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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?

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    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)?2011-09-14
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    Choose a random integer and check if it is divisible by a square.2011-09-14
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    @Dan: ?? $ $ $ $2011-09-14
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    Didier, throw a needle!2011-09-14
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    @Didier: it's a rather long-winded gag... :)2011-09-14
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    @J.M. I had guessed so and this is precisely my point.2011-09-14
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    Is the fact that the function being integrated is the generating function of the Bernoulli numbers of interest here?2011-09-14
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    @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.2011-09-14
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    If I recall correctly, the $n$th cumulant of the uniform distribution on the interval $[-1,0]$ is $B_n/n$. Does it make sense to ask why the $n$th cumulant of that distribution should be $-\zeta(1-n)$?2011-09-14
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    Generally speaking, $$\int_0^\infty\frac{x^n}{e^x-1}dx=n!\cdot\zeta(n+1)$$2013-11-04

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