Evaluating $\int_0^{\infty} \int_0^{\infty} \frac{\sin(x)\sin(x+y)}{x(x+y)}\,\mathrm dx\,\mathrm dy$ $$\int_0^{\infty} \int_0^{\infty} \frac{\sin(x)\sin(x+y)}{x(x+y)}\, \mathrm dx\,\mathrm dy$$
I'm having some difficulty evaluating this Improper Double Integral with standard techniques taught in Multivariable Calculus. I opted to use the Residue Theorem and obtained an answer of $\pi^2/2$. I was wondering if this is correct and if my work is on the right track?
Edit: Converted $\sin(x)$ to exponential form, converted $\sin(x+y)$ to Im(cis(x+y)) = Im(e^i(x+y)) and then used residue theorem from there.
My progress:

 A: I'll use the sine integral function: $$\operatorname{Si}(x) = \int_0^x \frac{\sin t}{t}dt$$
We will use the properties:
$$\lim_{x\rightarrow\infty} \operatorname{Si}(x) = \frac{\pi}{2}$$
$$\operatorname{Si}(0)=0$$
$$\frac{d}{dx}\operatorname{Si}(x)=\frac{\sin x}{x}$$
We may now proceed to the solution. Firstly, let us notice that the integrand is measurable so we may treat is as an iterated integral:
$$\int_0^\infty \frac{\sin{x}}{x}\Big(\int_0^\infty \frac{\sin(x+y)}{x+y}dy\Big)dx = (*)$$
Let us integrate the inside integral by substitution $t=x+y$.
$$\int_0^\infty \frac{\sin(x+y)}{x+y}dy = \int_x^\infty\frac{\sin t}{t}dt = \int_0^\infty\frac{\sin t}{t}dt - \int_0^x\frac{\sin t}{t}dt = \frac{\pi}{2}-\operatorname{Si}(x)$$
We then have:
$$(*) = \int_0^\infty \frac{\sin x}{x} \Big(\frac{\pi}{2}-\operatorname{Si}(x)\Big)dx=\frac{\pi}{2}\int_0^\infty \frac{\sin x}{x} dx - \int_0^\infty \frac{\sin x}{x} \operatorname{Si}(x)dx$$
From the properties of the sine integral we have $\frac{\pi}{2}\int_0^\infty \frac{\sin x}{x} dx=\frac{\pi^2}{4}$.
The second integral may be found using the derivative property of the sine integral inserted into the formula for integral of multiplication. For some functions $f$ and $g$ we have:
$$fg = \int f'g + \int fg'$$
We are able to calculate the indefinite integral $\int \frac{\sin x}{x} \operatorname{Si}(x)dx$ by putting $f=g=\operatorname{Si}(x)$:
$$\operatorname{Si}^2(x) =2\int \frac{\sin x}{x} \operatorname{Si}(x)dx $$
$$\int \frac{\sin x}{x} \operatorname{Si}(x)dx = \frac{\operatorname{Si}^2(x)}{2}$$
We can now calculate the value of the definite integral:
$$\int_0^\infty \frac{\sin x}{x} \operatorname{Si}(x)dx = \frac{\Big(\lim_{x\rightarrow\infty} \operatorname{Si}(x)\Big)^2}{2}-0=\frac{\pi^2}{8}$$
Eventually:
$$(*) = \frac{\pi^2}{4}-\frac{\pi^2}{8} = \frac{\pi^2}{8}$$
A: Denoting $ \mathrm{Si}:x\mapsto\int_{0}^{x}{\frac{\sin{y}}{y}\,\mathrm{d}y} $, we have the following : \begin{aligned} \int_{0}^{+\infty}{\frac{\sin{x}}{x}\int_{0}^{x}{\frac{\sin{y}}{y}\,\mathrm{d}y}\,\mathrm{d}x}=\int_{0}^{+\infty}{\mathrm{Si}\,{x}\,\mathrm{Si}'\,{x}\,\mathrm{d}x}&=\left[\frac{\mathrm{Si}^{2}\,{x}}{2}\right]_{0}^{+\infty}\\&=\frac{1}{2}\left(\int_{0}^{+\infty}{\frac{\sin{x}}{x}\,\mathrm{d}x}\right)^{2}\\&=\frac{\pi^{2}}{8} \end{aligned}
Thus : \begin{aligned}\int_{0}^{+\infty}{\int_{0}^{+\infty}{\frac{\sin{x}\sin{\left(x+y\right)}}{x\left(x+y\right)}\,\mathrm{d}y}\,\mathrm{d}x}&=\int_{0}^{+\infty}{\frac{\sin{x}}{x}\int_{x}^{+\infty}{\frac{\sin{y}}{y}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\int_{0}^{+\infty}{\frac{\sin{x}}{x}\left(\frac{\pi}{2}-\int_{0}^{x}{\frac{\sin{y}}{y}\,\mathrm{d}y}\right)\mathrm{d}x}\\ &=\frac{\pi^{2}}{4}-\int_{0}^{+\infty}{\frac{\sin{x}}{x}\int_{0}^{x}{\frac{\sin{y}}{y}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\frac{\pi^{2}}{4}-\frac{\pi^{2}}{8}\\&=\frac{\pi^{2}}{8}\end{aligned}
