Differentiation under the integral sign (one complex variable) Let $u(z), u'(z)$ be complex-analytic functions on an open neighborhood $\Omega \subseteq \mathbb{C}$ of the origin. Also, let $f(X)$ be a complex-analytic function. For $s \in [0,1],$ define
$$g(s,z) = f((1-s)u(z) + su'(z))$$
and 
$$h(z) = \int_0^1 g(s,z) ds. $$
How do we compute $h'(z)$? 


*

*Is it possible to differentiate under the integral sign? (Most of the results I have seen regarding this assumes that $z$ is also real.) If yes, I would appreciate being pointed to a proof of this result.

*In particular, can we estimate $|h'(z)|$ as follows?
$$ |h'(z)| \leq \int_0^1 |\partial_z g(s,z)| ds. $$


Since $g$ is smooth, we see that ${\partial g(s,z) \over \partial z}$ is bounded on $[0,1] \times U$, where $U$ is a bounded neighbourhood of $\Omega$.
The mean value theorem gives  $|g(s,z+h)-g(s,z)| \le \left( \sup_{(s,z) \in [0,1] \times U}|{\partial g(s,z) \over \partial z}| \right) | h| $.
Hence ${1 \over h} |g(s,z+h)-g(s,z)| \le \left( \sup_{(s,z) \in [0,1] \times U}|{\partial g(s,z) \over \partial z}| \right)$ and so the left hand side is uniformly bounded. Since $\lim_{ h \to 0} {1 \over h} (g(s,z+h)-g(s,z)) = {\partial g(s,z) \over \partial z}$, the bounded convergence theorem gives
$h'(z) = \lim_{h \to 0} \int_0^1 {1 \over h} (g(s,z+h)-g(s,z)) ds= \int_0^1{\partial g(s,z) \over \partial z} ds $.
The bound $|h'(z)| \le \int_0^1 |{\partial g(s,z) \over \partial z}| ds $ follows.
Addendum: Suppose $|\phi_n(t)| \le K$, $\phi_n(t) \to \phi(t)$ and $A$ has finite measure. Then $|\operatorname{re} \phi_n(t)| \le K$, and 
$\operatorname{re} \phi_n(t) \to \operatorname{re} \phi(t)$, so
$\int_A \operatorname{re} \phi_n(t)dt \to \int_A \operatorname{re} \phi(t) dt$. Similarly for the imaginary part. Then linearity of the integral gives
$\int_A \phi_n(t)dt \to \int_A \phi(t) dt$.
In the above case, take any sequence $h_n \to 0$ ($h_n \neq0$), and let
$\phi_n(t) = {1 \over h_n} (g(s,z+h_n)-g(s,z))$.
