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Define: $f_a(x)=\frac{a}{\sqrt \pi} \frac{1}{a^2+x^2}$. Prove that $$f_a *f_b=f_{a+b}$$

My Approach

$$(f_a *f_b)(x)=\int_{-\infty}^{\infty}\frac{ab}{\pi}\frac{a}{a^2+y^2} \frac{b}{(x-y)^2+b^2}\,\mathrm dy$$

Mark Viola
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Atul Anurag Sharma
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  • Ok so using partial fractions one can get the horrible anti-derivative $$\frac{b \left(-a^2+b^2+x^2\right) \tan ^{-1}\left(\frac{y}{a}\right)+a b x \left(\log \left(a^2+y^2\right)-\log \left(b^2+(x-y)^2\right)\right)+a \left(a^2+x^2\right) \tan ^{-1}\left(\frac{y-x}{b}\right)+a b^2 \tan ^{-1}\left(\frac{x-y}{b}\right)}{a b \left(a^4-2 a^2 \left(b^2-x^2\right)+\left(b^2+x^2\right)^2\right)}$$ Now take $y\to\pm\infty$. (There must be an easier way!) – Maximilian Janisch Mar 09 '20 at 17:17
  • Are you sure you mean to have a square root over the $\pi$? Without the square root $f_a$ is the pdf of a Cauchy distribution with scale $a$. The result would then be just the fact that the pdf of the sum of independent random variables is the convolution of their pdfs combined with the fact that a sum of Cauchy distributions is Cauchy (obviously you then have to have a separate way of checking that result, which you can do via characteristic functions). But then you are off by a factor of $\pi$. – Rhys Steele Mar 09 '20 at 17:45

1 Answers1

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The problem is a little exercise on the Fourier transform. But there seems to be a little misprint in the formulation.

So let $f_a$ be $$ f_a(x) = \frac{a}{\pi}\frac{1}{a^2+x^2}. $$ I claim that $f_a * f_b = f_{a+b}$ for all $a,b>0$. To prove this I will use two facts.

First: the Fourier transform of $f_a(x)$, $a>0$, is $$ \mathcal{F}_{x \to \xi}[f_a(x)] = \frac{a}{\pi} \int e^{-ix\xi}\biggl[\frac{1}{a^2+x^2}\biggr]\,dx = e^{-a|\xi|}. $$ (See the computation here.)

Second: for the convolution of two functions $f(x)$, $g(x)$, we have $$ \mathcal{F}[f * g] = \mathcal{F}[f] \, \mathcal{F}[g]. $$

Now $$ f_a * f_b = \mathcal{F}^{-1}_{\xi\to x} \, \mathcal{F}_{x\to\xi}[f_a * f_b] = \mathcal{F}^{-1}_{\xi\to x}\Bigl[e^{-a|\xi|}\, e^{-b|\xi|}\Bigr] = \mathcal{F}^{-1}_{\xi\to x}\Bigl[e^{-(a+b)|\xi|} \Bigr] = \frac{1}{2\pi} \int e^{i x\xi}\,e^{-(a+b)|\xi|}\,d\xi = \frac{a+b}{\pi} \, \frac{1}{(a+b)^2+x^2} = f_{a+b}(x), $$ as desired.

dsw
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  • Can we prove this without using Fourier? – Atul Anurag Sharma Mar 09 '20 at 20:59
  • By explicit computation, I guess. I doubt there is an easy way. – dsw Mar 09 '20 at 21:05
  • I tried by taking partial fractions but I was getting a different answer. – Atul Anurag Sharma Mar 09 '20 at 21:08
  • I believe my argument is correct. For the integrals (for $\mathcal{F}$ and $\mathcal{F}^{-1}$), I used the computations from the accepted answer to this post. You can find both of them there. – dsw Mar 09 '20 at 21:13
  • well I don’t have good knowledge of solving this integral using Fourier transform that’s why I can’t question what you just say. I am looking for a solution that is Free from Fourier. – Atul Anurag Sharma Mar 09 '20 at 21:16
  • In my opinion, learning the Fourier transform would be easier (and more helpful). Attacking convolutions with the help of Fourier transform is the most natural method, because the Fourier transform turns convolution into multiplication. This is very powerful! – dsw Mar 09 '20 at 21:21