 Solution

ID: past-exam-of-the-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}

. Multiplying by

(lo g x)^{2}

x = e^{- t}

to obtain

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

where scaling followed by two integrations by parts evaluates the last integral. Hence

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

The finite value also establishes integrability at both endpoints.

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