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This integral came up in an exercise on the estimation of the specific heat of a 1-D solid and is probably a standard integral, possibly one that can be solved by contour integration:

\begin{equation} \int_0^{+\infty} \frac{x}{e^x-1} dx \end{equation}

I have some rudimentary knowledge of contour integrals, but I can't come up with a proper path, also because of the many singularities along the imaginary axis. Any suggestion?

2 Answers 2

16

Make a change or variables, $e^{-x}=u$, and write the integral as

$\int_0^\infty \frac{x e^{-x}}{1 - e^{-x}}dx$

Now substituting gives

$-\int_0^1 \frac{\log u}{1 - u}du $

This is a known integral that evaluates to $\dfrac{\pi^2}{6}$.

If you want to prove it, you can use the dilogarithm. Let $1-u=x$, so that

$-\int_0^1 \frac{\log(1 - x)}{x}dx$

Now, since we're working on $(0,1)$ it is legitimate to use

$\frac{{ - \log \left( {1 - x} \right)}}{x} = \sum_{n = 1}^\infty {\frac{{{x^{n - 1}}}}{n}} $

Integrating termwise gives

$\int_0^t \frac{-\log(1 - x)}{x} dx = \sum_{n = 1}^\infty \frac{t^n}{n^2} = \mathrm{Li}_2 (t)$

Evaluating at $t=1$ gives

$ -\int_0^1 \frac{\log(1 - x)}{x} dx = \sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$

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Any text book which dicusses the Riemann Zeta function will likely have this.

We have for $\Re(z) \gt 1$, that

$ \zeta(z) \Gamma(z) = \int_{0}^{\infty} \frac{t^{z-1}}{e^t - 1} \text{d}t$

Your integral is therefore

$\zeta(2)\Gamma(2) = \frac{\pi^2}{6}$

For an online discussion, see this: http://www.math.utah.edu/~milicic/zeta.pdf

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    @PeterT.off: Thank$s$!2012-04-10