I am trying to evaluate the following integral using the residue theorem:
$$\int_0^{2\pi} e^{ \cos(\theta)} \cos(n\theta) d\theta$$
I have already evaluated $\int_0^{2\pi} e^{e^{-i\theta}} e^{i n\theta} d\theta$ using the residue theorem by making the substitution $z = e^{i\theta}$ and then finding the coefficient of the $z^{-1}$ term in the Laurent series expansion. Now, however, this no longer works, so any help would be appreciated.
Thank you.
$$\int_0^{2\pi} e^{ \cos(\theta)} \cos(n\theta) d\theta=\operatorname{Re}\left(\int_0^{2\pi} e^{ \cos(\theta)+in\theta} d\theta\right)$$ Let $z=e^{i\theta}$. Then, $$\int_0^{2\pi} e^{ \cos(\theta)+in\theta} d\theta=\dfrac{1}{i}\oint_{|z|=1}\dfrac{dz}{z}e^{(z+z^{-1})/2}z^n=\dfrac{1}{i}\sum^\infty_{k=0}\dfrac{1}{2^kk!}\oint_{|z|=1}dz\dfrac{e^{z/2}}{z^{k+1-n}}$$ Use the fact that the integrand is anayltic in the unit circle when $0\leq k\leq n-1$. This will help you evaluate the integral. The last equation occurs since $$\sum^\infty_{k=0}\dfrac{1}{2^kk!}\dfrac{e^{z/2}}{z^{k}}={e^{(z+z^{-1})/2}}=e^{z/2}e^{z^{-1}/2}=e^{z/2}\sum^{\infty}_{k=0}\dfrac{z^{-k}}{2^kk!}$$