Let $U\subseteq\mathbb{C}$ be open, $\gamma$ a way in $\mathbb{C}$ which is picewise differentiable continiously and $f\colon rg(\gamma)\times U\to\mathbb{C}$ a continious function. Consider the parameter integral
$$
F(z):=\int_{\gamma}f(\omega,z), d\omega, z\in U.
$$
Show: Let the function $z\mapsto f(\omega,z)$ be holomorphic in $U$ for every $\omega\in rg(\gamma)$ with continious derivation $\frac{\partial}{\partial z}f(\omega,z)$ on $rg(\gamma)\times U$. Then $F(z)$ is holomorphic in $U$, and one can differentiate under the integral:
$$
F'(z)=\int_{\gamma}\frac{\partial}{\partial z}f(\omega,z)\, d\omega, z\in U
$$
Unfortunately I do not know how to show that.
I know the proof when $f$ is a function with values in $\mathbb{R}$ but cannot proof it here when $f$ has values in $\mathbb{C}$.
Best Answer
Firstly, you can prove it with real methods, splitting the function and integrals into real and imaginary parts, and using the characterisation of holomorphic functions by the Cauchy-Riemann equations.
We have, however, nicer methods at our disposal in complex analysis. One very nice tool is
Theorem (Morera):
Let $U \subset \mathbb{C}$ open, and $f \colon U \to \mathbb{C}$ a continuous function. If, for all closed triangles $\Delta \subset U$, you have
$$\int_{\partial \Delta} f(z)\, dz = 0,$$
then $f$ is holomorphic on $U$.
(Cauchy's integral theorem provides the converse.)
Armed with that, we let $\Delta$ an arbitrary closed triangle in $U$ and compute
$$\begin{align} \int_{\partial \Delta} F(z)\,dz &= \int_{\partial \Delta} \int_\gamma f(\omega,\,z) \, d\omega\,dz\\ &= \int_\gamma \int_{\partial \Delta} f(\omega,\, z)\,dz\, d\omega & \text{(Fubini)}\\ &= \int_\gamma 0\, d\omega & \text{(Cauchy)}\\ &= 0, \end{align}$$
hence conclude by Morera's theorem that $F$ is holomorphic in $U$.
Since the partial derivative $\frac{\partial f}{\partial z}$ is continuous by the premise, the function
$$G(z) = \int_\gamma \frac{\partial f}{\partial z}(\omega,\, z)\,d\omega,$$
is continuous on $U$ (even holomorphic, by Morera), and we can compute
$$\begin{align} \int_\alpha G(z)\,dz &= \int_\alpha \int_\gamma \frac{\partial f}{\partial z}(\omega,\, z)\,d\omega\,dz\\ &= \int_\gamma \int_\alpha \frac{\partial f}{\partial z}(\omega,\, z)\,dz\,d\omega & \text{(Fubini)}\\ &= \int_\gamma f(\omega,\, \alpha(1)) - f(\omega,\,\alpha(0))\,d\omega\\ &= F(\alpha(1)) - F(\alpha(0)) \end{align}$$
for each path of integration $\alpha \colon [0,\,1] \to U$. That implies that $F'(z) = G(z)$.
(To make it explicit, fix $z_0 \in U$ and choose $\alpha$ a straight line segment connecting $z_0$ and $z \in U$ - for $r := \lvert z-z_0\rvert$ small enough, the segment is contained in $U$. Then
$$\begin{align} \left\lvert\frac{F(z) - F(z_0)}{z-z_0} - G(z_0)\right\rvert &= \left\lvert\frac{1}{z-z_0}\int_\alpha G(\zeta) - G(z_0)\, d\zeta\right\rvert\\ &\leqslant \frac{1}{\lvert z-z_0\rvert}\int_0^1 \lvert G(z_0 + t(z-z_0)) - G(z_0)\rvert\cdot \lvert z-z_0\rvert\, dt\\ &\leqslant \max_{\lvert w-z_0\rvert \leqslant r} \lvert G(w) - G(z_0)\rvert, \end{align}$$
and the continuity of $G$ in $z_0$ shows the desired convergence.)