How calculate this double integral? I'm having trouble with this double integral:
$$ \int_{0}^{\infty}\int_{0}^{\infty}\frac{\sin(x)\sin (y)\sin(x+y)}{xy(x+y)}dxdy$$
I can't  solve it.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&\int_{0}^{\infty}\int_{0}^{\infty}{\sin\pars{x}\sin\pars{y}\sin\pars{x + y} \over xy\pars{x + y}}\,\dd x\,\dd y =
\int_{0}^{\infty}\int_{0}^{\infty}{\sin\pars{x}\sin\pars{y} \over xy}\,
{1 \over 2}\int_{-1}^{1}\expo{\ic\pars{x + y}k}\,\dd k\,\dd x\,\dd y
\\[5mm] = &\
{1 \over 2}\int_{-1}^{1}\bracks{\int_{0}^{\infty}
{\sin\pars{x} \over x}\,\expo{\ic kx}}^{2}\,\dd k =
{1 \over 2}\int_{-1}^{1}
\braces{{1 \over 2}\bracks{\pi + 2\ic\,\mrm{arctanh}\pars{k}}}^{2}\,\dd k
\\[5mm] = &\
{\pi^{2} \over 4} - \underbrace{\int_{0}^{1}\mrm{arctanh}^{2}\pars{k}\,\dd k}
_{\ds{\pi^{2} \over 12}}\ =\ \bbx{\ds{\pi^{2} \over 6}}
\end{align}
A: \begin{align}&\int_0^\infty\int_0^\infty\frac{\sin(x)\sin(y)\sin(x+y)}{xy(x+y)}{\rm~d}x{\rm~d}y\\&=\frac12\int_0^\infty\int_0^\infty\int_{-1}^1\frac{\sin(x)\sin(y)}{xy}e^{it(x+y)}{\rm~d}t{\rm~d}x{\rm~d}y\\&=\frac12\int_{-1}^1\left[\int_0^\infty\frac{\sin(x)}xe^{itx}{\rm~d}x\right]^2{\rm~d}t\\&=\frac12\int_{-1}^1\left[\int_0^\infty\frac{e^{i(1+t)x}-e^{-i(1-t)}}{2ix}{\rm~d}x\right]^2{\rm~d}t\\&=\frac12\int_{-1}^1\left[\frac1{2i}\ln\left(-\frac{1-t}{1+t}\right)\right]^2{\rm~d}t\tag1\\&=\frac12\int_{-1}^1\left[\frac\pi2+\frac1{2i}\ln\left(\frac{1-t}{1+t}\right)\right]^2{\rm~d}t\\&=\frac{\pi^2}4-\frac18\int_{-1}^1\ln^2\left(\frac{1-t}{1+t}\right){\rm~d}t\\&=\frac{\pi^2}4-\frac12\int_{-1}^1{\rm arctanh}^2(t){\rm~d}t\\&=\frac{\pi^2}4-2\int_{-\infty}^\infty\frac{u^2}{(e^u+e^{-u})^2}{\rm~d}u\tag{$t=\tanh(u)$}\\&=\frac{\pi^2}4-4\int_0^\infty\frac{u^2}{(e^u+e^{-u})^2}{\rm~d}u\\&=\frac{\pi^2}4-4\int_0^\infty\frac{u^2}{e^{2u}(1+e^{-2u})^2}{\rm~d}u\\&=\frac{\pi^2}4-4\sum_{n=1}^\infty(-1)^{n+1}n\int_0^\infty u^2e^{-2nu}{\rm~d}u\tag2\\&=\frac{\pi^2}4-\sum_{n=1}^\infty\frac{(-1)^{n+1}}{2n^2}\int_0^\infty v^2e^{-v}{\rm~d}v\tag{$2nu=v$}\\&=\frac{\pi^2}4-\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}\\&=\frac{\pi^2}4-\frac{\pi^2}{24}\tag3\\&=\frac{\pi^2}6\end{align}
$(1)$ is an application of the complex Frullani with $f(z)=e^{-z}$.
$(2)$ is an application of the derivative of the geometric series:
$$\frac x{(1-x)^2}=\sum_{n=1}^\infty nx^n$$
$(3)$ uses the Dirichlet eta function. Note that
$$S-\zeta(2)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}-\frac1{n^2}=\sum_{n=1}^\infty\frac2{(2n)^2}=\frac12\zeta(2)\\S=\frac12\zeta(2)=\frac{\pi^2}{12}$$
A: This is not a complete answer, but it is what I have managed arrive at, a slightly easier looking problem than the one you have written.
Use the Inverse Laplace Transform on $\displaystyle \frac{1}{x+y}$ and expand $\displaystyle \sin{(x+y)}$ to arrive at:
$\displaystyle \frac{\cos{y}\sin{y}}{y} e^{-ty} \int_0^\infty \frac{\sin^2{x}}{x}  e^{-t x}\, \text{d}x + \frac{\cos{x}\sin{x}}{x} e^{-tx} \int_0^\infty \frac{\sin^2{y}}{y}  e^{-t y}\, \text{d}y$
The variables of integration are insignificant, and so the two integrals are identical.
It can be shown (though I do not know how) that:
$\displaystyle \int_0^\infty \frac{\sin^2{x}}{x}e^{-tx} \, \text{d}x = \frac{\log{\left(1+\frac{4}{t^2}\right)}}{4}$
So it follows that:
$\displaystyle \int_{0}^{\infty}\int_{0}^{\infty}\frac{\sin(x)\sin (y)\sin(x+y)}{xy(x+y)}\text{d}x\text{d}y = 
\frac{1}{4}\int_0^\infty \int_0^\infty \frac{\sin{2y} \, e^{-ty} \log{\left(1+\frac{4}{t^2}\right)}}{y} \text{d}y \text{d}t$
And here my progress terminates. :)
