Zeros of Prime Zeta Function

Question

Let $P(s)=\sum_{p}\frac{1}{p^s}$ be the prime zeta function. It is clear that $P(s)$ converges absolutely for $\sigma>1$, where $s=\sigma+it$, and can be analytically continued to the strip $0<\sigma\leq 1$ (Fröberg 1968)https://link.springer.com/article/10.1007%2FBF01933420.

Now, I have two questions about the zeros of $P(s)$:

(1) Are there any zeros of $P(s)$ on the line $\Re(s)=1$?

(2) Is $s=1$ the pole or the essential sigularity of $P(s)$?

If $s=1$ is the pole of $P(s)$, what is the Laurent expansion of $P(s)$ at $s=1$?

In fact, if $P(s)$ has no zeros on the line $\Re s=1$ and $$P(s)=\frac{\alpha}{s-1}+\alpha_0+(s-1)h(s)$$ for some complex number $\alpha,\alpha_0$ and a holomorphic function $h(s)$, then we can prove by the theorem 3.4.2 of Jameson's book: The Prime Number Theorem, that $$\lim_{x\to \infty}\frac{\pi(x)}{x}=\alpha$$ which is absurd. So where it goes wrong? Here, $\pi(x)$ is the prime counting function.

Answer 1

Euler's product for the Riemann zeta function implies that $$\log \, \zeta(s) = \sum_{k=1}^{\infty} \frac{1}{k} P(ks).$$ The series $\sum_{k=2}^{\infty} \frac{1}{k} P(ks)$ remains bounded at $s = 1$, so the singularity of $P(s)$ is of the same kind as the singularity of $\log\, \zeta(s)$ at $s=1$.

This is neither a pole nor an essential singularity; it is a branch point. (The analytic continuation $$P(s) = \sum_{k=1}^{\infty} \frac{\mu(k)}{k} \log \, \zeta(ks),$$ which you get from the first expression by Möbius inversion, depends on branches of logarithms and has to be understood carefully). In particular the expression $$P(s) = \frac{\alpha}{s-1} + ...$$ is invalid.

I don't have an answer for question (1). (I do not think this is known.)

Answer 2

You are asking here about zeros of $P(s)$ at $\Re(s)=1$, but showing that $P(s)$ has no zeros for $\Re(s)>1$ is already not trivial, since the prime indicator function has no Dirichlet inverse so constructing $\frac{1}{P(s)}$ as a dirichlet series which is absolutely convergent for $\Re(s)>1$ is not possible. One "trivial" zero free zone from this Math.SE question is by noting that $|P(s)|$ will be at least $\frac{1}{2^{\sigma}}-\sum_{p\geq3 }\frac{1}{p^{\sigma}}$ where $s=\sigma+it$.

This leads us to conclude (from numerical computations) that $P(s)$ has no zeros for $\Re(s)>1.76$. If we look at the relation function $P^*(s)=\sum_{p^k}\frac{1}{p^{ks}}$ which is the "prime power zeta function", the situation is even worse. Using the same bounding method as before, we get that $|P^*(s)|>\frac{1}{2^{\sigma}}-\sum_{p^k\geq3}\frac{1}{p^{k\sigma}}$ which gives the much worse zero free region of $\Re(s)>2.21$. In this answer I will outline a procedure to reduce the zero free region of $P^*(s)$ to $\Re(s)>1.11$, along with proving that for $\Re(s)>\frac{1}{2}$ the function $P^*(s)$ will only have zeros if $\zeta(2s)=\zeta(s)$.

To begin the proof, I will examine the divisor sum $\sum_{d|n}\frac{\lambda(d)}{\Omega(d)+1}$ where $\Omega(n)$ counts the number of prime factors of $n$ (with multiplicity) and $\lambda(n)=(-1)^{\Omega(n)}$. Since $\frac{\lambda(d)}{\Omega(d)+1}$ only depends on the number of prime factors of $d$ and there are ${\Omega(n)\choose k}$ divisors of $d$ with exactly $k$ prime factors, we can split our sum over all these classes to get that

\begin{align*} \sum_{d|n}\frac{\lambda(d)}{\Omega(d)+1}&=\sum_{k=0}^{\Omega(n)}{\Omega(n)\choose k}\frac{(-1)^k}{k+1}\\ &=\int_{0}^{1}\sum_{k=0}^{\Omega(n)}{\Omega(n)\choose k}(-1)^kx^kdx\\ &=\int_{0}^{1}(1-x)^{\Omega(n)}dx\\ &=\frac{1}{\Omega(n)+1} \end{align*}

Thus, we can apply this formula to get that

\begin{align*} \left(\sum_{n=1}^{\infty}\frac{\frac{\lambda(n)}{\Omega(n)+1}}{n^s}\right)\left(\sum_{n=1}^{\infty}\frac{\Omega(n)}{n^s}\right)&=\sum_{n=1}^{\infty}\frac{\sum_{d|n}\frac{\lambda(d)\Omega(n/d)}{\Omega(d)+1}}{n^s}\\ &=\sum_{n=1}^{\infty}\frac{1-\sum_{d|n}\lambda(d)}{n^s} \end{align*}

Since $\sum_{d|n}\lambda(d)$ is the indicator function for squares, this means that

\begin{align*} \left(\sum_{n=1}^{\infty}\frac{\frac{\lambda(n)}{\Omega(n)+1}}{n^s}\right)\left(\sum_{n=1}^{\infty}\frac{\Omega(n)}{n^s}\right)&=\zeta(s)-\zeta(2s)\tag{1} \end{align*}

We now work on simplifying our first sum, namely we see that

\begin{align*} \sum_{n=1}^{\infty}\frac{\frac{\lambda(n)}{\Omega(n)+1}}{n^s}&=\int_{0}^{1}\sum_{n=1}^{\infty}\frac{\lambda(n)z^{\Omega(n)}}{n^s}dz\\ &=\int_{0}^{1}\prod_{p}\frac{1}{1+zp^{-s}}dz\\ &=\int_{0}^{1}\zeta^{-z}(s)\zeta^{z}(2s)\prod_{p}\frac{\left(1+p^{-s}\right)^z}{1+zp^{-s}}dz \end{align*}

We now let

$$G(s;1-z):=\zeta^{z}(2s)\prod_{p}\frac{\left(1+p^{-s}\right)^z}{1+zp^{-s}}$$

so that

\begin{align*} \sum_{n=1}^{\infty}\frac{\frac{\lambda(n)}{\Omega(n)+1}}{n^s}&=\int_{0}^{1}G(s;1-z)\zeta^{-z}(s)ds\tag{2} \end{align*}

Substituting (2) into (1) and noting that $\sum_{n=1}^{\infty}\frac{\Omega(n)}{n^s}=\zeta(s)P^*(s)$ gets us that

\begin{equation} \frac{1}{P^*(s)}=\frac{1}{\zeta(s)-\zeta(2s)}\int_{0}^{1}G(s;z)\zeta^z(s)ds \end{equation}

The definition of $G(s;z)$ is absolutely convergent for $\Re(s)>\frac{1}{2}$, and thus since $\zeta^z(s)$ is defined in any zero free region of the zeta function we get (under RH) that $P^*(s)$ can only be zero of $\zeta(s)=\zeta(2s)$ for $\Re(s)> \frac{1}{2}$. This result holds without RH or anything special for $\Re(s)>1$ which is what we will use in the rest of the proof.

What we will simply do is note that

\begin{align*} P(s)+P(2s)&=\sum_{k=1}^{\infty}\frac{\mu(k)}{k}\ln(\zeta(ks))+\sum_{k=1}^{\infty}\frac{\mu(k)}{k}\ln(\zeta(2ks))\\ &=\frac{3}{2}\ln(\zeta(2s))+\sum_{k=3}^{\infty}\frac{\mu(k)}{k}\ln(\zeta(2s))+\sum_{k=2}^{\infty}\frac{\mu(k)}{k}\ln(\zeta(2ks)) \end{align*}

We can now lower bound $\ln(\zeta(2s))=-\sum_{p}\ln\left(1-\frac{1}{p^{2s}}\right)$ by the same methods as in the first paragraph, and we will get a greater than zero result for values of $s$ even with $\Re(s)<1$ because we are working with the doubled $\zeta(2s)$, due to $\zeta(s)=\zeta(s)$ holding. Including more terms and numerically computing the prime zeta function, we arrive at the desired bound of $P^*(s)$ having no zeros for $\Re(s)>1.11$. The acutal computations are tedious, but I will share them if anyone is interested.

This same attack won't work for $P(s)$, since the associated functions $\mu(n)$ and $\omega(n)$ are not completely multiplicative/additive which is what we used in the crucial proof of (1).