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The Prime Zeta function is defined as $$\zeta_{P}(s) = \sum_{p\in \mathbb P} \frac 1{p^s}$$ where $\mathbb P$ is the set of primes.

In an answer to this question, TravorLZH shows $$\sum_{p>x}p^{-s}={x^{1-s}\over(s-1)\log x}+\mathcal O\left(|s|x^{1-\Re(s)}\over|1-s|\log^2x\right)$$ for large $x$.

I want to know whether there are known (tight) upper and lower bounds for $\zeta_P(s)$. In other words, do we know of a function $f$ and constants $A$ and $B$ such that $$Af(s) \le \zeta_P(s) \le Bf(s)$$ for all positive integers $s>1$?

A couple of bounds for $\zeta_P(2)$ has been discussed in this question (although as I pointed out in the comments, some of the references have not been shared properly). But I want to know about the general case.

Sayan Dutta
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1 Answers1

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By partial summation we have $$\sum_{p\leq x}\frac{1}{p^{s}}=\sum_{n\leq x}\frac{1_{p}}{n^{s}}=\frac{\pi\left(x\right)}{x^{s}}+s\int_{1}^{x}\pi\left(t\right)t^{-s-1}dt$$ so $$\sum_{p\in\mathbb{P}}\frac{1}{p^{s}}=s\int_{1}^{+\infty}\pi\left(t\right)t^{-s-1}dt.$$ By the prime number theorem we know that there exists $A,B>0$ such that $$A\text{li}(t)\leq\pi(t)\leq B\text{li}(t)$$ where $\text{li}(t)$ is the logarithmic integral function, hence the problem boils down to the evaluation of the Mellin transform $$s\int_{1}^{+\infty}\text{li}\left(t\right)t^{-s-1}dt=\log\left(\frac{1}{s-1}\right)$$ which is valid for $\text{Re}(s)>1$ (see this answer for a complete calculation).

Marco Cantarini
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