Prove that the Laplace transform of $f$ is represented by a power series

Question

Let $f:[0, +\infty) \to \mathbb{R}$ be continuous and bounded. Define $$F(s) = \int_0^{\infty} f(t) e^{-st} \; \mathrm{d}t$$Consider an arbitrary $s_0>0$. Prove that $F(s)$ is represented by a power series $F(s) = \sum_{n = 0}^{\infty} a_n (s-s_0)^n$ which has a radius of convergence $R \geq s_0$.

I know that the Laplace transform is essentially the continuous version of the discrete power series. How should I go about proving this statement? Do I need to find a closed form for $\{ a_n \}$ in terms of $f$? Any advice would be appreciated.

Answer 1

Wikipedia:

The Laplace transform is analytic in the region of absolute convergence: this is a consequence of Fubini's theorem and Morera's theorem.

Let be $\gamma$ a simple closed curve: $$ \int_\gamma F(s)\,ds = \int_\gamma\int_0^{\infty} f(t) e^{-st}\,dt\,ds = $$ $$ = \int_0^{\infty}f(t)\int_\gamma e^{-st}\,ds\,dt = 0. $$

For an alternative proof see Theorem 6.7 of this lecture notes.