Is $\int_{0}^\infty g(s,u)du$ is holomorphic when $s\longmapsto g(s,u)$ is holomorphic? Let $U\subset \mathbb C$ an open and $g:U\times [a,b]\longrightarrow \mathbb C$ continuous s.t. $s\longmapsto g(u,s)$ is holomorphic. Then, a theorem says that $$s\longmapsto \int_a^b g(s,u)du$$
is holomorphic. Is there such a theorem if $g:U\times \mathbb R\longrightarrow \mathbb C$ ? i.e. if $g:U\times \mathbb R\longrightarrow \mathbb C$ is continuous and $s\longmapsto g(u,s)$ holomorphic, does $$s\longmapsto \int_{\mathbb R}g(u,s)du$$
holomorphic ?
 A: In general no. Consider the function $g(u,s) = 1$. In that case
$$
\def\d{\textrm{d}}
\int_{\mathbb{R}} g(u,s) \,\d u = \int_{\mathbb{R}} 1 \textrm{d} u 
$$
does not exist. You can also create some more subtle examples like $g(u,s) = \exp( - u^2 s)$ which can be integrated for some values of $s$ but not all of them.
Some sufficient (but not necessary) conditions you need to add to make everything work is that first of all $u \mapsto g(u,s)$ can be integrated over $\mathbb{R}$ for all $s$. And that there exists some integrable function $G$ such that $\left|\frac{\d g(u,s)}{\d s}\right| \leq G(u)$ for all $s$ and almost all $u$. This allows you to use Leibniz's rule to conclude that
$$
\frac{\d}{\d s} \int_{\mathbb{R}} g(u,s) \, \d u =\int_{\mathbb{R}} \frac{\d  g(u,s)}{\d s} \, \d u
$$
and (by assumption) the right hand side exists, so $\int_{\mathbb{R}} g(u,s) \,\d u$ is complex differentiable and therefore holomorphic.
Now in practice you can often ignore the condition that there is some uniform bound $G$ for the derivatives, since you can just use the theorem for a small enough neighbourhood of the point you're interested in and prove that it is holomorphic there. You only run into problems when no neighbourhood has such a bound, but that often just means that the integral of $\frac{\d g(u,s)}{\d s}$ does not exist, which is easy enough to check.
