Integrating $e^{a\cos(\phi_1-\phi_2)+b\cos(\phi_1-\phi_3)+c\cos(\phi_2-\phi_3)}$ over $[0,2\pi]^3$ I am trying to integrate the following function. (it arises in channel modeling in wireless communications, Rayleigh random variables)..Any help is appreciated.Thanks
$$\int_0^{2\pi}\int_0^{2\pi}\int_0^{2\pi}e^{a\cos(\phi_1-\phi_2)+b\cos(\phi_1-\phi_3)+c\cos(\phi_2-\phi_3)}d\phi_1d\phi_2d\phi_3$$
 A: (1) Note that your integrand is periodic in all the integration variables. We will use this fact later. 
Let us introduce the new variables $x_1= \phi_1- \phi_2$, $x_2 = \phi_2 -\phi_3$ and $x_3 = \phi_3-\phi_1$. We immediately notice that $x_3 = - x_1 - x_2$. The goal is thus, to perform a change of variables from $(\phi_1, \phi_2 , \phi_3)$ to $(X, x_1, x_2)$ with $X=(x_1 + x_2 + x_3)/3$. The Jacobian for this change is given by $|\det J|=  1$.
It would be most convenient, if the new integration region would be again $[0,2\pi]^3$. I will argue that this is correct due to the fact (1).
Here, I have shown the analog situation for a change of variable $(\phi_1,\phi_2)\mapsto(X,x_1)$ with $X=(\phi_1+\phi_2)/2$ and $x_1 =\phi_1-\phi_2$ involving only two of the three variables.

The original integration region $(\phi_1,\phi_2)\in[0,2\pi]^2$ is shown in blue. The new region $(X,x_1) \in [0,2\pi]^2$ is in red. It can be immediately seen that due to the $2\pi$ periodicity of the integrand, integration over these two regions yields the same result.
A simular result can be obtained for all three variables (the corresponding plot would be in a dimension higher).
So we have that
$$I=\int_{[0,2\pi]^3} \!d\phi_1\,d\phi_2\,d\phi_3 e^{a \cos(\phi_1-\phi_2) + b \cos (\phi_2-\phi_3) + c\cos(\phi_3-\phi_1)}= \int_{[0,2\pi]^3}\!dX\,dx_1\,dx_2 e^{a \cos(x_1) + b \cos(x_2) + c\cos(x_1+x_2)}.$$
The integral over $X$ is trivial (yielding $2\pi$).
Next, we perform the integral over $x_1$. We need the result that
$$ \int_0^{2\pi}\!dx_1 e^{a \cos(x_1) + c \cos(x_1+x_2)} = 2 \pi I_0\left(\sqrt{a^2+c^2+2 a c \cos(x_2)}\right)$$ 
with $I_0$ the modified Bessel function of the first kind. The result can be proven using $\cos(x_1+x_2) = \cos(x_1) \cos(x_2) -\sin(x_1) \sin(x_2)$ and the Jacobi-Anger expansion.
So, we have that
$$I = (2\pi)^2 \int_0^{2\pi} \! dx_2\,e^{b \cos(x_2)} I_0\left(\sqrt{a^2+c^2+2 a c \cos(x_2)}\right).$$
This last integral, I don't know if it can be evaluated explicitly. I would guess that this could be another question ...
Edit:
Using Graf's addition theorem for Bessel function, I managed to bring the integral onto the alternative form
$$ I = (2\pi)^3 \sum_{k\in\mathbb{Z}} I_k(a) I_k(b) I_k(c).$$
I am not sure if this is simpler though.
