calculate: $\int_{0}^{2\pi}e^{\cos\theta}(\cos(n\theta-\sin\theta))d\theta$
my try:
$
\begin{array}{c}
\int_{0}^{2\pi}e^{\cos\theta}(\cos(n\theta-\sin\theta))d\theta\\
\int_{0}^{2\pi}e^{\cos\theta}(\frac{e^{-i(n\theta-\sin\theta)}}{2}+\frac{e^{i(n\theta-\sin\theta)}}{2})d\theta\\
\int_{0}^{2\pi}(\frac{e^{-i(n\theta \cos\theta-\sin\theta \cos\theta)}}{2}+\frac{e^{i(n\theta \cos\theta-\sin\theta \cos\theta)}}{2})d\theta\\
\frac12\int_{-2\pi}^{2\pi}(\frac{e^{-i(n\theta \cos\theta-\sin\theta \cos\theta)}}{2}+\frac{e^{i(n\theta \cos\theta-\sin\theta \cos\theta)}}{2})d\theta\\
\frac14\int_{-2\pi}^{2\pi}(e^{-i(n\theta \cos\theta-\sin\theta \cos\theta)}+e^{i(n\theta \cos\theta-\sin\theta \cos\theta)})d\theta
\end{array}$
I failed to find a path that will allow me to evaluate it. I thought about a semi circle but I wasn't able to show the arc tends to 0.
Best Answer
You seem to have incorrectly thought $e^xe^y=e^{xy}$ rather than $e^xe^y=e^{x+y}$. Your integral is$$\Re\int_0^{2\pi}e^{\cos\theta+in\theta-i\sin\theta}d\theta=\Re\int_0^{2\pi}e^{in\theta}e^{e^{-i\theta}}d\theta\stackrel{z=e^{-i\theta}}{=}\Re\oint_{|z|=1}\frac{e^zdz}{-iz^{n+1}}.$$If you want to evaluate this (you should find it's $2\pi/n!$), bear in mind the contour is clockwise.