<s>[INST] How to find integral $\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y \sin (x+y)}{xy(x+y)}\,dx\,dy$ Find the integral
$$I=\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y \sin (x+y)}{xy(x+y)}\,dx\,dy$$
My try: define
$$I(b):=\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y \sin (x+y)}{xy(x+y)}e^{-bxy(x+y)}\,dx\,dy$$
then
$$I'(b)=-\int_0^\infty\int_0^{\infty}\sin x \sin y \sin (x+y) e^{-bxy(x+y)}\,dx\,dy$$
But I can't find $I(b)$.
Thank you
 [/INST] Note that $I=J(1)$, where
$$J(a) =\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y \sin [a(x+y)]}{xy(x+y)} dx dy$$
and
\begin{align}
J’(a) &=\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y \cos [a(x+y)]}{xy} dx dy\\
& =\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y [\cos (ax)\cos(ay)- \sin (ax)\sin(ay)]}{xy} dx dy\\
 & =\left(\int_0^\infty\frac{\sin x \cos (ax)}xdx \right)^2
- \left(\int_0^\infty\frac{\sin x \sin (ax)}xdx \right)^2\\
& =\frac14\left(\int_0^\infty\frac{\sin (1-a)x + \sin (1+a)x }{x}  dx\right)^2\\
& \>\>\>\>\>\>\>-\frac14\left(\int_0^\infty\frac{\cos (1-a)x -\cos (1+a)x }{x} dx\right)^2\\
&=\frac14\left(\pi^2 - \ln^2 \frac{1+a}{1-a} \right)
\end{align}
Thus
$$I=\int_0^1 J’(a)da=\frac14 \int_0^1 \left(\pi^2 - \ln^2 \frac{1+a}{1-a} \right) da=\frac14 \left(\pi^2 - \frac{\pi^2}3 \right) 
=\frac{\pi^2}{6}
$$
</s>