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I wanted to find $$\lim_{s\to 1} (P(s)-\ln(\zeta(s)))$$ and here is my attempt:

So we know that $$M=\gamma +\sum_{n=2}^\infty \mu(n) \frac {\ln(\zeta(n))}{n}$$ and that $$P(s)=\sum_{n=1}^\infty \mu(n) \frac {\ln(\zeta(sn))}{n}$$ where $M$ is Mertens constant (here), $\gamma$ the Euler Mascheroni constant (here), $\mu(s)$ the Möbius function (here), $\zeta(s)$ the Riemann zeta function (here) and $P(s)$ the prime zeta function (here).

If we let s=1 then $P(1)=\infty=\sum_{n=1}^\infty \mu(n) \frac {\ln(\zeta(n))}{n}$

but if we subtract $\mu(1) \frac {\ln(\zeta(1))}{1}$ (which is also equal to $\infty$) then we have $P(1)-\mu(1) \frac {\ln(\zeta(1))}{1}=P(1)-\ln(\zeta(1))=\sum_{n=2}^\infty \mu(n) \frac {\ln(\zeta(n))}{n}=M-\gamma$

and we can write that as $$\lim_{s\to 1^+} (P(s)-\ln(\zeta(s)))=M-\gamma$$

If you type $P(1)-\ln(\zeta(1))$ into wolfram alpha, it yields $\infty$. But if you give it very small numbers near one it gives nearly perfect results

$P(1.001)-\ln(\zeta(1.001))=-0.31496...$

$M-\gamma=-0.31571...$

I think that some things here are not really legit, but is that correct? Are there better ways to prove this result? And I'm sorry if there are any grammatical or spelling mistakes.

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  • $$\zeta(s) = \prod_p \frac1{1-p^{-s}}, \qquad \Re(s) > 1$$ gives $$\log \zeta(s) =- \sum_p \log(1-p^{-s}) = \sum_{p^k} \frac{p^{-sk}}{k} = \sum_k \frac{P(sk)}{k}$$ so $$P(s) = \sum_k \frac{\mu(k)}{k} \log \zeta(sk)$$ at first for $\Re(s) > 1$ and by analytic continuation for $\Re(s) > 0$ so that $$ \lim_{s \to 1} P(s) - \log \zeta(s) = \sum_{k\ge2} \frac{\mu(k)}{k} \log \zeta(k)$$

  • At first for $\Re(s) > 1$ and by analytic continuation for $\Re(s) > 0$ $$\zeta(s)- \frac1{s-1} = \sum_n (n^{-s} -\int_n^{n+1} x^{-s}dx)$$ gives that $f(s)=(s-1)\zeta(s) $ is analytic at $s=1$ with $f(1)=1$ thus $F(s)=\log (s-1)\zeta(s)$ is analytic at $s=1$ with $F(1)=0$.

  • The $\gamma$ appearing in the Mertens constant $M$ comes from $\gamma = \sum_n (n^{-1} -\int_n^{n+1} x^{-1}dx)$

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Your argument is (nearly) correct . Because, $P(1+e)=\ln\frac{1}{e}+C+O(e)$

Where C is your M - $\gamma$

And for

$\ln\zeta(1+e)= \ln \frac {1}{e} + O(\ln(e))$ For small e

For more details Visit https://link.springer.com/article/10.1007%2FBF01933420

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    $\begingroup$ At $s=1$, $(s-1)\zeta(s)$ is analytic and non-zero and $\log \zeta(s) - P(s)$ is analytic thus $\log \zeta(s) = -\log(s-1)+C+O(s-1)$, $P(s) = \log \zeta(s)+A+O(s-1)=-\log(s-1)+A+C+O(s-1)$. It remains to show $C=0, A =\sum_{n=2}^\infty \mu(n) \frac {\ln(\zeta(n))}{n}$ which is not hard $\endgroup$ – reuns Aug 17 '19 at 4:21
  • $\begingroup$ @reuns but my question is how to show that. So if its not hard, could you show me how its done? $\endgroup$ – Kinheadpump Aug 17 '19 at 10:41
  • $\begingroup$ To show what ? $\lim_{s\to 1} (P(s)-\ln(\zeta(s)))$ is obvious from the formula you wrote for $P(s)$@Kinheadpump $\endgroup$ – reuns Aug 17 '19 at 10:49

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