Resummation of the Series

Question

I wonder if there is a way to get resummation of this series? By this way , i am trying to get the integral representation of this series, it could be by Gamma function.

$$\sum _{k=0}^{\infty} \left [ (-1)^{k}(2k)!\left(\frac {1} {(i+a)^{2k+1}}-\frac {1} {(-i+a)^{2k+1}}\right) \right ]$$

thank you.

Answer 1

One can start with the Euler asymptotic expansion $$ f(z)=\int_0^\infty\frac{e^{-t}}{z+t}\,dt=\sum_{k=0}^\infty(-1)^k\frac{{k!}}{z^{k+1}}. $$ Putting $$ g(z)=\frac{f(-iz)-f(iz)}{2i}= \sum _{k=0}^{\infty} (-1)^k\frac{(2k)!}{z^{2k+1}} $$ we have formally $$ g((i+a)^{-1})+g((-i+a)^{-1})= \sum _{k=0}^{\infty} \left [ (-1)^{k}(2k)!\left(\frac {1} {(i+a)^{2k+1}}-\frac {1} {(-i+a)^{2k+1}}\right) \right ]. $$

Answer 2

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: $$\begin{align} &\sum_{k = 0}^{\infty} (-1)^{k} (2 k)! \Bigl((a + i)^{(-2k - 1)} - (a - i)^{(-2k - 1)}\Bigr) = \\ &\qquad -i\operatorname{e} ^{i a - 1} \mathrm{Ei} (-ia + 1) + i \operatorname{e} ^{-ia + 1} \mathrm{Ei} (i a - 1) + i \operatorname{e} ^{i a + 1} \mathrm{Ei} (-ia - 1) - i \operatorname{e} ^{-ia - 1} \mathrm{Ei} (i a + 1) \end{align}$$