# Expansion of prime zeta function near singularities

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

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

Therefore,

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

• 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 – reuns Jul 5 at 14:25
• 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? – user76284 Jul 5 at 17:03
• 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. – user76284 Jul 5 at 17:19
• 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. – runway44 Jul 5 at 18:18
• 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 – reuns Jul 5 at 18:26