2. Limiting Behaviour of Integrals Flashcards
Why is it useful to approximate integrals?
- 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
∫ exp(-t²) dt
-evaluated between -∞ and +∞
∫ exp(-t²) dt = √π
∫ exp(-t)/√t dt
-evaluated between 0 and ∞
∫ 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
The Gamma Function
Γ(x) = ∫ t^(x-1) exp(-t) dt, x>0
-evaluated between 0 and ∞
Evaluating the Gamma Function for x∈R
Γ(x) = (x-1) Γ(x-1)
Evaluating the Gamma Function for n∈Z+
Γ(n) = (n-1)! for n∈Z+
Evaluating the Gamma Function for (n∈Z+)+1/2
Γ(1/2) = √π
-then use Γ(x) = (x-1) Γ(x-1) valid for any x∈R
Direct Expansion of Integrals
Two Types
∫ 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
Direct Expansion of Integrals
Type I - f(εt)
∫ f(εt) g(t) dt , ε «_space;1
- since ε is small, can replace the more complex function with an expansion
- evaluate integral and sub in limits
Direct Expansion of Integrals
Type II - x
∫ 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
A Complete Series Expansion
-over a small interval of integration, can perform a complete series expansion
Extension of the Complete Series Expansion to Wide Intervals
-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
Exponential Decay Integral
∫ f(t) exp(-λt) dt, λ->∞
-where the interval of integration is t=0 to t=T
Ideas Behind Dealing With Exponential Decay Integrals
∫ 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)
Watson’s Lemma
-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->∞
Sketch Proof of Watson’s Lemma
- 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
MacLaurin Expansion Radius of Convergence
-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
Does the result of Watson’s Lemma converge?
The Two Cases
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
Extension of Watson’s Lemma
-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
Integration By Parts
-for some integrals repeated integration by parts can produce a power series in increasing powers of a small term 0
