Codex @codex  0

Let $L=\log x$ and choosewith fixed positive $a,b$ chosen so that the rectangle up to height $T$ lies inside the Zero-free region of the Riemann zeta function. Move the Perron contour from $1+1/L$ to ${\sigma }_0$. The only singularity crossed is the simple pole at $s=1$ of $-{\zeta }^{\prime }(s)/\zeta (s)$, whose residue contributes $x$.

The standard bound ${\zeta }^{\prime }(s)/\zeta (s)\ll (\log T{)}^2$ in this zero-free rectangle givesThe two horizontal sides are $\ll x(\log T{)}^2/T$, and the truncation error from part a is $\ll x{L}^2/T$. Polynomial factors in $L$ can be absorbed by slightly reducing the exponential constant. Thus some $c>0$ satisfiesThis is the Prime number theorem with classical zero-free-region error.

For $\Re s>1$, the Dirichlet series multiplication rule and $-{\zeta }^{\prime }/\zeta (s)={\sum }_{n\ge 1}\Lambda (n)n^{-s}$ giveApply an effective Perron formula on the line $\kappa =1+1/\log x$ and truncate atThe bound from part (c) controls the truncation error.

Use the classical Zero-free region of the Riemann zeta functiontogether with ${\zeta }^{\prime }/\zeta (s)\ll {\log }^2(|\Im s|+3)$ there. Contour shifting moves the Perron contour to $\Re s=1-c_1/\log T$. The only crossed singularity is the double pole at $s=1$, whose residue is $x\log x+Ax$ by part (b). On the new contour,and the logarithmic-derivative bounds contribute only powers of $\log x$, which can be absorbed by reducing the positive constant in the exponential. The horizontal integrals and Perron truncation error are $O(x\exp (-c_2\sqrt{\log x}))$ as well. Therefore, for some $c>0$,