Chebyshev psi prime number theorem error

Under countable choice, the supplied finite explicit-formula bound applies. Put $u=\sqrt{\log x}$ and choose a fixed $A>0$. For sufficiently large x take $T=e^{Au}\ge 3$. Then $\log (T+2)=Au+O(e^{-Au})$, so the finite-zero term is $O(x{u}^2{e}^{-(c_0\mathrm{/}A)u+o(1)})$, the truncation term is $O(x{u}^4{e}^{-Au})$, and the remaining error is $O(u^2)$.

Choose $0<c<\min (A,c_0\mathrm{/}A)$. For any fixed k and positive epsilon, $u^k{e}^{-ϵu}$ is bounded; thus each error above is $O(x{e}^{-cu})$. Enlarging the constant over the initial compact x-range proves the assertion for all $x\ge 2$.

The contour interpretation is consistent with the same bound: take ${\sigma }_1=1-c_1\mathrm{/}\log (T+2)$ with a sufficiently small region constant and ${\sigma }_0=1+1\mathrm{/}\log x$. Throughout $\mathrm{\mid }t\mathrm{\mid }\le T$ the left edge stays in the proved region. At high heights the derivative is $O(\log T)$; integrating $1\mathrm{/}\mathrm{\mid }s\mathrm{\mid }$ gives a vertical contribution $O(x^{{\sigma }_1}{\log }^{2}T)$, and horizontal edges give $O(x{\log }^2(xT)\mathrm{/}T)$. At bounded height the pole-subtracted estimate bounds the derivative by $O(1+1\mathrm{/}\mathrm{\mid }s-1\mathrm{\mid })$; on the left edge its integral is $O(\log \log (T+2))$. Only the pole at one is crossed. These edge bounds explain the scale used in the finite-zero proof.