# Show that an integral of an entire function is an entire function

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.

• Have you heard of Morera's theorem? – Daniel Fischer Jan 30 '14 at 11:00
• yes, but it's not a closed contour – user107761 Jan 30 '14 at 11:02
• For Morera's theorem, you consider $$\int_{\partial \Delta} A(z)\,dz,$$ where $\Delta$ is a triangle in $\mathbb{C}$. – Daniel Fischer Jan 30 '14 at 11:07
• @DanielFischer "Where $\Delta$ is a triangle" --- very amusing. – Newb Jan 31 '14 at 9:09

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.