Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 25 / 2

Absolute convergence and the Fundamental theorem of arithmetic give the Euler productFor a finite set of primes, expand the geometric factors: their product sums $n^{-s}$ over integers whose prime factors lie in that set. Let the finite sets increase through all primes. Absolute convergence permits passage to the limit and recovers the full Dirichlet series. Moreover ${\sum }_{p,m\ge 1}|p^{-ms}|/m<\infty$, so the logarithm converges and the product has no zeros there.

The same absolutely convergent logarithm, and $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2\ge 0$, giveThis is the product version of the three-four-one zero-free-region argument. It also proves there are no zeros on $\Re s=1$: if $\zeta (1+i{t}_0)=0$ for $t_0\ne 0$, its factor has order at least four as $\sigma \downarrow 1$, while the real pole contributes only order minus three and the $2t_0$ factor remains bounded. The displayed left side would tend to zero, a contradiction. At $t_0=0$ there is a pole, not a zero.

For large $|t|$, put $L=\log (|t|+2)$. The Hardy-Littlewood approximation to the Riemann zeta function at $x\asymp |t|$ gives $|\zeta (\sigma +it)|\ll L$ for $1-2/L\le \sigma \le 3$; its finite sum is bounded by ${\sum }_{n\le x}{n}^{-1+2/L}\ll L$ and the integral term is bounded. The Cauchy estimate for derivatives on circles of radius comparable to $1/L$ consequently gives $|{\zeta }^{\prime }(\sigma +it)|\ll L^2$ for $1-L^{-9}\le \sigma \le 2$.

Take ${\sigma }_0=1+a{L}^{-9}$ with a small fixed $a>0$. The product inequality, $\zeta ({\sigma }_0)\ll L^9/a$, and $|\zeta ({\sigma }_0+2it)|\ll L$ implyIf $1-c{L}^{-9}\le \sigma \le {\sigma }_0$, integration of the derivative along the horizontal segment changes this value by at most $C(a+c)L^{-7}$. Choose $a$ sufficiently small that $Ca<c_0{a}^{3/4}/4$, then $c\le a$ sufficiently small. The lower bound remains a positive multiple of $L^{-7}$. For ${\sigma }_0\le \sigma \le 2$, the same product inequality, $\zeta (\sigma )\le \zeta ({\sigma }_0)$, and the near-one upper bound give that lower bound directly. For $\sigma \ge 2$, the reciprocal Euler product gives $|1/\zeta (s)|\le \zeta (2)$.

Finally the no-zero result on $\Re s=1$, compactness at bounded heights and the regular reciprocal at the pole allow a further fixed reduction of $c$ to include bounded $t$. We have proved the weak logarithmic zero-free region for the Riemann zeta functionThe reciprocal at $s=1$ is its holomorphic extension, equal to zero.

Write $\psi (x)={\sum }_{n\le x}\Lambda (n)$, where $\Lambda$ is the Von Mangoldt function. The Riemann–von Mangoldt explicit formula, in its symmetric limiting form for $x>1$, isHere nontrivial zeros are counted with multiplicity, the limit is taken symmetrically through admissible heights, and ${\psi }_0$ assigns half weight at a jump. Its difference from $\psi$ is at most $(\log x)/2$. The Euler product and part (a) exclude zeros with real part at least one. The Functional equation of the Riemann zeta function leaves only the trivial zeros of the Riemann zeta function at negative even integers outside $0<\Re \rho <1$; their already displayed logarithmic correction is $O(x^{-2})$ for large $x$. These terms and the constant are negligible in the requested asymptotic error, rather than literally absent from the exact formula.

A useful truncated explicit formula for the second Chebyshev function is, uniformly for $2\le T\le x$,One may first take a height in $[T,T+1]$ separated from zeros and then adjust to $T$ using the local count. The local zero count for the Riemann zeta function isIt includes multiplicity and is uniform in real $t$. It follows by subtracting the Riemann–von Mangoldt formula at endpoints, handling bounded heights separately and using conjugation for negative heights. Both closed endpoints change the count only by another local bound.

By part (a), every nontrivial zero $\rho =\beta +i\gamma$ with $|\gamma |\le T$ satisfies $\beta \le 1-c_1/{\log }^{9}T$ for large $T$, after reducing the positive constant. The local zero count for the Riemann zeta function givesThere are finitely many zeros at bounded height, none at zero or at one, so that part of the sum is bounded. The truncated explicit formula for the second Chebyshev function now yieldsBalance the exponent losses $c_1\log x/(\log T{)}^9$ and $\log T$ by choosing $\log T=(c_1\log x{)}^{1/10}$. This is the optimal order obtainable from these two errors: making either exponent larger forces the other smaller. Thus, for a positive constant $c_2$,The logarithmic prefactor can be absorbed by reducing $c_2$. This is the prime number theorem error from a logarithmic zero-free region with ninth-power width.

Under the Riemann hypothesis, $|x^{\rho }|=x^{1/2}$. The same reciprocal-zero sum bounds the zero contribution by $O(x^{1/2}{\log }^{2}T)$. Taking $T=x$ makes the truncation error $O({\log }^{2}x)$, so