TITLE: Proof that $g(z) = \int_0^1 f(t)e^{tz} \, dt$ is entire
QUESTION [5 upvotes]: Let $f(z)$ be a complex and continuous function on $[0,1]$. We'll define $g(z) = \int_0^1 f(t)e^{tz} \, dt $. Prove that $g(z)$ is an entire function.

REPLY [12 votes]: Morera's theorem implies that if for every circle $C$ in the complex plane,
$$
\int_C g(z) \; dz = 0
$$
then $g$ is entire.  To find that integral, first write
$$\int_C g(z) \; dz = \int_C\int_0^1 f(t)e^{tz} \; dt\; dz.$$
Now we're integrating continuous functions over compact sets, so everything in sight is Lebesgue-integrable and Fubini's theorem can be used to get
$$
\int_C\int_0^1 f(t)e^{tz} \; dt\; dz=\int_0^1 \int_C f(t) e^{tz} \; dz\;dt = \int_0^1\left(f(t)\int_C e^{tz} \; dz \right) \; dt.
$$
The inner integral is $0$ because it is the integral of an entire function around a circle.  So the whole thing is $0$ and the conclusion from Morera's theorem is just what is needed.