Evaluating a double integral I was trying to evaluate the following integral
$$\int_{x=0}^{\infty}\int_{y=0}^{\infty}\frac{y \ln y \ln x}{(x^2+ y^2)( 1+y^2)} dy dx.$$
I have a guess that the value of this integral is $\frac{\pi^4}{8}$. But I am unable to prove it.
Could someone please help me in evaluating this integral?
Or, can we show the following identity holds without much calculation?
$$\int_{x=0}^{\infty}\int_{y=0}^{\infty}\frac{2y \ln y \ln x}{(x^2+ y^2)( 1+y^2)} dy dx=   \int_{x=0}^{\infty}\int_{y=0}^{\infty}\frac{y (\ln y )^2}{(x^2+ y^2)( 1+y^2)} dy dx.$$
Any help or hint would be appreciated. Thanks in advance.

We shall use an approach via Mellin transforms:
We aim to find $$I=2\int_{0}^{\infty}\frac{y\ln y}{1+y^2}\int_{0}^{\infty}\frac{\ln x}{x^2+y^2}\, dx\, dy$$
We shall begin by evaluating the Mellin transform $$\int_{0}^{\infty}\frac{x^{s-1}}{x^2+y^2}\, dx\stackrel{x\to u y}{=}y^{s-2}\int_{0}^{\infty}\frac{u^{s-1}}{u^2+1}\, du=\frac{y^{s-2}\pi}{2}\csc\left(\frac{\pi s}{2}\right)$$ which is a classical result that can be obtained via the beta function or contour integration for example: see here.
Now \begin{align*}\int_{0}^{\infty}\frac{\ln x}{x^2+y^2}\, dx &=\frac{\partial}{\partial s} \left[\frac{y^{s-2}\pi}{s}\csc\left(\frac{\pi s}{2}\right)\right]_{s=1}\\ &=\lim_{s\to 1}\left[-\frac{\pi}{4} y^{s-2}\csc\left(\frac{\pi s}{2}\right)\left(\pi\cot\left(\frac{\pi s}{2}\right)-2\ln y\right)\right]\\&=\frac{\pi\ln y}{2y}\end{align*}
This means $$I=\pi\int_{0}^{\infty}\frac{\ln^2 y}{1+y^2}\, dy$$ and using the same method $$\int_{0}^{\infty}\frac{y^{s-1}}{1+y^2}\, dy=\frac{\pi}{2}\csc\left(\frac{\pi s}{2}\right)$$ so \begin{align*}I&=\pi\frac{d^2}{ds^2}\left[\frac{\pi}{2}\csc\left(\frac{\pi s}{2}\right)\right]_{s=1}\\&=\pi\lim_{s\to 1}\left[\frac{\pi^3}{16}(\cos(\pi s)+3)\csc^3\left(\frac{\pi s}{2}\right)
\right]\\&=\frac{\pi^4}{8}\end{align*}
as desired. $\square$
