How can I prove that the ROC (region of convergence) of the Laplace Transform of any finite-duration signal is the entire complex plane?
Lets think of a "friendly" function $f(t)$ such is zero outside the interval $t \in (a,\ b)$ with $0\leq a<b<\infty$ (so, the function is compact-supported in the time variable $t$).
I want to understand why its Laplace Transform's region of convergence (ROC) is the whole complex plane, and how it is proved (or disproved).
Suppose $f(t)$ is the function that represents the signal "in the time domain." The Laplace transform (by definition) is $$\int_0^\infty f(t)e^{-st}\,dt$$ If $f\in L^1[0,\infty)$, this exists for $\mathrm{Re}\,s\ge 0$. Now suppose that $f$, in addition to being $L^1$, is zero outside the interval $[0,a]$, which corresponds to the signal having finite duration, namely from $t=0$ up to $t=a$. This implies that in the above integral, $|e^{-st}|=e^{-t\mathrm{Re}\,s}$ $\le e^{|s|a}$ for $t$ with $f(t)\ne 0$, which in turn gives $$\int_0^\infty |f(t)e^{-st}|\,dt\le e^{|s|a}\int_0^\infty |f(t)|\,dt$$ This shows that the Laplace transform is finite for all $s$.
Note this answer uses some "mathematical lingo," which may not be the usual language of signals and systems. Just in case, $f\in L^1$ just means that $\int |f|<\infty$. If the function $f$ has finite duration and is bounded, then it's easy to see that it's $L^1$.