Use the Fourier inversion formula to compute h(x) when $\hat{h}(x)=\frac{1}{(1+y^2)^2}$ Use the Fourier inversion formula to compute $h(x)$ when $\hat{h}(x)=\frac{1}{(1+y^2)^2}$.
$$\hat{h}(x)=\frac{1}{(1+y^2)^2}=\frac{1}{1+y^2} \times \frac{1}{1+y^2}=\widehat{\frac{1}{2}e^{-|x|}} \times \widehat{\frac{1}{2}e^{-|x|}}.$$
Let $f(x)=\frac{1}{2}e^{-|x|}$.
So $\hat{h}(x)=\hat{f} \times \hat{f}=\hat{f*f}$
By the Fourier inversion formula, $h(x)=f*f$.
So $h(x)=f*f=\int_R f(y)f(x-y)dy=\frac{1}{4} \int_R e^{-|y|}e^{-|x-y|}dy$.
How am I suppose to finish this problem?
 A: There are some constants missing somewhere. However,
$$
         \frac{1}{1+y^{2}} = \frac{1}{2}\int_{-\infty}^{\infty}e^{-|x|}e^{-ixy}\,dx.
$$
Use the scaling property of the Fourier transform for $a > 0$:
$$
\begin{align}
    \frac{1}{a^{2}+y^{2}} & =\frac{1}{a^{2}}\frac{1}{1+(y/a)^{2}} \\
    & = \frac{1}{2a^{2}}\int_{-\infty}^{\infty}e^{-|x|}e^{-ix(y/a)}\,dx\;\;\; \mbox{(now set $x=ax'$)}\\
    & = \frac{1}{2a}\int_{-\infty}^{\infty}e^{-a|x'|}e^{-iyx'}dx'.
\end{align}
$$
Differentiate both sides with respect to $a$ and set $a=1$ afterwards to get what you want:
$$
      -\frac{2}{(1+y^{2})^{2}}=-\frac{1}{2}\int_{-\infty}^{\infty}(1+|x'|)e^{-|x'|}e^{-iyx'}\,dx'.
$$
A: Hint: Note that $|y| = y$ when $y\geq 0$, and $|y| = -y$ otherwise. Similarly, $|x-y| = x-y$ when $x-y \geq 0 \iff y \leq x$ and $|x-y| = -(x-y)$ otherwise. First, consider $x \geq 0$. Then, consider evaluating the integral
$\int_{-\infty}^{\infty} e^{-|y|} e^{-|x-y|}\mathrm{d}y$ via the three parts $\int_{-\infty}^{\infty}  = \int_{-\infty}^{0} + \int_{0}^{x} + \int_{x}^{\infty}.$ The reason for this partitioning is to get rid of the absolute values inside the integral. Next, consider $x<0$, and find a similarly appropriate partitioning of the integral.
