<s>[INST] 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.
 [/INST] $$I=\int\frac{y \log(y) }{(x^2+ y^2)( 1+y^2)} dy $$ is not  bad since
$$A=\frac{y  }{(x^2+ y^2)( 1+y^2)}=\frac{y}{(y-i) (y+i) (y-i x) (y+i x)}$$ Using partial fraction decomposition
$$A=\frac 1{2(x^2-1)}\left(\frac 1{y-i} +\frac 1{y+i}-\frac 1{y-ix}-\frac 1{y+ix}\right)$$ and
$$\int \frac {\log(y)}{y+a}\,dy=\text{Li}_2\left(-\frac{y}{a}\right)+\log (y) \log   \left(1+\frac{y}{a}\right)$$ Therefore
$$I=\frac 1{2(x^2-1)}\left(\frac{1}{2}
   \left(\text{Li}_2\left(-y^2\right)-\text{Li}_2\left(-\frac{y^2
   }{x^2}\right)\right)+\log (y) \log \left(\frac{x^2
   \left(y^2+1\right)}{x^2+y^2}\right)  \right)$$ and then
$$J=\int_0^\infty\frac{y \log(y) }{(x^2+ y^2)( 1+y^2)} dy =\frac{\log ^2(x)}{2 \left(x^2-1\right)}$$
Then, what remains is the computation of
$$K=\int \frac{\log ^3(x)}{x^2-1}\,dx=\frac 12\left(\int \frac{\log ^3(x)}{x-1}\,dx- \int \frac{\log ^3(x)}{x+1}\,dx\right)$$ which is not too bad using a few integrations by parts.
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