 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-2/14a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

f (k) = \int_{0}^{1} e^{- ik x} d x - \int_{- 1}^{0} e^{- ik x} d x = \frac{2 - e ^{ik} - e ^{- ik}}{ik} = - \frac{2 i ( 1 - cos k )}{k} .

The continuous value at

k = 0

is zero, in agreement with the vanishing integral of the odd function

f

. Thus

f (k) = - \frac{2 i ( 1 - cos k )}{k} .

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