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

Solution 1:

Asymptotic: For $k>-1$ we have $$\sum_{p\leq x}p^{k}=\text{li}\left(x^{k+1}\right)+O\left(x^{k+1}e^{-c\sqrt{\log x}}\right).$$

Proof: We want to sum $\sum_{p\leq x}p^{-s}.$ Write this as a Riemann Stieltjes integral and use partial integration. The infinite series converges absolutely if $\text{Re}(s)>1$, so we assume that $\text{Re}(s)< 1.$ Then this is

$$\sum_{p\leq x}p^{-s}=\int_{2}^{x}t^{-s}d\left(\pi(t)\right)=t^{-s}\pi(t)\biggr|_{2}^{x}+s\int_{2}^{x}t^{-s-1}\pi(t)dt.$$

We expect this to be close to $\int_{2}^{x}t^{-s}d\left(\text{li}(t)\right)$, so consider

$$\int_{2}^{x}t^{-s}d\left(\pi(t)\right)-\int_{2}^{x}t^{-s}d\left(\text{li}(t)\right)=t^{-s}\left(\pi(t)-\text{li}(t)\right)\biggr|_{2}^{x}+s\int_{2}^{x}t^{-s-1}\left(\pi(t)-\text{li}(t)\right)dt$$

which by the quantitative prime number theorem is

$$=O\left(|s|xe^{-c\sqrt{\log x}}\int_2^x t^{-\text{Re(s)}-1}dt\right)=O\left(\frac{|s|}{\text{Re}(s)}x^{1-\text{Re}(s)}e^{-c\sqrt{\log x}}\right).$$ Notice if rewritten for real $s$, it appears much nicer.

Hence

$$\sum_{p\leq x}p^{-s}=\int_{2}^{x}\frac{t^{-s}}{\log t}dt+O\left(\frac{|s|}{\text{Re}(s)}x^{1-\text{Re}(s)}e^{-c\sqrt{\log x}}\right).$$

If we let $t=u^{\alpha}$, the integral term becomes $\int_{2}^{x^\frac{1}{\alpha}}\frac{u^{-\alpha s}u^{\alpha-1}}{\log u}du+O(1).$ Because we want the exponent to be zero, we need $-\alpha s+\alpha-1=0$ so let $\alpha=\frac{1}{1-s}$. Then we see that

$$\int_{2}^{x}\frac{t^{-s}}{\log t}dt=\int_{2}^{x^{1-s}}\frac{1}{\log u}du=\text{li}\left(x^{1-s}\right)+O(1).$$

(The $O(1)$ comes from the starting point of the integral) Consequently, for $\text{Re}(s)\neq 1$, we have that

$$\sum_{p\leq x}p^{-s}=\text{li}\left(x^{1-s}\right)+O\left(\frac{|s|}{\text{Re}(s)}x^{1-\text{Re}(s)}e^{-c\sqrt{\log x}}\right).$$

In particular for fixed $s$,

$$\sum_{p\leq x}p^{-s}\sim\frac{x^{1-s}}{(1-s)\log x}.$$

When $\text{Re}(s)=1$, things are special, and only when $s=1$ do we get $\log\log x$. Also, when $s=-k$ is real, we obtain

$$\sum_{p\leq x}p^{k}=\text{li}\left(x^{k+1}\right)+O\left(x^{k+1}e^{-c\sqrt{\log x}}\right).$$

Hope that helps,

Edit: I edited as previously the answer only applied to real $s$. Now it applies to all $s$ in the complex plane we $\text{Re}(s)<1$.

Edit: This question gets asked a lot on math.stackexchange, here are just some of the duplicates:

• Finding an asymptotic for the sum $\sum_{p\leq x}p^m$
• Generalization of the Prime number theorem to $\sum_{p\leq x}p^{m+1}$
• What is the sum of the prime numbers up to a prime number $n$?
• Estimate for sum of negative powers of primes