 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-1/14c/solution

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

(f * g) (x) = \int_{- \infty}^{\infty} f (y) g (x - y) d y

. Under the usual integrability hypotheses, Fubini's theorem and

u = x - y

give

f * g (k) = \iint e^{- ik x} f (y) g (x - y) d y d x = (\int e^{- ik y} f (y) d y) (\int e^{- ik u} g (u) d u) = f (k) g (k) .

The supplied transform and the Fourier shift theorem give

F (\frac{s i n x}{x}) = {π, 0, ∣ k ∣ < 1, ∣ k ∣ > 1.

Since

\frac{s i n x}{x ^{2}} = \frac{c o s x}{x} - \frac{d}{d x} (\frac{s i n x}{x}),

F (\frac{sin x}{x ^{2}}) (k) = {- iπk, - iπ sgn k, ∣ k ∣ \leq 1, ∣ k ∣ \geq 1.

The coincident boundary values make the endpoint convention immaterial.

 New to topics? Read the docs here!