Is it true that the Laplace Transform of a real function with compact support is always entire?

Question

Is it true that the Laplace Transform of a real function with compact support is always entire (entire = complex derivative exists on the entire complex plane)?

Answer 1

One should assume that the real function in question is integrable, otherwise the Laplace transform isn't defined at $\lambda=0$. Here is a more general statement.

Claim. Suppose $(\Omega, \mu)$ is a finite measure space and for every $\omega\in\Omega$ we have an entire function $f_\omega$. Suppose further that the family $\{f_\omega\}$ is uniformly bounded on compact subsets of $\mathbb C$. Then the function $$F(z)=\int_\Omega f_\omega(z)\,d\mu(\omega) \tag1$$ is entire.

Proof. The integral (1) converges for all $z\in\mathbb C$, being the integral of a bounded function over a finite measure space. Since $f_\omega'$ are also uniformly bounded on compact subsets, the family $\{f_\omega\} $ is uniformly Lipschitz on compact subsets. This property passes to $F$.

Let $T$ be a triangle in the complex plane. The integral $\int_T F(z)\,dz$ makes sense by the above, and can be computed with Fubini's theorem: $$ \int_T F(z)\,dz = \int_\Omega \left(\int_T f_\omega(z)\,d\mu(\omega)\right)dz =\int_\Omega \left(\int_T f_\omega(z)\,dz\right) d\mu(\omega) =0 $$ By Morera's theorem, $F$ is holomorphic in $\mathbb C$. $\quad\Box$

In our situation, $\Omega$ is a finite interval in $\mathbb R$, $d\mu(\omega) = f(\omega)\,d\omega$, and $f_\omega(z) = e^{-\omega z} $. The assumptions in the claim are satisfied, and $F$ in (1) is the Laplace transform of $f$.

Answer 2

Really true?

But in fact most simple entire elementary functions their Laplace Transform are often unavoidable to have some singularities and not entire, and in fact finding the inverse Laplace Transform of entire functions are often much more difficult than that of the functions having some singularities. (e.g. headache in $\mathcal{B}^{-1}_{s\to x}\{e^{as^2+bs}\}$ and $\mathcal{L}^{-1}_{s\to x}\{e^{as^2+bs}\}$ , where $a\neq0$)