Proving the Convolution Theorem for Laplace Transform

[matematikuvol]

$$\mathcal{L}\{f(t)*g(t)\}=F(s)G(s)$$

Is there some relation between

$$F(s)*G(s)$$ and $$f(t)g(t)$$?

$*$ is convolution.

 

[matematikuvol]

Any answer?

 

[Lajka]

Yes, there is.

See here. Check out the 'multiplication' part under the 'Properties and theorems' section.

 

[matematikuvol]

Sorry but I see here only convolution of originals.

 

[Lajka]

Here, let me help you, I've taken a snapshot of it:

The convolution is being done over an imaginary line $Re{(\sigma)} = c$

 

[matematikuvol]

Ok. Tnx. Do you know how to prove?

$F(s)*G(s)=f(t)g(t)$?