It looks like Andrew beat me to it
There is no guarantee that the following makes sense... But...
There is this Borel summation $ \sum_{j = 0}^{\infty} \frac{(-1)^{j} j!}{x^{j + 1 }} = -\operatorname{e} ^{x} \mathrm{Ei} (-x) \tag{1} $ Here, $\mathrm{Ei}$ is the exponential integral. Put $-x$ for $x$, $ \sum_{j = 0}^{\infty} \frac{j!}{x^{j + 1}} = \operatorname{e} ^{-x} \mathrm{Ei} (x) \tag{2} $ Add (1) and (2) $ \sum_{k = 0}^{\infty} \frac{(2 k)!}{x^{2 k + 1}} = -\operatorname{e} ^{x} \mathrm{Ei} (-x) + \operatorname{e} ^{-x} \mathrm{Ei} (x) $ Put $x=iz$, $ \sum_{k = 0}^{\infty} \frac{(-1)^{k} (2 k)!}{z^{(2 k + 1)}} = -i\operatorname{e} ^{i z} \mathrm{Ei} (-iz) + i \operatorname{e} ^{-iz} \mathrm{Ei} (i z) $ Put $z=a+i$ and $z=a-i$ and subtract: $\begin{align} &\sum_{k = 0}^{\infty} (-1)^{k} (2 k)! \Bigl((a + i)^{(-2k - 1)} - (a - i)^{(-2k - 1)}\Bigr) = \\ &\qquad -i\operatorname{e} ^{i a - 1} \mathrm{Ei} (-ia + 1) + i \operatorname{e} ^{-ia + 1} \mathrm{Ei} (i a - 1) + i \operatorname{e} ^{i a + 1} \mathrm{Ei} (-ia - 1) - i \operatorname{e} ^{-ia - 1} \mathrm{Ei} (i a + 1) \end{align}$