 Solution

ID: past-exam-of-the-options-ii-for-mathematics-and-philosophy-course-of-the-university-of-oxford/2013/d3/b/solution

For

x > 0

\frac{x e ^{- αx}}{1 - e ^{- x}} = \sum_{n = 0}^{\infty} x e^{- (n + α) x} .

Every summand is nonnegative and has integral

(n + α)^{- 2}

by part (a). This numerical series converges by comparison with

\sum_{n \geq 1} n^{- 2}

. The series-integration rule proved above therefore yields

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

The value at the single endpoint zero does not affect this Lebesgue integral.

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