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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 14E / c / Solution

The nonzero poles are

t = \pm i a_{n}, a_{n} = \frac{π}{2} (2 n + 1), n \geq 0.

Their paired residues sum to

- (- 1)^{n} a_{n}^{s - 1} cos \frac{π s}{2} .

The paired series converges for

Re s < 1

by the alternating-series test, and absolutely for

Re s < 0

. Therefore

\sum Res = - cos \frac{π s}{2} (\frac{π}{2})^{s - 1} β (1 - s) .

Using the stated contour closure, the Hankel integral is

- 2 πi

times this sum. Substitution into part (a) gives

β (s) = Γ (1 - s) cos \frac{π s}{2} (\frac{π}{2})^{s - 1} β (1 - s) .

Now

1/Γ (1 - s) = Γ (s) sin (π s) / π

and

sin (π s) = 2 sin (π s /2) cos (π s /2)

, so

β (1 - s) = Γ (s) (\frac{π}{2})^{- s} sin \frac{π s}{2} β (s) .

It was derived on a nonempty half-plane. Both sides have analytic continuations, with the apparent gamma poles cancelled by the sine factor or the trivial zeros from part (b); the identity theorem therefore extends it to every

s \in C

.

Solved by gpt-5.6-sol high.

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