Evaluation of $\int_{0}^{1}{\int_{0}^{1}{\frac{\ln \left( 1-x \right)-\ln \left( 1-y \right)}{x-y-1}dx}dy}$ I trying to verify the following:
$$
\int_{0}^{1}{\int_{0}^{1}{\frac{\ln \left( 1-x \right)-\ln \left( 1-y \right)}{x-y-1}dx}dy}=\frac{{{\pi }^{2}}}{12}+{{\ln }^{2}}\left( 2 \right)
$$
I thought the solution may be similar to this problem but i can't handle the integral in a similar way?!?!?!
 A: This is not a complete answer, but too long for a comment. I managed to reduce the two dimensional integral into a single definite integral
$$\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-x\right)-\ln\left(1-y\right)}{x-y-1}{\rm d}x{\rm d}y=\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-x\right)}{x-y-1}{\rm d}x{\rm d}y-\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-y\right)}{x-y-1}{\rm d}x{\rm d}y=$$
$$=\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-x\right)}{x-y-1}{\rm d}x{\rm d}y-\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-x\right)}{y-x-1}{\rm d}x{\rm d}y=$$
$$=\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-x\right)}{x-y-1}{\rm d}x{\rm d}y+\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-x\right)}{x-y+1}{\rm d}x{\rm d}y=$$
$$=\int_{0}^{1}\ln\left(1-x\right)\left(\ln\left(1-x\right)-\ln\left(2-x\right)\right){\rm d}x+\int_{0}^{1}\ln\left(1-x\right)\left(\ln\left(1+x\right)-\ln x\right){\rm d}x=$$
$$=\int_{0}^{1}\ln\left(1-x\right)\left(\ln\left(1-x\right)-\ln\left(2-x\right)+\ln\left(1+x\right)-\ln x\right){\rm d}x=$$
$$=\int_{0}^{1}\ln\left(1-x\right)\ln\left(\dfrac{1-x^{2}}{2x-x^{2}}\right){\rm d}x$$
According to Wolfram this integral involves the polylogarithm ${\rm Li}_{2}\left(x\right)$ and can be evaluated. Note that
$${\rm Li}_{2}\left(x\right)=\sum_{n=1}^{\infty}\dfrac{x^{n}}{n^2}$$
so you can immediately see the connection to ${\rm Li}_{2}\left(1\right)=\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$.
A: Performing the change of variable $$ u=\frac{1-x}{1-y}\,\,,\,\,v=x-y$$ the domain $D=\{(x,y)\,|\,0<x<1\,,\,0<y<1\}$ change into $$D_1=\{(u,v)\,|\,0<u<1\,,\,0<v<1-u\}\cup D_2=\{(u,v)\,|\,1<u<\infty\,,\,0<v<\frac{1-u}{u}\}$$
and your integral reads
$$\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-x\right)-\ln\left(1-y\right)}{x-y-1}{\rm d}x{\rm d}y=$$
$$\int_{0}^{1}\int_{0}^{1-u}\dfrac{\ln\left(u\right)}{v-1}\frac{v}{(u-1)^2}{\rm d}v{\rm d}u-\int_{1}^{\infty}\int_{\frac{1-u}{u}}^0\dfrac{\ln\left(u\right)}{v-1}\frac{v}{(u-1)^2}{\rm d}v{\rm d}u=$$
Here, the negative sign becomes from de Jocobian. The first integral in $v$ (in both cases) is inmediatly and get
$$\int_0^1 \frac{(1-u+\log (u)) \log (u)}{(u-1)^2} \, du -\int_1^{\infty } \frac{\left(1-\frac{1}{u}-\log \left(2-\frac{1}{u}\right)\right) \log (u)}{(u-1)^2} \, du$$
and with the change $1/u\rightarrow u$ in the second integral we arrives to
$$\int_0^1 \frac{(2(1-u)+\log (u)-\log(2-u)) \log (u)}{(u-1)^2} \, du$$
Now:


*

*$\displaystyle \int_0^1 \frac{2(1-u)\log (u)}{(u-1)^2} \, du=-2\int_0^1 \frac{ \log (u)}{u-1} \, du=2\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\int_0^1(u-1)^{n-1}du=2\sum_{n=1}^{\infty}\frac{(-1)}{n^2}=\boxed{-\frac{\pi^2}{3}}$

*Integrating by parts we obtain $\displaystyle \int_0^1 \frac{2\log (u)}{u-1} \, du=\int_0^1 \frac{\log^2 (u)}{(u-1)^2} \, du=\boxed{\frac{\pi ^2}{3}}$
Using similar techniques (Taylor series of $\log(u)$ or $\log(2-u)$) you can arrive 
-  $\displaystyle \int_0^1 -\frac{\log (2-u)\log (u)}{(u-1)^2} \, du=\boxed{\frac{\pi ^2}{12}+\log^2(2)}$
and the desired result follows.
A: With 

$$\int\frac{\ln(x+z)}{x-a}dx = \text{Li}_2\left(\frac{x+z}{a+z}\right) - \ln(x+z)\,\text{Li}_1\left(\frac{x+z}{a+z}\right) + C $$ 

we get 
$$\int\limits_0^1\frac{\ln(1-x)}{x-y-1}dx = \int\limits_0^{-1}\frac{\ln(x+1)}{x+y+1}dx = -\text{Li}_2\left(-\frac{1}{y}\right)$$ 
and the other part is: 
$$\int\limits_0^1\frac{-\ln(1-y)}{x-y-1}dx = -\ln(1-y)\ln\left(\frac{y}{y+1}\right)$$
Now we need
$$\int -\text{Li}_2\left(-\frac{1}{y}\right)dy = -y\,\text{Li}_2\left(-\frac{1}{y}\right) + y\ln\left(1+\frac{1}{y}\right) + \ln(y+1) + C$$ and 
$$\int -\ln(1-y)\ln\left(\frac{y}{y+1}\right)dy = \\ 2\,\text{Li}_2\left(\frac{1-y}{2}\right) + \text{Li}_2(y) + (y + (1-y)\ln(1-y))\ln\left(\frac{y}{y+1}\right) \\ + 2\ln(1-y)\ln\left(\frac{y+1}{2}\right) - \ln(y+1) + C$$
which leads to: 
$$\int\limits_0^1\int\limits_0^1 \frac{\ln(1-x) - \ln(1-y)}{x-y-1} dxdy = \int\limits_0^1 -\text{Li}_2\left(-\frac{1}{y}\right)dy + \int\limits_0^1 -\ln(1-y)\ln\left(\frac{y}{y+1}\right)dy \\ = \left(\frac{\pi^2}{12} + \ln 4\right) + \left((\ln 2)^2 - \ln 4\right) = \frac{\pi^2}{12} + (\ln 2)^2$$
