I am trying to solve an integration in the form
$$\int_{0}^{2\pi} e^{a \cos{(\theta-b)} + c \cos{(2\theta)}} d\theta$$
where $a$, $b$, and $c$ are constants. I know that if $c=0$, the integration reduces to $2\pi I_0(a)$. However, I want to solve it for $c\ne 0$.
Any suggestions? even in the form of a series expansion?
Thanks
$\def\I{\operatorname I}$ A Jacobi Anger expansion uses Bessel I:
$$\int_{0}^{2\pi} e^{a \cos{(\theta-b)} + c \cos{(2\theta)}} d\theta=\sum_{n\in\Bbb Z}i^n(-1)^n \I_n(-a)\int_0^{2\pi}e^{i n (\theta-b)}e^{c\cos(2\theta)}d\theta$$
Substitute $2t\to t$, notice the odd terms cancel, and convert $e^{ix}\to\cos(x)$ since the sum is real:
$$\sum_{n\in\Bbb Z}i^n(-1)^n \I_n(a)\int_0^{2\pi}e^{i n (\theta-b)}e^{c\cos(2\theta)}d\theta=\frac12\sum_{n\in\Bbb Z}\cos(2bn)\I_{2n}(a)\int_0^{4\pi}\cos(n\theta)e^{c\cos(\theta)}d\theta$$
The integrand is symmetric, so we can integrate from $0$ to $\pi$ to get another Bessel I function. Since $\I_n(-x)=\I_n(x)$, we get a sum over the naturals. Therefore:
$$\boxed{\int_{0}^{2\pi} e^{a \cos{(\theta-b)} + c \cos{(2\theta)}} d\theta=2\pi\I_0(a)\I_0(c)+4\pi\sum_{n=1}^\infty\cos(2bn)\I_{2n}(a)\I_n(c)}$$
which relates to “two variable Bessel functions”. The result is shown here: