 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-1/3e/b/solution

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

For

0 < ε < e^{- 1}

u = lo g (1/ x)

\int_{ε}^{e^{- 1}} \frac{d x}{x ( l o g ( 1/ x ) ) ^{θ}} = \int_{1}^{l o g (1/ ε)} u^{- θ} d u .

Taking

ε ↓ 0

is exactly the defining limit of the improper integral. By the improper power integral, it converges for

θ > 1

, and

\int_{0}^{e^{- 1}} \frac{d x}{x ( lo g ( 1/ x ) ) ^{θ}} = \int_{1}^{\infty} u^{- θ} d u = \frac{1}{θ - 1} .

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