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Feb 20, 2017 at 6:38 comment added Lucian @OlivierOloa: For instance, $~\displaystyle\prod_{n\in\mathbf M}\bigg(1-\frac1{n^4}\bigg)~$ yields $~\dfrac1{\zeta(4)}~$ for $~\mathbf M=\mathbb P,~$ and $~\dfrac1{4\pi\text{ csch }\pi}~$ for $~\mathbf M=\mathbb N_{>1}.$
Feb 19, 2017 at 21:51 comment added Olivier Oloa @Lucian Even if I've some difficulty to grasp your definition, I would say keep on working on your intuition.
Feb 18, 2017 at 16:26 comment added Lucian @OlivierOloa: I've noticed your fondness for various generalizations of $\zeta$ functions, based on infinite series. In your opinion, would a generalization of the same functions, but based on infinite products rather than infinite series, hold any merit ?
Aug 1, 2015 at 0:13 answer added Olivier Oloa timeline score: 4
Feb 24, 2015 at 10:49 vote accept Olivier Oloa
Feb 24, 2015 at 4:12 answer added Antonio Vargas timeline score: 8
Feb 23, 2015 at 9:20 comment added Olivier Oloa @AntonioVargas Yes, I have the same conjecture, I've defined some Laurent-Stieltjes type constants by $$\zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\gamma_k(a,b) s^k \tag1$$ near $0$. Then we may extend $ζ(⋅∣a,b)$ to a meromorphic function on $\mathbb{C}$, as is the classic Hurwitz zeta function. The point is to prove $(1)$. Maybe the Euler–Maclaurin formula could be an interesting tool to prove $(1)$.
Feb 22, 2015 at 23:35 comment added Antonio Vargas Such a series exists, at least. There are coefficients $c_k(a,b)$ such that, for some $\delta > 0$, $$\zeta(s | a,b) = \frac{1}{s} + \sum_{k=0}^{\infty} c_k(a,b) s^k$$ for all $0 < s < \delta$. The series here can be used to analytically continue $\zeta(s|a,b)$ to the annulus $0 < |s| < \delta$.
Feb 22, 2015 at 20:47 comment added Antonio Vargas Are you looking for a closed form for the coefficients?
Feb 22, 2015 at 15:54 history edited Olivier Oloa CC BY-SA 3.0
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Feb 22, 2015 at 2:43 history edited Olivier Oloa CC BY-SA 3.0
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Feb 22, 2015 at 2:29 history asked Olivier Oloa CC BY-SA 3.0