# Prime Zeta Function

The **prime zeta function** is an expression similar to the Riemann zeta function. It has interesting properties that are related to the properties of the Riemann zeta function, as well as a connection to Artin's conjecture about primitive roots.

The prime zeta function $\zeta_{\mathbb{P}}(s)$, where $s$ is a complex number, is defined by the series $\zeta_{\mathbb{P}}(s) = \sum_{p \in \mathbb{P}} \dfrac{1}{p^{s}}$ where $\mathbb{P}$ is the set of prime numbers.

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## Divergence of $\zeta_{\mathbb{P}}(1)$

The value of $\zeta_{\mathbb{P}}(1)$ is the sum of the reciprocals of the primes. This series diverges, but very slowly:
$\sum_{\stackrel{p \in \mathbb{P}}{p < n}} \frac1{p} - \log\big(\log(n)\big) \to M\ \text{ as }\ n\to \infty,$
where $M = 0.261497\ldots$ is a constant (called the **Meissel-Mertens constant**). This is reminiscent of the definition of the Euler-Mascheroni constant $\gamma$.

Euler asserted that the sum of the reciprocals of the primes diverged, and derived a correct estimate for it, but his proof involved manipulations of divergent infinite series and products (for instance, the Euler product expansion for $\zeta(s)$ evaluated at $s=1$). As is often the case with Euler's arguments, it can be made rigorous with some extra work.

## Expression in terms of the Riemann Zeta Function

The Euler product for $\zeta(s)$ $\big($for $\text{Re}(s)>1\big)$ is $\zeta(s) = \prod_{p \text{ prime}} \left( 1-\frac1{p^s} \right)^{-1}$ and taking the natural log of both sides gives $\begin{aligned} \log\big(\zeta(s)\big) &= - \sum_{p \text{ prime}} \log\left( 1-\frac1{p^s} \right) \\ &= \sum_{p \text{ prime}} \sum_{k=1}^{\infty} \frac{\hspace{1.5mm} \frac{1}{p^{ks}}\hspace{1.5mm} }{k}, \end{aligned}$ using the Maclaurin series $-\log(1-x) = x+\frac{x^2}2 + \frac{x^3}3 + \cdots.$

Now switching the two sums gives $\begin{aligned} \log\big(\zeta(s)\big) &= \sum_{k=1}^{\infty} \sum_{p \text{ prime}} \frac{\hspace{1.5mm} \frac{1}{p^{ks}}\hspace{1.5mm} }{k} \\ &= \sum_{k=1}^{\infty} \frac1{k} \sum_{p \text{ prime}} \frac1{p^{ks}} \\ &= \sum_{k=1}^{\infty} \frac{\zeta_{\mathbb{P}}(ks)}{k}. \end{aligned}$ A generalization of Möbius inversion says that $f(x) = \sum_{k=1}^{\infty} \frac{g(kx)}{k} \Leftrightarrow g(x) = \sum_{k=1}^{\infty} \mu(k)\frac{f(kx)}{k},$ where $\mu$ is the Möbius function, as long as the sums are absolutely convergent (the proof is straightforward). Applying this gives $\zeta_{\mathbb{P}}(s) = \sum_{k=1}^{\infty} \mu(k) \frac{\log\big(\zeta(ks)\big)}{k}.$ Note that this gives an idea of why $\zeta_{\mathbb{P}}(1)$ diverges at the same speed as $\log\big(\log(n)\big)$, since the $k=1$ term is the only undefined term at $s= 1,$ and $\zeta(1)$ diverges like $\log(n)$. In fact, expanding near $s=1$ gives $($for $x > 0)$ $\zeta_{\mathbb{P}}(1+x) = -\log(x)+(M-\gamma) + O(x),$ where $\gamma$ is the Euler-Mascheroni constant. The $O(x)$ terms go to $0$ as $x \to 0^+ .$

## Connection with the Twin Prime Constant

The twin prime conjecture is that there are infinitely many primes $p$ such that $p+2$ is also prime. While this is still open, heuristics suggest that it is true and that in fact the function $\pi_2(x)$ that counts twin primes $\le x$ satisfies
$\pi_2(x) \sim 2\Pi_2 \int_2^x \frac{dx}{\big(\log(x)\big)^2},$
where $\Pi_2 =0.66016\ldots$ is the **twin prime constant**
$\sum_{p \text{ odd prime}} \left( 1-\frac1{(p-1)^2}\right).$

A computation involving Taylor expansions of logarithms (similar to the above one) shows that the constant $\Pi_2$ is related to the values of $P(s)$ as follows: $\log(\Pi_2) = - \sum_{k=2}^{\infty} \frac{2^k-2}{k} \big(\zeta_{\mathbb{P}}(k)-2^{-k}\big).$ The point is that the prime zeta function comes up in evaluating constants involving products over all primes.

## Connection with Artin's Conjecture

**Artin's conjecture** states that if $a$ is an integer that is not a perfect square or $-1$, then it is a primitive root modulo $p$ for infinitely many $p.$ As usual, there is a heuristic estimate for the probability that $a$ is a primitive root mod a given prime $p.$ If $a$ is squarefree and not congruent to $1$ mod $4$, this probability is conjectured to be **Artin's constant**
$C = \prod_{p \text{ prime}} \left( 1-\frac1{p(p-1)} \right) = 0.37395\ldots.$
(For other values of $a$, there is also a conjectural probability that is a rational multiple of Artin's constant.)

A calculation involving taking logs and expanding gives $\log(C) = \sum_{n=2}^{\infty} (1-L_n) \frac{\zeta_{\mathbb{P}}(n)}{n},$ where $L_n$ is the $n^\text{th}$ Lucas number $\big($defined by $L_1 = 1, L_2 = 3, L_n = L_{n-1}+L_{n-2}\big).$

## Generalization

An interesting generalization can be made by summing over inverse of positive integers raised to power $s$ which are a product of $k$ primes (not necessarily distinct).

The $k$ prime zeta function $\zeta_{\mathbb{P}}(k,s)$, where $s$ is a complex number and $k$ is a non-negative integer, is defined by the series $\zeta_{\mathbb{P}}(k,s) = \sum_{n : \Omega(n) = k} \dfrac{1}{n^{s}},$ where $\Omega$ denotes the number of prime factors.

## Relation with other Zeta Functions

Try proving these interesting identities involving generalized prime zeta function:

$\begin{aligned} \sum_{k=0}^{\infty}\zeta_{\mathbb{P}}(k,s) &= \zeta (s) \\\\ \lim_{s \to 1} (s-1) \zeta_{\mathbb{P}}(k,s) &= 0 \ \forall \ k \in \mathbb N \\\\ \zeta_{\mathbb{P}}(2,s) &= \dfrac{\zeta_{\mathbb{P}}(s)^2 + \zeta_{\mathbb{P}}(2s)}{2}. \end{aligned}$

**Cite as:**Prime Zeta Function.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/prime-zeta-function/