 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ii/paper-4/31c/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

This is endpoint asymptotics after odd Gaussian cancellation. All interior algebraic terms cancel because

sin t e^{- λ t^{2}}

is odd. Split the interval into

[- 1, 1]

and the remaining tail:

\int_{- 2}^{1} sin t e^{- λ t^{2}} d t = - \int_{1}^{2} sin t e^{- λ t^{2}} d t .

The right-hand integral has its minimum phase at its left endpoint, with

ϕ^{'} (1) = 2

and nonzero amplitude

sin 1

. Endpoint Laplace method therefore give

I (λ) \sim - \frac{sin 1}{2 λ} e^{- λ} .

The answer is exponentially small, rather than zero; applying only the interior Gaussian approximation misses the asymmetric boundary tail.

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