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

I know that $\sum_p p^{-s}$, $s>1$, converges. Now, I define $J(s) = \sum_p p^{-s}$. Are there any "well known" values for $J(2)$, $J(3)$, $J(4)$, etc? We all know that $\zeta(2)= \frac{\pi^2}{6}$, $\zeta(4)=\frac{\pi^4}{90}$, etc.

share|improve this question
@Yemon: True, I misunderstood the question. I thought, he was asking formula known for $\zeta(s)$. –  S.C. Jun 13 '11 at 6:26
As an aside, one way of showing that the sum over primes diverges is via the identity $\sum_{p}{p^{-s}} = \log \zeta(s) - \sum_{p}\sum^{\infty}_{n=2}{n^{-1} p^{-ns}}$, which is valid for all $\Re(s) > 1$. This shows the connection between special values of $\sum_{p}{p^{-s}}$ and of $\zeta(s)$, but I don't think it's possible to obtain nice closed-form values for $\sum_{p}\sum^{\infty}_{n=2}{n^{-1} p^{-ns}}$ (though it is of course easy to show that it is uniformly bounded as $s \to 1$). –  Peter Humphries Jun 13 '11 at 6:31
This page: mathworld.wolfram.com/RiemannZetaFunction.html answers some for values for $\zeta(s)$ –  S.C. Jun 13 '11 at 6:40
I believe $J(s)$ is sometimes called "the prime zeta function" and information about it can be found by using that search term. –  Gerry Myerson Jun 13 '11 at 12:12
From the identity Peter mentioned and Mobius inversion one has $J(s) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$. So for even $s$ one can get a series formula for $J$ this way, but it is unlikely to lead to any particularly compact closed form. –  Terry Tao Feb 22 at 19:58

1 Answer 1

No, in the sense that there are (for all I know) no identies along the lines of those for $\zeta(s)$, that you recalled, known.

Your function $J$ is sometimes called the prime zeta function.

You can find some information, some approciamte numerical values, plots, and pointers to the literature, e.g., at




Two related hand-waving/heuristic arguments for the difficulty (not sure how good/convincing they are):

  1. The values would 'encode' quite precise information on the set of primes.

  2. The arithmetic function you are summing, that is, $f(n) = n^{-s}$ if $n$ is prime, and $f(n)=0$ if $n$ is not prime, is not a 'nice' arithmetic function; for example it is not multiplicative.

A related note that might interest you, in case you are not aware of it:

As you say $\sum_p p^{-1}$ diverges. However, the rate of divergence is fairly precisely known. Namely, by Mertens's Second Theorem $$\lim_{n \to \infty} \left ( \sum_{p\le n} p^{-1}\right ) - \log \log n $$ exists, and is equal to (or perhaps, rather defines) the Meissel--Mertens constant, which is approxiamtely $0.2614972$.

share|improve this answer
There is one explicit special value concerning $J$ I can think of (as long as one allows analytic continuation and regularization): $J'(0)=-2\log (2\pi)$. See the 'Product over all primes' reference given in the Mathworld link. –  dke Jun 13 '11 at 13:06

Your Answer


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.