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

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.

$$F(s)$$ 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 $$c$$.
• And for similar reasons $f$ is a compactly supported distribution iff $F$ is entire and $O(s^k e^{r|\Re(s)|})$ Commented Jun 19, 2020 at 21:53
• Which function ? If $f\in L^1[-r,r]$ (or any compactly supported distribution) then its Laplace transform is entire Commented Jun 19, 2020 at 22:08