# Series representation of Bell numbers in terms of Bessel functions

The generating function of the Bell numbers $B_{n}$ is
$B(x) = \sum_{n\,=\,0}^{\infty} \frac{B_{n}x^{n}}{n!} = e^{e^{x}-1}$

Set $x=it$:
$B(it) = \sum_{n\,=\,0}^{\infty} \frac{B_{n}(it)^{n}}{n!} = e^{e^{(it)}-1}$

Factor out $e^{-1}$ and split exponential:

$e^{e^{(it)}-1} = e^{-1}e^{e^{it}} = e^{-1}e^{\cos t}e^{i \sin t}$

Replace with series from Jacobi-Anger expansions for each:

$e^{-1}e^{\cos t}e^{i \sin t} = e^{-1} \left[\sum_{m\in\mathbb{Z}} i^{m}J_{m}(-i)e^{imt} \right] \left[\sum_{n\in\mathbb{Z}} J_{n}(1)e^{int} \right]$

Collect and reorganize:

$e^{-1} \left[\sum_{m\in\mathbb{Z}} i^{m}J_{m}(-i)e^{imt} \right] \left[\sum_{n\in\mathbb{Z}} J_{n}(1)e^{int} \right] = e^{-1} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)e^{imt}J_{n}(1)e^{int}$

Let’s multiply those exponentials:

$e^{-1} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)e^{imt}J_{n}(1)e^{int} = e^{-1} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)J_{n}(1)e^{it(m+n)}$

Let’s expand that exponential as a power series:

$e^{-1} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)J_{n}(1)e^{it(m+n)} = e^{-1} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)J_{n}(1) \sum_{v\,=\,0}^{\infty} \frac{(it(m+n))^{v}}{v!}$

Shuffling things around a bit:
$e^{-1} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)J_{n}(1) \sum_{v\,=\,0}^{\infty} \frac{(it(m+n))^{v}}{v!} = e^{-1}\sum_{v\,=\,0}^{\infty} \frac{(it)^{v}}{v!} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)J_{n}(1) (m+n)^{v}$

Therefore:
$B_{a} = \frac{1}{e} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)J_{n}(1) (m+n)^{a}$