Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $f^2$?

share|cite|improve this question
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.

share|cite|improve this answer
Thank you Fabian. – Kumara Jan 22 '13 at 10:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.