Past exam of the options II course of the University of Oxford / 2008 / D / D3 / c / ii

Table of contents
- Solution ii

Solution

For

x > 1

, the geometric series gives

\frac{l o g x}{x ^{p} ( x - 1 )} = \sum_{k = 0}^{\infty} x^{- (p + k + 1)} lo g x .

All terms are nonnegative, so monotone convergence permits termwise integration. Part (i), with

α = p + k + 1 > 1

, gives

\int_{1}^{\infty} \frac{lo g x}{x ^{p} ( x - 1 )} d x = k = 0 \sum \infty \frac{1}{( p + k ) ^{2}} .

This series is finite for

p > 0

, which also proves integrability of the original nonnegative function.

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