TITLE: Show that an integral of an entire function is an entire function
QUESTION [8 upvotes]: Given $f:\Bbb C\times \Bbb R \rightarrow \Bbb C$ is a continous function,
and for each $t\in \Bbb R, f(t,z)$ is an entire function. We'll define
$$
A(z) = \int_0^1f(t,z)\,dt.
$$
Show that $A(z)$ is an entire function.

REPLY [5 votes]: Since $f$ is continuous, we know by general integration theory that
$$A(z) = \int_0^1 f(t,z)\,dt$$
defines a continuous function on all of $\mathbb{C}$, and it remains to see that $A$ is holomorphic.
By Morera's theorem, $A$ is holomorphic if and only if
$$\int_{\partial \Delta} A(z)\,dz = 0$$
for all triangles $\Delta \subset \mathbb{C}$. Now,
$$\begin{align}
\int_{\partial\Delta} A(z)\,dz &= \int_{\partial\Delta} \int_0^1 f(t,z)\,dt\,dz\\
&= \int_0^1 \int_{\partial\Delta} f(t,z)\,dz\,dt\\
&= \int_0^1 0\,dt\\
&= 0,
\end{align}$$
since the continuity of $f$ allows changing the order of integration over the compact domains of integration $[0,1]$ and $\partial\Delta$.
So $A$ is holomorphic on all of $\mathbb{C}$, i.e. entire.