Consider the double-integral
$$
F(x,y) := \int_0^\infty d a \int_0^\infty d b\ \frac{ \sin(x a) \sin( y b ) }{a+b} \ .
$$
By playing around in Mathematica I have come to believe that the above evaluates to
$$
F(x,y) \stackrel{?}{=} \frac{\pi}{2(x+y)} \ .
$$
How would one prove this result?
EDIT: I've been able to evaluate the $a$-integral, showing that the above is equal to
$$
F(x,y) = \int_0^\infty db\ \sin(by) \bigg[ \text{Ci}(bx) \sin(bx) + \frac{1}{2} \cos(bx) \big( \pi – \text{Si}(bx) \big) \bigg] \ ,
$$
where $\text{Ci}$ and $\text{Si}$ are the cosine integral and sine integral functions, respectively. From here I cannot make progress.
There is a symmetry in the order of integration (ie. the above would look the same if I did the $b$-integral first). It is curious that the function seems to depend on $x+y$: I don't see how this dependence falls out of the definition of $F$.
Best Answer
$$I=\int_{0}^{\infty} \int_{0}^{\infty} \frac{\sin ax \sin bx}{a+b} da db = \int_{0}^{\infty} \int_{0}^{\infty} \sin ax \sin by ~ da ~ db ~ e^{-(a+b)t}~dt$$ $$I=\int_{0}^{\infty} dt\left(\int_{0}^{\infty} \sin ax ~e^{-ta} da \int_{0}^{\infty}\sin by ~ db ~ e^{-tb}\right)$$ $$I=\int_{0}^{\infty} \frac{xy}{(t^2+x^2)(t^2+y^2)}dt= \frac{xy}{y^2-x^2} \int_{0}^{\infty}\left(\frac{1}{t^2+x^2}-\frac{1}{t^2+y^2}\right)=\frac{\pi}{2(x+y)}.$$ Note that: $$J=\int \sin ax ~e^{-ta} da=\Im \int_{0}^{\infty} e^{-(t-ix)}=\frac{x}{t^2+x^2}$$