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}

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