 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2009/d/d3/iii/solution

The integrand is nonnegative and measurable. It is bounded by

x

near zero and by

x e^{- x}

for

x > 0

, so its integral is finite. For every

x > 0

the geometric expansion gives

x / (e^{x} + 1) = \sum_{n \geq 1} (- 1)^{n - 1} x e^{- n x}

. To justify termwise integration, note that

\sum_{n \geq 1} \int_{0}^{\infty} ∣ (- 1)^{n - 1} x e^{- n x} ∣ d x = \sum_{n \geq 1} \frac{1}{n ^{2}} < \infty.

Tonelli gives an integrable sum of absolute values; dominated convergence then applies to the signed partial sums. Consequently

\int_{0}^{\infty} \frac{x}{e ^{x} + 1} d x = n \geq 1 \sum \frac{( - 1 ) ^{n - 1}}{n ^{2}} .

If evaluated further, this is

(1 - 2^{- 1}) ζ (2) = π^{2} /12

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