Fourier transform of product I would like to know the fourier transform of the product of the Cauchy probability distribution $f(x)=\frac{1}{\pi (1+x^2)}, -\infty<x<\infty$ with itself.
I know that the fourier transform of $f$ is $e^{-|t|}$. So the fourier transform of $f^2$ is the convolution of $e^{-|t|}$ with itself. 
The convolution is $\int_{-\infty}^{\infty}e^{-|y|}\cdot e^{-|t-y|}dy$.
Can someone help me in evaluating this integral? Or is there any other way to find the fourier transform of $f^2$?

Convolution is fine: for evaluating 
$$\hat{f^2}(t)= \int_{-\infty}^{\infty}e^{-|y|} e^{-|t-y|}dy$$
split the integral up in parts. For $t>0$, we have
$$\begin{align}\hat{f^2}(t)&= \left(\int_{-\infty}^0 + \int_0^t + \int_t^\infty\right) e^{-|y| -|t-y|}dy
=\int_{-\infty}^0 e^{y-(t-y)}dy
+ \int_0^t e^{-y-(t-y)} dy+ \int_t^\infty e^{-y-(y-t)}dy \\
&= \frac{e^{-t}}2 + t e^{-t} +\frac{e^{-t}}2 = (t+1)e^{-t}
\end{align}.$$
A similar calculation for $t<0$ yields the final result
$$\hat{f^2}(t)=(|t|+1)e^{-|t|}.$$
Alternatively, you could directly arrive at the final result by calculating the Fourier integral using, e.g., the residue theorem.
