 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ib/paper-1/14b/a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-05

h (k) = \int_{R} e^{- ik x} h (x) d x

, with inverse

(2 π)^{- 1} \int_{R} e^{ik x} h (k) d k

. Splitting at the origin, with

a > 0

to ensure convergence, gives

h (k) = \int_{- \infty}^{0} e^{(a - ik) x} d x + \int_{0}^{\infty} e^{- (a + ik) x} d x = \frac{1}{a - ik} + \frac{1}{a + ik} .

Thus

h (k) = \frac{2 a}{a ^{2} + k ^{2}} .

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