Show that, $\int_{0}^{1}\int_{0}^{1}\frac{\ln x-\ln y}{x-y}dxdy=2\zeta(2)$ i take firstly, $w=\frac{x}{y}$ to get
$\int_{0}^{1}\int_{0}^{1/y}\frac{\ln w}{w-1}dwdy$
Now how shall i continue this exercise ?
 A: By symmetry
$$ \int_{0}^{1}\int_{0}^{1}\frac{\log x-\log y}{x-y}\,dx \,dy = 2\int_{0}^{1}\int_{0}^{x}\frac{\log x -\log y}{x-y}\,dy\,dx = 2\int_{0}^{1}\int_{0}^{1}\frac{-\log z}{1-z}\,dz\,dx $$
and the last integral boils down to
$$ 2\sum_{n\geq 0}\int_{0}^{1}-\log(z)z^n\,dz = 2\sum_{n\geq 0}\frac{1}{(n+1)^2}=2\,\zeta(2)=\frac{\pi^2}{3}.$$
A: Let us transform the problem to polar coordinates $x=r\cos\phi$, $y=r\sin\phi$, $dxdy=r dr d\phi$:
$$
\int_{0}^{1}\int_{0}^{1}\frac{\log x-\log y}{x-y}dxdy\\
=\int_{0}^{\pi/4}\frac{\log\cos\phi-\log\sin\phi}{\cos\phi-\sin\phi}d\phi\int_{0}^{1/\cos\phi}dr
+\int_{\pi/4}^{\pi/2}\frac{\log\cos\phi-\log\sin\phi}{\cos\phi-\sin\phi}d\phi\int_{0}^{1/\sin\phi}dr\\
\stackrel{*}{=}2\int_{0}^{\pi/4}\frac{\log\cos\phi-\log\sin\phi}{\cos\phi(\cos\phi-\sin\phi)}d\phi\stackrel{t=\tan\phi}{=}2\int_{0}^{1}\frac{\log t}{t-1}dt
\stackrel{u=1-t}{=}-2\int_{0}^{1}\frac{\log (1-u)}{u}du\\
=2\int_{0}^{1}du\sum_{n=1}^\infty\frac{u^{n-1}}{n}=2\sum_{n=1}^\infty\left[\frac{u^{n}}{n^2}\right]_0^1=2\sum_{n=1}^\infty\frac{1}{n^2}=2\zeta(2).
$$
In $\stackrel{*}{=}$ we used the fact that both integrals in the previous line are equal, which can be seen by substitition $\phi\mapsto\frac{\pi}{2}-\phi$ into the second one.
A: Hint:


*

*Use that $(1-w)^{-1} = \sum_{k=0}^\infty w^k$

*And verify $\int \log(x) x^k \, \mathrm{dx} = (k+1)^{-2} x^{k+1} ((k+1) \log(x)-1) $

A: $$I=\int_{0}^{1}\int_{0}^{1}\frac{ln(x)-ln(y)}{x-y}dxdy\\
\\
\\$$
$$=\int_{0}^{1}\int_{0}^{1}\frac{ln(\frac{x}{y})}{x-y}dxdy\ \ \ , let\ \ \frac{x}{y}=t\\
\\$$
$$=\int_{0}^{1}\int_{x}^{\infty }\frac{ln(t)}{t(t-1)}dt=[\int_{x}^{\infty }\frac{xln(t)}{t(t-1)}dt]\tfrac{1}{0}+\int_{0}^{1}\frac{ln(x)}{x-1}dx\\
\\
=\int_{1}^{\infty }\frac{ln(t)}{t(t-1)}dt+\int_{0}^{1}\frac{ln(x)}{x-1}dx=\int_{0}^{1}\frac{ln(x)dx}{x-1}+\int_{0}^{1}\frac{ln(x)dx}{x-1}\\
\\
=2\int_{0}^{1}\frac{ln(x)}{x-1}dx\ \ \ \ \ ,\ \ \ let\ u=x-1\ \ \\
\\$$
$$\therefore I=2\int_{-1}^{0}\frac{ln(u-1)}{u}du=2\int_{-1}^{0}\sum_{n=0}^{\infty }\frac{(-1)^{n}}{n+1}.u^ndu\\
\\
\\
\therefore I=2\sum_{n=0}^{\infty }\frac{(-1)^{n}}{(n+1)^2}u^{n+1}\tfrac{0}{-1}\\
\\
\\
\therefore I=2\sum_{n=0}^{\infty }\frac{1}{(n+1)^2}=2.\frac{\pi ^{2}}{6}=\frac{\pi ^{2}}{3}=2\zeta (2)
\\$$
