In Ahlfors' Complex Analysis text, the Bernoulli numbers, $B_k$, are defined as the coefficients in a Laurent development: $$(e^z-1)^{-1}=\frac{1}{z}-\frac{1}{2}+ \sum_1^\infty (-1)^{k-1} \frac{B_k}{(2k)!} z^{2k-1}. $$ (I'm aware that this definition is different from the modern one on Wikipedia.)

Over the course of proving Stirling's formula, the author states that it can be shown that for all $\nu \geq1$ $$ (-1)^{\nu-1} \frac{1}{\pi} \int_0^\infty \eta^{2\nu-2} \log \left( \frac{1}{1-e^{-2 \pi \eta}} \right) \mathrm{d} \eta=(-1)^{\nu-1} \frac{1}{(2\nu-1)2\nu} B_\nu$$ "by means of residues".

I have tried showing that by reaching the function in the definition of the Bernoulli numbers, using integration by parts. I found that $$ \frac{1}{\pi} \int_0^\infty \eta^{2\nu-2} \log \left( \frac{1}{1-e^{-2 \pi \eta}} \right) \mathrm{d} \eta=\frac{2}{2\nu-1} \int_0^\infty \eta^{2\nu-1} (e^{2 \pi \eta}-1)^{-1} \mathrm{d} \eta $$ Sadly, I can't find any point $\eta$ with a residue containing $B_\nu$... Maybe if the powers of $\eta$ were in the denominator I could have done something.

Can anyone help me prove this identity?

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    $\begingroup$ I am not very good at math, but maybe you could multiply the integrand by $\eta/\eta$ and expand $- log(1 - e^{-2\pi\eta})$ as a series? Sorry if this is all wet. $\endgroup$ – user12802 Aug 21 '13 at 18:49
  • $\begingroup$ @Andrew thanks for the advice, but I can't see how will the Bernoulli numbers pop up that way. $\endgroup$ – user1337 Aug 21 '13 at 18:54
  • $\begingroup$ Your integral can be evaluated using the zeta function. See here. $\endgroup$ – Mhenni Benghorbal Aug 21 '13 at 20:41
  • $\begingroup$ @user1337 : I was stuck on this same problem. I understand the solution here - it's cool. In the text where Ahlfors does this, he says you can user residues to get it. Have you found a simple way to do that? $\endgroup$ – bryanj Apr 23 '14 at 23:34
  • $\begingroup$ @bryanj no, sorry. $\endgroup$ – user1337 Apr 24 '14 at 4:01

See this answer for starters. The latter integral is expressed in terms of a Riemann zeta function:

$$\frac{2}{2\nu-1} \int_0^\infty d\eta \, \eta^{2\nu-1} (e^{2 \pi \eta}-1)^{-1} = \frac{2}{2\nu-1} \frac{(2 \nu-1)!}{(2 \pi)^{2 \nu}} \zeta(2 \nu)$$

Use the relation

$$\zeta(2 \nu) = \frac{(-1)^{\nu+1} B_{2 \nu} (2 \pi)^{2 \nu}}{2 (2 \nu)!}$$

and the sought-after result follows.

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