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Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 2 / 14A / a / Solution

Direct evaluation of the integral gives

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} .

Solved by gpt-5.6-sol high.

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