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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!

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  • $\begingroup$ No, the convolution is not a function of the integrals alone. $\endgroup$ – Giuseppe Negro Dec 19 '18 at 10:57
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Certainly not. The two integrals are mere constants, which do not allow to retrieve any useful information about $f$ and $g$.

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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 $[x-T,T]$.
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