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The prime zeta function is defined for $\mathfrak{R}(s)>1$ as

$P(s)=\sum_{p} \ p^{-s}$, where $p \in \mathbb{P}$.

It is well-know this series converges whenever $\mathfrak{R}(s)>1$.

Now, consider the infinite sum

$\sum_{p \geq x} \ p^{-s}$, where $x \in \mathbb{R}$ and $x > 2$.

How can I find a lower bound close enough to this sum?

I mean if there exists some smooth function $f(y, z)$ on real domain such that

$\sum_{p \geq x} \ p^{-s} > f(x, s) > 0$.

Frank Vega
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  • It really would be $f(x,s),$ a function of two variables, right? An obvious lower bound is $(2x)^{-s}.$ – Thomas Andrews May 04 '24 at 02:59
  • What does “close enough to this sum” mean exactly? Certainly you could take $2^{-s}$ as a lower bound (though it’s trivial). I would also suggest treating it as a two-variable function $f(x,s)$ since any lower bound will necessarily depend on both of these quantities. – Clayton May 04 '24 at 03:00
  • @clayton $2^{-s}$ won't be a lower bound for large $x.$ – Thomas Andrews May 04 '24 at 03:02
  • Thank you, I put the function of two variables. When I mean "close enough", then I actually wish to say that the difference between the sum and the lower bound could be a smallest as possible. – Frank Vega May 04 '24 at 03:04
  • @ThomasAndrews I'm not sure that $2^{-s}$ could be a good lower bound. Consider the following: $\frac{1}{2} = \sum_{n = 2} \frac{1}{2^{n}}$ and so, we would have $2^{-s} = \left(\sum_{n = 2} \frac{1}{2^{n}}\right)^{-s} > 4^{-s}+8^{-s}+\ldots$. – Frank Vega May 04 '24 at 03:09
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    @FrankVega I didn't say $2^{-s}$ was a lower bound. It isn't a lower bound. The lower bound I gave, $(2x)^{-s},$ is definitely not a good lower bound. – Thomas Andrews May 04 '24 at 03:12
  • Does $P’(s) = \sum_{n=1}^{\infty} \mu(n)\frac{\zeta’(ns)}{\zeta(ns)}$ help? It lets you bound the derivative and with suitable initial data can bound the original function. – Sidharth Ghoshal May 04 '24 at 03:13
  • @ThomasAndrews sorry, I misunderstood your comment. Thank you for letting me know. – Frank Vega May 04 '24 at 03:13
  • @SidharthGhoshal it might help the following link (where they found the partial sum: however, this question is not the same): https://math.stackexchange.com/questions/3497053/stronger-asymptotics-of-the-partial-sums-of-the-prime-zeta-function – Frank Vega May 04 '24 at 03:20
  • the question is confusing as one hand you talk about $\Re s >1$ (so implying $s$ complex, while on the other had you talk about a positive $\sum p^{-s}$ which kind of assumes $s>1$ real; what is exactly you want to bound? – Conrad May 04 '24 at 03:31
  • @ThomasAndrews: You’re correct, I was too hasty in my response. I’ll leave my comment because it isn’t clear to me what “close enough” actually means. – Clayton May 04 '24 at 11:24
  • @Conrad I actually want to find the lower bound when $s$ is real and $1.5 < s < 1.6$. – Frank Vega May 04 '24 at 18:35

1 Answers1

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Let $\pi(x)$ be the prime counting function. Then the prime number theorem states that

$$ \pi(x)=\int_2^x{\mathrm du\over\log u}+O\left(x\over\log^2x\right), $$

so by Riemann-Stieltjes integration, we have for $\Re(s)>1$ that

$$ \sum_{p>x}p^{-s}=\int_x^{+\infty}u^{-s}\mathrm d\pi(u)=\int_x^{+\infty}{\mathrm du\over u^s\log u}+R(x,s), $$

where

\begin{aligned} R(x,s) &=\int_x^{+\infty}u^{-s}\mathrm d[O(u/\log^2u)]=O(x^{1-\Re(s)}/\log^2x)+s\int_x^{+\infty}O(u^{-\Re(s)}/\log^2u)\mathrm du \\ &=O\{|s|x^{1-\Re(s)}/\log^2x\}. \end{aligned}

By integration by parts, there is

$$ \int_x^{+\infty}{\mathrm du\over u^s\log u}={x^{1-s}\over(s-1)\log x}+O\left(|s|x^{1-\Re(s)}\over|1-s|\log^2x\right). $$

Conclusively, we have for large $x$ that

$$ \sum_{p>x}p^{-s}={x^{1-s}\over(s-1)\log x}+O\left(|s|x^{1-\Re(s)}\over|1-s|\log^2x\right). $$

TravorLZH
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  • $\int_x^{+\infty}u^{-s}\mathrm d[O(u/\log^2u)]$ is very abuisive notation –  May 05 '24 at 21:17