# Integration of exponential of a function of cosines [closed]

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

• Any reason in particular you're trying to solve this integral? Commented Aug 28, 2023 at 22:02
• It pops out from some fluid mechanics equations applied to wakes Commented Aug 29, 2023 at 10:22

$$\def\I{\operatorname I}$$ 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:

• This is great! thanks a lot Commented Aug 29, 2023 at 10:26