Integral of $e^{-i\theta}e^{e^{i\theta}} d\theta$ over the unit circle I need to evaluate $\int_{0}^{2\pi} e^{-i\theta}e^{e^{i\theta}} d\theta$. 
My attempt at a solution was to try and find the antiderivative of $e^{-i\theta}e^{e^{i\theta}}$ and use the antiderivative theorem but I have not been able to find one for this expression. Is this the right approach or is there some other approach I am not seeing?
 A: Remark: The first part was written before the OP changed the question.  The question (in the body) originally asked to compute $$\int_0^{2\pi}\,\,\exp(-\text{i}\theta)\,\exp\big(\exp(-\text{i}\theta)\big)\,\text{d}\theta\,.$$
Let $z:=\exp(-\text{i}\theta)$.  So, you have $$\begin{align}\int\,\exp(-\text{i}\theta)\,\exp\big(\exp(-\text{i}\theta)\big)\,\text{d}\theta&=\int\,z\,\exp(z)\,\left(\frac{1}{-\text{i}z}\right)\,\text{d}z
\\&=\text{i}\,\int\,\exp(z)\,\text{d}z
\\&=\text{i}\,\exp(z)+\text{constant}\\&=\text{i}\,\exp\big(\exp(-\text{i}\theta)\big)+\text{constant}\,.\end{align}$$
Therefore,
$$\int_0^{2\pi}\,\,\exp(-\text{i}\theta)\,\exp\big(\exp(-\text{i}\theta)\big)\,\text{d}\theta=\text{i}\,\exp\big(\exp(-\text{i}\theta)\big)\Big|_0^{2\pi}=0\,.$$

If you meant to compute $$I:=\int_0^{2\pi}\,\exp(-\text{i}\theta)\,\exp\big(\exp(+\text{i}\theta)\big)\,\text{d}\theta\,,$$
then note that
$$I=\oint_\gamma\,\frac{1}{w}\,\exp(w)\,\left(\frac{1}{\text{i}w}\right)\,\text{d}w\,,$$
where $w:=\exp(\text{i}\theta)$ and $\gamma:=\big\{w\in\mathbb{C}\,\big|\,|w|=1\big\}$.  Therefore,
$$I=2\pi\,\left(\frac{1}{2\pi\text{i}}\,\oint_\gamma\,\frac{\exp(w)}{w^2}\,\text{d}w\right)=2\pi\,\left(\left.\frac{\text{d}}{\text{d}w}\right|_{w=0}\,\exp(w)\right)=2\pi\,,$$
by Cauchy's Integral Formula.
A: Let $z=e^{i\theta}$.  The integral of interest is 
$$\int_0^{2\pi}e^{-i\theta}e^{e^{i\theta}}\,d\theta=\oint_{|z|=1}\frac{e^{z}}z \left(\frac1{iz}\right)\,dz$$
Using the Taylor series for $e^{z}$, we see that 
$$\begin{align}
\oint_{|z|=1}\frac{e^{z}}z \left(\frac1{iz}\right)\,dz&=\sum_{n=0}^\infty \frac{1}{n!}\oint_{|z|=1}\left(\frac1{iz}\right) \frac{z^n}{z}\\\\
&=2\pi i \sum_{n=0}^\infty \frac1{n!}\text{Res}\left(\frac{z^{n-2}}{i}\right)\\\\
&=2\pi  
\end{align}$$
A: Using the Taylor expansion of the exponential function, we see that $e^{e^{i\theta}}$ admits the complex Fourier series $e^{e^{i\theta}} = \sum_{n=0}^{\infty} \frac{1}{n!} e^{in\theta}$, which converges uniformly over $\mathbb{R}$. So the integral denotes $2\pi$ times the complex Fourier coefficient of $e^{i\theta}$, which is
\begin{align*}
\int_{0}^{2\pi} e^{-i\theta}e^{e^{i\theta}} \, d\theta
= \sum_{n=0}^{\infty} \frac{1}{n!} \int_{0}^{2\pi} e^{-i\theta}e^{in\theta} \, d\theta
= \sum_{n=0}^{\infty} \frac{1}{n!} \left( 2\pi \mathbf{1}_{\{ n = 1\}} \right)
= 2\pi.
\end{align*}
