Is there a closed form expression for $ \int_0^{2\pi}{\exp(A \cos(x) + B \sin(x)) \cos(n(y-x))dx}$? The $n$ in the integral is an integer, ranging from $0$ to $\infty$
$$ \int_0^{2\pi}{\exp(A \cos(x) + B \sin(x))(\cos(n(y-x))dx}$$
I am wondering if there is a closed form solution for this type of integral?
I am interested to know if there for some n's for which this integral goes to zero.
 A: This integral can be written in terms of special functions. Consider the following extension of the integral above
$$R_n(z,z^*)=e^{iny}\int_{0}^{2\pi}dx ~\exp(z^*e^{ix}+ze^{-ix}-inx)$$
Then it is clear that the integral we want is given by $$Q=\text{Re}\left(R_n\left(\frac{A+iB}{2}, \frac{A-iB}{2}\right)\right)$$
which is manifestly real. Now notice that
$$R_n(z,z^*)=e^{iny}\frac{\partial^n}{\partial z^n}\int_0^{2\pi}\exp(ze^{-ix}+z^*e^{-ix})dx=\\=e^{iny}\frac{\partial^n}{\partial z^n}\int_0^{2\pi}\exp(A\cos x+B\sin x)dx\\=e^{iny}\frac{\partial^n}{\partial z^n}\int_0^{2\pi}\exp(\sqrt{A^2+B^2}\cos(x-x_0))dx\\=2\pi e^{iny}\frac{\partial^n}{\partial z^n}I_0(4zz^*)\\=2\pi~ 2^n ~e^{in(y-\arg z)}(A^2+B^2)^{n/2}I_0^{(n)}(\sqrt{A^2+B^2})$$
and thus
$$Q=2\pi ~2^n(A^2+B^2)^{n/2} ~\cos(y-\arg(A+iB))I_0^{(n)}(\sqrt{A^2+B^2})$$
where $I_0^{(n)}(x)$ stands for the $n$-th derivative of the modified Bessel function $I_0(x)$. This integral can be identically zero only if
$$y-\arg(A+iB)=\frac{p\pi}{2}~,~ p\in \mathbb{Z}$$
One may be interested in a simpler formula of high derivatives of $I_0$. Indeed, one can show that it can be written in terms of higher-order modified Bessel functions as follows:
$$I_0^{(n)}(x)=\begin{Bmatrix}\frac{1}{2^{n-1}}\sum_{k=0}^{\lfloor n/2\rfloor}{n\choose k}I_{n-2k}(x)~~~ &\text{n odd}\\ 
-\frac{1}{2^{n}}{n\choose n/2}I_0(x)+\frac{1}{2^{n-1}}\sum_{k=0}^{\lfloor n/2\rfloor}{n\choose k}I_{n-2k}(x)~~~ &\text{n even}\end{Bmatrix}$$
