From Fourier transform to Laplace Transform It's well known that you can estimate the Fourier Transform $X(f)$ of a signal $x(t)$ via its Laplace Transform $X(s)$, just by setting $s = j2\pi f$ to the latter, as long as the region of convergence includes the imaginary axis. 

However, I do not have a clear view of how (and when) we can obtain the Laplace Transform via the Fourier Transform of a signal (which is the opposite of what I've stated before).

For example, $x(t) = e^{-at}u(t)$ has a Fourier Transform $X(f) = \frac{1}{a+j2\pi f}$ as long as $a &gt; 0$. Its Laplace Transform is $X(s) = \frac{1}{a+s}$, for any value of $a$, as long as $\mathrm{Re}\{s\} &gt; -a$. We can see that if we set $j2\pi f = s$ to the Fourier Transform, we can directly obtain the Laplace Transform. The same holds for any rational function of $j2\pi f$.

Is there a theorem or something that can be clearly stated about it? It looks to me that it has something to do with the convergence of the Fourier integral (if it does converge, then the Laplace Transform converges as well).
So it sounds like X(s) is the analytic continuation of X(f), given the proper restrictions, is that right?


that's what i would say.  be careful about expressing the function and its argument.
use a different "$X(\cdot)$" if you need to.  like: 
$$ X_\mathscr{L}(j 2 \pi f) = X_\mathscr{F}(f) $$
where 
$$ X_\mathscr{L}(s) \triangleq \mathscr{L}\Big\{ x(t) \Big\} = \int\limits_{-\infty}^{\infty} x(t) \, e^{-st} \, \mathrm{d}t $$ 
and 
$$ X_\mathscr{F}(f) \triangleq \mathscr{F}\Big\{ x(t) \Big\} = \int\limits_{-\infty}^{\infty} x(t) \, e^{-j2\pi f t} \, \mathrm{d}t $$