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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.

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@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: 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 $0.2614972$.

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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

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