What is the argument between compute the Fourier transform of a function and compute the integral? Which is the relationship between compute the Fourier transform of $f(x) = x e^{- |x|}$ and show that
$$
\int_{\mathbb{R}} \frac{x^2}{{(1 + x^2)}^4} \, dx = \frac{\pi}{16}\mbox{?}
$$
In other words, If I get $\hat{f}$, what must I use to show the last equality?
 A: Using the following normalization for defining the Fourier transform
$$ \widehat{f}(s) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} f(x) \,e^{-isx}\,dx $$
by the Plancherel theorem it happens that
$$ \int_{-\infty}^{+\infty}\left|\,f(x)\right|^2\,dx = \int_{-\infty}^{+\infty}\left|\,\widehat{f}(s)\right|^2\,ds \tag{1} $$
In our case, the Fourier transform of $f(x)=x e^{-|x|}$ is given by $\widehat{f}(s)=2\sqrt{\frac{2}{\pi}}\frac{is}{(1+s^2)^2}$ and the LHS of $(1)$ equals $\frac{1}{2}$. As a straightforward consequence,
$$ \int_{-\infty}^{+\infty}\frac{s^2\,ds}{(1+s^2)^4} =\frac{\pi}{16}.\tag{2}$$
The same can be proved in a variety of different ways, for instance by differentiation under the integral sign (Feynman's trick). For any $a>0$ we have $\int_{-\infty}^{+\infty}\frac{dx}{a+x^2}=\frac{\pi}{\sqrt{a}}$, hence by considering $\frac{d}{da}$ of both sides we also have $\int_{-\infty}^{+\infty}\frac{dx}{(a+s^2)^2}=\frac{\pi}{2a\sqrt{a}}$ and
$$ \int_{-\infty}^{+\infty}\frac{s^2}{(1+s^2)^2}\,ds = \int_{-\infty}^{+\infty}\frac{ds}{(1+s^2)}-\int_{-\infty}^{+\infty}\frac{ds}{(1+s^2)^2}=\pi-\frac{\pi}{2}.\tag{3}$$
If we differentiate twice more, we also get
$$ \forall a>0,\qquad \int_{-\infty}^{+\infty}\frac{s^2\,ds}{(a+s^2)^4} = \frac{\pi}{16 a^2\sqrt{a}}\tag{4} $$
as needed.
A: Hint: try Plancherel's theorem.
