I was reading the proof that Laplace transform maps the convolution of two functions to the multiplication of their transforms. Or mathematically $$\mathcal{L}[f*g]=\mathcal{L}[f]\,\mathcal{L}[g],$$ with $$\mathcal{L}[f](s)=\int_0^\infty e^{-st}f(t)dt$$ and $$f*g(t)=\int_0^t f(\tau)g(t-\tau)d\tau.$$
In the proof it is written $$\int_0^\infty \int_0^t e^{-st} f(\tau)g(t-\tau)d\tau dt=\int_0^\infty \int_\tau ^\infty e^{-st} f(\tau)g(t-\tau) dt d\tau.$$
The justification is 2 pictures that show that these integrals go through the same area. You can see this here, page 2.
I understand this but I believe the same could be proved just with changes in the integration variables. However I could not figure this out. Can this be done analytically?
Edit: I assume that $f$ and $g$ are analytic and $|f(t)|,|g(t)| \le M e^{a|t|}$. It is easy to prove that $|f*g(t)|\le M' e^{a|t|}$. Then $\mathcal{L}[f*g](s)$ is defined for $\mbox{Re}\, s> a$. So in particular I can't assume that $f(t)=g(t)=0$ when $t<0$.
My goal is to prove something similar but instead of integration over a line, I have integration over a path. That's why I want an analytic way for the change.