Can one solves this using contour integration ?
Indeed this look straightforward but having a close to it shows that we have some parity obstruction therein.
I have applied residues method the complex function $f(z) =\frac{e^{izb}}{z^2+a^2}$ but give the following
$$ \int_{\Bbb R}\frac{\sin(xb)}{x^2+a^2}dx = 0$$
Which is natural since the integrand is an odd function. However this does not help to get the value of $$\int_{0}^{\infty}\frac{\sin x}{x^2+a^2}\,dx$$
as opposed to the following cases
$$\int_{0}^{\infty}\frac{x\sin x}{x^2+a^2}\,dx~~~~\mbox{and}~~~~~\int_{0}^{\infty}\frac{\cos x}{x^2+a^2}\,dx$$ where the situation is more simpler. Where residues method gives of
$$\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+a^2}\,dx=2\int_{0}^{\infty}\frac{x\sin x}{x^2+a^2}\,dx~~~~\mbox{and}~~~~~\int_{-\infty}^{\infty}\frac{\cos x}{x^2+a^2}\,dx =2\int_{0}^{\infty}\frac{\cos x}{x^2+a^2}\,dx $$ and one decues the values of the aforementioned integral by parity. Finding a contour to evaluate$\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+a^2}\,dx$
Question: How can one compute the above integral using Contour integral?
Note I am also opened to any suggestion not using contour integration.