Note the $ p < x $ in the sum stands for all primes less than $ x $. I know that for $ s=1 $,
$$ \sum_{p Note again: I'm asking about asymptotics when $ \mathrm{Re}(s) < 1 $.
How does $ \sum_{p
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0The first term should be the prime zeta function itself, so I'm not sure your conjecture works (because I think it goes to $0$ as $x$ goes to $\infty$). – 2011-07-04
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0Sorry I may not have been clear earlier, I was talking about the first term of the asymptotic expansion of $\sum$. But since your series converges, shouldn't you have $\sum \sim P(s)$ (where $P$ denotes the prime zeta function) ? On the other hand, I think $\pi(x)^{1-s} \longrightarrow 0$. – 2011-07-04
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1@Joel: As I put in the title, $ \mathrm{Re}(s) < 1 $. – 2011-07-04
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0Oh sorry, I was mislead by the "$\textrm{Re}(s) > 1$" in the body of the question (to avoid further confusion, I suggest you mention it again at the end). Never mind then :). – 2011-07-04
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0@Anon: What you want is $$\pi\left(x^{1-s}\right)\ \ \text{rather than}\ \ \ \pi(x)^{1-s}.$$ – 2011-07-04
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0@EricNaslund I think I found a way to rewrite $P_\color{red}x(\color{blue}s)$, see [here](http://math.stackexchange.com/q/115230/19341). What do you think about my derivation...? – 2013-01-16
1 Answers
Asymptotic: For $k>-1$ we have $$\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.$$
We expect this to be close to $\int_{2}^{x}t^{-s}d\left(\text{li}(t)\right)$, so consider
$$\int_{2}^{x}t^{-s}d\left(\pi(t)\right)-\int_{2}^{x}t^{-s}d\left(\text{li}(t)\right)=t^{-s}\left(\pi(t)-\text{li}(t)\right)\biggr|_{2}^{x}+s\int_{2}^{x}t^{-s-1}\left(\pi(t)-\text{li}(t)\right)dt$$
which by the quantitative prime number theorem is
$$=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).$$ Notice if rewritten for real $s$, it appears much nicer.
Hence
$$\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:
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0Eric, while looking at your integral $\displaystyle \int_2^{x^{1-s}} 1/\log(u) du$, I wonder, why the integration path doesn't matter? – 2012-03-23
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0@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. – 2012-03-23
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0Why 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? – 2012-11-11
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1@draks: Well, $\pi(x)$ is supported on $[2,\infty]$, so they are the same. – 2012-11-11
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0Hi @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? – 2013-01-06
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0Relevant other questions: http://math.stackexchange.com/questions/95058/does-the-correctness-of-riemanns-hypothesis-imply-a-better-bound-on-sum-limi?rq=1, http://math.stackexchange.com/questions/623872/what-is-the-sum-of-the-prime-numbers-up-to-a-prime-number-n/624296#624296. – 2014-01-01
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0Duplicates of the question: http://math.stackexchange.com/questions/231380/a-conjecture-of-mine-about-primes?lq=1 and http://math.stackexchange.com/questions/72796/generalization-of-the-prime-number-theorem?lq=1 – 2014-01-01
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0Another duplicate: http://math.stackexchange.com/questions/1532561/estimate-for-sum-of-negative-powers-of-primes/1532577#1532577 – 2015-11-16