 .
For the curve γ3 we choose the parametrization in the form:
x(τ ) = τ,
y(τ ) = 2τ 2 ,
(6.1.21)
where τ ∈ [0, 1], which gives
d r
= ex + 4τ ey .
dτ
(6.1.22)
Let us now calculate
1
1
I3 =
ωA =
γ3
1
(4τ · 1 + τ · 4τ ) dτ =
3
0
8τ 3 dτ = 2,
2
(6.1.23)
0
again obtaining the same result.
This exercise will be of interest again in Sect. 6.3, in which the so-called Stokes’
theorem will be dealt with. Then it will be seen that these results are not accidental,
and they stem from the simple fact that the integral of a compact form over a closed
path vanishes.
6.2 Calculating Integrals over Surfaces
189
6.2 Calculating Integrals over Surfaces
Problem 1
The flux of the field B = x ex + y ey + zez through the surface of the upper
hemisphere of radius R centered at the origin and oriented “upward” will be found.
Solution
In Chap. 4, we were concerned with the calculation of unoriented integrals over
surfaces. We then learned that if a surface S is parametrized using the dependence r(u, v) in order to calculate the integral of a certain function f (r ), the
formula (4.2.3) can be used:
f (u, v) tu × tv dudv,
f dS =
S
(6.2.1)
D
where
f (u, v) = f (r (u, v))
and
u, v ∈ D.
This expression is useful if one, for example, looks for the area of a curved
surface (then f (r ) simply equals 1), the location of the center of mass or the moment
of inertia of the mass distributed on a surface. However, if one wants to calculate
the flux of a vector quantity through a surface, such as the flux of the magnetic field
or the weight of water flowing through the surface per unit
B or electric field E,
of time, one should integrate not the entire vector but only the component normal
to the surface. The tangential component is “sliding” on the surface and does not
contribute to the flux. In this case, in place of (6.2.1), one should write
r (u, v)) · n(u, v) tu × tv dudv,
B(
S =
Bd
S
(6.2.2)
D
n(u, v) being a unit vector perpendicular to the surface at the point specified by the
parameters u and v.
The normal vector can be obtained as the cross product of the tangent vectors
associated with both parameters (see Sect. 4.1):
tu =
∂ r
,
∂u
tv =
∂ r
,
∂v
(6.2.3)
190
6 Examining Oriented Curvilinear Integrals
as
tu × tv
.
n = tu × tv (6.2.4)
Thanks to the denominator it is normalized to unity. Upon plugging this expression
into (6.2.2), one obtains the formula that will be used later in the solution.
S =
Bd
S
D
r (u, v)) · tu × tv tu × tv dudv
B(
tu × tv %
$
r (u, v)) · tu × tv dudv.
B(
=
(6.2.5)
D
As opposed to the integral (6.2.1), the above expression is, however, sensitive
to the chosen direction of the normal vector: the change of n reverses the sign of
the flux. There is nothing strange with that. After all in physics the “outgoing” and
“incoming” fluxes differ just by their signs. Therefore, when calculating an integral
of the type (6.2.5), one has to establish beforehand the direction of the normal vector.
This will fix the orientation of the surface spoken of in the theoretical introduction.
Naturally, the reversal of the direction of n to the opposite can be obtained by
changing the order of the tangent vectors tu and tv in the cross product (6.2.4).
Hence the orientation of a surface can be fixed from the very beginning, by selecting
its parametrization and deciding which of the parameters is considered as the “first,”
and which as the “second.”
If a normal vector has been defined at some point of our surface and this surface
is smooth (with no edges), it can be moved to any other point, maintaining the
property of its perpendicularity. We will get in this way the field of normal vectors,
as it is shown in Fig. 6.4. However, the reader probably knows already from the
lecture of analysis that there exist surfaces for which it is not possible to introduce
the orientation. The so-called Möbius strip can serve as a well-known example. Any
normal vector selected at any point after having been moved along the appropriate
path reverses its direction to the opposite one. This means that the surface is not
orientable and one is unable to define anything like “flux through the Möbius strip”
(except where locally, on a small piece of the surface). For the rest of our discussion
we only deal with orientable surfaces.
Let us now transform the expression (6.2.5), using the previously introduced
Levi–Civita symbol together with the Einstein summation convention:
S =
Bd
S
Bi (u, v)ij k
drj drk
dudv
du dv
D
∂(y, z)
∂(z, x)
∂(x, y)
Bx (u, v)
+ By (u, v)
+ Bz (u, v)
dudv.
∂(u, v)
∂(u, v)
∂(u, v)
=
D
(6.2.6)
6.2 Calculating Integrals over Surfaces
191
Fig. 6.4 The field of normal
vectors defining the
orientation of the hemisphere
Note that this expression is identical to (6.0.4) for the case of k = 2, m = 3, if the
r ) is associated with the 2-form according to the formula (5.2.36):
vector field B(
2
ωB = Bx dy ∧ dz + By dz ∧ dx + Bz dx ∧ dy,
(6.2.7)
i.e.,
ω12 = Bz ,
ω13 = −By ,
ω23 = Bx .
(6.2.8)
One can, therefore, write
2
S =
Bd
S
ωB ,
(6.2.9)
S
recalling that for integrals involving differential forms one normally uses a single
symbol of integration.
We now return to the current problem, starting with the choice of the parametrization. The most natural one is that using the angles ϕ and θ (see (4.1.36)), where
ϕ ∈ [0, 2π [ and θ ∈ [0, π/2]. It was sufficient in the chapter concerned with
unoriented integrals. When calculating the flux, however, one needs to fix in addition
the surface orientation. So, if one applies the formula (6.2.6), the order of parameters
ϕ and θ or—as equivalent—the field of normal vectors must be determined, as
already explained above. The sphere, whether it is the hemisphere, is naturally an
192
6 Examining Oriented Curvilinear Integrals
orientable surface and, therefore, the orientation can be determined by selecting a
normal vector at an arbitrary point.
In accordance with the text of the exercise, the surface is to be oriented “upward,”
which means that the unit normal vector located at the “north pole” should be
identical to the versor of the z-axis. This special point is, however, singular from
the point of view of our parametrization, since it corresponds to θ = 0 and to the
undetermined value of ϕ. There the “parallels” become degenerated to one point
only and there is no tangent vector tϕ but an infinite number of independent tangent
vectors tθ . And that is just the tangent vectors, when calculating their cross product,
that serve to construct the normal vector. This problem is not due to the surface itself,
because there is nothing peculiar at the point under consideration, but is merely the
consequence of the specific parametrization chosen. To get out of this difficulty,
another parametrization can be used, but equally well a different point on the surface
may be picked out. One is able to do this, since the surface is orientable.
Looking at Fig. 6.4, we see that there are no obstacles to consider the normal
vector located at the point of parameters θ = π/2 and ϕ = 0. If it aligns with the
versor of the x-axis (i.e., if it “looks” outward), the orientation will be “upward” as
well. Naturally, equally well, one could take θ = π/2 and ϕ = π/2 or any other
point.
Let us now find the tangent vectors at the point (R, 0, 0), which corresponds to
θ = π/2 and ϕ = 0:
tϕ =
∂ r =
−R
sin
θ
sin
ϕ
e
+
R
sin
θ
cos
ϕ
e
= R ey ,
x
y
ϕ=0
∂ϕ ϕ=0
θ=π/2
θ=π/2
∂ r tθ =
= R cos θ cos ϕ ex + R cos θ sin ϕ ey − R sin θ ez = −R ez .
ϕ=0
∂θ ϕ=0
θ=π/2
θ=π/2
(6.2.10)
Calculating their cross product, we get the normal vector in the form:
tϕ × tθ = (R ey ) × (−R ez ) = −R 2 ex .
(6.2.11)
It points inward and not outward, which means that one is dealing with the opposite
parametrization (i.e., called “downward”). This indicates that the parameters should
appear in the reverse order, and in the formula (6.2.6) the role of u should be played
by the angle θ , and the role of v by the angle ϕ.
Instead of the formula (6.2.6), one can use the equivalent expression (6.2.5),
which requires calculating the cross product tθ × tϕ not only at one given point, but
for any parameters. In accordance with what has been established previously, the
vector tθ should appear in the first place. In this way, one gets
6.2 Calculating Integrals over Surfaces
193
tθ × tϕ = (R cos θ cos ϕ ex + R cos θ sin ϕ ey − R sin θ ez )
×(−R sin θ sin ϕ ex + R sin θ cos ϕ ey )
= R 2 sin2 θ cos ϕ ex + R 2 sin2 θ sin ϕ ey + R 2 sin θ cos θ ez ,
(6.2.12)
where the relations ex × ey = ez , ey × ez = ex , ez × ex = ey have been used.
The reader can easily check that the coefficients that occur in front of the versors
ex , ey , and ez are exactly Jacobian determinants ∂(y, z)/∂(θ, ϕ), ∂(z, x)/∂(θ, ϕ),
and ∂(x, y)/∂(θ, ϕ) from the formula (6.2.6). Using the formula (6.2.5) or (6.2.6),
we obtain the required value of the flux:
π/2
2π
S =
Bd
S
dθ [R sin θ cos ϕ ·R 2 sin2 θ cos ϕ
dϕ
0
0
Bx (θ,ϕ)
2
+ R sin θ sin ϕ ·R 2 sin2 θ sin ϕ + R cos
θ ·R sin θ cos θ ]
Bz (θ,ϕ)
By (θ,ϕ)
π/2
2π
=R
3
dθ sin θ = 2π R 3 .
dϕ
0
(6.2.13)
0
This exercise will appear again in the next section, dealing with Stokes’ theorem.
Problem 2
The flux of the vector field B = y 2 ex + z2 ey + x 2 ez through the surface of the
circle x 2 + y 2 ≤ 1 for z = 0, oriented “downward” will be found.
Solution
In order to calculate the flux, the formula (6.2.5) can be used. To this end two tangent
vectors are needed, which will be obtained upon differentiating the position vector
r with respect to the parameters. Because the surface being examined is contained
in the plane xy, the natural parametrization is
x(ϕ, ρ) = ρ cos ϕ,
y(ϕ, ρ) = ρ sin ϕ,
z(ϕ, ρ) = 0,
(6.2.14)
194
6 Examining Oriented Curvilinear Integrals
with 0 ≤ ϕ < 2π and 0 ≤ ρ ≤ 1. In accordance with the formulas (6.2.3) the
tangent vectors have the form:
tϕ =
∂ r
= −ρ sin ϕ ex + ρ cos ϕ ey ,
∂ϕ
tρ =
∂ r
= cos ϕ ex + sin ϕ ey ,
∂ρ
(6.2.15)
and their cross product gives
tϕ × tρ = (−ρ sin ϕ ex + ρ cos ϕ ey ) × (cos ϕ ex + sin ϕ ey ) = −ρ ez .
(6.2.16)
As one can see, this is a normal vector pointing downward, i.e., the orientation
which has been chosen by selecting the parameter ϕ as the first one, and ρ as the
second one (i.e., tϕ as the first tangent vector, and tρ as the second one), complies
with that given in the text of the exercise.
Now one can complete the formula (6.2.5), obtaining
2π
S =
Bd
S
1
dρ [y(ϕ, ρ)2 ex + z(ϕ, ρ)2 ey + x(ϕ, ρ)2 ez ] · [−ρ ez .]
dϕ
0
0
2π
=−
1
π
dρ ρ 3 = − .
4
0
dϕ cos2 ϕ
0
=π
(6.2.17)
=1/4
The negative value of the flux for the specified surface could have been easily
predicted. The vector B has a positive coefficient at ez , which means that it is
directed upward. In turn, the vector normal to the surface, in accordance with the
chosen orientation, “looks” downward. A scalar product of two vectors inclined at
an obtuse angle with respect to each other is obviously negative.
Problem 3
""
The integral S B · d S will be calculated, where B = (z − x) ex + (z − 2y) ez , and
S is the surface of the cube specified by the planes:
x = 0,
y = 0, z = 0,
x = 1, y = 1,
oriented “outward.”
z = 1,
(6.2.18)
6.2 Calculating Integrals over Surfaces
195
Solution
In Fig. 6.5, the faces of the cube in question are drawn. Since this surface is not
smooth (there are edges), for each of the walls S1 , . . . , S6 we have to choose the
parametrization and fix the orientation so that the normal vector “looks” outward.
This stays in line with the method of “smoothing edges,” mentioned in the previous
section.
Let us consider below all the faces one by one.
1. S1 . It is the bottom wall, lying in the plane xy. The simplest parametrization is,
therefore,
x(u, v) = u,
y(u, v) = v,
z(u, v) = 0,
(6.2.19)
where 0 ≤ u, v ≤ 1. In this case, the tangent vectors correspond to the versors of
the axes:
tu = ex ,
tv = ey ,
(6.2.20)
and their cross product provides
tu × tv = ex × ey = ez .
(6.2.21)
This vector points to the interior of the cube and, therefore, when calculating the
integral, one ought to change the order of tangent vectors or simply add a minus
sign in front of the normal vector:
%
$
r (u, v)) · tv × tu dudv
B(
S =
Bd
S1
D
1
=
1
dv[(0 − u) ex + (0 − 2v) ez ] · [−ez ] = 2
du
0
1
0
1
dv v = 1.
du
0
0
(6.2.22)
Fig. 6.5 The surface of the
cube (6.2.18). The invisible
walls are marked with arrows
196
6 Examining Oriented Curvilinear Integrals
2. S2 . This time one is dealing with the upper wall (the “lid”), located in the plane
z = 1. The parametrization that gets imposed is
x(u, v) = u,
y(u, v) = v,
z(u, v) = 1,
(6.2.23)
for 0 ≤ u, v ≤ 1. As before, the tangent vectors are the versors of the axes:
tu = ex ,
tv = ey ,
(6.2.24)
so that their vector product is identical as above:
tu × tv = ex × ey = ez ,
(6.2.25)
with this difference that presently it is correctly oriented, as the vector ez points
outward. Hence one has
%
$
r (u, v)) · tu × tv dudv
B(
S =
Bd
S2
D
1
=
1
dv[(1 − u) ex + (1 − 2v) ez ] · ez =
du
0
1
0
1
dv (1 − 2v) = 0.
du
0
0
(6.2.26)
3. S3 . The subsequent surface is the left wall, lying in the plane xz. We choose the
parametrization in the form:
x(u, v) = u,
y(u, v) = 0,
z(u, v) = v,
(6.2.27)
where again 0 ≤ u, v ≤ 1. The tangent vectors are:
tu = ex ,
tv = ez ,
(6.2.28)
and their cross product
tu × tv = ex × ez = −ey
(6.2.29)
is directed to the left, out of the cube. One has, therefore,
%
$
r (u, v)) · tu × tv dudv
B(
S =
Bd
S3
D
1
=
1
dv[(v − u) ex + (v − 0) ez ] · [−ey ] = 0,
du
0
0
(6.2.30)
6.2 Calculating Integrals over Surfaces
197
since B is tangential to the surface S3 (B does not have the component By ).
4. S4 . For the right wall, one must again get 0, since here B is tangential to this
surface as well.
5. S5 . Now we move on to the back wall lying in the plane yz. The parametrization
has the form:
x(u, v) = 0,
y(u, v) = u,
z(u, v) = v,
(6.2.31)
again with 0 ≤ u, v ≤ 1. The tangent vectors now are
tu = ey ,
tv = ez .
(6.2.32)
Their vector product
tu × tv = ey × ez = ex
(6.2.33)
is directed inward. This means that when calculating the integral, one has to plug
in the additional minus sign:
%
$
r (u, v)) · tv × tu dudv
B(
S =
Bd
S5
D
1
=
1
dv[(v − 0) ex + (v − 2u) ez ] · [−ex ] = −
du
0
1
0
1
1
dv v = − .
2
du
0
0
(6.2.34)
6. S6 . The front wall is still left, for which x = 1. We have
x(u, v) = 1,
y(u, v) = u,
z(u, v) = v,
(6.2.35)
where again 0 ≤ u, v ≤ 1. The tangent vectors are identical as above:
tu = ey ,
tv = ez ,
(6.2.36)
and their vector product:
tu × tv = ey × ez = ex
(6.2.37)
defines the correct orientation. One gets
%
$
r (u, v)) · tu × tv dudv
B(
S =
Bd
S6
D
1
=
1
dv[(v − 1) ex + (v − 2u) ez ] · ex =
du
0
1
0
1
1
dv(v − 1) = − .
2
du
0
0
(6.2.38)
198
6 Examining Oriented Curvilinear Integrals
The total flux is then equal to zero:
S =
Bd
S
S +
Bd
S1
S +
Bd
S2
S +
Bd
S3
S +
Bd
S4
S = 1 + 0 + 0 + 0 − 1 − 1 = 0.
Bd
2 2
+
S
Bd
S5
(6.2.39)
S6
This exercise will be again the subject of our interest in the next section. We will
see then that instead of calculating the fluxes through the subsequent walls, one can
easily and readily obtain the final result, with the use of Stokes’ theorem. The result
obtained (i.e., total flow equal to zero) will turn out not to be accidental. It will be
seen that it is simply much akin to the physical law, which states that the total flux of
a magnetic field by a closed surface always vanishes, due to the absence of magnetic
charges.
6.3 Using Stokes’ Theorem
Problem 1
The Green theorem in R2 will be verified, taking
F =
1
ex + ey
x+y
(6.3.1)
and the curve γ constituting the boundary of the closed area:
(
)
S2 = (x, y) ∈ R2 | 0 ≤ y ≤ x ∧ 1 ≤ x + y ≤ 5 .
(6.3.2)
Solution
Stokes’ theorem, certainly known to the reader from the lecture of analysis and
recalled in the theoretical introduction, can be written in terms of differential forms
as
k−1
k−1
d ω=
Sk
ω,
∂Sk
(6.3.3)
6.3 Using Stokes’ Theorem
199
where Sk stands for a smooth, orientable, compact, k-dimensional surface contained
k−1
in Rn (therefore k ≤ n), ∂Sk for its k − 1-dimensional boundary, and ω for a
differential form of the degree k − 1. The formula (6.3.3) means that instead of
integrating a form over the boundary of a surface, one can calculate the integral
of the exterior derivative of this form over the surface itself. In many cases, this
theorem constitutes the convenient tool for finding the integrals without their tedious
calculations.
This exercise is aimed at getting used to this theorem and gaining some
proficiency in applying it. The expressions on the left-hand and right-hand sides
of (6.3.3) will be independently calculated and one will be able to verify whether
the same results are obtained. In our case, the surface Sk lies on the plane xy, and
the vector F is contained in it as well. Thus one has k = n = 2. With the vector F ,
a 1-form is associated in the following known way:
1
ωF = Fx dx + Fy dy =
1
dx + dy,
x+y
(6.3.4)
and its exterior derivative equals
1
d ωF =
1
dx ∧ dy.
(x + y)2
(6.3.5)
As we know, in this particular case the two-dimensional Stokes’ theorem bears
the name of the Green theorem, often written without referring to differential forms
as:
∂Fy
∂Fx
−
dxdy = (Fx dx + Fy dy).
(6.3.6)
∂x
∂y
S2
∂S2
The surface S is shown in Fig. 6.6. Its boundary γ = ∂S, which is a piecewise
smooth curve, consists of four smooth fragments, drawn with thicker lines.
Let us calculate first the left-hand side of (6.3.3) or, as equivalent, of (6.3.6). The
natural parametrization of the surface S2 is
x(u, v) = u,
y(u, v) = v,
(6.3.7)
where, as easily deciphered from the drawing, it is convenient to split the range of
variation of u into 3 intervals. If u ∈ [1/2, 1[, then 1 − u ≤ v ≤ u. In turn for
u ∈ [1, 5/2[, one has 0 ≤ v ≤ u, and for u ∈ [5/2, 5] there occurs 0 ≤ v ≤ 5 − u.
For the selected parametrization, the corresponding Jacobian determinant is
equal to unity:
∂(x, y)
=1
∂(u, v)
(6.3.8)
200
6 Examining Oriented Curvilinear Integrals
Fig. 6.6 The surface S
drawn in gray with the edges
marked with thicker lines
and the formula (6.0.4) gives
1
d ωF
S2
1
=
5/2
u
du
1/2
1−u
1
=−
1/2
1
=−
1/2
1
dv
+
(u + v)2
1 u
du
−
u + v 1−u
1
− 1 du +
2u
5
u
1
dv
+
(u + v)2
du
1
0
5/2
1 u
du
−
u + v 0
5/2
1
5/2
5/2
5
1
du −
2u
1
5
5−u
du
dv
0
1
(u + v)2
1 5−u
du
u + v 0
1 1
1
−
du = log 5.
5 u
2
(6.3.9)
5/2
Now let us find the right-hand side of the formula (6.3.3) or (6.3.6). The boundary
curve is not smooth, as already discussed, but it breaks down in a natural way into
four smooth segments contained between the points A and B, B and C, C and
D, and D and A subsequently, denoted with symbols γAB , γBC , γCD , and γDA ,
respectively. The integrals will be calculated independently along each of them, but
before we start, the orientation of the curve has to be established. In order for the
conclusion of Stokes’ theorem to be true, the orientations of the interior and of the
boundary of the surface must agree with each other. Reversing the orientation of the
boundary changes the sign of the right-hand side of the expression (6.3.3), without
6.3 Using Stokes’ Theorem
201
modifying that on the left, so the equality can hold only for a concrete orientation.
The principle one should be guided by when fixing the orientation of the edge is
as follows: the orientations of the interior and of the boundary are consistent if
the external normal vector complements the orientation of the edge to that of the
interior. Below we will take a look at how this principle works in practice. When
calculating (6.3.9), it was assumed that the first parameter is u, and the second one
v. One can now, using the ordered tangent vectors:
tv = [0, 1],
tu = [1, 0],
construct the 2 × 2 matrix and calculate its determinant:
10
tu
= det
= 1.
det
tv
01
(6.3.10)
(6.3.11)
Since these vectors are independent, it is different from zero for sure. It could,
however, be either positive or negative. In particular, if the two parameters switched
their places, and thus the tangent vectors as well, the determinant would change
its sign. As one can see from (6.3.11), a positive value is obtained, so the selected
orientation will be called “positive” too.
Now let us have a look at the boundary and, for example, fix the orientation in
such a way that the curve γ is traveled in the counterclockwise direction. For this
purpose we choose for the segment AB the tangent vector in the form t = [1, 0]. As
we know, upon “smoothing the corners” it can be moved onto the other segments
of the curve γ . If now the normal external vector n is added at the forefront (e.g.,
in the form of n = [0, −1] for the segment γAB ) to the previously selected tangent
vector t, then—for consistent orientations—their determinant should be positive as
well. Let us see that in fact it is so:
n
0 −1
det
= det
= 1 > 0.
(6.3.12)
t
1 0
The chosen orientation of a curve, i.e., walking in the counterclockwise direction,
is always correct for the positive orientation of the interior and we accept this as a
rule for the future, without referring to the determinants.
One can now proceed to calculate the boundary integrals.
1. γAB . Let us choose the parametrization x(u) = u, y(u) = 0. The tangent vector
is t = ex . Therefore
5
1
du = log 5.
u
1
ωF =
γAB
(6.3.13)
1
2. γBC . Now the parametrization x(u) = u, y(u) = 5 − u is chosen. The tangent
vector has the form t = ex − ey . The curve γBC , however, travels in the direction
of the decreasing value of the parameter u (orientation!), so we have
202
6 Examining Oriented Curvilinear Integrals
5/2
1
ωF =
γBC
1
− 1 du = 2.
5
(6.3.14)
5
Here one could equally well change the tangent vector into one that is in the
direction of our motion on the curve, i.e., to choose t = −ex + ey , but integrate
over the parameter u from 5/2 to 5.
3. γCD . Choosing the parametrization x(u) = u, y(u) = u, one gets the tangent
vector as t = ex + ey . Therefore, one has
1/2
1
ωF =
γCD
1
1
+ 1 du = − log 5 − 2.
2u
2
(6.3.15)
5/2
4. γDA . Now we fix the parametrization as x(u) = u, y(u) = 1 − u. The tangent
vector appears to be in the form t = ex − ey . Hence
1
1
ωF =
γCD
(1 − 1) du = 0.
(6.3.16)
1/2
Upon collecting all the results, we finally obtain
1
1
ωF =
γ
ωF +
γAB
γBC
1
ωF +
γCD
1
ωF +
1
ωF =
1
log 5.
2
(6.3.17)
γDA
Thus one has also made sure that when calculating the expressions on both sides of
the Eq. (6.3.3), the same result is obtained.
Problem 2
Stokes’ theorem in R3 will be verified for the vector field
B = y 2 ex + x 2 ey
(6.3.18)
and the domain with boundary:
(
)
S2 = (x, y, z) ∈ R3 | 4x 2 + y 2 + z2 = 4 ∧ x, y, z ≥ 0 .
(6.3.19)
6.3 Using Stokes’ Theorem
203
Solution
In Fig. 6.7, the surface S2 is depicted: it is the eighth part of the surface of an
ellipsoid. The boundary constitutes a curve composed of three smooth parts: γ1 , γ2 ,
and γ3 . The direction of the motion when integrating, which indicates the orientation
of the curve, is marked with arrows. One has to consider now, which orientation
should be attributed to the surface S2 itself, for both orientations to be consistent.
In the previous problem, we formulated a principle valid for the flat case: it could
be called the “rule of bypassing the interior on one’s left” (then the orientation is
positive). It can be easily extended to two-dimensional surfaces immersed in R3
and called the “right-hand screw rule.” The reader knows it already very well from
physics classes at school, where the direction of the magnetic field produced by a
current flowing in a conductor was to be determined.
When encircling the interior in accordance with the specified direction, the rule
states that the normal vector n at the chosen point of the surface S2 and fixing its
orientation must be directed upward, as is shown in the figure. However, as we
know, the direction of the normal vector stems from the chosen sequence of tangent
vectors, i.e., the sequence of parameters. In the figure, the two tangent vectors are
drawn, one directed along the “parallel,” and the other along the “meridian.” It is
clear that the normal vector is set by the cross product tθ × tϕ , which means that
Fig. 6.7 The surface S2 with its boundary, drawn with thicker lines
204
6 Examining Oriented Curvilinear Integrals
the correct orientation of the interior will be given by the following sequence of
parameters: θ , ϕ. These angles, when parameterizing an ellipsoid surface, play the
same role as in the case of a spherical surface. Anyway, the parametrization of the
ellipsoid has already been encountered in the second part of this book series (see
formulas (13.3.23)). This time we are interested solely in the surface, so we have
x(θ, ϕ) = sin θ cos ϕ,
y(θ, ϕ) = 2 sin θ sin ϕ,
z(θ, ϕ) = 2 cos θ,
(6.3.20)
with 0 ≤ θ, ϕ ≤ π/2, and thus,
∂ r
= cos θ cos ϕ ex + 2 cos θ sin ϕ ey − 2 sin θ ez ,
∂θ
∂ r
tϕ =
= − sin θ sin ϕ ex + 2 sin θ cos ϕ ey .
∂ϕ
tθ =
(6.3.21)
By selecting some exemplary values of the angles θ and ϕ (e.g., θ = ϕ = π/4),
the reader can easily check that both vectors are correctly shown in the figure (they
“look” in the direction of increasing parameter values).
After having fixed the orientation, one can already begin to calculate the
the following 1-form is associated:
integrals. With the vector B,
1
ωB = y 2 dx + x 2 dy.
(6.3.22)
Next the curve ∂S2 is divided into three smooth fragments:
1
1
ωB =
γ1
∂S2
1
ωB +
1
ωB +
γ1
ωB
(6.3.23)
γ2
and the integrals over each of them can be calculated.
1. γ1 . The parametrization of the curve γ1 stems from (6.3.20) after having put
ϕ = 0:
x(θ ) = sin θ,
y(θ ) = 0,
z(θ ) = 2 cos θ,
(6.3.24)
(0 · (cos θ ) + cos2 θ · 0)dθ = 0.
(6.3.25)
which implies
0
1
ωB =
γ1
π/2
2. γ2 . This time we substitute θ = π/2 into (6.3.20):
x(ϕ) = cos ϕ,
y(ϕ) = 2 sin ϕ,
z(ϕ) = 0,
(6.3.26)
6.3 Using Stokes’ Theorem
205
which yields
π/2
2
ωB =
γ1
(4 sin2 ϕ · (− sin ϕ) + cos2 ϕ · 2 cos ϕ) dϕ
0
π/2
=
π/2
(2 cos ϕ − 4 sin ϕ) dϕ = −2
3
3
0
cos3 ϕ dϕ,
(6.3.27)
0
the latter step having been done with the use of the known fact that
" π/2 n
" π/2
sin ϕ dϕ = 0 cosn ϕ dϕ. If necessary, it can be easily demonstrated
0
upon substituting ϕ → π/2 − ϕ into the integral. The integral (6.3.27) can
be calculated in a simple way, with the use of the trigonometric Pythagorean
identity and upon substituting the new variable u = sin ϕ:
π/2
2
ωB = −2
γ1
1
4
(1 − u2 )du = − .
3
(1 − sin2 ϕ) cos ϕ dϕ = −2
0
(6.3.28)
0
3. γ3 . The last integral equals zero, as can be seen without any calculations. The
curve γ3 lies in the plane yz and, therefore, one constantly has x = 0 on it. The
derivative of x with respect to the parameter, that now is the angle θ , equals zero
as well and the two terms in (6.3.22) vanish, as it was for the curve γ1 .
Then the entire result is
4
1
ωB = − .
3
(6.3.29)
∂S2
Now let us calculate the left-hand side of (6.3.3) and make sure that it equals
1
−4/3 as well. The exterior derivative of the form ωB amounts to
1
d ωB = 2(x − y) dx ∧ dy.
(6.3.30)
The entire intricate formula (6.0.4) boils down to only one term, which is easy to
calculate (remember the appropriate order of parameters: θ and ϕ):
π/2
1
d ωB = 2
S2
π/2
dϕ (x(θ, ϕ) − y(θ, ϕ))
dθ
0
0
π/2
π/2
=2
=sin(2θ)
dϕ (sin θ cos ϕ − 2 sin θ sin ϕ) sin(2θ )
dθ
0
∂(x, y)
∂(θ, ϕ)
0
206
6 Examining Oriented Curvilinear Integrals
π/2
=2
π/2
(sin θ − 2 sin θ ) sin 2θ dθ = −4
0
sin2 θ cos θ dθ
0
1
4
u2 du = − .
3
= −4
(6.3.31)
0
The result identical to (6.3.29) has been obtained.
Problem 3
Using the Green theorem, the integral
"
V · d l will be calculated, where V =
γ
2xy ex + x 2 ey , if the curve γ , lying in the plane R2 , is:
(a) the segment of the straight line between the points A(0, 0) and B(1, 2),
(b) the polyline A(0, 0) → C(1, 0) → B(1, 2),
(c) the fragment of the parabola y = 2x 2 between the points A and B.
Solution
We have already seen this problem in Sect. 6.1. However, then, Stokes’ (Green’s)
theorem was not used, so we are now returning to this problem for a while. The idea
of solving this type of problem using Stokes’ theorem is as follows. From among
curves (in this exercise they are γ1 , γ2 , and γ3 ) we select that for which calculating
"
" 1
the integral γ ωV = γ V · d l from the definition is the simplest, and then find this
value. Then the invoked theorem allows us to find the integrals along other curves
(let us denote some other curve with the symbol γ ) connecting the same end-points
as γ without calculating them. For one has
1
1
ωV −
γ
1
ωA V =
γ
d ωV ,
(6.3.32)
S2
where S2 is the surface, whose boundary is the curve consisting of γ and γ , and
the minus sign comes from the fact that in order for γ ∪ γ to be a closed oriented
curve, one needs to reverse the orientation of γ .
The idea is, therefore, based on the hope that perhaps calculating the right-hand
" 1
side will turn out to be much easier than finding directly γ ωV . Then, one will be
able to find the last integral, using the formula:
6.3 Using Stokes’ Theorem
207
1
γ
1
1
ωV =
ωV −
γ
d ωV .
(6.3.33)
S2
With this approach, the current exercise becomes particularly simple. This is
because it has already been found (see (6.1.12)) that the 1-form associated with the
vector V is closed, which means that the right-hand side of (6.3.32) equals zero.
Thereby, we simply have
1
1
ωV =
γ
ωV
(6.3.34)
γ
and all integrals over arbitrary curves with common ends lying in a contractible
domain must be equal to one another. The result of Problem 3 from Sect. 6.1 is no
longer a surprise.
As we remember, we know already the value of the integral over γ3 , depicted in
Fig. 6.8:
1
ωV = 2.
(6.3.35)
γ3
Now, if one wants to find
"
γ1
1
ωV , it is sufficient to apply this Stokes’ theorem to the
surface marked with dark gray color, for which γ1 ∪ γ3 constitutes a boundary. In
that way, one immediately gets
Fig. 6.8 The domains
between the curves γ3 , γ1
(dark gray), and γ3 , γ2 (light
gray)
208
6 Examining Oriented Curvilinear Integrals
1
ωV = 2.
(6.3.36)
γ1
In order to find
"
1
ωV , the same idea is applied to the surface marked with light
γ2
gray with the obvious result:
1
ωV = 2.
(6.3.37)
γ2
It is clear that the outcome will be identical for any curve connecting the points A
and B. This fact can also be stated in the following manner: the integral of a closed
1-form, over a closed curve lying in a contractible domain, is equal to zero.
Problem 4
""
Using Stokes’ theorem, the integral ∂S3 B · d S will be found, where B = (z −
x) ex + (z − 2y) ez , and ∂S3 is the surface of the cube (i.e., S3 ) specified with the
inequalities:
0 ≤ x ≤ 1, 0 ≤ y ≤ 1,
0 ≤ z ≤ 1,
(6.3.38)
oriented “outward.”
Solution
The flux in question was already found in Problem 3 of the previous section. The
job required then a lot of patience because one had to calculate successively fluxes
through all 6 walls, first fixing their orientations. Now we are going to use Stokes’
theorem, which will make the exercise very easy. It has now the form:
2
2
d ωB =
S3
ωB ,
(6.3.39)
∂S3
2
where ωB is the 2-form associated with the vector field B (see Sect. 5.2):
2
ωB = Bx dy ∧ dz + By dz ∧ dx + Bz dx ∧ dy.
(6.3.40)
6.3 Using Stokes’ Theorem
209
As we know from the theoretical introduction, Stokes’ theorem in this case bears
the name of the Gauss–Ostrogradsky’s theorem (in short GO) and it can be given
the vector form:
B d S.
· B d 3 r =
∇
S3
(6.3.41)
∂S3
The right-hand side is just the flux to be found. Instead of being bothered with the
subsequent walls of the cube, let us calculate the expression on the left-hand side.
One has
· B = −1 + 0 + 1 = 0.
∇
(6.3.42)
Thus, the vector field B is sourceless (hence it is the curl of a certain field A,
which is called the vector potential). Consequently, the integral on the left-hand
side of (6.3.41) vanishes, which implies that the total flux equals zero as well. In
this simple way our earlier result has been restored.
Note that if instead of a cube one considered any other (but closed!) surface,
· B = 0 means, after all, that the
the total flux would vanish as well. The result ∇
left-hand side of (6.3.41) always equals zero, so the same must apply to the righthand side. If we for example recognized that B is a magnetic field (and it could be,
then the result would be well known
× A),
because there exists A such that B = ∇
for physicists. It would simply entail that no magnetic charges exist.
As the reader has certainly noticed, the question of fitting the interior orientation
has not been addressed yet. This is understandable: since the integral turns out to be
0, it is inessential whether it is +0, or −0. However, in some other examples, the
integral may prove to be nonvanishing (for example, if rather than magnetic flux,
one would calculate the electric one), so at the end of this exercise this issue will be
reviewed.
Let us start with the boundary, that is to say, with the surface of the cube. In
accordance with the content of the exercise, the normal vector n is to be oriented
“outward.” Thus, if the two parameters were chosen and in consequence at a given
point of the surface (forget the edges, which can always be smoothed) one had two
tangent vectors, t1 and t2 , it should occur
t1 × t2 = α n,
(6.3.43)
up to the certain positive constant α. This means that
n |2 > 0
n · (t1 × t2 ) = α|
(6.3.44)
210
6 Examining Oriented Curvilinear Integrals
or equivalently:
⎡
⎤
n
det ⎣ t1 ⎦ > 0.
t2
(6.3.45)
Thanks to the form of the inequality, this orientation can be called “positive.”
In order for the orientations of the interior and of the boundary to be consistent,
the external normal vector must complement the border orientation to that of the
interior (this rule has already been met in the flat case). Thereby, if the orientation
of the interior stays in line with (6.3.45), it must be positive as well, i.e.:
⎡
⎤
tu
det ⎣ tv ⎦ > 0,
tw
(6.3.46)
where tu , tv and tw are some vectors tangent to S3 , stemming from the selected
parameters:
x = x(u, v, w),
y = y(u, v, w),
z = z(u, v, w).
(6.3.47)
For the interior of the cube S3 , the simplest parametrization has the form:
x(u, v, w) = u,
y(u, v, w) = v,
z(u, v, w) = w,
(6.3.48)
which gives
tu = ex ,
tv = ey ,
tw = ez .
(6.3.49)
As a result,
⎡
⎤
⎡
⎤
1 00
tu
det ⎣ tv ⎦ = det ⎣ 0 1 0 ⎦ = 1 > 0
001
tw
(6.3.50)
and it is clear that the parametrization (6.3.48) defines the orientation of the interior
consistent with the “outward” orientation of the boundary.
Problem 5
Using Stokes’ theorem, the flux of the vector field B = x ex +y ey +zez through the
upper hemisphere of the radius R oriented upward, with the center at the coordinate
system origin, will be found.
6.3 Using Stokes’ Theorem
211
Solution
The flux in question was found in Problem 1 of the previous section, by the direct
calculation of the integral of the vector field B over the upper hemisphere. Now
we would like, for the same purpose, to make use of Stokes’ (GO) theorem. Let
us denote with the symbol S3 the upper half of the ball of radius R. Its boundary
is composed of two parts: the hemisphere denoted with Sa (see Fig. 6.4) and the
bottom, i.e., the circle of radius R lying in the plane xy, for which the symbol Sb is
reserved.
The surface Sa , admittedly, is not closed, i.e., it is not the boundary of S3 ;
however, it occurs
Sa ∪ Sb = ∂S3 ,
(6.3.51)
which allows us to write Stokes’ theorem as:
2
2
d ωB =
2
ωB +
S3
Sa
ωB ,
(6.3.52)
Sb
2
with the form ωB :
2
ωB = x dy ∧ dz + y dz ∧ dx + z dx ∧ dy.
(6.3.53)
The relation (6.3.52) allows us to find the needed flux from the equation:
2
2
ωB =
Sa
2
d ωB −
S3
ωB ,
(6.3.54)
§b
i.e.,
S =
Bd
Sa
S.
Bd
3r −
· Bd
∇
S3
(6.3.55)
Sb
The idea contained in these equations is the following: since we are to find the flux
through a surface that is not closed, let us for a while supplement it to a closed
one (for which the GO theorem may be applied), and then subtract what was added
from the outcome. Calculating the integral over the bottom should be, after all, much
easier than over the hemisphere since the surface Sb is flat. In turn, the divergence
of the field B is a constant:
· B = 1 + 1 + 1 = 3,
∇
(6.3.56)
212
6 Examining Oriented Curvilinear Integrals
which means that the first integration is simply three halves of the sphere volume:
3 r = +2π R 3 .
· Bd
∇
(6.3.57)
S3
With the symbol +, it was emphasized that the positive orientation of the interior
had been chosen, which, as we know from the previous exercise, stays in line with
the “outward” orientation of the boundary.
It remains to calculate the flux through the surface Sb oriented “downward” (i.e.,
“outward” with respect to S3 ). Note that on this surface one has z = 0, which entails
Bz = 0. Therefore, the contribution could come only from the x and y components
However, they both are perpendicular to the normal vector n = −ez .
of the field B.
Thereby, one gets
S = 0.
Bd
(6.3.58)
Sb
As a consequence, it stems from the Eq. (6.3.55) in this simple way that the flux
looked for is equal to
S = 2π R 3
Bd
(6.3.59)
Sa
in accordance with (6.2.13).
Problem 6
Stokes’ theorem will be verified for the form
2
ω= (x − y 2 ) dy ∧ dz + xyz dz ∧ dx + z2 dx ∧ dy
(6.3.60)
and the domain S3 with the boundary:
(
)
S3 = (x, y, z) ∈ R3 | x 2 + y 2 ≤ 1 ∧ 0 ≤ z ≤ 2 ∧ x ≥ 0 .
(6.3.61)
Solution
In this problem, one has to independently calculate the left-hand and right-hand
sides of the Eq. (6.3.3) and check if the results are identical. In the case of a
6.3 Using Stokes’ Theorem
213
three-dimensional surface S3 contained in R3 —as we know from the previous
examples—this equation might also be given the vector form, upon associating with
2
ω the vector E:
S,
Ed
(6.3.62)
∂S3 = Sa ∪ Sb ∪ Sc ∪ Sd .
(6.3.63)
3r =
· Ed
∇
S3
∂S3
the boundary of S3 being
However, below we prefer to use differential forms. Thus, we are required to make
sure that
2
2
S3
2
ω+
d ω=
Sa
2
ω+
Sb
2
ω+
Sc
ω.
(6.3.64)
Sd
In Fig. 6.9, the domain S3 is drawn together with its boundary consisting of four
parts: Sa , which will be called the “lid,” Sb —the “bottom,” Sc —the “back,” and
Sd —the “belly.” Let us fix the orientation of the boundary to be “outward” and
choose the positive orientation of the interior, i.e., consistent with the former.
First let us consider the left-hand side of the Eq. (6.3.64). The exterior derivative
2
of the form ω is
2
d ω= (1 + xz + 2z) dx ∧ dy ∧ dz.
Fig. 6.9 The domain D with
the boundary consisting of
four parts: Sa —“lid,”
Sb —“bottom,” Sc —“back,”
and Sd —“belly”
(6.3.65)
214
6 Examining Oriented Curvilinear Integrals
Then one has to choose the parametrization of S3 and the orientation related to it.
Because of the shape of the solid (a half of a cylinder), cylindrical variables get
imposed, for which
x(r, ϕ, u) = r cos ϕ,
y(r, ϕ, u) = r sin ϕ,
z(r, ϕ, u) = u,
(6.3.66)
where 0 < r < 1, −π/2 < ϕ < π/2, and 0 < u < 2. Assume, for a trial, the
following sequence of parameters: r, ϕ, u, and check whether
⎤
tr
det ⎣ tϕ ⎦ > 0.
tu
⎡
(6.3.67)
Then one would have the positive orientation, staying in line with the “outward”
orientation of the boundary.
The tangent vectors at some selected point have the form:
tr = cos ϕ ex + sin ϕ ey ,
tϕ = −r sin ϕ ex + r cos ϕ ey ,
tu = ez ,
(6.3.68)
so for the determinant (6.3.67), we have
⎡
⎡ ⎤
cos ϕ sin ϕ
tr
⎦
⎣
⎣
det tϕ = det −r sin ϕ r cos ϕ
0
0
tu
⎤
0
0 ⎦ = r > 0.
1
(6.3.69)
This result is also the value of the Jacobian determinant ∂(x, y, z)/∂(r, ϕ, u) to be
made use of later in this solution.
Thereby, the orientation is positive, so the left-hand side of (6.3.64) should be
calculated with the use of the formula (6.0.4) in the following way:
π/2
2
dω=
1
dϕ
−π/2
S3
0
π/2
=
0
dϕ
=r
du (1 + ru cos ϕ + 2u)r
0
π/2
1
dr (2+2r cos ϕ+4)r=
dϕ
0
∂(x, y, z)
∂(r, ϕ, u)
2
dr
0
π/2
−π/2
du (1 + x(r, ϕ, u)z(r, ϕ, u) + 2z(r, ϕ, u))
1
−π/2
=
2
dr
−π/2
2
4
3 + cos ϕ dϕ = 3π + .
3
3
(6.3.70)
6.3 Using Stokes’ Theorem
215
Now let us pass to the boundary, starting our discussion from the “lid.” Its
parametrization may be chosen in the form:
x(r, ϕ) = r cos ϕ,
y(r, ϕ) = r sin ϕ,
z(r, ϕ) = 2,
(6.3.71)
the range of variation of the parameters r, ϕ remaining the same as above. There are
two tangent vectors:
tr = cos ϕ ex + sin ϕ ey ,
tϕ = −r sin ϕ ex + r cos ϕ ey ,
(6.3.72)
and their vector product “looks” upward, which means outward, as it should be:
tr × tϕ = (cos ϕ ex + sin ϕ ey ) × (−r sin ϕ ex + r cos ϕ ey ) = r ez .
(6.3.73)
Thus, the first term of the sum on the right-hand side of (6.3.64) can be easily
calculated. It should be noted that in this case z is constant and, therefore, the term
containing dz will not appear at all. So one has
π/2
2
z dx ∧ dy = 4
2
ω=
Sa
dϕ
−π/2
Sa
π/2
1
0
1
∂(x, y)
dr
=4
dϕ dr r = 2π.
∂(r, ϕ)
−π/2
0
=r
(6.3.74)
The value coming from the “bottom” can be immediately estimated as equal to
2
zero. The first two terms of ω do not give any contribution for the same reason as it
was for the “lid,” and the last one vanishes since z = 0.
Now let us have a look at the flux passing through the “back.” It is a rectangle
lying in the plane yz. The simplest parametrization is:
x(u, v) = 0,
y(u, v) = u,
z(u, v) = v,
(6.3.75)
where −1 ≤ u ≤ 1 and 0 ≤ v ≤ 2. The tangent vectors:
tu = ey ,
tv = ez
(6.3.76)
define the “inward” orientation, since the vector
tu × tv = ey × ez = ex
(6.3.77)
is directed into the interior of the solid. Thereby, when writing down the expression
for the surface integral, one should (in the Jacobian determinant) take v as the first
parameter and u as the second one. In addition, it should be noted that the value of
216
6 Examining Oriented Curvilinear Integrals
x is constant on Sc , which means that the only contribution will come from the first
2
term of ω:
1
2
(x − y ) dy ∧ dz = −
2
ω=
Sc
2
du
−1
Sa
1
2
∂(y, z)
4
dv u
= du dv u2 = .
∂(v, u)
3
−1
0
2
0
=−1
(6.3.78)
As to the “belly,” its parametrization is:
x(ϕ, u) = cos ϕ,
y(ϕ, u) = sin ϕ,
z(ϕ, u) = u,
(6.3.79)
where −π/2 ≤ ϕ ≤ π/2 and 0 ≤ u ≤ 2. The tangent vectors are
tϕ = − sin ϕ ex + cos ϕ ey ,
tu = ez ,
(6.3.80)
and their cross product is oriented “outward”:
tϕ × tu = (− sin ϕ ex + cos ϕ ey ) × ez = cos ϕ ex + sin ϕ ey .
(6.3.81)
Choosing for example ϕ = 0, we see that tϕ = ex . For the other points on the
“belly,” the direction must be correct as well, because this surface is orientable.
From the formula (6.0.4), it can now be found:
2
ω=
Sd
[(x − y 2 ) dy ∧ dz + xyz dz ∧ dx + z2 dx ∧ dy]
Sd
π/2
=
2
dϕ
−π/2
0
∂(y, z)
∂(z, x)
du (cos ϕ − sin2 ϕ)
+u cos ϕ sin ϕ
∂(ϕ, u)
∂(ϕ, u)
=cos ϕ
=sin ϕ
π/2
2
∂(x, y)
=
+u
dϕ du[(cos ϕ − sin2 ϕ) cos ϕ + u cos ϕ sin2 ϕ]
∂(ϕ, u)
−π/2
0
2
=0
π/2
=2
−π/2
π/2
cos ϕ dϕ − 2
2
−π/2
π/2
sin ϕ cos ϕ dϕ + 2
2
sin2 ϕ cos ϕ dϕ = π.
−π/2
(6.3.82)
Now, collecting the contributions coming from all four terms present on the righthand side of (6.3.64), one gets the result 3π + 4/3, i.e., identical to (6.3.70).
6.4 Exercises for Independent Work
217
6.4 Exercises for Independent Work
Exercise 1 Calculate the work L of the force
(a) F = xy ex + yz ey + zx ez over the curve r(t) = t 2 ex + t ey + t 3 /2 ez oriented
pursuant to the increasing value of the parameter t for 0 ≤ t ≤ 1,
(b) F = y sin x ex + xcosy ey + ez over the segment r(t) = t (ex + ey + ez ) oriented
pursuant to the increasing value of the parameter t for −π ≤ t ≤ π ,
(c) F = x ex + y ey + z ez over the helix r(t) = cos t ex + sin t ey + t 2 /2 ez oriented
pursuant to the increasing value of the parameter t for 0 ≤ t ≤ 2π .
Answers
(a) L = 19/32.
(b) L = 4π .
(c) L = 2π 4 .
Exercise 2 Calculate the flux of the vector field
(a) B = (y − z) ex + (z − x) ey + (z − y) ez over the conical surface oriented
“outward” x 2 + y 2 = z2 , for 0 ≤ z ≤ 1,
x
1
y
(b) B = ex + ey + ez over the surface of the ellipsoid x 2 /4 + y 2 /9 + z2 = 1
x
y
z
oriented “outward,”
(c) B = x 2 ex + y 2 ey + z2 ez over the surface (oriented “outward”) constituting
the boundary of the domain defined with the inequalities: x > 0, y > 0, z > 0,
x + y + z < 1.
Answers
(a) = −2π/3.
(b) = 24π .
(c) = 1/4.
Exercise 3 Verify Stokes’ theorem for the differential form
2
ω= xy dy ∧ dz − xz dz ∧ dx + z dx ∧ dy
and the domain with the boundary, defined with the inequalities 0 ≤ z ≤ 4−x 2 −y 2 .
218
6 Examining Oriented Curvilinear Integrals
Exercise 4 Using Stokes’ theorem, calculate the integral I
1
(a) of the form ω= (x +y) dx +(x −y) dy over the upper semicircle x 2 = y 2 = R 2
oriented “clockwise,”
2
(b) of the form ω= 2x 2 dy ∧ dz + xy dz ∧ dx + xz2 dx ∧ dy over the surface
(oriented “outward”) constituting the boundary of the domain defined with the
inequalities: 0 < x < 1, 0 < y < 1, 0 < z < xy,
2
(c) of the form ω= x(z + 1) dy ∧ dz + z2 dz ∧ dy + (z + 1)2 dx ∧ dy over the
lateral surface of the cylinder (oriented “outward”) defined with the inequalities:
x 2 + y 2 = R 2 , −1 ≤ z ≤ 1.
Answers
(a) I = 0.
(b) I = 11/12.
(c) I = 2π R 2 .
Chapter 7
Studying Functions of Complex Variable
The following three chapters are concerned with functions of the complex variable
z = x + iy. The principal notion here is that of the holomorphicity of a function. It
is defined as follows.
Given a complex-valued function f : O → C where O ⊂ C is an open set. The
function f is said to be complex-differentiable at a certain point z0 ∈ O if there
exists the limit
f (z0 ) = lim
z→z0
f (z) − f (z0 )
.
z − z0
(7.0.1)
The value f (z0 ) is called the complex derivative of the function f at z0 . If this
derivative exists at every point of O, the function f is said to be holomorphic on O.
If the complex derivative exists at z0 and on some neighborhood of it, the function
is holomorphic at z0 . The notion of holomorphicity is explained in detail in the
first problem below.
Let the complex function be written as a function of two real variables x and y:
f (z) = u(x, y) + iv(x, y),
(7.0.2)
where u(x, y), v(x, y) are real functions. Then from the existence of (7.0.1), the
following Cauchy–Riemann conditions can be derived:
∂v
∂u
=
,
∂x
∂y
∂u
∂v
=− .
∂y
∂x
(7.0.3)
The opposite theorem is not true unless the partial derivatives of u and v with respect
to x and y are continuous (strictly speaking this requirement can be somewhat
mitigated).
© Springer Nature Switzerland AG 2020
T. Radożycki, Solving Problems in Mathematical Analysis, Part III, Problem Books
in Mathematics, https://doi.org/10.1007/978-3-030-38596-5_7
219
220
7 Studying Functions of Complex Variable
The complex power series
f (z) =
∞
!
an (z − z0 )n ,
(7.0.4)
n=0
in the ball defined with the inequality
|z − z0 | < R,
(7.0.5)
where
R=
lim
n→∞
n
|an |
−1
,
(7.0.6)
defines the holomorphic function f (z). The opposite is also true: a function f (z)
holomorphic on a certain open ball has the Taylor expansion of the form of (7.0.4)
in this ball. Therefore, the holomorphic functions are also called analytic ones.
The complex contour integral
(7.0.7)
f (z)dz
C
is performed according to the formula
b
f (z)dz =
f (z(t))
dz
dt
dt
(7.0.8)
a
C
upon the appropriate parametrization of the curve C with a real parameter a ≤ t ≤ b
through the dependence z(t).
Contour integrals are subject to the following important Cauchy theorem: Given
a holomorphic function f (z) in a simply connected domain O ⊂ C. Then the
integral over a closed contour C lying in O vanishes:
/
f (z) dz = 0.
(7.0.9)
C
7.1 Examining the Holomorphicity of Functions
Problem 1
It will be verified that the functions:
(a) f (z) = z2 ,
(b) g(z) = sin z
are holomorphic on the entire complex plane.
7.1 Examining the Holomorphicity of Functions
221
Solution
We assume that the reader has finished the full course of algebra and is fluent in
manipulation of complex numbers. Therefore, we will not devote time to basic facts
in this field but rather immediately go to the concept of holomorphicity of functions.
It will be appearing quite often in this and in the following chapters. As we know
from the theoretical introduction above, a holomorphic function is just a function
for which the derivative with respect to the complex variable exists. This definition
means that when talking about the holomorphicity of a function on a given set, we
always mean an open set.
Let us first explain why the property of the holomorphicity is so important that
it earned a special name. It turns out that the requirement of the existence of the
derivative for a certain function f (z) at some point or domain on the complex plane
is much stronger than the requirement of differentiability for a real function on the xaxis. Holomorphic functions have a number of important properties that allow to use
them to calculate sums of series, to evaluate definite integrals, to solve differential
equations, etc. A certain particular property of the families of curves generated by
the equation f (z) = const was encountered already in the first volume of this book
series (see Problem 2 in Sect. 10.3).
To explain the concept of holomorphicity, let us consider a certain function g
defined on a certain domain ⊂ R2 with the values in R2 :
g(x, y) =
g1 (x, y)
.
g2 (x, y)
(7.1.1)
Assume that it is of the class C 1 , i.e., it is a function for which there exists the socalled strong (Fréchet) derivative. With this notion, the reader became familiar in
Part II in Sect. 5. The Jacobian matrix of this function has the form:
⎡ ∂g ∂g ⎤
1
1
⎢ ∂x ∂y ⎥
g (x, y) = ⎣ ∂g ∂g ⎦ ,
2
2
∂x ∂y
(7.1.2)
and no special restrictions on the values of these partial derivatives, apart from their
continuity (as functions of two variables), are imposed.
The things are different for a holomorphic function f (z). If the argument of
this function is written in the combined form z = x + iy and the value of this
function, which also is a complex number, in the form f = u + iv (of course now
x, y, u, v ∈ R), then this function formally can be looked at as a function from R2
into R2 . There is, however, an important difference: it does not depend separately
on x and on y, as it was in the case of g, but on one variable z. The variables x and
y are bound (i.e., “complexed”) into one!
222
7 Studying Functions of Complex Variable
A function that would separately depend on x and y should in general depend on
the complex conjugate of z (i.e., z̄ = x − iy), as:
x=
1
(z + z̄),
2
y=
1
(z − z̄).
2i
(7.1.3)
Now—and it is a very strong restriction—a holomorphic function may not depend
on z̄. If the reader encounters an expression that in addition to z also depends on
z̄, it certainly does not represent any holomorphic function! This fact is due to
the requirement of the existence of the derivative. To find it out, let us create the
differential quotient, taking z = x + iy:
f (z + z) − f (z)
z
u(x + x, y + y) + iv(x + x, y + y) − u(x, y) − iv(x, y)
=
x + iy
(7.1.4)
and demand that the limit at z → 0 exists. The existence of such a limit entails
that one has to get the identical result, approaching the point z along the real axis,
imaginary axis, or any other path.
Assume at the beginning that z = x, which means that one is moving along
the “horizontal” line, i.e., with the imaginary part of z fixed: Im z = y. Then the
following value of the limit is obtained:
f (z + z) − f (z)
z→0
z
lim
= lim
x→0
∂u
∂v
u(x + x, y) + iv(x + x, y) − u(x, y) − iv(x, y)
=
+i
.
x
∂x
∂x
(7.1.5)
Now let us choose z = iy, i.e., the straight line is “vertical,” the real part of
z being constant: Rez = x. One has then
f (z + z) − f (z)
z→0
z
lim
= lim
y→0
∂u ∂v
u(x, y + y) + iv(x, y + y) − u(x, y) − iv(x, y)
= −i
+
.
iy
∂y
∂y
(7.1.6)
These two expressions must be equal for a holomorphic function. The value of
the derivative for such a function may not be dependent on the path, along which
7.1 Examining the Holomorphicity of Functions
223
one is approaching a given point z = x + iy. The equality of the expressions (7.1.5)
and (7.1.6) entails, therefore, the necessity of the compliance with the conditions:
∂v
∂u
=
,
∂x
∂y
∂u
∂v
=− ,
∂y
∂x
(7.1.7)
called the Cauchy–Riemann (CR) conditions.
As it will be seen below, if they are satisfied, the derivative of the function f with
respect to z̄ vanishes. Let us temporarily accept the opposite, i.e., that the function
f depends on two variables: z and z̄. One would then have:
∂f
∂f ∂z ∂f ∂ z̄
∂f
∂f
=
+
,
+
=
∂x
∂z ∂x
∂ z̄ ∂x
∂z
∂ z̄
=1
=1
∂f
∂f ∂z ∂f ∂ z̄
∂f
∂f
=
−i
.
+
=i
∂y
∂z ∂y
∂ z̄ ∂y
∂z
∂ z̄
=i
(7.1.8)
=−i
From these equations it immediately stems that
∂f
1
=
∂ z̄
2
∂f
∂f
+i
∂x
∂y
(7.1.9)
.
Similarly, although it is not required here but may be useful later, we would get that
∂f
1
=
∂z
2
∂f
∂f
−i
∂x
∂y
(7.1.10)
.
However, the obvious relations hold:
∂f
∂u
∂v
=
+i
,
∂x
∂x
∂x
∂f
∂u
∂v
=
+i
∂y
∂y
∂y
(7.1.11)
and after plugging them into (7.1.9) and using the CR conditions one obtains:
1
∂f
=
∂ z̄
2
∂u ∂v
−
∂x
∂y
+
i
2
∂u ∂v
+
∂y
∂x
= 0.
(7.1.12)
Thereby, it is seen that the holomorphicity of a function implies its independence
of z̄ and satisfying the Cauchy–Riemann conditions. The opposite is not true. It can,
however, be shown, and the reader knows it from the lecture of analysis, that if,
in addition to the equations (7.1.7), the continuity of partial derivatives appearing
in CR conditions is assumed, then it is sufficient to prove that the function is
holomorphic.
224
7 Studying Functions of Complex Variable
Let us come back to the function g(x, y) considered earlier. If our function f
was treated as the function of two variables x and y and written in the vector form,
i.e., as
f (x, y) =
u(x, y)
,
v(x, y)
(7.1.13)
then the Jacobian matrix would have the form
⎡ ∂u ∂u ⎤
⎢
∂y ⎥
f (z) = ⎣ ∂x
∂v ∂v ⎦ .
(7.1.14)
∂x ∂y
At this point, the reader probably notices a significant limitation imposed on the
form of this matrix by the property of holomorphicity. Of these 4 numbers, only 2
are independent! If one denotes: ∂u/∂x = a and ∂u/∂y = b, then the CR conditions
necessitate the following form of the matrix:
f (z) =
a b
.
−b a
(7.1.15)
So, as we see, the holomorphicity constitutes a strong requirement with respect
to a function, but on the other hand it should be emphasized that plenty of functions
one most often deals with, as polynomial functions, exponential functions, sine or
cosine are holomorphic, and others, as the logarithmic function, tangent, etc., have
large domains of holomorphicity.
After this recap one can now move on to solving the current problem.
(a) f (z) = z2 .
We are going to verify the differentiability from the definition, calculating the
limit of the differential quotient:
lim
z→0
f (z + z) − f (z)
(z + z)2 − z2
= lim
z→0
z
z
= lim
z→0
2zz + (z)2
= 2z.
z
(7.1.16)
This result does not depend on the way one is going with z to zero, hence the
function is differentiable in the complex sense, i.e., it is holomorphic. There are
no restrictions on the value of z either, which means that the whole complex
plane C constitutes the region of holomorphicity. Functions possessing this
property are called entire functions.
7.1 Examining the Holomorphicity of Functions
225
(b) g(z) = sin z.
This time the Cauchy–Riemann conditions will be used. To this goal, the
function g(z) needs to be rewritten as the sum of the real and imaginary parts.
Putting z = x + iy, one gets
1
2i
1
ei(x+iy) − e−i(x+iy)
2i
1 −y
1
e−y eix − ey e−ix =
e (cosx + i sin x)
=
2i
2i
1 −y
(e − ey )cosx + i(e−y + ey ) sin x
−ey (cosx − i sin x) =
2i
(7.1.17)
= sin x cosh y +i cos x sinh y .
g(z) = sin z =
u(x,y)
eiz − e−iz =
v(x,y)
Now we can calculate the partial derivatives:
∂u
∂v
= cos x cosh y =
,
∂x
∂y
∂v
∂u
= sin x sinh y = − .
∂y
∂x
(7.1.18)
Thus the CR conditions are satisfied. The continuity of all partial derivatives
raises no doubt either. One can conclude, therefore, that the function sin z is
holomorphic on the whole complex plane. It is, therefore, an entire function
(similarly as cos z).
Problem 2
The holomorphicity domains for the functions:
(a) f (z) = |z|,
1
(b) g(z) =
z−2
will be found.
Solution
In the previous problem the notion of holomorphicity was discussed in detail, so one
can immediately go to the examination of the functions f (z) and g(z).
226
7 Studying Functions of Complex Variable
(a) f (z) = |z|.
The conclusions already formulated should allow the reader to immediately
judge whether the function f (z) is holomorphic. Rewriting it in the following
way:
f (z) = |z| =
√
(7.1.19)
zz̄,
one can see that this function depends on both z and z̄. It is impossible to write
the modulus of the complex number z = x +iy without the involvement of z̄. As
a consequence, f (z) cannot be a holomorphic function. With this observation
we could finish our solution. As an exercise let us, however, consider the
Cauchy–Riemann conditions, to clearly see that they are not satisfied anywhere.
To this goal f (z) is rewritten as a function of x and y in order to find its real
part u(x, y) and imaginary part v(x, y):
f (z) =
x2 + y2
⇒ u(x, y) =
x2 + y2,
v(x, y) = 0.
(7.1.20)
It is now easy to calculate all partial derivatives:
∂u
y
,
=
∂y
x2 + y2
∂u
x
,
=
∂x
x2 + y2
∂v
∂v
=
= 0.
∂x
∂y
(7.1.21)
As one can see, they exist everywhere apart from the system origin, where, of
course, the function cannot be holomorphic.
Now suppose that (x, y) = (0, 0). The CR conditions (7.0.3) would require the
simultaneous satisfying of the equations:
x
x2
+ y2
= 0,
y
x2
+ y2
= 0.
(7.1.22)
It is obvious that these equations cannot hold. Consequently the function f (z)
does not have any areas of holomorphicity.
(b) g(z) = 1/(z − 2).
This time we are going to check whether the limit of the differential quotient
exists. We get
lim
z→0
1/(z + z − 2) − 1/(z − 2)
−z
= lim
z→0 z(z + z − 2)(z − 2)
z
=−
1
.
(z − 2)2
(7.1.23)
It is visible that this limit does exist (i.e., the complex derivative exists)
everywhere apart from z = 2. Thus, the domain of holomorphicity for the
function g(z) is the set C \ {2}.
7.1 Examining the Holomorphicity of Functions
227
Problem 3
The holomorphicity domains for the functions:
(a) f (z) = z log z,
(b) h(z) = arctan z
will be found.
Solution
(a) f (z) = z log z.
The notion of the logarithm appearing here has not been defined yet in the set
of complex numbers. Therefore, before one moves on to the holomorphicity of
this function, one needs to understand the meaning of this symbol. The most
natural definition of the logarithm is such that it constitutes an inverse of the
exponential function:
g(z) = ez
⇒ log z = g −1 (z).
(7.1.24)
The definition of the latter on the complex plane C does not cause any problems:
all one has to do is using the Taylor series:
g(z) = ez :=
∞ n
!
z
n=0
n!
.
(7.1.25)
The problem of raising the number e to a complex power (what would that
mean?) is boiled down to the multiplication of the number z by itself and then to
adding terms, hence the operations are well known even if z ∈ C. One can show,
which is certainly already known to the reader, that the function so defined
has all the properties that would be expected from exponential functions, for
example:
e0 = 1,
1
= e−z ,
ez
ez1 ez2 = ez1 +z2 ,
ez1
= ez1 −z2 .
ez2
[ez ] = ez .
(7.1.26)
There is a further question, whether the infinite sum on the right-hand side
of (7.1.25) makes sense and possibly for which arguments z. Such issues will
be addressed in detail in the following section, and here we only mention an
important fact: the series spoken of is convergent, and thus defines the function
g(z), for any value of the complex z.
If the exponential function is already available, one can think of its reversal, i.e.,
of the logarithm. There emerges here, however, an important question, whether
228
7 Studying Functions of Complex Variable
the function defined in this way is reversible at all. In the set of real numbers
the function ex was injective, so no problem appeared: the entire image of this
function, i.e., the set R+ , constituted the domain of the logarithm. However, in
the set C, this is not the case. We know from algebra that
e2π ki = cos(2π k) + i sin(2π k) = 1,
for k ∈ Z,
(7.1.27)
which means that there is an infinite number of arguments zk = z + 2π ki, for
which the exponential function has the same value. So, if one wants to invert
the function ez , its domain needs to be cut to a strip of width 2π , which would
contain only one number zk , and the function would become one-to-one. Let us
then accept (this is the most common choice, although not unique) that
− ∞ < Rez < ∞,
−π < Im z < π.
(7.1.28)
As to the local reversibility, this is no problem at all. The appropriate theorem,
in respect of the real numbers dealt with in Chap. 8 of Part II, requires the
derivative at a given point be nonvanishing. This condition is, of course,
fulfilled, as
|[ez ] | = |ez | = |ex+iy | = |ex eiy | = |ex | |eiy | = ex > 0.
(7.1.29)
=1
Thanks to the fact that the domain (7.1.28) is an open set (the boundary
points corresponding to Im z = ±π have been excluded), no problem with
the calculation of the derivative at any point arises, because this set is made up
of internal points only.
Thus, as can be seen, the exponential function cut in the above way can be easily
inverted, the rule for calculating its inverse being obvious. One simply has to
represent the number z as ew (remembering that −π < Im w < π ), and then
read off the value of w. If z is written in the so-called polar form, i.e.,
z = reiϕ ,
where − π < ϕ < π,
(7.1.30)
r being the modulus of the number z and ϕ its argument, i.e., the phase (often
denoted with arg z), then
z = eln r+iϕ
⇒ log z = ln r + iϕ.
(7.1.31)
Contrary to the other chapters of this book, the symbol “ln” for the ordinary
natural logarithm is used in order to avoid collision with the symbol “log” for
the complex logarithm.
What is the domain of the logarithm so defined? It is just the image of ez . The
condition (7.1.30) means that it is the complex plane without the nonpositive
real semiaxis: C\] − ∞, 0]. And on this set the function log z is actually
7.1 Examining the Holomorphicity of Functions
229
holomorphic, which can be easily found out by calculating the derivative. For
we have
log(ez ) = z
⇒ [log(ez )] = 1
⇒ ez log (ez ) = 1,
(7.1.32)
the symbol log denoting the differentiation with respect to the whole argument
w = ez and not over z. Consequently, one gets
[log w] =
1
w
(7.1.33)
in the entire domain (see also the discussion on the page 274 and following).
Let us now go back to the function specified in the text of the exercise. Because
any polynomial is an entire function, the additional factor z in the expression
f (z) = z log z does not affect the area of holomorphicity. As we know from the
lecture of analysis, the product (but also the sum, difference, and quotient apart
from zeros of the denominator) of holomorphic functions is again holomorphic.
The function f (z) is then holomorphic in the domain C\] − ∞, 0].
At the end, it should be noted that if in place of (7.1.28) some other strip of
the width 2π was chosen, as for instance
− ∞ < Rez < ∞,
−π + θ < Im z < π + θ,
(7.1.34)
with θ being any real number, the holomorphicity domain of the logarithmic
function would be the following: C\] − ∞, 0] × eiθ .
(b) h(z) = arctan z.
The properties of the function log z, discussed above, will now be used
to examine the holomorphicity domains of the function arctan z. Since by
definition it is the inverse function with respect to the tangent, the equation
arctan z = w entails tan w = z, i.e.,
eiw − e−iw
sin w
= −i iw
=z
cos w
e + e−iw
⇒ eiw (1 − iz) = e−iw (1 + iz).
(7.1.35)
i
i−z
⇒ w = arctan z = log
.
2
i+z
(7.1.36)
Thus one obtains
e
2iw
i−z
1 + iz
=
=
1 − iz
i+z
Let us consider now the holomorphicity domains of the right-hand side. Surely
the point z = −i has to be excluded because of the denominator i + z. In
addition, it stems from the item a), that one needs to remove all such numbers
z for which the argument is equal to the logarithm of a certain nonpositive real
number, i.e., those satisfying the equation
230
7 Studying Functions of Complex Variable
i−z
= −r
i+z
for r ≥ 0.
(7.1.37)
Let us now multiply both the numerator and denominator by the conjugate of
the denominator, i.e., by −i + z̄ (remember that the point z = −i has already
been excluded). One obtains then
(i − z)(−i + z̄)
= −r,
|i + z|2
i.e.,
1 + i(z + z̄) − zz̄ = −r|i + z|2 .
(7.1.38)
Note that apart from the expression i(z+ z̄), which is purely imaginary, all other
terms in the above equation are real. The equality of both sides shall entail that
z + z̄ = 2Re z = 0, which means that the number z must be imaginary: z = iy.
Let us plug z in this form into (7.1.37). This yields
1−y
= −r
1+y
⇒ y=−
r +1
.
r −1
(7.1.39)
It is very easy to examine the behavior of the right-hand side depending on
the variable r ∈ [0, ∞[. It is simply a section of a hyperbolic curve shown in
Fig. 7.1.
It follows from the drawing that for y ∈] − ∞, −1[∪[1, ∞[ one can always
find a nonnegative value of r for which the Eq. (7.1.37) will be satisfied, i.e., the
function arctan z will not be holomorphic. Since the point z = −i (i.e., y = −1)
has already been excluded from the holomorphicity domain, one can conclude
that the function arctan z is holomorphic on the set C \ (] − i∞, −i] ∪ [i, i∞[).
The symbols ±i∞ denote the infinities reached by moving up or down along
r
r
1
1
1
1
1
Fig. 7.1 The plot of the right-hand side of the second equation of (7.1.39)
r
7.1 Examining the Holomorphicity of Functions
231
Fig. 7.2 The holomorphicity
domain for the function
arctan z
the imaginary axis. This area is shown in Fig. 7.2. The reader is encouraged
to establish the set of holomorphicity if for the logarithm, the domain C\] −
∞, 0] × eiθ was chosen with θ = 0. In particular one can find out that if θ = π ,
the segment lying on the imaginary axis between the points z = −i and z = i
ought to be removed.
Problem 4
The form of a holomorphic function f (z) will be found if one knows that its real
part is given by the formula:
u(x, y) = ex [(x 2 − y 2 ) cos y − 2xy sin y].
(7.1.40)
Solution
Any holomorphic function has, among other things, the property that knowing its
real part, one can restore—up to an additive constant—the imaginary part (and
vice versa), i.e., reconstruct the whole function. This opportunity is created by the
Cauchy–Riemann conditions (7.1.7). However, before using them, it is reasonable
to make sure, whether this task makes sense, i.e., whether the given function u(x, y)
232
7 Studying Functions of Complex Variable
can at all constitute the real part of a certain holomorphic function. Not every
function of two real variables is allowed to be the real or imaginary part of a
holomorphic function. It is easy to demonstrate, by manipulating the CR conditions,
that they have to be harmonic functions and hence they must satisfy the twodimensional Laplace equation. Let us then see if u(x, y) provided in the text of
the exercise has the required property:
u=
∂ 2u ∂ 2u
x
(2 + 4x + x 2 − y 2 ) cos y − 2(2 + x)y sin y
+
=
e
∂x 2
∂y 2
(7.1.41)
+ex −(2 + 4x + x 2 − y 2 ) cos y + 2(2 + x)y sin y = 0.
Having ensured that the exercise makes sense, one can now move on to solving
the problem by the use of the CR conditions. Substituting the function u(x, y) into
the first of them, one gets the equation:
∂v
= ex (x 2 − y 2 ) cos y − 2xy sin y + 2x cos y − 2y sin y .
∂y
(7.1.42)
After having integrated both sides with respect to y, one obtains the following
expression for the sought after function:
v(x, y) = ex 2xy cos y + (x 2 − y 2 ) sin y + C(x),
(7.1.43)
where the integration “constant” C can depend, in general, on the variable x. The
following known integrals have been used here (up to some additive constants):
y sin y dy = −y cos y + sin y,
y 2 cos y dy = 2y cos y + (y 2 − 2) sin y.
(7.1.44)
If we now plug the known expression for u(x, y) and that obtained above for
v(x, y) into the second of the conditions (7.1.7), we will come to an ordinary
differential equation for the unknown function C(x). In our case, all terms cancel
and it turns out to be trivial:
C (x) = 0,
(7.1.45)
which means that C is actually a constant.
One can now assemble the complete expression for the function f (z):
u(x, y) + iv(x, y)
= ex [(x 2 − y 2 ) cos y − 2xy sin y] + iex 2xy cos y + (x 2 − y 2 ) sin y + iC
7.2 Finding Domains of Convergence of Complex Series
233
= ex x 2 (cos y + i sin y) − y 2 (cos y + i sin y) + 2ixy(cos y + i sin y) + iC
= ex eiy (x 2 − y 2 + 2ixy) = ex+iy (x + iy)2 + iC = z2 ez + iC =: f (z).
(7.1.46)
What is very important to notice is the final expression turns out to depend solely
on the combination x + iy, and not on both these variables separately. It had to be
so, because as we know, a holomorphic function may not depend on z̄. If eventually,
after having found v(x, y), it was not possible to combine the arguments x and y
into one argument x + iy, this would mean that some calculational mistake had been
made or the initial function u(x, y) could not be the real part of any holomorphic
function.
7.2 Finding Domains of Convergence of Complex Series
Problem 1
It will be examined in what domain the series:
∞ 2n
!
z
n=0
n!
(7.2.1)
defines a holomorphic function.
Solution
The holomorphicity of a function is a very strong requirement which was spoken of.
If in a given domain the first complex derivative of a function exists, then the second,
third, and all subsequent ones do as well. This situation is completely different
from what one was accustomed when considering real functions defined on R. The
existence of the first derivative did not imply the same for the second and further
ones at all.
Any holomorphic function in the neighborhood of a certain point in the complex
plane can be expanded there in the Taylor series, i.e., it is the analytical function.
In turn, a function given in the form of a certain power series is, within the area of
its convergence, a holomorphic function. For this reason, one often interchangeably
uses the notions “analytical function” and “holomorphic function.”
If one wants to determine the domain of holomorphicity for a function defined
with a power series
234
7 Studying Functions of Complex Variable
f (z) =
∞
!
an (z − z0 )n ,
n=0
it is enough to find the ball of convergence. As the reader knows from the lecture
and from the theoretical summary at the beginning of this chapter, for this purpose
one has to calculate the quantity:
R=
lim
n→∞
n
|an |
−1
(7.2.2)
,
which plays the role of the radius of this ball. If it is found to be greater than
zero, then the ball of holomorphicity of the function f (z) is the set of all complex
numbers z, for which
|z − z0 | < R.
(7.2.3)
It can, naturally, happen that R is infinite. This means that we have an entire function
which is holomorphic on the whole complex plane. In turn, if R = 0, the function
f (z) is not holomorphic even on an arbitrarily small neighborhood of the point z0 .
The formula (7.2.2) is due to the application of the Cauchy criterion to the power
series, dealt with in the first part of the book (see Problem 4 in the Sect. 13.2). It
entails that if
lim n |an (z − z0 )n | = q < 1,
(7.2.4)
n→∞
the series is convergent. This condition corresponds to (7.2.3) with the radius R
defined with the formula (7.2.2).
Let us then find the convergence radius for the series (7.2.1). After an obvious
substitution n → 2n stemming from the power of the variable z, one has
R=
lim
n→∞
1/n!
2n
−1
.
(7.2.5)
From the first part of this set of problems and from the lecture of analysis, we know
that instead of the limit
n
|bn |,
(7.2.6)
bn+1 ,
lim n→∞ b (7.2.7)
lim
n→∞
for some bn , one can equally well calculate
n
7.2 Finding Domains of Convergence of Complex Series
235
and if both these limits exist, they lead to the same result. Thus one can write
2n
lim
n→∞
1/n! =
lim
n→∞
=
n
1/n!
1/2
1/2
1
n→∞ n + 1
n!
= lim
n→∞ (n + 1)!
= 0.
lim
1/2
(7.2.8)
This result indicates that R = ∞ and the function given by the formula (7.2.1) is an
entire function. Its ball of convergence has the form (here z0 = 0):
|z| < ∞.
(7.2.9)
The similar result would be obtained for the function ez , sin z, or cos z, and is
owed to the presence of the factorial in the denominators of the general terms of the
series in all of these cases (see (7.1.25)).
Problem 2
It will be examined, in what domain the series:
f (z) :=
∞
!
(−1)n z2n
(7.2.10)
n=0
defines a holomorphic function.
Solution
After having analyzed the previous problem, the present one turns out to be very
easy. Simply use the formula (7.2.2), which in our case takes the form:
R=
lim
n→∞
|(−1)n |
2n
−1
=
lim
n→∞
√
2n
1
−1
= 1.
(7.2.11)
Since z0 = 0, the convergence ball of (7.2.10), and at the same time the
holomorphicity ball of the function f (z) defined by the series in question, boils
down to:
|z| < 1.
(7.2.12)
This time the considered function is not entire since it is not holomorphic on the
whole complex plane C. It is perfectly clear: the series (7.2.10) is just the geometric
236
7 Studying Functions of Complex Variable
series, and as such, it is convergent only when the common ratio is smaller (or rather
its absolute value) than 1.
Having observed that we have a geometric series, we can be tempted to find
explicitly its sum, using the following formula:
a + a · q + a · q2 + a · q3 + . . . =
a
,
1−q
(7.2.13)
for |q| < 1. In our case a = 1 and q = −z2 , so after having met the
condition (7.2.12), one gets
f (z) =
1
1
.
=
(z − i)(z + i)
1 + z2
(7.2.14)
The function defined with the above formula has singularities for z = ±i. It
is then obvious that it cannot be any entire function. This is why the radius of the
convergence ball around the point z0 = 0 has proved to be equal to 1. This number
is simply equal to the distance from z0 to the nearest singularity.
At the end, it is worth mentioning one more aspect. The function f (z) defined in
the text of the exercise is determined exclusively for |z| < 1. Therefore, only in this
domain the formula (7.2.14) may be used. If one would like to use the expression
beyond this set, where formally the expression 1/(1 + z2 ) still makes sense, then it
is not the function f itself that one is talking about, but its “analytical continuation.”
Problem 3
It will be examined, in what domain the series:
f (z) :=
∞
!
(z − 2)n
n=0
n
(7.2.15)
defines a holomorphic function.
Solution
Making use of the formula (7.2.2), one immediately gets
R=
lim
n→∞
n
1/n
−1
= 1,
(7.2.16)
from which it stems that the function f (z) is holomorphic in the ball:
|z − 2| < 1.
(7.2.17)
7.3 Calculating Contour Integrals
237
Fig. 7.3 The holomorphicity domain for the function defined with the power series (7.2.15)
One can again find out that the value R = 1 constitutes the distance between
the point z0 = 2 and the location of the nearest singularity. Applying the methods
exploited in Part I in Sect. 15.4, one is able to explicitly find the sum of (7.2.15),
obtaining
f (z) = log(3 − z)
(7.2.18)
(see also formula (11.2.21) in Part I).
The logarithmic function is already known to be holomorphic apart from the
nonpositive real semi-axis. Since the series (7.2.15) is “centered” around z0 = 2,
the distance to the nearest singularity, which is located at the point z = 3 (there
the argument of the logarithm becomes zero), equals 1 and hence the radius of the
convergence ball is 1. This is shown graphically in Fig. 7.3.
Outside the gray ball, the formula (7.2.18) still defines the holomorphic function
(apart from the half-line from z = 3 to infinity), but it constitutes only the analytical
continuation of the function (7.2.15).
7.3 Calculating Contour Integrals
Problem 1
The integrals of the function f (z) = sin z will be calculated, over the following
contours from the point z0 = 1 + i to the point z1 = 3 + 3i:
(a) the polyline of vertices: z = 1 + i, z = 3 + i, z = 3 + 3i,
(b) the polyline of vertices: z = 1 + i, z = 1 + 3i, z = 3 + 3i.
238
7 Studying Functions of Complex Variable
Solution
In this section, we are facing the problem of calculating integrals of functions of the
complex variable over a contour (curve) C lying on the complex plane C. For this
purpose, one has to parameterize the curve (it is assumed that the curve is smooth,
and if it had “corners,” it can be divided onto smooth pieces), by specifying the
relation z(t) = x(t) + iy(t) for t ∈ [a, b]. Then the contour integral is understood
in the following way:
b
f (z) dz =
f (z(t))
dz
dt.
dt
(7.3.1)
a
C
These are, in fact, two real integrals: one for the real, and the other for the imaginary
part. For, writing f = u + iv, we have
b
dy
dx
+i
[u(x(t), y(t)) + iv(x(t), y(t))]
dt
dt
f (z) dz =
dt
a
C
b
=
dx
dy
dt
u(x(t), y(t))
− v(x(t), y(t))
dt
dt
a
b
+i
dy
dx
u(x(t), y(t))
+ v(x(t), y(t))
dt.
dt
dt
(7.3.2)
a
The result of the integration will obviously depend on the orientation of the contour:
the opposite orientations lead to opposite signs.
In Fig. 7.4, the two contours, referred to in the content of the exercise, are shown.
Let us consider the first one. It is composed of two segments: the horizontal and
the vertical. On the first segment, one can choose the parametrization z(t) = t + i,
where t ∈ [1, 3]. In turn on the vertical segment one has z(t) = 3 + it, where again
t ∈ [1, 3]. Using Euler’s formula for z = x + iy, i.e., writing:
sin z =
1 iz
1
(e − e−iz ) = (eix e−y − e−ix ey ),
2i
2i
and applying (7.3.1), one finds
(7.3.3)
7.3 Calculating Contour Integrals
239
Fig. 7.4 The contours C1
(dashed line) and C2 (dotted
line)
3
1
f (z) dz =
2i
C1
3
(e
−1 it
1 −it
e −e e
1
) dt +
2i
1
(e3i e−t − e−3i et )i dt
1
3
−i it
1
−it e − ie e
−e3i e−t − e−3i et
+
e
2
1
1
=
2i
1
− (e3i−1 + e−3i+1 ) + (ei−1 + e−i+1 )
=
2
=2 cos(3+i)
− (e
3i−3
+e
−3i+3
=2 cos(3+3i)
=2 cos(1+i)
) + (e
3i−1
+e
−3i+1
=2 cos(3+i)
3
1
)
= cos(1 + i) − cos(3 + 3i).
(7.3.4)
On the contour C2 , the parametrization z(t) = 1 + it on the vertical segment and
z(t) = t + 3i on the horizontal one, where in both cases t ∈ [1, 3] can be chosen.
Then one gets
3
1
f (z) dz =
2i
3
i −t
(e e
1
− e e )i dt +
2i
−i t
1
C2
=
1
−ei e−t − e−i et
2
(e−3 eit − e3 e−it ) dt
1
3
+ 1 −ie−3 eit − ie3 e−it
2i
1
3
1
240
7 Studying Functions of Complex Variable
=
1
[− (ei−3 + e−i+3 ) + (ei−1 + e−i+1 )
2 =2 cos(1+3i)
=2 cos(1+i)
− (e3i−3 + e−3i+3 ) + (ei−3 + e−i+3 )]
=2 cos(3+3i)
=2 cos(1+3i)
= cos(1 + i) − cos(3 + 3i).
(7.3.5)
The identical result has been obtained for both contours. This is not a coincidence, since we have the holomorphic function f (z) = sin z. For this type of
function, each contour with fixed ends will give the same result. This fact can be
also expressed in the form of the following statement: the integral of a holomorphic
function over a closed contour vanishes. This observation is the subject of Cauchy’s
theorem, the next section is devoted to.
Problem 2
The integrals of the function f (z) = 1/z will be calculated over the following
contours running from the point z0 = −1 to the point z1 = 1:
(a) the lower semi-circle defined with the equation |z| = 1,
(b) the upper semi-circle defined with the equation |z| = 1.
Solution
The natural parametrization for the circle of radius R and center at z0 = x0 + iy0 is
z(ϕ) = z0 + Reiϕ ,
(7.3.6)
i.e.,
x(ϕ) = x0 + R cos ϕ,
y(ϕ) = y0 + R sin ϕ,
(7.3.7)
where ϕ ∈ [0, 2π [. In the case of the contours drawn in Fig. 7.5, one has R = 1,
z0 = 0, and then z(ϕ) = eiϕ , with ϕ ∈ [−π, 0] for C1 and ϕ ∈ [0, π ] for C2 . Since
there occurs dz/dϕ = ieiϕ , according to the formula (7.3.1), the first integration
takes the form (the integration limits stem from the contour orientation):
0
1
dz = i
z
I1 =
C1
−π
0
1
e iϕ dϕ = i
e
iϕ
−π
dϕ = π i.
(7.3.8)
7.3 Calculating Contour Integrals
241
Fig. 7.5 The contours C1
(dashed line) and C2 (dotted
line)
For the second contour, the integration is performed in the identical way. Only
the integration limits for the definite integral are modified, and one gets
0
1
dz = i
z
I2 =
C2
0
1
e iϕ dϕ = i
e
dϕ = −π i.
iϕ
π
(7.3.9)
π
As opposed to the previous example, this time different results are obtained. They
differ exactly by 2π i. This fact will find its explanation in the next chapter, in which
the so-called residue theorem will be discussed. Here we should only notice that
the function f (z) = 1/z is not holomorphic in the area that is surrounded (jointly)
by the contours C1 and C2 . It has a visible singularity at z = 0 connected with the
vanishing of the denominator. If the curve C1 was deformed so as to be converted
into C2 , one would inevitably run into this point. It is impossible to carry out any
continuous deformation so that the function remains constantly holomorphic at all
points of the curve being deformed.
Problem 3
The integral of the function f (z) = 1/(z2 − 1) over the entire imaginary axis will
be calculated.
242
7 Studying Functions of Complex Variable
Solution
The integral of a holomorphic function may not exist if the contour extends to
infinity, similarly as it can happen for the real improper integrals (see Chap. 2
in Part II). However, in this problem the integrand function goes to zero quickly
enough (as 1/z2 ). On the contour C itself (i.e., on the imaginary straight line),
the function is holomorphic, being the quotient of two holomorphic functions (a
constant polynomial in the numerator and the second-degree polynomial in the
denominator), with roots of the denominator lying at z = ±1, i.e., outside the
imaginary axis.
The natural parametrization for the considered contour has the form:
z(t) = it,
where − ∞ < t < ∞.
(7.3.10)
The orientation has been chosen arbitrarily, since in the text of the exercise none was
imposed. In the case of the opposite orientation, the final result will simply change
its sign. In accordance with the formula (7.3.1), one then has
∞
f (z) dz =
−∞
C
1
dz
dt = i
2
z(t) − 1 dt
∞
−∞
∞
1
dt = −i arctan t = −iπ.
2
−t − 1
−∞
(7.3.11)
This result can be very easily justified using the residue theorem.
Problem 4
The integral of the function f (z) = z2 over the entire edge of the square with the
vertices at −1 + i, −1 − i, 1 − i, 1 + i will be calculated.
Solution
Let us denote the sides of the square with the symbols CA , CB , CC , and CD , as it is
shown in Fig. 7.6. Then the following parametrization can be used:
CA : z(t) = −1 + it,
CB : z(t) = −i + t,
CC : z(t) = 1 + it,
CD : z(t) = i + t.
(7.3.12)
Now applying the formula (7.3.1) to the subsequent sides (and having regard to
their orientations), one gets
7.3 Calculating Contour Integrals
243
Fig. 7.6 The integration
contour
−1
IA =
f (z) dz =
CA
1
(−1 + it) i dt = −i
−1
1
1
IB =
f (z) dz =
4
(−1 − 2it + t 2 ) dt = − ,
3
(−i + t)2 dt =
−1
1
IC =
f (z) dz =
1
f (z) dz =
CD
4i
,
3
(−1 + 2it + t 2 ) dt =
4
.
3
−1
−1
ID =
(1 + 2it − t 2 ) dt =
(1 + it) i dt = i
2
−1
CC
1
(i + t) dt = −
2
1
4i
,
3
1
−1
CB
(1 − 2it − t 2 ) dt = −
2
−1
(7.3.13)
As a result for the closed contour (C), it is found
f (z) dz = IA + IB + IC + ID = −
4 4i
4
4i
− +
+ = 0.
3
3
3
3
(7.3.14)
C
The value of the integral equals zero. The reader should not be surprised by this
fact. It was already pointed out in Problem 1 that such a result should occur for the
holomorphic function, that, after all, each polynomial is.
244
7 Studying Functions of Complex Variable
Problem 5
The integral of the function f (z) = Re z will be calculated over the following
contours between the points z = 1 and z = i:
(a) the polyline of vertices at 1, 1 + i, i,
(b) the segment.
Solution
The function Re z clearly cannot be holomorphic, as it depends on both z and z̄ (see
Problems 1 and 2 in Sect. 7.1):
Re z =
1
(z + z̄).
2
(7.3.15)
The similar conclusion refers to the imaginary part, for which one would have
Im z =
1
(z − z̄).
2i
(7.3.16)
It is very easy to check that at any point of the complex plane the Cauchy–Riemann
conditions are not satisfied. For the first function one has
Re z = x
Fig. 7.7 The contours C1
(dashed line) and C2 (dotted
line)
⇒ u(x, y) = x, v(x, y) = 0
(7.3.17)
7.4 Using Cauchy’s Theorem
245
which entails
∂v
∂u
= 1,
=0
∂x
∂y
⇒
∂u
∂v
=
.
∂x
∂y
(7.3.18)
For the second one it occurs
Im z = y
⇒ u(x, y) = 0, v(x, y) = y,
(7.3.19)
and then
∂u
∂v
= 0,
=1
∂x
∂y
⇒
∂u
∂v
=
.
∂x
∂y
(7.3.20)
Since the function f (z) is not holomorphic in any domain, we expect that the
integrals calculated over the contours C1 and C2 can have different values. It is
actually the case which can be seen below.
The contour C1 consists of two segments. On the vertical segment we choose the
parametrization: z(t) = 1+it, and on the horizontal one: z(t) = i +t. Consequently,
1
f (z) dz =
0
Re(1 + it)i dt +
0
C1
1
Re(i + t) dt = i
1
1
1
t dt = − + i.
2
dt −
0
0
(7.3.21)
The contour C2 will be parametrized by choosing z(t) = 1 + t (i − 1). When
calculating the integral, one finds
1
f (z) dz =
C2
1
i
1
(1 − t) dt = − + .
2 2
Re(1 + t (i − 1))(i − 1) dt = (i − 1)
0
As can be seen,
0
"
f (z) dz =
C1
"
(7.3.22)
f (z) dz.
C2
7.4 Using Cauchy’s Theorem
Problem 1
The value of the integral:
∞
I=
0
will be found.
sin x
dx
x
(7.4.1)
246
7 Studying Functions of Complex Variable
Solution
The subject of this section is the direct application of Cauchy’s theorem announced
in the previous problems and formulated in the theoretical introduction. This is
an extremely powerful and useful statement, which says that the integral of a
holomorphic function f (z) over a closed contour C in a simply connected domain
(i.e., without “holes”) vanishes:
/
f (z) dz = 0.
(7.4.2)
C
This theorem turns out to be an exceptionally convenient tool to calculate certain
definite integrals even if the primitive function for the integrand expression cannot
be found. If our goal is to find a definite integral (including improper one), i.e.,
a number, then usually first the indefinite integral is calculated, i.e., a function,
and next the integration limits are substituted. There appears, however, a fairly
obvious question: is it necessary to calculate the complete function if one is
actually interested in one number only? Finding a function seems in general much
more challenging than a single value. Cauchy’s theorem and the residue theorem
discussed in the following section are just tools we need. They create the possibility
to find a definite integral without designating the primitive function. But there is an
important restriction: this method works only for very specific integration intervals.
Knowing the values of the definite integral for arbitrary intervals would in fact mean
knowing the indefinite integral.
Before proceeding with the calculations the reader should make sure that the
problem set before us makes sense and the integral (7.4.1) is actually convergent.
From Sect. 2.2 in the second part of this book series, where, among others, we were
dealing with Dirichlet’s test of convergence of indefinite integrals, we know that it
is actually the case.
If one wants to apply Cauchy’s theorem, stated with formula (7.4.2) in order
to calculate integrals of the type (7.4.1), two things have to be established at the
beginning: a certain function f (z) has to be chosen and a closed integration contour
C has to be proposed. Of course not just any, but one fit for the problem to be
solved. In the present section, as well as in Sect. 8.3, we would like to deal with
the question of how to choose an appropriate function together with an integration
contour in each specific case.
Let us start with the contour. Since we are interested in the integral from 0 to ∞,
there is no doubt that a fragment of it must run along the positive real semi-axis.
Such a contour would not, however, be closed and Cauchy’s theorem might not be
used with it. It has to be supplemented with other pieces, but such for which one
is able to find the integrals or to show, for example, that they tend to zero. It often
happens, however, that the task becomes much easier—and this is the situation of
the current exercise—if the original integral extends not from 0 to ∞, but from −∞
to ∞. Then one can think about closing the entire real axis with the semi-circle of
7.4 Using Cauchy’s Theorem
247
the infinite radius lying in the upper or lower half-plane. For the semi-axis, instead
of the entire axis, it would be more problematic. For this reason, relying on the
parity of the function, the lower limit of the integration is extended till −∞:
∞
I=
1
sin x
dx =
x
2
0
∞
−∞
sin x
dx.
x
(7.4.3)
It is convenient to begin with considering the integral from −R to R, and only
later execute the limit R → ∞. Suppose, in addition, that we have decided to close
the original integration contour with the upper semi-circle of the radius R denoted
with the symbol CR . If the function undergoing the integration was holomorphic on
the entire half-plane, Cauchy’s theorem could be applied, allowing to write
f (z) dz +
[−R,R]
f (z) dz = 0.
(7.4.4)
CR
It is obvious that if one wished to get the value of the first integral out of
this equation, the second one would have to be independently found. In general,
however, one is unable to calculate the integral over the semi-circle of radius R.
Fortunately quite often it can be demonstrated (with the appropriate choice of the
function f ) that this integral tends to zero when R → ∞. This occurs, for example,
when, together with the increase of R the integrand function goes to zero fast
enough. We are writing “fast enough,” because it has to compensate for the length
of the integration contour (i.e., of the semi-circle) increasing to infinity.
We have come to the point where one has to propose the form of the integrand
function. The choice that gets imposed is of course:
f (z) =
sin z
,
z
(7.4.5)
but, if one decided to go for it, one would immediately run into a serious problem:
this function not only does not aim at zero at CR , but exponentially grows to
(complex) infinity. This is because if one plugs z = x + iy into Euler’s formula
for sine, one obtains
sin z
eiz − e−iz
ei(x+iy) − e−i(x+iy)
=
=
z
2iz
2i(x + iy)
∼ (· · · )
large y
ey
.
y
(7.4.6)
Similarly,
sin z
z
∼
large −y
(· · · )
e−y
.
y
(7.4.7)
248
7 Studying Functions of Complex Variable
If the sine function contained only one component, there would be no problem.
One could always close the contour be it up or down. There are two components,
however, and one of them diverges when y → −∞, and the other when y → ∞.
The obvious idea is that, rather than considering the function (7.4.5), use
f (z) =
eiz
.
z
(7.4.8)
Surely at this stage the reader may be skeptical, whether this is really a good choice
for calculating the integral (7.4.1).
It would seem that the problem is already resolved: both the function and
the contour are fixed. However, that is not all. Due to the presence of z in
the denominator, the function proposed above has a singularity for z = 0, and
unfortunately the chosen integration contour passes through this point. In such a
situation, Cauchy’s theorem cannot be used. Ultimately, the entire contour, together
with the area encircled by it, has to lie in the holomorphicity domain of the function
f (z). For this reason, we are going make an additional minor modification of this
contour: the troublesome point will be bypassed up over a small semi-circle of radius
r (contracted to zero afterwards), as it is shown in Fig. 7.8. The entire integration
contour will then be composed of the following fragments: the segments [−R, −r]
and [r, R] lying on the real axis and the semi-circles Cr and CR . This contour in its
entirety lies, as it should be, in the holomorphicity domain of the function f (z).
The integration contour as well as the complex function f (z) have finally been
chosen, so Cauchy’s theorem may be given the form:
f (z) dz +
C1
f (z) dz +
Cr
I1
Fig. 7.8 The integration
contour for calculating the
integral (7.4.1). The dashed
line corresponds to the
integral looked for, and the
dotted ones are parts
supplemented in order to use
Cauchy’s theorem
Ir
f (z) dz +
C2
I2
f (z) dz = 0.
CR
IR
(7.4.9)
7.4 Using Cauchy’s Theorem
249
One should notice that although each of the four integrals separately depends on R
or r (or on both), the sum of all four terms equals zero. We can also formally write:
lim lim (I1 + I2 + IR + Ir ) = 0.
R→∞ r→0+
(7.4.10)
It ought to be clearly pointed out here that we do not have yet any certainty
that the limits for each of these integrations separately exist. We only know that
the double limit above does exist and equals zero. Therefore, below we are going
to study each of the integrations individually. Let us start with I1 and I2 . On the
contour C1 , one has z = x (as one can see, this time the integration parameter is
denoted with x) and one gets
−r
I1 =
−R
eix
dx =
x
−r
−R
cos x
dx + i
x
−r
−R
sin x
dx.
x
(7.4.11)
The first integration, i.e., the real part of I1 , will not be dealt with (the reader will see
in a moment why), while in the second one, the imaginary part of I1 , there appear
no obstacles to go with R and r to the appropriate limits (see Chap. 2 in Part II). We
have, therefore
−r
lim lim Im I1 = lim lim
R→∞ r→0+
R→∞ r→0+
−R
0
sin x
dx =
x
−∞
sin x
dx.
x
(7.4.12)
Similarly,
R
I2 =
R
eix
dx =
x
r
R
cos x
dx + i
x
r
sin x
dx
x
(7.4.13)
r
and for the second integral one gets
R
sin x
dx =
x
lim lim Im I2 = lim lim
R→∞ r→0+
R→∞ r→0+
r
∞
sin x
dx.
x
(7.4.14)
0
Finally one can write
∞
lim lim Im (I1 + I2 ) =
R→∞ r→0+
−∞
sin x
dx = 2I.
x
(7.4.15)
250
7 Studying Functions of Complex Variable
In this way, the expression containing the desired integral (7.4.1) has been
obtained. At this point, it becomes clear why one could modify the integrand
function by selecting it in the form of (7.4.8). The needed quantity is simply
obtained by taking only the imaginary part of the Eq. (7.4.9). It is why we did not
have to deal at all with the real contributions in (7.4.11) and (7.4.14).
Now it is time to go to the integrations over the semi-circles. On the contour CR ,
the natural parametrization is
z(ϕ) = Reiϕ
⇒
dz
= iReiϕ ,
dϕ
(7.4.16)
e−R sin ϕ eiR cos ϕ dϕ.
(7.4.17)
so one can write
π
π
1
iϕ
eiRe iReiϕ dϕ = i
Reiϕ
IR =
0
0
As it has been already discussed, we do not know how to calculate explicitly this
integral, but the following estimate can be made:
|IR | = π
0
e−R sin ϕ eiR cos ϕ dϕ ≤
−R sin ϕ iR cos ϕ e
e
dϕ
0
π/2
π
=
π
e
−R sin ϕ
dϕ = 2
0
π/2
e
−R sin ϕ
e−2Rϕ/π dϕ
dϕ ≤ 2
0
π −2Rϕ/π π/2
π
1 − e−R .
=− e
=
R
R
0
0
(7.4.18)
The following inequality, true for ϕ ∈ [0, π/2], has been used here:
sin ϕ ≥
2
ϕ,
π
(7.4.19)
which stems from the fact that the sine function is concave (i.e., its graph lies above
the intersecting chord). In turn this implies the conclusion we need:
e−R sin ϕ ≤ e−R·2ϕ/π .
(7.4.20)
As can be seen from this estimate, it occurs (the limit with respect to r is executed
in the trivial way)
lim lim |IR | = 0
R→∞ r→0+
⇒
lim lim IR = 0,
R→∞ r→0+
(7.4.21)
7.4 Using Cauchy’s Theorem
251
and the corresponding term disappears from the Eq. (7.4.10).
One is still left with the integral Ir . On the contour Cr , the similar parametrization
as before can be adopted:
⇒
z(ϕ) = reiϕ
dz
= ireiϕ ,
dϕ
(7.4.22)
e−r sin ϕ eir cos ϕ dϕ.
(7.4.23)
and then
0
π
1 ireiϕ iϕ
e
ire dϕ = −i
reiϕ
Ir =
π
0
The value of this integral is essential only for r → 0+ . There is no problem to
perform this limit because the integrand expression is continuous as the function of
two variables: ϕ and r. These issues have already been dealt with in Sect. 3.1, so
they should be well known to the reader. We obtain, therefore,
π
lim lim Ir = −i
dϕ = −iπ.
R→∞ r→0+
(7.4.24)
0
The result is purely imaginary, so it contributes only to the imaginary part of the
Eq. (7.4.10). We have already at our disposal all the elements and we get (from the
imaginary part) the required result:
∞
2I − π = 0
⇒ I=
π
sin x
dx = .
x
2
(7.4.25)
0
The problem was solved without having to find the primitive function, which, after
all, could not be explicitly written, unable to be expressed through elementary
functions.
The reader could still ask a question, what would be obtained from the real part of
the Eq. (7.4.10). In this case, passing with r to zero in (7.4.12) and (7.4.14) would be
ineligible because the expression cos x/x (as opposed to sin x/x) is not integrable
at zero. On the other hand, there would be no obstacle for performing the limit as
R → ∞. In turn the integrals IR and Ir in these limits would not contribute: the
former tends to zero, and the latter does not have the real part at all. As a result the
following equation would be obtained:
⎡
−r
lim ⎣
r→0+
−∞
cos x
dx +
x
∞
r
⎤
cos x ⎦
dx = 0.
x
(7.4.26)
252
7 Studying Functions of Complex Variable
The expression obtained on the left-hand side is called the principal value of the
integral and is conventionally denoted with P:
⎡ −r
⎤
∞
∞
cos x
cos
x
cos
x
P
dx = lim ⎣
dx +
dx ⎦ .
(7.4.27)
x
x
x
r→0+
−∞
−∞
r
The zero value of (7.4.26) is not surprising, since the integrand function is odd.
It should be noted that the principal value of an integral can exist (as it is just in
the present case), although the integral itself does not. The main difference between
the integral and its principal value is the symmetric (from above and from below)
limit as r → 0+ in (7.4.27). For an integrable function the way of taking this limit
would be inessential. One can take two different values r1 and r2 on both sides of
the singularity and independently let them go to zero. By contrast this is not possible
in the above formulas.
If a given function is integrable, the principal value is equal to the integral itself.
One has for example:
∞
P
−∞
sin x
dx =
x
∞
−∞
sin x
dx.
x
(7.4.28)
The principal values of integrals will be dealt with in the subsequent examples.
At the end, it should be noted that in this exercise instead of (7.4.8) one could
choose the function e−iz /z, and close the integration contour in the lower half-plane.
Then, in order to be allowed to use Cauchy’s theorem, the small semi-circle should
also be drawn from underneath (otherwise the singularity of the function would
appear inside the contour). This new contour would be the mirror image of that
shown in the figure. We encourage the reader to explore how the integrations would
be led in this case.
Problem 2
Making use of the known value of the Gaussian integral:
√
∞
π
−x 2
,
e dx =
2
0
(7.4.29)
the so-called Fresnel integrals:
∞
I1 =
sin(x )dx,
0
will be found.
∞
2
I1 =
cos(x 2 )dx
0
(7.4.30)
7.4 Using Cauchy’s Theorem
253
Solution
The convergence of the first integral (7.4.30) was dealt with in the second part of this
book series (see Problem 3 in Sect. 2.2). It was then found, with the use of Dirichlet’s
test, that this integral exists (for the second one the same conclusion would have
been obtained), although we were unable to find its value. As we will see in a
moment, they can be found relatively easily thanks to the application of Cauchy’s
theorem. After having examined the previous problem, the reader is definitely not
surprised that to this goal the following function is chosen:
2
f (z) = eiz .
(7.4.31)
The Fresnel integrals will then correspond to the real and imaginary parts of f (z)
integrated over the positive real semi-axis. However this contour has to be somehow
closed up. Extending it along the real axis to −∞ and adding the upper semicircle will not work, however, due to the presence of the square in the exponent.
Substituting z = x + iy, one would have
2
eiz = ei(x
2 −y 2 )
e−2xy .
(7.4.32)
The first factor has the modulus equal to 1 and is irrelevant. For the behavior of
the integral over the large semi-circle the second factor is crucial. Well, it is clear
that when y → ∞, one should also have x > 0, and in turn for y → −∞, x < 0
is needed. Otherwise, the function would grow without any bound. This means that
it is not possible to use the entire semi-circle: eventually a fragment of it in the first
(or possibly third) quadrant would fit. If so, to close the contour one also needs a
half-line outgoing from the origin. On such a straight line one has
z = teiα ,
(7.4.33)
with α being an angle in the range ]0, π/2[ (or eventually ]π, 3π/2[, but we continue
with the first option only). After plugging it into the expression for f (z), one gets
2
eiz = eit
2 e2iα
.
(7.4.34)
Now choosing α = π/4, which means that the straight line is inclined at this
angle with respect to the real axis, as shown in Fig. 7.9, it can be seen that the
additional integral to be calculated corresponds simply to (7.4.29), which is given
in the text of this exercise. For, one has: ie2iπ/4 = ieiπ/2 = i 2 = −1.
Finally the integration contour is composed of three parts shown in the figure:
C1 , CR , and C2 . As usual, at the beginning R is assumed to be finite, and then the
limit R → ∞ will be taken. The chosen contour often bears the customary name:
“a contour of the type a piece of cake” and lies in the holomorphicity domain of the
function f , which is, after all, the entire function. Thus Cauchy’s theorem has the
254
7 Studying Functions of Complex Variable
Fig. 7.9 The integration
contour for calculating
Fresnel integrals. The dashed
line corresponds to the
integral looked for, and the
dotted ones are parts
supplemented in order to use
Cauchy’s theorem
form:
2
2
eiz dz +
C1
I1
2
eiz dz +
CR
IR
eiz dz = 0.
(7.4.35)
C2
I2
Let us now focus on the first fragment. On C1 , one has z = x, and then
R
I1 =
e
dx =
0
∞
∞
R
ix 2
[cos(x ) + i sin(x )]dx −→
2
cos(x )dx + i
2
2
R→∞
0
0
sin(x 2 )dx.
0
(7.4.36)
In accordance with the results obtained in Part II, mentioned above, one could go to
the limit R → ∞ without any problems.
On C2 , we choose the parametrization z = Reiϕ = R(cos ϕ + i sin ϕ) and make
the following estimate:
π/4
iR 2 (cos ϕ+i sin ϕ)2
iϕ
|IR | = e
Rie dϕ 0
π/4
2 2
iR (cos ϕ−sin2 ϕ) −2R 2 sin ϕ cos ϕ iϕ e
e dϕ
e
≤R
0
7.4 Using Cauchy’s Theorem
255
π/4
=R
e
π/2
−R 2 sin(2ϕ)
R
dϕ =
β=2ϕ 2
e
0
π/2
−R 2 sin β
R
dβ ≤
2
0
e−2R
2 β/π
dβ,
0
(7.4.37)
where the inequality (7.4.20) has been used. This last integral can be easily carried
out and the result is
|IR | ≤
π
2
1 − e−R
4R
−→ 0.
(7.4.38)
R→∞
Thereby, the integral over CR vanishes after taking this infinite limit.
There remains still C2 , spoken of above. Here one gets
I2 =
0
R
e−t eiπ/4 dt = −eiπ/4
2
e−t dt −→ −
2
R→∞
0
R
π
(1 + i),
8
(7.4.39)
thanks to (7.4.29).
In the limit R → ∞, the thesis of Cauchy’s theorem may be written as:
∞
∞
cos(x )dx + i
sin(x 2 )dx =
2
0
π
(1 + i),
8
(7.4.40)
0
which leads to the following values of the Fresnel integrals:
∞
∞
cos(x )dx =
sin(x 2 )dx =
2
0
π
.
8
(7.4.41)
0
Problem 3
The integral:
∞
e−x cos x dx
2
I=
0
will be found, using the known value of the Gaussian integral (7.4.29).
(7.4.42)
256
7 Studying Functions of Complex Variable
Solution
It seems reasonable that in order to find the value of the integral (7.4.42) the function
f (z) is chosen in the form:
f (z) = e−z .
2
(7.4.43)
For, inserting z = x + i/2, one gets:
e−(x+i/2) = e−(x
2
2 −1/4+ix)
= e1/4 e−x (cos x − i sin x),
2
(7.4.44)
and it is visible that the required expression is obtained from the real part. In turn on
2
the positive real axis, one has z = x, so f (z) will reduce there to e−x and nothing
more than the known Gaussian integral occurs.
The equations z = x + i/2 and z = x parametrize the two horizontal sides of
a rectangle (its vertices are initially chosen at 0, R, R + i/2, i/2). To get a closed
contour, let us supplement them with two other, vertical sides, with the hope that
the values of integrals along them can be estimated or shown not to contribute to
the final result. The proposed contour is depicted in Fig. 7.10. Of course finally the
rectangle will be extended to infinity as R → ∞.
2
The function f (z) = e−z is obviously holomorphic, so one can use Cauchy’s
theorem, which gives
e−z dz +
e−z dz +
2
C1
C2
I1
Fig. 7.10 The integration
contour for calculating the
integral (7.4.42)
e−z dz +
2
I2
e−z dz = 0.
2
C3
I3
2
C4
I4
(7.4.45)
7.4 Using Cauchy’s Theorem
257
As previously, we are going to proceed with the subsequent calculation of the above
integrals. First, plugging z = x, one finds
∞
R
I1 =
e
−x 2
e−x dx =
2
dx −→
R→∞
0
√
π
.
2
(7.4.46)
0
On the contour denoted with C2 , one has z = R + it and the integral I2 will take
the form:
1/2
I2 =
1/2
e
−(R+it)2
i dt = ie
−R 2
0
e−2iRt et dt.
2
(7.4.47)
0
The quickly decreasing factor e−R suggests that this integral will tend to zero
with the increasing R. In fact, it is easy to be shown with the use of the following
estimate:
1/2
1/2
−R 2
−2iRt t 2 −2iRt t 2 −R 2
|I2 | = ie
e
e dt ≤ e
e dt
e
0
0
2
1/2
=e
−R 2
2
et dt −→ 0.
(7.4.48)
R→∞
0
The integral
1/2
"
2
et dt is, after all, a constant and finite number and does not depend
0
on R.
On the next side of the rectangle, i.e., on C3 , we take z = x + i/2, getting
0
I3 =
0
e
−(x+i/2)2
dx = e
R
e−x (cos x − i sin x) dx
2
1/4
R
R
= −e
1/4
R
e
−x 2
cos x dx + ie
0
e−x sin x dx
2
1/4
0
∞
−→ −e
1/4
∞
e
R→∞
0
−x 2
cos x dx + ie
e−x sin x dx.
2
1/4
0
(7.4.49)
258
7 Studying Functions of Complex Variable
Both improper integrals are convergent due to the presence of the factor e−x , so
one could easily go with R to infinity.
The integral over the fourth side of the rectangle is still left. There, one has z = it
and hence
2
1/2
0
I4 =
e
−(it)2
2
idt = −i
1/2
et dt.
(7.4.50)
0
This expression cannot be explicitly calculated, but it is clear that it is purely
imaginary. Thereby, it does not contribute to the real part of the Eq. (7.4.45), which
in the limit R → ∞ will take the form:
√
π
− e1/4
2
∞
e−x cos x dx = 0.
2
(7.4.51)
0
Finally it can be found that
∞
e−x cos x dx = e−1/4
2
√
π
.
2
(7.4.52)
0
It should be noted that the above integral could not be found in a traditional way.
The primitive function for the integrand expression cannot be obtained in an explicit
form, since it is expressed through special functions.
The imaginary part of Cauchy’s equation leads to the relation:
∞
1/2
e
−x 2
sin x dx = e
0
−1/4
2
et dt,
(7.4.53)
0
which can also become useful in some cases. The right-hand side of (7.4.53) cannot
be found explicitly either. It is expressed through the so-called error function of the
imaginary argument.
7.5 Looking for Images of Sets
Problem 1
The image of the area defined with the inequality Re z ≥ 1 will be found in the
mapping f (z) = 1/z.
7.5 Looking for Images of Sets
259
Solution
Functions of the complex variable z may be considered as mappings of the plane C
into itself, in accordance with the rule:
w = f (z).
(7.5.1)
In particular holomorphic functions have here interesting properties, not dealt with,
however, in this book. As a result of such mappings subsets of the plane of the
complex variable z are converted into other subsets of the complex plane in the
variable w. The present section is devoted to determining these images.
The set, the image of which is looked for in this exercise, defined with the
inequality Re z ≥ 1, constitutes the half-plane, for which x ≥ 1. Our goal is to
determine the condition to be satisfied by the new variable defined with the formula
w = 1/z. This task is not difficult. Since z = 1/w, it must be simply required that
Re
1
≥ 1,
w
(7.5.2)
which implies
1
2
1
1
+
w w̄
≥ 1.
(7.5.3)
Now both sides of this inequality may be multiplied by the quantity w w̄, which
is a positive real number. The reader should be warned against an error, occurring
sometimes, which consists of multiplying both sides of an inequality by a complex
number. An inequality obtained in this way would not make any sense. We must
remember that the set of complex numbers is not linearly organized by any order
relation. One cannot, therefore, say which of two complex numbers is smaller or
larger. On the other hand, one can compare the moduli of complex numbers, their
real or imaginary parts which are real numbers. Therefore, one must constantly keep
an eye on the inequalities so that on both sides always real numbers appear.
As a result of this multiplication, we get the inequality:
1
(w + w̄) > w w̄,
2
(7.5.4)
to be rewritten in the form:
1
1
1
w−
w̄ −
<
⇐⇒ w −
2
2
4
1 2 1
≤ .
2
4
(7.5.5)
260
7 Studying Functions of Complex Variable
Substituting w = x + iy, it is easy to see that this inequality defines a ball with the
center at (1/2, 0) and the radius of the length 1/2:
1
1
1 2
1
1
x − iy −
≤
⇐⇒
x−
x + iy −
+ y2 ≤ .
2
2
4
2
4
(7.5.6)
As it is seen, the image of the half-plane is a ball. The border of the area, in this
case the straight line x = 1, is transformed into a circle, which in turn constitutes
the border of the ball. It did not happen by accident. It can be demonstrated that the
so-called homographic mappings:
w=
az + b
,
cz + d
where ad = bd
(7.5.7)
(the supplementary condition simply meaning that a fraction cannot be shortened),
always transform straight lines into other straight lines or circles, and circles into
other circles or straight lines. The function given in the text of the exercise is a
specific case of this type of a mapping, for which a = d = 0 and b = c = 1.
Sometimes the definition of the homographic function is supplemented by the
additional condition of c = 0, which will be assumed in this section as well.
Homographic functions are often introduced on the so-called compactified
complex plane C̄ = C ∪ {∞}, also called Riemann’s sphere (see formula (7.5.9)).
One can then imagine the set C̄ as a kind of surface of the sphere (strictly speaking,
the so-called stereographic projection of it onto a plane), the south pole of which is
the point z = 0, and the north pole z = ∞. Then ∞ becomes a point on the sphere
as any other, and the line interpreted as straight on a plane, in fact, is a circle passing
through the north pole. Therefore, rather than distinguishing between straight lines
and circles, the concept of generalized circles is used with respect to both. Using
this language, we can say that a homography transforms generalized circles into
other generalized circles.
Riemann’s sphere will not be dealt with in the present book. With its detailed
construction, the reader will certainly be familiarized during lectures of analysis or
further course of study.
Problem 2
The image of the domain defined with the inequalities
|z| < 1
and
Im z > 0
will be found in the mapping f (z) = z/(z + i).
(7.5.8)
7.5 Looking for Images of Sets
261
Solution
The domain concerned with in this problem is the intersection of the interior of
a ball with a half-plane (see Fig. 7.11). Its edge is composed of a segment (i.e., a
fragment of a straight line) and a semi-circle. In turn the function f (z) is an example
of a homography defined with the formula (7.5.7), which, as we know, transforms
circles and straight lines into circles or straight lines, i.e., generalized circles into
generalized circles. We expect, therefore, that the needed image of the domain will
be the intersection of a half-plane with the interior of a ball.
In theory some other possibilities might come into play as well, which, however,
can be rejected after a while of reflection. Notice that the singular point of the
homography f (z) is z = −i, for which the denominator vanishes. Such a point
cannot be mapped onto any “finite” point lying on the C plane. Consequently the
situation when the image of this point is located on the circle is excluded. The image
of the circle |z| = 1, containing the singular point z = −i, inevitably must be a
straight line. On the other hand, on the straight line Im z = 0, which is the edge
of the half-plane, no singular point of the homography appears, which would be
transformed into ∞, so its image this time must be a circle.
All of these conclusions become clear, if the homographic mapping h(z) is
defined as transforming C̄ into C̄ in the following way:
⎧
az + b
d
⎪
⎪
for z = − i z = ∞,
⎪
⎪
c
⎨ cz + d
d
h(z) = ∞
for z = − ,
⎪
c
⎪
⎪
a
⎪
⎩
for z = ∞.
c
(7.5.9)
The conditions to be obeyed by the coefficients a, b, c, d were given in the previous
exercise.
We can now proceed with solving the present problem. Let us consider first the
inequality |z| < 1. When solving the equation w = z/(z + i) for the variable w, one
gets z = iw/(1 − w), leading to the inequality:
iw 1 − w < 1,
(7.5.10)
w w̄
< 1.
(1 − w)(1 − w̄)
(7.5.11)
which can also be given the form:
After having multiplied both sides by the denominator (real, positive number!), one
comes to
262
7 Studying Functions of Complex Variable
Fig. 7.11 The area indicated in the text of the problem and its image in the homography w =
z/(z + i)
w w̄ < 1 − w − w̄ + w w̄
⇒ w + w̄ > 1.
(7.5.12)
The obtained inequality constitutes simply the condition Re w > 1/2, i.e., x > 1/2.
In accordance with our prediction, the image of the ball turned out to be a half-plane.
Now consider the inequality Im z > 0. Expressing the variable z through w, we
obtain
iw
−i w̄
1
−
> 0.
(7.5.13)
2i 1 − w 1 − w̄
Let us now again multiply both sides of the inequality by (1 − w)(1 − w̄) and
additionally by 2 (multiplying by i is forbidden!), obtaining after some minor
transformations
1
1
1
w + w̄ − 2w w̄ > 0 ⇒
w−
w̄ −
< .
(7.5.14)
2
2
4
Once again our prediction is confirmed: the image of the half-plane is, in this case,
a ball. As one can see, the center of this ball is located at w = 1/2 and such is also
the value of its radius.
As a result, the image of the area specified in the text of the exercise is the semicircle without boundary, described with the inequalities:
Re w >
1
2
and
w−
1
2
1
1
w̄ −
< .
2
4
Both areas, the initial one and its image, are shown in Fig. 7.11.
(7.5.15)
7.5 Looking for Images of Sets
263
Problem 3
The image of the hyperbola xy = 1 in the mapping f (z) = z2 will be found.
Solution
Since w = z2 , and z = x + iy, one gets:
w = (x + iy)2 = x 2 − y 2 + 2ixy,
(7.5.16)
Re w = x 2 − y 2 , Im w = 2xy.
(7.5.17)
or
We are interested in the image of the hyperbola of the equation xy = 1, so
naturally there occurs Im w = 2. This means that the image is contained within
the horizontal straight line specified with this equation. However, it is still unclear
whether this image constitutes the entire straight line or only a subset of it. In order
to resolve this issue, one needs to consider the first of the relations (7.5.17). The
question to be answered, is, in fact, the question about the range of the function:
g(x) = x 2 −
1
,
x2
(7.5.18)
where in place of y it has already been put 1/x. Upon solving the equation:
x2 −
1
= ξ,
x2
(7.5.19)
ξ being an arbitrary real parameter, one obtains
(x ) − ξ x − 1 = 0
2 2
⇒ x =
2
2
ξ±
ξ2 + 4
.
2
(7.5.20)
The sign “−” has to be rejected because it would lead to the negative value of x 2 . In
turn, for the sign “+” there appear two solutions for x, regardless of the value of ξ :
0
x=±
ξ+
11/2
ξ2 + 4
.
2
(7.5.21)
264
7 Studying Functions of Complex Variable
This result indicates that no matter which point was chosen on the line Im w = 2
(i.e., regardless of the choice of ξ on the right-hand side of (7.5.19)), a point (and
even two points in accordance with the formula (7.5.21)) on the hyperbola can be
found, whose image is w = ξ + 2i. We conclude, therefore, that the image of the
hyperbola xy = 1 in the mapping w = f (z) is the entire straight line Im w = 2.
Problem 4
The image of the circle |z| = R, where R > 0, will be found in the mapping
f (z) = (1 + z)2 .
Solution
This problem will be solved in a slightly different way than the previous ones in this
section. Rather than to look for the condition for w (where w = f (z)), we will try to
parametrize the curve in the plane of the complex z and next obtain the parametric
formula for the curve in the complex w plane. The initial curve is a circle, for which
the natural parametrization is z(ϕ) = Reiϕ , where ϕ ∈ [0, 2π [. By substituting this
expression into the formula for w, one finds
w(ϕ) = (z(ϕ) + 1)2 = 1 + Reiϕ
2
= 1 + 2Reiϕ + R 2 e2iϕ .
(7.5.22)
Denoting with the symbols x and y, respectively, the real and imaginary parts of
the variable w, one gets the wanted curve in the parametric form:
x(ϕ) = 1 + 2R cos ϕ + R 2 cos(2ϕ),
y(ϕ) = 2R sin ϕ + R 2 sin(2ϕ).
(7.5.23)
Such a curve is called the cardioid due to its similarity in shape to the heart. It is
shown in Fig. 7.12.
7.6 Exercises for Independent Work
Exercise 1 Find the integrals of the functions f (z) = Im z, g(z) = z2 , and h(z) =
|z|2 over the contours drawn in Fig. 7.7.
7.6 Exercises for Independent Work
265
y
2R
1 R
x
2R
Fig. 7.12 The cardioid (7.5.23)
Answers
f (z) dz = −1 + i/2,
C1
f (z) dz = (i − 1)/2,
C2
g(z) dz =
C1
g(z) dz = −(1 + i)/3,
C2
h(z) dz = (4i − 4)/3,
C1
h(z) dz = 2(i − 1)/3.
C2
Exercise 2 Find the full expression for the holomorphic function, if
(a) its real part is given by the formula u(x, y) = x 3 − 3xy 2 ,
(b) its imaginary part is given by the formula v(x, y) = y cos x cosh y
− x sin x sinh y.
Answers
(a) f (z) = z3 .
(b) f (z) = z cos z.
266
7 Studying Functions of Complex Variable
Exercise 3 Using Cauchy’s theorem obtain the value of the integral:
/ 2
z (1 + z)
(a) I =
dz, where C is the positively oriented circle of the equation
(z4 − 16)2
C
|z| =/1,
(b) I =
C
1
dz, where C is the positively oriented triangle with
(z2 − z + 25/36)2
vertices at 0, 1, 1 + i.
Answers
(a) I = 0.
(b) I = 0.
Exercise 4 Find the image of the set
(a) defined with the inequality Re z ≥ 0 in the mapping w = (z − i)/(z + i),
(b) defined with the inequalities 0 < Im z < π in the mapping w = ez ,
(c) defined with the inequalities π/6 < arg z < π/3 in the mapping w = (z6 +
i)/(z6 − i).
Answers
(a) The half-plane Im w ≤ 0.
(b) The half-plane Im w > 0.
(c) The ball |w| < 1.
Chapter 8
Investigating Singularities of Complex
Functions
In the present chapter, the central role will be played by isolated singularities of
complex functions. By “singularity” we mean the point of non-holomorphicity of a
given function. The detailed classification of singularities is given in the solution of
the first problem. In the second section, examples of problems with singularities are
dealt with from the point of view of the Laurent series.
Given a function f (z) holomorphic in the ring r < |z − z0 | < R, where 0 ≤ r <
R ≤ ∞. Then this function can be (in this domain) expanded in the Laurent series,
which has the form
∞
!
f (z) =
an (z − z0 )n .
(8.0.1)
n=−∞
The terms corresponding to negative values of n constitute the principal part of the
series and all other the analytic (Taylorian) part. The expansion coefficients may
be found with the use of the following integral formula
an =
1
2π i
/
C
f (z)
dz,
(z − z0 )n+1
(8.0.2)
with the contour C lying entirely in the considered ring and encircling the point
z0 exactly once. This type of a contour, i.e., constituting one non-self-intersecting
continuous loop is sometimes called a Jordan curve or a simple closed curve.
In the case when z0 constitutes an isolated singular point of the function f (z)
(i.e., such that all other points lying in a certain neighborhood of z0 are regular ones),
the inner ring has the form: 0 < |z − z0 | < R. Then the following classification of
singularities can be made. Thefunction at z = z0 has
© Springer Nature Switzerland AG 2020
T. Radożycki, Solving Problems in Mathematical Analysis, Part III, Problem Books
in Mathematics, https://doi.org/10.1007/978-3-030-38596-5_8
267
268
8 Investigating Singularities of Complex Functions
1. a removable singularity, if ∀n<0 an = 0,
2. a pole of order k, if ∀n<−k an = 0, and a−k = 0,
3. an essential singularity, if the principal part has infinitely many nonzero terms.
The integrals over closed contours can be calculated according to the following
Cauchy residue theorem. Let an open set O ⊂ C be simply connected and let
the function f (z) be holomorphic on O except at the isolated points z1 , . . . , zm
belonging to O. For any closed, simple, positively oriented contour C ⊂ O evading
all the singular points z1 , . . . , zm , the following formula holds:
/
f (z) dz = 2π i
!
i
C
Res f (z)
(8.0.3)
z=zi
where the summation runs over all singular points encircled by C. The residue
of the function f (z) at a given pole z0 is the value of the coefficient a−1 in the
expansion (8.0.1) around z = z0 in the inner ring. Without expanding the function
into the Laurent series, the residue at a point z0 constituting the kth order pole can
be found according to the following formula:
a−1 = Res f (z) =
z=z0
d k−1 1
(z − z0 )k f (z)
(k − 1)! dzk−1
.
(8.0.4)
z=z0
8.1 Identifying the Types of Singularities
Problem 1
All singular points of the function given with the formula
f (z) =
z
sin(π z)
(8.1.1)
will be examined.
Solution
In the previous chapters the word “singularities” has often appeared with regard to
the non-holomorphicity points of functions. Above, in the theoretical introduction,
the initial classification of such points based on the behavior of the Laurent series
has been done. In the present section, we will discuss such points in more detail. At
the beginning we come back to their classification because their character is very
important for the behavior of functions. First of all, let us divide these points into
isolated ones and non-isolated ones.
8.1 Identifying the Types of Singularities
269
The notion of an isolated singular point for a certain function f (z) is quite clear.
Let us imagine that this function is defined and is holomorphic on a certain ball
K(z0 , r) except its very center z0 , which does not belong to the domain. This point
is singular from the viewpoint of the function, but at the same time it is isolated,
hence the name. In fact, one usually proceeds vice versa: first one identifies the
singular points of a function, and only then one considers whether a given point
can be surrounded with a ball (of an arbitrarily small radius, as long as such radius
exists), within which the function would be holomorphic (except the center of the
ball). If it is not possible, and the needed ball does not exist, this means that the
given singularity is non-isolated (i.e., is a branch point).
If such a ball does exist, which means that the singularity is isolated, one can
think about further classification. The precise identification of the nature of a given
singularity is very important for a variety of applications, which will be explored by
the reader when solving problems of this chapter and the next one. The following
important situations mentioned in the theoretical introduction can occur.
1. The removable singularity.
If the following condition is met:
lim f (z) = a,
z→z0
(8.1.2)
where a is a finite complex number, the singularity is removable. Formally z0
does not belong to the domain for some reason, but it could be included in it, if
the definition of the function f was supplemented with the condition:
f (z0 ) = a.
(8.1.3)
The point z0 = 0 for the function f (z) = z/z can serve as a trivial example of the
removable singularity. This function differs from that defined with the formula
g(z) = 1 only with the domain, since the former is defined on C \ {0}, and the
latter on C. However, one could supplement the definition of the function f with
the additional condition f (0) = 1 and get the function free of this singularity.
This kind of a singularity is, therefore, called removable.
2. The pole of the nth order.
Suppose now that the limit (8.1.2) does not exist, but one can find such n ∈ N
that
lim (z − z0 )n f (z) = b,
z→z0
(8.1.4)
with b = 0. The number n should, therefore, be so adjusted that b be neither
zero nor infinity. A function possessing such singularity can have for example
the following form f (z) = 1/(z − 2)3 . The singular point is undoubtedly z0 = 2.
It is, of course, the isolated point since the function is holomorphic everywhere
around it. If now the limit of the following expression was looked for
270
8 Investigating Singularities of Complex Functions
(z − 2)n
1
(z − 2)3
(8.1.5)
as z → 2, it is clear that the finite result is obtained only for n = 3. If n was
taken too large, then the limit (8.1.4) would be zero, if too small—it would not
exist at all. One can, therefore, say that the function in question has at z0 = 2 the
pole of the 3rd order.
3. The essential singularity.
If a given function has an isolated singular point, but it is neither a removable
singularity, nor a pole of the nth order (i.e., it is not possible to find any n which
would satisfy the condition (8.1.4)), one has an essential singularity. An example
of a function possessing this type of a singularity will be met in the next problem.
After having gone through this classification, we can move on to examine the
function (8.1.1) given in the text of the exercise. First of all, it should be noted that
it is the quotient of two entire functions: a polynomial and the sine function, so
the only points of non-holomorphicity constitute the zeros of the denominator, i.e.,
those satisfying the equation:
sin(π z) = 0.
(8.1.6)
The solutions of this equation are well known: they are all integer numbers n. It
simply stems from the consideration of the equation:
1 iπ z
(e − e−iπ z ) = 0
2i
⇒ e2iπ z = 1.
(8.1.7)
If one inserts z = x + iy, the following condition is obtained:
e2iπ x e−2πy = 1.
(8.1.8)
For a real x, one has |e2iπ x | = 1, so upon taking the modulus of both sides of (8.1.8)
one obtains
e−2πy = 1
⇒ y = 0.
(8.1.9)
Thus we come to the first conclusion: the roots of the function sin(π z) are real.
After using (8.1.9), the Eq. (8.1.8) leads to
e2iπ x = cos(2π x) + i sin(2π x) = 1
cos(2π x) = 1 ∧ sin(2π x) = 0
⇒
⇒ x = n ∈ Z.
(8.1.10)
8.1 Identifying the Types of Singularities
271
This result entails that the function f (z) has an infinite number of singularities
located at the points zn = n ∈ Z. Obviously all these singularities are isolated,
as inside the balls K(zn , 1/2), except centers there are no other roots of the function
sin(π z).
Now one needs to accurately determine the nature of the singularities having
been found. Let us examine first the point z = 0, calculating the limit:
1
z
= .
z→0 sin(π z)
π
(8.1.11)
lim
Some finite number has been obtained, which means that the singularity is removable. This should have been expected, since for z → 0 the numerator and the
denominator are going to zero equally fast.
Now let us pass to the point z = k ∈ Z, the condition k = 0 having already
been assumed. Because of the periodic nature of the sine function, we expect all the
subsequent zeros to have similar character, as for z = 0. If so, then multiplying the
numerator by the factor (z − k), the finite limits (8.1.4) should be obtained. Thus,
let us calculate:
lim
z→k
(z − k)z
z (z + k)
= lim
sin(π z) z=z +kz →0 sin(π(z + k))
z (z + k)
z →0 sin(π z ) cos(kπ ) + cos(π z ) sin(kπ )
= lim
=(−1)k
z (z + k)
k
= (−1)k .
π
z →0 sin(π z )
= (−1)k lim
=0
(8.1.12)
Since k = 0, the condition (8.1.4) is satisfied. When calculating the above limit, the
function f (z) has been multiplied by the factor (z − k), i.e., (z − k)1 , which means
that n = 1. One can, therefore, deduce that at all points z = k, except zero, the
function has poles of the first order.
Problem 2
All singular points of the function given with the formula
f (z) = e1/z
will be examined.
1
z−2
(8.1.13)
272
8 Investigating Singularities of Complex Functions
Solution
Let us first take a look at the form of the function (8.1.13). It is a quotient, the
denominator of which constitutes a holomorphic function (a polynomial). What
should be examined is the root of the denominator, i.e., the point z = 2. In the
numerator there occurs an exponential, which, as we know, is holomorphic on the
entire complex plane. It is, however, composed with 1/z, for which z = 0 is (the
only) point of non-holomorphicity. The function f (z) has, therefore, two isolated
singularities, which will be subsequently examined.
1. The point z = 2.
Let us find the limit:
lim (z − 2)e1/z
z→2
√
1
= lim e1/z = e.
z − 2 z→2
(8.1.14)
According to the condition (8.1.4), it is clear that the function f (z) has at z = 2
the pole of the first order.
2. The point z = 0.
Since the limit
lim e1/z
z→0
1
z−2
(8.1.15)
does not exist, the removable singularity is for sure not an option. One should
then verify whether one is dealing with a pole of a certain order k, by calculating
lim zk e1/z
z→0
1
.
z−2
(8.1.16)
The expression 1/(z − 2) is not an issue here: it has the limit equal to −1/2,
therefore one should focus on the product zk e1/z .
If the limit at z → 0 was to exist for a certain natural k, it could not depend on
how the transition of z to zero is done. So let us choose a particular path: let us
move along the positive real axis, substituting simply z = x. In addition let us
introduce the new variable t = 1/x, obtaining (the symbol ∞ below stands for
the “normal” positive infinity in the real sense)
lim x k e1/x = lim t −k et = ∞.
x→0+
t→∞
(8.1.17)
This result is well known to the reader from the first part of this book series: the
exponential behavior dominates over the power one, regardless of the value of k.
However, as the limit does not exist in this particular (i.e., real) case, it does not
exist at all. At the singular point z = 0 the function f (z) has no pole of any
order. What remains is only the third and last possibility: one is dealing with an
essential singularity.
8.1 Identifying the Types of Singularities
273
This conclusion can also be found out in another way, namely by using the wellknown Taylor expansion for the exponential function:
ez =
∞ n
!
z
n=0
n!
.
(8.1.18)
From this formula one can write:
zk e1/z =
∞
!
1
1
· n−k .
n! z
(8.1.19)
n=0
Since the sum over n extends to infinity, no matter the magnitude of k, for n >
k one will get the positive power of 1/z. Such expressions obviously have no
limit for z → 0. This is the typical situation, when a function has an essential
singularity at a point z0 : the series with negative powers of (z−z0 ) does not peter
away, so they cannot be compensated by any factor (z − z0 )k in the numerator.
Problem 3
All singular points of the function given with the formula
f (z) =
log z
z−1
(8.1.20)
will be examined.
Solution
When solving Problem 3 of Sect. 7.1, we were dealing in detail with the holomorphicity domain of the logarithmic function. It was then found that the largest area
constitutes the complex plane C, the half-line starting at z = 0 having been removed.
We decided to choose the real nonpositive semi-axis, specifying the domain of the
logarithm as C\] − ∞, 0]. Whatever way this half-line was chosen, the point z = 0
will always stay outside the holomorphicity region. Therefore, this point constitutes
the singularity of the function (8.1.20) for sure. It is easy to establish that this
singularity is non-isolated (i.e., it is a branch point) as not only the point itself,
but the entire half-line must be removed from the domain. Consequently, the points
belonging to this half-line cannot appear in any ball surrounding the point z = 0,
which consequently ceases to be a neighborhood.
274
8 Investigating Singularities of Complex Functions
That it is impossible to define the logarithm as a holomorphic function on a set
K(0, r) \ {0} even for arbitrarily small r, one can be convinced in the following
way. Let us choose an exemplary point z1 = −r/2 naturally located inside the ball.
If the function were to be holomorphic on this ball (apart from its center), then it
would have to be continuous. As it will be seen below, this is not possible. Let us
choose in the domain under consideration the second point, e.g., z0 = r/2 and two
paths, which lead from z0 to z1 . Both of them will be parametrized with the formula
z(ϕ) = r/2 eiϕ , but for the first one ϕ varies from 0 to π , and for the second one
from 0 to −π . Both are also entirely contained in the set K(0, r) \ {0}.
Now let us check, with the use of the formula (7.1.31), what values of the function
will be obtained by letting z do to z1 in both these situations. It would be
lim log(z(ϕ)) = lim (ln
ϕ→π
ϕ→π
r
r
+ iϕ) = ln + iπ
2
2
(8.1.21)
in the first case, and
lim log(z(ϕ)) = lim (ln
ϕ→−π
ϕ→−π
r
r
+ iϕ) = ln − iπ
2
2
(8.1.22)
in the second one. These values are different and, therefore, it is visible that the
function has a “fault.” Of course, the same result would be obtained by selecting
any other point (and other than z = 0), as long as the branch point is being encircled
from different sides. As a result, it becomes obvious that the function cannot be
holomorphic on the chosen set and, thereby, the point z = 0 constitutes a nonisolated singularity. It could be said that the number −r/2 appears in two copies
and at one of them the function acquires the value (8.1.21), and at the other (8.1.22).
What is more, the same refers to all other points beyond zero. In reality, these copies
are infinitely many, and they can be reached by encircling the point z = 0 the
desired number of times. This can be imagined as moving around the curly stairs
in a multi-storey (i.e., an infinite-storey) building, the earlier removed half-line ] −
∞, 0] being a kind of a “barrier” between the subsequent floors. When moving in
a circle on the staircase, one is going to a higher or lower floor, on which the new
copies of −r/2 can be found, but also copies of other numbers: 1, −1, 2, −2, i, −i,
etc. At the point z = 0, the plane branches off onto different floors and hence the
name: the branch point. With this construction, the logarithmic function becomes
holomorphic (beyond z = 0), but only being defined on this complex multisheet
surface. The price which is payed for this function becoming holomorphic is the
infinite multiplication of each complex number, and, when calculating the value of
log z, one must have the additional information: which of the infinitely many copies
of the number z one is talking about.
The second singular point of the function f (z) is the root of the denominator, i.e.,
z = 1. Naturally this is an isolated singularity. A ball around this point can be chosen
with such radius (r < 1) that the other troublesome point z = 0 associated with the
numerator will remain outside and no problems connected with the logarithm will
appear. In order to investigate the nature of this singularity, let us calculate
8.1 Identifying the Types of Singularities
lim f (z) = lim
z→1
z→1
275
log z
1/z
= lim
= 1.
z − 1 z→1 1
(8.1.23)
To obtain the final result, the complex version of l’Hospital’s rule has been used.
First of all, the functions in the numerator and denominator converge to zero when
z → 1. Second, both functions are holomorphic. In the case of the denominator,
it is obvious: after all, we are dealing with a polynomial. Some concerns can be
raised by the reader, however, as to the statement about the holomorphicity of the
logarithm. The essence of the issue here is the fact that we are interested in the area
around z = 1 and not around z = 0. If we limit ourselves to the small neighborhood
of z = 1, no problem with the non-holomorphicity of the logarithm will come into
play. It should be strongly emphasized that the singularity of the logarithm is related
to the point z = 0. Excluding some other points of the complex plane from the
domain (e.g., some half-line) is only a matter of agreement. In fact, nothing “bad”
happens around the point z = 1. If we find ourselves near some z = 0 on one of
the floors of our skyscraper, we can walk around it in circles, as long as the radius
is small enough not to take us to the staircase.
From (8.1.23) under the condition (8.1.2), one sees that the singularity at z = 1
is removable. This is because in the denominator of the function f (z) one finds the
root of the first order and identical is the root of the numerator. This is easy to see if
one uses the Taylor expansion of the logarithm around unity:
(z − 1)2
(z − 1)3
(z − 1)4
+
−
+ ...
2
3
4
(z − 1) (z − 1)2
(z − 1)3
= (z − 1) 1 −
+
−
+ ... .
(8.1.24)
2
3
4
log z = log(1 + z − 1) = z − 1 −
Finally:
log z
(z − 1) (z − 1)2
(z − 1)3
=1−
+
−
+ . . . −→ 1.
z→1
z−1
2
3
4
(8.1.25)
Problem 4
All singular points of the function given with the formula
f (z) =
will be examined.
√
3
z2
z
+1
(8.1.26)
276
8 Investigating Singularities of Complex Functions
Solution
First, let us focus on the denominator of the function f . Upon decomposing it into
linear factors, i.e., writing
z2 + 1 = (z − i)(z + i),
(8.1.27)
one can see that it has two roots: z = i and z = −i. Since they are first order roots,
and the numerator does not vanish at any of them, our suspicion is that the function
f (z) in both cases has poles of the first order. To find it out, let us calculate the
appropriate limits:
√
√
3
3
z
z
i
lim (z − i)f (z) = lim (z − i)
= lim
=
,
z→i
z→i
(z − i)(z + i) z→i z + i
2i
√
√
√
3
3
3
z
z
−i
= lim
=
.
lim (z + i)f (z) = lim (z + i)
z→−i
z→−i
(z − i)(z + i) z→−i z − i
−2i
√
3
(8.1.28)
Two finite expressions have been obtained, which, however,
require
some clarifica√
√
3
3
tion. There may appear a question: what do
the
symbols
i
and
−i
mean?
√
A simple answer to the question, what 3 i is, reads as follows: it is such number
that when raised to the third power becomes equal to i. The problem, however, is
that there are three different numbers of that kind, namely:
− i,
1 √
( 3 + i),
2
1 √
(− 3 + i).
2
(8.1.29)
A simple exercise for the reader is to check that each of them raised to the cube
gives i.
In the same way,
√ one would ascertain that there are also three numbers that can
be understood as 3 −i:
i,
1 √
( 3 − i),
2
1 √
(− 3 − i).
2
(8.1.30)
In order for the expressions (8.1.28) to have a unique meaning, one must
determine which of these numbers one is talking about. This issue stays beyond
the current chapter and is connected with the so-called multivalued functions. To
this subject, the entire next chapter is devoted. Here it should be noticed only that
clarifying the right-hand sides of (8.1.28) is possible, and if so, in accordance with
the criterion (8.1.4), at z = ±i one has poles of the first order.
Now one has to address the numerator, which can be saved in the form:
√
3
z = e1/3 log z .
(8.1.31)
8.1 Identifying the Types of Singularities
277
As we know, the exponential function is holomorphic on the entire complex plane,
so any point of non-holomorphicity must be associated with the presence of the
logarithm. Such a point does really exist, namely z = 0. This is, as we already know,
the branch point for the logarithmic function. One can easily see that this singularity
appears also in the case of the function (8.1.31), and consequently (8.1.26) as well—
although our “multi-storey building” in this case has only three floors. For this
purpose one can do the same as in the previous exercise: to encircle it with some
ball K(0, r), the radius r being as small as one wishes, and to choose an exemplary
point z0 = r/2 inside it. From both sides of 0, the two paths (contained in the
ball) to the second point, e.g., to z1 = −r/2, will be led. For example, they can be
parameterized with the formula z(ϕ) = r/2 eiϕ , only that once ϕ runs from 0 to π ,
and once from 0 to −π . Now, using the formula (7.1.31), let us calculate the values
that would be obtained when going from z to z1 in both situations:
lim
√
3
ϕ→π
r
ϕ→π
=
lim
ϕ→−π
√
3
r
z = lim e1/3 log z = lim e1/3 (ln 2 +iϕ) = e1/3 (ln 2 +iπ )
3
r iπ/3
e
=
2
ϕ→π
3
π
π
r
cos + i sin
2
3
3
=
1
2
3
√
r
(1 + i 3),
2
r
r
z = lim e1/3 log z = lim e1/3 (ln 2 +iϕ) = e1/3 (ln 2 −iπ )
ϕ→−π
=
3
r −iπ/3
=
e
2
ϕ→−π
3
−π
1
r
−π
cos
=
+ i sin
2
3
3
2
3
√
r
(1 − i 3).
2
(8.1.32)
The presence of 1/3 in the exponents shows what have been already signaled.
If we additionally encircled the origin of the coordinate system three times, i.e.,
ϕ would vary not from 0 to π , but to 6π + π = 7π , we would again get the value
1
2
3
√
r
(1 + i 3)
2
(and analogously for the lower formula). This would mean that if one walked up the
staircase to the fourth floor, he would find himself on the first floor again.
It should be emphasized that unlike (8.1.26) and subsequent expressions, the
symbols of a root used on the right-hand side of the formulas (8.1.32) represent
ordinary roots in the real sense familiar to the reader from school. This may cause
some confusion, but should become clear after studying the next chapter of this
book. A similar problem with the symbol of the logarithm has been solved by using
in real terms “ln” instead of “log.”
The results of (8.1.32) mean that the function f (z) cannot be holomorphic in any
ball (even removing the center) surrounding the system origin. The point z = 0
constitutes, therefore, a non-isolated singularity.
278
8 Investigating Singularities of Complex Functions
8.2 Expanding Functions into Laurent Series
Problem 1
The Laurent series for the function given with the formula
f (z) =
z
z2 − 4
(8.2.1)
will be found in all annuli around the point z0 = 0.
Solution
As it was spoken of in the theoretical introduction at the beginning of this chapter,
the Laurent series is the series of the form:
∞
!
an (z − z0 )n .
(8.2.2)
n=−∞
As one can see, it contains—contrary to the Taylor series—beside positive powers
of the expression (z − z0 )—the negative ones as well. The sum of these terms (for
n = −1, −2, −3, . . .) is called the principal part of the series, and the remaining
terms form the analytic (Taylorian) part.
For what functions, apart from the analytic part, does there occur also the
principal part of the expansion? Well, it happens in general when one has to do
with singularities. The function which has no singularities, i.e., the holomorphic
function, can be expanded around each point into a series, which is an ordinary
(although complex) Taylor series. There are no negative powers of (z − z0 ).
Such an expandable function is called analytical and often the alternative terms
“holomorphic” and “analytic” are used in this context. These issues were discussed
in Sect. 7.2.
However, if one liked to expand a function around a point z0 , which is its pole,
then in this expansion the principal part occurs as well and it becomes a true Laurent
series. From the lecture of analysis, the reader probably knows that if z0 constituted
a non-isolated singularity, then the Laurent series on a ball with its center at z0 could
not be defined at all. One could speak neither about analytic nor about principal
parts, since there is no series at all!
One can also encounter a situation where z0 is the point of holomorphicity of the
function under consideration, but there is a singularity at certain z1 located nearby.
It is then clear that the expansion around z0 (it will be a pure Taylorian form) can
only be done on the ball K(z0 , |z1 − z0 |). The maximal radius is the distance from
z0 to the nearest singularity, which by definition must remain outside. Inside the
8.2 Expanding Functions into Laurent Series
279
convergence ball, the function is holomorphic, and it could not be such if the point
z1 was there.
We are not going to continue these theoretical considerations and discuss possible
expansions in the so-called annuli, since it will be easiest to do this when discussing
a specific example. So let us move on to analyzing the function (8.2.1).
It is obvious that this function has two singularities—these are the poles of the
first order at the points z1 = 2 and z2 = −2—resulting from the roots of the
denominator. The point z0 = 0 itself is the point where the function is holomorphic,
so one expects a Taylorian expansion around it, however limited to the ball of the
radius r = 2. Otherwise one would catch in the interior the singular points z1 and
z2 , which is not allowed.
Now let us see how this works in practice. We start with rewriting the function
f (z) in the form that will allow us to use the known formula for the sum of terms
of the geometric series:
1
z
.
f (z) = − ·
4 1 − z2 /4
(8.2.3)
The second fraction is associated immediately with the expression:
1
= 1 + q + q2 + q3 + . . . ,
1−q
(8.2.4)
if one puts q = z2 /4. There occurs, however, a certain restriction: the formula (8.2.4) is only applicable when the common ratio of the geometric series
(strictly speaking its modulus) is smaller than 1. So if one wishes to use this
expansion, one is forced to assume that
2
z < 1,
4
i.e.
|z| < 2.
(8.2.5)
As one can see, the expansion obtained will be correct inside the ball of radius 2. It
is not surprising: we have already come to this conclusion. In this simple way, one
gets the expansion of the function f (z) in the form:
1
0
2 2 2 3
z
z
z
z2
f (z) = −
+
+
+ ...
1+
4
4
4
4
∞
! z2n+1
z3
z
z5
= − − 2 − 3 ... = −
.
4 4
4
4n+1
(8.2.6)
n=0
It is visible that there appear only positive powers of the expression (z − z0 ), which
in our case is simply (z − 0) = z. The ball on which (8.2.6) is “legal” is marked in
Fig. 8.1 with gray color.
280
8 Investigating Singularities of Complex Functions
Fig. 8.1 The ball of holomorphicity on which the expansion (8.2.6) holds, and singularities of the
function (8.2.1)
One can now ask how to expand the function f (z) beyond the above ball. As it
will be verified below, it is possible, but the principal part will appear as well. Using
again the formula for the sum of the geometric series, one gets
1
0
2 3
1
1
1
4
4
4
f (z) = ·
=
1 + 2 + 2 + 2 + ...
z 1 − 4/z2
z
z
z
z
∞
−1
!
!
1
4
4n
1 2n+1
42
= + 3 + 5 + ... =
=
z
.
n+1
z z
z
z2n+1
4
n=−∞
(8.2.7)
n=0
This time, however, the common ratio is the quantity 4/z2 , which leads to the
condition:
4
< 1, i.e. |z| > 2.
(8.2.8)
z2 Thereby the series (8.2.7) is convergent in the region outside the ball of radius 2.
This means that there are two domains, as shown in Fig. 8.1, which could be called
the “annuli”: the internal one, on which the function has the expansion (8.2.6), and
the external one, for which (8.2.7) holds. If there were further singularities at greater
distances from z0 , there would be even more different regions (and they would then
deserve even more the name “annuli”). This will be seen in the following problems.
For the time being, let us pay attention to some facts below, which will occur in our
further examples.
8.2 Expanding Functions into Laurent Series
281
1. Inside each annulus (in this case |z| < 2 or |z| > 2), the function is holomorphic.
2. Since the expansion is made around the point z0 , i.e., the point of holomorphicity,
in the inner annulus it has the purely Taylorian character.
3. From the presence of the principal part of the series in the outer annulus, i.e.,
terms with negative powers, one cannot claim the existence of a singularity at the
point z0 . This is determined by the form of expansion in the inner annulus only.
4. The expansion around the point z0 = 0 of an even function contains only even
powers of z and of an odd function—only odd ones. This is not true for z0 = 0.
Note that in order to obtain the formulas (8.2.6) and (8.2.7), the known form of
the sum of a geometric series was used. However, the reader might feel unsatisfied
here. After all, not every function we wish to expand must be of the type (8.2.1) that
is suitable for this. For other functions one would have to come up with a different
way. Fortunately, there exists a universal integral formula for the coefficients an
of the series (8.2.2) given in the theoretical introduction (see (8.0.2)). Its form is
recalled below:
/
f (z)
1
dz.
(8.2.9)
an =
2π i
(z − z0 )n+1
C
If the closed curve C, running around the point z0 , was entirely contained within
the ball |z| < 2, one would get, after calculating the integrals, the coefficients an
of the formula (8.2.6). On the other hand, if it were contained in the outer annulus,
i.e., for |z| > 2, one would get (8.2.7). Let us follow this procedure for this latter
annulus. The residue theorem is useful here, which will be dealt with only in the
next section. The reader probably knows it already from the lecture. If not, then
the following considerations can be omitted, and one can come back to them after
having worked through the present chapter to the end.
Let us choose a closed curve C lying entirely in the outer annulus, with the
positive orientation (this means that the interior of the enclosed area is passed on the
left hand). An example of such a curve could be a circle with the center at z = 0 and
of any radius greater than 2, but for another curve the same result would be obtained.
The positive orientation means anticlockwise direction. The formula (8.2.9) will
then take the form:
/
/ 1
1
1
1
z
1
1
an =
−
· n dz.
· n+1 dz =
(8.2.10)
2
2π i
8π i
z−2 z+2
z
z −4 z
C
C
g(z)
Within the region encircled by the curve C, there are three singularities of the
integrand expression (i.e., of the function g(z)): the poles of the first order at z = ±2
and the pole of the nth order (if n > 0) at z = 0. The residue theorem discussed
in the next section states that in this case the value of the integral is equal to the
sum of residues at all these singularities times 2π i. We would like to remind you
that the residue at a certain point z = ξ is the value of the factor multiplying the
282
8 Investigating Singularities of Complex Functions
term 1/(z − ξ ) read off from the Laurent expansion in the inner annulus around this
point. If at this moment it is still not clear how to find this factor, then one should
arm oneself with a bit of patience.
First let us assume that n > 0. According to the given recipe, one has
an =
=
=
=
=
2π i
Res g(z) + Res g(z) + Res g(z)
z=−2
z=0
8π i z=2
1
d n−1
1
1
1
1 1
·
−
−
+
n
n
n−1
4 2
(−2)
(n − 1)! dz
z − 2 z + 2 z=0
1 1
(−1)n
(−1)n−1
(−1)n−1 −
+
−
4 2n
2n
(z − 2)n
(z + 2)n z=0
(−1)n
(−1)n−1
(−1)n−1
1 1
−
+
−
4 2n
2n
(−2)n
2n
1 1
(−1)n
1
(−1)n
−
−
+
= 0.
(8.2.11)
4 2n
2n
2n
2n
All the coefficients of the Taylorian part have proved to vanish, which is consistent
with the form of the series (8.2.7), which has reduced to the principal part only.
When n ≤ 0, the function g(z) in (8.2.10) has no pole for z = 0. Inside the
integration contour there remain only poles at z = ±2, so we find
2π i
1 1
1
Res g(z) + Res g(z) =
−
z=−2
8π i z=2
4 2n
(−2)n
⎧
for n = 2k + 2,
⎨0
1
= n+2 [1 − (−1)n ] =
⎩ −k−1
2
for n = 2k + 1,
4
an =
where k = −1, −2, −3, . . . . Only odd powers of z “survive” (since f (z) is odd),
and the coefficients obtained are identical to those of the series (8.2.7).
Problem 2
The Laurent series for the function defined with the formula
f (z) =
z2
(z − 2)2 (z + 2)
will be found in all annuli around the point z0 = 2.
(8.2.12)
8.2 Expanding Functions into Laurent Series
283
Solution
In this problem the function f (z) will be expanded into the Laurent series around
the point z0 = 2, i.e., around its pole. According to Fig. 8.2, two annuli will occur:
the inner one with the radius of r = 4 (as this is the distance from z0 to the other
singularity at the point −2) and the outer one. In each of them, the Laurent series
will have a different form. In the inner area, we expect the presence of the principal
part of the expansion because the point z0 is not a point of holomorphicity of the
function f (z), but a pole.
In order to find the numerical coefficients of the series, let us transform the
expression for the function f (z) in a similar way as in the previous exercise.
Remember, however, that the expansion point is now z0 = 2, so that the series
will contain the powers of (z − 2)n . one has
f (z) =
=
1
(z − 2 + 2)2
z2
=
·
z−2+4
(z − 2)2 (z + 2)
(z − 2)2
(z − 2)2 + 4(z − 2) + 4
1
.
·
1 + (z − 2)/4
4(z − 2)2
(8.2.13)
In the expression 1/(1+(z−2)/4), the sum of the geometric series with the common
ratio q = −(z − 2)/4 is easily recognized:
Fig. 8.2 The ball on which the expansion (8.2.17) holds and the singularities of the function (8.2.12)
284
8 Investigating Singularities of Complex Functions
∞
!
z−2
z−2 2
1
(z − 2)n
z−2 3
= 1−
+
−
+... =
(−1)n
,
1 + (z − 2)/4
4
4
4
4n
n=0
(8.2.14)
where it must be assumed that
z − 2
(8.2.15)
4 < 1, i.e. |z − 2| < 4.
The expansion obtained in this way holds inside the annulus 0 < |z−2| < 4 marked
in the figure in gray. It has the form:
∞
n
!
1
2
n (z − 2)
(z
−
2)
+
4(z
−
2)
+
4
(−1)
4n
4(z − 2)2
f (z) =
n=0
∞
1
(z − 2)n
1
1 !
=
+
(−1)n
+
2
z−2 4
4n
(z − 2)
n=0
=
∞
n
!
1
1
3
n (z − 2)
·
+
+
(−1)
.
4 z−2
(z − 2)2
4n+2
n=0
principal part
(8.2.16)
analytic part
Note a very important fact: for negative values of n, the series terminates at n =
−2. This is due to the fact that at z0 = 2 the function has a second order pole. In
general, for a pole of the kth order at z0 , the main part of the series peters away
at (z − z0 )−k . On the other hand, an infinite number of terms in the principal part
would indicate that at z0 we are dealing with an essential singularity.
When |z − 2| > 4, another expansion has to be made. To do this, let us convert
the expression for f (z) in the following way:
f (z) =
=
1
(z − 2 + 2)2
z2
=
(z − 2)2 (z + 2)
(z − 2)2 z − 2 + 4
1
1
.
[(z − 2)2 + 4(z − 2) + 4)2 ]
1 + 4/(z − 2)
(z − 2)3
(8.2.17)
This time one has
2 3
n
∞
!
1
4
4
4
4
= 1−
+
−
+... =
(−1)n
1 + 4/(z − 2)
z−2
z−2
z−2
z−2
n=0
(8.2.18)
8.2 Expanding Functions into Laurent Series
285
and after plugging this into (8.2.17), one gets
∞
!
1
f (z) =
[(z − 2)2 + 4(z − 2) + 4]
(−1)n
3
(z − 2)
n=0
=
=
4
4
1
+
+
z−2
(z − 2)3
(z − 2)2
!
∞
(−1)n
n=0
−3
!
(z − 2)n
1
−
(−1)n n+2 .
z − 2 n=−∞
4
4
z−2
4
z−2
n
n
(8.2.19)
As can be seen, the obtained expansion has only the principal part. However, the
series does not break away and arbitrarily large negative powers of the expression
(z − 2) appear. We already know that this does not mean that we are dealing with
an essential singularity for z = 2, for which it would be the typical behavior. The
series (8.2.19) has this form only in the outer annulus, so one cannot infer as to a
singularity directly from it.
Problem 3
The Laurent series for the function defined with the formula
f (z) =
(z2
2
+ 1)2 (z + 1)
(8.2.20)
will be found in all annuli around the point z0 = i.
Solution
This time the function f (z) has three poles located at −i, i, −1. First of all, let
us consider what annuli can be expected.
√ The distance from z0 = i to the nearest
singularity located at z = −1 equals 2, so the inner annulus will have the form
(the left inequality is due to the fact that i is not a point of holomorphicity for f (z)):
0 < |z − i| <
√
2.
(8.2.21)
The second singularity is at distance of two units away, hence for the next annulus
the following conditions apply:
√
2 < |z − i| < 2.
(8.2.22)
286
8 Investigating Singularities of Complex Functions
Fig. 8.3 The annuli of holomorphicity and singularities of the function (8.2.20)
There are no more singularities, so the outer annulus
|z − i| > 2
(8.2.23)
is the last one to be dealt with. The system of obtained annuli is shown in Fig. 8.3.
In each of these areas, the Laurent series will have a different character.
Now, considering the inner area defined by the inequalities (8.2.21) and using
the decomposition into simple fractions, let us transform the function (8.2.20) as
follows:
2
2
=
(z2 + 1)2 (z + 1)
(z − i)2 (z + i)2 (z + 1)
z − 1 + 2i
−i
1
=
−
z+1
(z − i)2
(z + i)2
z − i − 1 + 3i
−i
1
=
−
z−i+1+i
(z − i)2 (z − i + 2i)2
z − i − 1 + 3i
−i
1
=
·
(z − i)2
(2i)2
(1 + (z − i)/2i)2
1
1
·
.
−
1 + i 1 + (z − i)/(1 + i)
f (z) =
(8.2.24)
8.2 Expanding Functions into Laurent Series
287
The formula for the sum of the geometric series and that given below
∞
!
1
= 1 − 2q + 3q 2 − 4q 3 + . . . =
(−1)n (n + 1)q n
2
(1 + q)
(8.2.25)
n=0
can now be used by writing
z−i 2
z−i
z−i 3
1
+
3
=
1
−
2
−
4
+ ...
2i
2i
2i
(1 + (z − i)/2i)2
∞
!
z−i n
n
=
(−1) (n + 1)
,
for |z − i| < 2,
2i
n=0
1
z−i
z−i 2
z−i 3
= 1−
+
−
+ ...
1 + (z − i)/(1 + i)
1+i
1+i
1+i
∞
!
√
z−i n
=
(−1)n
,
for |z − i| < |1 + i| = 2.
1+i
n=0
(8.2.26)
In order to be able to apply both the upper and lower formulas, one must use the
intersection of the areas in which they are justified, and hence the condition (8.2.21).
After inserting into (8.2.24) and rearranging the terms (followed by the appropriate
adjustment of the summation indices), one gets
f (z) =
i−1
1
1
1 + 2i
·
·
−
4
4
z−i
(z − i)2
principal part
∞
1 !
4(1 − i)
n n(1 + i) + 5 + 3i
(z − i)n .
+
(−1)
+
16
(2i)n
(1 + i)n
n=0
analytic part
(8.2.27)
Please note that the principal part terminates at 1/(z − i)2 . We already know that
this must have been the case, because at the point z0 = i the function has the pole
of the second order.
√
Now let us turn to the second annulus defined with the inequalities: 2 < |z −
i| < 2. Note that the first of the expansions (8.2.26) does not change, however, the
second one will now become
288
8 Investigating Singularities of Complex Functions
1
1
1
1
=
=
·
z+1
z−i+1+i
z − i 1 + (1 + i)/(z − i)
1
0
1+i 2
1
1+i 3
1+i
+
=
−
+ ...
1−
z−i
z−i
z−i
z−i
=
∞
1+i n
1 !
(−1)n
,
z−i
z−i
for |z − i| > |1 + i| =
√
2.
n=0
(8.2.28)
Plugging both expansions into the formula for the function f (z), we find:
∞
!
1
1
(1 + i)n
3+i
1 − 2i
·
·
(−1)n
−
+
n+3
2
4
4
z−i
(z − i)
(z − i)
n=0
f (z) = i
principal part
+
1
16
∞
!
(−1)n
n=0
n(1 + i) + 5 + 3i
(z − i)n .
(2i)n
(8.2.29)
analytic part
This time the principal part does not peter away, but we already know that this fact
is not related to the character of the singularity because the formula (8.2.29) does
not apply to the inner annulus.
In the outer area, for which |z| > 2, the formula (8.2.28) remains legitimate, but
the first expression (8.2.26) has to be modified into:
z − 1 + 2i
z − i − 1 + 3i
z − i − 1 + 3i
1
=
=
·
=
2
2
2
(z + i)
(z − i + 2i)
(z − i)
(1 + 2i/(z − i))2
∞
z − i − 1 + 3i !
2i n
n
=
(−1) (n + 1)
.
(8.2.30)
z−i
(z − i)2
n=0
Again (8.2.25) has been used here. By inserting both expansions into the formula
for the function f (z), one finally obtains
f (z) =
∞
!
(−1)n [(3n + 1 + i(n + 1)]
n=0
(2i)n
.
(z − i)n+4
(8.2.31)
As one can see in this annulus, the Laurent series consists exclusively of the
principal part. By writing out the above expression, a lot of calculating details have
been omitted. They are elementary, but quite lengthy and it makes no sense to write
out all of them. The reader is encouraged to make the necessary calculations on
8.3 Using the Residue Theorem to Calculate Definite Integrals
289
their own. In particular, one should remember to shift the summation limits for n, if
one wants to place the expressions containing 1/(z − i)n+4 and 1/(z − i)n+3 under
the symbol of the common sum, when (8.2.28) and (8.2.30) are inserted into the
formula for f (z).
8.3 Using the Residue Theorem to Calculate Definite
Integrals
Problem 1
Making use of the residue theorem, the integral
∞
I=
−∞
1
dx
x4 + 4
(8.3.1)
will be calculated.
Solution
In this section we are going to deal with the applications of the residue theorem
in order to calculate various definite integrals. We start with a detailed restatement
of the theorem itself. Consider the function f (z), which, in a certain domain U of
the complex plane, does not have any non-isolated singularities and is holomorphic
everywhere except for the points z1 , z2 , . . . , zn . All of them can be cut out of this
domain with small circles, centered at these points, as shown in Fig. 8.4. This is
possible because it has been assumed that the singularities are isolated.
The resulting area, denoted with the symbol U , is no longer simply connected
(it has “holes”), unless n = 0. However, the function is holomorphic on it and
one can still use Cauchy’s theorem (7.0.9). The edge of this area becomes now
quite complicated because it consists of the contour C as well as small circles
C1 , C2 , . . . , Cn . The detailed construction was certainly presented to the listeners
during a lecture of analysis.
The arrows in the figure are used to indicate the orientation of the curves chosen
in such a way that it is consistent for all fragments. The selected orientation is
positive: the interior of the area U is always passed on one’s left hand. The Cauchy
theorem will now take the form:
/
/
/
/
f (z) dz + f (z) dz + f (z) dz + . . . + f (z) dz = 0.
(8.3.2)
C
C1
C2
Cn
290
8 Investigating Singularities of Complex Functions
Fig. 8.4 An example of the
domain U for n = 3
The integral over a small circle Ck around the singularity at the point zk divided
by 2π i is called the residuum of the function f (z) (at this point) if the orientation is
positive in relation to the circle interior. Let us mark with the symbol C̆k the curve
Ck , but with the orientation opposite to that shown in the figure. Then one has
1
2π i
Res f (z) =
z=zk
/
(8.3.3)
f (z) dz.
C̆k
For example, for the function g(z) = 1/z, which has the pole at z = 0, according to
this definition one would obtain
1
Res g(z) =
z=0
2π i
/
2π
1
1
dz =
z
2π i
2π
1
1
ireiϕ dϕ =
iϕ
re
2π
0
C̆r
dϕ = 1,
(8.3.4)
0
with Cr being a circle of radius r surrounding the point z = 0. Equation (8.3.2) can
now be given the form:
/
/
f (z) dz = −
C
/
/
f (z) dz −
C1
/
f (z) dz +
=
C̆1
/
f (z) dz − . . . −
C2
/
f (z) dz + . . . +
C̆2
f (z) dz
Cn
f (z) dz
C̆n
= 2π i Res f (z) + Res f (z) + . . . + Res f (z) .
z=z1
z=z2
z=zn
(8.3.5)
This equation constitutes just the thesis of the residue theorem, to be used in a while.
8.3 Using the Residue Theorem to Calculate Definite Integrals
291
In general, as the reader surely already knows, the residuum of a function at a
certain point z0 is simply the coefficient in front of the expression 1/(z − z0 ) in
the Laurent expansion of the function around z0 , in the inner annulus. Thus, if one
knows these coefficients for each of the isolated singularities located in the area U
(i.e., inside the contour C), one immediately gets the value of the integral (8.3.5).
This constitutes the potency of the theorem and the convenience of its use.
It is still necessary to learn how to find this coefficient (which can be called a−1 )
from the respective Laurent series. Of course, they could be read off from the explicit
form of these series by expanding the function beforehand. However, this method
would be quite troublesome and tedious. Why should we need a whole series if we
are only interested in one number out of it? Fortunately, one can use a formula for
the coefficient a−1 , and it is easy to apply provided that one determines the order of
the pole in advance, which usually does not pose a problem.
Suppose one wants to find the residue of the function f (z) at the point z0 , where
it has the kth order pole. As we know, the Laurent series in the inner annulus must
then have a form:
∞
f (z) =
!
a−(k−1)
a−2
a−1
a−k
+
+ ... +
+
+
an (z − z0 )n ,
k
k−1
2
(z − z0 )
z − z0
(z − z0 )
(z − z0 )
↑
n=0
(8.3.6)
with the arrow pointing to the interesting term one wants to pick up. Now let us
multiply both sides by (z − z0 )k and get
(z − z0 )k f (z) = a−k + a−(k−1) (z − z0 ) + . . . + a−2 (z − z0 )k−2
+ a−1 (z − z0 )
↑
k−1
+
∞
!
an−k (z − z0 )n ,
(8.3.7)
n=k
while in the last term the summation index has been shifted (n + k → n). Note that
all terms, which are to the left of the marked one, constitute a polynomial of the
degree of at most k − 2, so they will disappear after k − 1-fold differentiation. In
this way, one will get rid of all the unnecessary principal part (except for the sought
after term) of the series (8.3.6). Thus one has
d k−1 (z − z0 )k f (z)
dzk−1
= (k − 1)! a−1 +
↑
∞
!
an−k n(n − 1) · . . . · (n − k + 2)(z − z0 )n−k+1 .
(8.3.8)
n=k
Because n − k + 1 > 0, all terms under the symbol of the sum have positive
powers of (z − z0 ). Therefore, if one puts z = z0 , they will amount to zeros and thus
one will get rid of the Taylorian part as well:
292
8 Investigating Singularities of Complex Functions
d k−1 (z − z0 )k f (z)
dzk−1
= (k − 1)! a−1 .
(8.3.9)
z=z0
As can be seen, thanks to these operations, it was possible to find the necessary
coefficient. We can save this procedure in the form of a formula, which will be used
repeatedly:
a−1 = Res f (z) =
z=z0
d k−1 1
(z − z0 )k f (z)
(k − 1)! dzk−1
.
(8.3.10)
z=z0
Remember, however, that before applying the above formula one has to determine
the order of the pole, i.e., the number k.
We already have all the necessary tools at our disposal, so we will now proceed
to find the integral (8.3.1). In principle, this integral could be calculated, albeit
with a lot of work, using the methods from Part I. The expression 1/(x 4 + 4) is
a rational function, and for these the primitive functions can be found by the method
of expanding into simple fractions. However, as a result, one is only interested in
the definite integral, i.e., a number, and not the whole function! The effort invested
in finding the indefinite integral is not necessary. It will be shown below that it is
much more effective to use the residue theorem. In order to do this, just like in the
exercises of the previous chapter in which Cauchy’s theorem was used, one has to
choose:
• the form of the function f (z), which will be integrated,
• the integration contour.
Since as a result of the use of the residue theorem one would like to get the value
of the integral (8.3.1), so it seems reasonable to simply choose
f (z) =
1
.
z4 + 4
(8.3.11)
We already know that there are exceptions in which the function f (z) should have
a slightly different form than the integrand expression in the real integral to be
calculated. The reader probably remembers that when calculating the Fresnel integrals instead of trigonometric functions the exponential one was chosen. Another
example of such an exception is to be found in the next chapter on the occasion of
integrating the logarithmic function. Generally, however, a good choice is simply a
form identical to the integrand expression.
As far as the contour selection is concerned, it obviously has to be closed, and
since one has to calculate the integral (8.3.1), its fragment must be the real axis or,
strictly speaking, the interval [−R, R]. At the end R will of course go to infinity.
Such a contour can be easily closed with an upper or lower semi-circle of the radius
R, but one must be sure that the integral over this added semi-circle vanishes for
R → ∞ or at least can explicitly be found. In the case of the function (8.3.11), one
8.3 Using the Residue Theorem to Calculate Definite Integrals
293
Fig. 8.5 The integration
contour for the
function (8.3.11)
can hope because of the relatively high power of z in the denominator. In a moment,
a strict estimation of this integral will be provided.
From the reasoning described above, it seems logical to choose the integration
contour shown in Fig. 8.5. The segment [−R, R] could equally well be closed with
the lower semi-circle (this contour would then have the negative orientation, as the
interior would be passed on the right hand and the additional minus sign would have
to appear on the right-hand side of the residue theorem). Such a choice is not always
available, however, as we will see in the next problem.
As shown in the figure, there are two (out of four) poles of the function f (z)
inside the selected contour: those located at the points ±1 + i. All singularities of
this function are easy to identify if its denominator is decomposed into linear factors:
f (z) =
z4
1
1
.
=
(z + 1 − i)(z + 1 + i)(z − 1 − i)(z − 1 + i)
+4
(8.3.12)
Upon multiplying them by one another, the reader can easily check that this
decomposition is correct. All singularities, that are roots of the denominator, are
poles of the first order. There are no other singularities of the function f (z) because
it is a rational function (and, therefore, the quotient of two polynomials, i.e.,
holomorphic functions).
The residue theorem (8.3.5) will take on in our case a relatively simple form:
/
f (z) dz =
C
f (z) dz +
C1
f (z) dz = 2π i
Res f (z) + Res f (z) .
z=−1+i
z=1+i
CR
(8.3.13)
First let us have a look at the left-hand side of the equation. On the contour C1 ,
which runs over the real axis, one can put z = x, obtaining
R
f (z) dz =
C1
−R
x4
1
dx.
+1
(8.3.14)
294
8 Investigating Singularities of Complex Functions
When R is let go into infinity, this integral will be equal to I , so it will constitute
our main unknown, which will be determined from the Eq. (8.3.13). The second
integral of (8.3.13) can be estimated below by selecting the parametrization on the
semi-circle using the angle ϕ (i.e., by putting z = Reiϕ ):
π
IR =
f (z) dz =
0
CR
1
iReiϕ dϕ.
R 4 e4iϕ + 1
(8.3.15)
Of course, it occurs
π
|IR | ≤
0
1
iϕ R 4 e4iϕ + 1 iRe dϕ = R
π
0
1
R 4 e4iϕ + 1 dϕ.
(8.3.16)
In turn,
1
1
1
=
R 4 e4iϕ + 1 = 4 4iϕ
8
4
4
−4iϕ
R + 2R cos(4ϕ) + 1
(R e + 1)(R e
+ 1)
≤ √
1
R8
− 2R 4
+1
=
|R 4
1
.
− 1|
(8.3.17)
Thus, we see that
π
R
|IR | ≤ 4
|R − 1|
dϕ =
0
πR
|R 4 − 1|
(8.3.18)
and, as a result, when R → ∞, one has also IR → 0. As pointed out above, this
result is owed to the high power of z in the formula for f (z).
One must now look at the right-hand side of this equation. As R grows, the only
singularities that will be found inside the contour are those shown in the figure
and located above the real axis. No new poles will appear there because the function
f (z) simply does not have any. In order to calculate the residues at the points ±1+i,
one has to use the formula (8.3.10), of course putting k = 1, as we are dealing with
the poles of the first order. So one has
Res f (z) = [(z + 1 − i)f (z)]|z=−1+i =
z=−1+i
Res f (z) = [(z − 1 − i)f (z)]|z=1+i =
z=1+i
1−i
1
=
,
2i(−2)(−2 + 2i)
16
1+i
1
=−
.
2(2 + 2i)2i
16
(8.3.19)
8.3 Using the Residue Theorem to Calculate Definite Integrals
295
Combining the whole equation in the limit R → ∞, we get the result:
∞
I=
−∞
1+i
π
1
1−i
−
= .
dx = 2π i
16
16
4
x4 + 4
(8.3.20)
Although there occurred complex numbers at intermediate stages, the final result
is real and positive. This must have been the case for the integral of the real positive
function.
Problem 2
Using the residue theorem, the integral:
∞
I=
0
cos x
dx,
x 2 + a2
(8.3.21)
where a > 0, will be calculated.
Solution
The first point we should pay attention to when analyzing the integral (8.3.21) is
that it extends only over the positive semi-axis of x. As we know, in this situation
there may appear a problem with closing the contour because next to the fragment
of the (infinite) circle one would need a half-line, which would connect this circle
with the origin of the coordinate system. Hence, if it was possible to extend the
integral (8.3.21) to the full real axis, these problems could be avoided and one might
try to use, for example, the contour shown in Fig. 8.6, which does not require any
integral to be calculated along an additional line. This possibility is indeed present,
as the function cos x/(x 2 + a 2 ) is even and the integral can, therefore, be written as:
1
I=
2
∞
−∞
cos x
dx.
x 2 + a2
(8.3.22)
Now it is time to choose the function for which the residue theorem will be
applied. The first choice that gets imposed:
f (z) =
cos z
+ a2
z2
(8.3.23)
296
8 Investigating Singularities of Complex Functions
Fig. 8.6 The integration
contour for the
function (8.3.25)
is not suitable because one cannot show that the integral over a large semi-circle
vanishes (nor does one know how to calculate it explicitly). The problem comes
from the fact that the cosine function represented by the formula:
cos z =
1 iz
1 ix −y
(e + e−iz ) =
(e e + e−ix ey )
z=x+iy 2
2
(8.3.24)
contains two exponential functions with different signs at y (see Problem 1 in
Sect. 7.4). On the contour CR , when y → ∞, the second term will explode,
and if one decides to supplement the contour with the lower semi-circle, where
y → −∞—the same happens with the first one. And the presence of an additional
factor of 1/(z2 +a 2 ), which actually converges to zero with the increasing y, will not
help in any case because the exponential behavior always dominates over the power
one. The solution proposed in the previous chapter was to choose the function f (z)
in such a form that would contain only one of the exponents in (8.3.24), for example:
f (z) =
eiz
.
z2 + a 2
(8.3.25)
If one decided to choose this second exponent, then instead of the contour CR , one
would have to complete the real axis with the lower semi-circle, which would also
lead us to the goal.
Thus we have the complete set: the function f (z), which will be integrated, and
the contour shown in Fig. 8.6. Consequently the residue theorem can be used. Note
that the only singularities of the function f (z) are the first order poles located at
the points ±ia, of which only one is placed inside the contour C. This allows us to
write:
/
f (z) dz = f (z) dz + f (z) dz = 2π i Res f (z).
(8.3.26)
z=ia
C
C1
CR
8.3 Using the Residue Theorem to Calculate Definite Integrals
297
If R → ∞, for the first integration one gets
∞
lim
R→∞
C1
f (z) dz =
−∞
cos x + i sin x
dx = 2I,
x 2 + a2
(8.3.27)
with the integral containing the sine vanishing due to the odd character of the
integrand function. For the second integration, the estimation will be done in a
similar way as in the previous problem. For this we insert z = Reiϕ and pass to
the integration over the parameter ϕ:
π
IR =
f (z) dz =
CR
0
1
iϕ
eiRe
iReiϕ dϕ
2
2iϕ
2 R
e
+
a
dz
f (z)
π
eiR cos ϕ e−R sin ϕ eiϕ
= iR
1
R 2 e2iϕ
0
+ a2
dϕ.
(8.3.28)
The modulus of the quantity IR can then be estimated from above in the following
way:
π
|IR | ≤ R
iR cos ϕ −R sin ϕ iϕ
e
e
e
0
1
R 2 e2iϕ
dϕ
+ a2 π
=R
π
e
−R sin ϕ
0
1
dϕ ≤ R
R 2 e2iϕ + a 2 0
1
dϕ
|R 2 − a 2 |
π
=
R
|R 2 − a 2 |
dϕ =
0
πR
−→ 0,
|R 2 − a 2 | R→∞
(8.3.29)
since e−R sin ϕ ≤ 1 for ϕ ∈ [0, π ], and the estimation of the fraction runs identically
as in the formula (8.3.17).
One sees, therefore, that after taking the limit R → ∞, only the quantity 2I
remains on the left-hand side of (8.3.26). Now let us calculate the right-hand side,
getting
eiz π
2π i Res f (z) = 2π i
= e−a .
z=ia
z + ia z=ia
a
(8.3.30)
298
8 Investigating Singularities of Complex Functions
In this simple way, the final result has been obtained:
2I =
π −a
e
a
⇒ I=
π −a
e ,
2a
(8.3.31)
which would be very hard to find by first calculating an indefinite integral. It would
moreover require the use of the knowledge of special functions.
Problem 3
Using the residue theorem, the integral:
∞
I =P
−∞
x
dx
(x − 1)(x 2 + 1)
(8.3.32)
will be calculated.
Solution
As we know from the second part of the book, the integral of the form
∞
−∞
x
dx
(x − 1)(x 2 + 1)
(8.3.33)
does not exist, due to the non-integrable singularity at x = 1. One can only calculate
it in the sense of the principal value and hence the symbol P in (8.3.32). The reader
has already familiarized themselves with its meaning in the previous chapter (see
formula (7.4.27)). For the time being, however, let us not worry about this issue and
choose to integrate the function of the form:
f (z) =
z
.
(z − 1)(z2 + 1)
(8.3.34)
The integration contour should consist of the real axis and the large semi-circle.
However, the presence of the singularity on this axis for z = 1 obliges us to bypass
it (for example above) over a semi-circle of the radius r, as we did in Problem 1 of
Sect. 7.4. For the residue theorem to be applied, there cannot be any singularities on
the integration contour. Therefore, the contour shown in Fig. 8.7 is chosen, and after
calculating the integral, r will be contracted to zero, and R extended to infinity. Due
to the high power of z in the denominator, we do not expect any difficulties with the
8.3 Using the Residue Theorem to Calculate Definite Integrals
299
Fig. 8.7 The integration
contour for (8.3.34)
latter limit. However, some attention should be paid to the former, because it is its
character that makes the integral exist only in the sense of the principal value.
There is only one pole (for z = i) of the function f (z) inside the selected
integration contour. Let us calculate the required residue immediately:
z
1
Res f (z) =
= − (1 + i),
z=i
(z − 1)(z + i) z=i
4
(8.3.35)
useful in a while. The residue theorem now has the form:
/
f (z) dz = f (z) dz + f (z) dz + f (z) dz +
C
C1
C2
I1
Cr
Ir
I2
π
1
= 2π i − (1 + i) = (1 − i).
4
2
f (z) dz
CR
IR
(8.3.36)
As will be seen in a moment, the integrals I1 and I2 should be kept together.
On the contours C1 and C2 , there is a natural parametrization: z = x, so it can be
written that
I1 + I2 =
f (z) dz +
C1
f (z) dz
C2
1−r
=
−R
R
x
dx +
(x − 1)(x 2 + 1)
1+r
x
dx.
(x − 1)(x 2 + 1)
(8.3.37)
Each of the above integrals has a well-defined limit when R → ∞, due to the fact
that for large values of |x| the integrand functions behave effectively as 1/x 2 . As
300
8 Investigating Singularities of Complex Functions
far as the limit r → 0+ is concerned, each of the integrals individually diverges
logarithmically. There exists neither the integral
1
−∞
x
dx,
(x − 1)(x 2 + 1)
nor
∞
1
x
dx,
(x − 1)(x 2 + 1)
and, therefore, the value of r is left temporarily finite. Note, however, that the righthand side of (8.3.36) does not depend on r, so the left-hand side cannot depend on
it either. This means that the limit r → 0+ can certainly be executed, although not
in each integral separately, but only in their sum!
Now let us turn to the integral over the small semi-circle on which the
parametrization z = 1 + reiϕ can be chosen. One gets
0
Ir =
f (z) dz =
Cr
π
1 + reiϕ
ireiϕ dϕ = −i
reiϕ [(1 + reiϕ )2 + 1]
π
0
1 + reiϕ
dϕ.
(1 + reiϕ )2 + 1
(8.3.38)
In fact, we are interested only in
π
lim Ir = −i lim
r→0+
r→0+
0
1 + reiϕ
dϕ,
(1 + reiϕ )2 + 1
(8.3.39)
but we do not know if one is allowed to enter with the limit with respect to r under
the integral. If it was possible, the integrand function would immediately become
1/2, and the integral would amount to −iπ/2. To ascertain whether these steps
are admissible, we will refer to the knowledge gained in Sect. 3.1. Considering
the formula (3.1.2), it was stated that if the integrand function is continuous as a
function of two variables on the set [a, b] × [A, B], then the integral is a continuous
function of the parameter t on the interval [A, B] and the necessary limit can be
performed. In our case, the role of t is played by r, and ϕ is the integration variable.
The integrand function is obviously continuous on the rectangle [0, π ] × [A, B],
where A and B are fixed and selected freely so that −1 < A < 0 < B < 1. The
above requirements are thus met and one can enter with the limit under the integral
in the expression (8.3.39), obtaining
lim Ir = −i
r→0+
π
.
2
(8.3.40)
8.3 Using the Residue Theorem to Calculate Definite Integrals
301
One is still to estimate the integral over the large semi-circle. As usual, we will
try to show that it aims at zero when R → ∞. As in the previous examples, we plug
in z = Reiϕ and estimate the expression in the following way:
|IR | = ≤
π
(Reiϕ
0
Reiϕ
iϕ
iRe dϕ ≤
− 1)[(Reiϕ )2 + 1]
R2
|(R − 1)(R 2 − 1)|
π
π
0
iR 2 e2iϕ
(Reiϕ − 1)(R 2 e2iϕ + 1) dϕ
π R2
−→ 0.
|R 2 − 1| |R − 1| R→∞
dϕ =
0
(8.3.41)
The subsequent obvious inequalities have been used here:
|Reiϕ − 1| ≥ |R − 1|
and
|R 2 e2iϕ + 1| ≥ |R 2 − 1|.
(8.3.42)
Upon collecting all partial results, inserting them into (8.3.36), and executing the
limit r → 0+ , one gets
lim (I1 + I2 ) − i
r→0+
π
π
= (1 − i).
2
2
(8.3.43)
The imaginary parts on both sides cancel, and as a result one gets
∞
P
−∞
x
dx
(x − 1)(x 2 + 1)
⎡ 1−r
= lim ⎣
r→0+
−∞
x
dx +
(x − 1)(x 2 + 1)
∞
1+r
⎤
π
x
dx ⎦ = .
2
(x − 1)(x 2 + 1)
(8.3.44)
Problem 4
Using the residue theorem, the integral:
2π
I=
0
will be calculated.
cos2 θ + 3 sin θ
dθ
sin θ + 2
(8.3.45)
302
8 Investigating Singularities of Complex Functions
Solution
The problem to be addressed now is different from the one previously considered,
although in order to find the integral (8.3.45) the residue theorem will still be
used. The reader probably remembers that this expression could be found in the
traditional way by finding the primitive function and then substituting the limits of
integration. This is a rather tedious procedure, but we are well acquainted with it
(see Sect. 14.4 of Part I), and as long as we avoid traps that can appear (see Sect. 1.5
of the second part), it will lead to the goal. However, one should ask a question
already posed in the previous problems: are we really obliged to look for the whole
primitive function, when actually we only need one number? Surely, one can avoid
this lengthy procedure if the residue theorem is used.
For this purpose, as we know, one has to choose a function and an integration
contour. Let us start with the latter. The limits of the integral I , which are 0 and 2π ,
clearly suggest choosing as a contour C a circle with the center at the point z = 0
and of a certain radius R. On such a (positively oriented) circle the parametrization
z = Reiθ can be used, yielding
/
2π
(. . .) dz −→
(. . .)iReiθ dθ,
(8.3.46)
0
C
but now one applies this formula from right to left.
So suppose that we decide to make such a choice for now and let us think
about the integration function. One can see that the integrand expression in (8.3.45)
contains trigonometric functions sin θ and cos θ , and if one wants to pass from
integrating with respect to θ to that over z, one must be able to express both of
these functions through z. The use of the following formulas gets imposed here:
sin θ =
1 iθ
e − e−iθ
2i
and
cos θ =
1 iθ
e + e−iθ ,
2
(8.3.47)
and
cos θ =
1
(z + z̄).
2R
(8.3.48)
i.e.
sin θ =
1
(z − z̄)
2iR
Unfortunately, this would not be a satisfactory way to proceed, and hopefully,
after reviewing Sect. 7.1, the reader immediately recognizes why. By inserting into
the integrand expression sin θ and cos θ in the form of (8.3.48), one would introduce
into it both z and z̄. For such a (non-holomorphic) function one would not be allowed
to use the residue theorem. It is worth recalling that this theorem requires from the
integrand function to have at most poles in the considered domain.
Fortunately, there is a way out of this problem. If a circle of the radius R = 1
is chosen as the integration contour, it is possible to write the expressions for the
trigonometric functions without having to introduce z̄, i.e.:
8.3 Using the Residue Theorem to Calculate Definite Integrals
1
sin θ =
2i
1
z−
z
and
1
1
cos θ =
z+
.
2
z
303
(8.3.49)
Thanks to it, one gets
/
I=
C
/ 4
(z + 1/z)2 /4 + 3(z − 1/z)/2i dz
z − 6iz3 + 2z2 + 6iz + 1
1
·
dz.
=
(z − 1/z)/2i + 2
iz
2
z2 (z2 + 4iz − 1)
C
dθ
(8.3.50)
Let us identify now the poles of the function
f (z) =
z4 − 6iz3 + 2z2 + 6iz + 1
.
z2 (z2 + 4iz − 1)
(8.3.51)
The functions in the numerator and denominator are holomorphic (they are polynomials), so we are only interested in the roots of the denominator. Upon decomposing
it onto elementary factors, one finds
z2 (z2 + 4iz − 1) = z2 [z + i(2 +
√
√
3)] [z + i(2 − 3)],
(8.3.52)
from which it is clear that the function f (z) has the second √
order pole at the point
z = 0 and the first order poles at the points z = i(−2 ± 3). Their position in
relation to the integration contour is shown in Fig. 8.8.
Only two poles are located inside the unit circle. We have, therefore,
1
0
1
I = 2π i
Res √ f (z) .
Res f (z) +
2 z=0
z=i(−2+ 3)
Fig. 8.8 The integration
contour for (8.3.51)
(8.3.53)
304
8 Investigating Singularities of Complex Functions
All that remains is to determine the values of the required residues. First one gets
d 2
d z4 − 6iz3 + 2z2 + 6iz + 1 =
= −10i,
Res f (z) =
[z f (z)]
z=0
z=0
dz
dz
z2 + 4iz − 1
z=0
(8.3.54)
and then
√
Res √ f (z) = [z + i(2 − 3)]f (z)
√
z=i(−2+ 3)
z=i(−2+ 3)
z4 − 6iz3 + 2z2 + 6iz + 1 =
√
z2 (z + i(2 + 3))
√
z=i(−2+ 3)
√
= 6i 3,
(8.3.55)
where elementary computations have been omitted.
By plugging the obtained results into (8.3.53), we finally find
2π
I=
0
√
cos2 θ + 3 sin θ
dθ = 2π(5 − 3 3).
sin θ + 2
(8.3.56)
8.4 Using Residue Theorem to Find Sums of Series
Problem 1
Using the residue theorem, the sum of the series:
S=
∞
!
1
n4
(8.4.1)
n=1
will be found.
Solution
To begin with, let us note a particular feature of the function g(z) = π cot(π z),
which will play an ancillary role in further calculations. If given the form of the
quotient of the cosine and sine functions:
g(z) =
π cos(π z)
,
sin(π z)
(8.4.2)
8.4 Using Residue Theorem to Find Sums of Series
305
it can be seen that in the numerator and denominator there are entire functions, so
the only singularities can be roots of the denominator. We are going to establish
what these points are. Thus let us demand that
sin(π z) = 0
⇒ eiπ z − e−iπ z = 0
⇒ e2iπ z = 1,
(8.4.3)
which implies that the latter exponent must be an integer multiple of 2π , i.e., z has
to be equal to an integer n. They are, naturally, isolated singularities. What is more,
it occurs that
lim (z − n)g(z) = lim cos(π z)
z→n
z→n
π(z − n)
= 1,
sin(π z)
(8.4.4)
since (w − π n)/ sin w −→ (−1)n and cos(π n) = (−1)n . Thus, we are dealing
w→π n
with first order poles, and the residues at all poles are identical and equal to 1. This
is of course nothing new to the reader, who has already studied the first exercise in
Sect. 8.1. In a moment, it will be found out where these properties will be useful for
us.
To find the sum of the series (8.4.1), the integration contour C in the form of a
rectangle shown in Fig. 8.9 will be chosen and we shall calculate the integral:
/
g(z)f (z) dz,
(8.4.5)
C
the function f (z) being adjusted to the form of the particular series, the sum of
which is to be found. Let us repeat: g(z) is the universal function for each series of
the type (8.4.1), and the form of f (z) will be established in each case. In particular
in this problem, we will later take
Fig. 8.9 The integration
contour for the
function (8.4.5)
306
8 Investigating Singularities of Complex Functions
f (z) =
1
,
z4
(8.4.6)
and for now we are going to assume only that f (z) has no singularities other than
(possibly a few) poles. Of course, |f (z)| has to decrease fast enough when z → ∞,
because otherwise the exercise would not make sense (one would be looking for the
sum of a divergent series).
According to the residue theorem, the value of the integral (8.4.5) gets expressed
by the residues of poles lying inside the contour C. These are poles originating
from the function g(z) and located, as we know, on the real axis at the points of
z = −N, −N + 1, . . . , 0, . . . , N − 1, N , as well as from the function f (z). It may
happen that a singularity of the function f (z) coincides with one of the singularities
of the function g(z) (such a situation occurs in this exercise for z = 0). Then such
poles cannot be counted twice and we agree to include them only in the list of
singularities of the function f (z). These points will be denoted with the symbols
z1 , z2 , . . . , zk . That means that one has
/
⎡
⎢
g(z)f (z) dz = 2π i ⎢
⎣
C
⎤
N
!
Res g(z)f (z) +
z=n
n=−N
n=zi , i=1,...,k
k
!
i=1
⎥
Res g(z)f (z)⎥
⎦.
z=zi
(8.4.7)
Since the residues of the function g(z) for integer values of z are equal to 1, for
the product g(z)f (z) we have
Res g(z)f (z) = lim (z − n)g(z)f (z) = f (n).
(8.4.8)
z→n
z=n
Of course, this result does not apply to the points where any merging of singularities coming from both functions occurred; these will be considered separately.
Equation (8.4.7) will now take the form:
/
C
⎡
⎢
g(z)f (z) dz = 2π i ⎢
⎣
⎤
N
!
f (n) +
n=−N
n=zi , i=1,...,k
k
!
i=1
⎥
Res g(z)f (z)⎥
⎦.
z=zi
(8.4.9)
The appearance on the right-hand side of the sum of expressions f (n) gives a
chance to find the sum of the series if we manage to execute the limit of N going to
infinity, which corresponds to moving, in Fig. 8.9, infinitely far the vertical sides of
the rectangle to the left and to the right (R will also go to infinity for convenience).
One must of course be able to calculate the left-hand side of the above equation in
this limit, and strictly speaking, it will be shown to vanish for the infinite rectangle.
We will now work on the estimation needed for this, for the chosen function (8.4.6).
Let us start by storing the function |cot (π z)| for z = x + iy in the following way:
8.4 Using Residue Theorem to Find Sums of Series
307
cos(π z) eiπ(x+iy) + e−iπ(x+iy) =
|cot (π z)| = sin(π z) eiπ(x+iy) − e−iπ(x+iy) πy
(e + e−πy ) cos(π x) − i(eπy − e−πy ) sin(π x) =
(−eπy + e−πy ) cos(π x) + i(eπy + e−πy ) sin(π x) cosh(2πy) + cos(2π x) 1/2
=
,
(8.4.10)
cosh(2πy) − cos(2π x)
some elementary transformations having been omitted.
First we address the vertical sides of the rectangle, i.e., C2 and C4 , selecting them
in such a way that they always run between consecutive integers: N − 1 and N on
the right and −N − 1 and −N on the left. Then one has the clarity as to which
singularities are inside the contour and which remain outside. However, it is still up
to us where exactly (i.e., for which value of Re z) the given part of the contour will
be located. For reasons that will become clear in a moment, let us agree that the
vertical sides always run exactly halfway between consecutive integers. Then we
have the following parametrizations:
1
+ iy, on C2 ,
2
1
z = −N − + iy, on C4 ,
2
z=N+
(8.4.11)
where −R ≤ y ≤ R. Now the integral over C2 can be examined. In this case one
has
cosh(2πy) + cos(2π(N + 1/2))
|cot (π z)| =
cosh(2πy) − cos(2π(N + 1/2))
1/2
cosh(2πy) − 1
=
cosh(2πy) + 1
1/2
≤ 1,
(8.4.12)
because the values of the hyperbolic cosine are always greater than or equal to unity,
and cos(2π N + π ) = −1. The estimate for the integral over C2 looks then as
follows:
R
1
1
1
dy.
π cot (π z) 4 dz ≤ π |cot (π z)| 4 |dz| ≤ π
z
|z |
(y 2 + (N + 1/2)2 )2
C2
C2
−R
(8.4.13)
One does not have to bother with calculating the integral with respect to y of the
fraction which has a fourth degree polynomial in the denominator, although it is
possible. For large values of N, certainly the following inequality holds:
(y 2
1
1
< 2
,
2
2
+ (N + 1/2) )
y + (N + 1/2)2
308
8 Investigating Singularities of Complex Functions
and the integral of the quadratic expression can be performed very easily. In this
way we find the estimate:
R
∞
1
1
1
<
π
π
cot
(π
z)
dz
dy
<
π
dy
z4 y 2 + (N + 1/2)2
y 2 + (N + 1/2)2
C2
−∞
−R
∞
y
π
π2
arcot
−→ 0.
=
=
(8.4.14)
N + 1/2
N + 1/2 −∞
N + 1/2 N →∞
The estimation of the integral over C4 proceeds identically.
Now let us have a look at the horizontal sides. On C1 we have the parametrization: z = x − iR. Note that in such a case, the following applies:
|cot (π z)| =
cosh(2π R) + cos(2π x)
cosh(2π R) − cos(2π x)
1/2
≤
cosh(2π R) + 1
cosh(2π R) − 1
1/2
−→ 1.
R→∞
(8.4.15)
For a sufficiently large R, therefore, the following inequality must occur:
|cot (π z)| < 2.
(8.4.16)
N has already gone into infinity, so the integral with respect to x now runs from −∞
to +∞. Using (8.4.16) and proceeding in the same way as with the formula (8.4.14),
one can write
∞
1
2
1
dx
π cot (π z) 4 dz ≤ π |cot (π z)| 4 |dz| ≤ π
z
|z |
(x 2 + R 2 )2
C1
−∞
C1
∞
< 2π
−∞
x ∞
1
2π
2π 2
arcot
−→ 0.
dx
=
=
R
R −∞
R R→∞
x 2 + R2
(8.4.17)
The same result is obtained for the integral over C3 .
Coming back now to the formula (8.4.9), it can be seen that after an infinite
stretching of the rectangle (in both directions, i.e., for N → ∞ and R → ∞)
the integral on the left-hand side will vanish and a simple formula for the sum of
the series will be obtained. In the case of our function f (z), only one term must be
excluded from this sum. The point z = 0 is at the same time the pole of the functions
f (z) and g(z). That is why we get
∞
!
1
1
π
cot
(π
z)
.
=
−Res
z=0
n4
z4
n=−∞
n=0
(8.4.18)
8.4 Using Residue Theorem to Find Sums of Series
309
The expression 1/n4 is even in n, so
∞
∞
!
!
1
1
=
2
4
4
n
n
n=−∞
n=1
n=0
and one sees that in order to find the searched sum of the series, it is enough to
calculate the residue of the function g(z)f (z) at z = 0 :
∞
!
1
1
1
= − Res π cot (π z) 4 .
2 z=0
n4
z
(8.4.19)
n=1
This is the fifth order pole (the four powers of z come from the function f (z),
and one from the function g(z)), so the following formula is necessary:
1
1 d4
1 π d4
5
Res π cot (π z) 4 =
z
π
cot
(π
z)
=
cot
(π
z)]
[z
.
z=0
z=0
4! dz4
24 dz4
z
z4 z=0
(8.4.20)
It is quite a tedious calculation because one has to compute the fourth derivative of
the function, but it is an elementary derivation. As a result, we get
1
π4
Res π cot (π z) 4 = − .
z=0
45
z
(8.4.21)
After plugging this result into (8.4.19), the final formula is found:
∞
!
1
π4
.
=
4
90
n
(8.4.22)
n=1
This result can be obtained independently by the method of expanding the function
into the so-called Fourier series as described in Problem 1 of Sect. 10.2.
Problem 2
Using the residue theorem, the sum of the series:
S=
∞
!
(−1)n
n=1
will be found.
n2
(8.4.23)
310
8 Investigating Singularities of Complex Functions
Solution
The reader has certainly noticed a significant change in the content of this exercise
compared to the previous one: this time we have to sum up the alternating series.
If the factor (−1)n had not been present, one would have followed the same
procedure as with the series (8.4.1), but this time selecting not the function 1/z4
to be integrated, but
f (z) =
1
.
z2
(8.4.24)
However, if one wishes to get the additional altering factor, one has to modify the
function g(z). As we remember, the function (8.4.2) had the first order poles on the
real axis at the points z = n, where n ∈ Z, and all of the residues were equal to 1.
Now one also needs a function that has poles for z = n, but it would be helpful if the
residues in these poles were equal to ±1 alternately. This condition will of course
be fulfilled if instead of (8.4.2) one takes
g(z) =
π
.
sin(π z)
(8.4.25)
The reader will easily agree with this by analyzing the formula (8.4.4). By selecting
again to integrate the contour shown in Fig. 8.9, one will get this time the equation:
/
⎡
⎢
g(z)f (z) dz = 2π i ⎢
⎣
C
⎤
N
!
(−1)n f (n) +
n=−N
n=zi , i=1,...,k
k
!
i=1
⎥
Res g(z)f (z)⎥
⎦.
z=zi
(8.4.26)
So we have in fact two “universal” functions g(z): one given with the formula (8.4.2)
for series with positive terms and the other defined by (8.4.25) for alternating series.
Evaluating the integrals over the sides of the rectangle when stretched to infinity
is very similar to, and even easier than, that of the previous problem, as currently
in the numerator there is no cosine function, which is unbounded for complex
arguments. For example, on the side C2 one has (see (8.4.12))
1/2
1 1
=
≤1
sin(π z) cosh(2πy) + 1
and, as a result,
(8.4.27)
8.4 Using Residue Theorem to Find Sums of Series
311
R
∞
1
π
1
1
dz < π
dy < π
dy
2
2
2
2
sin(π z) z
y + (N + 1/2)
y + (N + 1/2)2
C2
−∞
−R
∞
π
π2
y
=
=
(8.4.28)
arctan
−→ 0.
N + 1/2
N + 1/2 −∞
N + 1/2 N →∞
The identical estimation runs for the side C4 .
In turn for C1 , we find (see (8.4.15))
1/2 1/2
1 1
1
=
≤
−→ 0.
sin(π z) R→∞
cosh(2π R) − cos(2π x)
cosh(2π R) − 1
(8.4.29)
Thus, for the appropriately large values of R, the following inequality must hold:
1 sin(π z) < 1.
(8.4.30)
Consequently, one obtains
∞
1
π
1
dz ≤ π
dx
2
2
sin(π z) z
x + R2
C1
−∞
=
x ∞
π
π2
arctan −→ 0
=
R
R −∞
R R→∞
(8.4.31)
and similarly for C3 .
Hence, the left-hand side of the Eq. (8.4.26) vanishes for the infinite rectangle
and one gets
∞
!
π
1
(−1)n
,
=
−Res
z=0 sin(π z) z2
n2
n=−∞
(8.4.32)
n=0
the fact that the only singularity common to the functions f (z) and g(z) is the point
z = 0 having been taken into account. The expression (−1)n /n2 does not change
with the replacement n → −n, so we have
∞
∞
!
!
(−1)n
(−1)n
=
2
.
n2
n2
n=−∞
n=0
n=1
(8.4.33)
312
8 Investigating Singularities of Complex Functions
It remains to calculate the residue of the function g(z)f (z) for z = 0:
∞
!
(−1)n
n=1
n2
1
π
1
.
= − Res
2 z=0 sin(π z) z2
(8.4.34)
We are dealing with a third order pole, so the following formula is to be used:
π
1 d2
1
1
1 π
π d2
3
=
z
z
Res
=
,
z=0 sin(π z) z2
2! dz2
sin(π z) z2 z=0
2 dz2
sin(π z) z=0
(8.4.35)
from which it stems after simple calculations that
π
π2
1
=
.
Res
z=0 sin(π z) z2
6
(8.4.36)
Upon plugging this result into (8.4.34), one obtains
∞
!
(−1)n
n=1
n2
=−
π2
.
12
(8.4.37)
In Problem 1 of Sect. 10.2 we will try to get the same result by expanding the
function into the Fourier series.
8.5 Exercises for Independent Work
Exercise 1 Determine the nature of all singularities of the functions:
sin z
,
ez − 1
(b) f (z) = log(cos
z),
3 2
(c) f (z) = z + z − 2.
(a) f (z) =
Answers
(a) Removable singularity for z = 0, first order poles at the points z = 2π in,
where n = ±1, ±2, . . . .
(b) Branch points for z = π/2 + nπ , where n ∈ Z.
(c) Branch points for z = 1 and for z = −2.
8.5 Exercises for Independent Work
313
Exercise 2 Obtain again the result (7.3.11), using the residue theorem.
Exercise 3 Obtain formulas for the coefficients an in the Laurent series for the
functions (8.2.1), (8.2.12), and (8.2.20), using the formula (8.2.9) and the residue
theorem.
Exercise 4 Find the expansion into the Laurent series of the functions:
1
around z0 = 0,
z2 − 3z + 2
1
around z0 = 1.
(b) f (z) = 3
z − 3z + 2
(a) f (z) =
Answers
(a)
∞
−∞
∞
!
!
!
(1 − 2−n−1 )zn for |z| < 1, −
zn −
2−n−1 zn for 1 < |z| < 2,
n=0
1+
(b)
∞
!
n=−1
−∞
!
n=0
(1 + 2−n−1 )zn for |z| > 2.
n=−1
(−1)n 3−n−3 (z − 1)n for |z − 1| < 3,
n=−2
−∞
!
(−1)n 3−n (z − 1)n−3 for
n=0
|z − 1| > 3.
Exercise 5 Using the residue theorem find the integrals:
∞
(a) I =
−∞
∞
(b) I =
x2 + x − 2
dx,
(x 2 − 2x + 2)2
(x 2
0
2π
(c) I =
0
2π
(d) I =
0
x sin x
dx,
+ 4)(x 2 + 1)
cos x sin x + 1
dx,
sin x + 2 cos x + 3
cos2 x + 1
sin2 x − cos x + 5
dx.
314
8 Investigating Singularities of Complex Functions
Answers
(a)
(b)
(c)
(d)
= π/2.
= (e − 1)π/6e2 .
= 27π/25. √
√
= π(−2 + 2 + 2/ 3).
I
I
I
I
Exercise 6 Using the residue theorem find the sums of the series:
(a )
∞
!
1
,
n2
n=1
(b)
∞
!
n=1
(c)
1
,
(4n2 − 1)2
∞
!
(−1)n
n=1
n4
.
Answers
(a) π 2 /6.
(b) π 2 /16 − 1/2.
(c) −7π 4 /720.
Chapter 9
Dealing with Multivalued Functions
The present chapter is devoted to the so-called multivalued functions, their analytic
continuations, and applications of the residue theorem to integrals involving such
functions.
Suppose that z0 ∈ C is a singular point of a function f (z). As we know from
the previous chapter, if a certain ball K(z0 , ) can be found for some > 0, such
that the function f (z) is holomorphic on K(z0 , ) \ {z0 }, then the singularity is
called “isolated.” Otherwise, if no such ball exists, the singularity is non-isolated or
a branch point. Functions possessing branch points are multivalued in the sense
that the value at a given point of the complex plane depends not only on the point
itself, but also on the path along which the
√ point in question has been achieved.
Typical functions with branch points are z − z0 or log(z − z0 ), which will be
explained in detail when solving specific problems.
The multivalued functions can be analytically continued beyond their initial
analyticity domains in the following way. Assume that there are two complex
functions, f (z) and g(z), analytic on the domains D1 and D2 , respectively, and
also assume that D = D1 ∩ D2 is an open, connected set such that
∀z∈D f (z) = g(z).
(9.0.1)
Then one says that the function g(z) is the analytical continuation of f (z) onto the
set D2 , and vice versa: f (z) is the analytical continuation of g(z) onto the set D1 .
© Springer Nature Switzerland AG 2020
T. Radożycki, Solving Problems in Mathematical Analysis, Part III, Problem Books
in Mathematics, https://doi.org/10.1007/978-3-030-38596-5_9
315
316
9 Dealing with Multivalued Functions
9.1 Analytically Continuing Functions
Problem 1
It will be shown that the functions defined as the sums of the series:
f (z) =
∞
!
2n (z + 1)n and g(z) =
n=0
∞
!
(−1)n+1 2n zn
(9.1.1)
n=0
are analytical continuations of each other.
Solution
After having studied Sect. 7.2, we know that the domains of holomorphicity of
functions defined in the form of Taylor series are also balls of convergence for
these series. So let us assume that some function f1 (z) of the complex variable z
defined by a power series has the convergence ball K1 , and another function f2 (z)—
the convergence ball K2 . The function f1 (z) is, therefore, holomorphic on K1 , and
f2 (z) on K2 . Even if both functions could be written down in a compact way after
summing up the series, and the formulas obtained were well defined outside the
balls K1 and K2 , the domains for both functions would remain the indicated balls,
because the corresponding series converged only inside them, and the expressions
defined by them made sense only there.
If one wants to extend the domains of functions beyond the original convergence
balls, then one has to carry out a procedure called the analytical continuation
mentioned in the theoretical introduction. Let us assume now that the balls K1 and
K2 indicated above have a certain nonempty area of intersection K1 ∩ K2 . Both
functions f1 and f2 are simultaneously defined in this area. If additionally they are
equal to each other on this domain, i.e.,
f1 = f2 ,
(9.1.2)
K1 ∩K2
K1 ∩K2
we say that they are analytical continuations of each other. In this way, the function
f2 extends the function f1 onto the entire ball K2 , and the function f1 extends the
function f2 onto the entire ball K1 . One should remember that the key here is to find
the whole domain (and not only one point), where both functions are defined and
equal.
Now let us move on to the current exercise. According to the formula (7.2.2), the
radius of the convergence ball for the function f (z) amounts to:
R1 =
lim
n→∞
√
n
2n
−1
=
1
,
2
(9.1.3)
9.1 Analytically Continuing Functions
317
so the ball K1 is K(−1, 1/2). Inside it one can easily get a compact formula for the
function f (z) because the corresponding Taylor series is simply a geometric series
with the modulus of the common ratio smaller than unity:
f (z) =
∞
!
2n (z + 1)n =
n=0
−1
1
=
.
1 − 2(z + 1)
2z + 1
(9.1.4)
One has
R2 =
lim
n→∞
n
|(−1)n+1 2n |
−1
=
1
,
2
(9.1.5)
which gives K2 = K(0, 1/2). Here the summation of the series can be easily
performed as well, leading to
g(z) =
∞
!
(−1)n+1 2n zn =
n=0
−1
.
1 + 2z
(9.1.6)
As can be seen, we have obtained the identical expressions for the functions f (z)
and g(z) and one would like to write that f (z) = g(z), but. . . . the expression (9.1.4)
only applies within the ball K1 , and (9.1.6) only within the ball K2 . Unfortunately,
as shown in Fig. 9.1, they are disjoint and do not have any common area that could
be used to perform the analytical continuation.
If we were to terminate the solution here, we would have to write that we
were unable to demonstrate that f (z) and g(z) are the analytical continuations
of each other. Fortunately, one can get out of this troublesome situation by using
another function in the form of a Taylor series, the convergence ball of which
would have nonempty intersections with both K1 and K2 . To do this, let us expand
the expression −1/(1 + 2z) in a series around z = (−1 + i)/2. One cannot
Fig. 9.1 The holomorphicity
balls for the functions f (z),
g(z), and h(z)
318
9 Dealing with Multivalued Functions
select z = −1/2 because it is not the point of holomorphicity for the expression
−1/(1 + 2z). The following expansion is obtained as a result:
−
1
i
1
=−
=
1 + 2z
i + 2(z + 1/2 − i/2)
1 − 2i(z + 1/2 − i/2)
∞
n
!
1
i
=i
(2i)n z + −
.
2 2
(9.1.7)
n=0
If one now defines the function h(z) by the formula:
h(z) = i
∞
!
1
i n
(2i)n z + −
,
2 2
(9.1.8)
n=0
then first it is holomorphic on the ball K3 = K(−1/2 + i/2, 1/2), and second
it is equal to the functions f (z) and g(z) in the domains K1 ∩ K3 and K2 ∩ K3 ,
respectively.
In that way, we have made an indirect analytical continuation of the function f (z)
from the ball K1 to the function g(z) onto the ball K2 : the function h(z) continues
f (z) onto the ball K3 , and the function g(z) continues the function h(z) onto the
ball K2 .
Problem 2
The function
f (z) = z2 (z + 2)
1/4
(9.1.9)
will be analytically continued along the curve shown in Fig. 9.2, assuming that
f (2) = 2.
Fig. 9.2 The curve along
which the function (9.1.9) is
being analytically continued
9.1 Analytically Continuing Functions
319
Solution
The reader may have encountered the concept of the analytical continuation of a
function along a curve (path). The idea of this procedure is as follows. Imagine that
one has a given function f (z), whose value at a certain point z0 equals f0 , and one
wants to analytically continue it to a certain point zk . Let z0 and zk be respectively
the start and the end points of a curve C, which lies entirely in the holomorphicity
domain of the function. Applying repeatedly the procedure presented in the previous
problem, we can cover the whole curve with consecutive balls in the way shown in
Fig. 9.2, from the initial to the final point (in the figure only the first few balls have
been marked). We remember that, by definition, a “domain” is an open set. Since
the curve lies in the area of holomorphicity, the radii of these balls can be adjusted
so that none of the singularities are contained in any of them.
In each of the balls, the function has its expansion into the Taylor series, just like
in the previous exercise. If these Taylor series take the same values on the subsequent
intersections, then in the last of the balls, containing in its interior the endpoint zk ,
the value of the analytical continuation can be found.
Of course, for complicated curves, for which one would have to analyze a large
number of balls, this procedure would be quite troublesome, although feasible. It
will be carried out in detail in the next exercise. Often, however, it is possible to
facilitate the work, based on the experience we have already gained by analyzing
the behavior of various functions of the complex variable.
The function f (z) given in the content of this problem has two singularities,
both of the character of branch points. They are located at z = 0 and z = −2. When
analytically continuing the function f (z) from z0 = 2 to zk = 2 (although actually
the start and end points of the curve in this exercise are located on different “floors”),
one can deform the curve C as one wishes, provided no singularity cuts this curve
and none of them are found on the curve. In particular, instead of the contour C, that
denoted with C as shown in Fig. 9.3 can be selected.
Imagine now that we are going from the point z = 2 lying on the positive real
semi-axis (the endpoint, which will be reached after passing the whole curve C, is
called for distinguishing z = 2C ) and moving to the left. Until the first singularity,
which is located at the origin of the system, the curve lies in the holomorphicity
area of the function f (z) and inside the convergence ball of the corresponding
Taylor series (e.g., K(2, 2)). So here there is no need to perform any analytical
Fig. 9.3 The deformed
curve C
320
9 Dealing with Multivalued Functions
continuation. When approaching the point z = 0, one has to bypass it from above
(because such curve was chosen in the text of the exercise). It is achieved by moving
around along the semi-circle shown in Fig. 9.3. On this semi-circle, one surely has
z = reiϕ , where r is the radius of the circle and ϕ is altered from 0 to π . Then
r = |z|, so as a result of this circumvention, one has
z = |z|ei0 −→ z = |z| eiπ = (−z)eiπ .
(9.1.10)
=−z
We have benefited from the fact that on the negative real semi-axis |z| = −z. Thus,
after bypassing the point z = 0, one obtains
z2 (z + 2)
1/4
−→ eiπ/2 z2 (z + 2)
1/4
(9.1.11)
.
The same result would be derived by expanding the function f (z) into the Taylor
series, e.g., around the points z = 1, z = i, and z = −1.
The reader could ask here why one does not deal with the factor z + 2 at all, but
only with z2 . This is because the phase of the variable z relative to the point z = 0
varies from 0 to π and it had to be taken into account. On the other hand, the phase
relative to the point z = −2 when circumventing the system origin increases from
zero to a certain (small) value, but then decreases again to zero (after all, we do not
bypass the point z = −2 in this step).
Now we are going further to the left until the second singularity is reached, which
is circumvented over the whole circle (but now the phase measured relative to the
system origin remains unchanged). Here one has z = −2 + reiϕ where ϕ grows
from 0 to 2π . As a result
z + 2 −→ e2π i (z + 2),
which, after taking into account (9.1.11), entails
1/4
1/4
z2 (z + 2)
−→ eiπ/2 e2iπ/4 z2 (z + 2)
= eiπ z2 (z + 2)
(9.1.12)
1/4
.
(9.1.13)
Alternatively, this result could be obtained by expanding the function f (z) into the
Taylor series relative to the points −1, −2 + i, −3, −2 − i, −1. How it works, we
will see in detail in the next exercise.
It is still left to bypass from below the singularity at z = 0 and the point 2C will
be reached. This gives rise to another factor eiπ/2 . Now the value z = 2 can finally
be substituted:
f (2C ) = eiπ eiπ/2 22 (2 + 2)
1/4
= −2i.
(9.1.14)
As a result of the analytical continuation applied, the number 2 has been multiplied
by the factor −i. A similar result would be obtained for other nearby points.
9.1 Analytically Continuing Functions
321
The method of analytical continuation presented above will prove very useful in
the next section, when the integrals of the so-called multivalued functions will be
calculated.
Problem 3
Expanding the function f (z) = log z subsequently around the points: 1 = e0i ,
i = ei π/2 , −1 = ei π , −i = ei 3π/2 , and 1 = ei 2π , its analytical continuation along
the circle |z| = 1 positively oriented will be found.
Solution
We begin with recalling the form of the Taylor series for the logarithmic function:
log(1 + w) =
∞
!
n=1
(−1)n+1
wn
,
n
(9.1.15)
which is convergent for |w| < 1. It will be used to define the holomorphic function
in the first ball (i.e., K0 ) shown in Fig. 9.4, with the center point z0 = 1 and the
radius equal to 1:
log z = log[1 + (z − 1)] =
∞
!
(z − 1)n
.
(−1)n+1
n
(9.1.16)
n=1
Fig. 9.4 The method to
analytically continue the
function log z along the circle
from the point z0 = e0i to the
point z4 = e2π i
K1
z1
K0
K2 z 2
z0
z4
C
z3
K3
K4
322
9 Dealing with Multivalued Functions
The ball K0 coincides with K4 , so for purely graphic reasons they are shown in the
drawing slightly shifted in relation to each other.
In the second ball (K1 ), with the center at z1 , we derive in the same way:
log z = log[z1 + (z − z1 )] = log[z1 (1 − i(z − i))] = log z1 −
∞
!
[i(z − i)]n
n=1
.
n
(9.1.17)
Since z1 = eiπ/2 , using the formula (7.1.31), one could immediately write that
log z1 = iπ/2. However, the reasoning presented for the logarithm in the Sect. 7.1
was limited to the phases in the range ] − π, π [, so in a moment it will not suffice
for our needs. Therefore, we will proceed the other way. Let us agree that we do
not know the value of log z1 and we will attempt to determine it. However, before
going into this, let us note that z1 = i has already been put in the power part of
the formula (9.1.17), without specifying whether i = eiπ/2 , e−i3π/2 , ei5π/2 , etc.
The same will also be done in subsequent balls. This is correct because the power
series for all these values gives the same result. The distinction between these points
becomes important only when one is dealing with a so-called multivalued function,
such as logarithm or square root, that is to say functions that have singularities of
the branch-point type.
The convergence balls for the expansions (9.1.16) and (9.1.17) obviously have a
common domain for which both formulas remain valid and at the same time equal
each other. On this intersection, it must occur
log z1 =
∞
!
[i(z − i)]n
n=1
n
−
∞
!
[−(z − 1)]n
n=1
n
.
(9.1.18)
As mentioned above, the areas of convergence for these series are open balls,
but in reality they are also convergent at the boundary point 1 + i (this is one of
the intersection points of both balls), which is very easy to see by substituting this
value and using Leibniz’s test discussed in the first part of this set of problems (see
Sect. 13.2). Abel’s theorem, known from the lecture of analysis, states that in such a
situation the function defined by the power series is continuous not only inside the
ball, but also at this boundary point if it is approached from the interior. We will use
this fact below.
The expression log z1 is a constant, so one can add the symbol of the limit in
front of it:
0∞
1
∞
! [i(ζ − i)]n !
[−(ζ − 1)]n
−
log z1 = lim log z1 = lim
ζ →1+i
ζ →1+i
n
n
n=1
=
∞ !
n=1
in
(−i)
−
n
n
n=1
n
.
(9.1.19)
9.1 Analytically Continuing Functions
323
Here, the symbol “lim” stands for just such “Abel” limit. The expression i n − (−i)n
vanishes when n is an even number, while for n odd, i.e., for n = 2k + 1, one has
i 2k+1 − (−i)2k+1 = 2i 2k+1 = 2i(−1)k ,
and hence it can be written
log z1 = 2i
∞
!
π
(−1)k
π
= 2i = i .
2k + 1
4
2
k=0
(9.1.20)
=π/4
The known sum of the series 1 − 1/3 + 1/5 − 1/7 + . . . = π/4 can easily be
calculated using the methods of Chap. 15 of Part I (see also Problem 2 in Sect. 10.2
of the present part).
As a result, in K1 the following formula applies:
∞
log z = i
π ! [i(z − i)]n
−
.
2
n
(9.1.21)
n=1
Now let us focus on the ball K2 . We are going to rewrite the formula of the
function in the form:
log z = log[z2 + (z − z2 )] = log[z2 (1 − (z + 1))] = log z2 −
∞
!
(z + 1)n
n=1
,
n
(9.1.22)
and the value of log z2 will be determined. On the common part of K1 and K2 , the
formulas (9.1.21) and (9.1.22) apply. They must give the same result, hence
log z2 = i
∞
∞
n=1
n=1
π ! [i(z − i)]n ! (z + 1)n
−
+
.
2
n
n
(9.1.23)
Again, we are now using the trick of going to the limit, this time to the point of
−1 + i. We have
0 ∞
1
∞
! [i(ζ − i)]n !
(ζ + 1)n
π
log z2 = lim log z2 = i + lim
+
−
ζ →−1+i
2 ζ →−1+i
n
n
n=1
=i
π
+
2
∞ n
!
i
n=1
n
−
(−i)
n
n
=i
π
π
+ i = π i,
2
2
n=1
(9.1.24)
324
9 Dealing with Multivalued Functions
where the expression for the sum (9.1.19) has already been taken into account. As a
result, one can see that on the ball K2 the following formula holds true:
log z = π i −
∞
!
(z + 1)n
n
n=1
(9.1.25)
.
Now let us move on to the ball K3 and rewrite log z in the form:
log z = log[z3 +(z−z3 )] = log[z3 (1+i(z+i))] = log z3 +
∞
!
[i(z + i)]n
.
(−1)n+1
n
n=1
(9.1.26)
By matching the values of (9.1.25) and (9.1.26) in the same way as before, this time
for the point −1 − i, one finds
0∞
1
∞
! [−i(ζ + i)]n !
(ζ + 1)n
−
log z3 = lim log z3 = π i + lim
ζ →−1−i
ζ →−1−i
n
n
n=1
= iπ +
∞ !
n=1
in
(−i)
−
n
n
n
= iπ + i
n=1
π
3
= π i,
2
2
(9.1.27)
and, as a result, on this ball there occurs the expansion:
∞
!
[i(z + i)]n
3
log z = π i +
(−1)n+1
.
2
n
(9.1.28)
n=1
For the last of the balls it can be written:
∞
!
(z − 1)n
.
(−1)n+1
n
n=1
(9.1.29)
Comparing this to the value of (9.1.28), for z → 1 − i, one finds
log z = log[z4 + (z − z4 )] = log[z4 (1 + (z − 1))] = log z4 +
0 ∞
1
∞
! [−(ζ − 1)]n !
[−i(ζ + i)]n
3
log z4 = lim log z4 = π i − lim −
+
ζ →1−i
ζ →1−i
2
n
n
n=1
=
3
πi +
2
∞ n
!
i
n=1
n
−
(−i)
n
n
=
n=1
3
π
π i + i = 2π i,
2
2
(9.1.30)
and as a result, for K4 we get the analytical continuation we need:
log z = 2π i +
∞
!
(z − 1)n
.
(−1)n+1
n
n=1
(9.1.31)
9.1 Analytically Continuing Functions
325
With the symbols of the previous exercise and assuming that z ∈ K0 , and zC ∈
K4 , it is possible to write that
log zC = log z + 2π i.
(9.1.32)
Problem 4
The function f (z) = arctan z will be analytically continued along the path shown
in Fig. 9.5, assuming that f (0) = 0.
Solution
The function arctan z has two singularities on the complex plane. They are located
at the points z = ±i. It is easy to establish if the function is given the form:
arctan z =
i−z
1
log
.
2i
i+z
(9.1.33)
Thanks to the fact that we already know how to deal with the logarithmic function,
this formula can possibly be used to make an analytical continuation in the way
discussed in the previous problem. Along the curve, the consecutive holomorphicity
circles can be drawn, as shown in the figure, and then the joint values of the Taylor
series obtained for each circle can be agreed on their common parts. In this exercise,
however, we will proceed by other means, as we would like to familiarize the reader
with another way of the analytical continuation of a function, provided that it can be
z
Fig. 9.5 The path along
which the function arctan z is
to be analytically continued
C
i
1
-i
326
9 Dealing with Multivalued Functions
given the form of a contour integral. In the case of the function arcus tangent, one
can write
arctan zC =
C
1
dw,
2+1
w
(9.1.34)
g(w)
where we agree that the curve C extends from z = 0 to zC . With this choice, one is
guaranteed to have arctan 0 = 0.
Due to the fact that the only singularities of the integrand function are first-order
poles at the points w = ±i, the contour in Fig. 9.5, which is simultaneously the
integration contour for (9.1.34), can be deformed to make it consist of a circle
with the center at w = i, starting and ending at the system origin (this part of
the contour will be called C1 ), and of a straight segment running along the real axis
from w = 0 to w = 1 (called C2 ). Performing these integrations, one will find the
value of arctan 1C that is being looked for. Let us calculate them one by one. The
first one, i.e.,
/
/
1
1
dw,
(9.1.35)
I1 =
dw =
(w + i)(w − i)
w2 + 1
C1
C1
can be easily calculated using the residue theorem. The contour C1 runs around the
pole at the point w = i, so one has
I1 = 2π iRes g(w) = 2π i
w=i
1
= π.
2i
(9.1.36)
In turn, the integral over C2 can be found directly, upon choosing the natural
parametrization w = x and writing
1
I2 =
C2
1
dw =
w2 + 1
0
1
1
= π.
dx
=
arctan
x
2
4
x +1
0
(9.1.37)
In this latter expression, the symbol arctan occurs in the usual, real sense, known to
the reader from school. As a result, one gets
arctan 1C = π +
π
4
(9.1.38)
or, more generally,
arctan zC = arctan z +
π
.
4
(9.1.39)
9.2 Calculating Integrals Involving Functions with Branch Points
327
The reader is encouraged to obtain the same formula by applying methods of the
previous problems, with the use of the formula (9.1.33).
9.2 Calculating Integrals Involving Functions with Branch
Points
Problem 1
Using the residue theorem, the integral
∞
I=
0
x α−1
dx
x2 + 1
(9.2.1)
will be found, where 0 < α < 2.
Solution
First of all, we will agree that α is not an integer (in our case only the value of
1 would come into play). Otherwise, the function f (z) referred to below would
not have any branch points, so there would be no need to use the methods of the
present section. The integral (9.2.1) would simply become the integral of a rational
function, the calculation of which (both indefinite and definite) is not problematic
(see Sect. 14.3 of Part I). For the complex integration, we now choose the function
f (z) in the form identical to that of the integrand function:
f (z) =
zα−1
,
z2 + 1
(9.2.2)
and the integration contour is shown in Fig. 9.6. It is sometimes referred to as a
“keyhole”-type contour due to the visual similarity to a keyhole.
As can be seen, this contour consists of four fragments. That marked with the
symbol C1 , after contracting the small circle to zero (i.e., for r → 0) and expanding
the large one to infinity (i.e., for R → ∞), restores the integral I to be found. As
will be seen below, the integral over C2 will be expressed by I as well.
At this point, the reader could raise the following objection to the selected
contour: since the integration over C1 is in fact the integration over the positive real
semi-axis from 0 to ∞ and the integration over C2 inversely, both these integrals
should cancel out, and thus the unknown to be calculated would disappear entirely
from the equation. And that would be really the case, if α was an integer. Then z = 0
328
9 Dealing with Multivalued Functions
Fig. 9.6 The integration
contour for calculating the
integral (9.2.1)
would not represent a branch point for the function, and its values on C1 and on C2
would be identical. This would result in the deletion of both integrals. However, if α
is not an integer, then in order to get from C1 to C2 one has to bypass the singularity
at the origin, i.e., to analytically continue the function, as it was done in the previous
section. As a consequence, the function values on C2 will not be equal to those on
C1 and the deletion will not occur. All these steps will be discussed in detail below.
On the curve C1 we choose the parametrization z = x and it can be written
R
f (z) dz =
r
C1
x α−1
dx.
x2 + 1
(9.2.3)
For 0 < α < 2 there are no obstacles to perform the double limit, i.e., r → 0+ and
R → ∞ (see Chap. 2 of Part II), so one has
∞
f (z) dz =
lim lim
R→∞ r→0+
C1
0
x α−1
dx = I.
x2 + 1
(9.2.4)
In order to get on the curve C2 , one has to bypass the point z = 0 anticlockwise.
The direction of this circumvention has nothing to do with the contour orientation
shown in the figure. One simply has to answer the question: “How can I access C2
starting from C1 ?” The answer is clear: one can for example walk along a big circle
or along a small circle, but it will always be counterclockwise. This means that the
variable z will undergo the transformation:
z −→ ze2π i .
(9.2.5)
9.2 Calculating Integrals Involving Functions with Branch Points
329
For this reason we will choose on C2 the parametrization z = xe2π i . However, it
should be emphasized that e2π i = 1 and replacing x with xe2π i is only relevant if
the function “distinguishes” between these values. In the previous section we have
already encountered a situation where a function took different values for x and
different values for xC .
Looking at the expression (9.2.2), one can see that z = x can simply be
substituted into the denominator. The one, which genuinely takes a different value
for C1 and C2 , is the factor zα in the numerator (−1 in the exponent does not play
any role either, because it is an integer). As a result, the function value analytically
continued to C2 will be equal to
e2π iα
x α−1
x2 + 1
(9.2.6)
and one gets
0
f (z) dz = e
lim lim
R→∞ r→0+
2απ i
∞
C2
x α−1
dx = −e2απ i I.
x2 + 1
(9.2.7)
Next to the minus sign an additional phase factor has appeared, which is different
from unity, since α is not an integer. It plays a very important role because, thanks
to it, the integrals (9.2.4) and (9.2.7) will not be canceled, as was promised.
On the contour CR , the natural parametrization has a form: z = Reiϕ , with
0 < ϕ < 2π . Therefore, one has
2π
IR =
f (z) dz =
CR
0
(Reiϕ )α−1
iReiϕ dϕ = iR α
(Reiϕ )2 + 1
2π
0
eiαϕ
dϕ
R 2 e2iϕ + 1
(9.2.8)
and the following estimate can be done:
|IR | ≤ R
eiαϕ
α
R 2 e2iϕ + 1 dϕ = R
2π 2π
0
0
α
2π
≤ Rα
0
1
dϕ
R 2 e2iϕ + 1
Rα
1
dϕ = 2π −→ 0,
R 2 − 1
R 2 − 1 R→∞
(9.2.9)
since α < 2. Similarly, upon selecting z = reiϕ for the small circle, we get
|Ir | = as α > 0.
Cr
rα
−→ 0,
f (z) dz ≤ 2π 2
r − 1 r→0+
(9.2.10)
330
9 Dealing with Multivalued Functions
There are only two first-order poles inside the integration contour, at the points
z = ±i. A non-isolated singularity lying at z = 0 was left outside. This is very
important because otherwise one would not be allowed to use the residue theorem.
For the choice of the contour depicted in the figure, this theorem leads to the
equation:
/
f (z) dz =
C
f (z) dz +
C1
f (z) dz +
C2
f (z) dz +
CR
= 2π i Res f (z) + Res f (z) ,
z=i
z=−i
f (z) dz
Cr
(9.2.11)
and taking into account the results (9.2.4), (9.2.7), (9.2.9), and (9.2.10), we have
(1 − e2απ i )I = 2π i Res f (z) + Res f (z) .
z=i
z=−i
(9.2.12)
It remains to find both residues. One has to be very careful because we are dealing
with multivalued functions. As we know, for such functions different values can be
obtained if the analytical continuation to the same point is performed along different
paths. In particular, one has
Res f (z) =
z=i
zα−1 ,
z + i z=i
(9.2.13)
but the value in the numerator will depend on the curve along which the point z = i
has been reached. It is obvious, however, that it cannot lead us beyond the integration
contour! So, if we start from a point on the contour of C1 , where all phases were
assumed to be zero (z = xei0 ), then we will arrive at the point we are interested in
by bypassing the branch point at zero from above. Therefore, in (9.2.13), one has to
put i = eiπ/2 . Similarly, if the other pole is considered, one will have −i = e3iπ/2
and not −i = e−iπ/2 . In the picture it is indicated with arrows. If so, it occurs
Res f (z) =
z=i
1
(eiπ/2 )α−1
= − eiαπ/2
2i
2
(9.2.14)
1
(e3iπ/2 )α−1
= − e3iαπ/2 .
−2i
2
(9.2.15)
and similarly
Res f (z) =
z=−i
The expression (9.2.12) will now take the form:
(1 − e2απ i )I = 2π i
−1 iαπ/2
e
+ e3iαπ/2 ,
2
(9.2.16)
9.2 Calculating Integrals Involving Functions with Branch Points
331
i.e., after the multiplication by e−iαπ :
(e−iαπ − eiαπ )I = −π i(e−iαπ/2 + eiαπ/2 ).
(9.2.17)
Of course, the integral I must be real and such is precisely the value resulting
from our calculations. For we have
e−iαπ − eiαπ = −2i sin(απ )
e−iαπ/2 + eiαπ/2 = 2 cos(απ/2)
(9.2.18)
and we finally find
I =π
π
cos(απ/2)
=
.
sin(απ )
2 sin(απ/2)
(9.2.19)
Admittedly, the computations were made for a non-integer value of α, but the
result turns out to be correct also for α = 1. The formula (9.2.19) then has the value
of π/2 and the same result will be obtained by calculating the integral in a direct
way:
∞
0
∞
1
π
dx = arctan x = .
2
2
x +1
0
(9.2.20)
Problem 2
Using the residue theorem, the integral
∞
I=
(x 2
0
log x
dx
+ 1)(x + 4)
(9.2.21)
will be found.
Solution
When calculating the real integrals using the residue theorem or Cauchy’s theorem,
we were generally choosing the complex function f (z) in the form identical to the
integrand expression. The only exceptions so far were sine and cosine functions,
instead of which it was advisable to take an exponential function (see Sect. 8.3). In
332
9 Dealing with Multivalued Functions
Fig. 9.7 The integration
contour for calculating the
integral (9.2.21)
this exercise, another exception will be encountered. It will turn out that the right
choice for the calculation of the integral (9.2.21) will be
f (z) =
log2 z
.
(z2 + 1)(z + 4)
(9.2.22)
As we will see in a moment, in this way we avoid canceling the unknown I and,
thereby, preventing it from being eliminated from the equation. This can be treated
as a rule: for the integrals of the type (9.2.21) the power of the logarithm should be
raised by 1. There will be an exception to this rule, which will appear in Problem 6.
The selected function f (z) has a branch point for z = 0 and three first-order
poles located at the points z = −4, ±i. The analytical properties of this function
are, therefore, similar to that of the previous problem, and since the integral (9.2.21)
also extends along the positive real semi-axis, the “keyhole”-type contour shown in
Fig. 9.7 will be chosen.
As can be seen, all isolated singularities are located inside the contour C, so one
has
/
f (z) dz = f (z) dz + f (z) dz + f (z) dz + f (z) dz
C
C1
C2
CR
Cr
= 2π i Res f (z) + Res f (z) + Res f (z) .
z=i
z=−i
z=−4
(9.2.23)
We will calculate the components of the sum on the left-hand side, letting
simultaneously R go to infinity and r to zero. On C1 z = x so one has
R
f (z) dz =
C1
r
log2 x
dx −→
R→∞
(x 2 + 1)(x + 4)
+
r→0
∞
0
log2 x
dx.
(x 2 + 1)(x + 4)
(9.2.24)
9.2 Calculating Integrals Involving Functions with Branch Points
333
Similarly, for C2 , where z = xe2π i , one gets
r
f (z) dz =
C2
R
log2 (xe2π i )
dx =
(x 2 + 1)(x + 4)
∞
−→−
R→∞
r→0+ 0
∞
=−
0
r
R
(log x + 2π i)2
dx
(x 2 + 1)(x + 4)
log2 x + 4π i log x + 4π 2
dx
(x 2 + 1)(x + 4)
log2 x
dx − 4π iI + 4π 2
(x 2 + 1)(x + 4)
∞
(x 2
0
1
dx,
+ 1)(x + 4)
(9.2.25)
the phase e2π i having been included only where it is relevant and the relation (9.1.32) having been used.
A glance at (9.2.24) and (9.2.25) is enough to conclude that the expressions
containing log2 x under the integral will be canceled after taking the sum in (9.2.23),
while the integral containing log x (i.e., simply the unknown I ) will survive. It is
obvious that if the function f (z) was not modified by increasing the power of the
logarithm by 1, just the expressions we need would be deleted!
Now we are going to show that the integrals along the big and small circles vanish
(within the appropriate limits). For CR one has z = Reiϕ , which gives
2π
IR =
f (z) dz =
0
CR
2π
= iR
0
log2 (Reiϕ )
iReiϕ dϕ
[(Reiϕ )2 + 1][Reiϕ + 4]
(log R + iϕ)2
eiϕ dϕ,
[(Reiϕ )2 + 1][Reiϕ + 4]
(9.2.26)
and thus
2π 2π
0
0
|IR | ≤ R
(log R + iϕ)2
iϕ e
[(Reiϕ )2 + 1][Reiϕ + 4] dϕ ≤ R
R(log2 R + 4π 2 )
≤
|(R 2 − 1)(R − 4)|
2π
dϕ = 2π
0
log2 R + ϕ 2
dϕ
|(R 2 − 1)(R − 4)|
R(log2 R + 4π 2 )
−→ 0
|(R 2 − 1)(R − 4)| R→∞
due to the sufficiently high power of R in the denominator.
(9.2.27)
334
9 Dealing with Multivalued Functions
In turn on the small circle choosing the parametrization z = reiϕ , one obtains
|Ir | = Cr
r(log2 r + 4π 2 )
−→ 0,
f (z) dz ≤ 2π
|(r 2 − 1)(r − 4)| r→0+
(9.2.28)
thanks to the presence of r in the numerator.
In order to be able to use the Eq. (9.2.23), one still has to calculate the residues at
the points z = −4, ±i. From the previous exercise we already know how to obtain
the analytical continuation of the function to these points: it is not allowed to leave
the interior enclosed by the contour! That is why we will have
i = eiπ/2 ,
−i = e3iπ/2 ,
−4 = 4eiπ .
(9.2.29)
Consequently the following values of the residues are obtained
π 2 1 + 4i
log2 z
(iπ/2)2
=
·
,
=
z=i
(z + i)(z + 4) z=eiπ/2
2i(i + 4)
8
17
9π 2 1 − 4i
log2 z
(3iπ/2)2
Res f (z) =
=
·
,
=
z=−i
(z − i)(z + 4) z=e3iπ/2
−2i(−i + 4)
8
17
log2 z (log 4 + iπ )2
Res f (z) = 2
.
(9.2.30)
=
z=−4
17
z + 1 z=4eiπ
Res f (z) =
Upon inserting all partial results into the Eq. (9.2.23), one gets the relation:
∞
−4π iI + 4π 2
= 2π i
=
0
1
dx
(x 2 + 1)(x + 4)
π 2 1 + 4i
9π 2 1 − 4i
(log 4 + iπ )2
·
+
·
+
8
17
8
17
17
πi 2
8π 2
(π − log 2) +
(π + 16 log2 2),
17
34
(9.2.31)
out of which by the comparison of the real and imaginary parts on both sides the
following values of integrals can be found:
∞
0
π 2 + 16 log2 2
log x
,
dx
=
−
136
(x 2 + 1)(x + 4)
∞
(x 2
0
2
1
(π − log 2).
dx =
17
+ 1)(x + 4)
(9.2.32)
9.2 Calculating Integrals Involving Functions with Branch Points
335
The latter could be equally well obtained, using the method of simple-fractions
decomposition. Here it is got as a “by-product.”
Problem 3
Using the residue theorem, the integral
∞
I =P
0
x α−1
dx
x−1
(9.2.33)
will be found, where 0 < α < 1.
Solution
The reader, who has carefully studied the behavior of the improper integrals, will
notice at once that in the usual sense the integral (9.2.33) does not exist due to
the non-integrable singularity for x = 1. It can only be calculated in the sense of
the principal value and hence the symbol P in front of the integral (see p. 252). The
principal value will appear in the following calculations in a completely natural way
and will be a consequence of circumventing the singularity located on the positive
real semi-axis.
Let us choose the function to be integrated as follows:
f (z) =
zα−1
.
z−1
(9.2.34)
It has two singularities: the branch point at z = 0 and the pole at z = 1. However,
this pole is located on the positive real semi-axis, i.e., in the place through which
the integration in (9.2.33) is running. If we wish to apply the residue theorem or
Cauchy’s theorem, then the integration contour must be free of singularities. For
this reason, the point z = 1 will be bypassed from above along a semi-circle Ca
of the radius , as shown in Fig. 9.8. This troublesome point could be equally well
circumvented from below, but then it would appear inside the integration contour
and additional and unnecessary work would be required to calculate its residue.
After circumventing the branch point along the big circle, one will again come
across the singularity. This time it will be bypassed from below (for the same reason
as above) along the semi-circle Cb . Its radius does not have to be the same as that of
Ca , but for convenience we assume that it is also equal to . It should be clear to the
reader that encircling this point does not involve any change in the phase measured
relative to the origin of the coordinate system.
336
9 Dealing with Multivalued Functions
Fig. 9.8 The integration
contour for calculating the
integral (9.2.33)
The fragments marked as C1 and C2 can be handled together. On both of them,
the parametrization in the form of z = x is chosen, which leads to
1−
f (z) dz +
C1
f (z) dz =
r
C2
1−
−→
R→∞
r→0+ 0
x α−1
dx +
x−1
R
1+
x α−1
dx +
x−1
∞
1+
x α−1
dx
x−1
x α−1
dx.
x−1
(9.2.35)
For the time being, however, we will suspend performing the limit → 0+ .
To the pieces marked with C3 and C4 one gets by walking around the branch
point at zero, so similarly as in the first problem of this section, one has
r
f (z) dz +
C3
f (z) dz = e
C4
2π iα
1−
⎡ 1−
−→ −e2π iα ⎣
R→∞
r→0+
0
x α−1
dx + e2π iα
x−1
x α−1
x−1
∞
dx +
1+
1+
R
x α−1
dx
x−1
⎤
x α−1
x−1
dx ⎦ .
(9.2.36)
One still has to estimate the integrals over the circles and semi-circles. Only the
contributions from Ca and Cb will be dealt with in more detail, as it is very easy to
prove, in the same way as in the first exercise, that the following holds true:
lim
R→∞
CR
f (z) dz = lim
f (z) dz = 0.
r→0+
Cr
(9.2.37)
9.2 Calculating Integrals Involving Functions with Branch Points
337
On the semi-circle Ca , we have the parametrization z = 1 + eiϕ and then
0
f (z) dz =
$
%α−1
1 + eiϕ
ieiϕ dϕ = −i
eiϕ
π
Ca
π
1 + eiϕ
α−1
dϕ −→ −iπ.
→0+
0
(9.2.38)
Note that from the integration over the semi-circle around the first order pole at
z = 1, one has in the limit:
−2π i ×
1
residuum,
2
the minus originating from the negative orientation of the contour with respect to
the pole (when bypassing, one has it on the right-hand and not on the left-hand). In
turn on Cb one has to put z = e2π i (1 + eiϕ ), obtaining
π
$
f (z) dz =
Cb
e2π i (1 + eiϕ )
eiϕ
%α−1
ieiϕ dϕ
2π
2π
= −ie
1 + eiϕ
2π iα
α−1
dϕ −→ −iπ e2π iα .
→0+
(9.2.39)
π
The factors e2π i are omitted wherever they do not play role.
Looking at Fig. 9.8, it is clear that there are no singularities inside the contour.
The following equation must, therefore, be fulfilled:
f (z) dz+ f (z) dz +
C1
C2
f (z) dz +
C3
+ f (z) dz +
Ca
f (z) dz +
C4
f (z) dz +
Cr
f (z) dz
CR
f (z) dz = 0,
(9.2.40)
Cb
which is simply Cauchy’s theorem. Note that the right-hand side of this equation
depends neither on r, R nor on . Therefore, it is possible to perform the appropriate
limits on the left and the equation will remain correct. The limits r → 0+ and
R → ∞ have already been executed: as a result, the integrals (9.2.37) vanish. There
is only the limit over remaining, which must exist (since the Eq. (9.2.40) holds):
f (z) dz +
lim
→0+
C1
f (z) dz +
C2
+
f (z) dz +
C3
f (z) dz +
Ca
f (z) dz
C4
f (z) dz = 0.
Cb
(9.2.41)
338
9 Dealing with Multivalued Functions
After having used the results (9.2.35), (9.2.36), (9.2.38), and (9.2.39), we find
⎡ 1−
(1 − e2π iα ) lim ⎣
→0+
0
⎤
∞
x α−1
x α−1
dx +
dx ⎦ = iπ(1 + e2π iα ),
x−1
x−1
1+
=P
"∞ x α−1
0
x−1
(9.2.42)
dx
which implies that the principal value of the integral (9.2.33) is equal to
∞
P
0
x α−1
e−π iα + eπ iα
1 + e2π iα
=
iπ
dx = iπ
x−1
e−π iα − eπ iα
1 − e2π iα
= iπ
2 cos(π α)
= −π cot(π α).
−2i sin(π α)
(9.2.43)
As we see, none of the integrals on the left-hand side of (9.2.42) individually exist
for → 0+ , but this does not prevent their sum from having a well-defined limit.
This is precisely the sense of the principal value of an integral.
Problem 4
Using the residue theorem, the integral
∞
I=
0
xβ − 1
dx
x2 − 1
(9.2.44)
will be found, where −1 < β < 1.
Solution
Let the function f (z) be chosen in the form:
f (z) =
zβ − 1
.
z2 − 1
(9.2.45)
It has two isolated singularities at the points z = ±1 and the branch point for z = 0.
Since one of these singularities is located on the positive real semi-axis, we already
9.2 Calculating Integrals Involving Functions with Branch Points
339
Fig. 9.9 The integration
contour for calculating the
integral (9.2.44)
know that it has to be bypassed. For this purpose, let us choose the contour similar
to that already used, shown in Fig. 9.9.
After studying the previous exercise, the reader will surely agree that for the parts
C1 and C2 one has
1−
f (z) dz+ f (z) dz =
C1
r
C1
∞
−→
R→∞
r→0+ 0
→0+
xβ − 1
dx +
x2 − 1
R
1+
xβ − 1
dx
x2 − 1
xβ − 1
dx = I,
x2 − 1
(9.2.46)
where is the radius of the semi-circle Ca , but also Cb . Note that unlike the previous
problem, in (9.2.46) one can easily let go to zero, separately in each of the
integrals. This is due to the vanishing of the numerator for x = 1. This means
that the integral of I exists in the usual sense, and not only in the sense of Cauchy’s
principal value. It is related to the character of the singularity for z = 1, which in
fact is only a removable singularity.
In turn for the fragments of C3 and C4 , we get
r
f (z) dz+ f (z) dz =
C3
C4
1−
⎡
r
= e2π iβ ⎣
1−
(e2π i x)β − 1
dx +
x2 − 1
xβ − 1
dx +
x2 − 1
1+
R
1+
R
(e2π i x)β − 1
dx
x2 − 1
⎤
xβ − 1 ⎦
dx
x2 − 1
340
9 Dealing with Multivalued Functions
⎡
1+
r
+(e2π iβ − 1) ⎣
1−
1
dx +
x2 − 1
∞
−→−e
R→∞
r→0+
→0+
2π iβ
I − (e
2π iβ
− 1) P
0
R
⎤
1
dx ⎦
x2 − 1
1
dx.
x2 − 1
(9.2.47)
This time, however, the principal value of the integral appeared in the expression.
This is entirely understandable because after having continued the function f (z)
onto C3 and C4 the numerators of the integrand expression (i.e., the factors
e2π iβ x β − 1) ceased to vanish for x = 1 and the analytically continued function
f (z) does not have a removable singularity any more, but the first-order pole!
The integrals over the circles Cr and CR in the usual limits vanish:
lim
R→∞
CR
f (z) dz = lim
f (z) dz = 0,
r→0+
(9.2.48)
Cr
which will not be demonstrated any longer. For the semi-circles Ca and Cb , using
the parametrizations identical to those of the preceding exercise, we find
0
f (z) dz =
π
Ca
π
f (z) dz =
Cb
2π
(1 + eiϕ )β − 1
ieiϕ dϕ = −i
(2 + eiϕ )eiϕ
π
0
(1 + eiϕ )β − 1
dϕ −→ 0,
(2 + eiϕ )
→0+
(e2π i (1 + eiϕ ))β − 1
ieiϕ dϕ = −i
(2 + eiϕ )eiϕ
−→ −iπ
→0+
−1
e2π iβ
2
2π
π
e2π iβ (1 + eiϕ )β − 1
dϕ
(2 + eiϕ )
(9.2.49)
.
That the first integral turned out to be convergent to zero and the second one
to a constant is a further reflection of the previously noted fact that the singularity
at unity is removable, but after the analytical continuation it has the character of a
first-order pole.
As can be seen in Fig. 9.9, one pole is located (for z = −1) inside the integration
contour, so the residue theorem leads to the equation:
f (z) dz +
C1
f (z) dz +
C2
+
f (z) dz +
C3
f (z) dz +
Ca
f (z) dz +
C4
f (z) dz +
Cr
f (z) dz = 2π i Res f (z).
z=−1
Cb
f (z) dz
CR
(9.2.50)
9.2 Calculating Integrals Involving Functions with Branch Points
341
We should remember, of course, that in order to find the value of the residue one has
to insert −1 = eiπ (see p. 330):
zβ − 1 eiπβ − 1
.
(9.2.51)
Res f (z) =
=
−
z=−1
z − 1 z=eiπ
2
All the partial results have yet to be plugged into (9.2.50), naturally after the
relevant limits have been executed, and the result is the equation:
∞
(1 − e
2π iβ
)I + (1 − e
2π iβ
)P
x2
0
e2π iβ − 1
1
dx − iπ
= −π i(eiπβ − 1),
2
−1
(9.2.52)
which after some minor transformations implies
∞
I +P
0
eiπβ − 1
1
π 1 − cos(πβ)
1
+
.
dx
=
π
i
−
= ·
2 e2iπβ − 1
2
sin(πβ)
x2 − 1
(9.2.53)
In order to get the value of I from this formula, we still have to independently
determine the missing principal value on the left-hand side. This will be achieved
by referring directly to the definition:
∞
P
0
⎡ 1−
1
1
dx=lim ⎣
dx +
x 2 − 1 →0+
x2 − 1
0
∞
1+
⎡ 1−
1
1
1
=
lim ⎣
−
dx +
2 →0+
x−1 x+1
=
1
lim
2 →0+
0
0
⎤
1
dx ⎦
x2 − 1
∞
1+
⎤
1
1
−
dx ⎦
x−1 x+1
1
x − 1 ∞
2 + x − 1 1−
1
= 0.
lim log + log log =
x + 1 0
x + 1 1+
2 →0+
2−
(9.2.54)
Finally the following result is obtained:
I=
π 1 − cos(πβ)
π
πβ
·
=
tan
.
2
sin(πβ)
2
2
(9.2.55)
342
9 Dealing with Multivalued Functions
Problem 5
Using the residue theorem, the integral
∞
I=
0
log x
dx
(x − 1)(x 2 + a 2 )
(9.2.56)
will be found for a > 0.
Solution
As can be easily seen, despite the different form of the integrand function, the
current problem is very similar to the previous one. In the integration interval there
appears a root of the denominator so the integration contour has to bypass it. At the
same time, the fact that log 1 = 0 makes the integral (9.2.56) exist in the normal
sense and not only in the sense of the principal value. The integration contour will
then have the character of a modified “keyhole” and is shown in Fig. 9.10.
Regarding the choice of the function to be integrated, we must remember to raise
the power of the logarithm by 1 (see Problem 2). So we set it up as follows:
f (z) =
Fig. 9.10 The integration
contour for calculating the
integral (9.2.56)
log2 z
.
(z − 1)(z2 + a 2 )
(9.2.57)
9.2 Calculating Integrals Involving Functions with Branch Points
343
As can be seen, this function has four singularities: a branch point for z = 0 and
isolated singularities for z = 1 and z = ±ia. Inside the integration contour only the
latter ones are located, as shown in the figure.
We are now going to find the integrals over the subsequent pieces of the contour.
The contributions from Cr and CR , which converge to zero, will be omitted
straightaway, since their estimation is similar to that given in the formulas (9.2.27)
and (9.2.28):
lim
R→∞
CR
f (z) dz = lim
f (z) dz = 0.
r→0+
(9.2.58)
Cr
On C1 and C2 , one jointly has
1−
f (z) dz =
r
C1
log2 x
dx +
(x − 1)(x 2 + a 2 )
R
1+
∞
−→
R→∞
r→0+ 0
→0+
log2 x
dx
(x − 1)(x 2 + a 2 )
log2 x
dx = I.
(x − 1)(x 2 + a 2 )
(9.2.59)
As before, here too, stands for the radius of the semi-circles Ca and Cb . The
isolated singularity at z = 1 turned out to be removable, so there were no obstacles
in performing the limit: → 0+ . The situation will change after the analytical
continuation of the function onto C3 and C4 . There, instead of x, one will need to
insert xe2π i and the factor in the numerator will cease to vanish for x = 1:
2π i log(xe )
= (log x + 2π i)
= 2π i.
(9.2.60)
x=1
x=1
This will result in the first-order pole. As a consequence, the principal value of the
integral will occur in the following expression:
r
f (z) dz +
C3
⎡
f (z) dz =
C4
r
1−
log2 (xe2π i )
dx +
(x − 1)(x 2 + a 2 )
(log x + 2π i)2
dx +
(x − 1)(x 2 + a 2 )
=⎣
1−
⎡ 1−
= −⎣
r
2
1+
R
R
log2 (xe2π i )
dx
(x − 1)(x 2 + a 2 )
⎤
(log x + 2π i)2
dx ⎦
(x − 1)(x 2 + a 2 )
R
log x
dx +
(x − 1)(x 2 + a 2 )
1+
1+
⎤
2
log x
dx ⎦
(x − 1)(x 2 + a 2 )
344
9 Dealing with Multivalued Functions
⎡ 1−
R
log x
dx +
(x − 1)(x 2 + a 2 )
− 4π i ⎣
r
1+
⎡ 1−
− 4π 2 ⎣
r
∞
−→ −
R→∞
r→0+
→0+
0
R
1
dx +
(x − 1)(x 2 + a 2 )
1+
⎤
log x
dx ⎦
(x − 1)(x 2 + a 2 )
⎤
1
dx ⎦
(x − 1)(x 2 + a 2 )
log2 x
dx − 4π iI − 4π 2 P
(x − 1)(x 2 + a 2 )
∞
0
1
dx.
(x − 1)(x 2 + a 2 )
(9.2.61)
We still need to consider the semi-circles Ca and Cb . As before, we find:
0
log2 (1 + eiϕ )
ieiϕ dϕ
+ a2)
f (z) dz =
eiϕ ((1 + eiϕ )2
π
Ca
0
log2 (1 + eiϕ )
dϕ −→ 0,
(1 + eiϕ )2 + a 2
→0+
=i
π
π
f (z) dz =
2π
Cb
log2 [e2π i (1 + eiϕ )]
ieiϕ dϕ
eiϕ ((1 + eiϕ )2 + a 2 )
2π
= −i
π
−→
→0+
[log[(1 + eiϕ ) + 2π i]2
dϕ
(1 + eiϕ )2 + a 2
4iπ 3
.
1 + a2
(9.2.62)
As can be seen in the figure, inside the integration contour there are two poles at
the points z = ±ia, so from the residue theorem one gets:
f (z) dz +
C1
f (z) dz +
C2
+
f (z) dz +
C3
f (z) dz +
Ca
f (z) dz +
C4
Cr
CR
f (z) dz
f (z) dz = 2π i Res f (z) + Res f (z) . (9.2.63)
z=ia
Cb
f (z) dz +
z=−ia
9.2 Calculating Integrals Involving Functions with Branch Points
345
We already know that the points ±ia should be interpreted as
ia = eiπ/2 a,
−ia = e3iπ/2 a.
(9.2.64)
The elementary calculation gives
Res f (z) =
(iπ/2 + log a)2
,
2ia(ia − 1)
Res f (z) =
(3iπ/2 + log a)2
,
2ia(ia + 1)
z=ia
z=−ia
(9.2.65)
and after plugging all partial results into (9.2.63), the following equation is obtained:
∞
− 4π iI + 4π P
2
0
= 2π i
1
2ia
4iπ 3
1
dx +
2
2
(x − 1)(x + a )
1 + a2
(iπ/2 + log a)2
(3iπ/2 + log a)2
+
ia − 1
ia + 1
. (9.2.66)
This time one does not have to calculate on the side the principal value of the
integral, which appeared in our formulas, although one could do it quite easily (after
all, under the integral we have a rational function). For, it is enough to compare the
real and the imaginary parts on both sides of the equation in order to catch the
unknown I . After rearranging the right-hand side of (9.2.66), we finally find
∞
I=
0
∞
P
0
3π 2
a log2 a − π log a
log x
dx =
+
,
2
2
2
(x − 1)(x + a )
8(1 + a )
2a(1 + a 2 )
2a log a − π
1
dx =
.
(x − 1)(x 2 + a 2 )
2a(1 + a 2 )
(9.2.67)
At the end of this exercise, thanks to the knowledge we have gained by
solving the last few problems, we can try to simplify the integration contours of
Figs. 9.9 and 9.10, where a removable singularity appeared. Let us note that the
contribution from the upper semi-circle labeled as Ca always turned out to be zero
(see formulas (9.2.49) and (9.2.62)). Consequently, the removable singularity never
provided its contribution. Moreover, the integrals over the pieces C1 and C2 could be
combined into a single integral running from 0 to ∞, without the need to introduce
the principal value of the integral.
346
9 Dealing with Multivalued Functions
Fig. 9.11 Simplified
integration contour for
calculating the
integrals (9.2.44) and (9.2.56)
With this in mind, it is possible to simplify our procedures in such cases.
First of all, one needs to redefine the function at a singular point (thus removing
the singularity). For example, the definition of the function (9.2.45) should be
completed by the value f (1) = f (ei0 ) = β/2 and the definition (9.2.57) by
f (1) = f (ei0 ) = 0.
Second, the integration contour can be selected in a slightly simpler form, as
shown in Fig. 9.11. Thanks to these modifications, the solution of such problems
will become somewhat easier.
Problem 6
Using the residue theorem, the integral
∞√
I=
0
x log x
dx
(x + 2)2
(9.2.68)
will be found.
Solution
It has already been written many times that when looking for real integrals using
the residue theorem, or Cauchy’s theorem, the complex function f (z) is most often
chosen in the form identical to the integrand expression. However, two exceptions
9.2 Calculating Integrals Involving Functions with Branch Points
347
Fig. 9.12 The integration
contour for calculating the
integral (9.2.68)
have been recognized. The first one concerned the integration of trigonometric
functions, where instead of the sine or cosine function the exponential function must
be used. The second one was encountered on the occasion of Problems 2 and 5 of
this section, where it was found out that when integrating the logarithmic function
over a “keyhole” contour, its power had to be raised by one. Below we are going
to familiarize the reader with an “exception to the exception”: if the logarithm is
accompanied by a multivalued functions such as roots, there is no need to increase
the power of the logarithm. Therefore, the correct choice in this exercise is
√
z log z
.
(z + 2)2
f (z) =
(9.2.69)
If the reader remembers why the power of the logarithmic function was previously increased, it will also become clear why it is not needed now. The aim was
to avoid canceling the integrals along the positive real semi-axis we were interested
√
in, which run twice in different directions. Now, thanks to the presence of z, this
cancellation will not occur. After bypassing the branching point at zero, the function
will gain an additional phase factor in the form of eiπ .
Thus, the function f (z) is chosen in the form of (9.2.69), and the integration
contour as shown in Fig. 9.12. The only singularity captured inside it is the secondorder pole at the point z = −2. The residue theorem therefore demands that the
following equation is fulfilled:
/
f (z) dz =
C
f (z) dz +
C1
C2
f (z) dz +
CR
f (z) dz +
f (z) dz = 2π i Res f (z).
z=−2
Cr
(9.2.70)
348
9 Dealing with Multivalued Functions
Based on the results of previous exercises, one can immediately write
R
x log x
dx −→ I,
R→∞
(x + 2)2
+
f (z) dz =
r→0
r
C1
√
eiπ x (log x + 2π i)
dx −→ I + 2π i
R→∞
(x + 2)2
+
r
f (z) dz =
C2
√
r→0
R
∞
0
√
x
dx.
(x + 2)2
(9.2.71)
The latter result comes from the fact that the point z = 0 is the branch point for
both the root√and the logarithm. After it is circumvented, the root is continued to
the value eiπ x and the logarithm to log x + 2π i. On the other hand, the change of
the sign resulting from the reversal of the integration limits is compensated by the
factor eiπ = −1.
It is easy to see that the relations (9.2.58) hold again. It remains to find the residue
of the function at z = −2 = 2eiπ . Here the formula for the second-order pole must
be used:
√
z log z d
Res f (z) =
(z + 2)2
z=−2
dz
(z + 2)2 z=2eiπ
% i
log 2
π
d $√
+
1
. (9.2.72)
−
=
z log z √
√
z=2eiπ 2 2
dz
2
2
Upon inserting the obtained results into the Eq. (9.2.70), one gets
∞
2I + 2π i
0
√
i
x
log 2
π
+
1
−
dx
=
2π
i
√
√
2
(x + 2)2
2 2
2
=
√
2π
π 2i
log 2
+1 + √ .
2
2
(9.2.73)
Thus, by comparing the real and imaginary parts on both sides, we finally find
∞√
I=
0
∞
0
x log x
π
dx = √
(x + 2)2
2
log 2
+1 ,
2
√
x
π
dx = √ .
2
(x + 2)
2 2
(9.2.74)
9.2 Calculating Integrals Involving Functions with Branch Points
349
This second integral
√ could also be calculated in a direct way, using the substitution
in the form of t = x.
Problem 7
Using the residue theorem, the integral
2
(x + 2)1/3 (2 − x)2/3 dx
I=
(9.2.75)
−2
will be found.
Solution
We are now going to learn how to integrate over the contour shown in Fig. 9.13,
sometimes referred to as the “bone”-type contour (where this name comes from, the
reader can easily guess by looking at the figure). Guided by the integrand expression
in (9.2.75), let us choose the complex function in the form:
f (z) = (z + 2)1/3 (z − 2)2/3 .
(9.2.76)
It is clear that the horizontal segments running along the real axis marked with C1
and C2 should give us the opportunity to find the unknown I .
An attentive reader probably observed that when choosing the function f (z) the
order of terms in the second brackets was reversed: instead of the factor (2 − z)2/3
we chose (z − 2)2/3 . This was not necessary, but it facilitates the tracing of the
phases relative to both branch points, located at z = ±2. This is because we are
accustomed to measure phases with respect to the positive real semi-axis. For the
point z = 2, it is natural to assume that the positive numbers (z = x) located to the
right of 2 have phases equal to zero relative to it, i.e., (z − 2) = |z − 2|ei0 = (x − 2).
After all, we prefer to write that (z − 2) = (x − 2) takes place here rather than,
Fig. 9.13 The integration
contour for calculating the
integral (9.2.75)
350
9 Dealing with Multivalued Functions
for example, (z − 2) = (2 − x)e−iπ . Similarly, the phase relative to −2 vanishes
as well, since one has (z + 2) = |z + 2|ei0 = (x + 2). However, the reader should
be aware that (9.2.76) constitutes our arbitrary choice, although it seems reasonable
and helps avoiding accidental mistakes.
Before moving on to concrete calculations, we need to briefly stop on two
aspects. The first one concerns the contour itself. That shown in the figure represents
a closed line. This is nothing new: one always looks for such a contour because only
then the residue and Cauchy’s theorems can be used. However, the question arises
whether the contour C is really closed. Of course one can draw everything, but is
the character of f (z) consistent with our drawing? This question is really about
continuing the analytical function along our curve. So let us make sure we are doing
the right thing. Imagine that initially we stay on the contour at the point indicated
with the arrow under the symbol C1 . Then we walk around both singularities in
any direction (let us assume opposite to the arrows), so that ultimately we return
to the starting point. The function will gain the following phase factors related to
bypassing each singularity along a full circle: e2π i/3 e4π i/3 = e2π i = 1. This result
means that the function f (z) at the start and end points has the same value. One can
say that we have come back to the starting point, so the selected contour is closed.
For the sake of the subsequent problems, it is worth considering whether this will
always be the case if we are dealing with a factor of the type:
(z − z1 )α (z − z2 )β .
After the full circumvention of both singularities along the “bone,” this factor will
provide the phase in the form: e2π iα e2π iβ = e2π i(α+β) . In order for it to be equal
to unity, i.e., for the contour to be closed, the following condition would have to be
met:
α + β = n ∈ Z.
(9.2.77)
This is a criterion that is easy to apply, and thanks to which it can be easily
determined whether one has a chance to calculate a given integral by integrating
a complex function along a “bone”-type contour.
The second aspect is related to the use of the residue theorem. It requires that
only isolated singularities be found inside the contour. It is not so in our case
because there are branch points at z = ±2. However, if we recall the construction
of Riemann’s sphere (see p. 260), we can easily come to the conclusion that it is
a matter of convention: what is meant by the interior of the contour, and what is
its exterior? Just like an animal in a very large cage at a zoo may feel that it has
freedom and that the rest of the world is surrounded by the cage. Consequently, the
residue theorem will not be applied relative to the interior, but to the exterior of the
contour, naturally ensuring that there are no non-isolated singularities outside. At
the same time, one has to pay attention to the orientation of the contour because the
positive orientation with respect to the interior is negative to the exterior and it is
easy to make a sign mistake. However, as can be seen from Fig. 9.13, the contour
9.2 Calculating Integrals Involving Functions with Branch Points
351
depicted there is positively oriented towards the exterior, because it is what passed
on the left-hand when moving as indicated by the arrows.
One can now proceed with the calculation of integrals on individual pieces of the
contour. Naturally, the radii of circles (they will be denoted by a and b respectively)
will be contracted to zero at the end.
Imagine that we stay on the positive real semi-axis to the right on the point z = 2
and the phase calculated in relation to this point is equal to zero. In order to get
to the segment C1 , this point has to be circumvented from above, which gives the
following analytical continuation (of course, at the end z = x is set):
(z + 2)1/3 (z − 2)2/3 −→ (x + 2)1/3 ((2 − x)eiπ )2/3 = e2π i/3 (x + 2)1/3 (2 − x)2/3 .
(9.2.78)
Similarly, in order to get to C2 , we are wandering below:
(z + 2)1/3 (z − 2)2/3 −→ (x + 2)1/3 ((2 − x)e−iπ )2/3 = e−2π i/3 (x + 2)1/3 (2 − x)2/3 .
(9.2.79)
When continuing the function, one should not worry about the orientation of the
contour, i.e., of the arrows that have been drawn. This orientation is needed only to
determine the sign of the right-hand side in the residue theorem.
On the other hand, the reader could note at this point that one might get to
the fragment C2 from a different way. Since in the first step the point z = 2 was
circumvented from above, we could now continue and circumvent the point z = −2
by the full angle of 2π . Would such a continuation be compatible with (9.2.79)? Let
us check it:
(z + 2)1/3 (z − 2)2/3 −→ ((x + 2)e2π i )1/3 ((2 − x)eiπ )2/3
= e4π i/3 (x + 2)1/3 (2 − x)2/3 = e2π i e−2π i/3 (x + 2)1/3 (2 − x)2/3 .
(9.2.80)
Of course, the factor e2π i can be omitted and one can see that the same phase is
obtained as in (9.2.79). It is not a coincidence. This result is due to the fulfillment
of the condition (9.2.77), or in other words, that the contour is “closed.”
Taking these results into account, one can write
2−b
f (z) dz +
C1
f (z) dz = (e
2π i/3
−e
−2π i/3
)
(x + 2)1/3 (2 − x)2/3 dx
−2+a
C2
√
−→2i sin(2π/3)I = i 3I.
a→0+
b→0+
(9.2.81)
352
9 Dealing with Multivalued Functions
We are left to estimate the integrals over the small circles. On Ca , we put z =
−2 + a eiϕ , which yields
0
Ia =
f (z) dz =
(a eiϕ )1/3 ((4 − a eiϕ )eiπ )2/3 ia eiϕ dϕ.
(9.2.82)
2π
Ca
This in turn entails
2π
|Ia | ≤ a
4/3
iϕ 1/3
(e ) ((4 − a eiϕ )eiπ )2/3 eiϕ dϕ ≤ a 4/3
0
= 2π a
2π
(4 + a)2/3 dϕ
0
4/3
(4 + a)
2/3
−→ 0.
(9.2.83)
a→0+
Similarly on Cb , one has z = 2 + b eiϕ , which gives
−π
Ib =
f (z) dz =
(4 + b eiϕ )1/3 (beiϕ )2/3 ib eiϕ dϕ.
(9.2.84)
π
Cb
Finally,
π
|Ib | ≤ b
5/3
(4 + b eiϕ )1/3 (eiϕ )2/3 eiϕ dϕ ≤ b5/3
−π
= 2π b
5/3
π
(4 + b)1/3 dϕ
−π
(2 + b)
1/3
−→ 0.
(9.2.85)
b→0+
In the limit a, b → 0+ , the residue theorem (as we remember, applied relative to
the exterior) will, therefore, take the form:
!
√
Res f (z).
i 3I = 2π i
(9.2.86)
outside
Now one has to ask what singularities lie outside the contour. From the form of the
function f (z), one can see that the only point that comes into play is z = ∞. The
residue of the function at infinity (times 2π i) should be equal to the integral over a
circle of a large radius R (to be denoted by the symbol CR ), which surrounds the
infinity:
/
2π i Res f (z) =
f (z) dz,
z=∞
CR
(9.2.87)
9.2 Calculating Integrals Involving Functions with Branch Points
353
with the clockwise orientation. If the substitution of w = 1/z is made, this circle
gets transformed into a circle of radius 1/R positively oriented with respect to the
origin of the coordinate system and we get
/
2π i Res f (z) =
z=∞
−f (1/w)
−f (1/w)
.
dw
=
2π
i
Res
w=0
w2
w2
(9.2.88)
C1/R
This formula can be considered as the rule for calculating the residue at infinity in
the following examples. By inserting the function f in the form of (9.2.76), we see
that the singularity at w = 0 has the character of a third-order pole and hence
Res f (z) = Res
z=∞
w=0
−1
(1 + 2w)1/3 (1 − 2w)2/3
w3
1 d2
16
1/3
2/3 .
=− ·
[(1
+
2w)
(1
−
2w)
]
=
2
2 dw
9
w=0
(9.2.89)
Here the elementary calculation of the second derivative and after plugging w = 0
have been omitted.
Equation (9.2.86) will now take the form:
√
16
i 3I = 2π i ,
9
(9.2.90)
which entails
2
I=
−2
32π
(x + 2)1/3 (2 − x)2/3 dx = √ .
9 3
(9.2.91)
Problem 8
Using the residue theorem, the integral
1
I=
−1
will be found, where c > 0.
x2
1
1
·√
dx
2
+c
1 − x2
(9.2.92)
354
9 Dealing with Multivalued Functions
Fig. 9.14 The integration
contour for calculating the
integral (9.2.92)
Solution
Following the example of the previous problem, we choose the complex function in
the form:
f (z) =
1
1
·√
z2 + c 2
(z − 1)(z + 1)
(9.2.93)
and the “bone”-type contour as shown in Fig. 9.14.
When going on to calculate the integrals over the individual parts of the contour,
one has to make the analytical continuation from the positive real semi-axis (i.e.,
from the points x > 1), bypassing the singularity at z = 1 above (for C1 ) and below
(for C2 ):
1
1
1
1
−→ 2
·√
·
iπ
z2 + c 2
x + c2
(z − 1)(z + 1)
e (1 − x)(1 + x)
1
1
·√
,
x 2 + c2
1 − x2
1
1
1
1
→ 2
−
·√
·
2
2
2
z +c
x +c
(z − 1)(z + 1)
e−iπ (1 − x)(1 + x)
= −i
=i
x2
1
1
·√
.
2
+c
1 − x2
(9.2.94)
Thus, one has
1−b
f (z) dz +
C1
f (z) dz = (−i − i)
−1+a
C2
−→−2iI.
a→0+
b→0+
1
1
·√
dx
x 2 + c2
1 − x2
(9.2.95)
9.2 Calculating Integrals Involving Functions with Branch Points
355
The integrals over small circles are estimated as in the previous exercise. On Ca
there occurs z = −1 + a eiϕ and one gets
0
Ia =
f (z) dz =
2π
Ca
1
1
·
ia eiϕ dϕ,
iϕ
2
2
(−1 + a e ) + c
a eiϕ eiπ (2 − a eiϕ )
(9.2.96)
and, as a result,
2π |Ia | ≤ a 1/2
0
2π
≤a
1/2
0
1
(−1 + a eiϕ )2 + c2 ·
1
iϕ i
e
√
dϕ
eiϕ/2 eiπ/2 2 − a eiϕ
√
1
a
1
1
·√
·√
dϕ = 2π
−→ 0.
2
2
2
2
(1 − a) + c
(1 − a) + c
2−a
2 − a a→0+
(9.2.97)
On Cb , where z = 1 + b eiϕ , one obtains in the same way
Ib = −→ 0.
b→0+
(9.2.98)
After having executed the limits a, b → 0+ , the residue theorem (applied to the
exterior) leads to the equation:
− 2iI = 2π i
!
Res f (z).
(9.2.99)
outside
Let us now consider what singularities are located outside the contour. Certainly,
at the points z = ±ic the function has first-order poles, but z = ∞ may also come
into play. In order to find the contribution of the latter, formula (9.2.88) will be used:
0
1
−1
−f (1/w)
= Res
Res f (z) = Res
= 0.
z=∞
w=0
w=0 (c2 w 2 + 1) 1/w 2 − 1
w2
(9.2.100)
This is because one can see that on a suitably small ball around the point w = 0
(excluding the very point), the function is holomorphic (the singularities for w =
±1 are then left outside), so only an isolated singularity could come into play, and
it also occurs (see (8.1.2)):
lim
w→0 (c2 w 2
−1
= 0.
+ 1) 1/w 2 − 1
(9.2.101)
356
9 Dealing with Multivalued Functions
Fig. 9.15 The phases needed
for calculating the residue at
the point ia
In order to find the residue at the point w = ic, Fig. 9.15 will be used, in which
the phases of this point relative to both branch
points have been marked. Note that
√
an isosceles triangle arises (d1 = d2 = 1 + c2 ), so θ1 + θ2 = π . This result will
be of use to us in a moment.
As it has already been written, for z = ic we are dealing with the first-order pole
and, therefore,
1
1
Res f (z) =
·√
.
(9.2.102)
z=ic
z + ic
(z − 1)(z + 1) z=ic
As illustrated in the figure, ic − 1 = d1 eiθ1 , ic + 1 = d2 eiθ2 , so
(z − 1)(z + 1) =
z=ic
ei(θ1 +θ2 ) d1 d2 =
eiπ d1 d2 = eiπ/2 d1 d2
= i d1 d2 = i 1 + c2
(9.2.103)
and finally
Res f (z) = −
z=ic
1
1
·√
.
2c
1 + c2
(9.2.104)
For the pole at z = −ic, the same result will be obtained in the same way. By
virtue of the residue theorem, it can now be written
− 2iI = −2π i
1
2
·√
,
2c
1 + c2
(9.2.105)
which leads to the final result:
π
.
I= √
c 1 + c2
(9.2.106)
9.3 Exercises for Independent Work
357
9.3 Exercises for Independent Work
Exercise 1 Continue analytically the function log z along a clockwise oriented
circle using the method of expansion into the Taylor series described in Problem 3
of Sect. 9.1.
Exercise 2 Find f (−1C ) if f (z) = (z − 1)1/3 log z and the contour C runs from
z = 2 to z = 1, bypassing below the singularity at z = 1 and above that at z = 0.
We assume that f (2) = ln 2.
Answers
Exercise 1. log zC = log z − 2π i.
√
√
Exercise 2. f (−1C ) = ( 3 + i)π/ 3 4.
Exercise 3 Using the residue theorem, calculate the integrals:
∞
(a) I =
0
∞
(b) I =
√
x
dx,
(x 2 + 4)2
√
3
x−1
dx,
− 1)
(x 2
0
∞
(c) I =
0
√
x ln2 x
dx,
(x + 1)(x + 4)
∞
(d) I = P
0
2
(e) I =
−2
3
(f) I =
−3
ln x
dx,
−4
x2
3x + 1
dx,
(x + 2)1/3 (2 − x)2/3
√
9 − x2
dx.
x 2 − 6x + 10
358
9 Dealing with Multivalued Functions
Answers
(a)
(b)
(c)
(d)
(e)
(f)
I
I
I
I
I
I
= π/32.
√
= π/2 3.
= π(π 2 + 8 ln2 2)/3.
2 /8.
= π√
= 2 3π .
= 27π/2.
Chapter 10
Studying Fourier Series
The last chapter of the book is concerned with the Fourier series, i.e., series of the
form:
∞
f (x) =
a0 !
+
[an cos(nωx) + bn sin(nωx)],
2
(10.0.1)
n=1
where ω is a positive parameter, and an and bn bear the name of the Fourier
coefficients. The presence of 1/2 in the first term has no meaning and is simply
a result of a convention. If this sum is convergent, the so defined function f (x) is
clearly periodic, with the period equal to T = 2π/ω.
If a function f (x) is not periodic, but defined on a finite interval, say
[−T /2, T /2], it can easily be extended beyond this interval to the periodic one,
according to the rule:
f (x) = f (x0 + nT ),
where x0 ∈ [−T /2, T /2], and n ∈ Z.
(10.0.2)
Then, if its values at the boundary points of the interval [−T /2, T /2] are the
arithmetic mean of the limits, namely:
f (−T /2) = f (T /2) =
1
2
lim
x→−T /2+
f (x) +
lim
x→T /2−
f (x) ,
(10.0.3)
the function may be treated on an equal footing with the truly periodic function.
In order for a periodic function f (x) to have the expansion in the form
of (10.0.1), the following conditions, called the Dirichlet conditions, have to be
satisfied (the reader can sometimes meet another set of conditions bearing the same
name):
© Springer Nature Switzerland AG 2020
T. Radożycki, Solving Problems in Mathematical Analysis, Part III, Problem Books
in Mathematics, https://doi.org/10.1007/978-3-030-38596-5_10
359
360
10 Studying Fourier Series
1. the function has to be absolutely integrable on the interval of the length T , which
means that
T /2
|f (x)|dx < ∞;
(10.0.4)
−T /2
2. within one period, it may have at most a finite number of local extremes;
3. within one period, it may have at most a finite number of discontinuities of the
first type (i.e., for which the left and right limits exist).
These conditions are “sufficient” ones, and, therefore, they will rule on the
existence of the Fourier series only if they are met. The Fourier coefficients of the
formula (10.0.1) can be found with the use of the following integral formulas:
T /2
2
an =
T
g(ξ ) cos(nωξ )dξ, n = 0, 1, 2 . . . ,
−T /2
T /2
2
bn =
T
g(ξ ) sin(nωξ )dξ, n = 1, 2 . . . .
(10.0.5)
−T /2
Then the function defined with the sum (10.0.1) satisfies (10.0.3).
10.1 Examining Expandability of Functions into Fourier
Series
Problem 1
It will be examined whether the function
x + 1 for x ∈ [−2, 0[,
f (x) =
1 − x for x ∈ [0, 2]
(10.1.1)
satisfies the Dirichlet conditions.
Solution
As already mentioned, we deal below with the following problem: given a periodic
function, we will try to represent it in the form of (10.0.1), i.e., simply find all the
Fourier coefficients an and bn . In the first place, however, one has to check if this
10.1 Examining Expandability of Functions into Fourier Series
361
is possible at all. The Dirichlet conditions referred to in the content of this exercise
and formulated in the theoretical introduction above allow us to determine whether
the function defined on the set R has a representation in the form of a Fourier series.
The function f (x) given by the formula (10.1.1) is not periodic, but since it
is defined on a bounded, connected, and closed interval [−2, 2], it can easily be
extended to a periodic function. For this purpose, it is enough to assume that the
period is T = 4, and the value at any point x outside the interval [−2, 2] is chosen
according to the rule (10.0.2). This rule simply means that in order to find the value
of the function at a given point x ∈ R it is enough to select in the interval [−2, 2] a
point x0 , such that x − x0 is an integer multiple of the period, and to find f (x0 )
according to (10.1.1). In the remainder of this chapter, functions defined on an
interval [a, b] will be identified with periodic functions, obtained in this way. In
particular, for our procedure to be successful, one must also have f (−2) = f (2),
i.e., f (−T /2) = f (T /2), as T /2 = −T /2 + T , and, as mentioned in the theoretical
introduction, the condition (10.0.3) is to be met.
It should be added that the considered function does not always have to be defined
on a symmetric interval [−T /2, T /2]. Everything that has been written above (and
below) remains true and can be easily transcribed for the interval [a, a + T ], where
a is a constant. For example, in place of (10.0.3), we would have
1
lim f (x) + lim f (x) .
f (a) = f (a + T ) =
(10.1.2)
2 x→a +
x→a+T −
As we will discover below, the Dirichlet conditions are fulfilled by the function
defined by the formula (10.1.1). First of all, we notice that it is bounded, so without
performing any calculations it is clear that
2
|f (x)|dx < ∞.
(10.1.3)
−2
This value can, moreover, be found explicitly in a trivial way:
2
0
|f (x)|dx =
−2
−1
2
|x + 1|dx +
−2
|1 − x|dx =
1
0
(x + 1)dx
−1
2
(−x + 1)dx +
+
(−x − 1)dx +
−2
0
0
(x − 1)dx = 2.
(10.1.4)
1
Second, it is easy to observe that the only local extremes in the interval ] − 2, 2[
are located at the points of x = −1, 0, 1. The third condition is also fulfilled because
the function f (x) is continuous throughout the entire interval [−2, 2]. Thus, we
conclude that the function f (x) can be represented in the form of an appropriate
Fourier series.
362
10 Studying Fourier Series
Problem 2
It will be verified if the function
f (x) =
sin(π/x) for x ∈]0, 1],
0
for x = 0
(10.1.5)
satisfies the Dirichlet conditions.
Solution
According to what was written in the previous problem, first of all the function
f (x) defined by the formula (10.1.5) is extended periodically to the entire set of
real numbers. The period of the new function will be of course the number T = 1.
However, since the right-hand limit of the function f (x) at zero does not exist, there
is an immediate problem with satisfying the requirement (10.1.2).
Let us now move on to the Dirichlet conditions. First of all, one has to think
about the very existence of the integral:
1
π dx.
sin
x
0
Since the integrand function is continuous on ]0, 1] and | sin(π/x)| ≤ 1, the above
integral does exist and takes the value from within the interval of ]0, 1[.
As far as the second condition is concerned, it can be seen that it is not fulfilled.
From Problem 1 in Sect. 8.3 of the first part of this book series, we know that in
the interval of ]0, 1] infinitely many maxima and minima of our function appear.
The maxima are found everywhere where sin(π/x) takes the value of 1, i.e., at the
points:
π
π
= + 2π n,
x
2
i.e., x =
2
, n = 1, 2, 3, . . . .
1 + 4n
(10.1.6)
Similarly the minima are located at
x=
2
, n = 1, 2, 3, . . . .
−1 + 4n
(10.1.7)
As regards to the third of the Dirichlet conditions, the function f (x) is continuous
on ]0, 1]. However the condition (10.1.2) cannot be fulfilled since the limit at x →
0+ does not exist.
10.2 Finding Fourier Coefficients
363
10.2 Finding Fourier Coefficients
Problem 1
The expansion coefficients of the function f (x) = x 2 into the Fourier series on the
interval [−π, π ] will be found and on this basis the sums of the following series will
be calculated:
∞
!
1
,
n2
∞
!
(−1)n
n=1
n2
n=1
.
(10.2.1)
Solution
For the function from the text of the problem, it can immediately be seen that all
coefficients bn vanish. This is due to the fact that the function is even (f (ξ ) = ξ 2 ),
so after having been multiplied by sin(nωξ ) it becomes odd and the integral of
an odd function over the interval [−T /2, T /2] equals zero. On the other hand, the
coefficients an are obtained by integrating an even function, so there is no reason
why they should amount to zero. As a consequence, one can make a significant
observation that an even function gets expanded into the series of even functions
only (i.e., cosines). Needless to say, an odd function would be similarly expanded
into odd functions only (i.e., sines).
In our case, T = 2π and consequently ω = 1, so one has the following formula
π
1
an =
π
−π
π
1
f (ξ ) cos(nξ )dξ =
π
ξ 2 cos(nξ )dξ.
(10.2.2)
−π
For n = 0, in place of the cosine there appears a unity, so the integral can be
calculated mentally and the result is
a0 =
2 2
π .
3
(10.2.3)
For n = 1, 2, . . ., it is necessary to apply twice the integration by parts. This is an
elementary calculation, as a result of which we get
an =
4(−1)n
.
n2
(10.2.4)
364
10 Studying Fourier Series
y
f x
f x
f x
9
f
x
f
x
5
2
x
Fig. 10.1 The approximation of the function f (x) = x 2 on the interval [−π, π ], when taking into
account the successive terms of the Fourier series
The Fourier formula (10.0.1) to be found can now be written in the form:
∞
x2 =
! (−1)2
π2
+4
cos(nx).
3
n2
(10.2.5)
n=1
In order to see clearly that the following terms of the obtained series are more
and more accurately approximating the function f (x) in the interval [−π, π ], in the
joint Fig. 10.1 the function itself and some first, increasingly accurate expressions
have been plotted, with the following denotation:
! (−1)2
π2
and fk (x) =
+4
cos(nx), for k = 1, 2, . . . .
3
n2
n=1
(10.2.6)
There still remains the second part of the problem, in which, using the obtained
expression (10.2.5), we are to find explicitly the sums of the series (10.2.1). In order
to do this, one needs to select some specific values of the argument x in the interval
[−π, π ]. First let us put x = 0 on both sides of the Eq. (10.2.5). As a result one gets:
k
π2
f0 (x) =
3
∞
0=
! (−1)n
π2
+4
,
3
n2
n=1
(10.2.7)
10.2 Finding Fourier Coefficients
365
which implies that
∞
!
(−1)n
n=1
n2
=−
π2
.
12
(10.2.8)
Note that equally well, instead of x = 0, one might put x = π/2. Then the
following equation would be obtained:
∞
! (−1)n
π2
nπ
π2
=
+4
.
cos
4
3
2
n2
(10.2.9)
n=1
For half-values of π (i.e., for n = 2k + 1), the cosine equals zero, and for integer
values of π (i.e., for n = 2k), the cosine amounts alternately to ±1 and hence
∞
−
∞
! (−1)2k
! (−1)k
π2
cos(kπ
)
=
.
=4
12
(2k)2 k2
k=1
=(−1)k
(10.2.10)
k=1
As can be seen, the second of the series (10.2.1) has the value of −π 2 /12. In order
to find the sum of the first series in (10.2.5), one has to plug in x = π . The following
relation is then obtained:
∞
π2 =
∞
! (−1)n
! 1
π2
π2
+4
+4
cos(nπ ) =
,
2
3
3
n
n2
n=1
(10.2.11)
n=1
and hence
∞
!
1
π2
.
=
2
6
n
(10.2.12)
n=1
Problem 2
The expansion coefficients of the function f (x) = (π − x)/2 into the Fourier series
on the interval [−π, π ] will be found and on this basis the value of the sum of the
following series:
∞
!
(−1)n
2n + 1
n=0
will be calculated.
(10.2.13)
366
10 Studying Fourier Series
Solution
At the beginning, according to the condition (10.0.3), the function f (x) will be
supplemented by the values at the ends of the interval: f (−π ) = f (π ) = π/2.
Now the formulas (10.0.5) can be used, with T = 2π . At first glance, the function
to be expanded does not have a definite parity, so the expansion should be a mixed
one, i.e., it ought to contain both cosines and sines. However, as one can easily find
out, it occurs that
π
1
a0 =
π
−π
π
1
an =
π
−π
1
π −ξ
dξ =
2
π
πξ
ξ2
−
2
4
π −ξ
1
cos(nξ )dξ =
2
π
π
1
+
2n
sin(nξ )dξ
−π
π
=π
−π
π
π −ξ 1
· sin(nξ )
2
n
−π
π
1
= 0 − 2 cos(nξ ) = 0, for n = 1, 2, . . . .
2n
−π
(10.2.14)
Apart from the first one (or actually the zeroth), all coefficients at the cosines
turned out to vanish. Looking more closely at the form of the function, it is easy to
find the answer why this is the case. After subtracting from the expression for f (x)
the constant π/2 (i.e., de facto a0 /2), the function becomes odd. No wonder that
from among all of the coefficients an only a0 survives.
For the coefficients bn we find
π
1
bn =
π
−π
1
π −ξ
sin(nξ )dξ =
2
π
π
1
−
2n
cos(nξ )dξ
−π
π
π −ξ 1
−
· cos(nξ )
2
n
−π
π
1
(−1)n
(−1)n
− 2 sin(nξ ) =
,
=
n
n
2n
−π
(10.2.15)
and as a result the following formula is obtained:
∞
π −x
sin(nx)
π !
(−1)n
= +
.
2
2
n
n=1
(10.2.16)
10.2 Finding Fourier Coefficients
367
Fig. 10.2 The approximation of the function f (x) = (π − x)/2 on the interval [−π, π ] when
taking into account the subsequent terms of the Fourier series
In Fig. 10.2, the first few functions fk (x) are presented, which correspond to the
sums of terms on the right-hand side up to n = k. It is obvious that they are getting
increasingly closer to the function f (x).
In order to find the sum of the series (10.2.13), all one has to do now is put
x = π/2 on both sides. Since the sine of an integer multiple of π vanishes, and for
the half-values one obtains alternately ±1, this substitution will lead to the relation:
∞
∞
n=1
k=0
π !
π !
π
sin(nπ/2)
(−1)k
= +
= +
.
(−1)n
(−1)2k+1
2k + 1
4
2
n
2
(10.2.17)
=−1
After the simple transformation, we finally find
∞
!
1 1 1
π
(−1)k
= 1 − + − + ... = .
2k + 1
3 5 7
4
(10.2.18)
k=0
Problem 3
The expansion coefficients for the function f (x) = eαx into the Fourier series on
the interval [−π, π ] will be found, where α ∈ R, and then the value of the sum:
∞
!
(−1)n
α 2 + n2
n=1
will be calculated.
(10.2.19)
368
10 Studying Fourier Series
Solution
According to (10.0.3), the values of the function at the ends of the interval are fixed:
f (−π ) = f (π ) =
%
1 $ −aπ
e
+ eaπ = cosh(aπ ).
2
(10.2.20)
To find the Fourier coefficients, as always, the formulas (10.0.5) will be used. The
coefficients an are found by calculating the integrals:
π
1
a0 =
π
eaξ dξ =
−π
%
2
1 $ aπ
e − e−aπ =
sinh(aπ ),
aπ
aπ
π
1
an =
π
eaξ cos(nξ )dξ =
−π
2a
1
·
(−1)n sinh(aπ ), for n = 1, 2, . . . ,
π a 2 + n2
(10.2.21)
the latter ones having been obtained by integrating by parts twice as described in
Problem 3 of Sect. 14.1 of Part I. The reader is surely familiar with this procedure,
so the calculational details are omitted here.
In the same way, the coefficients bn are found as
π
1
bn =
π
eaξ sin(nξ )dξ =
−π
a2
2n
(−1)n−1 sinh(aπ ).
+ n2
(10.2.22)
Consequently, the Fourier formula is obtained in the form:
e
ax
2 sinh(aπ )
=
π
0
1
∞
!
1
n a cos(nx) − n sin(nx)
+
(−1)
.
2a
a 2 + n2
(10.2.23)
n=1
In the following graphs, in Fig. 10.3, some initial functions fk (x) (for k =
0, 1, 2, 3, 4, 5) are depicted, defined by the formulas:
f0 (x) =
sinh aπ
,
aπ
2 sinh(aπ )
fk (x) =
π
0
1
k
!
1
a cos(nx) − n sin(nx)
+
,
2a
a 2 + n2
(10.2.24)
n=1
including f (x) (for a particular value of a = 1/2). In order to better illustrate the
progress of all functions, their domains have been extended periodically beyond the
10.2 Finding Fourier Coefficients
369
Fig. 10.3 The graph of the function ex/2 , on which the graphs of f0 (x), f1 (x), f2 (x), f3 (x),
f4 (x), and f5 (x) have been placed and are plotted consecutively. The fragments of the oscillating
curve, which protrude downwards, are typical for the behavior of a function being the sum of the
finite number of terms of the Fourier series near the points of the discontinuity of the original curve.
Their appearance is called the Gibbs’ effect
interval [−π, π ]. The function f (x) is often given a “saw-shaped” character, by
connecting the ends of the charts at the “gluing” points.
We still have to find the sum of the series (10.2.19). It will be obtained by
plugging x = 0 into (10.2.23). Of course, then all cosines will become unities and
sines zeros, so the following equation is immediately obtained:
2 sinh(aπ )
1=
π
0
1
∞
!
1
a(−1)n )
+
.
2a
a 2 + n2
n=1
(10.2.25)
370
10 Studying Fourier Series
After some elementary transformations, one obtains the required formula
∞
!
(−1)n
π
1
=
− 2.
2
2
2a sinh(aπ ) 2a
a +n
(10.2.26)
n=1
In turn, putting a = 0 (or rather executing the appropriate limit), the reader can
easily obtain once again the result which has already been found:
∞
!
(−1)n
n=1
n2
=−
π2
.
12
(10.2.27)
Problem 4
The function f (x) = sin x will be expanded on the interval [0, π ] into the series of
cosines.
Solution
When reading the text of this problem, the reader could object immediately. After
all, we have written earlier that odd functions (which is, after all, the sine) can only
be expanded into odd functions (i.e., the sines). So, how can one expand the function
f (x) = sin x into a series of cosines? In order to understand that this is possible, it
is enough to realize that the interval [0, π ] is not symmetric in relation to the point
x = 0, so one cannot talk about the parity of the function at all.
It is natural to imagine that the function f (x) originally defined for x ∈ [0, π ] is
extended to the entire interval [−π, π ] and then it actually becomes an odd function.
As such it would be expanded into sines only. But one could do otherwise: extend
the function f (x) to an even function by taking f (x) = | sin x|. Such a function will
only be expanded into cosines, which is exactly what we are concerned about in this
exercise. Note that in both cases the form of f (x) does not change in the interval
[0, π ], and only some other extension to the interval [−π, 0[ is chosen. However,
this does not affect the values we are interested in.
Of course, there is no obligation to extend the function to an odd or even one.
The function f (x) can be extended in some other way and then it will be expanded
into a mixed series of both sines and cosines. This means that on the interval of
[0, π ] we have different Fourier series at our disposal. This is a consequence of the
fact, probably known to the reader from the lecture, that the functions sin(nωx) and
cos(nωx) do not constitute an orthogonal system in the interval of [0, T /2]. They
would constitute a new orthogonal system, however, if one puts ω → 2ω, i.e., one
would treat T /2 as a new period in place of T .
10.2 Finding Fourier Coefficients
371
As mentioned above, we choose the function f (x) = | sin x| and expand it on
the interval [−π, π ], although at the end we are only interested in [0, π ]. Naturally,
then, T = 2π and ω = 1. The coefficients bn do not need to be calculated: they
are all equal to zero, thanks to the parity of the function f (x), i.e., the oddity of the
function f (x) sin(nx). In turn, for the coefficients an one gets
π
1
a0 =
π
−π
π
2
| sin ξ |dξ =
π
sin ξ dξ =
0
π
1
an =
π
−π
4
,
π
π
2
| sin ξ | cos(nξ )dξ =
π
sin ξ cos(nξ )dξ
0
π
1
2
· 2
[n sin ξ sin(nξ + cos ξ cos(nξ )] π n −1
0
2
1
0 for n = 1, 3, 5, . . . ,
(−1)n+1 − 1 =
= · 2
4/(π(1 − n2 )) for n = 2, 4, 6, . . . .
π n −1
=
(10.2.28)
Here the relation derived from the first part of this book series is used (see formula
(14.1.24)):
1
[b sin ax sin bx + a cos ax cos bx] .
a 2 − b2
(10.2.29)
Since it follows from (10.2.28) that only even values of n are involved, one can
now put n = 2k and construct the Fourier series in the form:
sin ax cos bx dx = −
∞
2
4 ! cos(2kx)
.
| sin x| = −
π
π
4k 2 − 1
(10.2.30)
k=1
For x ∈ [0, π ], the symbol of the absolute value disappears and thus the right-hand
side is the needed Fourier series for the sine function.
As in previous exercises, in Fig. 10.4 some approximations of the function sin x
have been plotted, using the successive terms of the series (10.2.30).
Note that if one puts x = π/2 on both sides of (10.2.30), one will get a known
relation:
∞
∞
! (−1)k+1
π
1 ! (−1)k
1
= −
=
− ,
2
4
2
2
4k − 1
4k 2 − 1
k=1
(10.2.31)
k=0
which makes it possible to find an approximate value of the number π (similarly
as (10.2.18)).
372
10 Studying Fourier Series
y
y
1
1
x
π
π
2
y
π
2
π
x
π
2
π
x
y
1
1
π
2
x
π
Fig. 10.4 The approximations of the function sin x in the interval [0, π ] with successively one,
two, three, and four terms on the right-hand side of (10.2.30) taken into account
Problem 5
The function f (x) = eiz sin x , where z is a parameter, will be expanded into the
complex Fourier series on the interval [−π, π ].
Solution
Instead of representing a given function g(x) in the form of a series of sines and/or
cosines, it is possible to use the expansion into functions of the type eiωx , which also
constitute a complete system on the interval [−T /2, T /2]. The appropriate Fourier
series will then have the following form:
g(x) =
∞
!
cn eiωnx ,
(10.2.32)
n=−∞
with the expansion coefficients found from the formula:
cn =
1
T
T /2
−T /2
e−inωξ g(ξ )dξ.
(10.2.33)
Unlike an and bn , these coefficients are generally complex numbers even for the
real function g(x).
10.3 Exercises for Independent Work
373
The function given in the content of this exercise is periodic with the period equal
to 2π , so our considerations may be limited to the interval [−π, π ]. Thus, we again
have T = 2π and ω = 1. The formula (10.2.33) will now take the form:
cn =
π
1
2π
−π
e−inξ eiz sin ξ dξ.
(10.2.34)
Unfortunately, this integral cannot be calculated in an explicit way. It defines a
certain function of the parameter z (and of course n), which is very well known
in the theory of special functions and bears the name of the Bessel function, for
which the designation Jn (z) is customarily used:
cn = Jn (z).
(10.2.35)
The properties of the Bessel functions are well known and discussed in many
studies, so the coefficients of cn can be treated as known.
The Fourier series to be found has, therefore, the following form:
∞
!
eiz sin x =
Jn (z)einx .
(10.2.36)
n=−∞
It can be used to obtain some interesting and useful identities. For example, by
putting x = 0 in (10.2.36), one finds
∞
!
Jn (z) = 1.
(10.2.37)
n=−∞
In turn, by differentiating (10.2.36) on both sides with respect to x and putting x = 0
again, we get the relation:
∞
!
nJn (z) = z.
(10.2.38)
n=−∞
It can be said that we have obtained a decomposition of the monomial z into Bessel
functions Jn (z). In a similar way, by performing subsequent differentiations, one
can get decompositions for higher monomials.
10.3 Exercises for Independent Work
Exercise 1 Expanding into Fourier series on the interval [−π, π ] the functions
f (x) = x 4 and g(x) = x 6 correspondingly, find the sums of the series:
374
10 Studying Fourier Series
S1 =
∞
!
1
,
n4
S2 =
n=1
∞
!
1
.
n6
n=1
Answers
S1 = π 4 /90.
S2 = π 6 /945.
Exercise 2 Find the expansions of the functions:
(a)
(b)
(c)
(d)
f (x) = |x| into the Fourier series on the interval [−π, π ],
f (x) = x 4 into the complex Fourier series on the interval [−π, π ],
f (x) = cos x into the series of sines on the interval [0, π ],
f (x) = (π − 2x)/4 into the series of cosines on the interval [0, π ].
Answers
∞
(a) |x| =
! (−1)n − 1
π
+2
cos(nx).
2
n2
(b) x 4 = 4
∞
!
n=1
(−1)n
−∞
∞
!
(c) cos x = 2
n=2
∞
n2 π 2 − 6
.
n4
(1 + (−1)n )n
sin(nx).
n2 − 1
! 1 − (−1)n
π − 2x
(d)
=
cos(nx).
4
π n2
n=1
Exercise 3 Using the results of the last problem in this chapter, find an expansion
into the series of Bessel functions for the following functions: z2 , cos z, and sin z.
Answers
z =
2
∞
!
2
n Jn (z).
n=−∞
∞
!
sin z =
n=−∞
cos z =
∞
!
n=−∞
n
(−1) J2n+1 (z).
(−1)n J2n (z).
Index
A
Acceleration, 2, 15, 17
Alternating series, 310
Analytical continuation, 236, 315–321, 324,
326, 330, 334, 340, 343, 351, 354
Analytical function, 233, 350
Analytic (Taylorian) part of a Laurent series,
267, 278
Antisymmetric symbol, 92, 123, 176
Arc, 4, 5, 15, 17
Atlas, 179
B
Ball, 59, 211, 220, 234–237, 260–262, 266,
269, 271, 273, 274, 277–281, 283,
315–319, 321–324, 355
Base curve, 2, 32–35
Bessel function, 373, 374
Binormal vector, 1, 7, 8, 11
Boundary point, 228, 322, 359
Bounded set, 99
Branch point, 269, 273, 274, 277, 312, 315,
319, 322, 327–356
C
Cardioid, 264, 265
Cartesian coordinates, 41, 103, 129
Cartesian product, 123, 125
(Cauchy) residue theorem, 268, 289–314, 331,
335, 338, 340, 342, 344, 346, 347,
349–353, 355–357
Cauchy-Riemann conditions, 219, 223, 225,
226, 231, 244
Cauchy’s test (criterion) for sequences, 234
Cauchy’s test (criterion) for series, 234
Cauchy’s theorem, 245–258, 266, 289, 292,
331, 335, 337, 346, 350
Center of mass, 103, 117, 122, 181, 189
Center-of-mass location, 103, 122, 181
Charge, 104, 141, 142, 198, 209
Closed ball, 59
Closed form, 125, 150
Closed set, 59, 61
Compactified complex plane, ix, 260
Compact set, 47, 56, 83
Complex derivative, 219, 226, 233
Complex-differentiable function, 219
Complex plane, ix, 220, 221, 225, 227, 228,
234, 235, 238, 244, 259, 260, 272, 273,
275, 277, 289, 315, 325
Concave function, 250
Conditional extreme, 41–66
Cone, 31, 32, 38, 54
Connected set, 125
Continuity, 18, 67–81, 84, 86, 87, 90, 91, 93,
99–101, 221, 223, 225, 369
Continuity theorem, 67, 69–76
Continuous function, 1, 83, 125
Contour integral, 220, 237–245, 326
Contractible set, 157
Convergent sequence, 77, 78, 81, 86
Convergent series, 227, 235, 280, 322
Critical point, 48, 51, 55–57, 59, 61, 62
Curl (rotation), 166, 168
© Springer Nature Switzerland AG 2020
T. Radożycki, Solving Problems in Mathematical Analysis, Part III, Problem Books
in Mathematics, https://doi.org/10.1007/978-3-030-38596-5
375
376
Curvature, 1–17, 37, 38
Curve, 1–39, 43–45, 51, 53, 54, 62, 63, 99,
103, 105, 109–111, 113, 127, 141, 150,
152, 154, 155, 157, 158, 181–189,
198–208, 217, 220, 221, 230, 238, 241,
264, 267, 281, 289, 290, 318–320, 325,
326, 328, 330, 350, 369
Cylinder, 13, 39, 109, 119, 121, 122, 214, 218
D
Decreasing function, 12, 94
Definite integral, 184, 221, 241, 246, 289–304
Derivative, ix, 3, 5, 6, 9–12, 15, 17, 18, 21,
25–27, 29, 41–43, 45, 47, 49, 52, 53,
55, 59, 61, 68, 77, 78, 81, 82, 90, 91,
93, 95–97, 109, 123, 124, 127, 142,
145–150, 154, 158, 161, 165–169, 172,
176, 177, 199, 205, 213, 219, 221–223,
225, 226, 228, 229, 233, 309, 353
Differentiable curve, 1, 103
Differentiable function, 103
Differential form, v, 123–178, 180, 183, 191,
198, 199, 213, 217
Dirichlet’s conditions, 359–362
Dirichlet’s function, 361, 362
Dirichlet’s test, 246, 253
Divergence, 164, 166, 168, 170, 171, 173, 175,
211
Divergent series, 306
Domain, 43, 45, 51, 54, 55, 103, 109, 110, 132,
152, 155–157, 185, 202, 207, 208, 212,
213, 217, 218, 220, 221, 224–231, 233,
235–237, 245, 246, 248, 253, 260, 261,
267, 269, 273–275, 280, 289, 290, 302,
315, 316, 318, 319, 322, 368
Doubly ruled surface, 32–34
Dual basis, 129
E
Eigenvalue, 46, 50, 53
Einstein, 174, 190
Electric field, 141, 189
Electrostatic field, 185
Electrostatic potential, 186
Elimination of variables, 43
Ellipsoid, 61, 203, 204, 217
Essential singularity, 268, 270, 272, 284, 285
Euler’s Gamma function, 73, 76
Euler’s substitution, 112
Even function, 281, 363, 370
Exact form, 125, 127, 132, 133, 150
Index
Exterior derivative, 124, 142, 145–150, 154,
158, 161, 165–169, 177, 199, 205, 213
Exterior product, 123, 124, 128, 129, 133, 134,
137, 142, 145, 146
F
Flux, 141, 142, 189–191, 193, 194, 198,
208–212, 215, 217
Fourier series, 309, 312, 359–374
Four-potential, 144, 177
Fréchet-differentiable function, 221
Frenet frame, 1, 12
Frenet’s equations, 1, 9
Frenet trihedron, 1
Fresnel’s integral, 252–255, 292
Froullani’s integral, 89
Fubini theorem, 68
Function (mapping), 23, 259
G
Gauge transformation, 162
Gaussian integral, 84, 86, 252, 255, 256
Gauss-Ostrogradsky’s theorem, 181, 209
Generalized circle, 260, 261
Gibb’s effect, 369
Global extreme, 56–66
Global maximum, 60, 66
Global minimum, 59, 60, 66
Gradient, 24, 41–44, 51, 58, 164, 165,
167–169, 171, 172, 186
Gravitational potential, 186
Green’s theorem, 181, 198, 199, 206
H
Hodge-star operation, 142
Holomorphic (analytic) function, 220, 233,
237, 278
Holomorphicity, 219–235, 237, 248, 253, 273,
275, 278, 280, 281, 283, 285, 286,
316–319, 325
Homography, 260–262
Hyperboloid, 18, 28, 33, 34, 36, 37, 50
I
Improper integral, 68–70, 73, 80, 81, 86, 242,
258, 335
Increasing function, 335, 364, 367
Increasing sequence, 42, 187
Indefinite integral (antiderivative, primitive
function), 89, 91, 95, 246, 292, 298
Index
Integer numbers, 270
Integrand, 68, 71, 74, 77, 80–82, 84–86, 89,
90, 93–95, 98–100, 116, 242, 246, 247,
250–252, 258, 281, 292, 297, 299, 300,
302, 326, 327, 331, 340, 342, 346, 349,
362
Integration by parts, 96, 108, 170, 363
Integration by substitution, 117
Interior, 56–59, 61, 70, 195, 200, 201, 203,
204, 209, 210, 212, 213, 215, 261, 279,
281, 289, 290, 293, 319, 322, 334, 350
Internal point, 228
Intersection of sets, 45, 62, 63
Interval, 1, 70–77, 81–84, 86, 91, 93, 95, 97,
99, 101, 184, 199, 246, 292, 300, 342,
359–374
Inverse function, 229
Irrotational field, 185
Isolated singular point (singularity), 267, 269,
270
Iterated integral, 68, 98, 99
J
Jacobian determinant, 139, 140, 144, 193, 199,
214, 215
Jacobian matrix, 2, 19, 20, 25–27, 29, 30, 138,
221, 224
Jordan (simple closed) curve, 267
K
k-form, 123, 124, 126, 137, 138, 141, 145, 148,
149, 159, 160, 180
k-surface, 2, 18–30, 141
L
Lagrange function, 41
Lagrange multiplier, 41–56, 58, 60, 61
Laurent series, 267, 268, 278–289, 291, 313
Lebesgue integral, v
Left limit, 360
Leibniz rule, 68, 76, 77, 80, 322
Length of a curve, 181
Length parametrization, 15, 17
l’Hospital’s rule, 275
Linear form, 125, 126
Local extreme, 42, 43, 360, 361
Local maximum, 65
Local minimum, 65
Lorentz force, 142
Lorentz transformations, 143, 178
Lower bound, 75
377
M
Magnetic fields, 141, 142, 169, 177, 189, 198,
203, 209
Majorant, 75, 83, 84, 86, 87, 91, 94, 96, 100
Maxwell’s equations, 142, 178
Metric (distance), 104, 111, 113, 116, 118, 121
Metric tensor, 104, 111, 113, 116, 118, 121
Minors, 21–23, 27–29, 64, 248, 262, 341
Moment of inertia, 119–122, 189
Multivalued function, 315–358
N
Natural numbers, 73, 76
Negative-definite matrix, 46, 53, 56
Neighborhood, 2, 18–20, 22, 23, 26, 30, 70,
71, 167, 219, 233, 234, 267, 273, 275
Non-holomorphic function, 267, 268, 270,
272, 275, 277, 302
Non-isolated singular point (singularity), 277,
278, 289, 330, 350
Normal plane, 1, 13
Normal vector, 1, 4, 24, 106, 189–192, 194,
195, 201, 203, 209, 210, 212
Numerical series, 283
O
Odd function, 281, 363, 370
Open set, 41, 124, 125, 137, 179, 219, 221,
228, 268, 319
Ordinary differential equation, 232
Orientable surface, 180, 190, 192
Orientation of a surface, 179, 180, 190, 192,
200, 203, 206
Orientation of a linear space, 179
Oriented curvilinear integral, 179–218
Orthogonal vectors, 4, 7, 9, 13, 15, 17, 24, 106,
131
P
Paraboloid, 35, 36, 119, 122, 184, 185
Parametrization, 2, 13–15, 20–22, 27, 30–37,
39, 113, 116, 118–120, 183, 184, 187,
188, 190–193, 195–197, 199, 201, 202,
204, 210, 214–216, 220, 238–240, 242,
245, 250, 251, 254, 264, 294, 299, 300,
302, 307, 326, 328, 329, 334, 336, 337,
340
Parity, 247, 366, 370, 371
Partial derivative, ix, 26, 27, 41–43, 59, 61, 68,
77, 78, 81, 91, 93, 219, 221, 223, 226,
227
378
Piecewise differentiable curve, 1
Piecewise smooth curve, 1, 18, 182, 199
Poincaré lemma, 125, 150, 152, 155, 158
Pole, 22, 113, 192, 260, 268–272, 276, 278,
279, 281–285, 287, 290–294, 296, 299,
302, 303, 305, 306, 308–310, 312, 326,
330, 332, 335, 337, 340, 343, 344, 347,
348, 353, 355, 356
Polynomial, 89, 126, 224, 229, 242, 243, 270,
272, 275, 291, 293, 307
Positive-definite matrix, 47, 49, 50
Power series, 219, 233, 237, 316, 322
Predictions, 262
Primitive forms, 123, 125, 149–164, 167, 169
Principal part of a Laurent series, 267, 278,
283, 288
Pullback of a form, 124
Pythagorean trigonometric identity, 10, 21
R
Radius of convergence, 236, 237, 316
Rank of a matrix, 19, 21, 27, 29, 30
Rational function, 89, 112, 292, 293, 345
Real numbers, 41, 228, 229, 259, 362
Regular set, 31
Remainder, 361
Removable (apparent) singularity, 268–270,
272, 275, 312, 339, 340, 343, 345
Residue, 241, 242, 246, 268, 281, 289–315,
326, 327, 330, 331, 334, 335, 338,
340–342, 344, 346–353, 355–357
Riemann’s sphere, 251, 260
Right limit, 360
Ruled surface, 2, 31–37, 39
S
Scalar potential, 164, 167, 172, 173, 175, 176,
178
Second derivative, 47–49, 52, 53, 55, 148, 150,
172, 353
Segment, 58, 186, 187, 200, 201, 206, 217,
231, 238, 239, 244, 245, 248, 261, 293,
326, 349, 351
Series of positive terms, 310
Set, 7, 18–21, 23, 24, 26, 27, 29–31, 36, 38,
39, 41, 42, 45, 47, 50–63, 65–68, 70,
74, 77, 78, 80–82, 84, 86, 87, 90–93,
96, 98–100, 124, 125, 128–130, 133,
Index
137–139, 157, 159, 160, 179, 180, 184,
203, 219, 221, 226–228, 230, 231, 234,
236, 246, 258–261, 263, 266, 268, 274,
296, 300, 315, 319, 322, 342, 351, 359,
361, 362
Singularity, 236, 237, 241, 248, 252, 267–275,
277, 278, 281, 283–285, 288, 290, 298,
306, 311, 312, 315, 319, 320, 322, 325,
326, 328, 330, 332, 335, 337–340, 343,
345–347, 350, 352–355, 357
Smooth curve, 1, 182, 199
Smooth function, 105, 123, 126
Smooth surface, 18, 41, 109, 123, 126, 127
Stationary point, 42
Stokes’ theorem, 180, 181, 193, 198–217
Straightening plane, 1, 13
Strong (Fréchet) derivative, 221
Subset, 2, 18, 20, 22, 123, 128, 259, 263
Summation convention, 174, 190
Surface area, 103, 104, 106, 111, 118, 121
T
Tangential vector, 4, 5, 20, 53, 182
Tangent plane, 1, 13, 22, 49
Tangent space, 24, 25, 28, 123, 126, 127, 180
Taylor series, 227, 233, 278, 316, 317,
319–321, 325, 357
Taylor’s formula, 267, 317
Torsion, 2–17, 37, 38
Torus, 26, 27
Twice differentiable function, 149
U
Uniform continuity, 75
Uniform convergence, 68, 75, 83, 86, 91, 93,
100
Unoriented curvilinear integral, 103–122, 181,
189, 191
Upper bound, 211
V
Vector field, 41, 164–168, 170, 171, 173, 178,
186, 191, 193, 202, 208–211, 217
Vector potential, 166, 168–171, 173, 178, 209
Velocity, 3, 15, 17, 109, 141–143, 177
Viviani curve, 13, 14, 16