Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 22 (Analytic Number Theory)

Write $\vartheta (x)={\sum }_{p\le x}\log p$ for the Chebyshev theta function and $\psi (x)={\sum }_{p^j\le x}\log p$ for the Second Chebyshev function. Throughout, $p$ denotes a prime and all logarithms are natural. Each prime $n<p\le 2n$ divides the central binomial coefficient, soApply this at $n=1,2,4,\dots ,2^{r-1}$ and telescope: $\vartheta (2^r)\le (2^{r+1}-2)\log 2$. Bounding a general $x$ by the next power of two proves $\vartheta (x)=O(x)$. Splitting the prime-counting function at $\sqrt{x}$ now givesThis is the required Chebyshev estimate from central binomial coefficients, with no use of a prime asymptotic.

The higher prime powers also satisfyFor an integer $N$, the prime factorization of a factorial givesThe Stirling formula, or integral comparison for $\log t$, says $\log N!=N\log N-N+O(\log N)$. After division by $N$ we obtain ${\sum }_{p^j\le N}(\log p)/p^j=\log N+O(1)$. MoreoverSubtract this bounded contribution and replace $N$ by $\lfloor x\rfloor$. The factorial proof of Mertens first theorem therefore yields

For fixed $\lambda >1$, apply partial summation to $A(t)$ and $f(t)=(\log t{)}^{\lambda -1}$:The main contribution isThe bounded error in $A(t)$ contributes $O((\log x{)}^{\lambda -1})$ at the endpoint and in the integral, since ${\int }_2^x(\log t{)}^{\lambda -2}dt/t=O_{\lambda }((\log x{)}^{\lambda -1})$. Thus the positive logarithmic moments of reciprocal primes satisfy

First continue the Riemann zeta function into $\Re s>0$. Summing the integral of the integer-part function gives, initially for $\Re s>1$,The last integral defines a holomorphic function for $\Re s>0$, so this continuation has only a simple pole at one, of residue one.

For $\sigma >1$, logarithmic expansion of the Euler product and $3+4\cos v+\cos 2v=2(1+\cos v{)}^2\ge 0$ giveThis proves the three-four-one product proof of zeta boundary nonvanishing inequalityIf $t\ne 0$ and $\zeta (1+it)=0$ to order $r\ge 1$, the middle factor is $O((\sigma -1{)}^{4r})$, the first is $O((\sigma -1{)}^{-3})$, and the last remains bounded. Their product tends to zero, a contradiction. At $t=0$ there is a pole, not a zero. The zeta function has no zero on $\Re s=1$.

Put $I(x)={\int }_0^x\psi (u)du={\sum }_{n\le x}(x-n)\Lambda (n)$, where $\Lambda$ is the Von Mangoldt function. Logarithmic differentiation of the Euler product gives $-{\zeta }^{\prime }(s)/\zeta (s)={\sum }_n\Lambda (n)n^{-s}$ for $\Re s>1$. The first integral of the Chebyshev function is consequentlyTo check its kernel, evaluate the contour integral of $y^s/(s(s+1))$ by closing to the left for $y>1$ and to the right for $0<y<1$. The residues at zero and minus one give $1-y^{-1}$ in the first case and zero in the second. Multiplication by $x$ and substitution $y=x/n$ therefore produce $(x-n{)}_{+}$. On the line $c>1$, the Dirichlet series and vertical integral have absolute convergence, justifying their interchange; at $x=n$ the kernel is zero on either side.

Here is the smoothing step that establishes the requested asymptotic. Shift the line of this contour integral to the left. The pole of $-{\zeta }^{\prime }/\zeta$ at one has residue $+1$; at each Nontrivial zero of the Riemann zeta function $\rho$ of zero multiplicity $m_{\rho }$ its residue is $-m_{\rho }$. The denominators also contribute at zero and minus one, and the trivial zeros of the Riemann zeta function occur at $-2,-4,\dots$. Summing these residues yields the smoothed explicit formula for the Chebyshev functionZeros are repeated with zero multiplicity. For the limits of the contour integral, choose horizontal heights avoiding zero ordinates, where the logarithmic derivative on each fixed strip is $O((\log T{)}^2)$. The factor $s(s+1)$ makes their integrals tend to zero. On left edges chosen a fixed distance from the trivial zeros, the functional equation of the Riemann zeta function controls the logarithmic derivative by $O(\log (|s|+2))$; for fixed $x>1$ the remaining vertical integral vanishes as that edge moves left. This is why the extra integration of $\psi$ is useful.

The Riemann–von Mangoldt formula, stated in Question 3, gives $N(T)=O(T\log T)$, henceIndeed each dyadic height block contributes $O((\log T)/T)$. The zero-free line just proved, the nonvanishing Euler product on its right, and the functional equation of the Riemann zeta function put every Nontrivial zero of the Riemann zeta function in $0<\Re \rho <1$. Therefore each $x^{\rho -1}$ tends to zero, with absolute value at most one for $x\ge 1$. Divide the explicit formula by $x^2$ and use dominated convergence on the zero terms with absolute convergence. The remaining displayed terms are $o(x^2)$, and we obtain

Monotonicity of $\psi$ removes the smoothing. For $0<h<1$,After division by $x$, the limits of the two bounds are $1-h/2$ and $1+h/2$. Let $h\downarrow 0$ to get $\psi (x)\sim x$. Question 1 gives $\psi (x)-\vartheta (x)=O(\sqrt{x}\log x)=o(x)$, so $\vartheta (x)\sim x$. Partial summation now givesThe integral is $O(x/(\log x{)}^2)$, by splitting at $\sqrt{x}$ and using $\vartheta (t)=O(t)$. Inverting this at $x=p_n$, we have $n\sim p_n/\log p_n$ and hence $\log n\sim \log p_n$. Thus the $n$th prime satisfies

Use the Jacobi theta function $\Theta (t)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}t}$, $t>0$. Its transformation law follows directly from the Gaussian Fourier transform. With the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(u)e^{-2\pi i\xi u}du$,For example, differentiating the transform and using integration by parts gives ${\hat{f}}^{\prime }(\xi )=-(2\pi \xi /t)\hat{f}(\xi )$; its value at zero is the Gaussian integral $t^{-1/2}$. Periodize the Gaussian. Its Fourier coefficients are these transforms at integer frequencies, and both series converge absolutely, so evaluating its Fourier series at zero proves the Poisson summation formula here:

For $\Re s>1$, integrating the exponentially convergent series termwise gives the Mellin representation of the completed Riemann zeta functionSplit at one and substitute $u=1/t$ in the lower part, using the transformation just proved. The contribution of $t^{-1/2}-1$ is $1/(s-1)-1/s$. We obtain the pole-subtracted theta integral for the completed zeta functionThe integral is entire: $\Theta (t)-1$ decays exponentially, uniformly dominating the integrand and all its $s$ derivatives on compact sets. Both the rational term and the integral are invariant under $s\mapsto 1-s$. This proves the meromorphic identityEquivalently the Riemann xi functionis entire and satisfies $\xi (s)=\xi (1-s)$. The reflection and duplication formulas for the Gamma function turn the same identity intounderstood by meromorphic continuation at removable exceptional points. This establishes the functional equation of the Riemann zeta function, rather than assuming it.

If $N(T)$ counts zeros $\rho =\beta +i\gamma$ in the critical strip with $0<\gamma \le T$, including zero multiplicity, the Riemann–von Mangoldt formula isAt a zero ordinate one may instead use the conventional half-weight boundary count; that only alters the stated error. In particular $N(T)\sim T\log T/(2\pi )$.

To invert without assuming simple zeros, fix $0<\epsilon <1$ and set $T_n^{\pm }=(1\pm \epsilon )2\pi n/\log n$. Then $\log T_n^{\pm }/\log n\to 1$, soFor large $n$, $N(T_n^{-})<n<N(T_n^{+})$, and hence $T_n^{-}<{\gamma }_n\le T_n^{+}$. Let $\epsilon \downarrow 0$. The asymptotic inversion of the zeta zero count givesAs in the zero-counting formula, the sequence lists ordinates with zero multiplicity; “increasing” means nondecreasing here. The argument does not require the Riemann hypothesis or simplicity of zeros.

The Selberg upper-bound sieve replaces an exact but long inclusion-exclusion expansion by a short nonnegative square, and then minimizes its main term. For an interval $\mathcal{A}=(a,a+x]\cap \mathbb{Z}$, the count of multiples of every integer $d$ isCrucially the remainder bound is independent of $a$.

Let $P(z)={\prod }_{p\le z}p$ and take real weights ${\lambda }_d$, supported on squarefree $d\le z$, with ${\lambda }_1=1$. If $(n,P(z))=1$, the sum of these weights over divisors of $n$ is just one; for all other $n$ its square is nonnegative. ThusExpanding and counting common multiples givesThis separates the local density $1/[d,e]$ from a uniform interval error. The weights will be chosen to minimize $Q$.

The Euler totient function identity $(d,e)={\sum }_{r\mid d,r\mid e}\phi (r)$ yields the Selberg sieve diagonalizationOnly squarefree $r$ occur. Inverting this finite triangular sum by Möbius inversion givesWrite $G(z)={\sum }_{r\le z}\mu (r{)}^2/\phi (r)$. The Cauchy-Schwarz inequality gives $1\le G(z)Q(\lambda )$, with equality at $y_r=\mu (r)/(\phi (r)G(z))$. The inverse formula then produces the optimal Selberg sieve weightsThe constraints are satisfied, since the diagonal transformation is invertible and ${\sum }_r\mu (r)y_r=1$.

We need the leading constant, not just $G(z)\gg \log z$. Here is an elementary proof of the Selberg sieve denominator asymptotic. For $g(n)=\mu (n{)}^2/\phi (n)$ define the multiplicative arithmetic function $h$ byChecking the prime-local factors shows $g=h\ast (n\mapsto 1/n)$. Also ${\sum }_n|h(n)|\log (2n)<\infty$: the local factors for absolute values differ from one by $O(p^{-2})$, and their logarithmic first moments sum $O((\log p)/p^2)$. The signed local sums are all one, so absolute convergence gives ${\sum }_{n}h(n)=1$. Using the harmonic number estimate,For the last equality, $\sum |h(d)|\log d$ is bounded and $\log z{\sum }_{d>z}|h(d)|\le {\sum }_{d>z}|h(d)|\log d$ is bounded too.

The optimal Selberg weights have modulus at most one. To prove this here, for squarefree $d$ use $d/\phi (d)={\sum }_{e\mid d}1/\phi (e)$. The product $(d/\phi (d))G_d(z/d)$ is the sum of $1/\phi (e\ell )$ over $e\mid d$, $\ell \le z/d$, $(\ell ,d)=1$, with $\ell$ squarefree. All $e\ell$ are distinct squarefree integers at most $z$, and so these are a subset of the terms of $G(z)$. The displayed weight formula now gives $|{\lambda }_d|\le 1$. ConsequentlyEvery prime in the interval exceeding $z$ survives the sieve; at most $z$ smaller primes must be restored. Choose $z=\lfloor \sqrt{x}/(\log x{)}^2\rfloor$. Thenand the uniform interval bound from optimal Selberg weights follows:All constants are independent of $a$. Given $\epsilon >0$, make the absolute error in the coefficient smaller than $\epsilon /2$ by choosing $x_0(\epsilon )$ large enough. Uniformly for every $a>0$ and $x>x_0(\epsilon )$,

Fix a modulus $q$ and a residue class $a$ with $(a,q)=1$. The Dirichlet characters modulo $q$ are the one-dimensional characters of a representation of $(\mathbb{Z}/q\mathbb{Z}{)}^{\times }$, extended by zero on nonunits. Let ${\chi }_0$ be the principal Dirichlet character.

For every nonprincipal Dirichlet character the partial sums ${\sum }_{n\le u}\chi (n)$ are bounded: the sum over a full period is zero, and a remaining incomplete period has bounded length. Partial summation therefore continues its Dirichlet L-function holomorphically to $\Re s>0$ byThis shows in particular that it has no pole at one. By contrasthas a simple pole there. We use the stipulated nonvanishing $L(1,\chi )\ne 0$ for every real nonprincipal Dirichlet character, and establish the remaining nonvanishing as follows.

For real $\sigma >1$, take the product over all Dirichlet characters. The Euler product logarithms and Orthogonality of Dirichlet characters giveThus this product is real and at least one. If a nonreal Dirichlet character had $L(1,\chi )=0$, complex conjugation would also give $L(1,\overline{\chi })=0$. These are distinct factors. Two zeros and only one possible simple pole force the whole product to tend to zero as $\sigma \downarrow 1$, contradicting the lower bound. Hence all nonprincipal Dirichlet characters have $L(1,\chi )\ne 0$, under the given real-character assumption.

For $\sigma >1$, absolute convergence of each Euler product giveswith bounded remainder as $\sigma \downarrow 1$: the terms of prime-power exponent at least two have total modulus at most ${\sum }_{n\ge 2}1/(n(n-1))$. Since a nonprincipal $L$ defines a holomorphic function that is nonzero near one, its Euler-product logarithm differs from a local holomorphic logarithm by a fixed integer multiple of $2\pi i$ on a short real interval. It is therefore bounded there. For the principal Dirichlet character the pole instead givesApply the Dirichlet character orthogonality relation at each prime. The prime-character sum near one yieldsThis tends to infinity. A finite set of primes would have a bounded sum as $\sigma \downarrow 1$, so every coprime residue class contains infinitely many primes. This proves the Dirichlet theorem on primes in arithmetic progressions from the specified assumption. In fact it proves divergence of the corresponding reciprocal-prime series and gives Dirichlet density $1/\phi (q)$.

Finally, for real $s>1$ each prime satisfiesThe integrands are nonnegative, so Tonelli theorem permits summation inside the integral. At a fixed $t$, exactly $\pi (t,q,a)$ primes from the residue class have entered. ThereforeBoth sides are finite, since they are bounded respectively by ${\sum }_{n\ge 2}{n}^{-s}$ and $s{\int }_1^{\infty }{t}^{-s}dt$. The values of a counting function at its isolated jump endpoints do not change this integral.