Gergő Nemes
Asymptotic Expansions
for Integrals
M. Sc. Thesis
Loránd Eötvös University
May 23, 2012
Loránd Eötvös University
Faculty of Science
M. Sc. Thesis
Asymptotic Expansions
for Integrals
Author:
Gergő Nemes
M. Sc. Pure Mathematics
Supervisor:
Dr. Árpád Tóth
associate professor
Examination Committee:
Committee Chairman:
Prof. Dr. László Simon
........................
Committee Members:
Dr. Balázs Csikós
........................
Dr. Alice Fialowski
........................
Prof. Dr. Tibor Jordán
........................
Dr. Tamás Móri
........................
Dr. Péter Sziklai
........................
Dr. János Tóth
........................
May 23, 2012
Preface
The asymptotic theory of integrals is an important subject of applied mathematics
and physics. Although it is an old topic, with origins dating back to Laplace, new
methods, applications and formulations continue to appear in the literature. The
purpose of this thesis is to review some of the classical procedures for obtaining
asymptotic expansions of integrals, especially focusing on Laplace-type integrals.
As a contribution to the topic, we give a new method for computing the coefficients
of these asymptotic series with several illustrating examples. This method is a
generalization of the one given in my paper about the Stirling Coefficients (J.
Integer Seqs. 13 (6), Article 10.6.6, pp. 5 (2010)).
Chapter 1 covers the asymptotic theory of real Laplace-type integrals. The
computation of the coefficients appearing in the asymptotic expansions are described completely in this chapter. In Chapter 2, we discuss two methods from
the asymptotic theory of complex Laplace-type integrals: the Method of Steepest
Descents and Perron’s Method. Each chapter contains examples to demonstrate
the application of the results. In Appendix A, we collect the basic properties of
some combinatorial quantities we use in this thesis.
Budapest, Hungary, May 23, 2012
Gergő Nemes
i
Table of Contents
Preface
i
Table of Contents
ii
Acknowledgements
iii
1 Real Laplace-type integrals
1.1 Fundamental concepts and results . . . . . . . . . . . .
1.1.1 Asymptotic expansions . . . . . . . . . . . . . .
1.1.2 Watson’s Lemma . . . . . . . . . . . . . . . . .
1.2 Laplace’s Method . . . . . . . . . . . . . . . . . . . . .
1.2.1 Asymptotic expansion of Laplace-type integrals
1.2.2 Generalization of Perron’s Formula . . . . . . .
1.2.3 The CFWW Formula . . . . . . . . . . . . . . .
1.2.4 Another method . . . . . . . . . . . . . . . . .
1.2.5 Examples . . . . . . . . . . . . . . . . . . . . .
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1
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2 Complex Laplace-type integrals
21
2.1 The Method of Steepest Descents . . . . . . . . . . . . . . . . . . . 21
2.2 Perron’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
A Combinatorial objects
33
A.1 Ordinary Potential Polynomials . . . . . . . . . . . . . . . . . . . . 33
A.2 The r-associated Stirling Numbers . . . . . . . . . . . . . . . . . . . 34
Bibliography
35
ii
Acknowledgements
I am deeply grateful to my supervisor, Professor Árpád Tóth for his detailed and
constructive comments, and for his important support throughout this work. I owe
my loving thanks to Dorottya Sziráki. Without her support it would have been
impossible for me to finish this work. My special gratitude is due to my parents
and my brother for their loving support.
iii
Chapter 1
Real Laplace-type integrals
In this chapter, we investigate the asymptotic behavior of certain parametric integrals. The origins of the method date back to Pierre-Simon de Laplace (1749
– 1827), who studied the estimation of integrals arising in probability theory of
the form
Z b
In =
e−nf (x) g (x) dx (n → +∞) .
(1.1)
a
Here the functions f and g are real continuous functions defined on the real (finite
or infinite) interval [a, b]. Hereinafter, we call integrals of the type (1.1), Laplacetype integrals. Laplace made the observation that the major contribution to the
integral In should come from the neighborhood of the point where f attains its
smallest value. If f has its minimum value only at the point x0 in (a, b) where
f ′ (x0 ) = 0 and f ′′ (x0 ) > 0, then Laplace’s result is
s
Z b
2π
e−nf (x) g (x) dx ∼ g (x0 ) e−nf (x0 )
.
′′
nf (x0 )
a
The sign ∼ is used to mean that the quotient of the left-hand side by the righthand side approaches 1 as n → +∞. This formula is now known as Laplace’s
approximation. A heuristic proof of this formula may proceed as follows. First,
we replace f and g by the leading terms in their Taylor series expansions around
x = x0 , and then extend the integration limits to −∞ and +∞. Hence,
Z
b
a
e−nf (x) g (x) dx ≈
Z
b
e
f ′′ (x )
−n f (x0 )+ 2 0 (x−x0 )2
a
≈ g (x0 ) e
−nf (x0 )
= g (x0 ) e−nf (x0 )
Z
+∞
e−n
g (x0 ) dx
f ′′ (x0 )
(x−x0 )2
2
dx
−∞
s
2π
.
nf ′′ (x0 )
We divided this chapter into two parts. In Section 1.1, we deal with the fundamental theorem of the topic, called Watson’s Lemma. Section 1.2 presents a general
theorem on asymptotic expansions of Laplace-type integrals, and investigates several of their properties.
1
1.1. Fundamental concepts and results
1.1
Fundamental concepts and results
This section is divided into two parts. In the first part, we collect all the basic
concepts which we will use hereinafter. In the second part we state and prove an
important theorem of the topic, known as Watson’s Lemma.
1.1.1
Asymptotic expansions
Let f and g be two continuous complex functions defined on a subset H of the
complex plane. Let z0 be a limit point of H.
Definition 1.1.1. We write f (z) = O(g(z)) (”f is big-oh g”), as z → z0 , to
mean that there is a constant C > 0 and a neighborhood U of z0 such that |f (z)| ≤
C |g(z)| for all z ∈ U ∩ H.
Definition 1.1.2. We write f (z) = o(g(z)) (”f is little-oh g”), as z → z0 , to mean
that for every ε > 0, there exists a neighborhood Uε of z0 such that |f (z)| ≤ ε |g(z)|
for all z ∈ Uε ∩ H.
To define the asymptotic expansion of a function we need the concept of an
asymptotic sequence.
Definition 1.1.3. Let {ϕn }n≥0 be a sequence of continuous complex functions
defined on a subset H of the complex plane. Let z0 be a limit point of H. We
say that {ϕn }n≥0 is an asymptotic sequence as z → z0 in H if, for all n ≥ 0,
ϕn+1 (z) = o(ϕn (z)), as z → z0 .
Definition 1.1.4. If {ϕn }n≥0 is an asymptotic sequence as z → z0 , we say that
∞
X
an ϕn (z),
n=1
where the an ’s are constants, is an asymptotic expansion of the function f if for
each N ≥ 0
N
X
f (z) =
an ϕn (z) + o(ϕN (z)) as z → z0 .
(1.2)
n=1
If a function possesses an asymptotic expansion, we write
f (z) ∼
∞
X
n=1
an ϕn (z) as z → z0 .
Note that (1.2) may be written as
f (z) =
N
−1
X
n=1
an ϕn (z) + O(ϕN (z)),
which implies that the error is of the same order of magnitude as the first term
omitted. Usually, we use ϕn (z) = z −ρn with 0 < Re (ρ0 ) < Re (ρ1 ) < Re (ρ2 ) < . . .,
as z → ∞ in some sector of the complex plane.
2
1.1. Fundamental concepts and results
1.1.2
Watson’s Lemma
Consider the integral
I (λ) =
Z
+∞
0
e−λx
dx
1+x
where λ > 0. Using the well-known identity
N
−1
X
1
xN
=
(−1)k xk + (−1)N
1+x
1+x
k=0
(N ≥ 0) ,
term-by-term integration gives
N
−1
X
k!
I (λ) =
(−1) k+1 + (−1)N
λ
k=0
where
Z
N
(−1)
+∞
0
k
Z
+∞
0
xN −λx
e dx,
1+x
Z +∞ N
Z +∞
x
N!
xN −λx −λx
xN e−λx dx = N +1 .
e dx =
e dx ≤
1+x
1+x
λ
0
0
And hence, by definition,
I (λ) ∼
∞
X
k=0
(−1)k
k!
, as λ → +∞.
λk+1
This is a special case of a much more general result, now known as Watson’s
Lemma. The theorem is due to George Neville Watson (1886 – 1965). It is a
significant result in the theory of asymptotic expansions of Laplace-type integrals.
In view of its importance, the proof of the result is reproduced below.
Theorem 1.1.1 (Watson’s Lemma). Let f (x) be a complex valued function of a
real variable x such that
(i) f is continuous on (0, ∞);
(ii) as x → 0+ ,
f (x) ∼
∞
X
ak xρk −1
k=0
with 0 < Re (ρ0 ) < Re (ρ1 ) < Re (ρ2 ) < . . ., limk→+∞ Re (ρk ) = +∞; and
(iii) for some fixed c > 0, f (x) = O (ecx ) as x → +∞.
Then we have
Z
+∞
0
e−λx f (x) dx ∼
as λ → +∞.
3
∞
X
ak Γ (ρk )
,
ρk
λ
k=0
1.2. Laplace’s Method
Proof. By the conditions (i), (ii) and (iii); the integral
Z +∞
e−λx f (x) dx
0
converges for λ > c. Conditions (ii) and (iii) imply that
N
−1
X
ρk −1 ak x
f (x) −
≤ KN ecx xρN −1 for x > 0,
k=0
for every N ≥ 0 with some constant KN . Thus we have
Z
Z +∞
Z +∞
N
−1
+∞
X
−λx
−λx ρk −1
e f (x) dx −
ak
e x
dx ≤ KN
e−(λ−c)x xρN −1 dx.
0
0
0
k=0
Note that we have for λ > 0
Z +∞
Z +∞
Γ (ρn )
1
−λx ρk −1
.
e−t tρk −1 dt =
e x
dx = ρ
k
λ
λ ρk
0
0
Hence we have
Z
Z +∞
N
−1
+∞
X
ak Γ (ρn ) −(λ−c)x ρN −1 −λx
dx
x
≤
K
e
e f (x) dx −
N
0
λ ρk 0
k=0
= KN
that is
Z
+∞
e
−λx
f (x) dx =
0
N
−1
X
k=0
Γ (Re (ρN ))
,
|(λ − c)ρN |
ak Γ (ρn )
+O
λ ρk
1
λ ρN
as λ → +∞, which proves Watson’s Lemma.
1.2
Laplace’s Method
In this section, we revisit the classical Laplace Method. In the first subsection, we
prove the fundamental theorem on asymptotic expansion of Laplace-type integrals,
an extension of the formula mentioned in the introduction of the chapter. When
the functions appearing in the integral are holomorphic, the coefficients of the
asymptotic expansion can be given explicitly by Perron’s Formula. In the second
part, we extend this formula to the case when the power series of the functions are
not necessarily convergent but asymptotic. In the third subsection, we reproduce
the proof of the CFWW Formula which gives another explicit form of the coefficients in the asymptotic expansion. The fourth part of the section is about a new
and simpler explicit formula for these coefficients. Finally, in the fifth subsection,
we give three examples to demonstrate the application of the results.
4
1.2. Laplace’s Method
1.2.1
Asymptotic expansion of Laplace-type integrals
Laplace’s Method is one of the best-known techniques developing asymptotic expansions for integrals. Here we present an extension of the formula mentioned in
the introduction of this chapter. Consider the integral of the form
Z b
e−λf (x) g (x) dx,
I (λ) =
a
where (a, b) is a real (finite or infinite) interval, λ is a large positive parameter
and the functions f and g are continuous. We observe that by subdividing the
range of integration at the minima and maxima of f , and by reversing the sign of
x whenever necessary, we may assume, without loss of generality, that f has only
one minimum in [a, b] which occurs at x = a. Next, we assume that, as x → a+ ,
f (x) ∼ f (a) +
and
g (x) ∼
∞
X
k=0
∞
X
k=0
ak (x − a)k+α ,
bk (x − a)k+β−1
(1.3)
(1.4)
with α > 0, Re (β) > 0; and that the expansion of f can be term wise differentiated, that is,
∞
X
′
f (x) ∼
ak (k + α) (x − a)k+α−1
(1.5)
k=0
as x → a+ . Also, we suppose, without loss of generality, that a0 6= 0 and b0 6= 0.
The following theorem is due to Arthur Erdélyi (1908 – 1977) (see, e.g., [8, p.
85], [14, p. 57]).
Theorem 1.2.1. For the integral
I (λ) =
Z
b
e−λf (x) g (x) dx,
a
we assume that
(i) f (x) > f (a) for all x ∈ (a, b), and for every δ > 0 the infimum of f (x) − f (a)
in [a + δ, b) is positive;
(ii) f ′ (x) and g(x) are continuous in a neighborhood of x = a, expect possibly at
a;
(iii) the expansions (1.3), (1.4) and (1.5) hold; and
(iv) the integral I (λ) converges absolutely for all sufficiently large λ.
Then
I (λ) ∼ e
−λf (a)
∞
X
Γ
n=0
n+β
α
cn
,
(n+β)/α
λ
as λ → +∞, where the coefficients cn are expressible in terms of an and bn .
5
(1.6)
1.2. Laplace’s Method
The first three coefficients cn are given explicitly by
b0
b1 (β + 1) a1 b0
1
c0 =
,
−
,
c1 = (β+1)/α
β/α
α
α 2 a0
αa0
a0
and
c2 =
1
(β+2)/α
a0
(β + 1) b0
b2 (β + 2) a1 b1
−
+ (β + α + 2) a21 − 2αa0 a2
2
α
α a0
2α2 a20
.
Proof. By conditions (ii) and (iii) there exists a number c ∈ (a, b) such that
f ′ (x) and g(x) are continuous in (a, c], and f ′ (x) is also positive there. Let T =
f (c) − f (a), and introduce the variable t as
t = f (x) − f (a).
Since f (x) is increasing in (a, c), we can write
e
λf (a)
Z
c
e
−λf (x)
g (x) dx =
a
Z
T
e−λt h (t) dt
(1.7)
0
with h(t) being the continuous function in (0, T ] given by
h(t) = g(x)
By assumption
t∼
∞
X
k=0
g(x)
dx
= ′ .
dt
f (x)
ak (x − a)k+α
(1.8)
as x → a+ ,
and thus, by series reversion we obtain
x−a∼
∞
X
dk tk/α
k=1
as t → 0+ .
Substituting this into (1.8) yields
h (t) ∼
∞
X
ck t(k+β)/α−1
k=0
as t → 0+ , where the coefficients ck are expressible in terms of ak and bk . We now
apply Watson’s Lemma to the integral on the right-hand side of (1.7) to obtain
Z
c
e
a
−λf (x)
g (x) dx ∼ e
−λf (a)
∞
X
n=0
Γ
n+β
α
cn
,
(n+β)/α
λ
as λ → +∞. All that remains is to show that the integral on the remaining range
(c, b) is negligible. Define
def
ε = inf (f (x) − f (a)) .
c≤x<b
6
1.2. Laplace’s Method
This is positive by condition (i). Let λ0 be a value of λ for which I(λ) is absolutely
convergent. Assume that λ ≥ λ0 , then
λ (f (x) − f (a)) = (λ − λ0 ) (f (x) − f (a)) + λ0 (f (x) − f (a))
≥ (λ − λ0 ) ε + λ0 (f (x) − f (a))
and
Z
λf (a) b −λf (x)
e
≤ Ke−ελ ,
e
g
(x)
dx
c
where
K=e
λ0 (ε+f (a))
is a constant.
1.2.2
Z
b
e−λ0 f (x) |g (x)| dx
c
Generalization of Perron’s Formula
In this subsection, we derive an explicit formula for the coefficients cn appearing
in the asymptotic expansion (1.6). The notations are the same as in the previous
subsection.
Theorem 1.2.2. The coefficients cn appearing in (1.6) are given explicitly by
"
(
(n+β)/α )#
1
dn
(x − a)α
cn =
G (x)
αn! dxn
f (x) − f (a)
x=a
"
(1.9)
(n+β)/α #
n
α
k
X
1
bn−k d
(x − a)
=
,
α k=0 k! dxk f (x) − f (a)
x=a
P∞
k
where G (x) is the (formal) power series k=0 bk (x − a) and f (x) should be identified with its (formal) expansion (1.3).
Remark. When (f (x) − f (a)) (x − a)−α and g (x) (x − a)1−β are holomorphic
functions, the representation (1.9) is known as Perron’s Formula [10] [14, p. 103]
(see also Section 2.2). But here we show that this formula holds also when the
expansions (1.3), (1.4) are merely asymptotic. An alternative proof of essentially
the same formula has been given previously by Wojdylo [13].
Proof. Let ℓ > 0 be the index of the first nonvanishing coefficient in the expansion
(1.3) of f apart from a0 . Define
def
fℓ (x) =
f (x) − f (a) − a0 (x − a)α
(x − a)
α+ℓ
Then
I (λ) =
Z
b
e
−λf (x)
g (x) dx = e
−λf (a)
a
= e−λf (a)
=
e
−λf (a)
λ1/α
Z
Z
∼
∞
X
k=0
b
a
ak+ℓ (x − a)k
α
as x → a+ .
α+ℓ
e−a0 λ(x−a) e−λ(x−a) fℓ (x) g (x) dx
|
{z
}
h(λ,x)
b
α
e−a0 λ(x−a) h (λ, x) dx
a
Z
λ1/α (b−a)
0
7
α
e−a0 t h λ, λ−1/α t + a dt.
(1.10)
1.2. Laplace’s Method
Set s = λ−1/α t; then the function h λ, λ−1/α t + a reads
h (λ, s + a) = e−t
α sℓ f
ℓ (s+a)
g (s + a) .
We have
e
−tα sℓ fℓ (s+a)
∞
X
1 dk −tα wℓ fℓ (w+a)
∼
e
sk
k
k! dw
w=0
k=0
as s → 0+
and hence,
!
∞
n
X
X
bn−k dk −tα wℓ fℓ (w+a)
h (λ, s + a) ∼
sn+β−1
e
k
k!
dw
w=0
n=0
k=0
as s → 0+ .
Introducing this expansion in the last integral of (1.10) we find
Z 1/α
e−λf (a) λ (b−a) −a0 tα
I (λ) = 1/α
e
h λ, λ−1/α t + a dt
λ
0
"
!# !
Z λ1/α (b−a)
∞
n
−λf (a) X
X
ℓ
α
bn−k dk
e
e−(a0 +w fℓ (w+a))t sn+β−1 dt
∼ 1/α
k
λ
k!
dw
0
n=0
k=0
"
!# w=0!
Z
∞
n
λ1/α (b−a)
X X bn−k dk
1
ℓ
α
= e−λf (a)
e−(a0 +w fℓ (w+a))t tn+β−1 dt
k
(n+β)/α
k! dw
λ
0
n=0
k=0
w=0
!
Z +∞
∞
n
X
X
bn−k dk
1
−(a0 +wℓ fℓ (w+a))tα n+β−1
−λf (a)
∼e
e
t
dt
k
(n+β)/α
k! dw
λ
0
w=0
n=0
k=0
"
!#
!
n
∞
n+β X
k
X
Γ
b
d
1
1
n−k
α
= e−λf (a)
(n+β)/α
α
k! dwk (a0 + wℓ fℓ (w + a))
λ(n+β)/α
n=0
k=0
w=0
and using the definition of fℓ (w + a):
I (λ) ∼ e−λf (a)
= e−λf (a)
∞
X
n=0
∞
X
n=0
(n+β)/α # !
1
wα
(n+β)/α
α
k!
f (w + a) − f (a)
λ
k=0
w=0
"
#
!
(n+β)/α
n
X
Γ n+β
bn−k dk
(x − a)α
1
α
k
(n+β)/α
α
k! dx
f (x) − f (a)
λ
k=0
Γ
n+β
α
n
X
bn−k
"
dk
dwk
x=a
as λ → +∞. Formula (1.9) now follows from the asymptotic expansion (1.6) and
the uniqueness theorem on asymptotic series.
1.2.3
The CFWW Formula
In the previous subsection, we introduced formula (1.9) for the coefficients cn
apperaing in the asymptotic expansion of certain Laplace-type integrals. In this
subsection, we shall extract the higher order derivatives in formula (1.9) by means
8
1.2. Laplace’s Method
of the Partial Ordinary Bell Polynomials (see Appendix A.1). Using the series
expansion (1.3) of f (x) we find
"
(n+β)/α #
n
1 X bn−k dk
(x − a)α
cn =
α k=0 k! dxk f (x) − f (a)
x=a
#
"
n
(n+β)/α
α
k
X
bn−k d
1
a0 (x − a)
=
(n+β)/α
k
k! dx
f (x) − f (a)
αa0
k=0
x=a
#
"
n
−(n+β)/α
X bn−k dk f (x) − f (a)
1
=
(n+β)/α
k! dxk
a0 (x − a)α
αa0
k=0
x=a


!
−(n+β)/α
∞
n
k
X
X
1
ak
bn−k  d
 .
=
(x − a)k
1+
(n+β)/α
k
k!
dx
a
0
αa0
k=1
k=0
x=a
From the definition of the Ordinary Potential Polynomials and the Partial Ordinary Bell Polynomials we obtain
cn =
=
=
1
(n+β)/α
αa0
1
(n+β)/α
αa0
1
αΓ
n+β
α
n
X
ak
a1 a2
, ,...,
bn−k A−(n+β)/α,k
a
a0
0 a0
k=0
n
k n+β X
X
1
− α
bn−k
j Bk,j (a1 , a2 , . . . , ak−j+1 )
j
a
0
j=0
k=0
n
X
bn−k
k
X
j=0
k=0
(−1)j
Bk,j (a1 , a2 , . . . , ak−j+1 )
Γ
(n+β)/α+j
j!
a0
n+β
+j .
α
(1.11)
This is the Campbell–Fröman–Walles–Wojdylo Formula (from now on the CFWW
Formula) [1, 12, 13]. Sometimes it is possible to find the Partial Ordinary Bell
Polynomials explicitly, in other cases the recurrence (A.4) could be used.
1.2.4
Another method
In the previous subsection, we showed an explicit formula for the coefficients cn
appearing in the asymptotic expansion of Laplace-type integrals. This formula,
that we called the CFWW Formula, followed from the generalization of Perron’s
Formula (1.9) using the Partial Ordinary Bell Polynomials. In this subsection,
we give a new explicit formula using Lagrange Interpolation and the Ordinary
Potential Polynomials. We write formula (1.9) in the form
"
(n+β)/α #
n
X
bn−k dk
1
a0 (x − a)α
cn =
.
(1.12)
(n+β)/α
k
k!
dx
f
(x)
−
f
(a)
αa0
k=0
x=a
Define the sequence of functions (Fk (z))k≥0 by the generating function
a0 (x − a)α
f (x) − f (a)
z
=
∞
X
k=0
9
Fk (z) (x − a)k .
1.2. Laplace’s Method
Now, equation (1.12) can be written as
cn =
1
(n+β)/α
αa0
n
X
bn−k Fk
k=0
n+β
.
α
(1.13)
Since
a0 (x − a)α
f (x) − f (a)
z
=
=
1+
∞
X
k=0
∞
X
ak
a0
k=1
k X
j=0
(x − a)k
!−z
!
−z 1
k
j Bk,j (a1 , a2 , . . . , ak−j+1 ) (x − a) ,
j a0
the Fk is a polynomial of degree at most k. By Lagrange Interpolation
k
Γ (z + k + 1) X
j k Fk (−j)
Fk (z) =
,
(−1)
j
k!Γ (z)
z
+
j
j=0
(1.14)
where the Fk (−j)’s can be computed as follows
∞
X
k=0
k
Fk (−j) (x − a) =
f (x) − f (a)
a0 (x − a)α
j
=
1+
∞
X
k=0
Fk (−j) = Aj,k
a0
k=1
=
that is
∞
X
ak
Aj,k
a1 a2
ak
, ,...,
a0 a0
a0
(x − a)
k
!j
ak
a1 a2
, ,...,
a0 a0
a0
.
(x − a)k ,
(1.15)
From (1.14) and (1.15) we deduce
Fk
n+β
α
=
Γ
k
+ k + 1 X (−1)j k
a1 a2
ak
Aj,k
, ,...,
.
n+β
a0 a0
a0
+j j
k!Γ n+β
α
j=0 α
n+β
α
Plugging this into (1.13) yields
Theorem 1.2.3. The coefficients cn appearing in (1.6) are given explicitly by
cn =
1
αΓ
n+β
α
n
X
Γ
k=0
k
+ k + 1 bn−k X (−1)j k
a1 a2
ak
Aj,k
.
, ,...,
n+β
(n+β)/α
a0 a0
a0
+j j
k!a0
j=0 α
(1.16)
n+β
α
As we will see in the following subsection, it is possible, in several cases, to
derive an explicit formula for the Ordinary Potential Polynomials due to the somewhat simple generating function
!j j X
∞
∞
X
ak
ak
f (x) − f (a)
a1 a2
k
(x − a)k ,
=
=
Aj,k
(x − a)
, ,...,
1+
α
a
a
(x
−
a)
a
a
a
0
0
0
0
0
k=0
k=1
10
1.2. Laplace’s Method
whereas in the case of the CFWW formula, the corresponding Partial Ordinary
Bell Polynomials have more complicated generating functions, such as
!
∞
X
f (x) − f (a) − a0 (x − a)α
k
exp y
ak (x − a) = exp y
(x − a)α
k=1
!
k
∞
X
X
Bk,j (a1 , a2 , . . . , ak−j+1 ) j
=
y (x − a)k .
j!
j=0
k=0
1.2.5
Examples
Example 1.2.1. As a first example we take the Gamma Function
Z +∞
e−t tλ dt
Γ (λ + 1) =
(1.17)
0
for λ > 0. If we put t = λ(1 + x), we obtain
Z +∞
λ+1 −λ
e−λ(x−log(1+x)) dx
Γ (λ + 1) = λ e
−1
and hence,
Γ (λ)
=
λλ e−λ
Z
+∞
e
−λ(x−log(1+x))
dx +
0
Z
1
e−λ(−x−log(1−x)) dx.
0
Using Theorem 1.2.1 with f (x) = x − log (1 + x) (and f (x) = −x − log (1 − x)),
g(x) ≡ 1, α = 2 and β = 1; one finds that
Z +∞
∞
X
cn
n+1
−λ(x−log(1+x))
e
dx ∼
Γ
(n+1)/2
2
λ
0
n=0
and
Z
1
e
−λ(−x−log(1−x))
0
dx ∼
∞
X
Γ
n=0
n+1
2
(−1)n cn
,
λ(n+1)/2
as λ → +∞, where, by Theorem 1.2.2,
"
(n+1)/2 #
dn
x2
1
cn =
2n! dxn x − log (1 + x)
Finally,
Γ (λ) ∼
as λ → +∞, where
√
2πλ
λ−1/2 −λ
e
∞
X
γn
n=0
λn
.
x=0
,
(1.18)
r
1
2
Γ n+
c2n
γn =
π
2
"
n+1/2 #
1
d2n 1
x2
= n
2 n! dx2n 2 x − log (1 + x)
11
(1.19)
x=0
1.2. Laplace’s Method
are the so-called Stirling Coefficients. The first few are γ0 = 1 and
γ1 =
1
1
139
571
, γ2 =
, γ3 = −
, γ4 = −
.
12
288
51840
2488320
We shall give a more explicit formula for the γn ’s using the CFWW Formula (1.11).
Since
∞
X
(−1)k k+2
f (x) = x − log (1 + x) =
x , g(x) ≡ 1,
k+2
k=0
(−1)k−j+1
1 1
we have to compute the Partial Ordinary Bell Polynomials Bk,j − 3 , 4 , . . . , k−j+3 .
We have the exponential generating function
!
∞
X
x2
(−1)k k
y
x−
x = exp
− log (1 + x)
exp y
2
k
+
2
x
2
k=1
! !
k
∞
X
X
1 1
(−1)k−j+1 y j
xk .
Bk,j − , , . . . ,
=
3
4
k
−
j
+
3
j!
j=0
k=0
From the generating functions of the Hermite Polynomials and the Stirling Numbers of the First Kind we have


X
⌊k/2⌋
∞
j
2
X
x
(−1)

exp t x −
=
tk−j  xk
j
2
j!
(k
−
2j)!2
j=0
k=0
and
exp (−t log (1 + x)) =
∞
X
k=0
(−1)
k
X
k
j=0
s (k, j) tj
!
xk
.
k!
Performing the Cauchy product of the series and rearranging the terms in the
coefficients yields
x2
− log (1 + x)
exp t x −
2


⌊k/3⌋ k−j
j−i
∞
k+r
X X (−1)i X
X
(−1) s (j − i − r, k − 2i − r) j  k

=
t x .
i
2 i! r=0
r! (k − 2i − r)!
j=0 i=0
k=0
Plugging t =
y
x2
gives
x2
exp
x−
− log (1 + x)
2


⌊k/3⌋ k−j
j−i
∞
k+r
j
X X (−1)i X
X
(−1) s (j − i − r, k − 2i − r) y  k

x
=
i
2j
2
i!
r!
(k
−
2i
−
r)!
x
r=0
i=0
j=0
k=0
y
x2
j
j−i
k X
∞
X
X
(−1)i X (−1)r s (k + 2j − 2i − r, j − i − r) j
=
y
i i!
2
r!
(k
+
2j
−
2i
−
r)!
r=0
j=0 i=0
k=0
12
!
xk ,
1.2. Laplace’s Method
that is
Bk,j
1 1
(−1)k−j+1
− , ,...,
3 4
k−j+3
!
j
j−i
X
(−1)i X (−1)r s (k + 2j − 2i − r, j − i − r)
= j!
.
i i!
2
r!
(k
+
2j
−
2i
−
r)!
r=0
i=0
Finally, formula (1.11) produces
j
j−i
n
X 1 X
X
2(n−1)/2+j Γ n+1
+
j
(−1)j+i+r s (n + 2j − 2i − r, j − i − r)
2
.
cn =
2i i! r=0
r! (n + 2j − 2i − r)!
Γ n+1
2
i=0
j=0
From this and the expression (1.19) we deduce
j
j−i
2n
X
2n+j Γ n + j + 21 X 1 X (−1)j+i+r s (2n + 2j − 2i − r, j − i − r)
√
γn =
.
i i!
2
r!
(2n
+
2j
−
2i
−
r)!
π
r=0
j=0
i=0
(1.20)
This is the result of López, Pagola and Sinusı́a [6].
Now we show how our new method gives a much simpler result than (1.20) with
less calculation. To
have to compute the Ordinary Potential
use formula (1.16) we
k 2
2 1
Polynomials Aj,k − 3 , 2 , . . . , (−1) k+2 . This time the generating function is
∞
X
2
1+
(−1)k
xk
k
+
2
k=1
!j
j
x − log (1 + x)
= 2
x2
∞
X
2
2 1
k
xk .
=
Aj,k − , , . . . , (−1)
3 2
k+2
k=0
Using the generating function of the Stirling Numbers of the First Kind yields
j X
j j−i
j
x (− log (1 + x))i
(x − log (1 + x)) =
i
i=0
j ∞
X
xk
j j−i X
x
(−1)k i!s (k, i)
=
k!
i
i=0
k=0
j ∞ X
X
xk+j−i
j
(−1)k i!s (k, i)
=
i
k!
k=0 i=0
!
j ∞
X
X
j
s
(k
−
j
+
i,
i)
=
xk ,
(−1)k−j+i i!
(k
−
j
+
i)!
i
i=0
k=0
which gives
!
j X
j ∞
X
s (k − j + i, i)
j
x − log (1 + x)
(−1)k−j+i i!
xk−2j
2j
2
=
2
i
x
(k − j + i)!
i=0
k=0
!
j ∞
X
X
s
(k
+
j
+
i,
i)
j
k−j+i
(−1)
i!
=
2j
xk ,
i
(k
+
j
+
i)!
i=0
k=0
13
1.2. Laplace’s Method
that is
Aj,k
2 1
2
− , , . . . , (−1)k
3 2
k+2
=2
j
j X
j
i
i=0
(−1)k−j+i i!
s (k + j + i, i)
.
(k + j + i)!
Finally, by formula (1.16) we find
cn =
Γ
3n+1
+1
2
n+1
Γ 2
n
X
j=0
j
X (−1)n+i s (n + j + i, i)
2n/2+j−1/2
n+1
+ j (n − j)! i=0 (j − i)! (n + j + i)!
2
n
j
X
+ 23 X
Γ 3n
(−1)n+j−i s (n + 2j − i, j − i)
2n/2+j+1/2
2
.
=
(n + 2j + 1) (n − j)! i=0
i! (n + 2j − i)!
Γ n+1
2
j=0
From this and the expression (1.19) it follows that
j
X
2n+j+1 Γ 3n + 23
(−1)j−i s (2n + 2j − i, j − i)
√
γn =
i! (2n + 2j − i)!
π (2n + 2j + 1) (2n − j)! i=0
j=0
j−n
3n
X
(−1)n 2j+1 Γ 3n + 23 X (−1)j−i s (2j − i, j − n − i)
√
,
=
i!
(2j
−
i)!
π
(2j
+
1)
(3n
−
j)!
i=0
j=n
2n
X
♣
which is much simpler than (1.20).
Remark. The substitution x = log(t/λ) in (1.17) leads to the form
Z +∞
Z +∞
Γ (λ)
−x
−λ(ex −x−1)
=
e
dx +
e−λ(e +x−1) dx.
λ
−λ
λ e
0
0
Similar procedures to that described above give (1.18) with
"
n+1/2 #
1
d2n 1
x2
γn = n
2 n! dx2n 2 ex − x − 1
x=0
j
j−i
2n
1 X
n+j
X
X
2 Γ n+j+ 2
(−1)j+i+r S (2n + 2j − 2i − r, j − i − r)
1
√
=
2i i! r=0
r! (2n + 2j − 2i − r)!
π
i=0
j=0
j−n
3n
X
(−1)n 2j+1 Γ 3n + 23 X (−1)j−i S (2j − i, j − n − i)
√
,
=
i! (2j − i)!
π (2j + 1) (3n − j)! i=0
j=n
using formula (1.9), (1.11) and (1.16) respectively. Here S(n, k) denotes the Stirling
Numbers of the Second Kind.
Example 1.2.2. Our second example is the family of the Modified Bessel Functions of the Second Kind Kν . Suppose that ν, t > 0, then we have the integral
representation
Z
1 +∞ −ν(t cosh s−s)
Kν (νt) =
e
ds.
2 −∞
14
1.2. Laplace’s Method
The substitution s = sinh−1 (1/t) + x gives
!ν Z
√
+∞
√
2
1
t +1+1
2
Kν (νt) =
e−ν ( t +1 cosh x+sinh x−x) dx
2
t
−∞
!ν
√
Z +∞
√
√
t2 + 1 + 1
1
2
−ν t2 +1
e−ν ( t +1(cosh x−1)+sinh x−x) dx,
=
e
2
t
−∞
and by splitting up the integral
2
√
!−ν
√
t2 + 1 + 1
2
eν t +1 Kν (νt)
t
Z +∞
Z
√
−ν ( t2 +1(cosh x−1)+sinh x−x)
e
=
dx +
0
+∞
e−ν (
√
t2 +1(cosh x−1)−sinh x+x)
dx.
0
√
Using
Theorem
1.2.1
with
f
(x)
=
t2 + 1 (cosh x − 1) + sinh x − x (and f (x) =
√
2
t + 1 (cosh x − 1) − sinh x + x), g(x) ≡ 1, α = 2 and β = 1; one finds that
Z +∞
∞
√
X
n+1
cn
−ν ( t2 +1(cosh x−1)+sinh x−x)
dx ∼
Γ
e
(n+1)/2
2
ν
0
n=0
and
Z
+∞
e
√
−ν ( t2 +1(cosh x−1)−sinh x+x)
0
dx ∼
∞
X
Γ
n=0
n+1
2
(−1)n cn
,
ν (n+1)/2
as ν → +∞, where, by Theorem 1.2.2,
"
(n+1)/2 #
dn
x2
1
√
cn =
2n! dxn
t2 + 1 (cosh x − 1) − sinh x + x
.
x=0
Finally,
Kν (νt) ∼
√
t2 + 1 + 1
t
!ν
e
√
−ν t2 +1
r
∞
X (−1)n Un (τ )
π
√
,
νn
2ν t2 + 1 n=0
(1.21)
as ν → +∞, where
r
1
2√ 2
n
c2n
(1.22)
Un (τ ) = (−1)
t + 1Γ n +
π
2
"
n+1/2 #
1/4
2
2n
2
(t
+
1)
d
x
√
= (−1)n 2n+1/2
2
n! dx2n
t2 + 1 (cosh x − 1) − sinh x + x
x=0
"
n+1/2 #
n
2n
2
(−1)
d
x
= 2n+1/2 1/2
,
2n
−1
2
n!τ
dx
τ (cosh x − 1) − sinh x + x
x=0
and τ = (t2 + 1)−1/2 . It is known that the Un (τ )’s are polynomials in τ of degree
3n. The first few are given by U0 (τ ) = 1 and
U1 (τ ) = −
5 3 1
τ + τ,
24
8
15
1.2. Laplace’s Method
385 6
77 4
9 2
τ −
τ +
τ ,
1152
192
128
75 3
85085 9 17017 7 4563 5
τ +
τ −
τ +
τ .
U3 (τ ) = −
82944
9216
5120
1024
Thus, the expansion (1.21) holds uniformly for t > 0. Since
U2 (τ ) =
√
t2
+ 1 (cosh x − 1) − sinh x + x =
we have
∞
X
k=1
∞
X
1
1
2k
x −
x2k+1 ,
τ (2k)!
(2k
+
1)!
k=1
1
1
, a2k+1 = −
for k ≥ 0.
τ (2k + 2)!
(2k + 3)!
From the CFWW Formula (1.11) we obtain
n
X
(−1)j (2τ )(n+1)/2+j
1
1 1
n+1
cn =
Bn,j − ,
, . . . , an−j+1 Γ
+j ,
j!
6 24τ
2
Γ n+1
2
j=0
a2k =
from which it follows by (1.22) that
Un (τ ) =
2n
X
j=0
(−1)
n+j
2n+j+1 τ n+j
√
Γ
j! π
1
n+j+
2
B2n,j .
1
, . . . , an−j+1 , and from the recurrence of the Partial
Here Bn,j = Bn,j − 61 , 24τ
Ordinary Bell Polynomials, Bn,0 = 0, Bn,1 = an and
1
1
Bn,j = − Bn−1,j−1 +
Bn−2,j−1 + · · · + an−j+1 Bj−1,j−1 .
6
24τ
In this case, it is quite complicated to give any explicit formula for the Partial
Ordinary Bell Polynomials. Nevertheless, formula (1.16) enables us to derive a
fairly explicit expression for the polynomials Un (τ ). To use formula (1.16)
we
τ 1
have to compute the Ordinary Potential Polynomials Aj,k − 3 , 12 , . . . , 2τ ak . The
generating function is
j X
∞
τ −1 (cosh x − 1) − sinh x + x
τ 1
2τ
=
Aj,k − , , . . . , 2τ ak xk .
x2
3 12
k=0
Algebraic manipulation and the generating function of the 1-associated Stirling
Numbers of the Second Kind (see expression (A.5)) yields
j
τ −1 (cosh x − 1) − sinh x + x
j
1
1
1
1
x
−x
= (e − x − 1)
+ e +x−1
−
+
2τ
2
2τ
2
j i
j−i
X j
j−i
1
1
1
1
−
+
(ex − x − 1)i e−x + x − 1
=
2τ
2
2τ
2
i
i=0
i j−i X
j ∞
∞
X
1
1
xm
xk X
1
1
m
−
+
(−1) S1 (m, j − i)
S1 (k, i)
= j!
2τ
2
2τ
2
k! m=0
m!
i=0
k=0
!
i j−i
j k−2j+2i
∞
X
X
X
1
1
1
1
S1 (m, i) S1 (k − m, j − i)
k−m
(−1)
= j!
−
+
xk ,
2τ
2
2τ
2
m!
(k
−
m)!
i=0 m=2i
k=0
16
1.2. Laplace’s Method
which gives
j
τ −1 (cosh x − 1) − sinh x + x
2τ
x2
= j!
j k−2j+2i
∞
X
X
X
k=0
= j!
∞
X
k=0
(−1)
i=0 m=2i
j k+2i
X
X
(−1)
k−m
k−m
i=0 m=2i
i
(1 − τ ) (1 + τ )
i
(1 − τ ) (1 + τ )
j−i
j−i
S1 (m, i) S1 (k − m, j − i)
m! (k − m)!
!
xk−2j
S1 (m, i) S1 (k + 2j − m, j − i)
m! (k + 2j − m)!
!
xk ,
that is
τ 1
Aj,k − , , . . . , 2τ ak
3 12
j k+2i
X
X
S1 (m, i) S1 (k + 2j − m, j − i)
.
= j!
(−1)k−m (1 − τ )i (1 + τ )j−i
m!
(k
+
2j
−
m)!
i=0 m=2i
Finally, from formula (1.16), we find
n
1
3n+1
+ 1 X (−1)j Aj,n − τ3 , 12
, . . . , 2τ an
(n+1)/2 Γ
2
cn = (2τ )
n+1
j! (n − j)!
2Γ n+1
+
j
2
2
j=0
n
3n
1
+ 23 X
, . . . , 2τ an
Aj,n − τ3 , 12
(−1)j
(n+1)/2 Γ
2
= (2τ )
,
(n
+
2j
+
1)
j!
(n
−
j)!
Γ n+1
2
j=0
and thus by (1.22),
2 (2τ )n Γ 3n +
√
Un (τ ) = (−1)
π
n
where
def
un (i, j) =
2n+2i
X
m=2i
3
2
(−1)m
2n
X
(−1)
j=0
j
Pj
un (i, j) (1 − τ )i (1 + τ )j−i
,
(2n + 2j + 1) (2n − j)!
i=0
S1 (m, i) S1 (2n + 2j − m, j − i)
.
m! (2n + 2j − m)!
♣
Remark. It can be shown that the Modified Bessel Functions of the First Kind
Iν have a similar asymptotic expansion:
√
ν
∞
2
X
eν t +1
t
Un (τ )
p
,
Iν (νt) ∼ √
√
n
t2 + 1 + 1
2πν t2 + 1 n=0 ν
as ν → +∞, uniformly for t > 0. The coefficients Un (τ ) are the same as above. It
is known that a recurrence for the polynomials Un (τ ) is given as follows
Z
′
1 τ
1 2
2
5x2 − 1 Un (x) dx,
Un+1 (τ ) = τ 1 − τ Un (τ ) −
2
8 0
with U0 (τ ) = 1 [8, p. 376].
17
1.2. Laplace’s Method
Example 1.2.3. As a last example we take the Legendre Polynomials Pm . Suppose that t > 1, then we have the integral representation
Z
m
√
1 π
Pm (t) =
t + cos x t2 − 1 dx.
π 0
The substitution t = cosh θ (θ > 0) and a little algebraic manipulation gives
Z
emθ π −m(− log(1−sin2 ( x2 )(1−e−2θ )))
e
dx.
Pm (cosh θ) =
π 0
Using Theorem 1.2.1 with f (x) = − log 1 − sin2 x2 1 − e−2θ , g(x) ≡ 1, α = 2
and β = 1; one finds that
Z π
∞
X
cn
n+1
−m(− log(1−sin2 ( x2 )(1−e−2θ )))
dx ∼
Γ
e
(n+1)/2
2
m
0
n=0
as m → +∞, where, by Theorem 1.2.2,

!(n+1)/2 
n
2
1 d
x

cn =
2 x
n
2n! dx
− log 1 − sin 2 (1 − e−2θ )
.
x=0
It is not hard to see that cn = (−1)n cn , thus
e(m+1)θ
as m → +∞, where
Pm (cosh θ) ∼ p
πm (e2θ − 1)
∞
X
ρn (θ)
n=0
mn
,
1
1 p
−2θ
c2n
ρn (θ) = √
1−e Γ n+
2
π

!n+1/2 
√
2n
2
−2θ
1−e
x
d

= 2n+1
2 x
2n
2
n!
dx
− log 1 − sin 2 (1 − e−2θ )
(1.23)
(1.24)
.
x=0
The first few are ρ0 (θ) = 1 and
ρ1 (θ) = −
e2θ − 3
5e6θ + 35e4θ + 455e2θ + 105
e4θ + 10e2θ + 25
,
ρ
(θ)
=
.
,
ρ
(θ)
=
3
2
8 (e2θ − 1)
128 (e2θ − 1)2
1024 (e2θ − 1)3
Since
k
∞
X
1 − e−2θ
2 x
2k x
−2θ
− log 1 − sin
sin
=
1−e
2
k
2
k=1
!
k ∞
∞
k
X
X
1 − e−2θ X
(−1)k+j
2k
i
2j+2k
2 k+j
i
=
x
(−1)
k−i
k
22k−1 (2k + 2j)! i=1
j=0
k=1
j+1 j+1
!
∞
k−1
X
X
X
(−1)k 1 − e−2θ
2j
+
2
(−1)i
i2k x2k ,
=
2j+1
j−i+1
2
(j + 1) (2k)! i=1
j=0
k=1
18
1.2. Laplace’s Method
we have
a2k
j+1 j+1
k
X
X
(−1)k+1 1 − e−2θ
2j + 2
i
=
i2k+2
(−1)
2j+1
j−i+1
2
(j + 1) (2k + 2)! i=1
j=0
and a2k+1 = 0 for k ≥ 0. From the CFWW Formula (1.11) we obtain
n
X
1
(−1)j 4(n+1)/2+j
n+1
cn =
+ j Bn,j ,
Γ
−2θ )(n+1)/2+j
2
2Γ n+1
2
j=0 j! (1 − e
with Bn,j
3e−4θ −4e−2θ +1
= Bn,j 0,
, . . . , an−j+1 . From this and (1.24) it follows that
96
2n
X
4n+j
ρn (θ) =
(−1) √
Γ
j! π (1 − e−2θ )n+j
j=0
j
1
n+j+
2
B2n,j .
By the recurrence of the Partial Ordinary Bell Polynomials, B2n,0 = 0, B2n,1 = a2n
and
3e−4θ − 4e−2θ + 1
B2n−2,2j−1 + · · · + a2n−2j B2j,2j ,
B2n,2j =
96
3e−4θ − 4e−2θ + 1
B2n−2,2j + · · · + a2n−2j B2j,2j
B2n,2j+1 =
96
for j ≥ 0. If one thinks of the generating function of these Partial Ordinary Bell
Polynomials, it is clear that giving an explicit formula for them would be quite
complicated. On the other hand, formula (1.16) enables us to give an explicit
expression for the coefficients ρn (θ) in terms of the Stirling Numbers of the First
Kind. To use formula (1.16) wehave to compute the Ordinary Potential Polyno−2θ
4ak
mials Aj,k 0, 1−3e
, . . . , 1−e
. The generating function is
−2θ
24
!j
∞
X
− log 1 − sin2 x2 1 − e−2θ
1 − 3e−2θ
4ak
xk .
=
Aj,k 0,
,...,
4
2
−2θ
−2θ
x (1 − e )
24
1−e
k=0
To compute the corresponding Ordinary Potential Polynomials, note that
k
∞
x
j X
1 − e−2θ
2 x
−2θ
− log 1 − sin
sin2k
1−e
=
j!s (k, j)
2
k!
2
k=1
!
k ∞
k
∞
k+m
X
X
1 − e−2θ X
2k
(−1)
i
2k+2m
2 k+m
i
x
(−1)
=
j!s (k, j)
k−i
k!
22k−1 (2k + 2m)! i=1
m=0
k=1
!
m m
k
∞
X
X
(−1)k j!s (m, j) 1 − e−2θ X
2m
i
i2k x2k ,
=
(−1)
2m−1 m! (2k)!
m
−
i
2
m=0
i=1
k=1
for j ≥ 1, which gives
4a2k
1 − 3e−2θ
Aj,2k 0,
,...,
24
1 − e−2θ
k+j
m
X
2m
(−1)k+j 22j−2m+1 j!s (m, j) X
i
i2k+2j ,
(−1)
=
−2θ )j−m m! (2k + 2j)!
m
−
i
(1
−
e
m=0
i=1
19
1.2. Laplace’s Method
for k, j ≥ 1. Finally, formula (1.16) yields
cn =
=
2n+1 Γ
1
n+1
2
2Γ
n! (1
n+1
3
3n
+2
2
(n+1)/2
e−2θ )
2n+1 Γ
1
Γ
n+1
+n+1
2
− e−2θ )(n+1)/2
n
X
n
X
j=0
n
(−1)j
Aj,n
n+1
+j j
2
Aj,n
(−1)j
,
(n + 2j + 1) (n − j)!j!
(1 −
j=0
−2θ
4an
= Aj,n 0, 1−3e
,
.
.
.
,
. From this and the expression
24
1−e−2θ
2
for n ≥ 1 with Aj,n
(1.24) we deduce
2n
n+j
m
X
Γ 3n + 32 X (−1)n X
22n+2j−2m+2 s (m, j)
2m
i
√
ρn (θ) =
i2n+2j
(−1)
n+j−m
−2θ
m−i
(2n − j)! m=0 (1 − e )
π
m! (2n + 2j + 1)! i=1
j=0
3n
j
m
X
Γ 3n + 23 X (−1)n X
4j−m+1 s (m, j − n)
2m
i
√
=
i2j ,
(−1)
j−m
−2θ
m
−
i
(3n
−
j)!
π
)
m! (2j + 1)! i=1
m=0 (1 − e
j=n
(1.25)
for n ≥ 1, with ρ0 (θ) = 1.
♣
Remark. Another possible way to derive the asymptotic expansion (1.23) is to
use the generating function
∞
X
1
√
=
Pm (cosh θ) xm , θ > 0.
1 − 2x cosh θ + x2 m=0
A simple algebraic manipulation yields
√
∞
X
1
eθ
√
=
e−mθ Pm (cosh θ) z m .
2θ
1 − z e − z m=0
This generating function has an algebraic singularity at z = 1, the full asymptotic
expansion can now be carried out by Darboux’s Method (see, e.g., [8, p. 309], [14, p.
116]). The coefficients ρn (θ) this time take the form
2 n
X
2n − 2k
n − k + 1/2
(−1)k (2n − 4k + 1)
(n−k+1/2)
Bk
,
ρn (θ) =
n−k
4n−4k
2θ
n−k
k
2
(2n − 2k + 1) (e − 1)
k=0
(µ)
where Bn denote the Generalized Bernoulli Numbers. Using the explicit formula
[5]
k+µ k−µ
k
X
k+j
k−j
(µ)
S (k + j, j),
Bk =
k−µ
j=0
k
we get the following expression involving the Stirling Numbers of the Second Kind:
k 2 n
X
(−1)k (2n − 4k + 1) 2n − 2k
2k (−1)j S (k + j, j)
n + 1/2 X
,
ρn (θ) =
k − j 2n − 2k + 2j + 1
2k
n−k
24n−4k (e2θ − 1)n−k
j=0
k=0
which is even simpler than (1.25).
20
Chapter 2
Complex Laplace-type integrals
In this chapter, we investigate the asymptotic behavior of Laplace-type integrals
with complex parameter. The integrals under consideration are of the form
Z
I (λ) =
eλf (z) g (z) dz,
(2.1)
C
where the path of integration C is a contour in the complex plane and λ is a large
real parameter. The functions f and g are independent of λ and holomorphic in a
domain containing the path of integration.
Chapter 2 is organized as follows. In the first section, we revisit the Method of
Steepest Descents, a well-known procedure in the asymptotic theory of complex
Laplace-type integrals. In the second part, we present Perron’s Method that avoids
the computation of the path of steepest descent. Finally, in the third section, we
give three examples to demonstrate the application of these methods.
2.1
The Method of Steepest Descents
Consider the integral (2.1) with the assumptions we made. The basic idea of the
method is to deform the contour of integration C into a new path of integration
C ′ so that the following conditions hold:
(i) C ′ passes through one or more zeros of f ′ ;
(ii) the function Im(f ) is constant on C ′ .
The choice of a path with Im(f ) = constant has two major advantages. It removes
the rapid oscillations of the integrand and on such paths Re(f ) changes the most
rapidly (see, e.g., [9, p. 5]). Thus, the dominant contribution will arise from a
neighborhood of the point where Re(f ) is the greatest.
In order to obtain a geometrical interpretation of the method and the new path
of integration C ′ , we introduce the notations
f (z) = u(x, y) + iv(x, y)
with z = x + iy, and the functions u and v are real. Suppose that z0 = x0 + iy0
is a zero of f ′ . It follows easily that (x0 , y0 ) is a critical point of u, and since u is
harmonic, it must be a saddle point of u. For this reason, we call z0 a saddle point
of f .
21
2.1. The Method of Steepest Descents
Consider the surface F in the (x, y, u) space defined by u = u(x, y). The shape
of the surface F can also be represented on the (x, y) plane by drawing the level
curves on which u is constant. From the Cauchy–Riemann equations it follows
that the families of curves corresponding to constant values of u(x, y) and v(x, y)
are orthogonal at all their points of intersection [9, p. 6]. The regions where
u(x, y) > u(x0 , y0 ) are called hills and those where u(x, y) < u(x0 , y0 ) are called
valleys. The level curve through the saddle, u(x, y) = u(x0 , y0 ), separates the
neighborhood of the saddle point (x0 , y0 ) into a series of hills and valleys.
Suppose that z0 is a saddle point of order m − 1, m ≥ 2, i.e.,
f ′ (z0 ) = f ′′ (z0 ) = · · · = f (m−1) (z0 ) = 0,
with f (m) (z0 ) = aeiϕ , a > 0. Then, if z = z0 + reiθ , r > 0, we have
rm i(mθ+ϕ)
ae
+ ···
f (z) = f (z0 ) +
m!
and hence, near the saddle point z0 the level curves and steepest paths are roughly
the same as
rm
a cos (mθ + ϕ) ,
u (x, y) = u (x0 , y0 ) +
m!
rm
v (x, y) = v (x0 , y0 ) +
a sin (mθ + ϕ) .
m!
The directions of the level curves where u is constant, are given by the solutions
of the equation cos (mθ + ϕ) = 0, i.e.,
ϕ
1 π
θ=− + k+
, k = 0, 1, . . . , 2m − 1.
m
2 m
Similarly, the directions of the steepest paths satisfy sin (mθ + ϕ) = 0, i.e.,
θ=−
ϕ
π
+k ,
m
m
k = 0, 1, . . . , 2m − 1.
Therefore, there are 2m equally spaced steepest directions from z0 : m directions
of steepest descent and m directions of steepest ascent. In the neighborhood of z0 ,
the level curves u = u(x0 , y0 ) form the boundaries of m valleys surrounding the
saddle point, in which cos (mθ + ϕ) < 0, and m hills on which cos (mθ + ϕ) > 0.
The valleys and hills are situated respectively entirely below and above the saddle
point, and each has angular width equal to π/m.
Now suppose that the path of integration C in (2.1) is a steepest path through
the saddle point z0 of order m − 1. On this path we have
f (z) = f (z0 ) − τ,
where τ is non-negative and monotonically increasing as one progresses down the
path of steepest descent. Then (2.1) becomes
Z T
Z T′
dz
λf (z0 )
−λτ
λf (z0 )
e−λτ g (z) dτ ,
I (λ) = e
e g (z) dz = e
(2.2)
dτ
z0
0
where T denotes some point on the steepest descent path and T ′ > 0 is the map
of T in the τ -plane. Since, for large positive λ, the factor e−λτ decays rapidly, the
22
2.1. The Method of Steepest Descents
main contribution comes from the neighborhood of τ = 0. Thus, we can apply
Watson’s Lemma to the integral in the right-hand side of (2.2). To do this, we
dz
first require a series expansion for g (z) dτ
in ascending powers of τ . Near z0 ,
f (z) = f (z0 ) −
that is
τ=
∞
X
k=0
∞
X
k=0
m
ak (z − z0 )m+k ,
a0 6= 0,
ak (z − z0 )m+k
as z → z0 . If we put τ = υ , then
1/m
υ = a0
(z − z0 ) ϕ (z) ,
1/m
where ϕ is analytic around z0 with ϕ(z0 ) 6= 0 and a0 takes its principal value.
Hence, υ is a single-valued analytic function of z in a neighborhood of z0 and that
υ ′ (z0 ) 6= 0. Therefore, by the inverse function theorem (see [2, p. 121])
z − z0 =
∞
X
k
αk υ =
k=1
∞
X
αk τ k/m ,
k=1
where
α1 =
1
1/m
a0
,
α2 = −
a1
1+2/m
ma0
,
α3 =
(m + 3) a21 − 2ma0 a2
2+3/m
2m2 a0
,... .
dz
Furthermore, for small τ , g (z) dτ
has a convergent expansion of the form
∞
dz X
g (z)
=
cn τ (n−m+1)/m .
dτ
n=0
(2.3)
The first three coefficients cn are given explicitly by
b0
b1 2a1 b0
1
,
−
c0 =
,
c1 = 2/m
1/m
m m 2 a0
ma0
a0
and
c2 =
1
3/m
a0
b0
b2 3a1 b1
− 2 + (m + 3) a21 − 2ma0 a2
m m a0
m2 a20
.
Here n!bn = g (n) (z0 ). We are now in a position to derive the asymptotic expansion
of the integral (2.2) taken from the saddle point z0 of order m − 1 down one of
the m valleys. Depending on the path C , we choose one of the m valleys, labelled
by l, say. We replace τ by τ e2πil in the expansion (2.3) and substitute it into the
integral (2.2). The asymptotic expansion of I(λ) can now be obtained directly
from Watson’s Lemma:
∞
X
n + 1 cn e2πil(n+1)/m
λf (z0 )
,
(2.4)
I (λ) ∼ e
Γ
(n+1)/m
m
λ
n=0
23
2.2. Perron’s Method
as λ → +∞. To obtain an explicit formula for the coefficients cn we observe that
1−m dz
dz
with τ = υ m , so that dτ
= υ m dυ
, we find from (2.3)
g (z)
∞
X
dz
=m
cn υ n .
dυ
n=0
Cauchy’s formula then shows that
Z
Z
1
g (z)
1
dz dυ
cn =
=
g (z)
dz,
2πim γ
dυ υ n+1
2πim γ ′ (f (z0 ) − f (z))(n+1)/m
with an appropriate branch of (f (z0 ) − f (z))1/m . Here γ and γ ′ are simple closed
contours with positive orientation that enclose the points υ = 0 and z = z0 ,
respectively. This is Dingle’s Formula [3, p. 119]. It follows that
"
(
(n+1)/m )#
dn
(z − z0 )m
1
g (z)
cn =
mn! dz n
f (z0 ) − f (z)
"
# z=z0
(2.5)
n
(n+1)/m
m
(z − z0 )
1 X bn−k dk
,
=
m k=0 k! dz k f (z0 ) − f (z)
z=z0
which is Perron’s Formula. In the next subsection we derive this formula in more
general conditions.
Remark. It can be shown that the asymptotic expansion (2.4) is also valid when
|arg λ + arg a0 + mω − 2πl| ≤
and |λ| → +∞, where 0 < δ ≤
π
2
π
π
−δ <
2
2
is fixed and
ω = lim arg (z − a)
C ∋z→a
is the angle of slope of C at a.
2.2
Perron’s Method
In the previous subsection, we presented a method, namely, the Method of Steepest
Descents, to derive asymptotic expansions for integrals of the form
Z
I (λ) =
eλf (z) g (z) dz.
(2.6)
C
However, in many specific cases, the construction of steepest descent path can be
extremely complicated. In this subsection, we shall describe a method due to Perron that avoids the computation of the path of steepest descent. Our presentation
of Perron’s Method follows closely that given by Wong [14, p. 103].
In this subsection we allow λ to be complex. Let a be the starting point, and b
be the endpoint of the continuous curve C (b can by finite or infinite). We impose
the following four conditions:
24
2.2. Perron’s Method
(i) The path C lies in the sector
|arg(λ(f (a) − f (z)))| ≤
π
− δ,
2
where δ is a fixed positive number.
(ii) For each point c 6= a in C , there exists a number η = η(c), such that
|f (a) − f (z)| ≥ η > 0 for all z on the portion of C joining c to b.
(iii) In a neighborhood of a,
f (a) − f (z) =
∞
X
k=0
ak (z − a)k+α ,
(2.7)
where a0 6= 0 and α > 0. Since α is not necessarily a positive integer, f is
need not be analytic at a. We make (z − a)α single-valued by introducing a
cut in the z-plane from a to infinity along a convenient radial line.
(iv) The contour C must not cross this cut. Furthermore, suppose that there
exists a point c′ 6= a on C such that for any c on C with |ac| < |ac′ |, the
portion of C from a to c can be deformed into the straight line arg(z − a) =
arg(c − a).
Let
ω = lim arg (z − a)
C ∋z→a
be the angle of slope of C at a. In order to condition (i) holds, the set through
which |λ| → +∞ must be contained in the sector
1
1
2l −
π + δ ≤ arg λ + arg a0 + αω ≤ 2l +
π−δ
(2.8)
2
2
for some fixed δ in 0 < δ ≤ π2 , where l is a fixed integer.
Theorem 2.2.1. Consider the integral (2.6), and assume that it exists absolutely
for every fixed λ satisfying the inequalities in (2.8). Furthermore, assume that in
a neighborhood of a,
∞
X
g (z) =
bk (z − a)k+β−1 , b0 6= 0,
k=0
with some fixed complex number β, Re(β) > 0. Then, under the conditions (i) to
(iv), we have
∞
X
n + β cn e2πli(n+β)/α
λf (a)
I (λ) ∼ e
Γ
,
α
λ(n+β)/α
n=0
as |λ| → +∞, uniformly with respect to arg λ confined to the sector (2.8). The
coefficients cn are given by
"
(
(n+β)/α )#
dn
a0 (z − a)α
1
G (z)
cn =
(n+β)/α
n
f (a) − f (z)
αa0
n! dz
# z=a
"
(2.9)
n
(n+β)/α
α
k
X bn−k d
1
a0 (z − a)
=
,
(n+β)/α
k! dz k f (a) − f (z)
αa0
k=0
z=a
def
where G (z) =
P∞
k=0 bk
(z − a)k .
25
2.2. Perron’s Method
Proof. Let F be the function
∞
X
ak
def
F (z) = −
k=1
a0
(z − a)k
which is analytic at z = a and satisfies
f (z) = f (a) − a0 (z − a)α (1 − F (z))
(2.10)
in a neighborhood of a. Since the function G (z) exp (wF (z)) is, for each fixed w,
analytic at z = a, we have
G (z) exp (wF (z)) =
∞
X
k=0
Pk (w) (z − a)k ,
|z − a| < ρ,
for sufficiently small ρ, where Pk (w) is a polynomial in w whose degree does not
exceed k, and
1 dk
(G (z) exp (wF (z)))
Pk (w) =
.
(2.11)
k! dz k
z=a
Using (2.10), (2.11) can be written as
f (z) − f (a)
1 dk
−w
G (z) exp w
e Pk (w) =
.
k! dz k
a0 (z − a)α
z=a
(2.12)
By Cauchy’s inequality
|Pk (w)| ≤
1
max |G (z) exp (wF (z))| ,
rk |z−a|=r
for 0 < r < ρ. Since F vanishes at a, for every 0 < r < ρ,
|Pk (w)| ≤
1
M1 exp (M2 |w| r) ,
rk
where M1 and M2 are fixed numbers. Thus, for any fixed N > 0,
G (z) exp (wF (z)) =
N
X
k=0
Pk (w) (z − a)k + RN ,
(2.13)
where, for |z − a| ≤ r < ρ,
|RN | ≤ M3 |z − a|N +1 exp (M2 |w| r) ,
where M3 is a fixed number depending on N . Now, we put w = λa0 (z − a)α in
(2.13). It follows that
g (z) exp (λf (z) − λf (a)) =
N
X
k=0
e−w Pk (w) (z − a)k+β−1 + e−w (z − a)β−1 RN
(2.14)
and, for |z − a| ≤ r < ρ,
−w
β−1
RN ≤ M3 (z − a)N +β exp (−Re (w) + M2 |w| r) .
e (z − a)
26
(2.15)
2.2. Perron’s Method
We write the integral (2.6) in the form
I (λ) =
Z
c
e
λf (z)
g (z) dz +
a
Z
b
eλf (z) g (z) dz,
(2.16)
c
where c is a point on C such that the portion of C form a to c can be deformed
into the radial line arg(z − a) = arg(c − a) (see condition (iv)). Furthermore, we
require |c − a| < r so that the expansion (2.14)-(2.15) holds for all z on the new
path of integration from a to c. Inserting (2.14) in the first integral on the right
of (2.16) yields
" N
#
Z c
X
eλf (z) g (z) dz = eλf (a)
Ik (λ) + EN (λ) ,
(2.17)
a
k=0
where
def
Ik (λ) =
and
Z
c
a
def
EN (λ) =
e−w Pk (w) (z − a)k+β−1 dz
Z
c
a
(2.18)
e−w (z − a)β−1 RN dz.
By choosing c closer to a if necessary, we also have from (2.8)
1
1
2l −
π + δ0 ≤ arg λ + arg a0 + α arg (c − a) ≤ 2l +
π − δ0
2
2
for some 0 < δ0 < δ. This implies that
1
1
π + δ0 ≤ arg w ≤ 2l +
π − δ0 ,
2l −
2
2
(2.19)
and thus Re(w) ≥ |w| sin δ0 . Thus we obtain from (2.15)
−w
N +β β−1
RN ≤ M3 (z − a)
exp (− |w| sin δ0 + M2 |w| r)
e (z − a)
for the points on the path of integration. If we choose r such that r < sin δ0 /M2 ,
then there will exist a constant M4 > 0 such that for all z with arg(z − a) =
arg(c − a) and |z − a| ≤ r < ρ,
−w
N +β β−1
RN ≤ M3 (z − a)
exp (−M4 |w|) ,
e (z − a)
and therefore,
Z c
α
N +β |EN (λ)| ≤ M3
exp (−M4 |λa0 (z − a) |) |dz|.
(z − a)
a
If we let θ = arg(c − a) and z − a = ρeiθ , then
Z +∞
β+N exp (−M4 |λa0 | ρα ) dρ = O λ−(β+N +1)/α .
ρ
|EN (λ)| ≤ M3
0
27
(2.20)
2.2. Perron’s Method
We now turn our attention to the integrals Ik (λ) in (2.18). It is not hard to
show that
Z a+∞eiθ
e−w Pk (w) (z − a)k+β−1 dz = O e−ε|λ|
c
for some ε > 0. Next we note, from (2.19), that w = λa0 (z − a)α gives
−1/α −1/α
z − a = w1/α e2πli/α a0
and
λ
−(k+β−1)/α −(k+β−1)/α
(z − a)k+β−1 = w(k+β−1)/α e2πli(k+β−1)/α a0
Hence,
1 −(k+β)/α −(k+β)/α 2πli(k+β)/α
λ
e
Ik (λ) = a0
α
Z
∞eiθ
0
λ
.
′
e−w Pk (w) w(k+β)/α−1 dw+O e−ε|λ| ,
where, for fixed λ, |θ′ | < π2 . Now, we deform the path of integration into the
positive real axis. Therefore
Z
1 −(k+β)/α −(k+β)/α 2πli(k+β)/α +∞ −w
λ
e
Ik (λ) = a0
e Pk (w) w(k+β)/α−1 dw + O e−ε|λ| .
α
0
We now insert (2.12) into the integral, and then interchange the order of integration
and differentiation. This leads to
k + β ck e2πli(k+β)/α
−ε|λ|
Ik (λ) = Γ
.
+
O
e
α
λ(k+β)/α
Using this and (2.20) in (2.17) yields
Z
c
e
λf (z)
g (z) dz = e
λf (a)
a
N
X
k=0
Γ
k+β
α
ck e2πli(k+β)/α
+O
λ(k+β)/α
1
λ(N +β+1)/α
!
.
(2.21)
What remains is to consider the second integral on the right-hand side of (2.16).
We choose a point λ0 that satisfies the inequalities in (2.8) and write
Z b
Z b
λf (z)
e
g (z) dz =
e(λ−λ0 )f (z) eλ0 f (z) g (z) dz,
c
whence
c
Z b
Z
(λ−λ )f (z) b λ f (z)
λf (z)
0
e 0 g (z) |dz|,
e
g (z) dz ≤ max e
c
c
where the maximum is taken over all points on the portion of C joining c to b. By
assumption, the integral on the right exist, thus
Z b
λf (z)
≤ K max e(λ−λ0 )f (z) e
g
(z)
dz
c
for some constant K > 0. It follows that
Z b
λf
(z)
e
g (z) dz ≤ Ke−λ0 f (a) eλf (a) max e−(λ−λ0 )(f (a)−f (z)) .
c
28
(2.22)
2.3. Examples
Then
Re ((λ − λ0 ) (f (a) − f (z)))
= |f (a) − f (z)| {|λ| cos [arg (λ (f (a) − f (z)))] − |λ0 | cos [arg (λ0 (f (a) − f (z)))]}
= |λ| |f (a) − f (z)| {cos [arg (λ (f (a) − f (z)))] + o (1)} ,
as |λ| → +∞. Thus, by conditions (i) and (ii),
Re ((λ − λ0 ) (f (a) − f (z))) ≥ |λ| η (sin δ + o (1)) ,
as |λ| → +∞ in the sector (2.8), and consequently,
Re ((λ − λ0 ) (f (a) − f (z))) ≥ ε1 |λ|
for some ε1 > 0, uniformly in arg λ for all λ satisfying (2.8) and uniformly in z for
all z on C from c to b. This together with (2.22) implies that
Z b
λf (z)
= O eλf (a)−ε1 |λ| .
(2.23)
e
g
(z)
dz
c
The result now follows from (2.16), (2.21) and (2.23).
Remark. From (2.7) and (2.9) it can be seen that the representations (1.11) and
(1.16) hold in the complex case too.
2.3
Examples
Example 2.3.1. Our first example is the Reciprocal Gamma Function
Z
1
1
=
et t−λ dt,
Γ (λ)
2πi H
where the path of integration starts at ∞e−πi , goes round the origin once and ends
at ∞eπi . Suppose that λ is real an positive, then the substitution t = λz gives
Z
1
1
=
eλ(z−log z) dz,
λ−1
Γ (λ)
2πiλ
H
with the same path as before. This formula holds also when Re(λ) > 0, provided
that λλ means eλ log λ where log λ has its principal value. Using the notations of
Section 2.1, f (z) = z − log z and g(z) ≡ 1. The function f has only one saddle
point, namely z0 = 1. The steepest paths through z0 are described by the equation
y
y − tan−1
= 0.
x
The equation y = 0 represents the path of steepest ascent whereas the equation
x = y cot y (or rather the branch through (1, 0)) gives the path of steepest descent.
def
Then H can be deformed into C = {(x, y) : x = y cot y, |y| < π}. On the path
of steepest descent, we have
f (z) − f (1) = −1 + z − log z = −τ.
29
(2.24)
2.3. Examples
As z → 1,
f (z) − f (1) =
(z − 1)2 (z − 1)3
−
+ ··· ,
2
3
thus
z± = 1 +
dz±
=
dτ
∞
X
∞
X
(±1)n αn τ n/2 ,
(2.25)
n=0
(±1)n+1 cn τ (n−1)/2 ,
n=0
cn =
n+1
αn+1 ,
2
(2.26)
where z± denotes the upper and lower path of C through the saddle point z0 = 1.
Equation (2.24) defines a many valued function z(τ ) of a complex variable τ with
branch points at τ = 2πin with n ∈ Z \ {0}. Hence, the expansions (2.25)-(2.26)
are converget in the disc |τ | < 2π. Since
dz±
z±
=
dτ
1 − z±
is bounded when τ > 0, Watson’s Lemma is applicable and gives
Z +∞
1
eλ
dz+ dz−
−λτ
dτ
=
−
e
Γ (λ)
2πiλλ−1 0
dτ
dτ
r
∞ r
1 −ic2n
λ X 2
λ −λ
Γ n+
,
∼e λ
2π n=0 π
2
λn
as λ → +∞. From Perron’s Formula (2.5) we find

!(n+1)/2 
"
(n+1)/2 #
2
n+1
n
n
2
i
1 d
d
(z − 1)
z

=
cn =
2n! dz n 1 − z + log z
2n! dz n z − log (1 + z)
z=0
z=1
and hence
"
r
r
n+1/2 #
1
1 (−1)n d2n
z2
2
2
Γ n+
Γ n+
(−ic2n ) =
π
2
π
2 2 (2n)! dz 2n z − log (1 + z)
z=0
n
= (−1) γn .
Here γn denotes the Stirling Coefficients given in Example 1.2.1. Therefore,
r
∞
1
λ X (−1)n γn
λ −λ
∼e λ
,
Γ (λ)
2π n=0
λn
as λ → +∞. This expansion is also valid when |arg λ| ≤
0 < δ ≤ π2 .
π
2
− δ < π2 , with fixed
♣
Example 2.3.2. As a second example we take the Airy Function
Z
t3
1
2
2
eµ t− 3 dt,
Ai µ =
2πi L
30
,
2.3. Examples
4
where µ > 0 and the path L consists of two rays, one with end points ∞e 3 πi and
2
0, and the other with end points 0 and ∞e 3 πi . The substitution t = µz yields
Z
3
µ
µ3 z− z3
2
dz,
Ai µ =
e
2πi L
where L can be taken as the same as in the original integral. Using the notations
3
of Section 2.1, f (z) = z − z3 , g(z) ≡ 1 and λ = µ3 . The saddle points of f are
z0 = −1 and z1 = 1, we choose the former one. The steepest paths through z0 are
described by the equation
y y 2 − 3x2 + 3 = 0.
The equation y = 0 represents the path of steepest ascent whereas the equation
y 2 − 3x2 + 3 = 0 (left branch of a hyperbola) gives the path of steepest descent. It
def
is clear that L can be deformed into C = {(x, y) : y 2 − 3x2 + 3 = 0, x < 0}. We
put
z3 2
+ = −τ.
f (z) − f (−1) = z −
3
3
We have
(z + 1)3
2
f (z) − f (−1) = (z + 1) −
3
as z → −1, thus
∞
X
z± = −1 +
(±1)n αn τ n/2 ,
(2.27)
n=0
dz±
=
dτ
∞
X
(±1)n+1 cn τ (n−1)/2 ,
n=0
cn =
n+1
αn+1 ,
2
(2.28)
where z± denotes the upper and lower path of C through the saddle point z0 = −1.
Since
1
dz±
,
= 2
dτ
z± − 1
and τ (1) = − 43 , the expansions (2.27)-(2.28) are converget in the disc |τ | < 43 .
Watson’s Lemma then gives
Z +∞
dz+ dz−
µ
− 32 µ3 −µ3 τ
2
dτ
e
−
Ai µ =
2πi 0
dτ
dτ
2 3
∞
e− 3 µ X
1 −2ic2n
√ 3n ,
∼ √
Γ n+
2 πµ n=0
2
πµ
as µ → +∞. From Perron’s Formula (2.5) we find
"
(n+1)/2 #
(−i)n+1 dn 1
dn
z −(n+1)/2
3
=
cn =
1−
2n! dz n z − 2
2n!
dz n
3
z=0
z=−1
(−i)n+1 Γ 3n+1
2
,
=
2
3n n!Γ n+1
2
31
2.3. Examples
and thus
Ai µ
2
2 3
∞
1
e− 3 µ X
1
n Γ 3n + 2
∼ √
as µ → +∞.
(−1) √ n
2 πµ n=0
π9 (2n)! µ3n
This expansion is also valid when |arg µ3 | ≤
substitution z = µ2 leads to the final form
π
2
− δ < π2 , with fixed 0 < δ ≤ π2 . The
2 3/2
∞
1
X
e− 3 z
1
n Γ 3n + 2
Ai (z) ∼ √ 1/4
,
(−1) √ n
3n/2
2 πz n=0
π9 (2n)! z
as |z| → +∞ in the sector |arg z| ≤
π
3
− ε < π3 , with fixed 0 < ε ≤ π3 .
♣
Example 2.3.3. Our last example is the integral
Z +∞
1
2
eλ(2z−z ) z ν dz, Re(ν) > −1.
I (λ, ν) = √
π 0
We shall use Perron’s Method to obtain the asymptotic expansion of I (λ, ν) for
fixed ν and |λ| → +∞ in an appropriate sector of the complex plane. We split up
the integral into two parts as follows
Z +∞
Z 1
1
1
2
ν
λ(1−z 2 )
I (λ, ν) = √
e
(1 + z) dz + √
eλ(1−z ) (1 − z)ν dz.
π 0
π 0
Using the notations of Section 2.2, f (z) = 1 − z 2 , g+ (z) = (1 + z)ν , g− (z) =
(1 − z)ν . The only saddle point of f is z0 = 0. We have f (0) − f (z) = z 2 and
∞
X
ν k
z ,
g± (z) =
(±1)
k
k=0
k
as z → 0. We can apply Theorem 2.2.1 with α = 2, β = 1:
Z +∞
∞
X
n+1
cn
ν
λ(1−z 2 )
λ
e
(1 + z) dz ∼ e
Γ
,
(n+1)/2
2
λ
0
n=0
Z
1
e
0
λ(1−z 2 )
ν
(1 − z) dz ∼ e
λ
∞
X
n=0
Γ
n+1
2
(−1)n cn
,
λ(n+1)/2
as |λ| → +∞ in the sector |arg λ| ≤ − δ < π2 , with fixed 0 < δ ≤ π2 . Here
n
1
d
1 ν
ν
.
cn =
(1 + z)
=
2n! dz n
2 n
z=0
π
2
After simplification, the final result is
∞ eλ X ν (2n)! 1
I (λ, ν) ∼ √
,
λ n=0 2n 22n n! λn
as |λ| → +∞ and |arg λ| ≤
π
2
− δ < π2 , with fixed 0 < δ ≤ π2 .
32
♣
Appendix A
Combinatorial objects
A.1
Ordinary Potential Polynomials
P
n
Let F (x) = 1 + ∞
n=1 fn x be a formal power series. For any complex number ρ,
we define the Ordinary Potential Polynomial Aρ,n (f1 , f2 , . . . , fn ) (associated to F )
by the generating function
!ρ
∞
∞
X
X
ρ
n
(F (x)) = 1 +
fn x
=
Aρ,n (f1 , f2 , . . . , fn ) xn .
n=1
n=0
The first few are Aρ,0 = 1, Aρ,1 = ρf1 , Aρ,2 = ρf2 + ρ2 f12 , and in general
X ρ
k!
f1k1 f2k2 · · · fnkn ,
Aρ,n (f1 , f2 , . . . , fn ) =
k k1 !k2 ! · · · kn !
(A.1)
where the sum extending over all sequences k1 , k2 , . . . , kn of non-negative integers
such that k1 + 2k2 + · · · + nkn = n and k1 + k2 + · · · + kn = k. We write
n X
ρ
Bn,k (f1 , f2 , . . . , fn−k+1 ),
Aρ,n (f1 , f2 , . . . , fn ) =
k
k=1
where the Bn,k (f1 , f2 , . . . , fn−k+1 )’s are called the Partial Ordinary Bell Polynomials. From (A.1) it follows that
Bn,k (f1 , f2 , . . . , fn−k+1 ) =
X
k!
kn−k+1
f1k1 f2k2 · · · fn−k+1
.
k1 !k2 ! · · · kn−k+1 !
(A.2)
Here the sum runs over all sequences k1 , k2 , . . . , kn of non-negative integers such
that k1 + 2k2 + · · · + (n − k + 1) kn−k+1 = n and k1 + k2 + · · · + kn−k+1 = k.
Since (F (x))ρ = (F (x))ρ−1 F (x), we have the recurrence
Aρ,n (f1 , f2 , . . . , fn ) = Aρ−1,n (f1 , f2 , . . . , fn−k ) +
n
X
fk Aρ−1,n−k (f1 , f2 , . . . , fn−k ).
k=1
Taking into account formula (A.2), one finds that
!k
∞
∞
X
X
f n xn
=
(F (x) − 1)k =
Bn,k (f1 , f2 , . . . , fn−k+1 ) xn ,
n=1
n=k
33
(A.3)
A.2. The r-associated Stirling Numbers
and therefore
Bn,k+1 (f1 , f2 , . . . , fn−k ) =
n−k
X
fk Bn−k,k (f1 , f2 , . . . , fn−2k+1 ).
(A.4)
k=1
P∞
gn xn is a formal power series, then by (A.3), we have
!
∞
n
X
X
G (y (F (x) − 1)) =
gk Bn,k (f1 , f2 , . . . , fn−k+1 ) y k xn .
If G (x) =
n=0
n=0
k=0
Specially,
exp y
∞
X
f n xn
n=1
!
∞
n
X
X
Bn,k (f1 , f2 , . . . , fn−k+1 ) k
y
=
k!
n=0
k=0
!
xn .
For more details see, e.g., Riordan’s book [11, p. 189].
A.2
The r-associated Stirling Numbers
For every nonnegative integers r and k we define the r-associated Stirling Numbers
of the First and Second Kind by the generating functions
!k
r
∞
X
X
1
xm
xn
− log (1 − x) −
=
sr (n, k) ,
k!
m
n!
m=1
n=(r+1)k
1
k!
r
X
xm
x
e −
m!
m=0
!k
∞
X
=
Sr (n, k)
n=(r+1)k
def
xn
.
n!
(A.5)
def
If r = 0 then s (n, k) = s0 (n, k) and S (n, k) = S0 (n, k) are the Stirling Numbers
of the First and Second Kind, respectively.
It follows that


!!
⌊n/r+1⌋
∞
r
n
m
X
X
X
x
j x

=
,
sr (n, j) y
exp y − log (1 − x) −
m
n!
n=0
m=1
j=0
r
X
xm
x
exp y e −
m!
m=0
!!
=
∞
X
n=0
It is known (see [4]) that
sr (n, k) = n!
⌊n/r+1⌋

X
j=0

Sr (n, j) y j 
k
X
(−1)j sr−1 (n − rj, k − j)
rj j! (n − rj)!
j=0
Sr (n, k) = n!

,
k
X
(−1)j Sr−1 (n − rj, k − j)
j=0
(r!)j j! (n − rj)!
For further identities see, e.g., Howard’s paper [4].
34
.
xn
.
n!
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