 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/27j/c/solution

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

For

f = 1_{[- 1, 1]}

, part (b) gives

f (u) = 2 sin u / u

. The Plancherel theorem in the stated Fourier convention gives

\int_{R} ∣ f (x) ∣^{2} d x = \frac{1}{2 π} \int_{R} ∣ f (u) ∣^{2} d u .

Since the left side is two,

2 = \frac{1}{2 π} \int_{R} 4 (\frac{s i n u}{u})^{2} d u .

The integrand is even, so the sinc-squared integral is

\int_{0}^{\infty} (\frac{sin x}{x})^{2} d x = \frac{π}{2} .

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