Help with finding the region where the function has an antiderivate... I'm having trouble in finding the region on wich $f(z) = \exp(1/z)$ has an antiderivative, by making this region as large as i can. And i want to know how that will compare with the real function $f(x) = \exp(1/x)$. 
Any help would be appreciated. Thanks.
 A: Your function has a Laurent series
$$
            e^{1/z}=\sum_{n=0}^{\infty}\frac{z^{-n}}{n!},\;\;\; 0 < |z| < \infty.
$$
You can see that $g(z)=e^{1/z}-1/z$ has an anti-derivative $h$ on $\mathbb{C}\setminus\{0\}$, where
$$
          h(z) = z - \sum_{n=2}^{\infty}\frac{z^{-n+1}}{(n-1)n!}.
$$
So $e^{1/z}$ has an antiderivative $k(z)$ on some region $\Omega\subseteq\mathbb{C}\setminus\{0\}$ iff $1/z$ does. This is because $k'(z)-h'(z)=(e^{1/z})-(e^{1/z}-1/z)=1/z$ must hold on $\Omega$ for any such $k$. So the problem comes down to finding a region in $\mathbb{C}\setminus\{0\}$ where there is an analytic logarithm.
Modify $\mathbb{C}\setminus\{0\}$ by removing a slit or a curve from $0$ to $\infty$ in such a way that the resulting open region $\Omega$ is simply connected. Then $e^{1/z}$ has an antiderivative $a$ which is unique up to an additive constant. Assuming $1\in\Omega$, and choosing $a(1)=0$, then
$$
                a(z) = \int_{C_{z}}e^{1/w}\,dw,
$$
where $C_{z}$ is any simple path from $0$ to $z$ which stays in $\Omega$. This function $a$ will be real on the entire positive real axis, unless the real axis crosses the boundary of $\Omega$ (you can choose non-straight-line curves for branch cuts.) When you look at the the value of $a(-1)$, then the path $C_{-1}$ will either circle the origin in a clockwise or a counter-clockwise path from $1$ to $-1$. Then $a(-1)-a(-r)$ will be real for all $r > 0$ unless the branch crosses the negative real axis. The value of $a(-1)-a(1)$ is 
$$
         \int_{0}^{\pm\pi}\sum_{n=0}^{\infty}\frac{e^{-in\theta}}{n!}ie^{i\theta}\,d\theta = \pm\pi i +\sum_{n=0\;(n\ne 1)}^{\infty}\frac{e^{\pm i(1-n)\pi}-1}{(1-n)n!}=\pm\pi i +\sum_{n=0\;(n\ne 1)}^{\infty}\frac{(-1)^{n-1}-1}{(1-n)n!}. 
$$
So $a(-1)-a(1)=\pm\pi i+r$, where $r$ is the same real number for both cases ($\pm$). So the antiderivative on the positive and negative real axes are easily related to the usual real antiderivatives on the positive and negative real axes; however the antiderivative on the negative real axis for $e^{1/z}$ is related to that on the positive real axis, with a relation that depends on the type of branch-cut you choose (Note: the integral definition of the antiderivative does not technically exist on the entire real axis where the branch cut crosses.)
