Past exam of the options II for mathematics and philosophy course of the University of Oxford / 2013 / D3 / b

Table of contents
- Solution b

Solution

For

x > 0

, the geometric series gives

\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.

 Ancestors (11)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version