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Can I evaluate this integral:

$$ \int_0^1\frac{\log(x)}{x-1}dx $$

without knowing the value of $\zeta(2)$? In particular can I use methods of complex analysis?

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  • $\begingroup$ Such integral is $\zeta(2)$, by the substitution $x=e^{-t}$ and the integral representation of the $\zeta$ function, for instance. What are you exactly looking for? A direct proof of $\int_{0}^{1}\frac{\log x}{x-1}\,dx=\frac{\pi^2}{6}$ which does not exploit the identity $\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, for instance? $\endgroup$ – Jack D'Aurizio Feb 1 '18 at 19:42
  • $\begingroup$ In such a case, you may have a look at the Pace-Ritelli approach outlined in this classical thread. $\endgroup$ – Jack D'Aurizio Feb 1 '18 at 19:48
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$\begin{align} \int_0^1 \frac{\ln x}{x-1}\,dx&=\Big[\ln(1-x)\ln x\Big]_0^1-\int_0^1 \frac{\ln(1-x)}{x}\,dx\\ &=-\int_0^1 \frac{\ln(1-x)}{x}\,dx\end{align}$

and read my answer, https://math.stackexchange.com/a/2632547/186817

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$$-\int^{1}_{0}\frac{\ln x}{1-x}dx=-\int^{1}_{0}\ln x\sum^{\infty}_{n=0}x^ndx$$

$$=-\sum^{\infty}_{n=0}\int^{1}_{0}x^n\ln(x)dx$$

By parts method

$$ \sum^{\infty}_{n=0}\frac{1}{(n+1)^2}=\frac{\pi^2}{6}=\zeta(2)$$

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    $\begingroup$ Still, you need to prove that $\sum_{n\geq 1}\frac{1}{n^2}=\frac{\pi^2}{6}$ separately, which is what I believe the OP wants to avoid. $\endgroup$ – Jack D'Aurizio Feb 1 '18 at 19:45

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