The <Truncated Perron formula> says that if $F(s)=\sum_{n\geq1}a_nn^{-s}$ converges absolutely for $\Re s>\sigma_a$, then for $c>\sigma_a$, $T\geq2$, and $x$ not an integer,
$$
\sum_{n\leq x}a_n
=\frac1{2\pi i}\int_{c-iT}^{c+iT}F(s)\frac{x^s}{s}\,ds
+O\left(\sum_{n\geq1}|a_n|\left(\frac xn\right)^c
\min\left\{1,\frac1{T|\log(x/n)|}\right\}\right).
$$
Take $a_n=\Lambda(n)$ and $c=1+1/\log x$. The <logarithmic derivative> identity gives $F(s)=-\zeta'(s)/\zeta(s)$. Since $x-1/2$ is an integer, $|x-n|\geq1/2$ for every integer $n$. In the range $x/2<n<2x$,
$$
|\log(x/n)|\asymp\frac{|x-n|}{x},
$$
and hence the contribution there is
$$
\ll\frac{x\log x}{T}\sum_{x/2<n<2x}\frac1{|x-n|}
\ll\frac{x(\log x)^2}{T}.
$$
The ranges $n\leq x/2$ and $n\geq2x$ are bounded by the same quantity using absolute convergence and $-\zeta'(c)/\zeta(c)\ll\log x$. Therefore
$$
\sum_{n\leq x}\Lambda(n)
=-\frac1{2\pi i}\int_{1+1/\log x-iT}^{1+1/\log x+iT}
\frac{\zeta'(s)}{\zeta(s)}\frac{x^s}{s}\,ds
+O\left(\frac{x(\log x)^2}{T}\right).
$$

Solved by gpt-5.6-sol high.

Solution

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

{c}

A truncated Perron integral over $c-iT$ to $c+iT$ recovers a summatory function with an error controlled by
$$
\sum_n |a_n|\left(\frac xn\right)^c
\min\left(1,\frac1{T|\log(x/n)|}\right).
$$


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