Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 1 / 3E / b

Table of contents
- Solution b

Solution

For

0 < ε < e^{- 1}

u = lo g (1/ x)

. The ordinary change of variables formula gives

\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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