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How to calculate this improper integral $\int_{0}^{+\infty}\frac{x}{\mathrm{e}^x-1}\mathrm{d}x$

liaoweichuan
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2 Answers2

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Hint. One may justify a termwise integration using $$ \frac{x}{e^x-1}=e^{-x} \cdot\frac{x}{1-e^{-x}}=\sum_{n=0}^\infty xe^{-(n+1)x} $$ and using an integration by parts to get $$ \int_0^\infty xe^{-(n+1)x}\:dx=\frac1{(n+1)^2}, \quad n\ge0. $$

Olivier Oloa
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Collect $e^x$ at the denominator to obtain:

$$\frac{x}{e^x-1} \to \frac{x}{e^x(1 - e^{-x})} = \frac{x e^{-x}}{1 - e^{-x}}$$

Since the range of the integral is $(0, +\infty)$ we are allowed to use the geometric series for

$$\frac{1}{1 - e^{-x}} = \sum_{k = 0}^{+\infty} (e^{-x})^k$$

Hence substituting in the integral:

$$\sum_{k = 0}^{+\infty}\int_0^{+\infty} x e^{-x} e^{-kx}\ \text{d}x$$

$$\sum_{k = 0}^{+\infty}\int_0^{+\infty} x e^{-x(1+k)}\ \text{d}x$$

The very last integral is trivial, and it is easily solvable (you can do it by parts for example):

$$\int_0^{+\infty} x e^{-x(1+k)}\ \text{d}x = \frac{1}{(1+k)^2}$$

Hence you get

$$\sum_{k = 0}^{+\infty}\frac{1}{(1+k)^2}$$

Which we can write it also by shifting the sum index from $k = 0$ to $k = 1$, modifying the function:

$$\sum_{k = 1}^{+\infty}\frac{1}{k^2}$$

Which is nothing but the Riemann Zeta Function of $2$:

$$\sum_{k = 1}^{+\infty}\frac{1}{k^2} = \zeta(2) = \frac{\pi^2}{6}$$

Which is the result of your integration.

Hence

$$\boxed{\int_0^{+\infty} \frac{x}{e^x-1}\ \text{d}x = \frac{\pi^2}{6}}$$

Riemann Zeta Function

This special function is defined as

$$\zeta(s) = \sum_{k = 1}^{+\infty} \frac{1}{k^s}$$

Enrico M.
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    Note that for $x=0 \in [0,\infty)$ you can't use $\frac{1}{1 - e^{-x}} = \sum_{k = 0}^{+\infty} (e^{-x})^k$... – Olivier Oloa Nov 29 '16 at 10:34
  • @OlivierOloa Of course one can, since the whole integral is regular at $x = 0$:

    $$\lim_{x\to 0} \frac{xe^{-x}}{1 - e^{-x}} = 1$$

     – Enrico M. Nov 29 '16 at 10:37
  • The integrand is integrable at $x=0$, but the identity $\frac{1}{1 - e^{-x}} = \sum_{k = 0}^{+\infty} (e^{-x})^k$ does not exist at $x=0$, isn't it? – Olivier Oloa Nov 29 '16 at 10:42
  • @OlivierOloa Oh, sorry, I misunderstood! Yes that is right actually :D – Enrico M. Nov 29 '16 at 10:42
  • The integral is integrable at x=0 . If we let x>0 , the proof works . On the other hand, the integral is the same with x =0. – liaoweichuan Nov 29 '16 at 11:23