I am reading Ahlfors' Complex Analysis. On page 205 he defined $$C_n=(-1)^{n-1}\frac{1}{\pi} \int_0^\infty \eta^{2n-2} \log(\frac{1}{1-e^{-2\pi \eta}})d\eta$$ and he said "It can be proved (for instance by means of residues) that the $C_n$ are connected with the Bernoulli numbers by $$C_n=(-1)^{n-1}\frac{1}{(2n-1)2n}B_n.$$" And he defined Bernoulli numbers $B_n$ to be the constant in the Laurent series $$\frac{1}{e^z-1}=\frac{1}{z}-\frac{1}{2}+\sum_1^\infty (-1)^{k-1} \frac{B_k}{(2k)!} z^{2k-1}$$
I don't know how he got this. The definition of $C_n$ seems too complicated to simplyfy to me.
Integrate by parts. This gives: $$ \int_0^\infty \eta^{2n-2} \log(\frac{1}{1-e^{-2\pi \eta}})d\eta=\frac{2\pi}{2n-1}\int_0^{\infty}\frac{\eta^{2n-1}d\eta}{e^{2\pi\eta}-1}. $$ Now, use this
and you are there.