 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-1/13a/b/solution

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

Integrate

z e^{iπ z} / (z^{2} + a^{2})

over the upper semicircle. Jordan lemma removes the arc, and the pole at

z = ia

e^{- πa} /2

. Hence

\int_{- \infty}^{\infty} \frac{x e ^{iπ x}}{x ^{2} + a ^{2}} d x = πi e^{- πa} .

The cosine contribution is odd, while the sine contribution is even. Therefore

\int_{0}^{\infty} \frac{x sin ( π x )}{x ^{2} + a ^{2}} d x = \frac{π}{2} e^{- πa} .

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