 Solution

Assume first the Riemann hypothesis. Replace any $x\ge 2$ by $y=\lfloor x\rfloor +1/2$; then $\psi (x)=\psi (y)$ and $⟨y⟩\ge 1/2$. Take $T=y$ in part a. Every nontrivial zero has real part $1/2$, and the Riemann–von Mangoldt formula impliesConsequentlywhile both truncation errors in part a are $O((\log y{)}^2)$. Thuswhich implies the stated $O_{ϵ}(x^{1/2+ϵ})$ estimate.

Conversely, suppose that estimate holds for every $ϵ>0$. For $\Re s>1$, partial summation givesGiven any $s$ with $\Re s>1/2$, choose $ϵ<\Re s-1/2$. The error hypothesis makes the last integral locally uniformly convergent there, so it supplies a holomorphic continuation ofto the half-plane $\Re s>1/2$. A zero of $\zeta$ in that half-plane would create a pole of its logarithmic derivative, so none exists. The Functional equation of the Riemann zeta function reflects every nontrivial zero with real part below $1/2$ to one above $1/2$. All nontrivial zeros must therefore lie on the critical line, proving the Riemann hypothesis equivalence for the second Chebyshev function.