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I am trying to prove that $p(l,x)*p(u,x)=p(l+u,x)$ where * denotes the convolution of two functions and $p(l,x)=\dfrac{l}{\pi(l^2+x^2)}$. I am having trouble in integrating the left hand side.

$$\int_{-\infty}^\infty \frac{l}{l^2+(x-y)^2}.\frac{u}{u^2+y^2}dy,$$

this is the definition of convolution and is the left hand side of the above equation. Putting the command

Integrate[(l/(l^2+(x-y)^2))*(u/(u^2+y^2)),{y,-infinity,infinity}]

into Mathetica gives me $p(l-u,x)$, instead of $p(l+u,x)$. please help!

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  • $\begingroup$ Hi sami, welcome to Math.SE! Thanks for bringing your questions here, I hope to stay around and contribute to the site. $\endgroup$ – gt6989b Jun 17 '13 at 17:00
  • $\begingroup$ It is very helpful to edit your code in MathJAX (LaTeX). I will do it this time, please take a look at the edits... $\endgroup$ – gt6989b Jun 17 '13 at 17:00
  • $\begingroup$ Small nitpick: avoid using $*$ for multiplication, especially when you've already defined it to mean convolution. $\endgroup$ – icurays1 Jun 17 '13 at 17:09
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It's interesting, but this equality can be proved with the help of characteristic functions from the theory of probability.

Let's take 2-sided normal distribution,its density is expressed with a function

$$p(x)=\frac{a}{2}e^{-a\left|x\right|} $$

Now we found characteristic function of distribution (or, as in analysis, Fourier transformation).

$$g(t)=\int_{-\infty}^{+\infty}e^{itx}p(x)dx =\frac{a^2}{a^2+t^2} $$

Then we use the formula of inverse transformation

$$\frac{a}{2}e^{-a\left|t\right|}= \frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{-itx}\frac{a^2}{a^2+t^2}dt $$ $$e^{-a\left|t\right|} =\frac{1}{\pi}\int_{-\infty}^{+\infty}e^{itx}\frac{a}{a^2+t^2}dt$$

That means, that $g(t)=e^{-a\left|t\right|}$ is characteristic function of Cauchy distribution. Because of the property of characteristic functions

$$g_{a*b}=g_{a}g_{b} $$ $$g_{p_{l}(x)*p_{u}(x)}(t)=e^{-l\left|t\right|}e^{-u\left|t\right|}=e^{-(l+u)\left|t\right|}, $$ which means that $$p_{l}(x)*p_{u}(x)=p_{u+l}(x) $$

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  • $\begingroup$ Thanks Paul, I cannot use this approach. I did end up solving the problem using Fourier Inversion. $\endgroup$ – sami Jun 17 '13 at 21:43
  • $\begingroup$ No problem. You can rewrite this solutions in terms of Fourier transformation. Characteristic function=Fourier transformation Distribution -connected with absolutely integrated function Density of distribution = density of absolutely integrated function. $\endgroup$ – cool Jun 18 '13 at 13:22

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