 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-2/5/a/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

Choose

γ > 1/2

. The inverse integrand is

λ^{- s} / [2 (s - 1/2)]

. Because

- lo g λ > 0

, the exponential factor decays towards the left. Closing the Bromwich contour to the left and using the residue theorem picks up just the simple pole at

s = 1/2

, with residue

λ^{- 1/2} /2

. Hence

h (λ) = \frac{1}{2 λ}, 0 < λ < 1.

Its defining integral checks both the answer and its convergence condition:

\int_{0}^{1} λ^{s - 1} \frac{1}{2 λ} d λ = \frac{1}{2 ( s - 1/2 )} = \frac{1}{2 s - 1}, Re s > \frac{1}{2} .

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