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

What could I use to prove the following conjecture?

$ \sum_{p \le x} p^{m} \sim \operatorname{Li}(x^{m+1}) $

For $ m=0 $ this is just the prime number theorem, but would it be true for other numbers 'm' ? Or even for negative m?

Also if the function can be expanded in power series $ f(x)= \sum_{n=0}^{\infty} a(n)x^{n} $,

I also think that $ \sum _{p \le x}f(x) \sim \sum_{n=0}^{\infty}a(n)\operatorname{Li}(x^{n+1}) $.

Of course $ \operatorname{Li}(x)= \int_{2}^{x}\frac{du}{\ln(u)} $.

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marked as duplicate by Eric Naslund Nov 13 '12 at 17:03

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Did you test that computationally at least? Why do you think this is true? –  Pedro Tamaroff Nov 6 '12 at 14:16
2  
It seems your question mark key is stuck? –  joriki Nov 6 '12 at 14:25
2  
I think you are interested in this: How does $ \sum_{p<x} p^{-s} $ grow asymptotically for $ \text{Re}(s) < 1 $? and the answer given therein... –  draks ... Nov 6 '12 at 15:03
    
@Jose, where'd you get this from? –  akkkk Nov 6 '12 at 15:44
    
i got it intuitively for example formthe prime number theorem $ \sum_{n\le x}f(n) \sim \int_{2}^{x} f(x)/ln(x) $ , the probability of finding a prime is about $ \frac {1}{ln(x)} $ and i replace the sum by an integral, which can be obtained in a closed form in terms of the $ Li(x) $ integral at least when $ f(x)= x^{m} $ –  Jose Garcia Nov 6 '12 at 15:57
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1 Answer

up vote 10 down vote accepted

This is not an answer but a plot for $m=1,2,3,10,20$ because some peopel asked for that.

EDIT: This is somewhat an answer now, see below.

I entered the following code in Mathematica:

m = 1
maxx = 100
p1 = DiscretePlot[Sum[Prime[n]^m, {n, PrimePi[x]}], {x, maxx}];
p2 = DiscretePlot[LogIntegral[x], {x, maxx}, PlotStyle -> Red];
Show[{p1, p2}]

It seems alright for small $m$, as you can see below.


Okay, so in this paper we find that: $$\sum_{p\leq x\text{ prime}}f(p)\approx\int_2^x \frac{f(y)dy}{\ln y}$$

Substituting $f(p)=p^m$ gives, according to wolfram alpha: $$\sum_{p\leq x\text{ prime}}p^m\approx \text{Ei}((n+1)\ln x)= \text{Ei}(\ln x^{n+1})=\text{Li}(x^{n+1})$$


$m=1$ m=1 $m=2$ m=2 $m=3$ m=3 $m=10$ m=10 $m=20$, $maxx=10000$ m=20

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