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2 votes
0 answers
451 views

Analytic continuation of "composite" zeta function

Let us define the Dirichlet series $$\mathcal C(s):=\sum_{n\text{ composite}}\frac{1}{n^s},\quad P(s):=\sum_{p\text{ prime}}\frac{1}{p^s}.$$ They are absolutely convergent in the half-plane $\sigma>...
user219023's user avatar
2 votes
1 answer
561 views

On infinite sum containing logarithmic derivative of Zeta function and Möbius function:

Consider the following function: $$F(s)= \sum_m \mu(m) \sum_n \frac{e^{-n/2}\zeta^\prime (mns)}{n \zeta(mns)}$$ Now, we can see, that function has simple poles ${\left[\frac{1}{n}\right]}_{n=1}^\...
Zaza's user avatar
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1 vote
1 answer
170 views

Are there variations of Ramaswami's formula for the analytic continuation of the Riemann zeta function?

On p. 286 of Borwein's paper entitled "Computational Strategies for the Riemann zeta function", the author mentions a formula due to Ramaswami: $$(1-2^{1-s})\zeta(s) = \sum_{n=1}^{\infty} \...
Max Lonysa Muller's user avatar
0 votes
1 answer
169 views

Analytic extension of the Hurwitz ζ function

For the purpose of formalisation in a theorem prover, I am looking for a simple definition of the analytic extension of the Hurwitz ζ function $\zeta(s,q)$ valid for all $s\in\mathbb{C}\setminus\{1\}$ ...
Manuel Eberl's user avatar
  • 1,241
0 votes
1 answer
1k views

Analytical continuation of the reciprocal of the Zeta function [closed]

Is the reciprocal of the Zeta function analytically continuable? As $1/\zeta(n) = \sum_{n=1}^\infty \mu(n)/{n^s}$, this does not look obvious.
pat porto's user avatar