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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3/30c/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 30C / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

The endpoint

t = 0

is treated by rotating onto its method of steepest descent ray. It contributes

Γ (3/2) (- i x)^{- 3/2} = \frac{π}{2} (- i x)^{- 3/2} .

At

t = 1

, put

s = 1 - t

and expand

(1 - s)^{1/2} = \sum_{n = 0}^{\infty} (- 1)^{n} (n 1/2) s^{n} .

Termwise endpoint integration, using

\int_{0}^{\infty} s^{n} e^{- i x s} d s = n! (i x)^{- n - 1}

on the rotated ray, gives

I (x) \sim \frac{π}{2} (- i x)^{- 3/2} - e^{i x} \sum_{n = 0}^{\infty} a_{n} (- i x)^{- n - 1},

where

a_{n} = n! (n 1/2) .

The two displayed contributions come from the two endpoints; the remaining deformed contour is exponentially or algebraically smaller after any fixed truncation.

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