# Evaluate the integral $\int_0^\infty \frac{t}{e^t-1}\mathrm dt$

An integral related to the zeta function at the point $2$ is given by

$$\zeta(2) = \int\nolimits_0^\infty \dfrac{t}{e^t - 1}\mathrm dt$$

How to calculate this integral?

• Well... you already did (compute this integral), didn't you? Since you know its value. Or do you want a proof that it is indeed zeta(2)? – Did Sep 14 '11 at 13:12
• @Dan: ??   – Did Sep 14 '11 at 13:15
• Didier, throw a needle! – Dan Brumleve Sep 14 '11 at 13:17
• Is the fact that the function being integrated is the generating function of the Bernoulli numbers of interest here? – Michael Hardy Sep 14 '11 at 14:04
• @Michael Hardy: Yes it is of interest. We can see this directly with the "notion" of a negative Bernoulli number. (Non rigorous, it might be possible to make it rigorous though) The generating series fact can be reworded as $$B_n =\lim_{x\rightarrow 0} \frac{d^n}{dx^n} \frac{x}{e^x-1}.$$ In some "sense" the integral of $\frac{t}{e^t-1}$ should be like $B_{-1}$ since we would want to take $n=-1$. Recall as well that $-n\zeta(1-n)=B_n$ for positive $n$. If we "define" the negative Bernoulli numbers by that formula as well, then $B_{-1}=\zeta(2)$. Voilà. – Eric Naslund Sep 14 '11 at 15:02

The integrand can be expressed as a geometric series with first term $te^{-t}$ and common ratio $e^{-t}$. Integrate term-by-term (after justifying it, of course) and see if you don't recognize the result as $\zeta(2)$.
Somewhat equivalent to Gerry's answer: let $t=-\log(1-u)$, giving the integral
$$-\int_0^1 \frac{\log(1-u)}{u}\mathrm du$$
• Or notice that this integral is the dilogarithm evaluated at $1$. Wait, I guess you prove that $\text{Li}_2 (1)=\frac{\pi^2}{6}$ by expanding the integral anyway so nevermind.... – Eric Naslund Sep 14 '11 at 13:33