convolution of functions whose integral is known

Question

Is there a general approach to solve the convolution \begin{align} (f*g)(x) & = \int_{-\infty}^\infty f(t)g(x-t)\,dt \end{align} if the individual integrals $\int_{-\infty}^\infty f(t)\,dt, \; \int_{-\infty}^\infty g(t)\,dt $ are known?

(In my particular case $f$ and $g$ have support on $[0, T]$ only.)

Thanks!

Answer 1

It is just a matter of finding the integral on RHS and knowledge of the integrals you mention is not useful.

If $f$ and $g$ have support $[0,T]$ then the result will be the same if $f$ is replaced by $f\mathbf1_{[0,T]}$ and $g$ by $g\mathbf1_{[0,T]}$.

Now observe that: $$f(x)\mathbf1_{[0,T]}(x)g(x-t)\mathbf1_{[0,T]}(x-t)\neq0\implies$$$$ t\in[0,T]\cap[x-T,x]=[\max(0,x-T),\min(T,x)]$$ showing that it is handsome to discern cases:

• If $x<0$ or $x>2T$ then the integrand is the zero function so that $(f*g)(x)=0$.
• If $x\in[0,T]$ then it is enough to integrate over $[0,x]$.
• If $x\in[T,2T]$ then it is enough to integrate over $[x-T,T]$.

Answer 2

Certainly not. The two integrals are mere constants, which do not allow to retrieve any useful information about $f$ and $g$.