I want to evaluate $$ \int_{-\infty}^0\int_1^{\infty}\frac{1}{2}\log\left(\frac{x(1-y)}{y(1-x)}\right)(y-x)^{-2}dxdy $$ which Mathematica tells me is $\pi^2/6$. I tried a few change of variables, series expansions, etc., but didn't get anywhere.
[For context, the measure $(y-x)^{-2}dxdy$ is isometry invariant on the space of geodesics $(y,x)\in\mathbb{R}^2$ (upper half-plane model of two-dimensional hyperbolic space) and $\frac{1}{2}\log\left(\frac{x(1-y)}{y(1-x)}\right)$ is the length of the geodesic segment on the geodesic $(y,x)$ between the geodesics $(\infty,0)$ and $(\infty,1)$.]
The given integral can be expressed, after the change of variables $x-y\mapsto t, y\mapsto s$, as $$\int _1^{+\infty }\frac{1}{2t^2}\left(\int_{1-t}^0 \log\left[\frac{(1-s)(t+s)}{s(1-(t+s))}\right] \, ds\right)dt,$$ and through $t\mapsto \frac{1}{u}$ we obtain $$ \int_{1}^{+\infty}\frac{t\log t-(t-1)\log(t-1)}{t^2}\,dt =\int_{0}^{1}\frac{(u-1)\log(1-u)-u\log u}{u}\,du,$$ or, taking into account $\int_{0}^{1}\log(u)\,du = \int_{0}^{1}\log(1-u)\,du$, $$ -\int_{0}^{1}\frac{\log(1-u)}{u}\,du = \int_{0}^{1}\sum_{n\geq 1}\frac{u^{n-1}}{n}\,du = \sum_{n\geq 1}\frac{1}{n^2}={\zeta(2)}.$$