I need to compute $$I:=\int_{-\infty}^{\infty} \frac{\sin(ax)}{x(\pi^2-a^2x^2)}dx$$
($a>0$)
Probably there is a way to compute it with residue theorem.
My thoughts:
- The singularity at $x=0$ is removable
- $I=\displaystyle \DeclareMathOperator{Im}{Im} \Im \left( \int_{-\infty}^{\infty} \frac{e^{iax}}{x(\pi^2-a^2x^2)}dx \right)$ . I usually solve those integrals by using the residue theorem to integrate over the upper semicircle (then the integral over the arc goes to 0 as the radius goes to $\infty$ and we are left with I) but this doesn't work since we have non-removable singularities. I also thought about integrating over a circular sector whose border doesn't contain the singularities (the integrand is an even function) but then the integral over the arc does not go to $0$ anymore and I don't get the expression for $I$ anyway.
Best Answer
$$I=\int\frac{\sin (a x)}{x \left(\pi ^2-a^2 x^2\right)}\,dx=\int \frac{\sin (t)}{(t-\pi)t(t+\pi) }\,dt$$ $$\frac{1}{(t-\pi)t(t+\pi) }=\frac 1 {\pi^2}\Bigg[\frac{1}{2 (t-\pi )}+\frac{1}{2 (t+\pi )}-\frac{1}{t} \Bigg]$$
Consider $$\int \frac {\sin(t)}{t+k}\,dt=\int\frac {\sin(u-k)}{u}\,du=\cos (k)\int\frac{ \sin (u)}{u}\,du-\sin (k)\int\frac{ \cos (u)}{u}\,du$$ So now, you just face sine and cosine integrals.
Back to $x$, you will have $$I=\frac 1{2\pi^2}\Big[2 \text{Si}(a x)+\text{Si}(ax-\pi )+\text{Si}(a x+\pi ) \Big]$$ $$J=\int_{-p}^{+p}\frac{\sin (a x)}{x \left(\pi ^2-a^2 x^2\right)}\,dx=\frac 1{\pi^2}\Big[2 \text{Si}(a p)+\text{Si}(ap-\pi )+\text{Si}(a p+\pi ) \Big]\sim \frac 4{\pi^2}\text{Si}(a p)$$ Now, using the asymptotics $$K=\int_{-\infty}^{+\infty}\frac{\sin (a x)}{x \left(\pi ^2-a^2 x^2\right)}\,dx=\frac{2 |a|}{\pi a}$$