TITLE: Show that $\exp(\sin z)$ has an antiderivative on $\mathbb{C}$
QUESTION [5 upvotes]: How can we show that $\exp(\sin z)$ has an antiderivative on $\mathbb{C}$?

REPLY [2 votes]: $\exp{w}$ is an analytic function in $\mathbb{C}$ and therefore has a Maclurin series valid for all $w \in \mathbb{C}$:
$$\exp{w} = \sum_{k=0}^{\infty} \frac{w^k}{k!}$$
$w = \sin{z}$ is also an analytic function in $\mathbb{C}$:
$$w = \sin{z} = \sum_{n=0}^{\infty} (-1)^n \frac{z^{2 n+1}}{(2 n+1)!}$$
One may derive a single series for the composite function, valid for all $z \in \mathbb{C}$.  
$$\exp{(\sin{z})} = \sum_{m=0}^{\infty} a_m z^m$$
This series may be integrated term by term because it represents an analytic function for all $z \in \mathbb{C}$:
$$\int dz \, \exp{(\sin{z})} = \sum_{m=0}^{\infty} \frac{a_m}{m+1} z^{m+1}$$
The series on the right represents the antiderivative of $\exp{(\sin{z})}$ for all $z \in \mathbb{C}$.