Sums of powers of primes. II (Q6624947)

For a real number \(k\), let \(\pi_k(x) = \sum_{p\le x} p^k\). More precisely, let \(\pi(x)=\pi_0(x)\) be the number of primes less than or equal to \(x\). In the paper under review, the author explores the difference\N\[\N\Delta_k(x):=\pi_k(x) - \pi(x^{k+1}).\N\]\NLetting \(\theta_0\) to be the supremum of the real parts of the zeros of \(\zeta(s)\), he shows that as \(x\to\infty\),\N\[\N\Delta_k(x)= \begin{cases} \Omega_{\pm} (x^{\theta_0+k-\varepsilon}) & \text{if } k>0,\\\N\Omega_{\pm}( x^{(k+1)(\theta_0-\varepsilon)}) & \text{if } -1< k < 0 \text{ and } \theta_0<1, \end{cases}\N\]\Nwhere \(\varepsilon>0\) is arbitrary small. The author quantifies \(\varepsilon\) under assuming the Riemann Hypothesis (RH), by showing that if RH is true, then\N\[\N\Delta_k(x)= \begin{cases} \Omega_{\pm}(x^{k+\frac12}\frac{\log\log\log x}{\log x}) &\text{if } k>0,\\\N\Omega_{\pm}(x^{\frac{k+1}2}\frac{\log\log\log x}{\log x}) &\text{if } -1<k<0. \end{cases}\N\]\NFurthermore, he shows that\N\[\N\int_1^{\infty} \frac{\Delta_k(t)}{t^{k+2}} dt = \frac{-1}{k+1}\log(k+1) \quad\N\begin{cases} <0 &\text{if } k>0, \\\N>0 &\text{if } -1 < k < 0,\end{cases}\N\]\Nand proves that for \(0<k\le 2.3434\), the statement that\N\[\N\int_1^x \Delta_k(t)dt < 0\N\]\Nholds for all \(x\) sufficiently large, is equivalent with RH to be true.\N\NFor Part I see [\textit{J. Gerard} and the author, Ramanujan J. 45, No. 1, 171--180 (2018; Zbl 1428.11160)].