Convolution of $(f*f)(t)$

Question

I encounter an exercise where $f(t) = \frac{2}{1+t^2}$, and asked to find the convolution of $(f*f)(t)$. I'm also given the following $(f*g)(t) = \int_0^t f(u)g(t-u)du$, but not I'm not sure if it's relevant. I tried looking for some example with full solutions but haven't found any, I'd very much like to know the basic steps on solving this problem so I know how to tackle this kind of problem in the future! Thanks ahead.

Answer 1

Here's a more "Fourier" method.

First off, I don't think your definition of convolution is quite right. The convolution of two general measurable functions $f$ and $g$ is $$ (f * g)(t):=\int_{-\infty}^{\infty} f(u) g(t-u) \,du \tag{1} $$

The form you give in your question is equivalent to $(1)$ when $f$ and $g$ are supported on only $[0,\infty)$ which is not the case of your $f$ here.

Now, note that in your case it is very well known that $$ f = \hat{g} \tag{2} $$ where $g(t):=e^{-|t|}$.

By the convolution theorem, $$ \widehat{f*f}(\xi) = \hat{f}(\xi) \hat{f}(\xi)\quad\quad \forall \xi\in\Bbb{R} \tag{3} $$

In view of $(2)$, we have that $(3)$ is equivalent to $$ \widehat{f*f}(\xi) = \hat{\hat{g}}(\xi) \hat{\hat{g}}(\xi) \quad\quad \forall \xi\in\Bbb{R} \tag{4} $$

By the inversion theorem, and by the fact that $g$ is even, $(4)$ is equivalent to $$ \widehat{f*f}(\xi) = (2\pi)^2 g^2(\xi) \quad\quad \forall \xi\in\Bbb{R} \tag{5} $$

Taking the Fourier transform of each side of $(5)$, we get, applying the inversion theorem again, $$ (f*f)(t) = 2\pi\widehat{g^2}(t) \quad\quad \forall t \in\Bbb{R} \tag{6} $$

But here $g^2(\xi)=g(2\xi)$ hence $\widehat{g^2}(t) = \frac{1}{2}\hat{g}(t/2)$ by a well known property of the Fourier transform and finally $(6)$ reduces to $$ (f*f)(t) = \frac{8\pi}{4+t^2} \quad\quad \forall t \in\Bbb{R} $$

Answer 2

We may use partial fractions in this example. Let $$\frac{1}{1+u^2}\cdot\frac{1}{1+(t-u)^2}=\frac{Au+B}{1+u^2}+\frac{Cu+D}{1+(t-u)^2}.$$ Thus we have $$\begin{aligned}1&=(Au+B)(1+(t-u)^2)+(Cu+D)(1+u^2)\\ &=(A+C)u^3+(-2tA+B+D)u^2+(A-2tB+C)u+B(1+t^2)+D. \end{aligned}$$ This yields $$A=(2t)^{-1},\quad B=0,\quad C=-(2t)^{-1},\quad D=1,$$ provided that $t\neq 0$. Then you can find the integral easily.