2. Limiting Behaviour of Integrals

Question 1

Why is it useful to approximate integrals?

Answer 1

• integrals are often used to define functions
• many solutions of initial value problems can be expressed as integrals
• higher order ordinary differential equations or partial differential equations are often formally solved using Laplace or Fourier transforms

Question 2

∫ exp(-t²) dt

-evaluated between -∞ and +∞

Answer 2

∫ exp(-t²) dt = √π

Question 3

∫ exp(-t)/√t dt

-evaluated between 0 and ∞

Answer 3

∫ exp(-t)/√t dt
-substitute s =√t
= 2 ∫ exp(-s²) ds
-evaluated between 0 and ∞
-since the integrand is symmetric about 0, can divide by two and double the integration range so:
= ∫ exp(-s²) ds
-evaluated between -∞ and ∞
= √π

Question 4

The Gamma Function

Answer 4

Γ(x) = ∫ t^(x-1) exp(-t) dt, x>0

-evaluated between 0 and ∞

Question 5

Evaluating the Gamma Function for x∈R

Answer 5

Γ(x) = (x-1) Γ(x-1)

Question 6

Evaluating the Gamma Function for n∈Z+

Answer 6

Γ(n) = (n-1)! for n∈Z+

Question 7

Evaluating the Gamma Function for (n∈Z+)+1/2

Answer 7

Γ(1/2) = √π

-then use Γ(x) = (x-1) Γ(x-1) valid for any x∈R

Question 8

Direct Expansion of Integrals

Two Types

Answer 8

∫ f(εt) g(t) dt , ε«1
-evaluated between 0 and a with upper limit a not ‘too large’

∫ f(t) g(t) dt , |x|«1
-evaluated between 0 and x

-where f(t) is an arbitrary function which would be tricky to integrate if left alone but which has a known Taylor expansion around 0 and g(t) is a relatively simple function such that t^n g(t) can be integrated explicitly with n∈Z

Question 9

Direct Expansion of Integrals

Type I - f(εt)

Answer 9

∫ f(εt) g(t) dt , ε &laquo_space;1

• since ε is small, can replace the more complex function with an expansion
• evaluate integral and sub in limits

Question 10

Direct Expansion of Integrals

Type II - x

Answer 10

∫ f(t) g(t) dt , |x|«1

• since the range of the integral is small, can replace the more complicated expression f(t) with an expansion
• evaluate the integral

Question 11

A Complete Series Expansion

Answer 11

-over a small interval of integration, can perform a complete series expansion

Question 12

Extension of the Complete Series Expansion to Wide Intervals

Answer 12

-sometimes similar ideas can be applied to wider intervals:
∫ f(t) dt
-where the integration is between t=0 and t=x with x»1
-provided the integral ∫f(t)dt between 0 and ∞ exists we can write:
[0,x] ∫ f(t) dt = [0,∞] ∫ f(t) dt - [x,∞] ∫ f(t) dt
-make a substitution, s=1/t
= [0,∞] ∫ f(t) dt - [1/x,0] ∫ f(1/s)(-ds/s²)
= [0,∞] ∫ f(t) dt - [0,1/x] ∫1/s²f(1/s)ds
-the second integral is now over a small interval and so is amenable to series expansion

Question 13

Exponential Decay Integral

Answer 13

∫ f(t) exp(-λt) dt, λ->∞

-where the interval of integration is t=0 to t=T

Question 14

Ideas Behind Dealing With Exponential Decay Integrals

Answer 14

∫ f(t) exp(-λt) dt, λ->∞
-where the interval of integration is t=0 to t=T
-this can be split into the integral between 0 and ∞ minus the integral between T and ∞, the second of which can be discarded as it is so small
-only behaviour of f(t) close to t=0 should matter so we use a Taylor series for f around t=0 (MacLaurin Series)
-need to integrate terms like:
[0,∞] ∫ t^m e^(-λt) dt = Γ(m+1)

Question 15

Watson’s Lemma

Answer 15

-let:
I(λ) = [0,T] ∫ e^(-λt) t^α g(t) dt, T>0
-here α>-1 for the integral to exist and g(t) has a MacLaurin expansion:
g(t) = Σ gn t^n = Σ g'n(0)/n! t^n
-if T=+∞ then g(t) must be exponentially bounded i.e. 
|g(t)| < Ke^(ct)
-then:
I(λ) ~ Σ [gn Γ(α+n+1)] / λ^(α+n+1)
as λ->∞

Question 16

Sketch Proof of Watson’s Lemma

Answer 16

• separate the integral as the the integral between 0 & ∞ minus the integral between T & ∞
• substitute s=t-T in the second integral, it becomes an exponentially small term
• argue that the largest contributions come from near t=0 so replace g(t) by its MacLaurin series
• interchange order of integration and summation
• substitute s=λt and evaluate integral
• regard I as an asymptotic series in 1/λ

Question 17

MacLaurin Expansion Radius of Convergence

Answer 17

-for the function g to be used in Watson’s lemma, the radius of convergence of the MacLaurin expansion can be found using the ratio test which guarantees convergence when:
lim |[gn+1x^(n+1)]/[gnx^n]| < 1
-where the limit is n->∞
=>
|x| < R = lim |gn/gn+1|
-where the limit is n->∞
-there is divergence when |x|>R and behaviour at |x|=R must be determined on a case by case basis

Question 18

Does the result of Watson’s Lemma converge?

The Two Cases

Answer 18

1) if the MacLaurin Series g has a finite radius of convergence R, then this implies λ>∞ for convergence i.e. the series does not converge for a finite value of λ
2) if the MacLauring Series for g has an infinite radius of convergence then this can balance the growth resulting in convergence for sufficiently large λ

Question 19

Extension of Watson’s Lemma

Answer 19

-integrals can sometimes be transformed to correct form for Watson's Lemma using a change of variables e.g.
I(λ) = ∫ exp(-λ sins) ds
-between s=0 and s=π/4
-sub in t=sin(s)
=>
I(λ) = ∫ exp(-λt) / √[1-t²] dt
-between t=0 and t=1/√2

Question 20

Integration By Parts

Answer 20

-for some integrals repeated integration by parts can produce a power series in increasing powers of a small term 0