 Solution

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

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

Because

e^{- λ ∣ k ∣}

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} )} .

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