Closed form of series $\sum_{n=\frac {1}{2}}^{\infty}\Gamma (1-s,xn)e^{xn}n^{s-1}$ I'm trying to get some closed form solution of $\sum_{n=\frac {1}{2}}^{\infty}\Gamma (1-s,xn)e^{xn}n^{s-1}$. To be precise gotten closed form should be analytic continuation for any given $x$. There are no specified function of closed form. For example for $x=0$ it's just $\Gamma (1-s)\zeta (1-s,\frac {1}{2}) $ and this is absolutely satyfying result, because if given series is divergent I can just use analytic continuation of Hurwitz zeta function.
PS:What I mean by sum from  $n=\frac {1}{2} $ is for example anagously  $\sum_{n=0}^a n=\frac {a (a+1)}{2} $. So for $a=\frac {1}{2}$ we get $\frac {3}{8} $
 A: Here are some first thoughts.
We have that
$$\Gamma(1-s,xn) = \int_{xn}^\infty t^{-s} e^{-t}dt$$
It is much more convenient to turn this into
$$\Gamma(1-s,xn) = \int_{0}^\infty t^{-s} e^{-t}dt-\int_{0}^{xn} t^{-s} e^{-t}dt$$
Then we change variables so that we have for the first part
$$\int_{0}^{xn} t^{-s} e^{-t}dt = nx \int_0^1 (nxt)^{-s} e^{-nxt} dt$$
Now, to deal with the original series, we have
$$\sum_{n=1}^\infty \Gamma(1-s,xn) e^{xn} n^{s-1} = \int_0^\infty t^{-s} e^{-t} \sum_{n=1}^\infty e^{xn} n^{s-1} dt = \Gamma(1-s)Li_{-s+1}(e^x)$$
For the second piece
$$\int_0^1 (tx)^{-s} \sum_{n=1}^\infty n^{-s} e^{-nxt}e^{xn} n^{s-1} dt = \int_0^1 (tx)^{-s} \sum_{n=1}^\infty \frac{1}{n} (e^{x(1-t)})^n = x^{-s}\int_0^1 t^{-s} Li_1(e^{x(1-t)}) dt$$
So the final result is that the series equals
$$\Gamma(1-s)Li_{-s+1}(e^x)- x^{-s}\int_0^1 t^{-s} Li_1(e^{x(1-t)}) dt$$
Probably, there is extra analysis needed to understand exactly what the effect of replacing the divergent series in the inner summation with $Li$ does to the function. I suspect there is a branch cut somewhere, probably at negative values of $s$.
Is there a specific context you have in mind for working with this series? I could tailor my analysis a bit if you have something particular you're wanting to do this series.
