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):

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 $0.2614972$.

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):

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 $0.2614972$.

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):

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 $0.2614972$.

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):

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 $0.2614972$.