 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/14b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

Because

Re a > 0

, the endpoint factor

(1 + t^{2})^{a}

vanishes at the branch points

t = \pm i

. A finite contour may therefore run from

- i

i

on either side of a chosen branch cut; the diagram shows contours bowing through the left and right half-planes.

For

a = 1

f (t) = 1

is entire, so path independence makes either contour give, up to orientation,

\int_{- i}^{i} e^{z t} d t = \frac{e ^{i z} - e ^{- i z}}{z} = \frac{2 i s i n z}{z} .

Hence a nonzero constant multiple gives the solution

w_{1} (z) = \frac{sin z}{z} .

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