 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-3/15a/f/solution

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

Since the odd sine part of the complex exponential integrates to zero,

\frac{1}{2 π} \int_{- n}^{n} e^{ik x} d k = \frac{1}{π} \int_{0}^{n} cos (k x) d k .

Part (e)(ii) therefore yields

n \to \infty lim \frac{1}{2 π} \int_{- n}^{n} e^{ik x} d k = δ (x)

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