Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 150 (Analytic Number Theory)

Begin with the Von Mangoldt divisor identitySumming it for $n\le x$ and reversing the order givesThe given bound $\psi (x)\ll x$ and the Stirling formula therefore imply

Apply partial summation with the weight $1/\log n$. Writing $A(t)=\log t+E(t)$, where $E(t)=O(1)$, givesIndeed, the integral of $E(t)/(t(\log t{)}^2)$ converges, and its tail from $x$ to infinity is $O(1/\log x)$.

Grouping the left side by prime powers yieldsThe full double series over $k\ge 2$ converges. Its tail beyond $x$ is $O(x^{-1/2})$: split at $p=\sqrt{x}$, use a geometric series for $p\le \sqrt{x}$, and compare ${\sum }_{p>\sqrt{x}}{p}^{-2}$ with the corresponding sum over integers. Absorbing its limit into the constant proves the Mertens theorem for reciprocal primes

Putby part a. Double-counting divisibility gives the first moment of the prime omega function:Moreover,soExpanding the square and using the given bound $\pi (x)\ll x/\log x$ now gives the Turán normal-order theorem for distinct prime divisors estimate

By the Chebyshev inequality, the number of $n\le x$ for whichis $O(x/L^{1/2})=o(x)$. Discard the $O(\sqrt{x})$ integers below $\sqrt{x}$. For $\sqrt{x}<n\le x$,and $(\log \log n{)}^{3/4}\sim L^{3/4}$. Hence, for all sufficiently large $x$, every remaining integer counted in the question also satisfies the preceding inequality. Therefore

For $\Re s>1$, the absolutely convergent Dirichlet seriesdefines the Riemann zeta function. To continue it, use partial summation in Stieltjes form:Since $0\le \{x\}<1$, the final integral converges locally uniformly for $\Re s>0$ and is holomorphic there. The displayed expression is consequently a meromorphic function on that half-plane, with its only pole at $s=1$. Since $s/(s-1)$ has residue one there, so does $\zeta$. Agreement in $\Re s>1$ makes this continuation unique by the identity theorem.

The functional equation isEquivalently,

For a proof, letThe Poisson summation formula applied to a Gaussian function gives the theta transformationThe standard Gamma function integral and termwise integration initially give, for $\Re s>1$,Split the integral at one, substitute $u\mapsto 1/u$ in the lower half, and use the theta transformation. The result isThe integral is an entire function of $s$ because $\Theta (u)-1$ decays exponentially. The right side is visibly invariant under $s\mapsto 1-s$, proving both the analytic continuation and the Functional equation of the Riemann zeta function. This is the Mellin representation of the completed Riemann zeta function.

Write the second form of the functional equation asFor $-2\le \sigma \le 2$, the elementary exponential formula for the sine gives, uniformly for $t\ge 4$,The stated Stirling formula givesuniformly on the same strip. The bounded factors $2^{\sigma }{\pi }^{\sigma -1}$ and the cancelling exponentials therefore show thatTaking absolute values in the functional equation proves the Vertical-strip factor in the Riemann zeta functional equation:

For $\sigma >1$, unique prime factorization and absolute convergence give the Euler productEvery factor is nonzero and the product converges to a nonzero limit. Equivalently, the absolutely convergent identityprovides a reciprocal. This proves the Euler-product nonvanishing of the Riemann zeta function.

For $\sigma >1$, put $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$. Its absolutely convergent Dirichlet series andgive the three-four-one zero-free-region argumentThe pole of $\zeta$ at one gives

Suppose $\rho =\beta +i\gamma$ is a zero with $\gamma \ge 4$ and $\beta$ close to one. Apply the supplied Local partial-fraction expansion of the Riemann zeta logarithmic derivative at $\sigma +i\gamma$. Every term has positive real part, so retaining the term belonging to $\rho$ givesAt $\sigma +2i\gamma$ the same expansion gives merely $\Re F(\sigma +2i\gamma )\le O(\log \gamma )$. The zero is included in the supplied disk whenever $1-\beta$ and $\sigma -1$ are sufficiently small. Hence

Set $L=\log \gamma$ and $\sigma =1+a/L$, where $a>0$ is a sufficiently small fixed constant. If $(1-\beta )L$ were smaller than a sufficiently small constant $c>0$, division by $L$ would givea contradiction. Conjugation handles negative $\gamma$. Reducing $c$ to absorb the bounded range proves the classical Zero-free region of the Riemann zeta function

Shrink the constant $c$ from part b if necessary. Put $L=\log |t|$. Zeros in the disk appearing in the supplied partial-fraction formula have $\log |\gamma |=L+O(1/|t|)$, so part b ensuresfor every such zero.

If $\sigma \ge 1+c/(2L)$, absolute convergence of the logarithmic derivative giveswith the region $\sigma \ge 2$ even easier.

It remains to take $1-c/(2L)<\sigma <1+c/(2L)$. SetFor every local zero, both $\Re (s-\rho )$ and $\Re (s_0-\rho )$ are positive and comparable, while $|s-s_0|\ll 1/L$. The partial-fraction formula at $s_0$, together with the preceding Euler-product bound, givesSince $\Re (s_0-\rho )\gg 1/L$, it follows thatSubtracting the partial-fraction formulas at $s$ and $s_0$ now yieldsTherefore the logarithmic derivative inside the zeta zero-free region satisfiesthroughout the required half-width region.

Let $⟨x⟩$ denote the distance from $x$ to the nearest prime power. For $x\ge 2$ not an integer and $T\ge 2$, the truncated Riemann–von Mangoldt explicit formula isZeros are counted with multiplicity. The constant term comes from $s=0$, the logarithm collects the trivial zeros $-2,-4,\dots$, and the finite sum contains the nontrivial zeros. Enlarging the implied constant covers $1\le T<2$.

Assume first the Riemann hypothesis. Replace any $x\ge 2$ by $y=\lfloor x\rfloor +1/2$; then $\psi (x)=\psi (y)$ and $⟨y⟩\ge 1/2$. Take $T=y$ in part a. Every nontrivial zero has real part $1/2$, and the Riemann–von Mangoldt formula impliesConsequentlywhile both truncation errors in part a are $O((\log y{)}^2)$. Thuswhich implies the stated $O_{ϵ}(x^{1/2+ϵ})$ estimate.

Conversely, suppose that estimate holds for every $ϵ>0$. For $\Re s>1$, partial summation givesGiven any $s$ with $\Re s>1/2$, choose $ϵ<\Re s-1/2$. The error hypothesis makes the last integral locally uniformly convergent there, so it supplies a holomorphic continuation ofto the half-plane $\Re s>1/2$. A zero of $\zeta$ in that half-plane would create a pole of its logarithmic derivative, so none exists. The Functional equation of the Riemann zeta function reflects every nontrivial zero with real part below $1/2$ to one above $1/2$. All nontrivial zeros must therefore lie on the critical line, proving the Riemann hypothesis equivalence for the second Chebyshev function.