Simplifying modified Bessel function of the first kind The modified Bessel function of the first kind shows up in the normalizing constant of a lot of random variables (e.g. the normal product distribution, the noncentral chi-square distribution, the Skellam distribution, and the von Mises distribution). This is its formula (NB: this is false...see answer below):
$$
J_k(x) = \sum_{m=0}^{\infty} \frac{(-1)^m}{m!\Gamma(m+k+1)}\left(\frac{x}{2}\right)^{2m+k}
$$
How can we simplify this thing? Because I am interested in Skellam distributions in particular, I am interested in the case where $k$ is a possibly-negative integer, and $x$ is positive. 
This thing appears to depend on evaluations of the Gamma function at negative values, which can take on the value complex infinity. However, the wiki page for the Skellam distribution mentions that the Bessel function with a negative integer $k$ will be equal to the Bessel function with the absolute value of that integer as a subscript. Why is this?
 A: Actually, that formula above is not for the modified Bessel function of the first kind, it is for the (non-modified) Bessel function of the first kind! Furthermore, the Encyclopedia of Mathematics has an entry on this (non-modified) Bessel functions, and points out that, for $n \in \mathbb{N}$ and $z \in \mathbb{C}$
$$
J_{-n}(z) = (-1)^nJ_n(z) \tag{1}
$$
even though that the formula typically given is not technically correct:

...the formulas in (1) are not "strictly-speaking" well-defined
  because the $\Gamma$ function has poles in all negative integers.
  However, given the Laurent series of $\Gamma$ at such poles and the
  fact that the function appears in the denominators of fractions, the
  formula in (1) can be reinterpreted...

The correct formula is
$$
I_n(x) = i^{-n} J_{n}(ix) \tag{2}
$$
for $n \in \mathbb{N}$ and $x \in \mathbb{R}$, and it doesn't have that alternating sign situation. 
Using (1) and (2), and assuming $n \ge 0$ for the time being without loss of generality,
\begin{align*}
I_{-n}(x) &= i^n J_{-n}(ix) \\
&= (-i)^n J_n(ix) \\
&= (i)^{-n} J_n(ix) \\
&= I_n(x).
\end{align*}
This first clicked for me when I read an answer on another stackexchange cite, which can be found by clicking on this.
