real analysis - Showing that $int_0^{infty} e^{-ax} frac {sin x}{x} dx = frac {pi}{2} - arctan (a)$ for any $a >0$

$\displaystyle \int_0^{\infty} e^{-ax} \dfrac {\sin x}{x} dx = \dfrac {\pi}{2} - \arctan (a)$ for any $a >0$

I am really not sure how to go about doing this. I checked Wolfram Alpha and the function does not have an elementary antidericative. I do not have that many special tools (mostly just Fubini's Theorem). My best guess is that we have to do a substitution of some kind, and maybe integration by parts. Then hopefully we will get a "known" integral. However, I don't really know in what direction to go.