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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.

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  • $\begingroup$ The integral $F(a)=\int_{0}^{+\infty}\frac{\sin x}{x^2+a^2}\,dx $ equals $\frac{1}{a}\int_{0}^{+\infty}\frac{\sin(as)}{1+s^2}\,ds $, hence it depends on $\text{Si}$ and $\text{Ci}$. $\endgroup$ – Jack D'Aurizio Jan 18 '18 at 18:51
  • $\begingroup$ @JackD'Aurizio can I know why this phenomenon holds. because with cosine we have $$\int_0^\infty\frac{\cos ax}{x^2+1}\,dx =\frac{\pi}{2}e^{-|a|}$$ which does not depends one SI and Ci see here $\endgroup$ – Guy Fsone Jan 18 '18 at 18:54
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    $\begingroup$ But such integral is just half the integral over the whole real line, so it is not surprising. We have a parity obstruction in tackling the integral involving $\sin x$ through the same methods. $\endgroup$ – Jack D'Aurizio Jan 18 '18 at 18:57

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