Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let \begin{equation} R(x) = \sum_{k=1}^{\infty}\frac{\mu(k)}{k}li(x^{1/k}) \end{equation} where $\mu$ is the Mobius function and \begin{equation} li(x) = \int_0^x \frac{dt}{\log t} \end{equation} Is there a proof in the literature of \begin{equation} \pi(x)=R(x)-\sum_{\rho}R(x^{\rho}) \end{equation} where $\pi$ is prime counting function and the sum is over all complex zeros of $\zeta(s)$. The literature seems to treat it as fact while stating no proof is available - a strange situation.

Thanks in advance.

share|improve this question
1  
This link is on the Wiki article jstor.org/pss/2004630 –  Gjergji Zaimi Sep 13 '10 at 14:45
1  
I think Edwards' book <i>The Zeta Function</i> has a proof of this, but I don't have it available to check. –  David Speyer Sep 13 '10 at 14:54
    
I think there is an $O(1)$ missing. This is a formula stated by Riemann in "the paper" and proved by von Mangoldt. As David says, it's in chapter 3 of Edwards. –  Felipe Voloch Sep 14 '10 at 18:30
add comment

2 Answers

up vote 2 down vote accepted

Stopple, A Primer of Analytic Number Theory, proves a theorem which looks something like the one under discussion. On page 248, he has $$\pi(x)=R(x)+\sum_{\rho}R(x^{\rho})+\sum_{n=1}^{\infty}{\mu(n)\over n}\int_{x^{1/n}}^{\infty}{dt\over t(t^2-1)\log t}$$

You say that the literature treats your formula as a fact, but you give no citation. Where in the literature do you find your formula?

share|improve this answer
    
Borwein, Computational Strategies for the Riemann Zeta Function. This has the formula I quote and claims that it is exact. –  alext87 Sep 14 '10 at 11:19
1  
If Borwein is interested in computing $\pi (x)$ then he might be taking advantage of the fact that it is an integer and the error term that he is omitting is small. –  Felipe Voloch Sep 14 '10 at 18:33
2  
alext in the original question refers to a sum over 'all complex zeros'; it's not clear if he's referring to the trivial zeros or not. If he is, I think what he wrote is correct. The formula in my book that Gerry references has a sum over the nontrivial zeros only; the other infinite sum on n is the contribution of the trivial zeros, written in an explicit form to make clear that it is very small (decreasing in x). –  Stopple Sep 14 '10 at 20:14
4  
I've looked at the Borwein (and Bradley and Crandall) paper - a preprint is freely available at docserver.carma.newcastle.edu.au/211/2/… and where alext87 has equality Borwein et al. have a tilde. They say, "This relation has been called exact [by Ribenboim, in The New Book of Prime Number Records], yet we could not locate a proof in the literature." alext87, if you had access to the reference and the full quote, it would have saved us a lot of work had you included the information in your original question. –  Gerry Myerson Sep 14 '10 at 23:29
1  
@Gerry - I think the plus signs in my book are a typo; they should in fact be minus signs! The corresponding formula for $\Pi(x)$, (10.11) on p. 244, has minus. –  Stopple Oct 4 '10 at 20:52
show 3 more comments

It may be useful to read Section 10 of Chapter V of Ingham's "The Distribution of Prime Numbers."

Let $\Pi(x)=\pi(x)+\frac{1}{2}\pi(x^{1/2})+...$, then Moebius proved that

$$ \pi(x) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \Pi(x^{1/n}).$$

However, this isn't overly illuminating because it shows that

$$\pi(x) = \Pi(x) + O(\sqrt{x}/\log x ). $$

and Littlewood showed that

$$ \Pi(x) - \ell i(x) = \Omega_\pm(\sqrt{x} \log\log\log x/\log x).$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.