calculate: $\int_{0}^{2\pi}e^{\cos\theta}(\cos(n\theta-\sin\theta))d\theta$

Question

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.

Answer 1

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.

Answer 2

$$I_n(a)=\int_{0}^{2\pi}e^{a\cos\theta}\cos(n\theta-a\sin\theta)d\theta $$ Note that \begin{align} &I_{0}’(a)=\frac1a\int_{0}^{2\pi}d(e^{a\cos\theta}\sin(a\sin\theta))=0\implies I_0(a)= I_0(0)=2\pi \end{align}

Also note $I_n’(a)=I_{n-1}(a)$, $I_{n>0}(0)=0$, and integrate $I_n(a)=\int_0^a I_{n-1}(s)ds $ successively to get

$$ I_1(a) = 2\pi a,\>\>\> I_2(a)=\frac{2\pi a^2}{2!},\>\>\>...\>\>\> I_n(a)=\frac{2\pi a^n}{n!} $$ Thus

$$\int_{0}^{2\pi}e^{\cos\theta}\cos(n\theta-\sin\theta)d\theta =I_n(1)= \frac{2\pi}{n!}$$