I was wondering what happens when evaluating an improper integral involving a trigonometric function where the denominator is a rational function with a zero at $x=0$.
The example I have in mind is
$$\int_{-\infty}^{\infty}\frac{sin(ax)}{x(x-i)(x+i)} dx$$
If I rewrite the sine in terms of the exponential and then evaluate two integrals, one for the upper half-plane and one for the lower I have a problem because of the singularity at $z=0$ which seems to be in both halves of the plane. Do I treat it in both integrals then? Any suggestions/comments are welcome.
Best Answer
One way of avoiding to deal with the (redundant) singularity after spliting the integrals is to first differentiate w.r.t. $a$ and use Eulers identity
$$ I'(a)=\Re \int_{\mathbb{R}} \frac{e^{i a x}}{(x-i)(x+i)} $$
Assuming that $a>0$ we have to close our contour of integration, which is a big semicircle , in the upper halfplane(for $a<0$ it is the other way round)
We get
$$ I'(a)=\Re(2\pi i \text{Res}(x=i))=\pi e^{-a} $$
integrating back w.r.t. $a$ yields $$ I(a)=-\pi e^{-a}+C $$
and the constant of integration is fixed by the requirement that $I(0)=0$
which brings us to our final result
I leave the case $a<0$ to you