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I've been looking at the following integral, and trying to solve it through series:

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

I tried expanding out $e^x-1$:

$$\int_0^\infty \frac{x}{x+x^2/2!+x^3/3!+x^4/4!+\cdots} dx$$

but I do not know how to continue from there; I feel like it would have something to do with long division, but I don't know how.

I also tried the substitution

$$t=e^x-1$$ $$dt=e^xdx$$ $$x=\ln(1+t)$$

So after multiplying the integrand by $1 = e^x/e^x$ and substituting I got

$$\int_0^\infty \frac{\ln(1+t)}{t(1+t)}dt$$

I tried using series as well, but I was stuck after

$$\int_0^\infty \frac{t-t^2/2+t^3/3-t^4/4+\cdots}{t(1+t)}dt$$ $$\int_0^\infty \frac{1}{t+1}-\frac{t}{2(t+1)}+\frac{t^2}{3(t+1)}-\frac{t^3}{4(t+1)}+\cdots dt$$

I know that it is supposed to equal $\frac {\pi^2}{6}$, so I know that after doing everything it should come out to $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots$, but I do not know how to get there.

Thanks!

P.s. Wolfram Alpha gives a closed form for the antiderivative of $\frac{x}{e^x-1}$ as $\operatorname{Li}_2(e^x)-\frac{x^2}{2}+x\ln(1-e^x)+c$

Marco Cantarini
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Tdonut
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    We have $\frac{1}{e^x - 1} = \frac{1}{e^x} \frac{1}{1-e^{-x}}$ and $y = e^{-x} < 1$ if $x>0$ so we can instead try to expand $\frac{1}{1-e^{-x}} = \frac{1}{1-y} = 1+y+y^2+\ldots$ using the geometrical series. – Winther Oct 11 '15 at 19:24
  • In general, $$\int_0^\infty\frac{x^n}{e^x-1}~dx ~=~ n!~\zeta(n+1),$$ and $$\int_0^\infty\frac{x^n}{e^x+1}~dx ~=~ n!~\eta(n+1).$$ See the Riemann $\zeta$ and Dirichlet $\eta$ function for more information. – Lucian Oct 12 '15 at 01:56

2 Answers2

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Instead of trying to expand the denominator, try instead to expand the full fraction $\frac{1}{e^x-1}$ in a series. We have

$$\frac{1}{e^x-1} = e^{-x}\frac{1}{1-e^{-x}}$$

Now if $x>0$ then $e^{-x}<1$ so we can expand in a geometrical series

$$\frac{1}{e^x-1} = e^{-x}\sum_{k=0}^\infty e^{-kx} = \sum_{k=1}^\infty e^{-kx}$$

This gives the integral

$$\int_0^\infty\frac{x}{e^x-1}\,{\rm d}x = \int_0^\infty \sum_{k=1}^\infty xe^{-kx}\,{\rm d}x \\= \sum_{k=1}^\infty \int_0^\infty xe^{-kx}\,{\rm d}x = \sum_{k=1}^\infty \frac{1}{k^2}\int_0^\infty (kx)e^{-kx}\,{\rm d}(kx) = \sum_{k=1}^\infty\frac{1}{k^2}\int_0^\infty ye^{-y}\,{\rm d}y $$

where the switching of the summation and integration is justified by Fubini's theorem; see this answer for more details. The final integral can be evaluated using integration by parts.

The same series approach can be used to compute the integral $\int_0^\infty\frac{x^{s-1}}{e^x-1}{\rm d}x = \Gamma(s)\zeta(s)$ where $\Gamma$ is the $\Gamma$-function and $\zeta$ is the Riemann $\zeta$-function.

For an alternative way of deriving the result using contour integration, see this answer.

Community
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Winther
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    You don't really need to use the Gamma function, you just need to integrate by parts to show that $\int_0^\infty xe^{-x},\mathrm{d}x=1$. Then each integral, by substitution is $\frac1{k^2}$. – robjohn Oct 11 '15 at 20:06
  • @robjohn Yes of course. I guess I was thinking about the general zeta integral $\frac{x^s}{e^x-1}$ when writing it. Thanks for pointing it out. – Winther Oct 11 '15 at 20:09
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$$ \begin{align} \int_0^\infty\frac{x}{e^x-1}\,\mathrm{d}x &=\int_1^\infty\frac{\log(x)}{x(x-1)}\,\mathrm{d}x\tag{1}\\ &=-\int_0^1\frac{\log(x)}{1-x}\,\mathrm{d}x\tag{2}\\ &=-\int_0^1\frac{\log(1-x)}{x}\,\mathrm{d}x\tag{3}\\ &=\sum_{k=1}^\infty\int_0^1\frac{x^{k-1}}{k}\,\mathrm{d}x\tag{4}\\ &=\sum_{k=1}^\infty\frac1{k^2}\tag{5}\\ &=\frac{\pi^2}6\tag{6} \end{align} $$ Explanation:
$(1)$: Substitute $x\mapsto\log(x)$
$(2)$: Substitute $x\mapsto\frac1x$
$(3)$: Substitute $x\mapsto1-x$
$(4)$: Expand the series for $\log(1-x)$
$(5)$: Integrate each term
$(6)$: $\zeta(2)=\frac{\pi^2}6$

robjohn
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