Integrating sines

Let $g(x)=\int\limits_0^\infty \dfrac{exp(-2yx)}{1+y^2}\; dy$ and $f(x)=\large\int\limits_x^\infty\frac{sin(2(y-x))}{y}\ dy$ . We have $g(x)=f(x)$ ( We can start by showing that $f$ and $g$ are two functions of class $C_{2}$ then $f"+f=g"+g$ and $f(0)=g(0)$ .)

We have: $\large\int_{0}^{\infty}\int_{0}^{\infty} \frac{\sin x \sin y \sin (x+y)}{xy(x+y)} dy dx =\pi ^2 /4 - \int_{0}^{\infty}\int_{0}^{\infty} \frac{\ sin^2(x)\sin(2y)}{x^2 (x+y)}dy dx =\pi^2 /4 - \int_{0}^{\infty}\int_{x}^{\infty} \frac{\ sin^2(x)\sin(2(y-x))}{x^2 y}dy dx$

Now we can replace $f$ by $g$ and we get :

$\large\int_{0}^{\infty}\int_{0}^{\infty} \frac{\sin x \sin y \sin (x+y)}{xy(x+y)} dy dx =\pi ^2 /4 - \int_{0}^{\infty} \frac{f(x) \sin^2(x) }{x^2} dx = \pi ^2 /4 - \int_{0}^{\infty} \frac{g(x) \sin^2(x) }{x^2} dx = \pi ^2 /4 - \int_{0}^{\infty}\int_{0}^{\infty} \frac{\sin^2(x) exp(-2xy) }{(1+y^2)x^2} dx dy = \pi ^2 /4 -\int_{0}^{\infty} \frac{1}{y^2+1} \int_{0}^{\infty} \frac{\sin^2(x) exp(-2xy) }{x^2} dx dy$

We can now use Laplace transform of $\frac{\sin^2(x)}{x^2}$ which I will write it directly: $\int_{0}^{\infty} \frac{\sin^2(x) exp(-2xy) }{x^2} dx = arctan(\frac{1}{y}) -\frac{ylog(1+\frac{1}{y^2})}{2}$

Then we get :

$\pi ^2 /4 -\int_{0}^{\infty} \frac{1}{y^2+1} \int_{0}^{\infty} \frac{\sin^2(x) exp(-2xy) }{x^2} dx dy =\pi^2/4 - \int_{0}^{\infty} \frac{2arctan(\frac{1}{y}) -ylog(1+\frac{1}{y^2})}{2(1+y^2)} dy$

The remaining integral is equal to $\pi^2/12$ and we get $\pi^2/6$