How to recognize the Laplace transform of a function with compact support?

Question

The question is pretty much self-contained in the title: is there some criterion for recognizing the Laplace transforms of compact-supported functions, other than the explicit computation of $\mathcal{L}^{-1}$?

The question arises in a peculiar context: some integrals of oscillating functions can be converted into integrals of monotonic functions by exploiting the self-adjointness of the Laplace transform, for instance $$ \int_{0}^{+\infty}\frac{\sin(s)}{\sqrt{s}}\,ds = \int_{0}^{+\infty}\frac{dx}{\sqrt{\pi x}(1+x^2)} $$ and for numerical purposes the latter form is clearly more manageable than the former. On the other hand integrals of compact-supported functions are easier to handle through interpolation and quadrature, so it would be a nice thing to recognize in $\frac{1+e^{-\pi s}}{1+s^2}$ the Laplace transform of the chunk of the sine wave supported on $[0,\pi]$, in order to compute $$ \int_{0}^{+\infty}\frac{1+e^{-\pi s}}{\sqrt{s}(1+s^2)}\,ds $$ by applying a quadrature scheme (as done here) to $$ \int_{0}^{\pi}\frac{\sin(s)}{\sqrt{s}}\,ds. $$ The essence of the question is to understand which kinds of functions allow this trick.

Answer 1

$F(s)$ is the Laplace transform of a $L^2[-r,r]$ function iff $F(s)$ is entire, uniformly $L^2$ on vertical strips (*), and $F(s) = O(e^{r |\Re(s)|})$.

Proof : for $|t|> r+|a|$ let $c\to -sign(t) \infty$ in $$2i\pi f(t) \ast 1_{[0,a]}=\int_{c-i\infty}^{c+i\infty} \frac{1-e^{-a s}}{s} F(s)e^{st}ds\tag{1}$$

(*) this means $\int_{|y|>T} |F(x+iy)|^2dy,x\in [u,v]$ tends to $0$ uniformly as $T\to \infty$ so that $(1)$ doesn't depend on $c$.