 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-3/30a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Complete the square:

z^{2} - 2 i z = (z - i)^{2} + 1

. The pole of

(1 - i z)^{- 1}

is at

z = - i

, so the contour may be shifted upward through the saddle point

z = i

without crossing a singularity. Setting

z = t + i

gives the exact steepest-descent representation

I (x) = e^{- x} \int_{- \infty}^{\infty} \frac{e ^{- x t^{2}}}{2 - i t} d t .

Near the saddle,

\frac{1}{2 - i t} = \sum_{n \geq 0} \frac{( i t ) ^{n}}{2 ^{n + 1}} .

Odd terms integrate to zero, while

\int_{- \infty}^{\infty} t^{2 m} e^{- x t^{2}} d t = Γ (m + 1/2) x^{- m - 1/2} .

I (x) \sim e^{- x} m = 0 \sum \infty \frac{( - 1 ) ^{m} Γ ( m + 1/2 )}{2 ^{2 m + 1} x ^{m + 1/2}} .

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