 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-3/3a/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

With the stated Fourier transform convention, Fourier inversion gives

f (x) = \frac{1}{2 π} \int_{- \infty}^{\infty} \frac{- 2 ik}{p ^{2} + k ^{2}} e^{ik x}, d k .

For

x > 0

, close the contour in the upper half-plane. Jordan lemma removes the semicircle contribution, and the only enclosed pole is

k = i p

. Its residue is

Res_{k = i p} \frac{- 2 ik e ^{ik x}}{( k - i p ) ( k + i p )} = - i e^{- p x} .

The residue theorem therefore yields

f (x) = e^{- p x}, x > 0 .

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