The prime zeta function has the following expansion near its singularity at 1:

$$P(1+\varepsilon) = -\ln \varepsilon + C + O(\varepsilon)$$

It also has a singularity at the reciprocal of every square-free positive integer. What is its expansion near each of these singularities? Is it

$$P(1/n+\varepsilon) = -\frac{\mu(n)}{n} \ln \varepsilon + C + O(\varepsilon) $$

where $\mu$ is the Möbius function?


There are two main ingredients here. The first is $\zeta$ near its singularity $s=1$:

$$ \zeta(1+\varepsilon)=\frac{1}{\varepsilon}+\gamma+\mathcal{O}(\varepsilon) $$

(where $\gamma$ is the Euler-Mascheroni constant). The other is the Mobius inversion formula:

$$ P(s)=\sum_{k=1}^\infty \frac{\mu(k)}{k}\ln \zeta(ks). $$

This is mentioned on the MathWorld page you linked for the prime zeta function $P(s)$. Notice the only nonzero terms occur when $k$ is squarefree. Setting $s=1/n+\epsilon$ yields

$$ P(1/n+\epsilon)=\frac{\mu(n)}{n}\ln\zeta(1+n\varepsilon)+\sum_{\substack{k=1\\ k\ne n}}^{\infty} \frac{\mu(k)}{k}\ln\zeta(k(1/n+\varepsilon)). $$

Letting $\varepsilon\to0$ in this case, and using the $\mathcal{O}$ expansion of $\zeta$ and $\ln$ yields

$$ \begin{array}{ll} \ln\zeta(1+n\varepsilon) & =\ln\left(\frac{1}{n\varepsilon}+\gamma+\mathcal{O}(\varepsilon)\right) \\ & = -\ln(n\epsilon)+\ln(1+n\gamma\epsilon+\mathcal{O}(\varepsilon^2)) \\ & = -\ln(n\varepsilon)+\mathcal{O}(\varepsilon). \end{array} $$


$$ P(1/n+\varepsilon)=-\frac{\mu(n)}{n}\ln(\varepsilon)+\left[-\frac{\mu(n)}{n}\ln(n)+\sum_{\substack{k=1\\k\ne n}}^\infty \frac{\mu(k)}{k}\ln\zeta(k/n)\right]+\mathcal{O}(\varepsilon). $$

  • 1
    $\begingroup$ Your methods works the same way for the non-trivial zeros of $\zeta(s)$ : around $s=\rho/n$ or $1/n$, assuming the RH, $P(s) - \frac{\mu(n)}{n} \log\zeta(ns)$ is analytic thus $P(s) - \frac{\mu(n)}{n} \log (s-\rho/n),P(s) + \frac{\mu(n)}{n} \log (s-1/n)$ is analytic. Due to the infinitely many non-trivial zeros, those $\log$ singularities accumulate to the right of $\Re(s)=0$ which is $P(s)$'s natural boundary $\endgroup$ – reuns Jul 5 at 14:25
  • $\begingroup$ Thank you. If one subtracts these logarithmic singularities from the prime zeta function as in imgur.com/a/KqHbxWM, it seems like it should be possible to extend the real part of the result to $s < 0$ in a smooth, natural way. Is this possible? $\endgroup$ – user76284 Jul 5 at 17:03
  • $\begingroup$ More generally I wonder whether it is possible to extend a complex function past a natural boundary by subtracting the singularities on that boundary, performing analytic (?) continuation, and re-adding the singularities. $\endgroup$ – user76284 Jul 5 at 17:19
  • $\begingroup$ Can you really subtract a logarithmic function for every singularity though? The singularities may be a discrete set on Re(s)>0, but I suspect every pure imaginary is a limit point of the singularities (including the ones from the nontrivial zeros of $\zeta$ which @reuns mentions), which I feel will be a serious obstruction. $\endgroup$ – runway44 Jul 5 at 18:18
  • $\begingroup$ That's why I said $\Re(s)=0$ is a natural boundary : $P(s)$ isn't analytic anywhere there $\implies$ no possible analytic continuation beyond $\endgroup$ – reuns Jul 5 at 18:26

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