Proof of holomorphicity of an "integral function" Let $\Omega \subset \mathbb{C}$ an open set and $K \subset \mathbb{R}^n$ a compact set. Let $v: \Omega \times K \to \mathbb{C}$ be a continuous function with continuous partial derivatives $v_x, v_y$. Suppose that for all $y \in K$we have  $v(\cdot, y) \in H(\Omega)$. 
Consider $$u(z) = \int_K v(z,y)dy$$ for $z \in \Omega$. 


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*How do I prove that $u$ is holomorphic? 

*Can we relax the assumptions (on the sets or on $v$) and obtain the same result?

 A: Assume that $v : \Omega \times K \to \Bbb{C}$ is a jointly measurable function such that


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*$v$ is locally bounded, or equivalently, for each compact $L \subset \Omega$ we have $\sup_{L\times K}|v| < \infty$,

*$v(\cdot, y)$ is holomorphic on $\Omega$ for each $y \in K$.


(Notice that your assumption implies all these conditions.) Then
$$ u(z) = \int_{K} v(z, y) \, dy $$
is holomorphic. This is essentially because $u(z)$ is a linear combination of holomorphic functions. And all the conditions above are cooked up so that the integral is regular enough to be interpreted as linear combination of $v(\cdot, y)$'s.
Indeed, let $\gamma : [0,1] \to \Omega$ be any piecewise $C^1$-loop on $\Omega$. Then the image $\gamma([0,1])$ is a compact subset of $\Omega$ so that the function $v$ is bounded on $\gamma([0,1]) \times K$. Since $\gamma'(t)$ is integrable, we can apply the Fubini's theorem to get:
\begin{align*}
\int_{\gamma} u(z) \, dz
&= \int_{0}^{1} u(\gamma(t))\gamma'(t) \, dt \\
&= \int_{0}^{1} \left( \int_{K} v(\gamma(t), y) \gamma'(t) \, dy \right) \, dt \\
&= \int_{K} \left( \int_{0}^{1} v(\gamma(t), y) \gamma'(t) \, dt \right) \, dy \\
&= \int_{K} \left( \int_{\gamma} v(z, y) \, dz \right) \, dy = 0
\end{align*}
since $\int_{\gamma} v(z, y) \, dz = 0$ by analyticity of each $v(\cdot, y)$. Therefore by the Morera's theorem, $u$ is holomorphic on $\Omega$.
