how to prove this extended prime number theorem?

A Generalized Prime Number Theorem?

Conjecture

Let $n$ and $k$ be positive integers with $n - 50 > k^2 > 0$ and $n$ sufficiently large. Then for the odd primes we have, when $p$ is the biggest odd prime $\le n$, $$ 3^k + 5^k + 7^k + 11^k + ... + p^k \sim \frac{n^{k+1}}{(k+1) ( \log(n) - \log(k) ) } $$ I wonder if you guys have seen it before ?

How to prove it ?

Any useful references for $k > 1$ ?

Solution 1:

For $k<e^{c\sqrt{\log x}}$, you can prove the above using partial summation along with the prime number theorem, but it is provably false for $k\sim x^{u}$ when we assume RH. For a complete solution, take a look at this past answer. Specifically, I prove that $$\sum_{p\leq x}p^{k}=\text{li}\left(x^{k+1}\right)+O\left(x^{k+1}E(x)\right),$$ where $\text{li}(x)=\int_2^x \frac{1}{\log t}dt$ is the logarithmic integral, and where $E(x)$ is any positive increasing function which bounds the error term $\pi(x)-\text{li}(x)$. This gives the asymptotic $$\sum_{p\leq x}p^{k}\sim \frac{x^{k+1}}{(k+1)\log x}$$ uniformely for all $k<e^{c\sqrt{\log x}}$, and if you use the Walfisz bound it can be taken slightly further to $e^{c(\log x)^{3/5}}$ with some doubly logarithmic terms in the exponent.

If you assume the Riemann Hypothesis, this shows that your above expressions is not correct for $k\approx x^u$ where $0<u<1$. On RH we have that $E(x)\ll x^{\frac{1}{2}+\epsilon}$, and so the asymptotic holds for all $k<x^{\frac{1}{2}-\epsilon}$. Then for $k=x^u$ we have $\log x-\log k=(1-u)\log x$, which yields the asymptotic $$(1-u)\frac{x^{k+1}}{(k+1)},$$ which is off by the constant factor $(1-u)$.

There may be a simpler way to prove that your expression is incorrect for $k\sim x^u$ that does not require RH. I think you need to be clever though.

Hope that helps,

Solution 2:

A slightly weaker form is the direct application of:

Theorem: Let $p_k$ be the $k$-th prime and let $f$ be a continuous function Riemann integrable in $(0,1)$ then,

$$ \lim_{m \to \infty}\frac{1}{m}\sum_{r = 1}^{m}f\Big(\frac{p_r}{p_m}\Big) = \int_{0}^{1}f(x)dx. $$

For $f(x) = x^k$ we get the more elegant form

$$ 2^k + 3^k + 5^k + \ldots + p_m^k \sim \frac{mp_m^k}{k+1} \sim \frac{m^{k+1} \log^k m}{m+1} $$

and then express it in terms $n$ using $m \log m \sim n$.