why is the Laplace transform of local integrable function with support on $[0,\infty)$ analytic?

Question

There is a proposition about Laplace transform, but I don't know how to prove it.

Let $f \in L^1_{loc}(\mathbb{R})$, $\operatorname{supp}(f) \subset[0, \infty)$, such that $a$ is the abscissa of absolute convergence of the Laplace transform $F$ of $f$. Then $z \mapsto F(z)$ is analytic in the half plane $\operatorname{Re}(z)>a$.

A point $a $ is called the abscissa of absolute convergence of the Laplace transform $F$ of $f$, if $a$ is the minimum real number such that $ \int_0^\infty |f(t)| e^{-Re(z)t} dt $ exists for any $z$ with $Re(z) > a$.

Answer 1

Morera's theorem and Fubini's theorem provide a quick solution. Let $\gamma:[0,1]\to \{z:\Re(z)>a\}$ be a piecewise $C^1$ closed curve. Then, let $\theta_*$ be such that $a<\Re(\gamma(\theta_*)=\min_{0\leq \theta\leq 1} \Re(\gamma(\theta))$, then $$ |e^{-\gamma(\theta)t}|\leq e^{-Re(\gamma(\theta_*)t} $$ and $$ \int_0^1\int_0^\infty |e^{-\gamma(\theta)t}||\gamma'(\theta)||f(t)|\mathrm dt\mathrm d\theta\leq \int_0^1\int_0^\infty e^{-\Re(\gamma(\theta_*))t}|\gamma'(\theta)||f(t)|\mathrm dt\mathrm d\theta<\infty $$ so that by Fubini-Tonelli, $$ \int_\gamma F(z)\mathrm dz=\int_0^\infty f(t)\int_\gamma e^{-zt}\mathrm dz\mathrm dt=0 $$ and $F$ is holomorphic by Morera's theorem.