TITLE: How to evaluate $\int_{0}^{\infty} \int_{0}^{\infty} \frac{\sin(x) \sin(y) \sin(x+y)}{x y(x+y)} ~{\rm d}x ~{\rm d}y$?
QUESTION [9 upvotes]: 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.

REPLY [2 votes]: 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 $$I(r)=\int_0^\infty\frac{\sin(r x)e^{-\frac{a}{i}x}}{x}\ dx, \quad I(0)=0, \quad I(1)=A$$
$$\Longrightarrow I'(r)=\int_0^\infty \cos(r x)e^{-\frac{a}{i}x}\ dx=\frac{a/i}{r^2+(a/i)^2}=-\frac{ia}{r^2-a^2}$$
$$\Longrightarrow A=i\int_0^1\frac{a}{a^2-r^2}\ dr=i\ \text{arccoth}(a)=i\ \text{arctanh}(1/a)$$