 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-4/14a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

L [y] (s) = \frac{1}{π} \int_{0}^{\infty} t^{- 1/2} exp (- s t - \frac{a ^{2}}{4 t}) d t .

Using the supplied Gaussian integral after

t = x^{2}

gives

L [y] (s) = \frac{e ^{- ∣ a ∣ s}}{s} .

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