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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 $f^2$?

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up vote 2 down vote accepted

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.

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Thank you Fabian. –  Kumara Jan 22 '13 at 10:24
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