 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ib/paper-4/17d/iv/solution

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

Evaluate the absolutely convergent complex Fourier series at

x = π

. The factors

(- 1)^{n} e^{inπ}

equal one, and the

n = 0

term is

1/ α^{2}

. Pairing positive and negative integers gives

cos (π α) = \frac{α s i n ( π α )}{π} [\frac{1}{α ^{2}} - 2 \sum_{n = 1}^{\infty} \frac{1}{n ^{2} - α ^{2}}] .

Since

sin (π α) \neq = 0

for nonintegral complex

α

, rearranging proves

n = 1 \sum \infty \frac{1}{n ^{2} - α ^{2}} = \frac{1}{2 α ^{2}} - \frac{π}{2 α tan ( π α )} .

The last term means

π cot (π α) / (2 α)

, so half-integer arguments are included by the finite cotangent value zero.

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