Power series for $\sum_{n=0}^\infty(-1)^n/n!^s$ (around $s=0$)

Question

I'm looking for ways to compute the coefficients of the power series $$ \sum_{n=0}^\infty\frac{(-1)^n}{n!^s}=\sum_{k=0}^{\infty}c_k s^k $$ (a prior version of the question asked whether such an expansion exists at all, at least in the asymptotic sense as $s\to0^+$; thanks to the answer by @Svyatoslav, now I know that it does).

For now, I know that $\color{blue}{c_0=\frac12}$ and $\color{blue}{c_1=\frac14\log\frac2\pi}$. The value of $c_0$ is obtained e.g. in this answer, and I've just used this approach to get the value of $c_1$. Roughly, for $f(s)=\sum_{n=0}^\infty(-1)^n e^{-\lambda_n s}$, we define $$ f_0(t)=\sum_{n=0}^\infty\mathbf{1}_{(\lambda_{2n},\lambda_{2n+1})}(t),\qquad f_{n+1}(t)=\int_0^t f_n(x)\,dx $$ and, under certain conditions, if we have $f(s)\asymp\sum_{k=0}^{(\infty)}c_k s^k$ as $s\to 0^+$, then $$ c_n=\lim_{t\to\infty}\left(\frac{f_{n+1}(t)}{t}-\sum_{k=0}^{n-1}\frac{c_k t^{n-k}}{(n-k+1)!}\right). $$ We replace $t$ by $\lambda_m$ (and $m\to\infty$) and use $f_n(\lambda_m)=(1/n!)\sum_{k=0}^{m-1}(-1)^k(\lambda_m-\lambda_k)^n$.

With $\lambda_n=\log n!$, (even) the computation of $c_1$ this way results in a limit with sums of logarithmic terms, which is not too hard, but a little bit tedious, to evaluate. And $c_2$ is out of my patience for now ;)

Answer 1

Denoting $\,S(s)=\displaystyle \sum_{n=0}^\infty\frac{(-1)^n}{n!^s}=\sum_{n=0}^\infty(-1)^ne^{-s\ln\Gamma(n+1)}$ we can use the Lindelöf summation formula for alternating series (also, for example, here, formula 46). Decomposing the numerator $$S(s)=\int_{-\infty}^\infty\frac{e^{-s\ln\Gamma\big(\frac12+ix\big)}}{2\cosh\pi x}dx=\int_{-\infty}^\infty\frac{dx}{2\cosh\pi x}-\frac s2\int_{-\infty}^\infty\frac{\ln\Gamma\big(\frac12+ix\big)}{\cosh\pi x}dx+O(s^2)$$ $$=\frac12-\frac s4\int_{-\infty}^\infty\frac{\ln\big|\Gamma\big(\frac12+ix\big)\big|^2}{\cosh\pi x}dx+O(s^2)=\frac12-\frac s4\int_{-\infty}^\infty\frac{\ln\frac\pi{\cosh\pi x}}{\cosh\pi x}dx+O(s^2)$$ $$=\frac12-\frac14\ln\Big(\frac\pi2\Big)\,s+O(s^2)$$ I'm not sure whether the next asymptotic term can be found in a closed form: $$c_2=\frac14\int_{-\infty}^\infty\frac{\ln^2\Gamma\big(\frac12+ix\big)}{\cosh\pi x}dx=\frac14\Re\,\int_{-\infty}^\infty\frac{\ln^2\Gamma\big(\frac12+ix\big)}{\cosh\pi x}dx$$