# A 3-dimensional calculus

1. Feb 25, 2009

### leverrier

I'm a researcher of telecommunications. I was stumped by calculation of an integral when doing some proof. Would anyone have any idea how to give a close form of the following 3-dimensional integral? Your help will be highly appreciated! Here, A, B, C are all positive constants.

$$\int_0^{2\pi}\int_0^{2\pi}\int_0^{2\pi}\exp\left(A\cos(y-x)\right)\exp\left(B\cos(z-y)\right)\exp\left(C\cos(z-x)\right)dxdydz.$$

BTW, I've tried the 2-dimensional case before which is converted to the Modified Bessel function as follow,

$$\int_0^{2\pi}\int_0^{2\pi}\exp\left(A\cos(x-y)\right)dxdy=(2\pi)^2I_0\{A\}$$

where

$$I_0\{A\}=\frac{1}{2\pi}\int_0^{2\pi}\exp(A\cos\theta)d\theta.$$

Last edited: Feb 25, 2009