Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2}  \right) ^2 dx =  \lim_{ A \to \infty }  \int_{0}^{A} \left( \frac{x}{1+x^2}  \right) ^2 dx $$
I think that firstly I should compute $ \int \left( \frac{x}{1+x^2}  \right) ^2 dx $ but I don't have idea.

We can use Parseval's theorem!
If $F$ is the Fourier transform of $f$, and $G$ is the Fourier transform of $g$, then
$$
\int_{-\infty}^\infty\overline{f(t)}g(t)\,dt=
\frac{1}{2\pi}\int_{-\infty}^\infty\overline{F(x)}G(x)\,dx.
$$
We have the Fourier transform pairs


*

*$\displaystyle e^{-|t|}\longmapsto\frac{2}{1+\omega^2}$

*$\displaystyle \frac{d}{dt}f(t) \longmapsto i\omega F(\omega)$


It is clear from those that
$$
f(t)=\frac{1}{2}\frac{d}{dt}\left(e^{-|t|}\right)\longmapsto
\frac{ix}{1+x^2}=F(x),
$$
and conveniently $\overline{F(x)}F(x)=\frac{x^2}{(1+x^2)^2}$ is the integrand that we want.
Using this and Parseval's theorem, we see that
$$
\int_{-\infty}^\infty\frac{x^2}{(1+x^2)^2}dx=
2\pi\int_{-\infty}^\infty\left(\frac{1}{2}\frac{d}{dt}\left(e^{-|t|}\right)\right)^2dt,
$$
and since the integrand is even, one can write
$$
\int_{-\infty}^\infty\frac{x^2}{(1+x^2)^2}dx=
\pi\int_0^\infty\left(\frac{d}{dt}\left(e^{-t}\right)\right)^2dt=
\pi\int_0^\infty e^{-2t}dt=\frac{\pi}{2}.
$$
