What is the equation for a Bessel function of order zero? I am currently submitting a paper in a distantly related field (experimental psychology) in which we are using a von Mises distribution to model certain aspects of perceptually-driven behavior.  One of our reviewers has requested that we write out the full equation to the Bessel function of zero order, which is a component of the former.  I'm happy to oblige, but I've been unable to find a clear example of this equation.  Would any of you folks be so kind as to point me in the right direction?
Thanks!
 A: Your question should be more specific. First, don't confuse "Bessel functions" and "Modified Bessel functions": they are different. In one of each of whose two sets of functions, they can be of the "first kind" or of the "second kind" : again different sub-sets of functions. And in each one of these sub-sets, they are different Bessel functions of various order.
For examples: 
The modified Bessel function of the first kind and order $0$ is $I_0(x)$. One integral definition is :
$$I_0(x)=\frac{1}{\pi}\int_0^\pi \exp\left(x\cos(t)\right)dt$$
The modified Bessel function of the second kind and order $0$ is $K_0(x)$. One integral definition is :
$$K_0(x)=\int_0^\infty \cos\left(x \sinh(t) \right)dt$$
Series expressions can be found in :
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheSecondKind.html
and related differential equation :
http://mathworld.wolfram.com/ModifiedBesselDifferentialEquation.html
A: Perhaps is is best to give both informations. The modified Bessel functions $I_n$ of the first kind of order $n$ are solutions of the modified Bessel differential equation:
$$x^2y''(x) + xy'(x)-(x^2+n^2) y(x)=0$$
Their Taylor series are
$$I_n(x) = (\tfrac{1}{2}x)^n \sum\limits_{k=0}^{\infty} \frac{(\tfrac{1}{4}x^2)^k}{k!(n+k)!}$$
Note that in https://en.wikipedia.org/wiki/Von_Mises_distribution#Definition not only $I_0$ appears but all functions $I_n$. 
But this is only the formal part, I have no clue why the $I_n$ appear in the von Mises distribution.
