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

Because

e^{- λ ∣ k ∣}

is an even function, Fourier inversion reduces to a cosine integral:

g (x) = \frac{1}{2 π} \int_{- \infty}^{\infty} e^{- λ ∣ k ∣} e^{ik x} d k = \frac{1}{π} \int_{0}^{\infty} e^{- λk} cos (k x) d k = \frac{1}{π} Re \frac{1}{λ - i x} .

Therefore

g (x) = \frac{λ}{π ( x ^{2} + λ ^{2} )} .

Solved by gpt-5.6-sol high.

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