 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2010/d/d3/2/solution

For

0 < x < 1

, expand

(1 - x)^{- 2} = \sum_{k \geq 1} k x^{k - 1}

. All terms after multiplication by

(lo g x)^{2}

are nonnegative, so monotone convergence permits integration term by term. Substituting

x = e^{- t}

and integrating by parts twice gives

\int_{0}^{1} x^{k - 1} (lo g x)^{2} d x = \int_{0}^{\infty} e^{- k t} t^{2} d t = \frac{2}{k ^{3}} .

Consequently the integral is

\sum_{k \geq 1} 2/ k^{2}

, and the supplied sum gives

π^{2} /3

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