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Can one analytically continue the function (Not equal to the Zeta function)

$$Z(s)=\prod_{p}\frac{1}{1+p^{-s}}=\sum_{k=1}^{\infty}\frac{(-1)^{\Omega(k)}}{k^s}$$ Where $\Omega(k)$ is the number of distinct prime factors of $k$.

Specifically, can one find $Z(0)$?

Elie Bergman
  • 3,917
  • You can go from the infinite product to the infinite sum, and then analytically continue it. – PPP May 28 '14 at 21:03
  • Well, I started with the infinite sum, didn't know how to continue it, so I converted it to the product and posted it here. How would you continue the infinite Dirichlet series? – Elie Bergman May 28 '14 at 21:04
  • I asked this a time ago, here: http://math.stackexchange.com/questions/378803/zeta-function-zeros-and-analytic-continuation I think it will be useful – PPP May 28 '14 at 21:06
  • Oh, you must've seen the "+" sign as a "-". This is NOT the Zeta function – Elie Bergman May 28 '14 at 21:09
  • Hmm, yeap, Sorry! – PPP May 28 '14 at 21:13
  • In fact, $\Omega(k)$ is not the number of distinct prime factors of $k$; it is the total number of prime factors. E.g., $\Omega(9)=2$. With this adjustment, the OP's identity is correct, as is the answer below. BTW, $(-1)^{\Omega(k)}$ is called Liouville's function, denoted $\lambda(k)$. – stopple May 29 '14 at 17:39

1 Answers1

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$$Z(s)=\frac{\zeta(2s)}{\zeta(s)}$$

Elie Bergman
  • 3,917