On a certain identity involving the Gamma function

Abstract

The goal of this paper is to prove the identity \begin{align}\sum \limits_{j=0}^{\lfloor s\rfloor}\frac{(-1)^j}{s^j}η_s(j)+\frac{1}{e^{s-1}s^s}\sum \limits_{j=0}^{\lfloor s\rfloor}(-1)^{j+1}α_s(j)+\bigg(\frac{1-((-1)^{s-\lfloor s\rfloor +2})^{1/(s-\lfloor s\rfloor +2)}}{2}\bigg)\nonumber \\ \bigg(\sum \limits_{j=\lfloor s\rfloor +1}^{\infty}\frac{(-1)^j}{s^j}η_s(j)+\frac{1}{e^{s-1}s^s}\sum \limits_{j=\lfloor s\rfloor +1}^{\infty}(-1)^{j+1}α_s(j)\bigg)=\frac{1}{Γ(s+1)},\nonumber \end{align}where \begin{align}η_s(j):=\bigg(e^{γ(s-j)}\prod \limits_{m=1}^{\infty}\bigg(1+\frac{s-j}{m}\bigg)\nonumber \\e^{-(s-j)/m}\bigg)\bigg(2+\log s-\frac{j}{s}+\sum \limits_{m=1}^{\infty}\frac{s}{m(s+m)}-\sum \limits_{m=1}^{\infty}\frac{s-j}{m(s-j+m)}\bigg), \nonumber \end{align}and \begin{align}α_s(j):=\bigg(e^{γ(s-j)}\prod \limits_{m=1}^{\infty}\bigg(1+\frac{s-j}{m}\bigg)e^{-(s-j)/m}\bigg)\bigg(\sum \limits_{m=1}^{\infty}\frac{s}{m(s+m)}-\sum \limits_{m=1}^{\infty}\frac{s-j}{m(s-j+m)}\bigg),\nonumber \end{align}where $\Gamma (s+1)$ is the Gamma function defined by $\Gamma (s):=\underset{0}{\overset{\infty }{\int }}{e}^{-t}{t}^{s-1}dt$ and $\gamma =\underset{n⟶\infty }{\lim }(\sum _{k=1}^n\frac{1}{k}-\log n)=0.577215664\cdots$ is the Euler-Mascheroni constant.

arXiv commenttypos corrected; a missing factor incorporated; one corollary added