Note the $ p < x $ in the sum stands for all primes less than $ x $. I know that for $ s=1 $, $$ \sum_{p<x} \frac{1}{p} \sim \ln \ln x , $$ and for $ \mathrm{Re}(s) > 1 $, the partial sums actually converge to a finite limit called the prime zeta function, which has an analytic continuation to the whole right-half plane but the actual series diverges in the critical strip. So anyway, I'm wondering what the asymptotic behavior of the partial sums are in the limit as $ x \to \infty $ for a given value of $ s $ with $ \mathrm{Re}(s) < 1 $. At first I intuitively conjectured it might be something vaguely like the following $$ \sum_{p<x} \frac1{p^s} \sim f(s) \pi(x)^{1-s} , \quad f(s) = \lim_{n\to\infty} \int_0^1 g_n(u) u^{-s} du $$ but after some thought I'm not sure if this kind of formula will work after all. Any ideas?

Note again: I'm asking about asymptotics when $ \mathrm{Re}(s) < 1 $.

  • $\begingroup$ The first term should be the prime zeta function itself, so I'm not sure your conjecture works (because I think it goes to $0$ as $x$ goes to $\infty$). $\endgroup$
    – Joel Cohen
    Jul 4 '11 at 11:36
  • $\begingroup$ Sorry I may not have been clear earlier, I was talking about the first term of the asymptotic expansion of $\sum$. But since your series converges, shouldn't you have $\sum \sim P(s)$ (where $P$ denotes the prime zeta function) ? On the other hand, I think $\pi(x)^{1-s} \longrightarrow 0$. $\endgroup$
    – Joel Cohen
    Jul 4 '11 at 11:56
  • 1
    $\begingroup$ @Joel: As I put in the title, $ \mathrm{Re}(s) < 1 $. $\endgroup$
    – anon
    Jul 4 '11 at 12:02
  • $\begingroup$ Oh sorry, I was mislead by the "$\textrm{Re}(s) > 1$" in the body of the question (to avoid further confusion, I suggest you mention it again at the end). Never mind then :). $\endgroup$
    – Joel Cohen
    Jul 4 '11 at 12:07
  • $\begingroup$ @Anon: What you want is $$\pi\left(x^{1-s}\right)\ \ \text{rather than}\ \ \ \pi(x)^{1-s}.$$ $\endgroup$ Jul 4 '11 at 17:04

Asymptotic: For $k>-1$ we have $$\sum_{p\leq x}p^{k}=\text{li}\left(x^{k+1}\right)+O\left(x^{k+1}e^{-c\sqrt{\log x}}\right).$$

Proof: We want to sum $\sum_{p\leq x}p^{-s}.$ Write this as a Riemann Stieltjes integral and use partial integration. The infinite series converges absolutely if $\text{Re}(s)>1$, so we assume that $\text{Re}(s)< 1.$ Then this is

$$\sum_{p\leq x}p^{-s}=\int_{2}^{x}t^{-s}d\left(\pi(t)\right)=t^{-s}\pi(t)\biggr|_{2}^{x}+s\int_{2}^{x}t^{-s-1}\pi(t)dt.$$

We expect this to be close to $\int_{2}^{x}t^{-s}d\left(\text{li}(t)\right)$, so consider


which by the quantitative prime number theorem is

$$=O\left(|s|xe^{-c\sqrt{\log x}}\int_2^x t^{-\text{Re(s)}-1}dt\right)=O\left(\frac{|s|}{\text{Re}(s)}x^{1-\text{Re}(s)}e^{-c\sqrt{\log x}}\right).$$ Notice if rewritten for real $s$, it appears much nicer.


$$\sum_{p\leq x}p^{-s}=\int_{2}^{x}\frac{t^{-s}}{\log t}dt+O\left(\frac{|s|}{\text{Re}(s)}x^{1-\text{Re}(s)}e^{-c\sqrt{\log x}}\right).$$

If we let $t=u^{\alpha}$, the integral term becomes $\int_{2}^{x^\frac{1}{\alpha}}\frac{u^{-\alpha s}u^{\alpha-1}}{\log u}du+O(1).$ Because we want the exponent to be zero, we need $-\alpha s+\alpha-1=0$ so let $\alpha=\frac{1}{1-s}$. Then we see that

$$\int_{2}^{x}\frac{t^{-s}}{\log t}dt=\int_{2}^{x^{1-s}}\frac{1}{\log u}du=\text{li}\left(x^{1-s}\right)+O(1).$$

(The $O(1)$ comes from the starting point of the integral) Consequently, for $\text{Re}(s)\neq 1$, we have that

$$\sum_{p\leq x}p^{-s}=\text{li}\left(x^{1-s}\right)+O\left(\frac{|s|}{\text{Re}(s)}x^{1-\text{Re}(s)}e^{-c\sqrt{\log x}}\right).$$

In particular for fixed $s$,

$$\sum_{p\leq x}p^{-s}\sim\frac{x^{1-s}}{(1-s)\log x}.$$

When $\text{Re}(s)=1$, things are special, and only when $s=1$ do we get $\log\log x$. Also, when $s=-k$ is real, we obtain

$$\sum_{p\leq x}p^{k}=\text{li}\left(x^{k+1}\right)+O\left(x^{k+1}e^{-c\sqrt{\log x}}\right).$$

Hope that helps,

Edit: I edited as previously the answer only applied to real $s$. Now it applies to all $s$ in the complex plane we $\text{Re}(s)<1$.

Edit: This question gets asked a lot on math.stackexchange, here are just some of the duplicates:

  • $\begingroup$ Eric, while looking at your integral $\displaystyle \int_2^{x^{1-s}} 1/\log(u) du$, I wonder, why the integration path doesn't matter? $\endgroup$
    – draks ...
    Mar 23 '12 at 11:30
  • $\begingroup$ @draks I don't think I ever said it doesn't matter. You have to take the straight line, I think some crazy contour would make it false. $\endgroup$ Mar 23 '12 at 20:07
  • $\begingroup$ Why do you change $\int_\color{red}1$ in the first equation to $\int_\color{red}2$ in the rest of the post? What is correct? $\endgroup$
    – draks ...
    Nov 11 '12 at 16:49
  • 1
    $\begingroup$ @draks: Well, $\pi(x)$ is supported on $[2,\infty]$, so they are the same. $\endgroup$ Nov 11 '12 at 17:06
  • $\begingroup$ Hi @Eric ,one more question: Does $\int_{2}^{x}t^{-s-1}\pi(t)dt$ relate to a Mellin Transform $M_f(s) := \int \limits_{0}^\infty f(t)t^{s-1}\mathrm{d}t$? If so, what does that mean? $\endgroup$
    – draks ...
    Jan 6 '13 at 23:31

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