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How does $ \sum_{p\lt x} p^{-s} $ grow asymptotically for $ \mathrm{Re}(s) \lt 1 $?

from the prime number theorem is it possible to have

$$ \sum_{p \le x}\: x^{m} = \frac{ \text{Li}\: (x^{m+1})}{m+1} $$ ?

here 'Li' is the logarithmic integral $ \int_{2}^{\infty} \frac{dt}{logt} $

valid for m > -1 in the case m=0 we recover the usual prime number theorem.

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marked as duplicate by Eric Naslund, anon, J. M., Chandrasekhar, Ilya Oct 15 '11 at 11:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I guess you mean asymptotically equal there. – lhf Oct 15 '11 at 8:33
@Anon: That is really funny, I remembered typing this up before, and was about to go find that answer! It is an exact duplicate, so I am voting to close. – Eric Naslund Oct 15 '11 at 8:45

I assume you meant $\sum_{p\leq x} p^m$ since $\sum_{p\leq x} x^m=x^m \pi(x)$. In this case when $\text{Re}(s)>-1$ we have $$\sum_{p\leq x}p^{s}=\text{li}\left(x^{1+s}\right)+O\left(\frac{x^{1+s}}{1+s}e^{-c\sqrt{\log x}}\right)$$ where the error is uniform over $s$. This follows from partial summation and the prime number theorem, please see this answer.

Edit: As mentioned in the comments, the real part of $s$ should be greater the $-1$ since the sum converges when it is less then $-1$, and acts different when it equals $1$. (Specfically we need that $1+s>0$ so the $\text{li}(x^{1+s})$ term contributes.

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This is true if the real part of $s$ exceeds $-1$, as in the original question; otherwise the sum converges except at $s=-1$ itself, so the main term would change. – Greg Martin Oct 15 '11 at 9:10

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