The question is: prove that
$$\int^{\infty}_0\frac{x\sin(rx)}{a^2+x^2}dx=\frac{\pi}{2}e^{-ar}$$
This is what I've got so far:
Let $I(r)=\int^{\infty}_0\frac{x\sin(rx)}{a^2+x^2}dx$
$I'(r)=\int^{\infty}_0\frac{x^2\cos(rx)}{a^2+x^2}dx$ = $\int^{\infty}_0\cos(rx)dx-\int^{\infty}_0\frac{a^2\cos(rx)}{a^2+x^2}dx$
$I''(r)=\int^{\infty}_0-x\sin(rx)dx+\int^{\infty}_0\frac{xa^2\sin(rx)}{a^2+x^2}dx$
$I''(r)=\int^{\infty}_0-x\sin(rx)dx+a^2I(r)$
I think I'm supposed to form a differential equation and solve it by letting $I(r)=c_1e^r+c_2e^{-r}$, but the problematic part is this: $\int^{\infty}_0-x\sin(rx)dx$. I don't believe it simplifies to any constant and the differential equation doesn't solve nicely.
Hints/suggestions appreciated!
