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

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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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.

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For $\Re s>1$, absolute convergence permits a shift of the complex variable:

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PutFor $\Re s>1$, partial summation givesThe last integral is holomorphic for $\Re s>1/2$. Thus the logarithmic derivative on the left continues meromorphically to that half-plane with no pole except $s=1+iu$.

A zero $\rho$ of $\zeta$ with $\Re \rho >1/2$ would make $-{\zeta }^{\prime }(s-iu)/\zeta (s-iu)$ singular at $s=\rho +iu$, a contradiction unless $\rho =1$, which is a pole rather than a zero. Therefore no nontrivial zero lies to the right of the critical line. The Functional equation of the Riemann zeta function reflects zeros across that line, so none lies to its left either. Every nontrivial zero lies on the critical line, proving the Riemann hypothesis. This is the Twisted Von Mangoldt estimate implying the Riemann hypothesis.

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