Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 28 / 3

Put $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$. The Euler product gives $F(s)={\sum }_{n\ge 1}\Lambda (n)n^{-s}$ for $\Re s>1$. The elementary nonnegative polynomialtherefore gives the three-four-one zero-free-region argument:The coefficients $\Lambda (n)$ are nonnegative; applying the identity with $u=t\log n$ justifies the inequality term by term.

Take real parts in the global partial-fraction expansion of the zeta logarithmic derivative. For $1<\sigma \le 2$, the digamma function bounds yieldAll terms in the zero sum are nonnegative, because the Euler product and completed functional equation locate the nontrivial zeros in the closed critical strip. The constant terms are bounded: in particular ${\sum }_{\rho }\Re (1/\rho )={\sum }_{\rho }\beta /|\rho {|}^2$ converges. Near the real pole, $F(\sigma )=1/(\sigma -1)+O(1)$.

First exclude zeros on the line one. If $1+it$, $t\ne 0$, were a zero of multiplicity $m\ge 1$, the selected zero term would giveInserted into the positivity inequality, this yields $0\le (3-4m)/(\sigma -1)+O_t(1)$, impossible as $\sigma \downarrow 1$. At $t=0$, zeta has a pole, with a punctured neighbourhood containing no zeros. Thus the line one has no zeros.

Now let $\rho =\beta +i\gamma$ with $|\gamma |\ge 1$ and put $L=\log (|\gamma |+2)$. Select this zero in the formula at $t=\gamma$ and discard every other nonnegative zero kernel. The pole terms at heights $\gamma ,2\gamma$ are bounded. ConsequentlyChoose a fixed positive $a$ small enough that $C_{1}a\le 1/2$ and $1+a/L\le 2$, and set $\sigma =1+a/L$, $h=1-\beta$. Rearranging givesThis is the required logarithmic gap at large height.

For $|\gamma |\le 1$, zeros with $\beta \le 1/2$ already have a fixed gap. The remaining compact region contains finitely many zeros, none on the line one; the pole at one cannot be an accumulation point of zeros. Hence those zeros also have a fixed positive distance from that line. Reduce the absolute constant to cover this bounded-height region and to make the preceding weak inequality strict. The zero-free region of the Riemann zeta function is thereforeThis proof has established boundary nonvanishing as well as the quantitative gap, rather than assuming the former implicitly.

The permitted zero-count bound $N(T)\ll T\log (T+2)$ implies, by partial summation,Indeed, the zeros with $|\rho |\le 1$ contribute a bounded amount since $\xi (0)\ne 0$; the rest contribute $N(T)/T+{\int }_1^{T}N(u)u^{-2}du$. For every zero in the sum, the preceding zero-free region of the Riemann zeta function givesThe supplied truncated explicit formula for the second Chebyshev function therefore impliesChoose $T=\exp (a\sqrt{\log x})$, with any fixed positive $a$ and then, for example, $a=\sqrt{c}$. Both terms have exponential decay in $\sqrt{\log x}$, and the logarithmic factors can be absorbed by slightly decreasing its coefficient. Hence, for some absolute $b>0$,If the truncated formula is taken only at admissible heights and non-prime-power endpoints, choose such heights of the same size and first apply it at half-integers. Monotonicity and the $O(\log x)$ jump of $\psi$ at an integer transfer this bound to all $x$; these convention choices do not change the conclusion.

To pass from prime powers to primes, let $\vartheta (x)={\sum }_{p\le x}\log p$. The higher prime-power contribution satisfiesThus $\vartheta (x)\sim x$. Apply partial summation again:The integral is $O(x/{\log }^{2}x)$: use $\vartheta (t)\ll t$ and split at $\sqrt{x}$. It is therefore smaller than the first term, which is asymptotic to $x/\log x$. This proves the Prime number theorem in the requested form

A fixed zero-free strip changes the zero contribution qualitatively. If every nontrivial zero satisfies $\beta <1-c$, with fixed $0<c<1/2$, the same zero-count estimate and partial summation giveThe truncated explicit formula for the second Chebyshev function then yieldsTake $T=x^c$, or an admissible height of this size. The two contributions are bounded by the same order, proving the prime number theorem error from a fixed zero-free strip:In particular this is $O_{\epsilon }(x^{1-c+\epsilon })$ for every $\epsilon >0$. It is a power saving, stronger than the decay obtained from the logarithmic zero-free region. The triangle-inequality estimate of the zero sum explains the two logarithms; removing them would require additional information or cancellation, not merely the asserted zero-free strip.