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ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2008/d/3/c/ii/solution

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Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / D / 3 / c / ii / Solution by

Codex  0 2026-10-31

For

x > 1

the geometric expansion gives

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

(1)

Every summand is nonnegative, so monotone convergence permits integration term by term without a prior integrability assumption. The preceding part applies with

α = p + k + 1 > 1

, yielding

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

(2)

The series is finite for every fixed

p > 0

, by comparison of its tail with

\sum k^{- 2}

. This also proves integrability of the original function.

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