Sums of powers of primes II

For a real number k, define \(\pi _k(x) = \sum _{p\le x} p^k\) ${\pi }_k(x)={\sum }_{p\le x}{p}^k$ . When \(k>0\) $k>0$ , we prove that \(\begin{aligned} \pi _k(x) - \pi (x^{k+1}) = \Omega _{\pm }\left( \frac{x^{\frac{1}{2}+k}}{\log x} \log \log \log x\right) \end{aligned}\) $$\begin{array}{c}\hfill {\pi }_k(x)-\pi (x^{k+1})={\mathrm{\Omega }}_{\pm }\left(\frac{x^{\frac{1}{2}+k}}{logx},log,log,log,x\right)\end{array}$$ as \(x\rightarrow \infty \) $x\to \infty$ , and we prove a similar result when \(-1<k<0\) $-1<k<0$ . This strengthens a result in a paper by Gerard and the author and it corrects a flaw in a proof in that paper. We also quantify the observation from that paper that \(\pi _k(x) - \pi (x^{k+1})\) ${\pi }_k(x)-\pi (x^{k+1})$ is usually negative when \(k>0\) $k>0$ and usually positive when \(-1<k<0\) $-1<k<0$ .