How to calculate $\int_{-\infty}^{\infty}\frac{x \sin x}{x^2+a^2}dx=\pi e^{-a}$ using residue theory.

Question

I want to compute the following: $$\int_{-\infty}^{\infty}\frac{x \sin x}{x^2+a^2}dx=\pi e^{-a}$$

so I chose the following function $f(z)=\frac{-ize^{iz}}{z^2+a^2}$ and I decided to integrate over the real line and semicircle ($C$), but then when I try to bound over the semicircle I've got this:

$$\left\lvert \int _{C}f(z)dz \right\rvert \le \pi R· \frac{R}{R^2+a^2}$$

and that thing doesn't go to zero, then I tried to do this using now a square as a contour but the integral doesn't seem very friendly there, So Can someone help me with this issue please?

Thanks a lot in advance.

Answer 1

On the semicircular contour $C_{R}(\theta)=Re^{i\theta}$ for $0 \le \theta\le \pi$, and $R > |a|$, $$ \left|\frac{ze^{iz}}{z^2+a^2}\right| \le \frac{Re^{-R\sin\theta}}{R^2-|a|^2} $$ Therefore, by the Lebesgue dominated convergence theorem, $$ \lim_{R\rightarrow\infty}\left|\int_{C_{R}}\frac{ze^{iz}}{z^2+a^2}dz\right| \le \lim_{R\rightarrow\infty}\int_{0}^{\pi}\frac{Re^{-R\sin\theta}}{R^2-|a|^2}Rd\theta =0. $$