TITLE: Prove: $\int_0^1 \frac{\ln x }{x-1} d x=\sum_1^\infty \frac{1}{n^2}$
QUESTION [29 upvotes]: I'd like your help with proving that
$$\int_0^1 \frac{\ln x }{x-1}d x=\sum_{n=1}^\infty \frac{1}{n^2}.$$
I tried to use Fourier series, or to use a power series and integrate it twice but it didn't work out for me.
Any suggestions?
Thanks!

REPLY [10 votes]: Related problem: I, II. Using the change of variables $u=-\ln(x)$ and the identity

$$ \int_{0}^{\infty}\frac{u^{s-1}}{e^u -1}=\zeta{(s)}\Gamma{(s)} $$

we reach to the deisred result
$$ \int_0^1 \frac{\ln x }{x-1}= \int_{0}^{\infty}\frac{u}{e^u -1}=\zeta{(2)}\Gamma{(2)} =\sum_{n=1}^\infty \frac{1}{n^2}. $$
Added: Note that,

$$ \int_{0}^{\infty}\frac{u^{s-1}}{e^u - 1}=\int_{0}^{\infty}\frac{u^{s-1}}{e^u}(1-e^{-u})^{-1}= \sum_{n=0}^{\infty} \int_{0}^{\infty}{u^{s-1}e^{-(n+1)u}}$$
$$= \sum_{n=0}^{\infty}\frac{1}{(n+1)^s} \int_{0}^{\infty}{y^{s-1}e^{-y}}= \sum_{n=1}^{\infty}\frac{1}{n^s} \Gamma(s)= \zeta(s) \Gamma(s).$$