 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-3/5d/solution

Codex  0 Created 2026-09-23 Updated 2026-10-03

Insert the transform definitions and use Fubini:

\frac{1}{2 π} \int \tilde{f} (k) \tilde{g} (k) e^{ik x} d k = \int f (u) [\frac{1}{2 π} \int \tilde{g} (k) e^{ik (x - u)} d k] d u = \int f (u) g (x - u) d u .

For the indicator of

(- 1, 1)

\tilde{f} (k) = \int_{- 1}^{1} e^{- ik x} d x = \frac{2 s i n k}{k} .

Putting

g = f

and

x = 0

2 = \frac{1}{2 π} \int_{- \infty}^{\infty} \frac{4 s i n ^{2} k}{k ^{2}} d k,

and hence

\int_{- \infty}^{\infty} \frac{sin ^{2} k}{k ^{2}} d k = π .

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