Compute $\int_{\mathbb {R}}\sin \left(\frac {\pi x}{x^2+1}\right)\frac{1}{x^2+1} dx$ How do I compute this integral?
$$\int_{\mathbb {R}}\sin \left(\frac {\pi x}{x^2+1}\right)\frac{1}{x^2+1} dx$$
 A: This integral doesn't have a nice antiderivative; so, we need to think a little bit more.
Let
$$
f(x)=\sin\left(\frac{\pi x}{x^2+1}\right)\frac{1}{x^2+1}.
$$
Notice that
$$
f(-x)=\sin\left(-\frac{\pi x}{x^2+1}\right)\frac{1}{x^2+1}=-\sin\left(\frac{\pi x}{x^2+1}\right)\frac{1}{x^2+1}=-f(x),
$$
for all $x\in\mathbb{R}$. In other words, your integrand is an odd function. As a result, for any $a\in\mathbb{R}$, it must be the case that
$$
\int_{-a}^{a}f(x)\,dx=0.
$$
This is not, on its own, enough to guarantee that $\int_\mathbb{R} f(x)\,dx=0$; however, this tells us that if the improper integral converges, then it must be $0$. So, it then suffices to show that the improper integral must converge.  Can you show that this integral is absolutely convergent?
A: $$\left|\;\frac{\sin\frac{\pi x}{x^2+1}}{x^2+1}\;\right|\le\frac1{1+x^2}$$
and since 
$$\left.\int\limits_{-\infty}^\infty\frac{dx}{1+x^2}=\arctan x\right|_{-\infty}^\infty=\pi$$
our integral converges absolutely.
But since the integrand is an odd function then the integral's value must be zero.
