 Solution

By part (a), every nontrivial zero $\rho =\beta +i\gamma$ with $|\gamma |\le T$ satisfies $\beta \le 1-c_1/{\log }^{9}T$ for large $T$, after reducing the positive constant. The local zero count for the Riemann zeta function givesThere are finitely many zeros at bounded height, none at zero or at one, so that part of the sum is bounded. The truncated explicit formula for the second Chebyshev function now yieldsBalance the exponent losses $c_1\log x/(\log T{)}^9$ and $\log T$ by choosing $\log T=(c_1\log x{)}^{1/10}$. This is the optimal order obtainable from these two errors: making either exponent larger forces the other smaller. Thus, for a positive constant $c_2$,The logarithmic prefactor can be absorbed by reducing $c_2$. This is the prime number theorem error from a logarithmic zero-free region with ninth-power width.

Under the Riemann hypothesis, $|x^{\rho }|=x^{1/2}$. The same reciprocal-zero sum bounds the zero contribution by $O(x^{1/2}{\log }^{2}T)$. Taking $T=x$ makes the truncation error $O({\log }^{2}x)$, so