Vrftsjtryk

Suppose:
$$f(t)=e^{at}+e^{bt};quad a>b>0$$

Its Laplace transform is:
$$mathbf{L}[e^{at}+e^{bt}]=frac{1}{s-a}+frac{1}{s-b}$$
for $Re(s)>a$ where $Re$ stands for the real part; for $Re(s)<a$ Laplace transform of $e^{at}+e^{bt}$ doesn't exist because the transform integral isn't finite.

Now suppose we invert $$F(s)=frac{1}{s-a}+frac{1}{s-b}$$
using Bromwich integral. $F(s)$ has simple poles at $a,b$, with corresponding residues equal to $1$. Therefore the inverse transform using Bromwich integral would be:
$$mathbf{L}^{-1}[F(s)]=frac{1}{2pi i}int_Cds~e^{st}(frac{1}{s-a}+frac{1}{s-b})\
=Res_{s=a}[e^{st}(frac{1}{s-a}+frac{1}{s-b})]+Res_{s=b}[e^{st}(frac{1}{s-a}+frac{1}{s-b})]\
=e^{at}+e^{bt}$$

Even though it matches with the $f(t)$ we began with, why is this procedure correct? $F(s)$ is not defined when $Re(s)<a$ and the pole at $b$ lies in the region where $F(s)$ isn't defined. But using the residue at that pole nevertheless gives me $f(t)$ correctly.

My question: Does a transformed function $F(s)$ whose poles lie outside its domain of validity (in the sense that the Laplace transform of the inverted function $f(t)$ has a limited domain of validity in the complex $s$-plane) always give the "correct answer" $f(t)$ when Bromwich integral is evaluated by summing residues?

Since I am not a mathematician I don't how to better phrase my question, but I hope that my example above makes my question clear. Thanks in advance for your help.

Assume we're considering only the one-sided Laplace transform. If $F$ is a Laplace transform, $F$ will be regular to the right of a vertical line, because if the Laplace transform integral converges for $operatorname{Re} s = gamma$, it will converge uniformly on any compact subset of $operatorname{Re} s > gamma$, giving an analytic function in the half-plane $operatorname{Re} s > gamma$.

The function obtained in this way is not defined for $operatorname{Re} s < gamma$, so you're correct that there is an intermediate step of constructing an analytic continuation of $F$ before applying the residue theorem.

$1/sqrt s$ with the branch cut lying in the left half-plane is an example where the inverse transform is well-defined but the residue theorem is not applicable.

If $F$ is meromorphic with finite number of poles and is regular at infinity, it necessarily has a zero at infinity (otherwise the Bromwich integral would diverge), therefore Jordan's lemma applies and the inverse transform can be computed as the sum of the residues of $F$.

If we start with an arbitrary $F$, we do not necessarily get $f$ s.t. $mathcal L[f] = F$. For instance,
$$mathcal L^{-1} !{left[ frac {e^s - 1} s right]} = 0,$$
where the Bromwich integral can be taken over any vertical line.