The integral of the convolution between functions f

[muzialis]

Hello there,

I am really struggling to prove that
"The integral of the convolution between functions f and gequals the product of their integrals", http://en.wikipedia.org/wiki/Convolution#Integration
Can anybody give me a hint?

Many thanks

 

[micromass]

What did you try?? Where are you stuck?

 

[mathman]

∫{∫f(y)g(x-y)dy}dx = ∫f(y){∫g(y-x)dx}dy (Fubini)
= ∫f(y){∫g(u)du)}dy = ∫f(y)dy∫g(u)du

 

[muzialis]

Thanks very muhc for your help.
I was following the line given by Mathman, but did not realize that the variable translation would not affcet the value of the integral as the integration domain is the whole real line, many thanks

 

[CellCoree]

for your question. The integral of the convolution between two functions f and g can be proven using the definition of convolution and the properties of integrals. Here are some steps you can follow to prove this statement:

1. Start with the definition of convolution: (f * g)(x) = ∫f(x-y)g(y)dy.

2. Substitute the definition into the integral: ∫(f * g)(x)dx = ∫∫f(x-y)g(y)dydx.

3. Use the properties of integrals to rearrange the order of integration: ∫∫f(x-y)g(y)dydx = ∫∫f(x-y)g(y)dxdy.

4. Use a change of variables to simplify the integral: Let u = x-y, then du = dx. This gives us: ∫∫f(u)g(y)dudx.

5. Use the properties of integrals again to simplify the integral: ∫∫f(u)g(y)dudx = ∫f(u)du ∫g(y)dy.

6. Substitute back the original variables: ∫f(u)du ∫g(y)dy = ∫f(x-y)dx ∫g(y)dy.

7. Use the properties of integrals once more to simplify the integrals: ∫f(x-y)dx ∫g(y)dy = ∫f(x)dx ∫g(y)dy.

8. Finally, we have: ∫(f * g)(x)dx = ∫f(x)dx ∫g(y)dy, which proves the statement that the integral of the convolution between functions f and g equals the product of their integrals.

I hope this helps! If you have any further questions or need clarification, please don't hesitate to ask. Good luck with your proof!