# convolution of functions whose integral is known

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!

• No, the convolution is not a function of the integrals alone. – Giuseppe Negro Dec 19 '18 at 10:57

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