Section 3.2 Asymptotic Series, Laplace’s Method , Gamma ...home.ustc.edu.cn/~ifs84252/Applied_Maths/Chapter 3-2.pdf · • In practical problems , it is difficult to calculate enough

Section 3.2

Asymptotic Series, Laplace’s Method,

Gamma Function, Stirling’s Formula

( ) ( )!,

2 2 ! !

NpN N

C Nw m Np N p

= =−

• the exact form of 𝑤𝑤 𝑚𝑚,𝑁𝑁 was obtained. m: position. N: total steps.

p = (N + m)/2

• The calculation is tedious and heavy.• James Stirling (Scottish mathematician, 1692-1770)

presented a way to estimate it with enough accuracy.

3.2.0 Simplify probability function by Stirling’s formula

Stirling’s formula

3

1 1 1ln !~ ln ln 22 12 360

n n n nn n

π + − + + − +

Dominant term

( )1/2!~ 2 n nn n n eπ −


will be proved later

2N mp +

=( ) ( )!,

2 ! !N

Nw m Np N p

=−


( ) ( )( )

2

!ln , ln2 ! !

ln ! ln ! ln ! ln 2

2 1ln ln 1 ln 1 ln 12 2 2

N

Nw m Np N p

N p N p N

N m m m m mN N N Nπ

=−

= − − − −

+ = − − + − − +

( )1

2 2 2 22, 1 1 1

N m m

m m mw m NN N N Nπ

+− − = − − +

1ln !~ ln 2 ln2

n n n n nπ + −

2N mp +

=( ) ( )!,

2 ! !N

Nw m Np N p

=−

1ln !~ ln 2 ln2

n n n n nπ + −


( ) ( )( )

( ) ( ) ( ) ( )

!ln , ln2 ! !

ln ! ln ! ln ! ln 21 1 1 1ln 2 ln ln ln 2 ln ln2 2 2 21 1ln 2 ln ln ln 22 21 1 1 1ln 2 ln ln ln 2 ln ln2 2 2 2 2 2 2 21 1ln 2 ln ln2 2 2 2 2 2

N

Nw m Np N p

N p N p N

N N N N p p p p

N p N p N p N p N

N m N m N m N mN N N N

N m N m N m N m N

π π

π

π π

π

=−

= − − − −

= + + − − − − +

− − − − − − + − −

+ + + += + + − − − − +

− − − −− − − + −

( ) ( )( )

2 2 2 2

22

ln 2

1 1 1 1ln ln ln ln ln ln ln 2 ln 22 2 2 2 2 2 2 2 2 21 1 1ln ln ln ln ln ln 2 ln 22 2 4 2 4 2 2

1 / 1 /1 1 1ln ln ln ln ln 2 ln 22 2 4 2 1 / 2

1 1ln ln ln 2 ln2 2

N m N m N m N m N m N mN N N N

N m N N m m N mN N N NN m

m N m NN mN N N N Nm N

N NN N N N

π

π

π

+ + + − − −= + − − − − − −

− − −= + − − + − −

+

− −+= + − + − −

+

+= + − − ( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )

22

2

1 1ln 2 ln 1 / ln 1 / ln 1 /4 2 2 2 2

2 1ln ln 1 / ln 1 / ln 1 /2 2 2

N m mm N m N m N

N m mm N m N m NN

π

π

+− − − + − − +

+= − − + − − +

( )

( )

12 2 2 2

2

2

2

2

2, 1 1 1

2 1exp exp exp2 2 2

12 exp2

2 exp2

N m m

N

m m mw m NN N N N

m N m m m mN N N N

N mN N

mN N

π

π

π

π

+− −

→∞

= − − + →

+ = − − −

= −

≈ −

1lim 1x

xe

x→∞

+ =


( )22, exp

2mw m N

N Nπ

≈ −

N=5 N=100


( )22, exp

2mw m N

N Nπ

≈ −

( )2

2 2

,

em

NN

w m N

π

−

m m


3 5

1 1 1 1ln !~ ln ln 22 12 360 1260

n n n nn n n

π + − + + − + −

In this book, via gamma function

Stirling’s formula• 2 heuristic & 8 rigorous approaches to derive

• Diverge for any value of n.

• Not a series in rigorous mathematical sense.


( )1/2!

2 n n

nn n e

απ −

=

Because it is an asymptotic series.

• But ! The dominant term works!

n

5 1.0167810 1.0083750 1.00167100 1.000837

n α

( )1/2!~ 2 n nn n n eπ −

3.2.1 Examples of asymptotics

Jules Henri Poincaré introduced asymptotic expansion

in 1886. This concept enables one to

• manipulate a large class of divergent series

• obtain numerical as well as qualitative results for

many problems.

for large x.

we have a solution in the form

this divergent series is useful for numerical

calculations, and called an asymptotic series

Example -1: ODE1dy y

dx x+ =

( )2 3

1 !1 2! 3!n

ny

x x x x−

= + + + +


Example-2: regular quadratic

ε : a small constant, say ε =0.0000001.

2 1 0x xε+ − =

2 42

x ε ε− ± +=Exact solutions

1x = ±

As ε=0, we have the unperturbed solution


2 1 0x xε+ − =

converge if

2ε <

( )

( )

2 46

2 46

12 8 128

12 8 128

Ox

O

ε ε ε ε

ε ε ε ε

− + − +=

− − − + +

Series method

2 31 2 31x a a aε ε ε= ± + + + +

power series expansion around 𝑥𝑥 = ±1

The same Taylor expansions can be reproduced.

Taylor expansion of the exact solution:


ε is a small constant, say ε =0.0000001.

Example-3: singular quadratic

Exact solutions 1 1 42

x εε

− ± +=

As ε=0, we have the unperturbed solution

1x =Only one root !


Taylor expansion of the exact solution:

( )( )

2 3 4

2 3 4

1 2 5

1 1 2 5

Ox

O

ε ε ε ε

ε ε ε εε

− + − += − − + − + +

converge if

1/ 4ε <

Expansion method

210 1 2

ax a a aε εε−= + + + +

Assuming power series

Substituting into

Why? be patient


( ) ( ) ( )1 2 0 21 1 1 0 0 1 1 0 12 1 2

0

a a a a a a a a aε ε ε−− − − −+ + + − + + +

+ =

Comparing coefficients of ε of same order

ε-1 : 21 1 0a a− −+ = 1 1a− = −

Regular root

ε0 : 1 0 02 1 0a a a− + − = 0 1a = −

ε1 : 21 1 0 12 0a a a a− + + = 1 1a =

1 1x εε

= − − + +


1or 0a− =

0 1a =

1 1a = −

2

12

x ε

ε

= −

+

Singular root

balance the three terms 2 1 0x xε + − =

Rescaling method

(1) x and −1 is comparable, assuming ε x2 is smaller than other two terms.

2 1 0 1x x xε + − = ⇒ ≈

( ) ( )2~ 1 ~ 1x O x oε


Good guess

(2) ε x2 and −1 is comparable, assuming x is smaller than other two terms.

Bad guess

( ) ( )2 ~ 1 ~ 1x O x oε

2 11 0 1x x xεε

+ − = ⇒ ≈ >>


(3) ε x2 and x is comparable, assuming both terms >> 1.

( )2 ~ 1x O xε >>

Self consistent

2 11 0 1x x xεε

+ − = ⇒ ≈ >>

When ε x2 and x balance, x is very larger 𝑥𝑥~𝑂𝑂 𝜀𝜀−1

Xxε

=rescaling x, ( )with ~ 1X O


2 0X X ε+ − =

We get a regular looking problem

20 1 2X a a aε ε= + + +

1 1x εε

= − − + +

Using regular expansion

Comparing coefficients of ε of same order, we get


Example-4: Asymptotic series by parts integration3.2.1 Examples of asymptotics

Error function

• In the 19th century, error function from the theory of errors appeared in several contexts unrelated to probability, e.g. refraction and heat conduction.

• In 1871, J. W. Glaisher wrote that "Erf(x) may fairly claim at present to rank in importance next to the trigonometrical and logarithmic functions."

• Glaisher introduced the symbol Erf and the name it error function.

( ) 2

0

2Erfz

tz e dtπ

−= ∫

This series is convergence.

( ) ( )( )

22 1

00

3 5 7 9 11

12 2Erf2 1 !

23 10 42 216 1320

n nzt

n

zz e dt

n n

z z z z zz

π π

π

+∞−

=

−= =

+

= − + − + − +

∑∫

D'Alembert's ratio test:

( )( )( )

( )( )( )

2 3 21

2 1

2 1 !z 2 1lim lim lim 0

2 3 1 !z 2 3 1

nn

nn n nn

n n n zaa n n n n

++

+→∞ →∞ →∞

+ += = =

+ + + +


Consider an alternative form of the error function

Integrating by parts

and 3 more times

( )

( )

2 2

2

0 0

2 2Erf

21 1 Erfc

z zt t

t

z

z e dt e dt

e dt z

π π

π

∞− −

∞

∞−

= = +

= − = −

∫ ∫ ∫

∫complementary error function

( ) 2 2 22

2

Erfc2 2 22 /

t z tt

z z z

z de e ee dt dtt z tπ

∞ ∞ ∞− − −−= = − = −∫ ∫ ∫

( )( ) ( ) ( )

2 2

2 3 42 82 2 2

Erfc 1 1 3 1 3 5 1 3 5 7 10512 2 162 / 2 2 2

z t

z

z e e dtz z tz z zπ

∞− − ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = − + − + +

∫

udv uv vdu= −∫ ∫


the remainder can be bounded by

5 5R a<

2 2 22

5 8 9 59 9

105 105 105 10516 32 32 32

t t zt

z z z

e de eR dt det t z z

a∞ ∞ ∞− − −

−= = < = =∫ ∫ ∫

Thus we have proven that as z → ∞

( ) ( ) ( ) ( )( )

( )2

1 2 1122

2 3 !!Erf 1 Erfc 1 1 1

2

t NN N

Nn

Nez z O zz zπ

−+ − +

−=

− = − ≈ − + − +

∑

• This expansion for the error function diverges• However, the truncated series, is useful

double factorial


This expansion has some important properties:• the leading term is roughly correct • further terms are corrections of decreasing size. This property is called asymptoticness.

z = 2.5

( )Erf 0.999593z =

3 terms give0.999599

z = 3

( )Erf 0.999978z =

3 terms give0.9999778

3 terms

z( )Erf z

1 −e−𝑧𝑧2

𝑧𝑧 𝜋𝜋1 −

12𝑧𝑧2


To evaluate the integral

Example-5: from our book

( ) 1 x t

x

f x t e dt∞

− −= ∫ as x >>1

Notice the incomplete gamma function reads

( ) 1, s t

x

s x t e dt∞

− −Γ = ∫

( ) ( )1 0,x t x

x

f x e t e dt e x∞

− −= = Γ∫


integration by parts, udv uv vdu= −∫ ∫( ) 1 1 1 2

2 32

|

1 1 1 2

x t x t x t x tx

x x x

x t x t

x x

f x t e dt t de t e e t dt

e t dt e t dtx x x

∞ ∞ ∞− − − − − − ∞ − −

∞ ∞− − − −

= = − = − +

= + = − +

∫ ∫ ∫

∫ ∫by successive integration by parts,

( ) ( ) ( )n nf x S x R x= +

where ( ) ( ) ( )1

2 3

1 1 !1 1 2!n

n n

nS x

x x x x

−− −= − + − +

( ) ( ) ( ) ( ) ( )11 ! 1 ! ,n nn x t xn x

R x n t e dt n e -n x∞ − + −= − = − Γ∫


• as n→∞, Sn(x) diverges for a fixed x.

( )1

1

!lim lim lim1 !

nn

nn n nn

a n x na n x x+

+→∞ →∞ →∞= = = ∞

−

( ) ( ) ( )1 11! !n nx t

n nxR x n x e dt n x a

∞− + − +−+< = =∫

• as x→∞, Rn(x) →0 , for a fixed n.

• The ratio of the remainder to the last term approaches zero, as x→∞


We have some observations

• As x ≥ 2n

which is small even for moderate values of n.

for example, as n=3, x =2n ≥ 6

( ) 4 2

1 1 0.0072 3 144nR x < = ≈


( )f x

n=3

( )3 2 3

1 1 2!S xx x x

= − +

( ) ( ) ( )11 !n n x tn x

R x n t e dt∞ − + −= − ∫

n=3


x

x

• As x ≥ 2n

which is small even for moderate values of n.

for example, as n=3, x =2n ≥ 6

( ) 4 2

1 1 0.0072 3 144nR x < = ≈


n=10 n=50

n=100

x x

x

This convergent series is less useful in practice.


( ) 2

0

2Erfz

tz e dtπ

−= ∫

More asymptotic series0-order Bessel functions


Laplace integral

ODE

More asymptotic series3.2.1 Examples of asymptotics

Consider function f(ε) , there are three possibilities

• the speed at which f(ε) →∞ or f(ε) →0 can be expressed by comparing gauge functions.

• gauge functions may be

( ) 0, , as 0, 0f A Aε ε= ∞ → < < ∞

( ) 11/ 1 11, , , log , log log , , sin , sinh ,n n eεε ε ε ε ε ε−± ± − −

( )1/20

lim 0f

ε

εε→

=

e.g.This indicates 𝑓𝑓 𝜀𝜀 goes to 0 in a speed faster than that of 𝜀𝜀1/2

Appendix: Landau order symbols: O or o.


The Symbol O

Definition: if there exists a positive constant M such that |𝜙𝜙 𝑥𝑥 | ≤M |𝜑𝜑 𝑥𝑥 |, as 𝑥𝑥 → 𝑥𝑥0, we say

e.g. as x → 0

( ) ( ) ( ) ( )or ~x O x x xφ ϕ φ ϕ= 𝜙𝜙 is big-oh of 𝜑𝜑

( ) ( )2 3cos 1 , sinx O x x O x= =

( )( )0

limx x

xC

xφϕ→

=

𝜙𝜙 is asymptotic to 𝜑𝜑

Meaning 𝜙𝜙 is of the same order of 𝜑𝜑, or

is a constantC

The Symbol o

( ) ( )2sin 1 , sinx o x o x= =

Definition: if for every positive constant M such that |𝜙𝜙 𝑥𝑥 | ≤M |𝜑𝜑 𝑥𝑥 |, as 𝑥𝑥 → 𝑥𝑥0, we say

e.g., as x → 0

( ) ( ) ( ) ( ) or x o x x xφ ϕ φ ϕ= <<

Meaning 𝜙𝜙 is of lower order of 𝜑𝜑, or( )( )0

lim 0x x

xx

φϕ→

=

( ) ( ) ( ) ( ), , ln , lnln lnx a a x ae o x x o e x o x x o x− −= = = =

e.g., as x → ∞ & any a > 0

𝜙𝜙 is small-oh of 𝜑𝜑 𝜙𝜙 is much less than 𝜑𝜑


( ) ( )!,

2 ! !N

Nw m Np N p

=−

Review

Stirling’s formula

22 exp2m

N Nπ −

3

1 1 1ln !~ ln ln 22 12 360

n n n nn n

π + − + + − +

2 1 0x xε+ − =( )

( )

2 46

2 46

12 8 128

12 8 128

Ox

O

ε ε ε ε

ε ε ε ε

− + − +=

− − − + +

( )( )

2 3 4

2 3 4

1 2 5

1 1 2 5

Ox

O

ε ε ε ε

ε ε ε εε

− + − += − − + − + +

( ) ( )( )

22 1

00

12 2Erf2 1 !

n nzt

n

zz e dt

n nπ π

+∞−

=

−= =

+∑∫

convergent series

asymptotic series( ) ( )

( ) ( )( )

( )2

1 2 1122

Erf 1 Erfc

2 3 !!1 1 1

2

t NN N

Nn

z z

Ne O zz zπ

−+ − +

−=

= −

− = − + − +

∑

• as n→∞, Sn(x) diverges for a fixed x.

• as x→∞, Rn(x) →0 , for a fixed n.

• as x→∞, |𝑅𝑅𝑛𝑛| < 𝑎𝑎𝑛𝑛

( ) ( ) ( )n nf x S x R x= +

Review

Review

( ) ( ) ( ) ( ) or x o x x xφ ϕ φ ϕ= << ( )( )0

lim 0x x

xx

φϕ→

=

( ) ( ) ( ) ( )or ~x O x x xφ ϕ φ ϕ= ( )( )0

limx x

xC

xφϕ→

=

• in the example of 𝜀𝜀𝑥𝑥2 + 𝑥𝑥 − 1 = 0, we see

( )2 21 2x oε ε ε= − + +with asymptotic sequence {1, ε, ε2, …}

( )1 2 21 2x oε ε ε ε−= − − + − +with asymptotic sequence {𝜀𝜀−1, 1, 𝜀𝜀, 𝜀𝜀2,…}

• One can use a general sequence of gauge functions {φn(ε)} as asymptotic sequence as ε → 0

e.g. ( ) ( ) ( ), log , sin ...n nnnϕ ε ε ε ε−=

3.2.2 Definition asymptotic expansion

• Usually, a function may depend on x and a small parameter ε. we may look for an expansion in the form

( ) ( ) ( )0

, ~ as 0N

n nn

f x a xε φ ε ε=

→∑

• In terms of asymptotic sequences {φn(ε)}, we can expand functions f(𝜀𝜀) in asymptotic expansion if constants an exist,

( ) ( )0

~ as 0N

n nn

f aε φ ε ε=

→∑or,

( ) ( ) ( )0

as 0N

n n nn

f a oε φ ε φ ε ε=

= + → ∑


Consider an expansion

Sn(x) is an asymptotic expansion of f(x) , if

( ) ( )lim 0 with fixednnx

n

f x S xn

A x−→∞

−=

( ) ( ) ( ) with , fixednnf x S x o x x n−− = →∞

or

( )1 11 20 2 1 1

n n nnn n n

A A AA AA S xx x x x x

+ ++ ++ + + + + + ≡ + +

we then say ( )0

~ ,n

ii

if x A x x−

=

→∞∑asymptotic to


|𝑅𝑅𝑛𝑛| < 𝑎𝑎𝑛𝑛

( )0

~ ,n

ii

if x A x x−

=

→∞∑If for any n, we always have

( )0

~ ,ii

if x A x x

∞−

=

→∞∑

then we say f(x) has an asymptotic power series

This asymptotic series is usually divergent for any fixed x.

3.2.2 Definitions in asymptotic expansion

( ) 0for ,N

nM

f x M N Nε< >∑

An expansion ∑𝑁𝑁 𝑓𝑓𝑛𝑛 𝑥𝑥 is said to converge at a fixed value of x if given an arbitrary ε > 0 it is possible to find a number N0(x, ε) such that

This property of convergence is less useful in practice.

Convergence3.2.2 Definitions in asymptotic expansion

( )( )

( ) ( )

( )0 as 0

n

nn

n n

f fRf f

ε εεε

ε ε

−= → →

∑

The sum ∑𝑁𝑁 𝑓𝑓𝑛𝑛 𝜀𝜀 is said to be an asymptotic approximation to 𝑓𝑓 ε as ε →0, if for each n ≤ N

the remainder 𝑅𝑅𝑛𝑛 is smaller than the last term 𝑓𝑓𝑛𝑛.

• If the sum has this asymptotic property, one writes

( ) ( )~ as 0N

nf fε ε ε →∑

Asymptoticness3.2.2 Definitions in asymptotic expansion

More comments on convergence and divergenceLet’s look at Taylor series

• the sum converges to f(x) as n→∞. • Taylor series are de facto asymptotic expansions

( ) ( ) ( )( ) ( )( ) ( )( )2 20 0 0 0 0 0

1' ''2

f x f x f x x x f x x x o x x= + − + − + −

There are two limiting processes for an asymptotic expansion, n→∞ and ε → 0.• An asymptotic expansion provide an accurate

approximation as ε → 0 for each n• many useful expansions diverge as n→∞.


A famous example is the incomplete exponential integral

( ) ( )1

1 1 !1 1

!

nn

nn

na na n

εε

ε

++ += = + > 0

1nε =

increase as n>𝑛𝑛0

( ) ( )0 0n nR o aε =

Optimal truncation3.2.2 Definitions in asymptotic expansion

( ) ( ) ( )11

2 3

1 1 !1 1 2!~n

x tn

x

nf x t e dt

x x x x

−∞− − − −

= − + − +∫

Optimal truncation

• In practical problems, it is difficult to calculate enough terms to decide whether the asymptotic series is divergent, as opposed to mathematical asymptotic problem.

• impossible to prove that the remainder after even one or two terms is small enough.

• Luckily, one or two terms is enough for most problems encountered in applied math.


Laplace’s method :• to obtain the asymptotic expansion of certain

integrals containing a large parameter.

3.2.3 Laplace’s method

( ) ( ) ( ) 1f tF g t e dtβ

λ

α

λ λ−= >>∫idea:• the dominant contribution comes from a small

portion。

Consider


( ) ( )0.5 cos cos 2g t x x= + ( ) ( )2expf t x= −

( ) ( )( ) ( )20.5 cos cos 2 exp

g t f t

x x x= + −( ) ( ) ( )f tF g t e dt

βλ

α

λ −= ∫


Let’s stare at the integrand

( ) ( ) ( )f tF g t e dtβ

λ

α

λ −= ∫

( ) ( ) ( )200

''2

f tf t t t≈ −

a minimum at 𝑡𝑡0

( ) ( ) ( )0 2

0

0

ta t t

t

F g t e dtδ

δ

λ+

− −

−

≈ ∫

𝑡𝑡0 + 𝛿𝛿𝑡𝑡0 − 𝛿𝛿 𝑡𝑡0α β

10a =

1a =

100a =

( ) ( )20expQ t a t t = − −

( )0''2

f ta

λ−=

Taylor expansion about t0 .

Assuming

( ) ( ) ( )( ) ( )( ) ( )( )( ) ( )( )

2 20 0 0 0 0 0

20 0 0

1' ''2

1 ''2

f t f t f t t t f t t t o t t

f t f t t t

= + − + − + −

≈ + −

( ) ( )0 0' 0 '' 0f t f t= >


thus

( ) ( ) ( )

( ) ( ) ( ) ( )0

20 0

0

''~ exp

2

f t

f t

F g t e dt

f t t tg t e dt

βλ

α

λ

λ

λ

−

∞−

−∞

=

− −

∫

∫

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( )

( )

0

0

0

20 0

0

1/2

20

0

1/2

00

''~ exp

2

2 exp''

2''

f t

f t

f t

f t

F g t e dt

f t t tg t e dt

g t e u duf t

g t ef t

βλ

α

λ

λ

λ

λ

λ

λ

πλ

−

∞−

−∞

∞−

−∞

−

=

− −

= −

=

∫

∫

∫( ) ( )0

0

''2

f tu t t= −


( )2exp u du π∞

−∞

− =∫

Gauss integral

3.3.4 Asymptotic Stirling series for Gamma function

( ) ( )1

0

Re 0s xs x e dx s∞

− −Γ = >∫Gamma function

( ) ( )1 1 !2

n nπ Γ = Γ = −

for s >>1, the gamma function can be reformed as

( ) ( )1 ln

0

0

s x x

sF

s e e dx

e dx

∞− −

∞−

Γ =

=

∫

∫

( ) ( )1 1, 1 lnF x s s x s x− −= − +

where

10s =

( )1 1 lns x− − 1s x−

F

x

00 1F x sx

∂= ⇒ = −

∂

0x( )ln ax o x=


The minimal 𝑥𝑥0 = 𝑠𝑠 − 1 shifts with s . Let1

xts

=−

So that the minimal point 𝑥𝑥0= 𝑠𝑠 − 1 corresponds to

t0 = 1. This leads to

( ) ( ) ( )0

1 ssFs e dx s J λ∞

−Γ = = −∫

1 1sλ = −

where

ln t− t

0 1

0t

dfdt =

= ( )0 1f t =

( ) ( )

0

f tJ e dtλλ∞

−= ∫

( ) lnf t t t= −

( )f t

t0 1t =


In order to use Laplace’s method, we modify the integrand

( ) ( ) ( ) ( )

( ) ( )

0 0

20

0 0

0

f t f t f tf t

f t w t

J e dt e dt

e e dt

λλ

λ λ

∞ ∞ − + −−

∞− −

= =

=

∫ ∫

∫where

( ) ( ) ( )20 ln 1w t f t f t t t= − = − −


define w(t) as continuous and monotonic

( ) ( ) ( )( ) ( ) ( )

0 0

0 0

0w t f t f t t t

w t f t f t t t

= − − ≤ ≤

= − >

( ) ( ), 0,w t∈ −∞ ∞ ↔ ∈ ∞

( ) ( )20f t w t dtJ e e dw

dwλ λ

∞− −

−∞

= ∫

Rewrite

?dtdw

=

( ) ( ) ( )f tF g t e dtβ

λ

α

λ −= ∫


We now try to expand w(t) around t0=1

( )( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

2

2 3 4

2 3 4

1 ln

1 ln 1 1

1 11 1 1 1 1 ...2 3

1 11 1 12 3

w t t t

t t

t t t t O t

t t O t

= − −

= − − + − = − − − − − + − + −

= − − − + −

w: a small number


( ) ( )22 1 1 1 22

w t t t w= − ⇒ − =

Using the first order

0 10, 2u u= =

( ) ( ) ( ) ( )2 3 42 1 11 1 12 3

w t t t O t = − − − + −

assume an asymptotic series

20 1 21 ...t u u w u w− = + + +


( )21 21 2 1 ...t w a w a w− = + +rewrite

substituting into the Taylor expansion

( ) ( ) ( ) ( )( )2 3 32 1 11 1 12 3

w t t t o t= − − − + −

comparing term by term, we have

1 2 32 1 11 2, , , ....

3 18 27a a a= = = −

( )21 2

02 1 ... m

mm

dt d w a w a w b wdw dw

∞

=

= + + = ∑Then

0 2b =


( ) ( ) ( ) ( ) ( )2

0 0

0

f t f tw tm m

m

dtJ e e dw e b Idw

λ λλλ λ∞

− −−

=−∞

= = ∑∫

If term-by-term integration is valid,

where

( ) ( )/2 /2

1 !!, is even

2m m m

mI mπλ

λ λ−

=

( ) 0, is oddmI mλ =

0

mm

m

dt b wdw

∞

=

= ∑

( ) 2w mmI e w dwλλ

∞−

−∞

= ∫


0mI =

( ) 2 2

11 | 0

2w wI e wdw eλ λλ

λ

∞− − ∞

−∞−∞

−= = =∫

2 33 1 0w dI e w dw I

dλ

λ

∞−

−∞

= = − =∫

As m is odd

2 55 3 0w dI e w dw I

dλ

λ

∞−

−∞

= = − =∫


2

0 /wI e dwλ π λ∞

−

−∞

= =∫

2 22 0

w dI e w dw Id

λ

λ

∞−

−∞

= = −∫

( )22

244 2 021w d dI e w dw I I

d dλ

λ λ

∞−

−∞

= = − = −∫

( ) ( ) ( )/2/22 0

/2 /2 /2

1 !!1 1

2

mmm

m m m m

mdI d IId d

πλ λ λ λ− −

= − = − =

As m is even


To the first order approximation

1s λ= +

( ) ( ) ( ) ( ) ( ) ( )

( )

0 0 3/20 0

0

12 / 1

f t f tm m

mJ e b I e b I O

e O

λ λ

λ

λ λ λ λ

π λ λ

− − −

=

− −

= ≈ +

= +

∑

( ) ( ) ( )1 ss s J λΓ = −( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

1 1

1

ln ln 1

2ln 1 ln 112ln 1 1 ln ln 1

11 1 1ln 1 1 ln 22 2 1

s

s

s s J

s s e Os

s s s Os

s s s Os

λ

π λ

π λ

π

− − −

−

Γ = −

= − + + −

= − − − + + + − = − − − − + + −

0 /I π λ=0 2b =


Stirling’s approximation □

In the case of integer, i.e. s = n + 1

( )

( )1

ln ! ln 1

1 1ln ln 22 2

n n

n n n O nπ −

= Γ +

= + − + +

( )1 !n nΓ + =

( ) ( ) ( )1 1 1ln ln 1 1 ln 22 2 1

s s s s Os

π Γ = − − − − + + −

Documents

Section 3.2 Asymptotic Series, Laplace’s Method , Gamma ...home.ustc.edu.cn/~ifs84252/Applied_Maths/Chapter 3-2.pdf · • In practical problems , it is difficult to calculate enough