how to perform laplace transform of a convolution?

Question

Can i say that the laplace transform of a function convoluted with itself $\left[\,\mathrm{f}\left(\,t\,\right)*\mathrm{f}\left(\,t\,\right)\,\right]$ is the square of the Laplace Transform of that function $\left[\,\mathrm{F}\left(\,s\,\right)\,\right]^{\,2}$ ?.

Answer 1

Suppose $f$ and $g$ admit Laplace transforms. Then their convolution

$$ (f * g)(t) = \int_0 ^\infty f(\tau)g(t-\tau) \ d\tau $$

has Laplace transform $\mathcal{L}(f * g) = \mathcal{L}(f) \mathcal{L}(g)$. Of course, if $f = g$ then $\mathcal{L}(f * f) = (\mathcal{L}(f))^2 $. If you wish to evaluate the transform at a specific value then you can do so.

I suggest you expand on @user1952009's comment and prove the convolution theorem, that the transform of a convolution is the product of transforms.