Let $L=\log x$ and choose
$$
T=\exp(a\sqrt L),
\qquad
\sigma_0=1-\frac b{\sqrt L},
$$
with 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 $-\zeta'(s)/\zeta(s)$, whose residue contributes $x$.

The standard bound $\zeta'(s)/\zeta(s)\ll(\log T)^2$ in this zero-free rectangle gives
$$
\int_{\sigma_0-iT}^{\sigma_0+iT}
\frac{\zeta'(s)}{\zeta(s)}\frac{x^s}{s}\,ds
\ll x^{\sigma_0}(\log T)^3
\ll x\exp(-b\sqrt L)L^{3/2}.
$$
The two horizontal sides are $\ll x(\log T)^2/T$, and the truncation error from part a is $\ll xL^2/T$. Polynomial factors in $L$ can be absorbed by slightly reducing the exponential constant. Thus some $c>0$ satisfies
$$
\sum_{n\leq x}\Lambda(n)
=x+O\left(x\exp(-c\sqrt{\log x})\right).
$$
This is the <Prime number theorem with classical zero-free-region error>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 150 / 2 / b / Solution

{c}

The classical zero-free region and a truncated Perron contour give
$$
\sum_{n\leq x}\Lambda(n)
=x+O\left(xe^{-c\sqrt{\log x}}\right)
$$
for some constant $c>0$.


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 Incoming links: Prime number theorem with classical zero-free-region error