Evaluation of Real Integral Given the following definition:$$I=\int\limits_{0}^{2\pi}e^{-i\theta n}\left(\frac{1}{n}\right)^{\rho e^{i\theta}}d\theta$$ 
Is there an analytic method for evaluating this integral?
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Let $z=e^{i\theta}$, so that $dz=ie^{i\theta}d\theta$ and the integral $I$ corresponds to the contour integral
$$
\oint_C \frac{1}{z^{n}e^{z\rho\log n}}\frac{dz}{iz}=\frac{1}{i}\oint_C\frac{dz}{z^{n+1}e^{z\rho\log n}}
$$
where $C$ is the unit circle. The function $f(z)=\frac{1}{z^{n+1}e^{z\rho\log n}}$ is meromorphic with a pole of order $n+1$ at the origin, where it has a residue of 
$$
\text{res}_{0}f(z)=\frac{1}{n!}\lim_{z\to 0}\left(\frac{d}{dz}\right)^n\left(e^{-z\rho\log n}\right)=\frac{1}{n!}\lim_{z\to 0}(-1)^n(\rho \log n)^n e^{-z\rho \log n}=\frac{(-1)^n(\rho\log n)^n}{n!}
$$
Finally, we apply the residue formula to get
$$
I=\frac{1}{i}2\pi i\cdot\text{res}_0 f(z)=\frac{2\pi (-1)^n(\rho \log n)^n}{n!}
$$
