Suppose $f:[0,\infty)\rightarrow \mathbb{R}$ is lebesgue integrable with compact support. Define the laplace transform by $$F(t) = \int_0^\infty f(x) e^{-tx}dx.$$
Show that $F$ is continuously differentiable on $[0,\infty)$.
The best I've been able to do with this one is to show that it's a decreasing function, and thus is differentiable almost everywhere. And that's about it.
I found another solution here, but I don't really understand it.
Any thoughts on how I could make a statement about differentiability (and the subsequent continuity) for this problem would be greatly appreciated.
Thanks in advance.
Hint: You can simply "do it" without relying on any results on differentiation under the integral:
$$\frac{F(t+h)-F(t)}{h}=\int_0^\infty f(x)e^{-xt}\frac{e^{-xh}-1}{h}\,dx$$
and note that the fraction in the last integral converges uniformly on the (compact) support of $f$.