 Solution

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

The function is continuous and nonnegative, bounded above by

x

(0, 1)

and by

x e^{- x}

(1, \infty)

, so it is integrable. For

x > 0

\frac{x}{e ^{x} + 1} = \sum_{n = 1}^{\infty} (- 1)^{n - 1} x e^{- n x} .

The series is absolutely summable in integral because

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

Each integral follows by substituting

t = n x

and integrating

t e^{- t}

by parts. Tonelli's theorem applied to the absolute terms, followed by Fubini's theorem, therefore justifies termwise integration and gives

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

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