How to evaluate $\int_{0}^{\infty} \int_{0}^{\infty} \frac{\sin(x) \sin(y) \sin(x+y)}{x y(x+y)} ~{\rm d}x ~{\rm d}y$? I came across the problem of evaluating the double integral:
$$\int_{0}^{\infty} \int_{0}^{\infty} \frac{\sin(x) \sin(y) \sin(x+y)}{x y(x+y)} ~{\rm d}x ~{\rm d}y$$
My attempt at this was to substitute $\sin(x) \cos(y) + \cos(x) \sin(y)$ for $\sin(x+y)$, although I had no clue where to go from there. My second attempt was to use trigonometric identities to rewrite the integral as:
$$\frac{1}{4} \int_{0}^{\infty} \int_{0}^{\infty} \frac{\sin(2x) + \sin(2y) - \sin(2x + 2y)}{x^2 y + x y^2} ~{\rm d}x ~{\rm d}y$$
Which didn't seem to help, so I continued to get:
$$\frac{1}{2} \int_{0}^{\infty} \int_{0}^{\infty} \frac{\sin(2x) \sin^2 (y) + \sin(2y) \sin^2 (x)}{x^2 y + x y^2} ~{\rm d}x ~{\rm d}y$$
I am sure I am missing something, but I am unsure how to integrate this. Any help or advice is appreciated.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{0}^{\infty}\int_{0}^{\infty}{\sin\pars{x}\sin\pars{y}\sin\pars{x + y}
\over xy\pars{x + y}}\,\dd x\,\dd y}}
\\[5mm] = &\
\int_{0}^{\infty}\int_{0}^{\infty}{\sin\pars{x}\sin\pars{y}\over xy}\
\overbrace{\bracks{{1 \over 2}\int_{-1}^{1}\expo{\ic k\pars{x + y}}\dd k}}
^{\ds{\sin\pars{x + y} \over x + y}}\ \,\dd x\,\dd y
\\[5mm] = &\
{1 \over 2}\int_{-1}^{1}\bracks{\int_{0}^{\infty}{\sin\pars{x} \over x}
\,\expo{\ic k x}\,\dd x}^{2}\dd k\label{1}\tag{1}
=
{1 \over 2}\int_{-1}^{1}
\bracks{\pi + 2\ic\,\mrm{arctanh}\pars{k} \over 2}^{2}\dd k
\\[5mm] = &\
{1 \over 4}\int_{0}^{1}
\bracks{\pi^{2} - 4\,\mrm{arctanh}^{2}\pars{k}}\dd k
=
{\pi^{2} \over 4} -
{1 \over 4}\int_{0}^{1}\ln^{2}\pars{1 - k \over 1 + k}\,\dd k
\\[5mm] \stackrel{t\ =\ \pars{1 - k}/\pars{1 + k}}{=}\,\,\,&
{\pi^{2} \over 4} - {1 \over 2}\
\underbrace{\int_{0}^{1}{\ln^{2}\pars{t} \over \pars{1 + t}^{2}}\,\dd t}
_{\ds{\pi^{2} \over 6}}
=
{\pi^{2} \over 4} - {1 \over 2}\,{\pi^{2} \over 6} =
 \bbx{\pi^{2} \over 6}\label{2}\tag{2}
\end{align}
A: Let $$I=\int_0^\infty\int_0^\infty\frac{\sin(x)\cos(x)\ln(x+y)}{xy(x+y)}\ dx\ dy$$
and
$$I(a)=\int_0^\infty\int_0^\infty\frac{\sin(x)\cos(y)\sin(a(x+y))}{xy(x+y)}\ dx\ dy,\quad I(0)=0,\quad I(1)=I$$
$$I'(a)=\int_0^\infty\int_0^\infty\frac{\sin(x)\cos(y)\cos(a(x+y))}{xy}\ dx\ dy$$
$$=\Re\int_0^\infty\int_0^\infty\frac{\sin(x)\cos(y)e^{ia(x+y)}}{xy}\ dx\ dy$$
$$=\Re\int_0^\infty\int_0^\infty\frac{\sin(x)e^{iax}}{x}\cdot\frac{\sin(y)e^{iay}}{y}\ dx\ dy=\Re\left(\color{blue}{\int_0^\infty\frac{\sin(x)e^{iax}}{x}\ dx}\right)^2$$
$$=\Re\left(i\ \text{arctanh}(1/a)\right)^2=-\Re\ \text{arctanh}^2{(1/a)}$$
$$\Longrightarrow I=-\Re\int_0^1\text{arctanh}(1/a)^2\ da=-\frac14\Re\int_0^1\ln^2\left(-\frac{1-a}{1+a}\right)\ da=-\frac12\Re\int_0^1\frac{\ln^2(-x)}{(1+x)^2}\ dx$$
$$=-\frac12\int_0^1\frac{\ln^2(x)-\pi^2}{(1+x)^2}\ dx=-\frac12\int_0^1\frac{\ln^2(x)}{(1+x)^2}\ dx+\frac{\pi^2}2\underbrace{\int_0^1\frac{1}{(1+x)^2}\ dx}_{1/2}$$
$$=\frac12\sum_{n=1}^\infty(-1)^nn\int_0^1 x^{n-1}\ln^2(x)\ dx+\frac{\pi^2}{4}=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}+\frac{\pi^2}{4}=-\frac{\pi^2}{12}+\frac{\pi^2}{4}=\frac{\pi^2}{6}$$

Proving the blue integral;
Let $$A=\int_0^\infty\frac{\sin(x)e^{iax}}{x}\ dx=\int_0^\infty\frac{\sin(x)e^{-\frac{a}{i}x}}{x}\ dx$$
and $$I(r)=\int_0^\infty\frac{\sin(r x)e^{-\frac{a}{i}x}}{x}\ dx, \quad I(0)=0, \quad I(1)=A$$
$$\Longrightarrow I'(r)=\int_0^\infty \cos(r x)e^{-\frac{a}{i}x}\ dx=\frac{a/i}{r^2+(a/i)^2}=-\frac{ia}{r^2-a^2}$$
$$\Longrightarrow A=i\int_0^1\frac{a}{a^2-r^2}\ dr=i\ \text{arccoth}(a)=i\ \text{arctanh}(1/a)$$
A: This integral appeared as problem 11953 in the American Mathematical Monthly, Vol.124, January 2017 (proposed C. I. Valean (Romania)). The result is $\zeta(2)$.
This is the solution that 
I submitted a few months. For more solutions you can take a look at this page.
We have that
$$\frac{|\sin(x) \sin(y) \sin(x+y)|}{xy(x+y)}\leq 
\left\{\begin{array}{ll}
        \frac{1}{xy(x+y)}  
        & \text{for $(x,y)\in [1,+\infty)\times [1,+\infty)$}\\
        \frac{1}{x(x+y)}  
        & \text{for $(x,y)\in [1,+\infty)\times [0,1]$}\\
        \frac{1}{y(x+y)}  
        & \text{for $(x,y)\in [0,1]\times [1,+\infty)$}\\
        1  
        & \text{for $(x,y)\in [0,1]\times [0,1]$}\\
        \end{array}\right. 
$$
Moreover
$$\int_{1}^{\infty}\left(\int_{1}^{\infty}\frac{dx}{xy(x+y)}\right)dy=\int_{1}^{\infty}\frac{\ln(y+1)}{y^2}dy=2\ln(2),$$
and
$$
\int_{0}^{1}\left(\int_{1}^{\infty}\frac{dx}{x(x+y)}\right)dy=
\int_{0}^{1}\frac{\ln(y+1)}{y}dy=
\frac{\pi^2}{12}.$$
Therefore, by Fubini-Tonelli Theorem, the given double integral can be written as iterated integrals
\begin{align*}
I&:=\int_{0}^{\infty}\int_{0}^{\infty}\frac{\sin(x) \sin(y) \sin(x+y)}{xy(x+y)}\,dx\,dy=\lim_{R\to +\infty} 
\int_{x=0}^{R}\left(\int_{y=0}^{R}\frac{\sin(x) \sin(y) \sin(x+y)}{xy(x+y)}\,dx\right)\,dy\\
&=\lim_{R\to +\infty} 2\int_{x=0}^{R}\left(\int_{y=0}^{x}\frac{\sin(x) \sin(y) \sin(x+y)}{xy(x+y)}\,dy\right)\,dx\\
&=\lim_{R\to +\infty}\frac{1}{2}\int_{x=0}^{R}\left(\int_{y=0}^{x}\frac{\sin(2x)+\sin(2y)-\sin(2x+2y)}{xy(x+y)}\,dy\right)\,dx.
\end{align*}
Then, after letting $u=2x$, $v=2y=tu$,
\begin{align*}
I&=\lim_{R\to +\infty}\int_{u=0}^{R}\left(\int_{v=0}^{u}\frac{\sin(u)+\sin(v)-
\sin(u+v)}{uv(u+v)}\,dv\right)\,du\\
&=\lim_{R\to +\infty}\int_{u=0}^{R}\left(\int_{t=0}^{1}\frac{\sin(u)+\sin(tu)-
\sin(u+tu)}{t(1+t)u^2}\,dt\right)\,du
=\lim_{R\to +\infty}\int_{t=0}^{1}\frac{f_R(t)}{t(1+t)}\,dt,
\end{align*}
where
\begin{align*}
f_R(t):=\int_{u=0}^{R}&\frac{\sin(u)+\sin(tu)-\sin(u+tu)}{u^2}\,du
=\left[-\frac{\sin(u)+\sin(tu)-\sin(u+tu)}{u}\right]_{u=0}^{R}\\
&\qquad+\int_{u=0}^{R}\frac{\cos(u)-\cos((1+t)u)}{u}\,du
+t\int_{u=0}^{R}\frac{\cos(tu)-\cos((1+t)u)}{u}\,du.
\end{align*}
By using the Frullani's integral,
$$\int_0^{+\infty}\frac{\cos(au)-\cos(bu)}{u}\,dt=\ln(b/a)
\qquad\mbox{for $a,b>0$,}$$
it follows that
$$\lim_{R\to +\infty} f_R(t)=f(t):=\ln(1+t)+t\ln((1+t)/t)=(1+t)\ln(1+t)-t\ln(t).$$
Note that for any $t\in [0,1]$,
$$|f_R(t)-f(t)|\leq \int_{R}^{\infty}\frac{3}{u^2}\,du=\frac{3}{R}.$$
Hence, by the Dominated Convergence Theorem, we finally obtain
\begin{align*}
I&=\int_{0}^{1}\frac{f(t)}{t(1+t)}\,dt
=\int_0^1 \frac{\ln(1+t)}{t}dt-\left[\ln(t)\ln(1+t)\right]_0^1+\int_0^1 \frac{\ln(1+t)}{t}dt\\
&=2\int_0^1 \frac{\ln(1+t)}{t}dt=
2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\int_0^1 t^{n-1}dt
=2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}=\zeta(2)=\frac{\pi^2}{6}.
\end{align*}
