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.
 A: 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. $\square$
A: We can evaluate a more general integral. Let's denote
$$I(k)=\int_0^{\infty}\int_0^{\infty}\frac{y \ln ^ky \ln x}{(x^2+ y^2)( 1+y^2)} dy dx$$
where $k=0,1,2, ...$
Then
$$I(k)\overset{t=y^2}{=}\frac{1}{2^{k+1}}\int_0^\infty\int_0^\infty\frac{\ln^kt\ln x}{(x^2+t)(1+t)}dxdt\overset{x=\sqrt t \,z}{=}\frac{1}{2^{k+1}}\int_0^\infty\frac{\sqrt t\,\ln^kt}{t(1+t)}dt\int_0^\infty\frac{\ln(\sqrt t\,z)}{1+z^2}dz$$
Given that $\displaystyle \int_0^\infty\frac{\ln z}{1+z^2}dz=0$ and $\displaystyle \int_0^\infty\frac{dz}{1+z^2}dz=\frac{\pi}{2}$
$$I(k)=\frac{\pi}{2^{k+3}}\int_0^\infty\frac{\ln^{k+1}t}{\sqrt t(1+t)}dt\overset{t=x^2}{=}\frac{\pi}{2}\int_0^\infty\frac{\ln^{k+1}x}{1+x^2}dx\overset{x=e^{\pi t}}{=}\frac{\pi^{k+3}}{4}\int_{-\infty}^\infty\frac{x^{k+1}}{\cosh\pi x}dx\tag{1}$$
Let's denote $\displaystyle J(\beta)=\int_{-\infty}^\infty\frac{e^{\beta x}}{\cosh\pi x}dx$, then $\displaystyle\int_{-\infty}^\infty\frac{x^k}{\cosh\pi x}dx=\frac{d^k}{d\beta^k}J(\beta)\,\bigg|_{\beta=0}$
A straightforward evaluation of $J(\beta)$ via integration in the complex plane along a rectangular contour gives
$$J(\beta)=\frac{1}{\cos\frac{\beta}{2}};\quad\int_{-\infty}^\infty\frac{x^k}{\cosh\pi x}dx=\frac{1}{2^k}\frac{d^k}{dx^k}\frac{1}{\cos x}\,\bigg|_{x=0}\tag{2}$$
Putting (2) into (1)
$$I(k)=\Big(\frac{\pi}{2}\Big)^{k+3}\frac{d^{k+1}}{dx^{k+1}}\frac{1}{\cos x}\,\bigg|_{x=0}\tag{3}$$
$$I(k=1)=\Big(\frac{\pi}{2}\Big)^4$$
Using the presentation $\displaystyle \frac{1}{\cos x}=\sum_{n=0}^\infty(-1)^n\frac{E_{2n}}{(2n)!}x^{2n}$, where $E_{2n}$ denotes Euler' numbers,
$$I(2n-1)=(-1)^nE_{2n}\Big(\frac{\pi}{2}\Big)^{2(n+1)};\quad E_2=-1; \,E_4=5,...$$
For even $k=2n$ we get zero.
A: $$I=\int\frac{y \log(y) }{(x^2+ y^2)( 1+y^2)} dy $$ 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.
A: To show your identity holds, make the one-dimensional change of variables $x=ty$ to get
\begin{align*}
\int_{0}^\infty \int_0^\infty \frac{y \ln(y) \ln (x)}{(y^2+1)(x^2+y^2)} \ dy \ dx &= \int_{0}^\infty \int_0^\infty \frac{y^2 \ln(y) \ln (ty)}{(y^2+1)(t^2+1) y^2} \ dy \ dt \\
&=\int_{0}^\infty \int_0^\infty \frac{ \ln(y) \ln (t)}{(y^2+1)(t^2+1)} \ dy \ dt \\
&\qquad + \int_{0}^\infty \int_0^\infty \frac{ \ln^2(y)}{(y^2+1)(t^2+1)} \ dt \ dy \\
&= \left(\int_0^\infty \frac{ \ln (t)}{t^2+1} \ dt  \right)^2 + \frac{\pi}{2} \int_0^\infty \frac{\ln^2(y)}{y^2+1} \ dy.
\end{align*}
Note upon a change of variables $t=1/u,$ we have
\begin{align*}
\int_0^1\frac{ \ln (t)}{t^2+1} \ dt &= -\int_1^\infty \frac{ \ln (t)}{t^2+1} \ dt,
\end{align*}
which implies $\int_0^\infty\frac{ \ln (t)}{t^2+1} \ dt =0.$ Hence,
\begin{align*}
\int_{0}^\infty \int_0^\infty \frac{y \ln(y) \ln (x)}{(y^2+1)(x^2+y^2)} \ dy \ dx &=  \frac{\pi}{2} \int_0^\infty \frac{\ln^2(y)}{y^2+1} \ dy.
\end{align*}
In order to evaluate the right hand side, consider the triple integral
$$I=\int_0^\infty \int_0^\infty \int_0^\infty \frac{xz}{(x^2+1)(z^2+1)(x^2y^2z^2+1)} \ dy \ dx \ dz.$$ By carrying out the integration directly, we can find $I=\frac{\pi^3}{8}.$ On the other hand, using Fubini's Theorem to integrate with respect to $z$ first, we get by partial fractions (or Mathematica)
\begin{align*}
I &=\int_0^\infty \int_0^\infty \int_0^\infty \frac{xz}{(x^2+1)(z^2+1)(x^2y^2z^2+1)} \ dz \ dx \ dy \\
&= \int_0^\infty \int_0^\infty \frac{x \ln \left(xy\right)}{\left(x^2+1\right) \left( x^2 y^2-1\right)} \ dx \ dy  \\
&= \int_0^\infty \int_0^\infty \frac{x \ln \left(x\right)}{\left(x^2+1\right) \left( x^2 y^2-1\right)} \ dy \ dx + \int_0^\infty \int_0^\infty \frac{x \ln \left(y\right)}{\left(x^2+1\right) \left( x^2 y^2-1\right)} \ dx \ dy \\
&= \int_0^\infty \int_0^\infty \frac{x \ln \left(y\right)}{\left(x^2+1\right) \left( x^2 y^2-1\right)} \ dx \ dy \\
&= \int_0^\infty \frac{\ln^2(y)}{y^2+1} \ dy,
\end{align*}
in which the first double integral in the third to last equality has Cauchy Principal Value $0$ and thus vanishes, and we carried out the double integral with respect to $x$ in the second to last equality using partial fractions once again.
Hence, $$I=\int_0^\infty \frac{\ln^2(y)}{y^2+1} \ dy=\frac{\pi^3}{8},$$ and so the desired integral is equal to $$\frac{\pi}{2} I=\frac{\pi^4}{16}.$$
Remark: These type of integral calculations are prominent in my joint AMS paper with Daniele Ritelli. See here: https://www.ams.org/journals/qam/2018-76-03/S0033-569X-2018-01499-3/
