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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.

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

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