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Explicit bounds for the Riemann zeta function
and a new zero free region

Chiara Bellotti School of Science
The University of New South Wales, Canberra, Australia
c.bellotti@adfa.edu.au
Abstract.

We prove that |ζ(σ+it)|70.7|t|4.438(1σ)3/2log2/3|t||\zeta(\sigma+it)|\leq 70.7|t|^{4.438(1-\sigma)^{3/2}}\log^{2/3}|t| for 1/2σ11/2\leq\sigma\leq 1 and |t|3|t|\geq 3. As a consequence, we improve the explicit zero-free region for ζ(s)\zeta(s), showing that ζ(σ+it)\zeta(\sigma+it) has no zeros in the region σ11/(54.004(log|t|)2/3(loglog|t|)1/3)\sigma\geq 1-1/\left(54.004(\log|t|)^{2/3}(\log\log|t|)^{1/3}\right) for |t|3|t|\geq 3 and asymptotically in the region σ11/(48.0718(log|t|)2/3(loglog|t|)1/3)\sigma\geq 1-1/\left(48.0718(\log|t|)^{2/3}(\log\log|t|)^{1/3}\right) for |t||t| sufficiently large.

Key words and phrases:
Riemann zeta function, explicit bounds, zero-free region, Vinogradov integral
2020 Mathematics Subject Classification:
Primary 11M06, 11N05, 11L15; Secondary 11D72, 11M35.

1. Introduction

Let ζ(s)\zeta(s) be the Riemann zeta function, with s=σ+its=\sigma+it a complex variable. It is known that all the non trivial zeros of ζ(s)\zeta(s) have real part σ(0,1)\sigma\in(0,1). Detecting zero free regions for ζ(s)\zeta(s) inside the critical strip 0<σ<10<\sigma<1 is an open problem that has always caught much interest in analytic number theory. Great effort has been put in trying to find both asymptotically and explicit regions inside the critical strip where there are no zeros of ζ(s)\zeta(s). The classical zero-free region is of the form σ>11/(R0log|t|)\sigma>1-1/(R_{0}\log|t|), where R0R_{0} is a positive constant. The best known result of this form is due to Mossinghoff, Trudgian and Yang [MTY22] with R0=5.558691R_{0}=5.558691 for every |t|2|t|\geq 2 (see [Ste70, RS75, Kon77, Kad05, JK14, MT14] for previous results). Littlewood zero-free region [Lit22] is instead of the form of σ>1loglog|t|/C1log|t|\sigma>1-\log\log|t|/C_{1}\log|t|, where C1C_{1} is a positive constant. It has been made first explicit by Yang [Yan23], who found C1=21.432C_{1}=21.432 for |t|3|t|\geq 3.
Asymptotically larger zero-free regions for the Riemann zeta function ζ(s)\zeta(s), known as Korobov–Vinogradov zero-free regions, are of the form

σ>11C2(log|t|)2/3(loglog|t|)1/3,\sigma>1-\frac{1}{C_{2}(\log|t|)^{2/3}(\log\log|t|)^{1/3}}, (1.1)

where C2C_{2} is a positive constant and are due to the method of Korobov [Kor58] and Vinogradov [Vin58], in which the main tool is an upper bound on |ζ(σ+it)||\zeta(\sigma+it)| when σ\sigma is near to the line σ=1\sigma=1. Upper bounds of this form were first made explicit by Richert [Ric67], who used Korobov-Vinogradov method to prove that

|ζ(σ+it)|A|t|B(1σ)3/2log2/3|t||t|2,12σ1|\zeta(\sigma+it)|\leq A|t|^{B(1-\sigma)^{3/2}}\log^{2/3}|t|\qquad|t|\geq 2,\ \frac{1}{2}\leq\sigma\leq 1 (1.2)

with B=100B=100 and AA a certain absolute constant. Although smaller values for BB were already found (see [Kul99]), the first completely explicit bound of the form (1.2) is due to Cheng [Che99], with A=175A=175 and B=46B=46. This estimate was further improved in 2002 by Ford [For02a] to A=76.2A=76.2 and B=4.45B=4.45 for every |t|3|t|\geq 3 and 1/2σ11/2\leq\sigma\leq 1.
Explicit bounds of the form (1.2) play a fundamental role in detecting explicit Korobov–Vinogradov zero-free regions for ζ(s)\zeta(s). There are several results regarding the value of the constant C2C_{2} in (1.1) and the current best known estimate is due to Mossinghoff, Trudgian and Yang [MTY22], with C2=55.241C_{2}=55.241 for every |t|3|t|\geq 3 (see [Che00, For02, For22] for previous results).
In this paper, we will improve the values for both the constants A,BA,B in (1.2) and, as a consequence, we find an improved Korobov-Vinogradov zero-free region for ζ(s)\zeta(s). More precisely, denoting with ζ(s,u)\zeta(s,u) the Hurwitz zeta function ζ(s,u)=n=0(n+u)s\zeta(s,u)=\sum_{n=0}^{\infty}(n+u)^{-s} defined for every s>1\Re s>1 and 0<u10<u\leq 1, we will prove the following result for both ζ(s)=ζ(s,1)\zeta(s)=\zeta(s,1) and a generic Hurwitz function ζ(s,u)\zeta(s,u).

Theorem 1.1.

The following estimate holds for every |t|3|t|\geq 3 and 12σ1\frac{1}{2}\leq\sigma\leq 1:

|ζ(σ+it)|\displaystyle|\zeta(\sigma+it)| A|t|B(1σ)3/2log2/3|t|\displaystyle\leq A|t|^{B(1-\sigma)^{3/2}}\log^{2/3}|t| (1.3)
|ζ(σ+it,u)us|\displaystyle\left|\zeta(\sigma+it,u)-u^{-s}\right| A|t|B(1σ)3/2log2/3|t|,0<u1,\displaystyle\leq A|t|^{B(1-\sigma)^{3/2}}\log^{2/3}|t|,\qquad 0<u\leq 1,

with A=70.6995A=70.6995 and B=4.43795B=4.43795.

The bound on ζ(s,u)\zeta(s,u) found in Theorem 1.1 might be useful to bound Dirichlet LL-functions due to the relation L(s,χ)=qsm=1qχ(m)ζ(s,m/q)L(s,\chi)=q^{-s}\sum_{m=1}^{q}\chi(m)\zeta(s,m/q), where χ\chi is a Dirichlet character modulo qq.
Although the new values A=70.6995A=70.6995 and B=4.43795B=4.43795 found in Theorem 1.1 are modest improvements on those found by Ford in [For02a] (A=76.2A=76.2 and B=4.45B=4.45 respectively), their importance relies on the fact that just a small improvement for both AA and BB can lead to improvements in several other results in analytic number theory. In particular, they have many applications in finding improved estimates for Korobov–Vinogrado zero-free region and in estimating the error term in the prime number theorem, both in an effective and ineffective way.
Furthermore, improvements on AA and BB only of the size found in Theorem 1.1 were already expected. In [For02a] (Section 8, point 11), Ford already predicted that an optimization of the argument involving the Vinogradov integral would have lower BB by less than 0.020.02, which is consistent with our new value for B=4.43795B=4.43795 found in Theorem 1.1.

As already mentioned, an immediate consequence of Theorem 1.1 is a new explicit zero free region for the Riemann zeta function.

Theorem 1.2.

There are no zeros of ζ(σ+it)\zeta(\sigma+it) for |t|3|t|\geq 3 and

σ1154.004(log|t|)2/3(loglog|t|)1/3.\sigma\geq 1-\frac{1}{54.004(\log|t|)^{2/3}(\log\log|t|)^{1/3}}.

Theorem 1.1 has also an influence on asymptotically Korobov-Vinogradov zero-free regions of the form

σ11c(log|t|)2/3(loglog|t|)1/3,\sigma\geq 1-\frac{1}{c(\log|t|)^{2/3}(\log\log|t|)^{1/3}}, (1.4)

where cc is a positive constant for |t||t| sufficiently large. More precisely, we get c=48.0718c=48.0718, improving cc on the current best value of 48.158848.1588 due to Mossinghoff, Trudgian and Yang [MTY22] (see [For02] for previous results).

Theorem 1.3.

For sufficiently large |t|,|t|, there are no zeros of ζ(σ+it)\zeta(\sigma+it) with

σ1148.0718(log|t|)2/3(loglog|t|)1/3.\sigma\geq 1-\frac{1}{48.0718(\log|t|)^{2/3}(\log\log|t|)^{1/3}}. (1.5)

As in [MTY22], the proofs of Theorem 1.2 and Theorem 1.3 rely on some non-negative trigonometric polynomials. We recall that, for every K2K\geq 2, a KK-th degree non-negative trigonometric polynomial PK(x)P_{K}(x) is defined as

PK(x)=k=0Kbkcos(kx)P_{K}(x)=\sum_{k=0}^{K}b_{k}\cos(kx) (1.6)

where bkb_{k} are constants such that bk0,b1>b0b_{k}\geq 0,b_{1}>b_{0} and PK(x)0P_{K}(x)\geq 0 for all real xx.
Finally, one can use Theorem 1.3 to improve on the error term in the prime number theorem. As per Ford [For02a], we get the following estimate for the error term in the prime number theorem

π(x)li(x)xexp{d(logx)3/5(loglogx)1/5},\pi(x)-\operatorname{li}(x)\ll x\exp\left\{-d(\log x)^{3/5}(\log\log x)^{-1/5}\right\},

with

d=(562234c3)1/5,d=\left(\frac{5^{6}}{2^{2}\cdot 3^{4}\cdot c^{3}}\right)^{1/5},

where cc is the constant in (1.4). Using the new value for cc found in Theorem 1.3 we obtain d=0.212579d=0.212579, which is a slightly improvement on d=0.2123d=0.2123 found by Mossinghoff, Trudgian and Yang [MTY22].

As in [For02a], Theorem 1.1 follows from a uniform upper bound on the sum

S(N,t):=max0<u1maxN<R2N|NnR1(n+u)it|,S(N,t):=\max_{0<u\leq 1}\max_{N<R\leq 2N}\left|\sum_{N\leq n\leq R}\frac{1}{(n+u)^{it}}\right|, (1.7)

where NN is a positive integer and tNt\geq N. This upper bound is of the form

S(N,t)cN11/(uλ2),S(N,t)\leq cN^{1-1/(u\lambda^{2})},

with u,c>0u,c>0 and λ=logt/logN.\lambda=\log t/\log N. The strength of Ford’s argument in [For02a] relies on some explicit estimates for both the Vinogradov integral and a quantity that counts the number of solutions of incomplete Diophantine systems that, combined with estimates for the exponential sum S(N,t)S(N,t), give a completely explicit uniform upper bound on S(N,t)S(N,t). Explicit bounds for the Vinogradov integral were further improved by Preobrazhenskiĭ [Pre11] in 2011 and Steiner in 2019 [Ste19] (see [Hua49, Ste70, Tyr87, ACK04] for previous results).
We recall that the Vinogradov integral is defined as

Js,k(P)=[0,1]k|1xPe(α1x++αkxk)|2s𝑑𝜶J_{s,k}(P)=\int_{[0,1]^{k}}\left|\sum_{1\leq x\leq P}e\left(\alpha_{1}x+\cdots+\alpha_{k}x^{k}\right)\right|^{2s}d\boldsymbol{\alpha} (1.8)

where 𝜶=(α1,,αk)\boldsymbol{\alpha}=\left(\alpha_{1},\ldots,\alpha_{k}\right) and e(z)=e2πize(z)=e^{2\pi iz}, or equivalently, Js,k(P)J_{s,k}(P) is defined as the number of solutions of the simultaneous equations

i=1s(xijyij)=0(1jk);1xi,yiP.\sum_{i=1}^{s}\left(x_{i}^{j}-y_{i}^{j}\right)=0\quad(1\leq j\leq k);\quad 1\leq x_{i},y_{i}\leq P.

Regarding incomplete systems, we denote with Js,k,h()J_{s,k,h}(\mathscr{B}) the number of solutions of the system

i=1s(xijyij)=0(hjk);xi,yi,\sum_{i=1}^{s}\left(x_{i}^{j}-y_{i}^{j}\right)=0\quad(h\leq j\leq k);\quad x_{i},y_{i}\in\mathscr{B},

where \mathscr{B} is a suitable set. For our purpose, we will use the definition given by Ford in [For02a] =𝒞(P,R)\mathscr{B}=\mathscr{C}(P,R) where 𝒞(P,R)\mathscr{C}(P,R) is the set of integers P\leq P composed only of prime factors in (R,R](\sqrt{R},R]. Hence, Js,k,h(𝒞(P,R))J_{s,k,h}(\mathscr{C}(P,R)) is defined as

Js,k,h(𝒞(P,R))=[0,1]t|f(𝜶)|2s𝑑𝜶,J_{s,k,h}(\mathscr{C}(P,R))=\int_{[0,1]^{t}}|f(\boldsymbol{\alpha})|^{2s}d\boldsymbol{\alpha}, (1.9)

where

f(𝜶)=f(𝜶;P,R)=x𝒞(P,R)e(αhxh++αkxk)f(\boldsymbol{\alpha})=f(\boldsymbol{\alpha};P,R)=\sum_{x\in\mathscr{C}(P,R)}e\left(\alpha_{h}x^{h}+\cdots+\alpha_{k}x^{k}\right)

and 𝜶=(αh,,αk)\boldsymbol{\alpha}=\left(\alpha_{h},\cdots,\alpha_{k}\right).
In his paper [For02a], Ford estimated S(N,t)S(N,t) working separately and with different techniques on three different ranges of λ\lambda. More precisely, he considered the ranges λ87\lambda\leq 87, 87λ22087\leq\lambda\leq 220 and λ220\lambda\geq 220, where the critical case happens to be around λ=87\lambda=87. However, it is possible to shift the critical case to λ=84\lambda=84, with a consequent improved upper bound on S(N,t)S(N,t) and so improved values of AA and BB. Indeed, in his paper [For02a] (p. 590), Ford stated a variant of his argument to find even better estimates for the Vinogradov integral when kk is small. These results, combined with Preobrazhenskiĭ’s argument for k90000k\geq 90000 in [Pre11] and some explicit bounds for the Vinogradov integral when sk2s\ll k^{2} due to Tyrina [Tyr87], will give the following explicit bounds for Js,kJ_{s,k}.

Theorem 1.4.

Let kk and ss be integers with k90000k\geq 90000 and

0.138128k2s12k2(0.6494+log(8k25D)(1D/k2D/k)Dk+2.051k),0.138128k^{2}\leq s\leq\frac{1}{2}k^{2}\left(0.6494+\log\left(\frac{8k}{25D}\right)-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+\frac{2.051}{k}\right),

with 1D0.4k1\leq D\leq 0.4k. Then

Js,k(P)k2.055k30.414k2+3nk1.06nk2+2(n2n)k+0.099912k3P2sk(k+1)/2+Δs(P1),J_{s,k}(P)\leq k^{2.055k^{3}-0.414k^{2}+3nk}1.06^{nk^{2}+2\left(n^{2}-n\right)k+0.099912k^{3}}P^{2s-k(k+1)/2+\Delta_{s}}\quad(P\geq 1),

where, denoting with (smodk)(s\bmod k) the unique integer uu such that 0u<k0\leq u<k,

Δs\displaystyle\Delta_{s}\leq max(825k2exp(0.64942(sk)k2(1D/k2D/k)Dk+2.051k+2(smodk)k3),\displaystyle\max\left(\frac{8}{25}k^{2}\exp\left(0.6494-\frac{2(s-k)}{k^{2}}-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+\frac{2.051}{k}+\frac{2(s\bmod k)}{k^{3}}\right),\right. (1.10)
Dkexp(2(smodk)k2(1+1k))).\displaystyle\ \ \quad\ \ \left.Dk\exp\left(\frac{2(s\bmod k)}{k^{2}}\left(-1+\frac{1}{k}\right)\right)\right).

Furthermore, if k129k\geq 129, there is an integer sρk2s\leq\rho k^{2} such that for P1P\geq 1,

Js,k(P)kθk3P2s12k(k+1)+0.001k2,J_{s,k}(P)\leq k^{\theta k^{3}}P^{2s-\frac{1}{2}k(k+1)+0.001k^{2}},

with

{ρ=3.177207,θ=2.40930,if 129k137ρ=3.177527,θ=2.39529,if 138k139ρ=3.178551,θ=2.39167,if 140k146ρ=3.178871,θ=2.38259,if 147k148ρ=3.181869,θ=2.37929,if 149k170ρ=3.184127,θ=2.35334,if 171k190ρ=3.192950,θ=2.33313,if 191k339ρ=3.196497,θ=2.24352,if 340k499ρ=3.205502,θ=1.77775,if 500k<90000ρ=3.208630,θ=2.17720,if k90000.\begin{cases}\rho=3.177207,\quad\theta=2.40930,&\text{if }129\leq k\leq 137\\ \\ \rho=3.177527,\quad\theta=2.39529,&\text{if }138\leq k\leq 139\\ \\ \rho=3.178551,\quad\theta=2.39167,&\text{if }140\leq k\leq 146\\ \\ \rho=3.178871,\quad\theta=2.38259,&\text{if }147\leq k\leq 148\\ \\ \rho=3.181869,\quad\theta=2.37929,&\text{if }149\leq k\leq 170\\ \\ \rho=3.184127,\quad\theta=2.35334,&\text{if }171\leq k\leq 190\\ \\ \rho=3.192950,\quad\theta=2.33313,&\text{if }191\leq k\leq 339\\ \\ \rho=3.196497,\quad\theta=2.24352,&\text{if }340\leq k\leq 499\\ \\ \rho=3.205502,\quad\theta=1.77775,&\text{if }500\leq k<90000\\ \\ \rho=3.208630,\quad\theta=2.17720,&\text{if }k\geq 90000.\end{cases}

Using these new bounds for the Vinogradov integral we will prove the following result.

Theorem 1.5.

Suppose N2N\geq 2 is a positive integer, NtN\leq t and set λ=logtlogN\lambda=\frac{\log t}{\log N}. Then

S(N,t)8.7979N11/(132.94357λ2).S(N,t)\leq 8.7979N^{1-1/\left(132.94357\lambda^{2}\right)}.

In proving both Theorem 1.4 and Theorem 1.5, we tried to make near-optimal choices for all parameters that are used in the argument. Hence, it seems unlikely that a substantial improvement in the BB constant in Theorem 1.1 can be obtained via better choices of parameters alone.

An immediate corollary of Theorem 1.5 involving Dirichlet characters is the following one.

Corollary 1.6.

Suppose χ\chi is a Dirichlet character modulo qq, where qNq\leq N and 2Nqt2\leq N\leq qt. Then

maxN<R2N|N<nRχ(n)nit|9.7979ϕ(q)qNelog3(N/q)132.94357log2t.\max_{N<R\leq 2N}\left|\sum_{N<n\leq R}\chi(n)n^{-it}\right|\leq 9.7979\frac{\phi(q)}{q}Ne^{-\frac{\log^{3}(N/q)}{132.94357\log^{2}t}}.

The proof is the same as that of Corollary 2A in [For02a].

2. Background

We recall some preliminary results we will use later in the proof of the theorems.
First of all, given a fixed kk, suppose 0dk10\leq d\leq k-1 and TT is a positive integer. The kk-tuple of polynomials 𝚿=(Ψ1,,Ψk)[x]k\boldsymbol{\Psi}=\left(\Psi_{1},\ldots,\Psi_{k}\right)\in\mathbb{Z}[x]^{k} is said to be of type (d,T)(d,T) if Ψj\Psi_{j} is identically zero for jdj\leq d, and for some integer m0m\geq 0, when j>d,Ψjj>d,\ \Psi_{j} has degree jdj-d with leading coefficient j!(jd)!2mT\frac{j!}{(j-d)!}2^{m}T.
Then, let Ks,d(P,Q;𝚿;q)K_{s,d}(P,Q;\boldsymbol{\Psi};q) be the number of solutions of

i=1kd(Ψj(zi)Ψj(wi))+qji=1s(xijyij)=0(1jk),\sum_{i=1}^{k-d}\left(\Psi_{j}\left(z_{i}\right)-\Psi_{j}\left(w_{i}\right)\right)+q^{j}\sum_{i=1}^{s}\left(x_{i}^{j}-y_{i}^{j}\right)=0\quad(1\leq j\leq k),

with 1zi,wiP1\leq z_{i},w_{i}\leq P and 1xi,yiQ1\leq x_{i},y_{i}\leq Q. Also, let Ls,d(P,Q;𝚿;p,q,r)L_{s,d}(P,Q;\boldsymbol{\Psi};p,q,r) be the number of solutions of

i=1kd(Ψj(zi)Ψj(wi))+(pq)ji=1s(uijvij)=0(1jk)\sum_{i=1}^{k-d}\left(\Psi_{j}\left(z_{i}\right)-\Psi_{j}\left(w_{i}\right)\right)+(pq)^{j}\sum_{i=1}^{s}\left(u_{i}^{j}-v_{i}^{j}\right)=0\quad(1\leq j\leq k)

with 1zi,wiP1\leq z_{i},w_{i}\leq P, ziwi(modpr)z_{i}\equiv w_{i}\left(\bmod\ p^{r}\right) and 1ui,viQ1\leq u_{i},v_{i}\leq Q.

Lemma 2.1 ([For02a] Lemma 3.2’).

Suppose k,r,dk,r,d and ss are integers with

k4;2rk;0dr1;sd+1.k\geq 4;\quad 2\leq r\leq k;\quad 0\leq d\leq r-1;\quad s\geq d+1.

Let M,PM,P and QQ be real numbers with

P1/(k+1)MP1/r;32s2M<QP,Mk.P^{1/(k+1)}\leq M\leq P^{1/r};\quad 32s^{2}M<Q\leq P,\quad M\geq k.

Suppose qq is a positive integer and 𝚿\boldsymbol{\Psi} is a system of polynomials of type (d,T)(d,T) with TPdT\leq P^{d}. Denote by 𝒫\mathscr{P} the set of the k3k^{3} smallest primes greater than MM, and suppose 𝒫(M,2M]\mathscr{P}\subset(M,2M]. Then there are a system of polynomials 𝚽\boldsymbol{\Phi} of type (d,T)(d,T) and a prime p𝒫p\in\mathscr{P} such that

Ks,d(P,Q;𝚿;q)4k3k!p2s+(rd)(rd+1)/2Ls,d(P,Q;𝚽;p,q,r).K_{s,d}(P,Q;\boldsymbol{\Psi};q)\leq 4k^{3}k!p^{2s+(r-d)(r-d+1)/2}L_{s,d}(P,Q;\boldsymbol{\Phi};p,q,r).
Lemma 2.2 ([For02a] Lemma 3.3’).

Suppose that sd,kr2,dk2,q1,s\geq d,k\geq r\geq 2,d\leq k-2,q\geq 1, pp is a prime and 𝚽\boldsymbol{\Phi} is a system of polynomials of type (d,T)(d,T). Then there is a system of polynomials 𝚼\boldsymbol{\Upsilon} of type (d+1,T)\left(d+1,T^{\prime}\right) with TTPTT\leq T^{\prime}\leq PT such that

Ls,d(P;Q;𝚽;p,q,r)\displaystyle L_{s,d}(P;Q;\boldsymbol{\Phi};p,q,r) (2P)kdmax[kkdJs,k(Q),2pr(kd)Js,k(Q)(kd2)/(2(kd1))\displaystyle\leq(2P)^{k-d}\max\left[k^{k-d}J_{s,k}(Q),2p^{-r(k-d)}J_{s,k}(Q)^{(k-d-2)/(2(k-d-1))}\right.
×Ks,d+1(P,Q;𝚼;pq)(kd)/(2(kd1))].\displaystyle\left.\times K_{s,d+1}(P,Q;\boldsymbol{\Upsilon};pq)^{(k-d)/(2(k-d-1))}\right].

As per Ford [For02a], Lemma 2.1 and Lemma 2.2 imply the following result.

Lemma 2.3.

Suppose k26,4rk,ksk3k\geq 26,4\leq r\leq k,k\leq s\leq k^{3} and

Js,k(Q)CQ2sk(k+1)/2+Δ(Q1).J_{s,k}(Q)\leq CQ^{2s-k(k+1)/2+\Delta}\quad(Q\geq 1).

Let jj be an integer satisfying

2j910r,(j1)(j2)2Δ(kr)(kr+1).2\leq j\leq\frac{9}{10}r,\quad(j-1)(j-2)\leq 2\Delta-(k-r)(k-r+1). (2.1)

Define

ϕJ=12r+k2+k+r2r2Δ2rJ4r(kJ)ϕJ+1(1Jj1),\phi_{J}=\frac{1}{2r}+\frac{k^{2}+k+r^{2}-r-2\Delta-2rJ}{4r(k-J)}\phi_{J+1}\quad(1\leq J\leq j-1),

and suppose rr and jj are chosen so that ϕi1/(k+1)\phi_{i}\geq 1/(k+1) for every ii. Suppose

13logkω12,η=1+ω,V=max(e1.5+1.5/ω,18ωk3logk).\frac{1}{3\log k}\leq\omega\leq\frac{1}{2},\quad\eta=1+\omega,\quad V=\max\left(e^{1.5+1.5/\omega},\frac{18}{\omega}k^{3}\log k\right).

If PVk+1P\geq V^{k+1}, then

Js+k,k(P)k3kη4s+k2CP2(s+k)k(k+1)/2+ΔJ_{s+k,k}(P)\leq k^{3k}\eta^{4s+k^{2}}CP^{2(s+k)-k(k+1)/2+\Delta^{\prime}}

where

Δ=Δ(1ϕ1)k+12ϕ1(k2+k+r2r).\Delta^{\prime}=\Delta\left(1-\phi_{1}\right)-k+\frac{1}{2}\phi_{1}\left(k^{2}+k+r^{2}-r\right).

We will use the definition of ϕJ\phi_{J} given in Lemma 2.3 for the range k<90000k<90000. However, when kk becomes large, the influence that the new funcitons Φj\Phi_{j} have on estimating the Vinogradov’s integral becomes negligible, as already pointed out by Ford in [For02a] (p. 590). Hence, for k90000k\geq 90000, we use the following result used by Ford in his work.

Lemma 2.4 ([For02a] Lemma 3.4).

Suppose k26,4rk,ksk3k\geq 26,4\leq r\leq k,k\leq s\leq k^{3} and

Js,k(Q)CQ2sk(k+1)/2+Δ(Q1).J_{s,k}(Q)\leq CQ^{2s-k(k+1)/2+\Delta}\quad(Q\geq 1).

Let jj be an integer satisfying the same relations as in (2.1). Then, define

ϕj=1r,ϕJ=12r+k2+k+r2r+J2J2Δ4krϕJ+1(1Jj1),\phi_{j}=\frac{1}{r},\quad\phi_{J}=\frac{1}{2r}+\frac{k^{2}+k+r^{2}-r+J^{2}-J-2\Delta}{4kr}\phi_{J+1}\quad(1\leq J\leq j-1),

and suppose rr and jj are chosen so that ϕi1/(k+1)\phi_{i}\geq 1/(k+1) for every ii. Suppose

13logkω12,η=1+ω,V=max(e1.5+1.5/ω,18ωk3logk).\frac{1}{3\log k}\leq\omega\leq\frac{1}{2},\quad\eta=1+\omega,\quad V=\max\left(e^{1.5+1.5/\omega},\frac{18}{\omega}k^{3}\log k\right).

If PVk+1P\geq V^{k+1}, then

Js+k,k(P)k3kη4s+k2CP2(s+k)k(k+1)/2+ΔJ_{s+k,k}(P)\leq k^{3k}\eta^{4s+k^{2}}CP^{2(s+k)-k(k+1)/2+\Delta^{\prime}}

where

Δ=Δ(1ϕ1)k+12ϕ1(k2+k+r2r).\Delta^{\prime}=\Delta\left(1-\phi_{1}\right)-k+\frac{1}{2}\phi_{1}\left(k^{2}+k+r^{2}-r\right).

For a given k,rk,r and Δ\Delta, we let δ0(k,r,Δ)\delta_{0}(k,r,\Delta) be the value of Δ\Delta^{\prime} coming from Lemma 2.3 or Lemma 2.4, where we take jj maximal satisfying (2.1).

Lemma 2.5 ([For02a] Lemma 3.5).

Let k26k\geq 26 and let ω,η\omega,\eta and VV be as in Lemma 2.3 or Lemma 2.4. Let Δ1=12k2(11/k)\Delta_{1}=\frac{1}{2}k^{2}(1-1/k) and for n1n\geq 1, let rnr_{n} be an integer in [4,k][4,k] satisfying

ϕ(k,rn,Δn):=2k2rnk+2Δn(krn)(krn+1)1k+1.\phi^{*}\left(k,r_{n},\Delta_{n}\right):=\frac{2k}{2r_{n}k+2\Delta_{n}-\left(k-r_{n}\right)\left(k-r_{n}+1\right)}\geq\frac{1}{k+1}.

Then set Δn+1=δ0(k,rn,Δn)\Delta_{n+1}=\delta_{0}\left(k,r_{n},\Delta_{n}\right). If nk2n\leq k^{2}, then

Jnk,k(P)CnP2nkk(k+1)/2+Δn(P1),J_{nk,k}(P)\leq C_{n}P^{2nk-k(k+1)/2+\Delta_{n}}\quad(P\geq 1),

where C1=k!C_{1}=k! and, for n2n\geq 2,

Cn=Cn1max[k3kη4k(n1)+k2,V(k+1)(Δn1Δn)].C_{n}=C_{n-1}\max\left[k^{3k}\eta^{4k(n-1)+k^{2}},V^{(k+1)\left(\Delta_{n-1}-\Delta_{n}\right)}\right].

As already mentioned before, some explicit bounds for the Vinogradov’s integral were found by Tyrina [Tyr87], and they give evident improvements when sk2s\ll k^{2}. In her argument, Tyrina defines recursively sequences rn,snr_{n},s_{n}, and Δsn\Delta_{s_{n}}, with s1=ks_{1}=k and Δs1=k\Delta_{s_{1}}=k. More precisely, for s1s\geq 1 we take rnr_{n} to be the integer nearest to the number

2Δsn+k(k+1)2k+1\frac{2\Delta_{s_{n}}+k(k+1)}{2k+1}

and we define recursively

Δsn+1=Δsn+rn(2rnk1)(2rnk)2rnΔsnrn\Delta_{s_{n+1}}=\Delta_{s_{n}}+r_{n}-\frac{\left(2r_{n}-k-1\right)\left(2r_{n}-k\right)}{2r_{n}}-\frac{\Delta_{s_{n}}}{r_{n}}

where

sn+1=k+r1++rn.s_{n+1}=k+r_{1}+\cdots+r_{n}.

Then, we consider the function x(y),ky<k(k+1)/2x(y),\ k\leq y<k(k+1)/2 :

x(y)=y+2k(k+1)2ln(12(yk)/(k2k)).x(y)=-y+2k-(k+1)^{2}\cdot\ln\left(1-2\cdot(y-k)/\left(k^{2}-k\right)\right).

This function increases monotonically and assumes the values from kk to infinity as yy increases from kk to k(k+1)/2k(k+1)/2. The function y=y(x)y=y(x) which is inverse to x(y)x(y) is also monotonically increasing. Finally, let l0l_{0} denote the integer x(k2/2)\lceil x\left(k^{2}/2\right)\rceil, where

x(k2/2)=k2/2+2k+(k+1)2ln(k1).x\left(k^{2}/2\right)=-k^{2}/2+2k+(k+1)^{2}\ln(k-1).

Tyrina proved the following result.

Theorem 2.6 ([Tyr87] Thm. 1).

The mean value Js,k(P)J_{s,k}(P) satisfies the estimate

Js,k(P)DsP2sκsJ_{s,k}(P)\leq D_{s}\cdot P^{2s-\kappa_{s}}

where Ds=22s(k+sk1)kk+4s2s2(sk)D_{s}=2^{2s\left(k+sk^{-1}\right)}k^{k+4s^{2}}s^{2(s-k)}. Here

  1. (1)

    if s=sn,n=1,2,,sn<l0s=s_{n},n=1,2,\ldots,s_{n}<l_{0}, then κsΔsy(s)\kappa_{s}\geq\Delta_{s}\geq y(s);

  2. (2)

    if s=lt=l0+kt,t=0,1,s=l_{t}=l_{0}+kt,t=0,1,\ldots, then

    κsΔsk(k+1)2k+22(11k);\kappa_{s}\geq\Delta_{s}\geq\frac{k(k+1)}{2}-\frac{k+2}{2}\left(1-\frac{1}{k}\right);
  3. (3)

    if sn<s<sn+1s_{n}<s<s_{n+1}, then κsΔsΔsn+1s/sn+1\kappa_{s}\geq\Delta_{s}\geq\Delta_{s_{n+1}}\cdot s/s_{n+1};

  4. (4)

    if lt1<s<lt,t=1,2,l_{t-1}<s<l_{t},t=1,2,\ldots, then κsΔsΔlts/lt\kappa_{s}\geq\Delta_{s}\geq\Delta_{l_{t}}\cdot s/l_{t}.

New explicit bounds for Js,kJ_{s,k} were found by Ford in 2002 [For02a].

Theorem 2.7 ([For02a] Thm. 3).

Let kk and ss be integers with k1000k\geq 1000 and 2k2sk22(12+log3k8)2k^{2}\leq s\leq\frac{k^{2}}{2}\left(\frac{1}{2}+\log\frac{3k}{8}\right). Then

Js,k(P)k2.055k35.91k2+3s1.06sk+2s2/k9.7278k3P2s12k(k+1)+Δs(P1)J_{s,k}(P)\leq k^{2.055k^{3}-5.91k^{2}+3s}1.06^{sk+2s^{2}/k-9.7278k^{3}}P^{2s-\frac{1}{2}k(k+1)+\Delta_{s}}\quad(P\geq 1)

where

Δs=38k2e1/22s/k2+1.7/k.\Delta_{s}=\frac{3}{8}k^{2}e^{1/2-2s/k^{2}+1.7/k}.

Further, if k129k\geq 129, there is an integer sρk2s\leq\rho k^{2} such that for P1P\geq 1,

Js,k(P)kθk3P2s12k(k+1)+0.001k2J_{s,k}(P)\leq k^{\theta k^{3}}P^{2s-\frac{1}{2}k(k+1)+0.001k^{2}}

with

(ρ,θ)={(3.21432,2.3291)(k200)(3.21734,2.3849)(150k199)(3.22313,2.4183)(129k149).(\rho,\theta)=\left\{\begin{array}[]{ll}(3.21432,2.3291)&(k\geq 200)\\ \\ (3.21734,2.3849)&(150\leq k\leq 199)\\ \\ (3.22313,2.4183)&(129\leq k\leq 149)\end{array}\right.. (2.2)

In 2011, Preobrazhenskiĭ [Pre11] improved the value of ρ\rho in Theorem 2.7 when k16000k\geq 16000 to ρ=3.213303\rho=3.213303. Some further explicit bounds for the Vinogradov’s integral were found also by Steiner [Ste19] using Wooley’s efficient congruencing method [Woo12, Woo13, Woo17]. He proved that for every k3k\geq 3, s52k2+ks\geq\frac{5}{2}k^{2}+k, and Ps10P\geq s^{10}, we have Js,k(P)CP2s12k(k+1)J_{s,k}(P)\leq CP^{2s-\frac{1}{2}k(k+1)}, where CC is roughly kkO(log(k)/log(λ))k^{k^{O(\log(k)/\log(\lambda))}}. However, although the exponent of PP is the optimal one, the constant CC is far too large compared to that one found by Ford in Theorem 2.7, which is of order kO(k3)k^{O(k^{3})}. In order to have a reasonable estimate for Js,kJ_{s,k}, one should find a constant that is at most of order kO(k4)k^{O(k^{4})}, since otherwise, after having taken the k4k^{4}-th root (this passage is required by the argument in Section 4, where the constant appearing in the upper bound for the Vinogradov integral will be elevated to the power of 1/(2rs)1/(2rs), with both rr and ss of order O(k2)O(k^{2})), the constant cannot be controlled. Hence, as already Steiner pointed out in his paper ([Ste19], p. 359), the estimate he found for Js,kJ_{s,k} cannot be applied to Ford’s argument.

In [For02a], Ford found also an upper bound for the incomplete systems defined by (1.9).

Theorem 2.8 ([For02a] Thm.4).

Let k60k\geq 60, h[0.9k,k2]h\in[0.9k,k-2], s[2t,h/2t]s\in[2t,\lfloor h/2\rfloor t] and Pexp(Dk2)P\geq\exp(Dk^{2}) for some D10D\geq 10. Assume that

η(2k3,(2k)1],4logkDk2η[18k1,0.4].\eta\in\left(2k^{-3},(2k)^{-1}\right],\qquad\frac{4\log k}{Dk^{2}\eta}\in\left[18k^{-1},0.4\right].

Then

Js,k,h(𝒞(P,Pη))APEJ_{s,k,h}(\mathscr{C}(P,P^{\eta}))\leq AP^{E}

where

A:=exp(s2t+10.5tlog2kDkη2+s((η1+h)(1h1)s/th)log(10η)),A:=\exp\left(\frac{s^{2}}{t}+\frac{10.5t\log^{2}k}{Dk\eta^{2}}+s\left(\left(\eta^{-1}+h\right)\left(1-h^{-1}\right)^{s/t}-h\right)\log(10\eta)\right),
E:=2st(h+k)2+t(t1)2+ηs22t+htexp(sht).E:=2s-\frac{t(h+k)}{2}+\frac{t(t-1)}{2}+\frac{\eta s^{2}}{2t}+ht\exp\left(-\frac{s}{ht}\right).

Finally, we recall a few results found in [For02a] involving both S(N,t)S(N,t) defined in (1.7) and some bounds for both ζ(s)\zeta(s) and ζ(s,u)\zeta(s,u) that we will use to prove Theorem 1.5 and Theorem 1.1 respectively.

Lemma 2.9 ([For02a] Lemma 5.1).

Suppose k,rk,r and ss are integers 2\geq 2, and hh and gg are integers satisfying 1hgk1\leq h\leq g\leq k. Let NN be a positive integer, and M1,M2M_{1},M_{2} be real numbers with 1MiN1\leq M_{i}\leq N. Let \mathscr{B} be a nonempty subset of the positive integers M2\leq M_{2}. Then

S(N,t)2M1M2+t(M1M2)k+1(k+1)Nk\displaystyle S(N,t)\leq 2M_{1}M_{2}+\frac{t(M_{1}M_{2})^{k+1}}{(k+1)N^{k}}
+N(M2||)1/r((5r)kM22sM12r+k(k+1)/2Jr,k(M1)Js,g,h()WhWg)1/2rs,\displaystyle+N\left(\frac{M_{2}}{|\mathscr{B}|}\right)^{1/r}\left((5r)^{k}M_{2}^{-2s}\lfloor M_{1}\rfloor^{-2r+k(k+1)/2}J_{r,k}(\lfloor M_{1}\rfloor)J_{s,g,h}(\mathscr{B})W_{h}\dots W_{g}\right)^{1/2rs},

where

Wj=min(2sM2j,2sM2jrM1j+stM2jπjNj+4πj(2N)jrtM1j+2)(j1).W_{j}=\min\left(2sM_{2}^{j},\frac{2sM_{2}^{j}}{r\left\lfloor M_{1}\right\rfloor^{j}}+\frac{stM_{2}^{j}}{\pi jN^{j}}+\frac{4\pi j(2N)^{j}}{rt\left\lfloor M_{1}\right\rfloor^{j}}+2\right)\quad(j\geq 1).
Lemma 2.10 ([For02a], Lemma 7.1).

Suppose 12σ1,0<u1,t3\frac{1}{2}\leq\sigma\leq 1,0<u\leq 1,t\geq 3 and s=σ+its=\sigma+it. If either σ1516\sigma\leq\frac{15}{16} or t10100t\leq 10^{100}, then

|ζ(s)|,|ζ(s,u)us|58.1t4(1σ)3/2log2/3t.|\zeta(s)|,\left|\zeta(s,u)-u^{-s}\right|\leq 58.1t^{4(1-\sigma)^{3/2}}\log^{2/3}t.
Lemma 2.11 ([For02a], Lemma 7.2).

If s=σ+it,1516σ1,t10100s=\sigma+it,\frac{15}{16}\leq\sigma\leq 1,t\geq 10^{100} and 0<u10<u\leq 1, then

|ζ(s,u)0nt(n+u)s|1080.\left|\zeta(s,u)-\sum_{0\leq n\leq t}(n+u)^{-s}\right|\leq 10^{-80}.
Lemma 2.12 ([For02a], Lemma 7.3).

Suppose that S(N,t)CN11/(Dλ2)(1Nt)S(N,t)\leq CN^{1-1/\left(D\lambda^{2}\right)}(1\leq N\leq t) for positive constants CC and DD, where λ=logtlogN\lambda=\frac{\log t}{\log N}. Let B=293DB=\frac{2}{9}\sqrt{3D}. Then, for 1516σ1,t10100\frac{15}{16}\leq\sigma\leq 1,t\geq 10^{100} and 0<u10<u\leq 1, we have

|ζ(s)|\displaystyle|\zeta(s)| (C+1+1080log2/3t+1.569CD1/3)tB(1σ)3/2log2/3t\displaystyle\leq\left(\frac{C+1+10^{-80}}{\log^{2/3}t}+1.569CD^{1/3}\right)t^{B(1-\sigma)^{3/2}}\log^{2/3}t
|ζ(s,u)us|\displaystyle\left|\zeta(s,u)-u^{-s}\right| (C+1+1080log2/3t+1.569CD1/3)tB(1σ)3/2log2/3t.\displaystyle\leq\left(\frac{C+1+10^{-80}}{\log^{2/3}t}+1.569CD^{1/3}\right)t^{B(1-\sigma)^{3/2}}\log^{2/3}t.

3. Proof of Theorem 1.4

Let sk2s\ll k^{2}. Due to the size of ss, we are in the first or third cases of Tyrina’s Theorem 2.6. However, since the third case gives an estimate for those ss between two consecutive sns_{n} for which we have an estimate in the first case, we can restrict ourselves to the first case. By Theorem 2.6, we know that

Js,kP2sy(s),J_{s,k}\ll P^{2s-y(s)},

where y(x)y(x) is the inverse function of x(y)x(y), with

x(y)=y+2k(k+1)2log(12(yk)(k2k)).x(y)=-y+2k-(k+1)^{2}\log\left(1-\frac{2(y-k)}{\left(k^{2}-k\right)}\right).

A direct computation gives the following explicit expression for y(x)y(x):

y(x)=k(k+1)2(k+1)2W(k(k1)2(k+1)2exp(k2k+2+x(k+1)22k(k+1)2)),y(x)=\frac{k(k+1)}{2}-(k+1)^{2}W\left(\frac{k(k-1)}{2(k+1)^{2}}\exp\left(\frac{k}{2k+2}+\frac{x}{(k+1)^{2}}-\frac{2k}{(k+1)^{2}}\right)\right),

where WW is the principal branch of the Lambert product-log function.
Furthermore, the function x(y)x(y) is monotonically increasing in the interval [n,n(n+1)2)\left[n,\frac{n(n+1)}{2}\right). Therefore, to lower-bound y(x)y(x), it suffices to upper-bound x(y)x(y). We want to find ss such that y(s)0.101k2y(s)\geq 0.101k^{2}. This means that sx(0.101k2)0.1247k2s\leq x(0.101k^{2})\leq 0.1247k^{2} for every k50k\geq 50.
However,

2s0.101k22sk(k+1)2+0.4k2,k500,s0.1247k2.2s-0.101k^{2}\leq 2s-\frac{k(k+1)}{2}+0.4k^{2},\qquad\forall k\geq 500,\quad s\geq 0.1247k^{2}. (3.1)

It follows that, given

Js,k(P)P2sk(k+1)2+Δs,J_{s,k}(P)\ll P^{2s-\frac{k(k+1)}{2}+\Delta_{s}},

we have

Δs0.4k2s0.1247k2,k500,\Delta_{s}\leq 0.4k^{2}\qquad\forall s\geq 0.1247k^{2},\quad k\geq 500, (3.2)

since Δs\Delta_{s} is decreasing in ss, for every fixed kk.
In order to prove the first part of Theorem 1.4, we follow Ford’s method in [For02a] and we find an estimate for Jnk,kJ_{nk,k}. By (3.2), we know that, given k90000k\geq 90000 fixed, we have Δs0.4k2\Delta_{s}\leq 0.4k^{2} for every s0.1247k2s\geq 0.1247k^{2}. Hence, in particular, if we denote n0=0.1247kn_{0}=\lceil 0.1247k\rceil, we have Δn0:=Δn0k0.4k2\Delta_{n_{0}}:=\Delta_{n_{0}k}\leq 0.4k^{2}. As a result, for the case Δn1>k\Delta_{n-1}>k, we follow Ford’s method in Lemma 3.6 of [For02a] but instead of starting from Δ1\Delta_{1}, we start from Δn0\Delta_{n_{0}}.

Lemma 3.1.

For every 1D0.4k1\leq D\leq 0.4k, k90000k\geq 90000 and

0.138128kn12k(0.6494+log(8k25D)(1D/k2D/k)Dk+2.051k)+10.138128k\leq n\leq\frac{1}{2}k\left(0.6494+\log\left(\frac{8k}{25D}\right)-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+\frac{2.051}{k}\right)+1

we have

Jnk,k(P)CnP2nkk(k+1)/2+ΔnJ_{nk,k}(P)\leq C_{n}P^{2nk-k(k+1)/2+\Delta_{n}}

with

Δnmax(825k2e0.64942n/k(1D/k2D/k)Dk+2.051/k,Dk)\Delta_{n}\leq\max\left(\frac{8}{25}k^{2}e^{0.6494-2n/k-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+2.051/k},Dk\right)

and

Cnk2.055k30.414k2+3nk1.06nk2+2(n2n)k+0.099912k3.C_{n}\leq k^{2.055k^{3}-0.414k^{2}+3nk}1.06^{nk^{2}+2\left(n^{2}-n\right)k+0.099912k^{3}}.
Proof.

We take

rn=k2+k2Δnr_{n}=\left\lfloor\sqrt{k^{2}+k-2\Delta_{n}}\right\rfloor

in Lemma 2.5, where ϕJ\phi_{J} are defined as in Lemma 2.4. Then, we define δn=Δn/k2\delta_{n}=\Delta_{n}/k^{2} for every nn. Now, we fix n2n\geq 2 and we define δ=δn1\delta=\delta_{n-1}, δ=δn\delta^{\prime}=\delta_{n}, Δ=Δn1\Delta=\Delta_{n-1}, Δ=Δn\Delta^{\prime}=\Delta_{n} and r=rn1r=r_{n-1}.
If Δn1k\Delta_{n-1}\leq k, then the upper bound for Δn\Delta_{n} follows trivially, since

ΔnΔn1kDk1D0.4k,k90000.\Delta_{n}\leq\Delta_{n-1}\leq k\leq Dk\qquad\forall 1\leq D\leq 0.4k,\ k\geq 90000.

Hence, from now on, we just consider Δn1>k\Delta_{n-1}>k.
By (3.2), we know that, given k90000k\geq 90000 fixed, we have Δs0.4k2\Delta_{s}\leq 0.4k^{2} for every s0.1247k2s\geq 0.1247k^{2}. Hence, in particular, if we denote n0=0.1247kn_{0}=\lceil 0.1247k\rceil, we have Δn0:=Δn0k0.4k2\Delta_{n_{0}}:=\Delta_{n_{0}k}\leq 0.4k^{2}. At this point, we follow Ford’s method in Lemma 3.6 of [For02a] but instead of starting from Δ1\Delta_{1}, we start from Δn0\Delta_{n_{0}}.
Let

y=2Δ(kr)(kr+1),ϕ=ϕ(k,r,Δ)=2k2rk+y.y=2\Delta-(k-r)(k-r+1),\qquad\phi^{*}=\phi^{*}(k,r,\Delta)=\frac{2k}{2rk+y}.

Using the definition of rnr_{n}, we have

k2+k2Δ1r=rn1=k2+k2Δk2+k2Δ.\sqrt{k^{2}+k-2\Delta}-1\leq r=r_{n-1}=\lfloor\sqrt{k^{2}+k-2\Delta}\rfloor\leq\sqrt{k^{2}+k-2\Delta}.

Hence,

ϕ=ϕ(k,r,Δ)\displaystyle\phi^{*}=\phi^{*}(k,r,\Delta) =2k2rk+y\displaystyle=\frac{2k}{2rk+y}
2k2kk2+k2δk2+δk(2k1+12δk)\displaystyle\geq\frac{2k}{2k\sqrt{k^{2}+k-2\delta k^{2}}+\delta k\left(2k-1+\frac{1}{2\delta k}\right)}
=2k2k21+1k2δ+δk(2k1+12δk)\displaystyle=\frac{2k}{2k^{2}\sqrt{1+\frac{1}{k}-2\delta}+\delta k\left(2k-1+\frac{1}{2\delta k}\right)}
2k2k2(1+12kδ)+δk(2k1+12δk)\displaystyle\geq\frac{2k}{2k^{2}\left(1+\frac{1}{2k}-\delta\right)+\delta k\left(2k-1+\frac{1}{2\delta k}\right)}
=22k+12δk+2δkδ+12k\displaystyle=\frac{2}{2k+1-2\delta k+2\delta k-\delta+\frac{1}{2k}}
=22k+1δ+12k\displaystyle=\frac{2}{2k+1-\delta+\frac{1}{2k}}
1k+1\displaystyle\geq\frac{1}{k+1}

and so the hypotheses of Lemma 2.5 are satisfied.
As per Ford [For02a], we have

θ121jr+2ϕkr0.071k4r+2ϕkr.\theta_{1}\leq\frac{2^{1-j}}{r}+\frac{2\phi^{*}}{kr}\leq\frac{0.071}{k^{4}r}+\frac{2\phi^{*}}{kr}. (3.3)

Since δ0.4\delta\leq 0.4, denoting with

α=k2+k2δk21,\alpha=\sqrt{k^{2}+k-2\delta k^{2}}-1,

we have

ϕ\displaystyle\phi^{*} =2k2rk+y\displaystyle=\frac{2k}{2rk+y} (3.4)
2k2kα+2δk2(kα)(kα+1)\displaystyle\leq\frac{2k}{2k\alpha+2\delta k^{2}-(k-\alpha)(k-\alpha+1)}
2(2δ2)kδ1k+2k2k(3+4k)+2(13kk2)\displaystyle\leq\frac{2}{\left(2-\delta^{2}\right)k-\delta}-\frac{1}{k}+\frac{2k}{\sqrt{2k}(3+4k)+2(-1-3k-k^{2})}
<1.08696k+0.2363k21k+2k2k(3+4k)+2(13kk2)\displaystyle<\frac{1.08696}{k}+\frac{0.2363}{k^{2}}-\frac{1}{k}+\frac{2k}{\sqrt{2k}(3+4k)+2(-1-3k-k^{2})}
1.08696k+0.2363k21k\displaystyle\leq\frac{1.08696}{k}+\frac{0.2363}{k^{2}}-\frac{1}{k}
=0.08696k+0.2363k2\displaystyle=\frac{0.08696}{k}+\frac{0.2363}{k^{2}}

as

2k2k(3+4k)+2(13kk2)<0k90000.\frac{2k}{\sqrt{2k}(3+4k)+2(-1-3k-k^{2})}<0\qquad\forall k\geq 90000.

It follows that

θ10.071k4r+0.08696k2r+0.2363k3r0.08697k2r.\theta_{1}\leq\frac{0.071}{k^{4}r}+\frac{0.08696}{k^{2}r}+\frac{0.2363}{k^{3}r}\leq\frac{0.08697}{k^{2}r}.

As per Ford [For02a],

Δ\displaystyle\Delta^{\prime} =Δk+ϕ+θ12(2kry)\displaystyle=\Delta-k+\frac{\phi^{*}+\theta_{1}}{2}(2kr-y) (3.5)
Δk+12ϕ(2kry)+0.043485k2(2kyr)\displaystyle\leq\Delta-k+\frac{1}{2}\phi^{*}(2kr-y)+\frac{0.043485}{k^{2}}\left(2k-\frac{y}{r}\right)
=Δk+12ϕ(2kry)+0.043485k2(2k1r(2Δ(kr)(kr+1)))\displaystyle=\Delta-k+\frac{1}{2}\phi^{*}(2kr-y)+\frac{0.043485}{k^{2}}\left(2k-\frac{1}{r}\left(2\Delta-(k-r)(k-r+1)\right)\right)
Δk+12ϕ(2kry)+0.043485k2(r1+k2+k2Δr)\displaystyle\leq\Delta-k+\frac{1}{2}\phi^{*}(2kr-y)+\frac{0.043485}{k^{2}}\left(r-1+\frac{k^{2}+k-2\Delta}{r}\right)
=Δ2ky2rk+y+0.043485k2(r1+k2+k2Δr)\displaystyle=\Delta-\frac{2ky}{2rk+y}+\frac{0.043485}{k^{2}}\left(r-1+\frac{k^{2}+k-2\Delta}{r}\right)
=Δ2k+4k2r2rk+y+0.043485k2(r1+k2+k2Δr)\displaystyle=\Delta-2k+4k^{2}\frac{r}{2rk+y}+\frac{0.043485}{k^{2}}\left(r-1+\frac{k^{2}+k-2\Delta}{r}\right)

because

k+12ϕ(2kry)\displaystyle-k+\frac{1}{2}\phi^{*}(2kr-y) =k+122k2rk+y(2kry)=k+k(12y2kr+y)=2ky2rk+y.\displaystyle=-k+\frac{1}{2}\frac{2k}{2rk+y}(2kr-y)=-k+k\left(1-\frac{2y}{2kr+y}\right)=-\frac{2ky}{2rk+y}.

As a function of the real variable rr,

r2rk+y\frac{r}{2rk+y}

has a positive second derivative and a minimum at

r0=k2+k2Δ.r_{0}=\sqrt{k^{2}+k-2\Delta}.

It follows that in the interval

k2+k2Δn11rk2+k2Δ\sqrt{k^{2}+k-2\Delta_{n-1}}-1\leq r\leq\sqrt{k^{2}+k-2\Delta}

the maximum occurs in

r1=k2+k2Δn11.r_{1}=\sqrt{k^{2}+k-2\Delta_{n-1}}-1.

Hence,

r2rk+yr02r0k+y\displaystyle\frac{r}{2rk+y}\geq\frac{r_{0}}{2r_{0}k+y} (3.6)
=k2+k2δk22k(k2+k2δk2)+2δk2(kk2+k2δk2)(kk2+k2δk2+1)\displaystyle=\frac{\sqrt{k^{2}+k-2\delta k^{2}}}{2k(\sqrt{k^{2}+k-2\delta k^{2}})+2\delta k^{2}-(k-\sqrt{k^{2}+k-2\delta k^{2}})(k-\sqrt{k^{2}+k-2\delta k^{2}}+1)}
1δ2k\displaystyle\geq\frac{1-\delta}{2k}

and

r2rk+yr12r1k+y\displaystyle\frac{r}{2rk+y}\leq\frac{r_{1}}{2r_{1}k+y}
=k2+k2δk212k(k2+k2δk21)+2δk2(kk2+k2δk2+1)(kk2+k2δk2+1+1).\displaystyle=\frac{\sqrt{k^{2}+k-2\delta k^{2}}-1}{2k(\sqrt{k^{2}+k-2\delta k^{2}}-1)+2\delta k^{2}-(k-\sqrt{k^{2}+k-2\delta k^{2}}+1)(k-\sqrt{k^{2}+k-2\delta k^{2}}+1+1)}.

We want to estimate the quantity

r12r1k+y\frac{r_{1}}{2r_{1}k+y}

using Taylor expansion for k+k\rightarrow+\infty. We observe that

limkkr12r1k+y=12δ2+412δ+4δ.\displaystyle\lim_{k\rightarrow\infty}k\cdot\frac{r_{1}}{2r_{1}k+y}=\frac{\sqrt{1-2\delta}}{-2+4\sqrt{1-2\delta}+4\delta}.

Furthermore, for all δ(0,0.4)\delta\in(0,0.4) we have

12δ2+412δ+4δ<1δ2δ2.\frac{\sqrt{1-2\delta}}{-2+4\sqrt{1-2\delta}+4\delta}<\frac{1-\delta}{2-\delta^{2}}.

Now,

12δ2+412δ+4δ=12δ2+δ24δ34+δ4163δ516+O(δ6)\frac{\sqrt{1-2\delta}}{-2+4\sqrt{1-2\delta}+4\delta}=\frac{1}{2}-\frac{\delta}{2}+\frac{\delta^{2}}{4}-\frac{\delta^{3}}{4}+\frac{\delta^{4}}{16}-\frac{3\delta^{5}}{16}+O(\delta^{6})

and

1δ2δ2=12δ2+δ24δ34+δ48δ58+O(δ6).\frac{1-\delta}{2-\delta^{2}}=\frac{1}{2}-\frac{\delta}{2}+\frac{\delta^{2}}{4}-\frac{\delta^{3}}{4}+\frac{\delta^{4}}{8}-\frac{\delta^{5}}{8}+O(\delta^{6}).

It follows that

12δ2+412δ+4δ1k(1δ2δ2δ416δ516)1k.\frac{\sqrt{1-2\delta}}{-2+4\sqrt{1-2\delta}+4\delta}\frac{1}{k}\leq\left(\frac{1-\delta}{2-\delta^{2}}-\frac{\delta^{4}}{16}-\frac{\delta^{5}}{16}\right)\frac{1}{k}. (3.7)

Now, we repeat the process for the term of second order. We have

limkk2(r12r1k+y12δ2+412δ+4δ1k)=1+12δ+2δ4(1+212δ+2δ)2.\displaystyle\lim_{k\rightarrow\infty}k^{2}\cdot\left(\frac{r_{1}}{2r_{1}k+y}-\frac{\sqrt{1-2\delta}}{-2+4\sqrt{1-2\delta}+4\delta}\frac{1}{k}\right)=\frac{-1+\sqrt{1-2\delta}+2\delta}{4(-1+2\sqrt{1-2\delta}+2\delta)^{2}}.

As before, using the Taylor expansion in δ=0\delta=0 (δ>1/k0\delta>1/k\rightarrow 0 for k+k\rightarrow+\infty), we have

1+12δ+2δ4(1+212δ+2δ)21k2(4δ(2δ2)23δ4)1k2.\frac{-1+\sqrt{1-2\delta}+2\delta}{4(-1+2\sqrt{1-2\delta}+2\delta)^{2}}\frac{1}{k^{2}}\leq\left(\frac{4\delta}{(2-\delta^{2})^{2}}-\frac{3\delta}{4}\right)\frac{1}{k^{2}}. (3.8)

However, in order to have an upper bound for

r12r1k+y\frac{r_{1}}{2r_{1}k+y}

a direct computation shows that we need to use the slightly weaker bound

1+12δ+2δ4(1+212δ+2δ)21k2(4δ(2δ2)214δ25)1k2,\frac{-1+\sqrt{1-2\delta}+2\delta}{4(-1+2\sqrt{1-2\delta}+2\delta)^{2}}\frac{1}{k^{2}}\leq\left(\frac{4\delta}{(2-\delta^{2})^{2}}-\frac{14\delta}{25}\right)\frac{1}{k^{2}}, (3.9)

where we have 14δ/25-14\delta/25 instead of 3δ/4-3\delta/4. Using the upper bounds found in (3.7) and (3.9) we obtain

r2rk+y(1δ2δ2δ416δ516)1k+(4δ(2δ2)214δ25)1k2.\frac{r}{2rk+y}\leq\left(\frac{1-\delta}{2-\delta^{2}}-\frac{\delta^{4}}{16}-\frac{\delta^{5}}{16}\right)\frac{1}{k}+\left(\frac{4\delta}{(2-\delta^{2})^{2}}-\frac{14\delta}{25}\right)\frac{1}{k^{2}}. (3.10)

From (3.5) and (3.10), it follows that

δ\displaystyle\delta^{\prime} δ2k+4(1δ2δ2δ416δ516)1k+(4δ(2δ2)214δ25)1k2\displaystyle\leq\delta-\frac{2}{k}+4\left(\frac{1-\delta}{2-\delta^{2}}-\frac{\delta^{4}}{16}-\frac{\delta^{5}}{16}\right)\frac{1}{k}+\left(\frac{4\delta}{(2-\delta^{2})^{2}}-\frac{14\delta}{25}\right)\frac{1}{k^{2}}
+0.043485k4(r1+k2+k2Δr)\displaystyle\ \ +\frac{0.043485}{k^{4}}\left(r-1+\frac{k^{2}+k-2\Delta}{r}\right)
=δ(142δ(2δ2)kδ34kδ44k+4(2δ2)2k21425k2)\displaystyle=\delta\left(1-\frac{4-2\delta}{(2-\delta^{2})k}-\frac{\delta^{3}}{4k}-\frac{\delta^{4}}{4k}+\frac{4}{(2-\delta^{2})^{2}k^{2}}-\frac{14}{25k^{2}}\right)
+0.043485k4(k2+k2δk21+k2+k2δk2k2+k2δk21)\displaystyle\ \ +\frac{0.043485}{k^{4}}\left(\sqrt{k^{2}+k-2\delta k^{2}}-1+\frac{k^{2}+k-2\delta k^{2}}{\sqrt{k^{2}+k-2\delta k^{2}}-1}\right)
δ(142δ(2δ2)kδ34kδ44k+4(2δ2)2k21425k2)+0.08697(1δ)k3\displaystyle\leq\delta\left(1-\frac{4-2\delta}{(2-\delta^{2})k}-\frac{\delta^{3}}{4k}-\frac{\delta^{4}}{4k}+\frac{4}{(2-\delta^{2})^{2}k^{2}}-\frac{14}{25k^{2}}\right)+\frac{0.08697(1-\delta)}{k^{3}}
δ(12δ2δ2(2k+δ34k+δ44k3221k2+1425k2))+0.08697k32δ2δ2\displaystyle\leq\delta\left(1-\frac{2-\delta}{2-\delta^{2}}\left(\frac{2}{k}+\frac{\delta^{3}}{4k}+\frac{\delta^{4}}{4k}-\frac{32}{21k^{2}}+\frac{14}{25k^{2}}\right)\right)+\frac{0.08697}{k^{3}}\frac{2-\delta}{2-\delta^{2}}
=δ(12δ2δ2(2k+δ34k+δ44k506525k20.08697k3δ))\displaystyle=\delta\left(1-\frac{2-\delta}{2-\delta^{2}}\left(\frac{2}{k}+\frac{\delta^{3}}{4k}+\frac{\delta^{4}}{4k}-\frac{506}{525k^{2}}-\frac{0.08697}{k^{3}\delta}\right)\right)
δ(12δ2δ2(2k+14k4+14k5506525k20.08697k3δ)),\displaystyle\leq\delta\left(1-\frac{2-\delta}{2-\delta^{2}}\left(\frac{2}{k}+\frac{1}{4k^{4}}+\frac{1}{4k^{5}}-\frac{506}{525k^{2}}-\frac{0.08697}{k^{3}\delta}\right)\right),

where in the last passage we used the fact that δ>1/k\delta>1/k.
Now we define

β=2k506525k2+14k4+14k5,β=βcδ,c=0.08697k3\beta=\frac{2}{k}-\frac{506}{525k^{2}}+\frac{1}{4k^{4}}+\frac{1}{4k^{5}},\qquad\beta^{\prime}=\beta-\frac{c}{\delta},\qquad c=\frac{0.08697}{k^{3}}

and let

δ′′=δ(12δ2δ2β).\delta^{\prime\prime}=\delta\left(1-\frac{2-\delta}{2-\delta^{2}}\beta^{\prime}\right).

Since y+logy+log(2y)y+\log y+\log(2-y) is increasing on (0,12]\left(0,\frac{1}{2}\right], and hence in (0,0.4]\left(0,0.4\right], we have

δ+logδ+log(2δ)\displaystyle\delta^{\prime}+\log\delta^{\prime}+\log\left(2-\delta^{\prime}\right)\leq δ′′+logδ′′+log(2δ′′)\displaystyle\delta^{\prime\prime}+\log\delta^{\prime\prime}+\log\left(2-\delta^{\prime\prime}\right)
=\displaystyle= δ+logδ+log(2δ)\displaystyle\delta+\log\delta+\log(2-\delta)
2δδ22δ2β+log[(12δ2δ2β)(2δ′′2δ)].\displaystyle-\frac{2\delta-\delta^{2}}{2-\delta^{2}}\beta^{\prime}+\log\left[\left(1-\frac{2-\delta}{2-\delta^{2}}\beta^{\prime}\right)\left(\frac{2-\delta^{\prime\prime}}{2-\delta}\right)\right].

As per Ford [For02a], given

T=2δδ22δ2β+log(12δ2δ2β)+log(2δ′′2δ),T=-\frac{2\delta-\delta^{2}}{2-\delta^{2}}\beta^{\prime}+\log\left(1-\frac{2-\delta}{2-\delta^{2}}\beta^{\prime}\right)+\log\left(\frac{2-\delta^{\prime\prime}}{2-\delta}\right),

we have

T\displaystyle T β(β)22(2δ2)2((2δ)2+δ2)+(β)33(2δ2)3((2δ)3+δ3)\displaystyle\leq-\beta^{\prime}-\frac{\left(\beta^{\prime}\right)^{2}}{2\left(2-\delta^{2}\right)^{2}}\left((2-\delta)^{2}+\delta^{2}\right)+\frac{\left(\beta^{\prime}\right)^{3}}{3\left(2-\delta^{2}\right)^{3}}\left(-(2-\delta)^{3}+\delta^{3}\right)
β25(β)2\displaystyle\leq-\beta^{\prime}-\frac{2}{5}\left(\beta^{\prime}\right)^{2}
β25β2+c(1+0.8β)δ.\displaystyle\leq-\beta-\frac{2}{5}\beta^{2}+\frac{c(1+0.8\beta)}{\delta}.

Hence, using an iterative argument we get

δn+logδn+log(2δn)\displaystyle\delta_{n}+\log\delta_{n}+\log\left(2-\delta_{n}\right)\leq δn0+logδn0+log(2δn0)(nn0)(β+0.4β2)\displaystyle\delta_{n_{0}}+\log\delta_{n_{0}}+\log\left(2-\delta_{n_{0}}\right)-(n-n_{0})\left(\beta+0.4\beta^{2}\right)
+c(1+1.6/k)(1δn0++1δn1).\displaystyle+c(1+1.6/k)\left(\frac{1}{\delta_{n_{0}}}+\ldots+\frac{1}{\delta_{n-1}}\right).

Since we are working with Δn>k\Delta_{n}>k, that is δ>1/k\delta>1/k, and we found the inequality

δδ(12δ2δ2(2k+14k4+14k5506525k20.08697k3δ)),\delta^{\prime}\leq\delta\left(1-\frac{2-\delta}{2-\delta^{2}}\left(\frac{2}{k}+\frac{1}{4k^{4}}+\frac{1}{4k^{5}}-\frac{506}{525k^{2}}-\frac{0.08697}{k^{3}\delta}\right)\right), (3.11)

we have

δi+1δi(1α),α=0.869565(βkc).\delta_{i+1}\leq\delta_{i}(1-\alpha),\quad\alpha=0.869565(\beta-kc).

It follows that

c(1+1.6/k)(1δn0++1δn1)c(1+1.6/k)αδn10.051k.c(1+1.6/k)\left(\frac{1}{\delta_{n_{0}}}+\ldots+\frac{1}{\delta_{n-1}}\right)\leq\frac{c(1+1.6/k)}{\alpha\delta_{n-1}}\leq\frac{0.051}{k}.

Therefore

δnδn0(2δn0)eδn0(2δn)eδne(nn0)(β+0.4β2)+0.051/k\delta_{n}\leq\frac{\delta_{n_{0}}\left(2-\delta_{n_{0}}\right)e^{\delta_{n_{0}}}}{\left(2-\delta_{n}\right)e^{\delta_{n}}}e^{-(n-n_{0})\left(\beta+0.4\beta^{2}\right)+0.051/k}

Now,

β+0.4β22k506525k2+14k4+14k5+0.4k2(250652590000)22k.\beta+0.4\beta^{2}\geq\frac{2}{k}-\frac{506}{525k^{2}}+\frac{1}{4k^{4}}+\frac{1}{4k^{5}}+\frac{0.4}{k^{2}}\left(2-\frac{506}{525\cdot 90000}\right)^{2}\geq\frac{2}{k}.

Furthermore

eδn2δn12eδn/(2δn)δn\frac{e^{-\delta_{n}}}{2-\delta_{n}}\leq\frac{1}{2}e^{\delta_{n}/\left(2-\delta_{n}\right)-\delta_{n}}

and, since δn=Δn/k2\delta_{n}=\Delta_{n}/k^{2}, and δn0.4\delta_{n}\leq 0.4 for k90000k\geq 90000, if we write δn=D/k\delta_{n}=D/k, where 1D0.4k1\leq D\leq 0.4k, we have

eδn2δn12eδn/(2δn)δn=12e(1D/k2D/k)Dk.\frac{e^{-\delta_{n}}}{2-\delta_{n}}\leq\frac{1}{2}e^{\delta_{n}/\left(2-\delta_{n}\right)-\delta_{n}}=\frac{1}{2}e^{-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}}.

Also, using the fact that δn00.4\delta_{n_{0}}\leq 0.4, we have

δn0(2δn0)eδn00.4(20.4)e0.4=1625e0.4.\delta_{n_{0}}\left(2-\delta_{n_{0}}\right)e^{\delta_{n_{0}}}\leq 0.4\cdot\left(2-0.4\right)e^{0.4}=\frac{16}{25}e^{0.4}.

Furthermore, for k90000k\geq 90000, we have

e2n0/ke0.2494+2/k.e^{2n_{0}/k}\leq e^{0.2494+2/k}.

It follows that

δn825e0.64942n/k(1D/k2D/k)Dk+2.051/k\delta_{n}\leq\frac{8}{25}e^{0.6494-2n/k-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+2.051/k}

and so

Δn825k2e0.64942n/k(1D/k2D/k)Dk+2.051/k.\Delta_{n}\leq\frac{8}{25}k^{2}e^{0.6494-2n/k-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+2.051/k}.

Now we shift our attention on the constant. As per Ford [For02a], to bound the constants CnC_{n}, we choose ω=0.06>1/(3logk)\omega=0.06>1/(3\log k) in Lemma 2.5, so that

Vk+1=(300k3logk)k+1k4.11k=:WV^{k+1}=\left(300k^{3}\log k\right)^{k+1}\leq k^{4.11k}=:W

for every k90000k\geq 90000. Then,

WΔn1Δn>k3k1.064k(n1)+k2(n0.138k+1),W^{\Delta_{n-1}-\Delta_{n}}>k^{3k}1.06^{4k(n-1)+k^{2}}\quad(n\leq 0.138k+1), (3.12)

since it was proved by Ford in [For02a] for the wider range n1.97k+1n\leq 1.97k+1.
Now, let n1=0.138k+1n_{1}=\lfloor 0.138k\rfloor+1. By (3.12) and Lemma 2.5,

Cn1WΔ1Δn1k!Wk2/2Δn1C_{n_{1}}\leq W^{\Delta_{1}-\Delta_{n_{1}}}k!\leq W^{k^{2}/2-\Delta_{n_{1}}}

and, for n>n1n>n_{1},

Cnk3k1.064k(n1)+k2WΔn1ΔnCn1.C_{n}\leq k^{3k}1.06^{4k(n-1)+k^{2}}W^{\Delta_{n-1}-\Delta_{n}}C_{n-1}.

Iterating this last inequality gives, for n>n1n>n_{1},

Cn\displaystyle C_{n} Wk2/2k3k(nn1)1.06(nn1)k2+4k(n1++n1)\displaystyle\leq W^{k^{2}/2}k^{3k\left(n-n_{1}\right)}1.06^{\left(n-n_{1}\right)k^{2}+4k\left(n_{1}+\ldots+n-1\right)}
Wk2/2k3k(n0.138k)1.06(n0.138k)k2+2k(n2n(0.138k)2+0.138k)\displaystyle\leq W^{k^{2}/2}k^{3k(n-0.138k)}1.06^{(n-0.138k)k^{2}+2k\left(n^{2}-n-(0.138k)^{2}+0.138k\right)}
k2.055k30.414k2+3nk1.06nk2+2(n2n)k+0.099912k3.\displaystyle\leq k^{2.055k^{3}-0.414k^{2}+3nk}1.06^{nk^{2}+2\left(n^{2}-n\right)k+0.099912k^{3}}.

This concludes the proof of the lemma. ∎

Remark.

In Lemma 3.1, contrary to Lemma 3.6 in [For02a], we are not allow to consider 2k2k as a lower bound for the range of nn for which the lemma holds. Indeed, the upper bound for nn, that is 12k(0.6494+log(8k25D)(1D/k2D/k)Dk+2.051k)+1\frac{1}{2}k\left(0.6494+\log\left(\frac{8k}{25D}\right)-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+\frac{2.051}{k}\right)+1, is a decreasing function in DD. As a result, this upper bound reaches the minimum when, for k90000k\geq 90000 fixed, D=0.4kD=0.4k. Substituting D=0.4kD=0.4k in the above expression, one gets n2.0255+0.138128kn\leq 2.0255+0.138128k, which is less than 2k2k for k90000k\geq 90000. Hence, in order to have an estimate for Δn\Delta_{n} and CnC_{n} which is uniform for every 1D0.4k1\leq D\leq 0.4k and k90000k\geq 90000, we took 0.138128k0.138128k as lower bound for nn, in order to ensure that for every DD and kk there exists always at least a value of nn in the range given in the hypotheses of Lemma 3.1. Also, the choice 0.138128k0.138128k as lower bound for nn is admissible, as in Lemma 3.1 the starting point is n0=0.1247kn_{0}=\lceil 0.1247k\rceil which is less than 0.138128k0.138128k for k90000k\geq 90000.

Now, we will use Lemma 3.1 to prove the first part of the theorem.
Given k90000k\geq 90000, every admissible ss such that s0(modk)s\not\equiv 0\ (\bmod\ k), is of the form s=nk+us=nk+u, where 0<u<k,usmodk0<u<k,\ u\equiv s\bmod k, and

0.138128kn12k(0.6494+log(8k25D)(1D/k2D/k)Dk+2.051k).0.138128k\leq n\leq\frac{1}{2}k\left(0.6494+\log\left(\frac{8k}{25D}\right)-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+\frac{2.051}{k}\right).

Using Hölder’s inequality, we get

Js,k(P)=[0,1]k|S|2nk+2u𝑑𝜶([0,1]k|S|2nk𝑑𝜶)1u/k([0,1]k|S|2nk|S|2k𝑑𝜶)u/k,J_{s,k}(P)=\int_{[0,1]^{k}}|S|^{2nk+2u}d\boldsymbol{\alpha}\leq\left(\int_{[0,1]^{k}}|S|^{2nk}d\boldsymbol{\alpha}\right)^{1-u/k}\left(\int_{[0,1]^{k}}|S|^{2nk}|S|^{2k}d\boldsymbol{\alpha}\right)^{u/k},

where 𝜶=(α1,,αk)\boldsymbol{\alpha}=(\alpha_{1},\dots,\alpha_{k}) and

S:=1xPe(α1x++αkxk).S:=\sum_{1\leq x\leq P}e\left(\alpha_{1}x+\cdots+\alpha_{k}x^{k}\right).

By Lemma 3.1, we have

Js,k(P)k2.055k30.414k2+3nk1.06nk2+2(n2n)k+0.099912k3P2sk(k+1)/2+ΔJ_{s,k}(P)\leq k^{2.055k^{3}-0.414k^{2}+3nk}1.06^{nk^{2}+2\left(n^{2}-n\right)k+0.099912k^{3}}P^{2s-k(k+1)/2+\Delta}

where

Δmax(825k2e0.64942n/k(1D/k2D/k)Dk+2.051/k,Dk)×(1uk+uke2/k).\Delta\leq\max\left(\frac{8}{25}k^{2}e^{0.6494-2n/k-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+2.051/k},Dk\right)\times\left(1-\frac{u}{k}+\frac{u}{k}e^{-2/k}\right).

Further,

1uk+uke2/k12uk2+2uk3e2u/k2+2u/k3.1-\frac{u}{k}+\frac{u}{k}e^{-2/k}\leq 1-\frac{2u}{k^{2}}+\frac{2u}{k^{3}}\leq e^{-2u/k^{2}+2u/k^{3}}.

Hence,

Δ\displaystyle\Delta\leq max(825k2exp(0.64942(nk+uk)k2(1D/k2D/k)Dk+2.051k+2uk3),\displaystyle\max\left(\frac{8}{25}k^{2}\exp\left(0.6494-\frac{2(nk+u-k)}{k^{2}}-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+\frac{2.051}{k}+\frac{2u}{k^{3}}\right),\right.
Dkexp(2uk2+2uk3))\displaystyle\ \ \quad\ \ \left.Dk\exp\left(-\frac{2u}{k^{2}}+\frac{2u}{k^{3}}\right)\right)
=\displaystyle= max(825k2exp(0.64942(sk)k2(1D/k2D/k)Dk+2.051k+2(smodk)k3),\displaystyle\max\left(\frac{8}{25}k^{2}\exp\left(0.6494-\frac{2(s-k)}{k^{2}}-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+\frac{2.051}{k}+\frac{2(s\bmod k)}{k^{3}}\right),\right.
Dkexp(2(smodk)k2(1+1k))).\displaystyle\ \ \quad\ \ \left.Dk\exp\left(\frac{2(s\bmod k)}{k^{2}}\left(-1+\frac{1}{k}\right)\right)\right).

This completes the first part of Theorem 1.4.
We now turn to the second part of the theorem. For k<90000k<90000, we follow Ford’s argument for proving Theorem 2.7 in [For02a]. Running Program 1 listed in Section 9 with ϕJ\phi_{J} defined as in Lemma 2.3, we get the desired bounds for kk in the ranges

[129,137],[138,139],[140,146],[147,148],[149,170],[171,190],[191,339],[340,499].[129,137],\ [138,139],\ [140,146],\ [147,148],\ [149,170],\ [171,190],\ [191,339],\ [340,499].

For 500k<90000500\leq k<90000 we use Program 2 in Section 9, with ϕJ\phi_{J} defined as in Lemma 2.3. Program 2 differs from Program 1 in the definition of the variable del0\operatorname{del0}, as, due to (3.2), we know that, for k500k\geq 500, Δn0.4k2\Delta_{n}\leq 0.4k^{2} for every nn0=0.1247kn\geq n_{0}=\lceil 0.1247k\rceil. Hence, in order to initialize del0\operatorname{del0}, we start from Δn00.4k2\Delta_{n_{0}}\leq 0.4k^{2} instead of starting from Δ112k2(11k)\Delta_{1}\leq\frac{1}{2}k^{2}(1-\frac{1}{k}).
For k90000k\geq 90000, we follow Preobrazhenskiĭ’s argument [Pre11]. Taking D=0.001kD=0.001k in Lemma 3.1, we observe that for every k90000k\geq 90000,

2kn12k(0.6494+log(8k25D)(1D/k2D/k)Dk+2.051k).2k\leq n\leq\frac{1}{2}k\left(0.6494+\log\left(\frac{8k}{25D}\right)-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+\frac{2.051}{k}\right).

Hence, since from now on we will work with D=0.001kD=0.001k, we will restrict the range of nn in Lemma 3.1 to

2kn12k(0.6494+log(8k25D)(1D/k2D/k)Dk+2.051k)+1.2k\leq n\leq\frac{1}{2}k\left(0.6494+\log\left(\frac{8k}{25D}\right)-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+\frac{2.051}{k}\right)+1. (3.13)

Choosing

n=12k(0.6494+log(8k25D)(1D/k2D/k)Dk+2.051k),n=\left\lceil\frac{1}{2}k\left(0.6494+\log\left(\frac{8k}{25D}\right)-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+\frac{2.051}{k}\right)\right\rceil, (3.14)

we have, by Lemma 3.1, that

Δn\displaystyle\Delta_{n} max(825k2e0.64942n/k(10.00120.001)0.001+2.051/k,0.001k2)\displaystyle\leq\max\left(\frac{8}{25}k^{2}e^{0.6494-2n/k-\left(\frac{1-0.001}{2-0.001}\right)0.001+2.051/k},0.001k^{2}\right)
max(825k2elog(8250.001),0.001k2)\displaystyle\leq\max\left(\frac{8}{25}k^{2}e^{-\log\left(\frac{8}{25\cdot 0.001}\right)},0.001k^{2}\right)
=0.001k2,\displaystyle=0.001k^{2},

where we used the fact that

2nk(0.6494+log(8k25D)(1D/k2D/k)Dk+2.051k).-\frac{2n}{k}\leq-\left(0.6494+\log\left(\frac{8k}{25D}\right)-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+\frac{2.051}{k}\right).

Hence, for every k90000k\geq 90000 fixed, there exists sρk2s\leq\rho k^{2} such that

Js,k(P)kθk3P2s12k(k+1)+0.001k2,J_{s,k}(P)\leq k^{\theta k^{3}}P^{2s-\frac{1}{2}k(k+1)+0.001k^{2}},

where

ρnkk212(0.6494+log(8k25D)(1D/k2D/k)Dk+2.051k)+1k3.20863\rho\leq\frac{nk}{k^{2}}\leq\frac{1}{2}\left(0.6494+\log\left(\frac{8k}{25D}\right)-\left(\frac{1-D/k}{2-D/k}\right)\frac{D}{k}+\frac{2.051}{k}\right)+\frac{1}{k}\leq 3.20863

and θ=2.17720\theta=2.17720 ( the value of θ\theta follows from a uniform upper bound for the constant

k2.055k30.414k2+3nk1.06nk2+2(n2n)k+0.099912k3k^{2.055k^{3}-0.414k^{2}+3nk}1.06^{nk^{2}+2\left(n^{2}-n\right)k+0.099912k^{3}}

found in the first part of Theorem 1.4 which holds for every k90000k\geq 90000, and nn in the range (3.13)). This estimate holds since, using our choice (3.14) for nn, we have s=nkρk2s=nk\leq\rho k^{2} and Δn0.001k2.\Delta_{n}\leq 0.001k^{2}.

Remark.

The starting point Δn00.4k2\Delta_{n_{0}}\leq 0.4k^{2} with n0=0.1247kn_{0}=\lceil 0.1247k\rceil is nearly the optimal choice. Indeed, taking Δn0ck2\Delta_{n_{0}}\leq ck^{2} with c<0.4c<0.4 would imply a greater value for n0n_{0}. A direct computation shows that, although the constant 825\frac{8}{25} in the estimate of Δn\Delta_{n} would decrease, the quantity e0.6494e^{0.6494} would increase, leading to an overall estimate for Δn\Delta_{n} which is worse than that one found by Ford in Lemma 3.6 of [For02a] when Δn>k\Delta_{n}>k.
Furthermore, the lower bound for kk in (3.1) is optimal, as the inequality in (3.1) is no longer valid for k<500k<500. Finally, also the value 0.101k20.101k^{2} in (3.1) is nearly optimal, as this inequality is satisfied only with 2sak22s-ak^{2} and a>0.1a>0.1.

4. Proof of Theorem 1.5

We will consider the cases λ84\lambda\leq 84 and λ84\lambda\geq 84 separately.

4.1. Case λ84\lambda\geq 84

We combine Ford’s method [For02a] with the new estimates for the Vinogradov’s integral found in Theorem 1.4 to obtain an improved upper bound for S(N,t)S(N,t) in Lemma 2.9.
As in [For02a], we make the following assumptions:

M1M2100g,s2g,r13g,rs,gh3.\left\lfloor M_{1}\right\rfloor\geq M_{2}\geq 100g,\quad s\leq 2^{g},\quad r\geq 13g,\quad r\geq s,\quad g\geq h\geq 3.

Also, we define

M1=Nμ1,M2=Nμ2,μ1>μ2M_{1}=N^{\mu_{1}},\quad M_{2}=N^{\mu_{2}},\quad\mu_{1}>\mu_{2}

and

ϕ=g/λ,γ=h/λ,1γ11μ2<11μ1ϕ11μ1μ2.\phi=g/\lambda,\quad\gamma=h/\lambda,\quad 1\leq\gamma\leq\frac{1}{1-\mu_{2}}<\frac{1}{1-\mu_{1}}\leq\phi\leq\frac{1}{1-\mu_{1}-\mu_{2}}. (4.1)

Following the proof of Theorem 2 in [For02a] for λ\lambda large, using the notation of Lemma 2.9, we have

WhWg2g2M2h+(h+1)++gNHW_{h}\cdots W_{g}\leq 2^{g^{2}}M_{2}^{h+(h+1)+\cdots+g}N^{-H} (4.2)

where

H\displaystyle H\geq λ2(ϕ+γγ221μ1μ22ϕ22μ1μ22(1μ1)(1μ2))\displaystyle\lambda^{2}\left(\phi+\gamma-\frac{\gamma^{2}}{2}-\frac{1-\mu_{1}-\mu_{2}}{2}\phi^{2}-\frac{2-\mu_{1}-\mu_{2}}{2\left(1-\mu_{1}\right)\left(1-\mu_{2}\right)}\right) (4.3)
+λ(γ2ϕ2(1μ1μ2))2μ1μ28\displaystyle\quad+\lambda\left(\frac{\gamma}{2}-\frac{\phi}{2}\left(1-\mu_{1}-\mu_{2}\right)\right)-\frac{2-\mu_{1}-\mu_{2}}{8}
=:H2λ2+H1λH0.\displaystyle=:H_{2}\lambda^{2}+H_{1}\lambda-H_{0}.

At this point, we shall take the near-optimal values

μ1=0.1905,μ2=0.1603,k=λ1μ1μ2+0.000003129,r=ρk2+1,\mu_{1}=0.1905,\quad\mu_{2}=0.1603,\quad k=\left\lfloor\frac{\lambda}{1-\mu_{1}-\mu_{2}}+0.000003\right\rfloor\geq 129,\quad r=\left\lfloor\rho k^{2}+1\right\rfloor, (4.4)

where ρ\rho is taken from Theorem 1.4.
Furthermore, we choose Y=288Y=288 and we want to estimate S(N,t)S(N,t) when NeYλ2N\geq e^{Y\lambda^{2}}, since otherwise for NeYλ2N\leq e^{Y\lambda^{2}} Theorem 1.5 follows trivially:

S(N,t)NeY/132.94357N11/132.94357λ28.7979N11/132.94357λ2.S(N,t)\leq N\leq e^{Y/132.94357}N^{1-1/132.94357\lambda^{2}}\leq 8.7979N^{1-1/132.94357\lambda^{2}}.

Finally, we consider

105g1.254λ105\leq g\leq 1.254\lambda (4.5)

and the following bounds for the quantity k/λk/\lambda:

k0:=10.64920.999997λkλ10.6492+0.000003λ=:k1k_{0}:=\frac{1}{0.6492}-\frac{0.999997}{\lambda}\leq\frac{k}{\lambda}\leq\frac{1}{0.6492}+\frac{0.000003}{\lambda}=:k_{1}\text{. }

First of all, by (4.4) and Theorem 1.4 we have

M12r+12k(k+1)Jr,k(M1)C1M10.001k2,\left\lfloor M_{1}\right\rfloor^{-2r+\frac{1}{2}k(k+1)}J_{r,k}\left(\left\lfloor M_{1}\right\rfloor\right)\leq C_{1}M_{1}^{0.001k^{2}}, (4.6)

where C1=kθk3C_{1}=k^{\theta k^{3}} and θ\theta comes from Theorem 1.4.
Since NeYλ2N\geq e^{Y\lambda^{2}}, from (4.4) and (4.5) we have M2eμ2Yλ2e0.1019Yg2M_{2}\geq e^{\mu_{2}Y\lambda^{2}}\geq e^{0.1019Yg^{2}}. Let D=0.1019Y=29.3472D=0.1019Y=29.3472 and η=1ξg3/2\eta=\frac{1}{\xi g^{3/2}}, where 3ξ63\leq\xi\leq 6. By the assumption on gg, (4.5), the hypotheses of Theorem 2.8 are satisfied if we take P=M2P=M_{2} and k=gk=g. Hence, by Theorem 2.8,

Js,g,h(𝒞(M2,M2η))C2P2st2(h+g)+E2,J_{s,g,h}\left(\mathscr{C}\left(M_{2},M_{2}^{\eta}\right)\right)\leq C_{2}P^{2s-\frac{t}{2}(h+g)+E_{2}}, (4.7)

where

E2\displaystyle E_{2} =12t(t1)+ηs22t+htexp{sht}\displaystyle=\frac{1}{2}t(t-1)+\frac{\eta s^{2}}{2t}+ht\exp\left\{-\frac{s}{ht}\right\}
logC2\displaystyle\log C_{2} =s2t+10.5ξ2tg2log2gDs((ξg3/2+h)(11/h)s/th)log(ξg3/2/10).\displaystyle=\frac{s^{2}}{t}+\frac{10.5\xi^{2}tg^{2}\log^{2}g}{D}-s\left(\left(\xi g^{3/2}+h\right)(1-1/h)^{s/t}-h\right)\log\left(\xi g^{3/2}/10\right).

Since by the hypotheses of Theorem 2.8 and the range (4.5) for gg we have

R=M2ηeDg2ηg10>625,R=M_{2}^{\eta}\geq e^{Dg^{2}\eta}\geq g^{10}>6^{25},

following exactly the argument used by Ford in the proof of Theorem 2 in [For02a] for λ\lambda large with δ=125\delta=\frac{1}{25} instead of δ=126\delta=\frac{1}{26} we get

M2|𝒞(M2,R)|\displaystyle\frac{M_{2}}{\left|\mathscr{C}\left(M_{2},R\right)\right|} (logR)(1.0417ξg3/2+1)(26.0417ξg3/22.5)1.0417ξg3/2\displaystyle\leq(\log R)\left(1.0417\xi g^{3/2}+1\right)\left(\frac{26.0417\xi g^{3/2}}{2.5}\right)^{1.0417\xi g^{3/2}}
(logN)C3C3NE3,\displaystyle\leq(\log N)C_{3}\leq C_{3}N^{E_{3}},

where

C3\displaystyle C_{3} =(10.4167ξg3/2)1.0417ξg3/2\displaystyle=\left(10.4167\xi g^{3/2}\right)^{1.0417\xi g^{3/2}}
E3\displaystyle E_{3} =log(Yλ2)Yλ2.\displaystyle=\frac{\log\left(Y\lambda^{2}\right)}{Y\lambda^{2}}.

By (4.4), we have

(5r)k(38.2λ2)1.55λλ4.65λ(5r)^{k}\leq\left(38.2\lambda^{2}\right)^{1.55\lambda}\leq\lambda^{4.65\lambda} (4.8)

and

r7.6λ2λ2.r\geq 7.6\lambda^{2}\geq\lambda^{2}.

It follows that

E3rlog(Yλ2)7.6Yλ4log(Yλ2)Yλ4.\frac{E_{3}}{r}\leq\frac{\log\left(Y\lambda^{2}\right)}{7.6Y\lambda^{4}}\leq\frac{\log\left(Y\lambda^{2}\right)}{Y\lambda^{4}}. (4.9)
Lemma 4.1.

For 84λ22084\leq\lambda\leq 220 and Ne288λ2N\geq e^{288\lambda^{2}} we have the following estimate

S(N,t)8.7979N11/132.94357λ2.S(N,t)\leq 8.7979N^{1-1/132.94357\lambda^{2}}.
Proof.

From Lemma 2.9, (4.2), (4.4), (4.6), (4.7) and (4.8) we have

S(N,t)\displaystyle S(N,t) (C31r(λ4.65λC1C2)12rs)N1+E+2N0.36+1kN10.0000019476\displaystyle\leq\left(C_{3}^{\frac{1}{r}}\left(\lambda^{4.65\lambda}C_{1}C_{2}\right)^{\frac{1}{2rs}}\right)N^{1+E}+2N^{0.36}+\frac{1}{k}N^{1-0.0000019476} (4.10)
E\displaystyle E =log(Yλ2)Yλ4+12rs(H+0.001μ1k2+μ2E2).\displaystyle=\frac{\log\left(Y\lambda^{2}\right)}{Y\lambda^{4}}+\frac{1}{2rs}\left(-H+0.001\mu_{1}k^{2}+\mu_{2}E_{2}\right).

Now, we choose

ξ=3.612381,σ=0.330201.\xi=3.612381,\qquad\sigma=0.330201.

Furthermore, taking s=σht+1s=\lfloor\sigma ht\rfloor+1, we make the same assumptions as in Lemma 5.3 of [For02a]

g=λ1μ1+1+a,h=λ1μ2b,t=gh+1,a,b{0,1},g=\left\lfloor\frac{\lambda}{1-\mu_{1}}\right\rfloor+1+a,\quad h=\left\lfloor\frac{\lambda}{1-\mu_{2}}\right\rfloor-b,\quad t=g-h+1,\quad a,b\in\{0,1\},

where gg satisfies (4.5). Then, we bound the exponent of NN in each interval λI=[λ1,λ2)\lambda\in I=\left[\lambda_{1},\lambda_{2}\right) where each of the quantities m1=λ1μ1,m2=λ1μ2m_{1}=\left\lfloor\frac{\lambda}{1-\mu_{1}}\right\rfloor,m_{2}=\left\lfloor\frac{\lambda}{1-\mu_{2}}\right\rfloor and kk defined in (4.4) is constant. We also take constant values of aa and bb in II, so that g,h,t,s,rg,h,t,s,r are also fixed.
As in [For02a], we define for λI\lambda\in I the quantities

H\displaystyle H =Z0+Z1λ,\displaystyle=Z_{0}+Z_{1}\lambda,
Z0\displaystyle Z_{0} =(m12+m1)(1μ1)+(m22+m2)(1μ2)h2+h(1μ1μ2)(g2+g)2,\displaystyle=\frac{\left(m_{1}^{2}+m_{1}\right)\left(1-\mu_{1}\right)+\left(m_{2}^{2}+m_{2}\right)\left(1-\mu_{2}\right)-h^{2}+h-\left(1-\mu_{1}-\mu_{2}\right)\left(g^{2}+g\right)}{2},
Z1\displaystyle Z_{1} =h+gm1m21=ab{1,0,1},\displaystyle=h+g-m_{1}-m_{2}-1=a-b\in\{-1,0,1\},

so that

HH:=Z0+{λ1Z1=10Z1=0λ2Z1=1.H\geq H^{\prime}:=Z_{0}+\left\{\begin{array}[]{ll}\lambda_{1}&Z_{1}=1\\ 0&Z_{1}=0\\ -\lambda_{2}&Z_{1}=-1\end{array}\right..

It follows that

Elog(Yλ12)Yλ14H0.001μ1k2μ2(t(t1)2+s2ξtg3/2+htes/(ht))2rs:=E.E\leq\frac{\log\left(Y\lambda_{1}^{2}\right)}{Y\lambda_{1}^{4}}-\frac{H^{\prime}-0.001\mu_{1}k^{2}-\mu_{2}\left(\frac{t(t-1)}{2}+\frac{s^{2}}{\xi tg^{3/2}}+hte^{-s/(ht)}\right)}{2rs}:=E^{\prime}.

At this point we use Program 3 in Section 9 to find the best values for CC and uu, under the condition u132.94357u\leq 132.94357, so that, for λI\lambda\in I,

S(N,t)CN11/(uλ2)+1kN11/132λ2.S(N,t)\leq CN^{1-1/\left(u\lambda^{2}\right)}+\frac{1}{k}N^{1-1/132\lambda^{2}}.

where u=1/(Eλ12)u=1/\left(E^{\prime}\lambda_{1}^{2}\right) and C=C31/r(λ4.65λC1C2)1/(2rs)C=C_{3}^{1/r}\left(\lambda^{4.65\lambda}C_{1}C_{2}\right)^{1/(2rs)}. Running Program 3 we can notice that in each interval II we have C8.7979C\leq 8.7979 and u132.94357u\leq 132.94357. The conclusion follows. ∎

Lemma 4.2.

For λ220\lambda\geq 220 and Ne288λ2N\geq e^{288\lambda^{2}} we have the following estimate

S(N,t)7.5N11/132.94357λ2.S(N,t)\leq 7.5N^{1-1/132.94357\lambda^{2}}.
Proof.

From Lemma 2.9, (4.2), (4.4), (4.6), (4.7) and (4.8) we have

S(N,t)\displaystyle S(N,t) (C31r(λ4.65λC1C2)12rs)N1+E+2N0.36+1kN10.0000019476\displaystyle\leq\left(C_{3}^{\frac{1}{r}}\left(\lambda^{4.65\lambda}C_{1}C_{2}\right)^{\frac{1}{2rs}}\right)N^{1+E}+2N^{0.36}+\frac{1}{k}N^{1-0.0000019476} (4.11)
E\displaystyle E =log(Yλ2)7.6Yλ4+12rs(H+0.001μ1k2+μ2E2).\displaystyle=\frac{\log\left(Y\lambda^{2}\right)}{7.6Y\lambda^{4}}+\frac{1}{2rs}\left(-H+0.001\mu_{1}k^{2}+\mu_{2}E_{2}\right).

Now, we assume

h=1.17928λ+12,g=1.24788λ+12,s=σh(t1)+1,h=\left\lfloor 1.17928\lambda+\frac{1}{2}\right\rfloor,\quad g=\left\lfloor 1.24788\lambda+\frac{1}{2}\right\rfloor,\quad s=\lfloor\sigma h(t-1)+1\rfloor, (4.12)

where

σ=0.3299,t=gh+1.\sigma=0.3299,\quad t=g-h+1. (4.13)

From (4.1), (4.12) and (4.13), the relation (4.5) holds and, furthermore, we have

|γ1.17928|12λ,|ϕ1.24788|12λ.|\gamma-1.17928|\leq\frac{1}{2\lambda},\quad|\phi-1.24788|\leq\frac{1}{2\lambda}.

From (4.4), (4.12) and (4.13) we have

g275,h259,t1713,k338,s0.02668λ20.02294λ2.g\geq 275,\quad h\geq 259,\quad t\geq 17\geq 13,\quad k\geq 338,\quad s\geq 0.02668\lambda^{2}\geq 0.02294\lambda^{2}.

Following exactly the proof of Lemma 5.2 in [For02a], with Y=288Y=288 instead of Y=300Y=300, with the adjusted value of D=0.1090Y=29.3472D=0.1090Y=29.3472 instead of D=30.57D=30.57, we still have

C31/r(λ4.65λC1C2)1/2rs7.5.C_{3}^{1/r}\left(\lambda^{4.65\lambda}C_{1}C_{2}\right)^{1/2rs}\leq 7.5.

It remains to estimate EE. We have

E\displaystyle E\leq log(Yλ2)7.6Yλ4+H+0.001μ1k22.00002ρσγ(ϕγ)λ2k2+μ2E22ρk2s\displaystyle\frac{\log\left(Y\lambda^{2}\right)}{7.6Y\lambda^{4}}+\frac{-H+0.001\mu_{1}k^{2}}{2.00002\rho\sigma\gamma(\phi-\gamma)\lambda^{2}k^{2}}+\frac{\mu_{2}E_{2}}{2\rho k^{2}s}
\displaystyle\leq 1.52×107λ2+λ2H2λH1+H02.00002ρσγ(ϕγ)λ2k2+0.001μ12.00002ρσγ(ϕγ)λ2\displaystyle\frac{1.52\times 10^{-7}}{\lambda^{2}}+\frac{-\lambda^{2}H_{2}-\lambda H_{1}+H_{0}}{2.00002\rho\sigma\gamma(\phi-\gamma)\lambda^{2}k^{2}}+\frac{0.001\mu_{1}}{2.00002\rho\sigma\gamma(\phi-\gamma)\lambda^{2}}
+μ22ρk2[ϕγ+1/λ2σγ+(t/(t1))eσ+σ/tσ+σhg3/212].\displaystyle+\frac{\mu_{2}}{2\rho k^{2}}\left[\frac{\phi-\gamma+1/\lambda}{2\sigma\gamma}+\frac{(t/(t-1))e^{-\sigma+\sigma/t}}{\sigma}+\frac{\sigma hg^{-3/2}}{12}\right].

Following Ford’s argument in [For02a], we have

λ2E1.52×107+f(γ,ϕ)+G1/λ1/2+G2/λρ\lambda^{2}E\leq 1.52\times 10^{-7}+\frac{f(\gamma,\phi)+G_{1}/\lambda^{1/2}+G_{2}/\lambda}{\rho}

where

f(γ,ϕ)\displaystyle f(\gamma,\phi) =12.00002σγ[0.001μ1ϕγ+1k12(H2ϕγ+1.00001μ2(12(ϕγ)+γeσ))],\displaystyle=\frac{1}{2.00002\sigma\gamma}\left[\frac{0.001\mu_{1}}{\phi-\gamma}+\frac{1}{k_{1}^{2}}\left(\frac{-H_{2}}{\phi-\gamma}+1.00001\mu_{2}\left(\frac{1}{2}(\phi-\gamma)+\gamma e^{-\sigma}\right)\right)\right],
G1\displaystyle G_{1} =μ2σγϕ3/224k020.0008,\displaystyle=\frac{\mu_{2}\sigma\gamma\phi^{-3/2}}{24k_{0}^{2}}\leq 0.0008,
G2\displaystyle G_{2} =12.00002σ(k/λ)2[H1+H0/λ+1.33547μ2γeσγ(ϕγ)+1.00001μ22γ].\displaystyle=\frac{1}{2.00002\sigma(k/\lambda)^{2}}\left[\frac{-H_{1}+H_{0}/\lambda+1.33547\mu_{2}\gamma e^{-\sigma}}{\gamma(\phi-\gamma)}+\frac{1.00001\mu_{2}}{2\gamma}\right].

Using Ford’s estimate for the expression inside the brackets in the definition of G2G_{2}, we have

G20.03922.00002σγk020.0213552.G_{2}\leq\frac{0.0392}{2.00002\sigma\gamma k_{0}^{2}}\leq 0.0213552.

It follows that

λ2E\displaystyle\lambda^{2}E 1.56×107+f(γ,ϕ)+0.0008λ1/2+0.0213552λ1ρ\displaystyle\leq 1.56\times 10^{-7}+\frac{f(\gamma,\phi)+0.0008\lambda^{-1/2}+0.0213552\lambda^{-1}}{\rho}
0.0000473+f(γ,ϕ)ρ.\displaystyle\leq 0.0000473+\frac{f(\gamma,\phi)}{\rho}.

In the range

|ϕ1.24788|1440,|γ1.17928|1440,|\phi-1.24788|\leq\frac{1}{440},\qquad|\gamma-1.17928|\leq\frac{1}{440},

f(γ,ϕ)f(\gamma,\phi) is increasing, hence the maximum occurring at γ=1.17928+1440,ϕ=1.247881440\gamma=1.17928+\frac{1}{440},\phi=1.24788-\frac{1}{440}, with

f(γ,ϕ)0.024287046496.f(\gamma,\phi)\leq-0.024287046496.

Using ρ=3.20863\rho=3.20863 from Theorem 1.4, it follows that

λ2E1132.94357.\lambda^{2}E\leq-\frac{1}{132.94357}.

Theorem 1.5 for λ84\lambda\geq 84 follows directly from Lemma 4.1 and Lemma 4.2.

Remark.

One might have considered the ranges 84λN84\leq\lambda\leq N and λ>N\lambda>N, with N>220N>220. However, a direct computation shows that even for really large values of NN, of the order of 10610^{6}, and hence, by definition of kk, for values of kk much greater than 9000090000, the improvement is negligible. Indeed, one can see that a new choice for NN would not influence the estimate for S(N,t)S(N,t) in the range 84λN84\leq\lambda\leq N we found in Lemma 4.1, while for the case λ>N\lambda>N, if NN, and hence kk, would be sufficiently large, the value of ρ\rho in Theorem 1.4 would be 3.208613.20861 for kk sufficiently large, which is slightly smaller that ρ=3.20863\rho=3.20863 we found for k90000k\geq 90000. However, a direct computation shows that this really small improvment on ρ\rho for kk sufficiently large would not lead to an improvement in Lemma 4.2, when λ>N\lambda>N and NN sufficently large.

4.2. Case λ84\lambda\leq 84

For λ84\lambda\leq 84, the values for the CC constant found in [For02a] hold, both for 1λ2.61\leq\lambda\leq 2.6 and 2.6λ842.6\leq\lambda\leq 84. Indeed, for the range 2.6λ842.6\leq\lambda\leq 84, the same values for CC still hold under the new constraint that S(N,t)CN11/uλ2S(N,t)\leq CN^{1-1/u\lambda^{2}} with u132.94357u\leq 132.94357. We can notice that the maximum is reached inside the interval λ[83,84]\lambda\in[83,84], where C=8.7979C=8.7979. This concludes the proof of Theorem 1.5 also for λ84\lambda\leq 84.

5. Proof of Theorem 1.1

First of all, since ζ(s¯,u)=ζ(s,u)¯\zeta(\bar{s},u)=\overline{\zeta(s,u)} and ζ(s)=ζ(s,1)\zeta(s)=\zeta(s,1), we restrict our attention to ss lying in the upper half-plane. Then, we consider separately the cases σ1516\sigma\leq\frac{15}{16} or t10108t\leq 10^{108} or t10108t\geq 10^{108} and 1516σ1\frac{15}{16}\geq\sigma\geq 1. The main contribution will come from the case t10108t\geq 10^{108} and 1516σ1\frac{15}{16}\geq\sigma\geq 1.

5.1. Case 1516σ1,t10108\frac{15}{16}\leq\sigma\leq 1,\ t\geq 10^{108}

Using Lemma 2.12 with the values C=8.7979C=8.7979 and D=132.94357D=132.94357 found in Theorem 1.5 and t10108t\geq 10^{108}, we have

(C+1+1080log2/3t+1.569CD1/3)(C+1+1080log2/3(10108)+1.569CD1/3)70.6995.\left(\frac{C+1+10^{-80}}{\log^{2/3}t}+1.569CD^{1/3}\right)\leq\left(\frac{C+1+10^{-80}}{\log^{2/3}(10^{108})}+1.569CD^{1/3}\right)\leq 70.6995.

It follows that

|ζ(σ+it)|70.6995t4.43795(1σ)3/2log2/3t|\zeta(\sigma+it)|\leq 70.6995t^{4.43795(1-\sigma)^{3/2}}\log^{2/3}t

and

|ζ(s,u)us|70.6995t4.43795(1σ)3/2log2/3t.\left|\zeta(s,u)-u^{-s}\right|\leq 70.6995t^{4.43795(1-\sigma)^{3/2}}\log^{2/3}t.

5.2. Case 1516σ1, 3t10108\frac{15}{16}\leq\sigma\leq 1,\ 3\leq t\leq 10^{108}

Following the proof of Lemma 2.10 in [For02a], one has

|ζ(s,u)us|(t+3/2)1σ(1+1/t+log(2t+1))\left|\zeta(s,u)-u^{-s}\right|\leq(t+3/2)^{1-\sigma}(1+1/t+\log(2t+1))

for 1516σ1\frac{15}{16}\leq\sigma\leq 1 and 3t101083\leq t\leq 10^{108}.
If 3t1063\leq t\leq 10^{6}, from [For02a] we know that

|ζ(s,u)us|36.8.\left|\zeta(s,u)-u^{-s}\right|\leq 36.8.

If 106t1010810^{6}\leq t\leq 10^{108}, from the argument of Lemma 2.10 in [For02a], one gets

|ζ(s,u)us|\displaystyle\left|\zeta(s,u)-u^{-s}\right| (t+3/2)1σ(1+1/t+log(2t+1))\displaystyle\leq(t+3/2)^{1-\sigma}(1+1/t+\log(2t+1))
1.123t1σlogt\displaystyle\leq 1.123t^{1-\sigma}\log t
=1.123(t4(1σ)3/2log2/3t)(t1σ4(1σ)3/2log1/3t)\displaystyle=1.123\left(t^{4(1-\sigma)^{3/2}}\log^{2/3}t\right)\left(t^{1-\sigma-4(1-\sigma)^{3/2}}\log^{1/3}t\right)
1.123(t4(1σ)3/2log2/3t)(t1108log1/3t)\displaystyle\leq 1.123\left(t^{4(1-\sigma)^{3/2}}\log^{2/3}t\right)\left(t^{\frac{1}{108}}\log^{1/3}t\right)
70.6199t4(1σ)3/2log2/3t\displaystyle\leq 70.6199t^{4(1-\sigma)^{3/2}}\log^{2/3}t
70.6995t4.43795(1σ)3/2log2/3t.\displaystyle\leq 70.6995t^{4.43795(1-\sigma)^{3/2}}\log^{2/3}t.

Hence, we also have

|ζ(σ+it)|70.6199t4(1σ)3/2log2/3t70.6995t4.43795(1σ)3/2log2/3t.|\zeta(\sigma+it)|\leq 70.6199t^{4(1-\sigma)^{3/2}}\log^{2/3}t\leq 70.6995t^{4.43795(1-\sigma)^{3/2}}\log^{2/3}t.
Remark.

One can notice that the estimate found for ζ(s)\zeta(s) and ζ(s,u)\zeta(s,u) in this range is much better than that one found for 1516σ1\frac{15}{16}\leq\sigma\leq 1 and t10108t\geq 10^{108}. Indeed, the BB constant we found in this current case is just 44, which is less than the final value 4.437954.43795.

5.3. Case 12σ1516,t3\frac{1}{2}\leq\sigma\leq\frac{15}{16},\ t\geq 3

From the proof of Lemma 2.10 in [For02a] we have

|ζ(σ+it)|21.3t4(1σ)3/270.6995t4.43795(1σ)3/2log2/3t.|\zeta(\sigma+it)|\leq 21.3t^{4(1-\sigma)^{3/2}}\leq 70.6995t^{4.43795(1-\sigma)^{3/2}}\log^{2/3}t. (5.1)
Remark.

The choice of 1516\frac{15}{16} comes from the proof of Lemma 7.1 in [For02a], where (1σ)4(1σ)3/2(1-\sigma)\leq 4(1-\sigma)^{3/2} for σ1516\sigma\leq\frac{15}{16}.
With (1σ)4.4(1σ)3/2(1-\sigma)\leq 4.4(1-\sigma)^{3/2} one should get σ459484\sigma\leq\frac{459}{484}. However, the influence of this second choice on the AA constant is negligible and it does not lead to any further improvements on AA.

6. Proof of Theorem 1.2

For texp(463388)t\leq\exp(463388) we use the best known explicit Littlewood zero-free region due to Yang in [Yan23]. For texp(463388)t\geq\exp(463388) we follow the argument used to prove Theorem 1.1 in [MTY22] which relies on a non-negative trigonometric polynomial P40(x)P_{40}(x) defined by (1.6) with degree 40 having

b0=1,b1=1.74600190914994,b=k=140bk=3.56453965437134.b_{0}=1,\quad b_{1}=1.74600190914994,\quad b=\sum_{k=1}^{40}b_{k}=3.56453965437134.

Using the new values A=70.6995A=70.6995 and B=4.43795B=4.43795 found in Theorem 1.1 and making the following new assumptions

T0:=exp(463388),M1:=0.050007,E=1.8008278,R=468T_{0}:=\exp(463388),\qquad M_{1}:=0.050007,\qquad E=1.8008278,\qquad R=468

in [MTY22], the proof of Theorem 1.2 is complete.

7. Proof of Theorem 1.3

The proof is exactly the same as in [MTY22], except for the new value B=4.43795B=4.43795 found in Theorem 1.1, and relies on a non-negative polynomial P46(x)P_{46}(x) with degree 46 where

b0=1,b1=1.74708744081848,b=k=146bk=3.57440943022073.b_{0}=1,\quad b_{1}=1.74708744081848,\quad b=\sum_{k=1}^{46}b_{k}=3.57440943022073.

8. Some possible improvements on AA

In this section we give some suggestions for some possible improvements on the constant AA in Theorem 1.1. We will provide some quite detailed proofs of some useful lemmas to help the reader follow the argument more easily.

Instead of the definition (1.7) for S(N,t)S(N,t), one could consider the following sum

S~(N,t):=max0<u1maxN<RmN|NnR1(n+u)it|\tilde{S}(N,t):=\max_{0<u\leq 1}\max_{N<R\leq mN}\left|\sum_{N\leq n\leq R}\frac{1}{(n+u)^{it}}\right| (8.1)

with 1<m21<m\leq 2. For λ84\lambda\geq 84, we can find sharper estimates.

Theorem 8.1.

For 1.001m21.001\leq m\leq 2 and λ84\lambda\geq 84 we have:

S~(N,t)(m1)8.7979N11/132.94357λ2.\tilde{S}(N,t)\leq(m-1)8.7979N^{1-1/132.94357\lambda^{2}}.

If one could find a bound of the form S~(N,t)(m1)cN11/(uλ2)\tilde{S}(N,t)\leq(m-1)cN^{1-1/(u\lambda^{2})} also for λ84\lambda\leq 84, then the constant AA in Theorem 1.1 might be reduced to around 4949. We briefly outline the proof of Theorem 8.1.
We start with a preliminary lemma that is a more general version of Lemma 2.9.

Lemma 8.2.

Suppose k,rk,r and ss are integers 2\geq 2, and hh and gg are integers satisfying 1hgk1\leq h\leq g\leq k. Let NN be a positive integer, and M1,M2M_{1},M_{2} be real numbers with 1MiN1\leq M_{i}\leq N. Let \mathscr{B} be a nonempty subset of the positive integers M2\leq M_{2}. Then

S~(N,t)2M1M2+(m1)t(M1M2)k+1(k+1)Nk\displaystyle\tilde{S}(N,t)\leq 2M_{1}M_{2}+\frac{(m-1)t(M_{1}M_{2})^{k+1}}{(k+1)N^{k}}
+(m1)N(M2||)1/r((5r)kM22sM12r+k(k+1)/2Jr,k(M1)Js,g,h()WhWg)1/2rs,\displaystyle+(m-1)N\left(\frac{M_{2}}{|\mathscr{B}|}\right)^{1/r}\left((5r)^{k}M_{2}^{-2s}\lfloor M_{1}\rfloor^{-2r+k(k+1)/2}J_{r,k}(\lfloor M_{1}\rfloor)J_{s,g,h}(\mathscr{B})W_{h}\dots W_{g}\right)^{1/2rs},

where

Wj=min(2sM2j,2sM2jrM1j+stM2jπjNj+4πj(2N)jrtM1j+2)(j1)W_{j}=\min\left(2sM_{2}^{j},\frac{2sM_{2}^{j}}{r\left\lfloor M_{1}\right\rfloor^{j}}+\frac{stM_{2}^{j}}{\pi jN^{j}}+\frac{4\pi j(2N)^{j}}{rt\left\lfloor M_{1}\right\rfloor^{j}}+2\right)\quad(j\geq 1)

and S~(N,t)\tilde{S}(N,t) is defined in (8.1).

Proof.

We define M=M1M=\left\lfloor M_{1}\right\rfloor. For N<RmNN<R\leq mN, 1<m21<m\leq 2 and 0<u10<u\leq 1, we have

|N<nR(n+u)it|\displaystyle\left|\sum_{N<n\leq R}(n+u)^{-it}\right| =1M|||aM1bN<n+abR(n+ab+u)it|\displaystyle=\frac{1}{M|\mathscr{B}|}\left|\sum_{\begin{subarray}{c}a\leq M_{1}\\ b\in\mathscr{B}\end{subarray}}\sum_{N<n+ab\leq R}(n+ab+u)^{-it}\right|
1M|||aM1bN<nR1(n+ab+u)it|+1M||aM1b(2ab1)\displaystyle\leq\frac{1}{M|\mathscr{B}|}\left|\sum_{\begin{subarray}{c}a\leq M_{1}\\ b\in\mathscr{B}\end{subarray}}\sum_{\begin{subarray}{c}N<n\leq R-1\end{subarray}}(n+ab+u)^{-it}\right|+\frac{1}{M|\mathscr{B}|}\sum_{\begin{subarray}{c}a\leq M_{1}\\ b\in\mathscr{B}\end{subarray}}(2ab-1)
NM||maxNzmN|aM1beitlog(1+ab/z)|+2M1M2.\displaystyle\leq\frac{N}{M|\mathscr{B}|}\max_{N\leq z\leq mN}\left|\sum_{\begin{subarray}{c}a\leq M_{1}\\ b\in\mathscr{B}\end{subarray}}e^{-it\log(1+ab/z)}\right|+2M_{1}M_{2}.

For 0x10\leq x\leq 1 we have

|log(1+x)(xx2/2++(1)k1xk/k)|xk+1k+1.\left|\log(1+x)-\left(x-x^{2}/2+\cdots+(-1)^{k-1}x^{k}/k\right)\right|\leq\frac{x^{k+1}}{k+1}.

Also |eiy1|y\left|e^{iy}-1\right|\leq y for real yy and ab/zM1M2/Nab/z\leq M_{1}M_{2}/N. Thus, for some z[N,mN]z\in[N,mN],

S~(N,t)(m1)NM|||U|+t(m1)(M1M2)k+1(k+1)Nk+2M1M2\tilde{S}(N,t)\leq\frac{(m-1)N}{M|\mathscr{B}|}|U|+\frac{t(m-1)\left(M_{1}M_{2}\right)^{k+1}}{(k+1)N^{k}}+2M_{1}M_{2}

where U=a,be(γ1(ab)++γk(ab)k)U=\sum_{a,b}e\left(\gamma_{1}(ab)+\cdots+\gamma_{k}(ab)^{k}\right) and γj=(1)jt/(2πjzj)\gamma_{j}=(-1)^{j}t/\left(2\pi jz^{j}\right).
At this point, from Ford’s argument in Lemma 5.1 of [For02a], we have

NM|||U|N(M2||)1r((5r)kM22sM12r+12k(k+1)Jr,k(M1)Js,g,h()WhWg)12rs.\frac{N}{M|\mathscr{B}|}|U|\leq N\left(\frac{M_{2}}{|\mathscr{B}|}\right)^{\frac{1}{r}}\left((5r)^{k}M_{2}^{-2s}\left\lfloor M_{1}\right\rfloor^{-2r+\frac{1}{2}k(k+1)}J_{r,k}\left(\left\lfloor M_{1}\right\rfloor\right)J_{s,g,h}(\mathscr{B})W_{h}\cdots W_{g}\right)^{\frac{1}{2rs}}.

The conclusion follows. ∎

At this point, from Ford’s argument in section 55 of [For02a] we have the following bounds:

t(M1M2)k+1(k+1)Nk1kN10.0000019476,\frac{t(M_{1}M_{2})^{k+1}}{(k+1)N^{k}}\leq\frac{1}{k}N^{1-0.0000019476},
2M1M22N0.362M_{1}M_{2}\leq 2N^{0.36}

and

N(M2||)1/r((5r)kM22sM12r+k(k+1)/2Jr,k(M1)Js,g,h()WhWg)1/2rs\displaystyle N\left(\frac{M_{2}}{|\mathscr{B}|}\right)^{1/r}\left((5r)^{k}M_{2}^{-2s}\lfloor M_{1}\rfloor^{-2r+k(k+1)/2}J_{r,k}(\lfloor M_{1}\rfloor)J_{s,g,h}(\mathscr{B})W_{h}\dots W_{g}\right)^{1/2rs}
(C31r(λ5λC1C2)12rs)N1+E\displaystyle\leq\left(C_{3}^{\frac{1}{r}}\left(\lambda^{5\lambda}C_{1}C_{2}\right)^{\frac{1}{2rs}}\right)N^{1+E}

where

E=log(Yλ2)7.6Yλ4+12rs(H+0.001μ1k2+μ2E2)E=\frac{\log\left(Y\lambda^{2}\right)}{7.6Y\lambda^{4}}+\frac{1}{2rs}\left(-H+0.001\mu_{1}k^{2}+\mu_{2}E_{2}\right)

and C1,C2,C3C_{1},C_{2},C_{3} are the same as for Theorem 1.5. If we use these bounds in Lemma 8.2 we get

S~(N,t)(m1)(C31r(λ4.65λC1C2)12rs)N1+E+2N0.36+(m1)kN10.0000019476.\tilde{S}(N,t)\leq(m-1)\left(C_{3}^{\frac{1}{r}}\left(\lambda^{4.65\lambda}C_{1}C_{2}\right)^{\frac{1}{2rs}}\right)N^{1+E}+2N^{0.36}+\frac{(m-1)}{k}N^{1-0.0000019476}. (8.2)

Now, we consider Ne288λ2N\geq e^{288\lambda^{2}}, since otherwise we have trivially

S~(N,t)(m1)N(m1)e288/132.94357N11/132.94357λ2(m1)8.7979N11/132.94357λ2.\tilde{S}(N,t)\leq(m-1)N\leq(m-1)e^{288/132.94357}N^{1-1/132.94357\lambda^{2}}\leq(m-1)8.7979N^{1-1/132.94357\lambda^{2}}.
Lemma 8.3.

For 84λ22084\leq\lambda\leq 220 and Ne288λ2N\geq e^{288\lambda^{2}} we have the following estimate:

S~(N,t)(m1)8.797901N11/132.94357λ2.\tilde{S}(N,t)\leq(m-1)8.797901N^{1-1/132.94357\lambda^{2}}.
Proof.

From Lemma 4.1 we know that

(C31r(λ4.65λC1C2)12rs)N1+E8.7979N11/132.94357λ2.\left(C_{3}^{\frac{1}{r}}\left(\lambda^{4.65\lambda}C_{1}C_{2}\right)^{\frac{1}{2rs}}\right)N^{1+E}\leq 8.7979N^{1-1/132.94357\lambda^{2}}.

Using this result in (8.2) we get

S~(N,t)\displaystyle\tilde{S}(N,t) (m1)8.7979N11/132.94357λ2+2N0.36+(m1)kN10.0000019476\displaystyle\leq(m-1)8.7979N^{1-1/132.94357\lambda^{2}}+2N^{0.36}+\frac{(m-1)}{k}N^{1-0.0000019476}
=(m1)(8.7979N11/(132.94357λ2)+2N0.36m1+1k+1N11/132λ2)\displaystyle=(m-1)\left(8.7979N^{1-1/(132.94357\lambda^{2})}+\frac{2N^{0.36}}{m-1}+\frac{1}{k+1}N^{1-1/132\lambda^{2}}\right)
(m1)8.797901N11/(132.94357λ2)\displaystyle\leq(m-1)8.797901N^{1-1/(132.94357\lambda^{2})}

where the last passage comes from the fact that, for m=1.001m=1.001, we have

(m1)1078.7979N11/(132.94357λ2)2N0.36.(m-1)\cdot 10^{-7}\cdot 8.7979N^{1-1/(132.94357\lambda^{2})}\geq 2N^{0.36}.

Indeed, for 84λ22084\leq\lambda\leq 220 and Ne288λ2N\geq e^{288\lambda^{2}}, we have

N11/(132.94357λ2)0.36\displaystyle N^{1-1/(132.94357\lambda^{2})-0.36} e288λ2(11/(132.94357λ2)0.36)\displaystyle\geq e^{288\lambda^{2}(1-1/(132.94357\lambda^{2})-0.36)}
e288842(11/(132.94357842)0.36)\displaystyle\geq e^{288\cdot 84^{2}(1-1/(132.94357\cdot 84^{2})-0.36)}
28.7979107(m1)\displaystyle\geq\frac{2}{8.7979\cdot 10^{-7}\cdot(m-1)}

if

m121078.7979e288842(11/(132.94357842)0.36)10560000,m-1\geq\frac{2\cdot 10^{7}}{8.7979}\cdot e^{-288\cdot 84^{2}(1-1/(132.94357\cdot 84^{2})-0.36)}\geq 10^{-560000},

which holds, since with our choice m1=103m-1=10^{-3}. ∎

Lemma 8.4.

For λ220\lambda\geq 220 and Ne288λ2N\geq e^{288\lambda^{2}} we have the following estimate:

S~(N,t)(m1)7.5001N11/132.94357λ2.\tilde{S}(N,t)\leq(m-1)7.5001N^{1-1/132.94357\lambda^{2}}.
Proof.

From Lemma 4.2 we know that

(C31r(λ4.65λC1C2)12rs)N1+E7.5N11/132.94357λ2.\left(C_{3}^{\frac{1}{r}}\left(\lambda^{4.65\lambda}C_{1}C_{2}\right)^{\frac{1}{2rs}}\right)N^{1+E}\leq 7.5N^{1-1/132.94357\lambda^{2}}.

Using this result in (8.2) we get

S~(N,t)\displaystyle\tilde{S}(N,t) (m1)7.5N11/132.94357λ2+2N0.36+(m1)kN10.0000019476\displaystyle\leq(m-1)7.5N^{1-1/132.94357\lambda^{2}}+2N^{0.36}+\frac{(m-1)}{k}N^{1-0.0000019476}
(m1)(7.5N11/(132.94357λ2)+2N0.36m1+1k+1N11/132.94357λ2)\displaystyle\leq(m-1)\left(7.5N^{1-1/(132.94357\lambda^{2})}+\frac{2N^{0.36}}{m-1}+\frac{1}{k+1}N^{1-1/132.94357\lambda^{2}}\right)
(m1)7.5001N11/(132.94357λ2)\displaystyle\leq(m-1)7.5001N^{1-1/(132.94357\lambda^{2})}

where the last passage comes from the fact that, for m=1.001m=1.001, we have

(m1)1057.5N11/(132.94357λ2)2N0.36.(m-1)\cdot 10^{-5}\cdot 7.5N^{1-1/(132.94357\lambda^{2})}\geq 2N^{0.36}.

Indeed, for λ220\lambda\geq 220 and Ne288λ2N\geq e^{288\lambda^{2}}, we have

N11/(132.94357λ2)0.36\displaystyle N^{1-1/(132.94357\lambda^{2})-0.36} e288λ2(11/(132.94357λ2)0.36)\displaystyle\geq e^{288\lambda^{2}(1-1/(132.94357\lambda^{2})-0.36)}
e2882202(11/(132.943572202)0.36)\displaystyle\geq e^{288\cdot 220^{2}(1-1/(132.94357\cdot 220^{2})-0.36)}
27.5105(m1)\displaystyle\geq\frac{2}{7.5\cdot 10^{-5}(m-1)}

if

m121057.5e2882202(11/(132.943572202)0.36)103800000,m-1\geq\frac{2\cdot 10^{5}}{7.5}\cdot e^{-288\cdot 220^{2}(1-1/(132.94357\cdot 220^{2})-0.36)}\geq 10^{-3800000},

which holds, since with our choice m1=103m-1=10^{-3}. ∎

Theorem 8.1 follows immediately from Lemma 8.3 and Lemma 8.4.
At this point, if one could find an estimate of the form S~(N,t)(m1)cN11/(uλ2)\tilde{S}(N,t)\leq(m-1)cN^{1-1/(u\lambda^{2})} also for λ84\lambda\leq 84, then we can use a more general version of Lemma 2.12.

Lemma 8.5.

If s=σ+it,1516σ1,t1090s=\sigma+it,\frac{15}{16}\leq\sigma\leq 1,t\geq 10^{90} and 0<u10<u\leq 1, then

|ζ(s,u)0nt(n+u)s|1077.\left|\zeta(s,u)-\sum_{0\leq n\leq t}(n+u)^{-s}\right|\leq 10^{-77}.
Proof.

The proof is the same as that one of Lemma 2.11 in [For02a], with t1090t\geq 10^{90} instead of t10100t\geq 10^{100}. ∎

Lemma 8.6.

Suppose that S~(N,t)(m1)CN11/(Dλ2)(1Nt)\tilde{S}(N,t)\leq(m-1)CN^{1-1/\left(D\lambda^{2}\right)}(1\leq N\leq t) for positive constants CC and DD and 1<m21<m\leq 2, where λ=logtlogN\lambda=\frac{\log t}{\log N}. Let B=293DB=\frac{2}{9}\sqrt{3D}. Then, for 1516σ1,t1090\frac{15}{16}\leq\sigma\leq 1,t\geq 10^{90} and 0<u10<u\leq 1, we have

|ζ(s)|((m1)C+1+11077log2/3t+1.0875034(m1)CD1/3log(m))tB(1σ)3/2log2/3t.|\zeta(s)|\leq\left(\frac{(m-1)C+1+\frac{1}{10^{77}}}{\log^{2/3}t}+\frac{1.0875034(m-1)CD^{1/3}}{\log(m)}\right)t^{B(1-\sigma)^{3/2}}\log^{2/3}t.
Proof.

Let

S1(u)=1nt(n+u)sS_{1}(u)=\sum_{1\leq n\leq t}(n+u)^{-s}

By Lemma 8.5, |ζ(s,u)us|1077+S1(u)\left|\zeta(s,u)-u^{-s}\right|\leq 10^{-77}+S_{1}(u). Put r=logtlogmr=\left\lceil\frac{\log t}{\log m}\right\rceil. By partial summation,

|S1(u)|\displaystyle\left|S_{1}(u)\right| 1+j=0r1|mj<nmin(t,mj+1)(n+u)σit|\displaystyle\leq 1+\sum_{j=0}^{r-1}\left|\sum_{m^{j}<n\leq\min\left(t,m^{j+1}\right)}(n+u)^{-\sigma-it}\right|
1+j=0r1(mj)σS(mj,t)\displaystyle\leq 1+\sum_{j=0}^{r-1}\left(m^{j}\right)^{-\sigma}S\left(m^{j},t\right)
1+(m1)Cj=0r1eg(j),\displaystyle\leq 1+(m-1)C\sum_{j=0}^{r-1}e^{g(j)},

where

g(j)=(1σ)(jlogm)(jlogm)3Dlog2t.g(j)=(1-\sigma)(j\log m)-\frac{(j\log m)^{3}}{D\log^{2}t}.

As a function of x,g(x)x,\ g(x) is increasing on [0,x0]\left[0,x_{0}\right] and decreasing on [x0,)\left[x_{0},\infty\right), where x0logm=D(1σ)/3logtx_{0}\log m=\sqrt{D(1-\sigma)/3}\log t. Thus

|S1(u)|1(m1)C\displaystyle\frac{\left|S_{1}(u)\right|-1}{(m-1)C} eg(x0)+0reg(x)𝑑x\displaystyle\leq e^{g\left(x_{0}\right)}+\int_{0}^{r}e^{g(x)}dx
tB(1σ)3/2+D1/3log2/3tlogm0e3y2uu3𝑑u,\displaystyle\leq t^{B(1-\sigma)^{3/2}}+\frac{D^{1/3}\log^{2/3}t}{\log m}\int_{0}^{\infty}e^{3y^{2}u-u^{3}}du,

where y=(1σ)/3D1/6log1/3ty=\sqrt{(1-\sigma)/3}D^{1/6}\log^{1/3}t.
To bound the last integral, we make use of the inequality

e2y30e3y2uu3𝑑u1.0875034(y0),e^{-2y^{3}}\int_{0}^{\infty}e^{3y^{2}u-u^{3}}du\leq 1.0875034\quad(y\geq 0),

where the maximum occurs near y=0.710y=0.710. Therefore

|S1(u)|1(m1)CtB(1σ)3/2(1+1.0875034logmD1/3log2/3t)\frac{\left|S_{1}(u)\right|-1}{(m-1)C}\leq t^{B(1-\sigma)^{3/2}}\left(1+\frac{1.0875034}{\log m}D^{1/3}\log^{2/3}t\right)

However, Ford’s method in [For02a] for λ87\lambda\leq 87 (or λ84\lambda\leq 84 as in our paper), cannot be modified to extract the factor (m1)(m-1). Indeed, following the proof of Lemma 6.3 in [For02a], one should estimates the following quantity

(N<nmN1|T(n)|2s)1/2s,\left(\sum_{N<n\leq mN-1}|T(n)|^{2s}\right)^{1/2s},

where

T(n)=vMe(t2πlog(1+vn+u)).T(n)=\sum_{v\leq M}e\left(-\frac{t}{2\pi}\log\left(1+\frac{v}{n+u}\right)\right).

However, since an estimate on the whole sum of T(n)T(n) over NnmNN\leq n\leq mN is required, instead of one for the maximum of T(n)T(n) over the interval NnmNN\leq n\leq mN, it is not possible to extract a factor (m1)(m-1) at this step. A clever argument which would overcome this problem could lead to a suitable estimate also for the case λ84\lambda\leq 84, with a consequent improved value for AA.

9. Code listings

  Program 1 for Theorem 1.4

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#define max(x,y) (((x)>(y))?(x):(y))
#define min(x,y) (((x)>(y))?(y):(x))
double newdel(double k, double r, double del)
{
double y,p,pf,tkr;
long j,jf,jj;
if ((r<4.0) && (r>k)) return (2.0*del);
tkr=2.0*k*r; y=2.0*del-(k-r)*(k-r+1.0);
if ((y<0.0) && (2.0*k/(tkr+y))<= 1.0/(k+1.0))
return (del*2.0);
j=floor(min(0.5*(3.0+sqrt(4.0*y+1.0)), 9.0*r/10.0));
p=1.0/r;
for (jj=j-1; jj>=1; jj–) {
p=0.5/r+0.5*(1.0-y/(tkr-2.0*r*jj))*p;
}
return(del-k+0.5*p*(tkr-y));
}
int main()
{
long j,k,k0,k1,r,r0,r1,n,bestr,s;
double kk,logk,del0,del1,bestdel,goal,maxs,eta,om;
double logH,logW,logC,k3,theta,thetamax;
printf(”enter k range :”); scanf(”%ld %ld”, &k0,&k1);
maxs=0.0; thetamax=0.0;
for (k=k0; k<=k1; k++) {
kk=(double)k;
logk=log(kk); k3=kk*kk*kk*logk;
om=0.5;
for (j=1;j<=10;j++) om=1.5/(log(18.0*k3/om)-1.5);
eta=1.0+om;
logW=(kk+1.0)*max(1.5+1.5/om,log(18.0/om*k3));
del0=0.5*kk*kk*(1.0-1.0/kk);
goal=0.001*kk*kk;
logH=3.0*kk*logk+(kk*kk-4.0*kk)*log(eta);
logC=kk*logk;
for (n=1; ;n++) {
r0=(long)(sqrt(kk*kk+kk-2.0*del0)+0.5)-2;
r1=r0+4;
bestdel=kk*kk; bestr=-1;
for (r=r0;r<=r1;r++) {
del1=newdel(kk,(double)r,del0);
if (del1<bestdel) { bestdel=del1; bestr=r;}
}
del1=bestdel; r=bestr;
if ((del1>=del0) && (r<r0)) exit(-1);
logC +=max(logH+4.0*kk*n*log(eta),logW*(del0-del1));
if (del1<=goal) {
s=(long)((n+(del0-goal)/(del0-del1))*kk+1);
theta=logC/k3;
printf(”%4d: s=%8.6f k^2 eta=%9.7f theta=%10.8f\n”,
k,s/kk/kk,eta,theta);
if ((s/kk/kk)>maxs) maxs=s/kk/kk;
if (theta>thetamax) thetamax=theta;
break;
}
del0=del1;
}
}
printf(”\n max s=%9.6fk^2 maxtheta=%10.8f\n”,maxs,thetamax);
system(”pause”);
}

  Program 2 for Theorem 1.4

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#define max(x,y) (((x)>(y))?(x):(y))
#define min(x,y) (((x)>(y))?(y):(x))
double newdel(double k, double r, double del)
{
double y,p,pf,tkr;
long j,jf,jj;
if ((r<4.0) && (r>k)) return (2.0*del);
tkr=2.0*k*r; y=2.0*del-(k-r)*(k-r+1.0);
if ((y<0.0) && (2.0*k/(tkr+y))<= 1.0/(k+1.0))
return (del*2.0);
j=floor(min(0.5*(3.0+sqrt(4.0*y+1.0)), 9.0*r/10.0));
p=1.0/r;
for (jj=j-1; jj>=1; jj–) {
p=0.5/r+0.5*(1.0-y/(tkr-2.0*r*jj))*p;
}
return(del-k+0.5*p*(tkr-y));
}
int main()
{
long j,k,k0,k1,r,r0,r1,n,bestr,s;
double kk,logk,del0,del1,bestdel,goal,maxs,eta,om;
double logH,logW,logC,k3,theta,thetamax;
printf(”enter k range :”); scanf(”%ld %ld”, &k0,&k1);
maxs=0.0; thetamax=0.0;
for (k=k0; k<=k1; k++) {
kk=(double)k;
logk=log(kk); k3=kk*kk*kk*logk;
om=0.5;
for (j=1;j<=10;j++) om=1.5/(log(18.0*k3/om)-1.5);
eta=1.0+om;
logW=(kk+1.0)*max(1.5+1.5/om,log(18.0/om*k3));
del0=0.4*kk*kk;
goal=0.001*kk*kk;
logH=3.0*kk*logk+(kk*kk-4.0*kk)*log(eta);
logC=kk*logk;
for (n=ceil(0.1247*kk); ;n++) {
r0=(long)(sqrt(kk*kk+kk-2.0*del0)+0.5)-2; r1=r0+4;
bestdel=kk*kk; bestr=-1;
for (r=r0;r<=r1;r++) {
del1=newdel(kk,(double)r,del0);
if (del1<bestdel) {bestdel=del1; bestr=r; }
}
del1=bestdel; r=bestr;
if ((del1>=del0) && (r<r0)) exit(-1);
logC +=max(logH+4.0*kk*n*log(eta),logW*(del0-del1));
if (del1<=goal) {
s=(long)((n+(del0-goal)/(del0-del1))*kk+1);
theta=logC/k3;
printf(”%4d: s=%8.6f k^2 eta=%9.7f theta=%10.8f\n”,
k,s/kk/kk,eta,theta);
if ((s/kk/kk)>maxs) maxs=s/kk/kk;
if (theta>thetamax) thetamax=theta;
break;
}
del0=del1;
}
}
printf(”\n max s=%9.6fk^2 maxtheta=%10.8f\n”,maxs,thetamax);
system(”pause”);
}

  Program 3 for Lemma 4.1

#include<stdio.h>
#include <math.h>
long k,g,h,s,r,t,g0,h0,g1,h1,flag;
double mu1,mu2,xi,lam,lam1,lam2,D,sigma,Y,goal;
void calc(ex,c,pr)
double *ex,*c; int pr;
{
double kk,logk,k2,log(),exp(),floor(),ceil();
double th,rr,ss,tt,gg,hh,rho,H,E1,E2,E3,m1,m2,Z0,Z1,
reta,logC1,logC2,logC3,logC,dc;
k=(long) (lam/(1.0-mu1-mu2)+0.000003);
kk=(double) k;
logk=log(kk); k2=kk*kk;
rho=3.20863; th=2.17720;
if (k<=89999) {rho=3.205502; th=1.77775;}
if (k<=499) {rho=3.196497; th=2.24352;}
if (k<=339) {rho=3.192950; th=2.33313;}
if (k<=190) {rho=3.184127; th=2.35334;}
if (k<=170) {rho=3.181869; th=2.37929;}
if (k<=148) {rho=3.178871; th=2.38259;}
if (k<=146) {rho=3.178551; th=2.39167;}
if (k<=139) {rho=3.177527; th=2.39529;}
if (k<=137) {rho=3.177207; th=2.40930;}
r=(long) (rho*k2+1.0);
rr=(double) r; ss=(double) s;
gg=(double) g; hh=(double) h; tt=(double) t;
m1=floor(lam/(1.0-mu1));
m2=floor(lam/(1.0-mu2));
Z0=0.5*((m1*m1+m1)*(1.0-mu1)+(m2*m2+m2)*(1.0-mu2)-
-hh*hh+hh-(1.0-mu1-mu2)*(gg*gg+gg));
Z1=hh+gg-m1-m2-1.0;
if (Z1<0.0) H=Z0+lam2*Z1;
else H=Z0+lam1*Z1;
reta=xi*pow(gg,1.5);
E1=0.001*k2;
E2=0.5*tt*(tt-1.0)+hh*tt*exp(-ss/(hh*tt))+ss*ss/(2.0*tt*reta);
E3=log(Y*lam1*lam1)/(1.0*Y*lam1*lam1*lam1*lam1);
*ex=(-E3+(1.0/(2.0*rr*ss))*(H-mu1*E1-mu2*E2))*lam1*lam1;
logC1=th*k2*kk*logk;
logC2=ss*ss/tt+10.5*xi*xi*tt*gg*gg*log(gg)*log(gg)/D;
logC2 -=(ss*log(0.1*reta)*((reta+hh)*pow(1.0-1.0/hh,ss/tt)-h));
logC3=1.0417*reta*log(10.4167*reta);
logC=logC3/rr+(4.65*lam2*log(lam2)+logC1+logC2)/(2.0*rr*ss);
*c=exp(logC)+1.0/kk;
if (pr==1){
printf(”%8.4f-%8.4f %4ld”,lam1,lam2,k);
if(g>0) printf(”%3ld %2ld %2ld %2ld %9.4f %7.4f\n”,
s,g-g0,h1-h,t,1.0/(*ex)+0.00005,*c+0.00005);
else printf(”\n”);
}
}
int main()
{
double E,lam8,lam9,r[9],tmp,maxex,con,maxcon
double bestth,bestcon,bp[5000];
long i,j,i0,w,n,m,maxm,bestg,besth,bests,s0,s1;
mu1=0.1905; mu2=0.1603;
goal=132.94357;
while (1) {
printf(”enter Y:”); scanf(”%lf”, &Y);
D=0.1019*Y;
printf(”enter xi: ”); scanf(”%lf”, &xi);
printf(”enter sigma :”); scanf(”%lf”, &sigma);
if (sigma<0.0) flag=1; else flag=0;
printf(”enter lambda range: ”);
scanf(”%lf %lf”, &lam8, &lam9);
if ((lam9<lam8)||(lam8<=80.0)||(lam9>=300.0)) continue;
printf(” approx. \n”);
printf(”lambda range k s a b t exp const\n”);
printf(”————– ———–\n”);
bp[1]=lam8; bp[2]=lam9; j=3;
i0=(long) (lam9/(1.0-mu1-mu2))+10;
for (i=1; i<=i0;i++){
w=(double) i;
r[1]=w*(1.0-mu1);
r[2]=w*(1.0-mu2);
r[3]=(w-0.000003)*(1.0-mu1-mu2);
for (m=1;m<=3;m++) if ((r[m]<lam9) && (r[m]>lam8))
bp[j++]=r[m];
}
n=j-1;
for (i=1;i<=n-1;i++) for(j=i+1;j<=n;j++)
if (bp[j]<bp[i]) {tmp=bp[i]; bp[i]=bp[j]; bp[j]=tmp;}
maxex=0.0;
maxcon=0.0;
for (j=1;j<=n-1;j++){
lam=0.5*(bp[j]+bp[j+1]);
lam1=bp[j]; lam2=bp[j+1];
g0=(long) (lam/(1.0-mu1)+1.0); g1=g0+1;
h1=(long) (lam/(1.0-mu2)); h0=h1-1;
bestg=-1; besth=-1; bestth=1.0e20; bestcon=1.0e40;
for (g=g0;g<=g1;g++) for (h=h0;h<=h1;h++){
t=g-h+1;
if ((g>=100) && ((double) g<=1.254*lam1)){
if (flag==0) {
s0=(long) (sigma*h*t+1.0); s1=s0;
}
else {
s0=h*(t-1)/4;
s1=h*t/2;
}
for(s=s0; s<=s1; s++) {
calc(&E,&con,0);
if((E>0.0)&&(1.0/E<goal)&&(con<bestcon)){
bestth=1.0/E; bestg=g; besth=h; bests=s;
}
}
}
}
g=bestg; h=besth; t=g-h+1;
s=bests;
calc(&E,&con,1);
if (1.0/E>maxex) maxex=1.0/E;
if (con>maxcon) maxcon=con;
}
printf(”max. ex: %10.6f max. const.: %10.6f\n”, maxex,maxcon);
}
}

Acknowledgements

I would like to thank my supervisor Timothy S. Trudgian for his support and helpful suggestions throughout the writing of this article.

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