Complex function defined by contour integral along a smoothly varying path Let $D$ be a domain in the complex plane. Consider the function $F: D\to \mathbb{C}$, defined by
$$
F\left( z \right) = \int_{\mathscr{C}\left( z \right)} {f\left( {z,t} \right)dt} .
$$
Suppose that the path $\mathscr{C}\left( z \right)$ is a continuous function of $z\in D$, $f\left( {z,t} \right)$ is an analytic function of $z\in D$ and $t\in \mathscr{C}\left( z \right)$, and that the integral converges absolutely for any $z\in D$. Is it true that $F$ is analytic in $D$?

Simple special case for intuitions sake: Let $D=\mathbb C$, $f(z,t)=1$, $\gamma_z(t)=g(z)t$, $0\leq t \leq 1$, where the bar denotes complex conjugation. Then $F(z)=g(z)$ so $F$ is analytic if and only if $g$ is analytic. 
Suppose now in addition to your stated hypothesis that $z\mapsto \gamma_z'(t)$ and $z\mapsto \gamma_z(t)$ are analytic for each $t$. Then $z\mapsto f(z,\gamma_z(t))\gamma_z'(t)$ is analytic for each $t$ and uniformly bounded in every compact set $C\subset D$. It follows that
$$
F(z)=\int _0^1 f(z,\gamma_z(t))\gamma_z '(t) \,dt
$$
is well-defined and it is a simple consequence of Morera's Theorem (as pointed out by Mhenni Benghorbal) coupled with Fubini's Theorem that $F$ is analytic.
