How can we show that $\exp(\sin z)$ has an antiderivative on $\mathbb{C}$?
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$\exp{w}$ 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 $z \in \mathbb{C}$.
Ron Gordon
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Conventionally, symbol $\mathbb C$ means the complex numbers without infinity. – GEdgar Apr 08 '13 at 21:08
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Maybe next time I discuss a real-variable problem, I will talk about $\mathbb R \setminus {i}$. – GEdgar Apr 08 '13 at 21:11
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@GEdgar: ha! I'd like to see you do that. – Ron Gordon Apr 08 '13 at 21:11
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There is no need to discuss the series, really: the integral you wrote defines an antiderivative! (One has to check that it is well defined, surely) – Mariano Suárez-Álvarez Apr 08 '13 at 21:16
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A function has an antiderivate when it is analytic on a simply connected region, and the composition of analytic functions is analytic. As $\mathbb{C}$ is simply connected and $\exp(\sin(z))$ is analytic it has an antiderivative.
Dominic Michaelis
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You need to be careful here. Every holomorphic function having a primitive is equivalent to the domain of definition being simply connected. Here we are fine because $\mathbb{C}$ is simply connected. – Alex Youcis Apr 08 '13 at 20:07
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