Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 25 (The Riemann Zeta Function)

For $\Re s>1$, the Riemann zeta function is the absolutely convergent Dirichlet series $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$. Apply Abel summation to the tail, with counting function $\lfloor w\rfloor -\lfloor x\rfloor$. The upper boundary term tends to zero, givingSubstitute $\lfloor w\rfloor =w-\{w\}$ and integrate the $w$ term. This proves the fractional-part continuation formula for the Riemann zeta functionThe endpoint convention is valid whether or not $x$ is an integer. Since $0\le \{w\}<1$, the last integral converges locally uniformly for $\Re s>0$, including after differentiation on compact subsets. It therefore defines a holomorphic function there. The other terms are entire except for $x^{1-s}/(s-1)$. By the identity theorem for holomorphic functions, the formula supplies a meromorphic continuation with exactly one pole in $\Re s>0$: a simple pole at $s=1$ of residue one.

We use the Van der Corput sum-integral lemma. Put $e(u)=e^{2\pi iu}$ and $I_h={\int }_a^{b}e(f(x)-hx)dx$. The Fourier series of the periodization of $1_{[a,b]}(x)e(f(x))$ giveswhere integer endpoints have half weight. This is the Dirichlet-Jordan convergence theorem for a piecewise smooth, or more generally bounded-variation, periodic function. Here $f$ is $C^1$, so the periodized function has bounded variation. Changing to the requested endpoint convention costs at most one.

Write $u=f^{\prime }$. For $h\ne 0$, $|u-h|\ge |h|-\delta >0$. Since $u$ is continuous and monotone, the reciprocal has bounded variation, and integration by parts in the Riemann-Stieltjes sense yieldsThe variation of the reciprocal is at most $2\delta /(h^2-{\delta }^2)$. Summing over $h\ne 0$ gives $O((1-\delta {)}^{-1})$, separating $|h|=1$ and using convergence of ${\sum }_{h\ge 2}{h}^{-2}$. For each endpoint, useThe symmetric partial sums of the first term are a constant multiple of ${\sum }_{h=1}^H\sin (2\pi hx)/h$, uniformly bounded in $H$ and $x$; this standard Fourier series bound follows by splitting at $h\asymp 1/\Vert x\Vert$ and applying Abel summation to the remaining sine sum. The second term is absolutely summable with bound $O((1-\delta {)}^{-1})$. The same bound therefore holds for the whole sum of the $h\ne 0$ integrals. Since $I_0$ is the ordinary integral,No second derivative is required; monotonicity supplies the needed variation estimate.

For the Hardy-Littlewood approximation to the Riemann zeta function, take $f(w)=-t\log w/(2\pi )$. On $w\ge x\ge |t|/\pi$, $|f^{\prime }(w)|\le 1/2$ and $f^{\prime }$ is monotone. The proved lemma says that the difference between the partial sum of $w^{-it}$ and its integral over $(x,Y]$ is $O(1)$ uniformly in $Y$. Weighted Abel summation with the decreasing weight $w^{-\sigma }$ then makes the weighted difference $O(x^{-\sigma })$, since its total variation on $[x,\infty )$ is $x^{-\sigma }$. Initially for $\sigma >1$, the tail integral is $x^{1-s}/(s-1)$. The bounded primitive of the discrepancy gives a locally uniformly convergent weighted discrepancy integral for every $\sigma >0$, continuing the identity to that region. Thus, away from the pole,If $t=0$ the ordinary sum-integral comparison supplies the same estimate. At $s=1$ the formula is understood meromorphically. It approximates the Riemann zeta function by a finite Dirichlet polynomial, transfers exponential sum estimates to bounds in the critical strip, yields elementary near-one bounds for $\zeta$ and its derivative, and supports estimates for the mean value of Dirichlet polynomials and numerical calculations.

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

Differentiate the locally uniformly convergent logarithm of the Euler product. Its logarithmic derivative isThe differentiated sum is absolutely convergent, since $\Lambda (n)\le \log n$. Taking the real parts at the three heights givesEach summand is nonnegative because its bracket is $2(1+\cos (t\log n){)}^2$. This proves the derivative form of the three-four-one zero-free-region argument.

Here is a quantitative local form of the Landau zero-free-region theorem. Let $t\ge 2$, $0<\eta \le 1/4$, $M\ge 2$, and suppose $|\zeta (z)|\le M$ on the two closed discs of radius $\eta$ centred at $1+\eta /4+it$ and $1+\eta /4+2it$. Then every zero $\beta +it$ satisfiesfor an absolute positive $c$. In particular, if upper bounds of this form hold locally for every large height, they give a zero-free region of this width. The discs are away from the pole, and the Euler product excludes zeros to the right of one.

We state precisely the permitted local logarithmic-derivative lemma. If $f$ is holomorphic on a neighborhood of $|z-z_0|\le R$, $f(z_0)\ne 0$, and $|f|\le M$, then for $|z-z_0|\le R/3$ away from zeros,The zeros are counted with multiplicity. This standard disc estimate, which may be assumed here, follows by factoring nearby zeros and applying a Cauchy estimate for derivatives to the remaining logarithm. When every zero has real part at most one and $\Re z>1$, the zero terms have nonnegative real parts, giving the required lower bound. The estimate with fixed radius ratios is also recorded as Lemma 24.17 in Montgomery and Vaughan's general treatment.

The reciprocal Euler product gives $|\zeta (1+\eta /4+ijt)|\ge 1/\zeta (1+\eta /4)\gg \eta$ for $j=1,2$. Put $B=1+\log (M/\eta )$. The lemma's error on both discs is therefore $O(B/\eta )$. Let $d=1-\beta$. If $d\ge \eta /24$, the desired conclusion already holds after reducing $c$. Otherwise $\beta +it$ is among the local zeros and, for $1<\sigma \le 1+\eta /4$,All other zero terms may be discarded because their real parts are nonnegative. The simple pole at one gives $-{\zeta }^{\prime }(\sigma )/\zeta (\sigma )=(\sigma -1{)}^{-1}+O(1)$. Insert these inequalities into part (a):There is no zero on the line one by the argument in Question 2(a), so $d>0$. Choose $\sigma =1+6d$, which lies in the indicated range. The left side is $(4/7-3/6)/d=1/(14d)$. Hence $d\gg \eta /B$, proving the theorem. The logarithm of the upper bound, rather than the upper bound itself, is what enters the zero-free width.

Put $L=\log t$ and $\ell =\log L$, for sufficiently large $t$. Apply the Landau zero-free-region theorem withwhere $a>0$ is fixed and small enough that $\eta \le 1/4$. On its two discs, the real part is at least $1-3\eta /4$ and the imaginary part is comparable to $t$. The given Richert bound for the Riemann zeta function therefore gives, on the part left of one,On the part right of one, the separately given $O({\log }^{2/3}t)$ bound gives the same conclusion. We may thus choose $M=L^A$ for one fixed sufficiently large $A$. Also $\log (1/\eta )=O(\ell )$, so the logarithmic term in the Landau zero-free-region theorem is $O(\ell )$. Its conclusion isfor large $t$ and a sufficiently small positive $c$. Complex conjugation supplies negative heights. This is the Vinogradov-Korobov zero-free region. Only the stated Richert upper bounds, the Euler product, the pole at one and the proved Landau zero-free-region theorem were used; no prior zero-free-region theorem was assumed.

Put $Z=\lfloor N^{2/5}\rfloor$ and $S={\sum }_{N<n\le N+M}{n}^{-it}$. For any positive integer shift $h\le Z^2$, translating the interval changes its sum by at most $2h$. Averaging the shifts $h=xy$ gives the bilinear shift averaging for a logarithmic phase identitySince $xy/n\le N^{-1/5}$, the alternating Taylor expansion of the logarithm has remainder at most $(xy/n{)}^{r+1}/(r+1)$. ThusFor $r=\lfloor 5.01\log t/\log N\rfloor$, we have $r+1>5.01\log t/\log N$, hence the error is at most $t^{-1/500}$. The exponential function on an imaginary argument changes by at most the change in that argument. ThereforeUse $Z^2\asymp N^{4/5}$ and $N<n\le N+M\le 2N$. Taking absolute values provesThe boundary error comes from integer shifts, so this argument also covers intervals shorter than a shift. Here $x,y$ range over positive integers; no zero term is needed.

The Vinogradov mean value isBy orthogonality of integer Fourier modes, it counts the ordered integer solutions of ${\sum }_{i=1}^k{x}_i^j={\sum }_{i=1}^k{y}_i^j$ for $1\le j\le r$, with every coordinate between one and $\lfloor Z\rfloor$. The diagonal solutions $y_i=x_i$ give $J_{k,r}(Z)\ge \lfloor Z{\rfloor }^k$.

Put $D=r(r+1)/2$. The moment vector $({\sum }_i{x}_i^j{)}_{j=1}^r$ has at most ${\prod }_{j=1}^r(k\lfloor Z{\rfloor }^j+1){\ll }_{k,r}{Z}^D$ possible values. If $R(v)$ counts the tuples with vector $v$, then ${\sum }_{v}R(v)=\lfloor Z{\rfloor }^k$ and $J_{k,r}(Z)={\sum }_{v}R(v{)}^2$. The Cauchy-Schwarz inequality gives $J_{k,r}(Z){\gg }_{k,r}{Z}^{2k-D}$. Combining the two lower bounds, with one common positive constant for large $Z$, yields

Here is an explicit way that upper bounds enter the Vinogradov mean-value method for a bilinear exponential sum. Write $U={\sum }_{x,y\le Z}e({\sum }_j{\alpha }_j{x}^j{y}^j)$ and $L_j=k\lfloor Z{\rfloor }^j$. Two applications of the Holder inequality, followed by grouping equal differences of moment vectors, giveAt integer ${\alpha }_{j}v$ the minimum is defined as $2L_j+1$; $\Vert u\Vert$ means distance to the nearest integer. To explain the mean-value factor, let $r(v)$ count pairs of $k$-tuples with prescribed moment difference. It is an autocorrelation of $R$, so $r(v)\le {\sum }_{w}R(w{)}^2=J_{k,r}(Z)$ by Cauchy-Schwarz inequality. The first Holder inequality groups the $y$ tuples, and the second groups the $x$ tuples, providing the two factors $J$. The remaining sums over moment differences are bounded by the displayed finite geometric series estimates. Good Vinogradov mean value upper bounds, together with rational approximation or spacing bounds for ${\alpha }_j$, therefore give cancellation in $U$. The mean-value estimate alone does not force cancellation for arbitrary coefficients: when all ${\alpha }_j$ are integers, $U=\lfloor Z{\rfloor }^2$. For the paper take $Z=N^{2/5}$ and the specified ${\alpha }_j$.

Set $\epsilon =1-\sigma \in [0,1)$ and $L=\log (t+2)$. By the Hardy-Littlewood approximation to the Riemann zeta function at $x=t$, it is enough to bound ${\sum }_{n\le t}{n}^{-\sigma -it}$: the integral term has size $O(t^{-\sigma })$ because $|s-1|\ge t$.

On a dyadic interval $(N,N^{\prime }]\subset (N,2N]$ with $N^{\prime }\le t$, the assumed estimate holds for every initial subinterval. Abel summation with $n^{-\sigma }$ therefore givesIndeed the weighted endpoint and integral of the term proportional to the subinterval length are $O(N^{1-\sigma })$, and those of the constant $N^{4/5}$ term are $O(N^{4/5-\sigma })$. The constants can be uniform in $0<\sigma \le 1$.

For the first term, write $u=j\log 2$ and complete the square:The sum of a shifted Gaussian function on a fixed-spaced lattice is $O(\sqrt{L})$, uniformly in the shift. Thus these dyadic contributions are $O(\sqrt{L}{e}^{{\epsilon }^{2}L/(4c)})$. This is the Gaussian dyadic summation bound.

For the second term, if $\epsilon \le 1/10$ its dyadic sum is bounded. If $\epsilon >1/10$, a crude bound is $O(L\exp (\max (\epsilon -1/5,0)L))$. The positive exponent obeys $\epsilon -1/5\le (5/4){\epsilon }^2$, and $L$ can be absorbed into $\sqrt{L}{e}^{C{\epsilon }^{2}L}$ uniformly on $\epsilon >1/10$ by increasing the fixed constant $C$. The finitely many initial terms cause no problem. We conclude, with one fixed sufficiently large $C$,In particular the endpoint $\sigma =1$ gives $O(\sqrt{\log t})$ under the assumed exponential-sum hypothesis. This conditional conclusion uses that hypothesis, not an unconditional improvement of the stated Richert bound for the Riemann zeta function.

Let $A(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ converge absolutely at a real $c>0$, and take $T\ge \max (2,c)$. DefineThe truncated Perron kernel estimate isFor $v>1$ close the contour to the left, collecting the residue one at zero; for $v<1$ close it to the right, collecting no residue. On the horizontal sides, integrating $v^{\sigma }$ bounds the error by $O(v^c/(T|\log v|))$, and the remote vertical side tends to zero. Near $v=1$ use the bounded transition estimate instead, giving $O(v^c)$. These are the contours and bounds underlying the kernel formula. The constants are uniform for $1\le c\le 2$, the range needed below.

Absolute convergence permits termwise integration. The truncated Perron formula is consequentlyHere the primed sum has half weight when $x$ is an integer, and $a_x=0$ otherwise. To obtain the inclusive sum add $a_x/2$ at an integer. This endpoint convention avoids a false uniform assertion about the kernel at $v=1$.

Apply this with $a_n=\Lambda (n)$, $A=-{\zeta }^{\prime }/\zeta$, $c=1+1/\log x$ and $2\le T\le x$, for sufficiently large $x$. For $n\notin [x/2,2x]$, $|\log (x/n)|$ is bounded below, andusing $-{\zeta }^{\prime }(c)/\zeta (c)\ll (c-1{)}^{-1}$. In the central range $(x/n{)}^c\ll 1$, $\Lambda (n)\ll \log x$, and $|\log (x/n)|\asymp |n-x|/x$. Separate the nearest integers, then sum the harmonic tail over distances $j\ge 1$: its contribution is $O((x/T){\log }^{2}x)$. The possible endpoint weight and nearest terms, of size $O(\log x)$, are absorbed because $T\le x$. ThereforeFor $x+y$, use the same vertical line $c=1+1/\log x$; the error estimate remains valid since $x\le x+y\le 2x$. Subtract the two formulas. The identityhas absolute value $O(y)$ on that line, because $(2x{)}^{c-1}$ is bounded. Taking absolute values givesThis is the short-interval Perron bound for the second Chebyshev function.

The Möbius function has $\mu (1)=1$, vanishes on integers divisible by a square of a prime, and equals $(-1{)}^r$ on a product of $r$ distinct primes. Factoring the divisor sum prime by prime givesThis is the Möbius divisor-sum identity.

For $T/2\le t\le T$ and $0<\sigma \le 1$, the Hardy-Littlewood approximation to the Riemann zeta function at cutoff $T$ givesIndeed $T\ge |t|/\pi$, and the omitted integral term has size at most $T^{1-\sigma }/|s-1|\ll T^{-\sigma }$. Multiply by ${\sum }_{m\le M}\mu (m)m^{-s}$. Its absolute value is at mostwhere the elementary inequality follows from $m^{1-\sigma }\le M^{1-\sigma }$. Reindexing the finite double sum yields coefficients $a_n={\sum }_{m\mid n,m\le M,n/m\le T}\mu (m)$, with no terms for $n>MT$. For $n\le \min (M,T)$ all divisors meet both restrictions, so the Möbius divisor-sum identity gives $a_1=1$ and $a_n=0$ for $1<n\le \min (M,T)$. Hence the truncated Möbius inverse identity for the Riemann zeta function isThe displayed error is uniform in $0<\sigma \le 1$ and the stated height interval.