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It is well known that $$\int_0^1\frac{\log{x}}{1 - x}\,\mathrm{d}x = -\frac{\pi^2}{6} $$

This is generally proved by expanding the geometric series.

My question is: can this be done in reverse?

Can we evaluate the integral $\int_0^1\frac{\log{x}}{1 - x}dx$ using other method, for example, differentiation under the integral sign, and thus prove $\zeta(2) = \frac{\pi^2}6$?

Quanto
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Ayesha
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    A popular method is to rewrite $\int_0^1 \frac{\log x,\mathrm{d}x}{x-1}=\int_0^1\int_0^1 \frac{\mathrm{d}y,\mathrm{d}x}{1−xy}$, letting $y=\alpha-\beta,;x=\alpha+\beta;\text{etc}.$ – ocg Jan 05 '14 at 19:17
  • I wonder if it is possible to do this by considering a contour integral $$\oint \frac{\log^2 z}{1-z} $$ over some contour. – Bennett Gardiner Jan 05 '14 at 23:45
  • Differentiate under integral sign https://math.stackexchange.com/a/3404200/686284 – Quanto Sep 05 '21 at 16:20

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Credits to Daniele Ritelli: $$\begin{eqnarray*}\zeta(2)&=&\frac{4}{3}\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\color{red}{\frac{4}{3}\int_{0}^{1}\frac{\log y}{y^2-1}dy}\\&=&\frac{2}{3}\int_{0}^{1}\frac{1}{y^2-1}\left[\log\left(\frac{1+x^2 y^2}{1+x^2}\right)\right]_{x=0}^{+\infty}dy\\&=&\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+x^2 y^2)}dx\,dy\\&=&\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{dx\, dz}{(1+x^2)(1+z^2)}=\frac{4}{3}\cdot\frac{\pi}{4}\cdot\frac{\pi}{2}=\color{red}{\frac{\pi^2}{6}}.\end{eqnarray*}$$

Jack D'Aurizio
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