TITLE: A solution for $\int^{2\pi}_0e^{\cos \theta}\cos(a\theta -\sin \theta)\,d \theta $
QUESTION [10 upvotes]: It can be proved using complex analysis that 
$$\tag{1}\int^{2\pi}_0e^{\cos \theta}\cos(n\theta -\sin \theta)=\frac{2\pi}{n!}$$
My initial thought that,  we use the Gamma function for non-integer values. But it seems that we cannot since 
$$\int^{2\pi}_0e^{\cos \theta}\cos\left(\left(\frac{1}{2} +n\right)\theta -\sin \theta \right)=0$$
My Question : Can we solve 
$$\int^{2\pi}_0e^{\cos \theta}\cos(a\theta -\sin \theta)\,d \theta $$
for specific values of $a$ that are not of the above cases ?
EDIT
As requested in the comments here is a proof of (1) using contour integeration
Consider the following function
$$f(z)=e^{z^{-1}}z^{n-1}$$
Now we integrate the function along a circle of radius 1
$$\oint_{|z|=1}e^{z^{-1}}z^{n-1} dz=\oint_{|z|=1}\sum_{k\geq 0}\frac{z^{n-k-1}}{\, k!} dz  $$
Now we need to find the residue which is the coefficient of $1/z$
$$\text{Assume that } n-k-1=-1 \to n=k$$
Hence we have 
$$\oint_{|z|=1}e^{z^{-1}}z^{n-1} dz= 2\pi i \frac{1}{n!}  $$
Using a parametirzation of the circle 
$$i\int_{0}^{2\pi i}e^{e^{-i\theta }}e^{i n\theta } d\theta = 2\pi i \frac{1}{n!}  $$
$$\int_{0}^{2\pi i}e^{e^{-i\theta }}e^{i n\theta } d\theta = 2\pi\frac{1}{n!}  $$
Hence we have 
$$\Re\int_{0}^{2\pi i}e^{e^{-i\theta }}e^{i n\theta } d\theta =\int^{2\pi i}_0 e^{\cos \theta} \cos(n\theta-\sin \theta)\, d\theta= \frac{2\pi}{n!}  $$

REPLY [7 votes]: You can obtain the solution for $a\in\mathbb{R}$ in terms of the incomplete Gamma function:
$$
\begin{aligned}
\int _{0}^{2\,\pi }\!{{\rm e}^{\cos \left( \theta \right) }}\cos
 \left( n\theta-\sin \left( \theta \right)  \right) {d\theta}&={\it Re}
 \left( \int _{0}^{2\,\pi }\!{{\rm e}^{{{\rm e}^{-i\theta}}}}{{\rm e}^
{in\theta}}{d\theta} \right) \\
&={\it Re}
 \left( \sum _{k=0}^{\infty } \left( {\frac {\int _{0}^{2\,\pi }\!{
{\rm e}^{-i\theta\,k}}{{\rm e}^{ia\theta}}{d\theta}}{k!}} \right) 
 \right) \\
&=-\sum _{k=0}^{\infty }\,{\frac {\sin \left( 2\,\pi \,a \right) }{
 \left( k-a \right) k!}}
 \\
&= \left( -1 \right) ^{a}\sin \left( 2\,\pi \,a \right) 
 \left[\Gamma  \left( -a,-1 \right)- \Gamma  \left( -a\right) \right]
\end{aligned}
$$
where $\Gamma(-a, -1)$ is the incomplete Gamma function