Integral of $\frac{1+\sin \pi x}{(x^2+4)^2}$ on the interval $(-\infty,+\infty)$

Question

$$ \int_{-\infty}^\infty \frac{1+\sin \pi x}{(x^2+4)^2}\,dx $$

Hi, I got this exercise to be solved but i got no idea on how am I suposed to do it. Can somebody show me how to proceed? Thank you!

Answer 1

Assuming that the integral of interest is given by

$$\int_{-\infty}^{\infty} \frac{1+\sin(\pi x)}{(x^2+4)^2}\,dx$$

and noting that

$$\frac{\sin \pi x}{(x ^2+4)^2}$$

is an odd function that is integtated over symmetric limits (in the Cauchy Principal Value sense)

we have

$$\begin{align} \int_{-\infty}^{\infty} \frac{1+\sin(\pi x)}{(x^2+4)^2}\,dx&=\lim_{L\to \infty}\int_{-L}^{L} \frac{1+\sin(\pi x)}{(x^2+4)^2}\,dx\\\\ &=\lim_{L\to \infty}\int_{-L}^{L} \frac{1}{(x^2+4)^2}\,dx\\\\ &=2\lim_{L\to \infty}\int_{0}^{L} \frac{1}{(x^2+4)^2}\,dx\\\\ &=2\lim_{L\to \infty}\left(\frac{x/8}{x^2+4}+\frac{1}{16}\arctan\left(\frac{x}{2}\right)\right)\\\\ &=\frac{\pi}{16} \end{align}$$

Note that since the integral $\int_0^{\infty}\frac{\sin \pi x}{(x ^2+4)^2}\,dx$ converges, we need not have interpreted the integral of interest in the sense of a Cauchy Principal Value. The value of the integral was unaffected by this effective artifice of using symmetric limits.