<s>[INST] Is possible to use "Feynman's trick" (differentiate under the integral or Leibniz integral rule) to calculate $\int_0^1 \frac{\ln(1-x)}{x}dx\:?$ I heard that the equivalent integral: $-\int_0^\infty \frac{x}{e^x-1}dx$ can be done using Contour integration (I never studied this). Also that sometimes "Leibniz integral rule" is used instead of Contour integration. So can "Feynman's trick" be used to show that $\int_0^1 \frac{\ln(1-x)}{x}dx = -\frac{\pi^2}{6}$   $\:\:?$
 [/INST] Let $\displaystyle J=\int_0^1 \frac{\ln(1-x)}{x}\,dx$
Let $f$ be a function defined on $\left[0;1\right]$,
$\displaystyle f(s)=\int_0^{\frac{\pi}{2}} \arctan\left(\frac{\cos t-s}{\sin t}\right)\,dt$
Observe that,
$\begin{align} f(0)&=\int_0^{\frac{\pi}{2}}\arctan\left(\frac{\cos t}{\sin t}\right)\,dt\\
&=\int_0^{\frac{\pi}{2}} \left(\frac{\pi}{2}-t\right)\,dt\\
&=\left[\frac{t(\pi-t)}{2}\right]_0^{\frac{\pi}{2}}\\
&=\frac{\pi^2}{8}
\end{align}$
$\begin{align} f(1)&=\int_0^{\frac{\pi}{2}}\arctan\left(\frac{\cos t-1}{\sin t}\right)\,dt\\
&=\int_0^{\frac{\pi}{2}}\arctan\left(-\tan\left(\frac{t}{2}\right)\right)\,dt\\
&=-\int_0^{\frac{\pi}{2}}\arctan\left(\tan\left(\frac{t}{2}\right)\right)\,dt\\
&=-\int_0^{\frac{\pi}{2}} \frac{t}{2}\,dt\\
&=-\frac{\pi^2}{16}
\end{align}$
For $0<s<1$,
$\begin{align}
f^\prime(s)&=-\int_0^{\frac{\pi}{2}}\frac{\sin t}{1-2s\cos t+s^2}\,dt\\
&=-\Big[\frac{\ln(1-2s\cos t+s^2)}{2s}\Big]_0^{\frac{\pi}{2}}\\
&=\frac{\ln(\left(1-s)^2\right)}{2s}-\frac{\ln(1+s^2)}{2s}\\
&=\frac{\ln(1-s)}{s}-\frac{\ln(1+s^2)}{2s}\\
\end{align}$
Therefore,
$\begin{align}
f(1)-f(0)&=\int_0^1 f^\prime(s)\,ds\\
&=\int_0^1 \left(\frac{\ln(1-s)}{s}-\frac{\ln(1+s^2)}{2s}\right)\,ds\\
-\frac{\pi^2}{16}-\frac{\pi^2}{8}&=J-\int_0^1 \frac{\ln(1+s^2)}{2s}\,ds\\
-\frac{3\pi^2}{16}&=J-\int_0^1 \frac{\ln(1+s^2)}{2s}\,ds\\
\end{align}$
In the latter integral perform the change of variable $y=s^2$,
$\begin{align}
-\frac{3\pi^2}{16}&=J-\frac{1}{4}\int_0^1 \frac{\ln(1+y)}{y}\,dy\\
&=J-\frac{1}{4}\int_0^1 \frac{\ln(1-y^2)-\ln(1-y)}{y}\,dy\\
&=J+\frac{1}{4}J-\frac{1}{4}\int_0^1 \frac{\ln(1-y^2)}{y}\,dy\\
\end{align}$
In the latter integral perform the change of variable $x=y^2$,
$\begin{align}
-\frac{3\pi^2}{16}&=J+\frac{1}{4}J-\frac{1}{4}\times \frac{1}{2}J\\
&=\frac{9}{8}J\\
\end{align}$
Therefore,
$\begin{align}J&=\frac{8}{9}\times -\frac{3}{16}\pi^2\\
&=\boxed{-\frac{\pi^2}{6}}\end{align}$
</s>