9. Limiting Behaviour of Integrals II

Question 1

Laplace’s Method

Description

Answer 1

-used to solve integrals of the form:
I(λ) = ∫ g(t) e^[λh(t)] dt , λ->∞
-integrate from a to b

Question 2

Laplace’s Method

End Point Maximum: h’≠0

Answer 2

-suppose that the maximum of h on [a,b] lies at a & h’(a)≠0, indeed h’(a)<0
-the integral is dominated by contributions from near t=a so we write:
g(t) ≈ g(a)
h(t) ≈ h(a) + h’(a) (t-a)
-sub into the integral and evaluate as normal
-if the maximum is at b then the expression is the same, just replace a for b

Question 3

Laplace’s Method

End Point Maximum: h’=0

Answer 3

-suppose that the maximum of h on [a,b] lies at a with h’(a)=0, h’‘(a)≠0 & g(a)≠0
-so:
h’(a)=0 & h’‘(a)<0
-we write:
g(t) ≈ g(a)
h(t) ≈ h(a) + h’‘(a)/2 (t-a)²
-sub in to the original equation
-substitute s = √[λ|h’‘(a)|/2] (t-a) to evaluate the integral
-rewrite the integral from 0 to ∞ minus the integral from s* to ∞ allowing some of it to be neglected

Question 4

Laplace’s Method
End Point Maximum: h’=0
Notes

Answer 4

• for a maximum at the right of [a,b] the formula is the same but with a replaced by b
• here, I(λ) ∝ exp[λh(a)]/√λ whereas for h’≠0, I(λ) ∝ exp[λh(a)]/λ
• there are extensions if h’‘(a)=0

Question 5

Laplace’s Method

Interior Maxima

Answer 5

• suppose that the maximum of h on [a,b] lies at c∈(a,b) and that h’‘(c)≠0 & g(c)≠0
• can rewrite the integral I(λ) between a and b as the sum of integrals I1 and I2 between a&c and c&b respectively
• each of I1 and I2 then have an end-point maximum
• so I(λ) can be approximated by 2 * the end-point maximum with h’=0 with a replaced by c, the location of the interior maximum

Question 6

Laplace’s Method

Competing Maxima

Answer 6

• of the interval [a,b] has multiple maxima of the same y-value, then the contribution from each of these maxima needs to be calculated and compared
• compare the size of the contributions and take the dominant one as the leading order term

Question 7

Laplace’s Method

Unbounded Intervals

Answer 7

• Laplace’s method cannot be applied to unbounded intervals without modification
• we rewrite the unbounded integral on [a,∞] as the sum of two integrals on [a,p] & [p,∞] where p>a is at our disposal
• apply Laplace’s method to the first integral
• show that p can be chosen such that the value of the second integral is significantly smaller than the Laplace approximation to the first integral
• then the Laplace approximation can be taken as the leading order solution to the original integral

Question 8

Laplace’s Method

Movable Maxima

Answer 8

• introduce new coordinate s=t/t* where t* is the location of the maximum in h
• rewrite the original integral in terms of s
• there is a maximum at s=1 and the interval of integration becomes unbounded
• follow steps for Laplace’s method for unbounded integrals
• convert back to original coordinate t