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Possible Duplicate:
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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    I guess you mean *asymptotically* equal there.2011-10-15
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    @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.2011-10-15

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