Square unit integrals This is quite a messy integral, I am trying to show that it has an answer of $2C-1$.
Where $C=0.915966...$ is a Catalan's constant.
$$\int_{0}^{1}\int_{0}^{1}\frac{(1-xy)^2}{1+x^2y^2}\cdot \frac{y}{x}\cdot \ln(xy)\mathrm dx \mathrm dy$$
I am not sure where to start or how to begin.
 A: Note first that the integrand is negative but we have $2 \rm{C} - 1 \approx 0.832 > 0$, so this cannot possibly be the correct answer. 
In fact, things are even worse: due to the factor of $\frac{1}{x}$ the integral over $x$ diverges at the lower limit. Therefore the correct result is $-\infty$.
However, there is a certain part of your integral that has the desired value. If instead of $ (1 - x y)^2 = 1 + x^2 y^2 - 2 x y$ we only consider $ - 2 x y$ in the numerator, we find
\begin{align}
\int \limits_0^1 \int \limits_0^1 \frac{- 2 x y}{1+ x^2 y^2} \frac{y}{x} \ln (x y) \, \mathrm{d} x \,  \mathrm{d} y &= 2 \int \limits_0^1 y^2 \int \limits_0^1 \frac{- \ln (x y)}{1 + x^2 y^2} \, \mathrm{d} x \, \mathrm{d} y \\
&= 2 \int \limits_0^1 y \int \limits_0^y \frac{- \ln (t)}{1 + t^2} \, \mathrm{d} t \, \mathrm{d} y \\
&= \int \limits_0^1  \frac{- \ln (t)}{1 + t^2} \int \limits_t^1 2 y  \, \mathrm{d} y \, \mathrm{d} t \\
&= \int \limits_0^1  \frac{1-t^2}{1 + t^2} (- \ln (t)) \, \mathrm{d} t \\
&= 2 \int \limits_0^1  \frac{- \ln (t)}{1 + t^2} \, \mathrm{d} t + \int \limits_0^1 \ln(t) \, \mathrm{d} t \\
&= 2 \mathrm{C} - 1 \, .
\end{align}
Here we have made the substitution $t = x y$ and then employed Tonelli's theorem to interchange the order of integration. 
The last step just uses an integral representation of Catalan's constant found by writing $\frac{1}{1+t^2}$ as a geometric series.
A: See e.g. Section 3 in the article "Double integrals and infinite products for some constants via analytic continuations of Lerch's transcendent" by J. Guillera and J. Sondow in Raman. J. 16 (2008) 247-270 doi:10.1007/s11139-007-9102-0 .
