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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$\begingroup$ @Yemon: True, I misunderstood the question. I thought, he was asking formula known for $\zeta(s)$. $\endgroup$– C.S.Commented Jun 13, 2011 at 6:26
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3$\begingroup$ 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 $s \to 1$). $\endgroup$– Peter HumphriesCommented Jun 13, 2011 at 6:31
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3$\begingroup$ This page: mathworld.wolfram.com/RiemannZetaFunction.html answers some for values for $\zeta(s)$ $\endgroup$– C.S.Commented Jun 13, 2011 at 6:40
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2$\begingroup$ I believe $J(s)$ is sometimes called "the prime zeta function" and information about it can be found by using that search term. $\endgroup$– Gerry MyersonCommented Jun 13, 2011 at 12:12
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2$\begingroup$ 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. $\endgroup$– Terry TaoCommented Feb 22, 2015 at 19:58
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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
http://mathworld.wolfram.com/PrimeZetaFunction.html
and
http://en.wikipedia.org/wiki/Prime_zeta_function
Two related hand-waving/heuristic arguments for the difficulty (not sure how good/convincing they are):
The values would 'encode' quite precise information on the set of primes.
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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7$\begingroup$ 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. $\endgroup$– dkeCommented Jun 13, 2011 at 13:06