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

The Truncated Perron formula says that if $F(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ converges absolutely for $\Re s>{\sigma }_a$, then for $c>{\sigma }_a$, $T\ge 2$, and $x$ not an integer,Take $a_n=\Lambda (n)$ and $c=1+1/\log x$. The logarithmic derivative identity gives $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$. Since $x-1/2$ is an integer, $|x-n|\ge 1/2$ for every integer $n$. In the range $x/2<n<2x$,and hence the contribution there isThe ranges $n\le x/2$ and $n\ge 2x$ are bounded by the same quantity using absolute convergence and $-{\zeta }^{\prime }(c)/\zeta (c)\ll \log x$. Therefore

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