# a conjecture (of mine ) about primes [duplicate]

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 NaslundNov 13 '12 at 17:03

Did you test that computationally at least? Why do you think this is true? –  Pedro Tamaroff Nov 6 '12 at 14:16
It seems your question mark key is stuck? –  joriki Nov 6 '12 at 14:25
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

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: $$\sum_{p\leq x\text{ prime}}p^m\approx \text{Ei}((n+1)\ln x)= \text{Ei}(\ln x^{n+1})=\text{Li}(x^{n+1})$$

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

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