 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-3/30d/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

u = lo g (1 + t), t = e^{u} - 1, d t = e^{u} d u .

Then

J (x) = \int_{0}^{l o g (1 + π)} g (u) e^{- xu} d u, g (u) = e^{u} [1 - cos (e^{u} - 1)] .

The Taylor series at the endpoint is

e^{u} - 1 = u + \frac{u ^{2}}{2} + \frac{u ^{3}}{6} + O (u^{4}),

and hence

g (u) = \frac{u ^{2}}{2} + u^{3} + O (u^{4}) .

The Watson lemma, together with

\int_{0}^{\infty} u^{n} e^{- xu} d u = \frac{n !}{x ^{n + 1}},

therefore yields

J (x) = \frac{1}{x ^{3}} + \frac{6}{x ^{4}} + O (x^{- 5}) .

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