Cfxtrjtrk

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):

$$Jequiv int_{0}^{1}{int_{0}^{1}{frac{ln x-ln y}{x-y}}}dxdy .$$

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

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

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

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,
$$sequiv 1-t,$$
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 $frac{{{pi }^{2}}}{3}.$

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 $sum 1/n^2$).