 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ia/paper-2/2a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

Start from

I_{0} (λ) = 1/ λ

I_{n}^{'} (λ) = - I_{n + 1} (λ), I_{n} (λ) = (- 1)^{n} \frac{d ^{n}}{d λ ^{n}} \frac{1}{λ} .

This interchange is justified locally around any

λ > 0

by the dominated convergence theorem: the differentiated integrands are bounded by an integrable polynomial times

e^{- λ x /2}

. Successive derivatives of

λ^{- 1}

give

I_{n} (λ) = \frac{n !}{λ ^{n + 1}} .

This is also the integer case of the Gamma function integral after scaling the integration variable.

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