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The von Mangoldt function: $$\Lambda(n) = \begin{cases} \log p &; \mbox{if }n=p^k \mbox{ for some prime } p \mbox{ and integer } k \ge 1, \\ 0 &; \mbox{otherwise.} \end{cases}$$ establishes a relation between the Riemann $\zeta$-function and prime powers $p^k$ via: $$\log \zeta(s)=-\sum_{n=2}^\infty \frac{\Lambda(n)}{\log(n)}\,\frac{1}{n^s}$$ this is however for $Re(s)>1$.

Is there a way to expand the above relation to $Re(s) \le 1$?

Or is there instead another formalism that would relate between $\zeta$ to $p^k$ in such clarity in $Re(s) \le 1$?

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You can differentiate both sides and get $\zeta'(s)/\zeta(s)$ = $-\sum_{n=1}^\infty \Lambda (n)/n^s$. Now $\zeta'(s)/(\zeta(s))^2$ is analytic for $Re(s)>1/2$ if you assume Riemann Hypothesis. So you can consider $\zeta'(s)/(\zeta(s))^2$ = $\sum_{n=1}^{\infty}f(n)/n^s$ where $f(n)=-\sum_{d|n}\mu(d)\Lambda(n/d)$.

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  • $\begingroup$ thanks for the step forward, does this hold also for $Re(s)=1/2$ if assuming RH? $\endgroup$ – al-Hwarizmi Jul 4 '13 at 10:08
  • $\begingroup$ Of course it doesn't. Because irrespective of truth of RH, $\zeta(s)$ has zeros on that line, so that function can not be analytic there. An edit: There would be a negative sign in the equation! $\endgroup$ – Kunnysan Jul 4 '13 at 13:32

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