Contour integral $\int_0^{2\pi}e^{-\cos\theta} \cos⁡(n\theta+\sin\theta)~d\theta$

Question

In the book it was asked to evaluate the integral using the contour integration. For $n\in\mathbb{Z}$,

$$\int_0^{2\pi}e^{-\cos\theta} \cos⁡(n\theta+\sin\theta)~d\theta$$

I tried by putting $e^{i(n\theta+\sin\theta)}$ didn't get the answer correctly. The answer given by the book is $$(-1)^n\frac{2\pi}{n!}$$

Answer 1

$$I=\int_0^{2\pi}e^{-\cos\theta} \cos⁡(n\theta+\sin\theta)~d\theta$$

$$I=\Re\int_0^{2\pi} e^{-\cos\theta-i(n\theta+\sin\theta)}~d\theta=\Re\int_0^{2\pi} e^{-e^{i\theta}-in\theta}~d\theta=\Re\int_0^{2\pi} \frac{e^{-e^{i\theta}}}{\left(e^{i\theta}\right)^n}~d\theta$$

Let $z=e^{i\theta}$

$$I=\Re\int_0^{2\pi} \frac{e^{-z}}{z^n}\frac{1}{iz}~dz=\Re\oint \frac{e^{-z}}{i\cdot z^{n+1}}~dz=\Re\left(2\pi i\cdot Res(z=0)\right)$$

Do Laurent series and take the coefficient for $z^{-1}$ term:

$$\frac{e^{-z}}{i\cdot z^{n+1}}=\frac{1}{i\cdot z^{n+1}}\sum_{k=1} \frac{(-1)^k}{n!}z^k$$

we get

$$Res(z=0)=\frac{(-1)^n}{i\cdot n!}$$

Therefore,

$$\int_0^{2\pi}e^{-\cos\theta} \cos⁡(n\theta+\sin\theta)~d\theta =\frac{2\pi\cdot (-1)^n}{ n!}$$

Answer 2

Alternatively \begin{align} &\int_0^{2\pi}e^{-\cos\theta} \cos⁡(n\theta+\sin\theta)~d\theta\\ =& \int_0^{2\pi}e^{in \theta}\cdot e^{-e^{-i\theta}} d\theta = \int_0^{2\pi}e^{in \theta}\sum_{k=0}^\infty\frac{(-e^{-i\theta})^k}{k!}\ d\theta \\ =& \ \sum_{k=0}^\infty\frac{(-1)^k}{k!} \int_0^{2\pi}e^{i(n-k)\theta}\ d\theta =\frac{2\pi(-1)^n}{ n!} \end{align} where only the $k=n$ term survives the integration.