calculus - Evaluating the integral $int_0^infty frac{x sin rx }{a^2+x^2} dx$ using only real analysis

Calculate the integral $$\int_0^\infty \frac{x \sin rx }{a^2+x^2} dx=\frac{1}{2}\int_{-\infty}^\infty \frac{x \sin rx }{a^2+x^2} dx,\quad a,r \in \mathbb{R}.$$
Edit: I was able to solve the integral using complex analysis, and now I want to try and solve it using only real analysis techniques.

Answer

It looks like I'm too late but still I wanna join the party. :D

Consider
$$\int_0^\infty \frac{\cos rx}{x^2+a^2}\ dx=\frac{\pi e^{-ar}}{a}.$$
Differentiating the both sides of equation above with respect to $r$ yields
$$\begin{align} \int_0^\infty \frac{d}{dr}\left(\frac{\cos rx}{x^2+a^2}\right)\ dx&=\frac{d}{dr}\left(\frac{\pi e^{-ar}}{a}\right)\\ -\int_0^\infty \frac{x\sin rx}{x^2+a^2}\ dx&=(-a)\frac{\pi e^{-ar}}{a}\\ \Large\int_0^\infty \frac{x\sin rx}{x^2+a^2}\ dx&=\Large\pi e^{-ar}. \end{align}$$
Done! :)