 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ib/paper-4/15d/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Integrating the interval indicator gives

\hat{f}_{1} (λ) = \frac{2 sin ( λ /2 )}{λ}, \hat{f}_{1} (0) = 1.

f_{2} (x)

is the length of overlap of two intervals of length one whose centres differ by

x

. Thus

f_{2} (x) = 1 - ∣ x ∣

for

∣ x ∣ \leq 1

, and zero otherwise. The convolution theorem gives

\hat{f}_{2} (λ) = (\frac{2 sin ( λ /2 )}{λ})^{2} .

For this Fourier transform normalization, the Parseval identity says

\int ∣ f ∣^{2} d x = (2 π)^{- 1} \int ∣ \hat{f} ∣^{2} d λ

. Since

\int ∣ f_{2} (x) ∣^{2} d x = 2 \int_{0}^{1} (1 - x)^{2} d x = \frac{2}{3},

we have

\int ∣ \hat{f}_{2} ∣^{2} d λ = 4 π /3

. Substituting

λ = 2 y

gives

2 \int_{- \infty}^{\infty} (\frac{s i n y}{y})^{4} d y = \frac{4 π}{3}, \int_{- \infty}^{\infty} (\frac{sin y}{y})^{4} d y = \frac{2 π}{3} .

The quotient at zero is understood by its continuous limit.

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