Double integral with logarithms [closed]

Question

Faster method to calculate the exact solution of the following integral (based on the ideas of Fedor Petrov: https://mathoverflow.net/users/4312/fedor-petrov, Double integral with logarithms, URL (version: 2019-04-15): https://mathoverflow.net/q/328126):

$$J\equiv \int_{0}^{1}{\int_{0}^{1}{\frac{\ln x-\ln y}{x-y}}}dxdy .$$

Since $$f\left( x,y \right)=\frac{\ln x-\ln y}{x-y}=f\left( y,x \right),$$ the surface $f\left( x,y \right) $ is symmetric with respect to the bisector plane $x = y$; so,

$$\frac{J}{2}=\int_{0}^{1}{dx\int_{0}^{x}{\frac{\ln x-\ln y}{x-y}}}dy.$$

With the change of variable$$y\equiv tx,\ t\in \left( 0,\ 1 \right),$$ the integral $$\int_{0}^{x}{\frac{\ln x-\ln y}{x-y}}dy,$$ is transformed into the following one that does not depend on $x$, $$ I\equiv -\int_{0}^{1}{\frac{\ln t}{1-t}\,}dt.$$

The integration on the unit square $(0, 0), (1, 0), (1, 1), (0,1)$ is reduced to the integration on the triangle $(0, 0), (1, 0), (1, 1).$

To solve the integral $I$ we will carry out the new change of variable, $$s\equiv 1-t,$$ by which $I$ is transformed into the integral that defines the dilogarithm, whose value for $s = 1$ coincides with the Riemann zeta $\zeta \left( 2 \right)$,whose value is well known:

$$I=\int_{1}^{0}{\frac{\ln \left( 1-s \right)}{s}\,}ds=\text{L}{{\text{i}}_{2}}\left( 1 \right)=\zeta \left( 2 \right)=\frac{{{\pi }^{2}}}{6}.$$ Therefore, the solution to the proposed integral is $$J=\frac{{{\pi }^{2}}}{3}.$$

Note. I've tried it by polylogarithmic transformations, but I couldn't get the result $\frac{{{\pi }^{2}}}{3}.$

Answer 1

By symmetry we have $J/2=\int_0^1 dx \int_0^x f(x, y) dy$ where $f(x, y) $ is your integrand. Integrating against $y$ for fixed $x$ we denote $y=tx$, $t$ varies from 0 to 1 and the integral against $y$ reads as $-\int_0^1 \frac{\log t} {1-t}dt$. It does not depend on $x$ and is well known to be equal to $\pi^2/6$ (you may use the geometric progression expansion $\frac{1}{1 - t} =\sum_{n>0 } t^{n-1}$ and integrate term-wise to get $\sum 1/n^2$).