Sums of powers of primes II

For a real number k, define \(\pi _k(x) = \sum _{p\le x} p^k\). When \(k>0\), we prove that

$$\begin{aligned} \pi _k(x) - \pi (x^{k+1}) = \Omega _{\pm }\left( \frac{x^{\frac{1}{2}+k}}{\log x} \log \log \log x\right) \end{aligned}$$

as \(x\rightarrow \infty \), and we prove a similar result when \(-1<k<0\). 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 \(\pi _k(x) - \pi (x^{k+1})\) is usually negative when \(k>0\) and usually positive when \(-1<k<0\).