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Explicit estimates for the Riemann zeta function close to the 11-line

Michaela Cully-Hugill111Address: School of Science, University of New South Wales (Canberra), ACT, Australia
Email: m.cully-hugill@unsw.edu.au
  and Nicol Leong222Address: School of Science, University of New South Wales (Canberra), ACT, Australia
Email: nicol.leong@unsw.edu.au
Abstract

We provide explicit upper bounds of the order logt/loglogt\log t/\log\log t for |ζ(s)/ζ(s)||\zeta^{\prime}(s)/\zeta(s)| and |1/ζ(s)||1/\zeta(s)| when σ\sigma is close to 11. These improve existing bounds for ζ(s)\zeta(s) on the 11-line.

1 Introduction

In the study of the Riemann zeta function ζ(s)\zeta(s) and related functions, one often requires upper bounds for the logarithmic derivative of ζ(s)\zeta(s) and the reciprocal of ζ(s)\zeta(s). The Chebyshev function ψ(x)=nxΛ(n)\psi(x)=\sum_{n\leq x}\Lambda(n), where Λ(n)\Lambda(n) is the von Mangoldt function, or M(x)=nxμ(n)M(x)=\sum_{n\leq x}\mu(n), where μ(n)\mu(n) is the Möbius function are particularly relevant examples of where these estimates are useful. To estimate either with Perron’s formula, we use the fact that

n1Λ(n)ns=ζ(s)ζ(s)andn1μ(n)ns=1ζ(s)\sum_{n\geq 1}\frac{\Lambda(n)}{n^{s}}=-\frac{\zeta^{\prime}(s)}{\zeta(s)}\qquad\text{and}\qquad\sum_{n\geq 1}\frac{\mu(n)}{n^{s}}=\frac{1}{\zeta(s)} (1)

for Re(s)>1{\textup{Re}}(s)>1 (see e.g. [18], [2], [3], [16, (3.14)] for more detail).

Trudgian [18] gave explicit upper bounds of the order O(logt)O(\log t) for both functions in (1) (see also [10, Theorem 6.7]). These estimates are especially notable because they are valid inside the critical strip (0<Re(s)<10<{\textup{Re}}(s)<1), but only within known zero-free regions. Because of this, the constants had to be sufficiently large to bound the functions as ζ(s)0\zeta(s)\rightarrow 0, which would happen as ss approaches the boundary of the zero-free region. As such, the problem of improving these explicit constants is extremely non-trivial, and explicit estimates heavily depend on explicit zero-free regions.

The recent work of Yang [20] establishes a Littlewood-type zero-free region (see Lemma 4), which is the largest known region for exp(209)texp(5105)\exp(209)\leq t\leq\exp(5\cdot 10^{5}). With this, Hiary, Leong and Yang [6] gave bounds on |ζ(s)/ζ(s)||\zeta^{\prime}(s)/\zeta(s)| and |1/ζ(s)||1/\zeta(s)| of the order logt/loglogt\log t/\log\log t on the 11-line, which are the first explicit bounds of this type. The goal of this article is to improve and extend their results inside the critical strip.

2 Main results

The following four corollaries are the new estimates we prove in Sections 4 and 5. They are the first explicit bounds of this order inside the critical strip, and also improve current bounds at σ=1\sigma=1. On the 11-line, Corollary 2 improves [6, Theorem 2], and Corollary 4 improves [6, Theorem 2] for t3t\geq 3, and [1, Proposition A.2] for t222116t\geq 222116.

Throughout this article, we denote the height to which the Riemann Hypothesis has been verified [12] by H:=3 000 175 332 800H:=3\,000\,175\,332\,800.

Corollary 1.

For tt0eet\geq t_{0}\geq e^{e} and

σ1loglogtWlogt,\sigma\geq 1-\frac{\log\log t}{W\log t},

we have

|ζ(σ+it)ζ(σ+it)|R1logtloglogt,\left|\frac{\zeta^{\prime}(\sigma+it)}{\zeta(\sigma+it)}\right|\leq R_{1}\frac{\log t}{\log\log t},

where (W,R1)=(22,5471)(W,R_{1})=(22,5471) is valid for t0=eet_{0}=e^{e}, and other values are given in Table 3.

Corollary 2.

For tt0t\geq t_{0} and σ1\sigma\geq 1, we have

|ζ(σ+it)ζ(σ+it)|K1logtloglogt,\left|\frac{\zeta^{\prime}(\sigma+it)}{\zeta(\sigma+it)}\right|\leq K_{1}\frac{\log t}{\log\log t},

where K1=113.3K_{1}=113.3 is valid for t0=500t_{0}=500, and other values are given in Table 4.

Corollary 3.

For tt0eet\geq t_{0}\geq e^{e} and

σ1loglogtWlogt,\sigma\geq 1-\frac{\log\log t}{W\log t},

we have

|1ζ(σ+it)|R2logtloglogt,\left|\frac{1}{\zeta(\sigma+it)}\right|\leq R_{2}\frac{\log t}{\log\log t},

where (W,R2)=(22,3438)(W,R_{2})=(22,3438) is valid for t0=500t_{0}=500, and other values are given in Table 5.

Corollary 4.

For tt0eet\geq t_{0}\geq e^{e} and σ1\sigma\geq 1 we have

|1ζ(σ+it)|K2logtloglogt,\left|\frac{1}{\zeta(\sigma+it)}\right|\leq K_{2}\frac{\log t}{\log\log t},

where K2=107.7K_{2}=107.7 is valid for t0=500t_{0}=500, and other values are given in Table 6.

Furthermore, with the restriction σ=1\sigma=1, the range of tt can be extended. In other words, for t3t\geq 3 we have

|1ζ(1+it)|107.7logtloglogt.\left|\frac{1}{\zeta(1+it)}\right|\leq 107.7\frac{\log t}{\log\log t}.

3 Some preliminaries

This section contains the lemmas and theorems used to prove our main results. The first two lemmas, which form the basis of our method, are built upon a theorem of Hadamard, Borel and Carathéodory, sometimes also referred to as the Borel–Carathéodory theorem. This theorem enables us to deduce an upper bound for the modulus of a function and its derivatives on a circle, using bounds for its real part on a larger concentric circle.

Theorem 1.

[6, Theorem 4][Borel–Carathéodory] Let s0s_{0} be a complex number, and let RR be a positive number possibly dependent on s0s_{0}. Suppose that the function f(s)f(s) is analytic in a region containing the disc |ss0|R|s-s_{0}|\leq R. Let MM denote the maximum of Ref(s)\textup{Re}\,f(s) on the boundary |ss0|=R|s-s_{0}|=R. Then, for any r(0,R)r\in(0,R) and any ss such that |ss0|r|s-s_{0}|\leq r,

|f(s)|2rRrM+R+rRr|f(s0)|.|f(s)|\leq\frac{2r}{R-r}M+\frac{R+r}{R-r}|f(s_{0})|.

If in addition f(s0)=0f(s_{0})=0, then for any r(0,R)r\in(0,R) and any ss such that |ss0|r|s-s_{0}|\leq r,

|f(s)|2R(Rr)2M.|f^{\prime}(s)|\leq\frac{2R}{(R-r)^{2}}M.
Lemma 1.

[6, Lemma 5] Let s0s_{0} be a complex number, and let rr and α\alpha be positive numbers (possibly depending on s0s_{0}) such that α<1/2\alpha<1/2. Suppose that the function f(s)f(s) is analytic in a region containing the disc |ss0|r|s-s_{0}|\leq r. Suppose further that there is a number A1A_{1} independent of ss such that

|f(s)f(s0)|A1\left|\frac{f(s)}{f(s_{0})}\right|\leq A_{1}

for |ss0|r|s-s_{0}|\leq r. Then, for any ss in the disc |ss0|αr|s-s_{0}|\leq\alpha r we have

|f(s)f(s)ρ1sρ|4logA1r(12α)2,\left|\frac{f^{\prime}(s)}{f(s)}-\sum_{\rho}\frac{1}{s-\rho}\right|\leq\frac{4\log A_{1}}{r(1-2\alpha)^{2}},

where ρ\rho runs through the zeros of f(s)f(s) in the disc |ss0|12r|s-s_{0}|\leq\tfrac{1}{2}r, counted with multiplicity.

Lemma 2.

[6, Lemma 6] Let ss and s0s_{0} be complex numbers with real parts σ\sigma and σ0\sigma_{0}, respectively. Let rr, α\alpha, β\beta, A1A_{1} and A2A_{2} be positive numbers, possibly depending on s0s_{0}, such that α<1/2\alpha<1/2 and β<1\beta<1. Suppose that the function f(s)f(s) satisfies the conditions of Lemma 1 with rr, α\alpha and A1A_{1}, and that

|f(s0)f(s0)|A2.\left|\frac{f^{\prime}(s_{0})}{f(s_{0})}\right|\leq A_{2}.

Suppose, in addition, that f(s)0f(s)\neq 0 for any ss in both the disc |ss0|r|s-s_{0}|\leq r and the right half-plane σσ0αr\sigma\geq\sigma_{0}-\alpha r. Then, for any ss in the disc |ss0|αβr|s-s_{0}|\leq\alpha\beta r,

|f(s)f(s)|8βlogA1r(1β)(12α)2+1+β1βA2.\left|\frac{f^{\prime}(s)}{f(s)}\right|\leq\frac{8\beta\log A_{1}}{r(1-\beta)(1-2\alpha)^{2}}+\frac{1+\beta}{1-\beta}A_{2}.
Theorem 2.

Let k3k\geq 3 be an integer and σk:=1k/(2k2)\sigma_{k}:=1-k/(2^{k}-2). Then for all t3t\geq 3,

|ζ(σk+it)|1.546t1/(2k2)logt.|\zeta(\sigma_{k}+it)|\leq 1.546t^{1/(2^{k}-2)}\log t.
Proof.

Yang [20, Theorem 1.1] proves the result for all integers k4k\geq 4, and [7, Theorem 1.1] implies the result for k=3k=3, from |ζ(1/2+it)|0.618t1/6logt|\zeta(1/2+it)|\leq 0.618t^{1/6}\log t for all t3t\geq 3. ∎

Lemma 3.

Let s=σ+its=\sigma+it for real σ,t\sigma,t. Choose real parameters ω1>0\omega_{1}>0 and t03t_{0}\geq 3 such that

ω1maxtt0(8loglogtlogt).\omega_{1}\geq\max_{t\geq t_{0}}\left(\frac{8\log\log t}{\log t}\right).

Also define

ω2:={1ω1log2(11e+logω1loglogt0)ifω1<11ω1log2(11e)ifω11.\omega_{2}:=\begin{cases}\frac{1}{\omega_{1}\log 2}\left(1-\frac{1}{e}+\frac{\log\omega_{1}}{\log\log t_{0}}\right)\quad&\text{if}\quad\omega_{1}<1\\ \frac{1}{\omega_{1}\log 2}\left(1-\frac{1}{e}\right)\quad&\text{if}\quad\omega_{1}\geq 1.\end{cases}

Then for every κ>0\kappa>0, all tt0t\geq t_{0}, and 1ω2(loglogt)2logtσ1+κ1-\frac{\omega_{2}(\log\log t)^{2}}{\log t}\leq\sigma\leq 1+\kappa we have

|ζ(s)|AκlogBt,|\zeta(s)|\leq A_{\kappa}\log^{B}t,

where B=1+8/3ω1B=1+8/3\omega_{1} and

Aκ:=1.546ζ(1+κ)(1+2+κt0)1/6(1+2+κt0logt0)(1+21+κt0).A_{\kappa}:=1.546\zeta(1+\kappa)\left(1+\frac{2+\kappa}{t_{0}}\right)^{1/6}\left(1+\frac{2+\kappa}{t_{0}\log t_{0}}\right)\left(1+\frac{2\sqrt{1+\kappa}}{t_{0}}\right). (2)

We give some admissible values of Aκ,B,κ,ω1,ω2,A_{\kappa},\,B,\,\kappa,\,\omega_{1},\,\omega_{2}, and t0t_{0} in Table 1.

Proof.

We use the bound for ζ(s)\zeta(s) given in Theorem 2. The Phragmén–Lindelöf Principle can be used to show that a bound on ζ(s)\zeta(s) of the form in Theorem 2 holds for σkσ1+κ\sigma_{k}\leq\sigma\leq 1+\kappa for any real κ>0\kappa>0 and integer k3k\geq 3. We refer to the statement of the principle in [17, Lemma 3] (or [19, Lemma 2.1]), in which we consider f(s)=(s1)ζ(s)f(s)=(s-1)\zeta(s). As the lower bound on kk implies σk[1/2,1]\sigma_{k}\in[1/2,1], we can take Q=1Q=1 to satisfy the condition in [17, (4.1)] with a=σka=\sigma_{k}, b=1+κb=1+\kappa, α1=1+1/(2k2)\alpha_{1}=1+1/(2^{k}-2), α2=1\alpha_{2}=1, β1=1\beta_{1}=1, β2=0\beta_{2}=0, A=1.546A=1.546, and B=ζ(1+κ)B=\zeta(1+\kappa). This choice of QQ comes from separately checking |t|3|t|\leq 3 and |t|3|t|\geq 3. First, we can numerically verify

|(s1)ζ(s)|{1.546|s+1|1+1/(2k2)log|s+1|forσ=σk,ζ(1+κ)|s+1|forσ=1+κ,\displaystyle|(s-1)\zeta(s)|\leq\begin{cases}1.546|s+1|^{1+1/(2^{k}-2)}\log|s+1|\quad&\text{for}\ \sigma=\sigma_{k},\\ \zeta(1+\kappa)|s+1|\quad&\text{for}\ \sigma=1+\kappa,\end{cases}

both for |t|3|t|\leq 3, noting that the bound for σ=σk\sigma=\sigma_{k} follows from

max1/2σ<1|t|3(|(s1)ζ(s)|1.546|s+1|log|s+1|)<0.21,\max_{\begin{subarray}{c}1/2\leq\sigma<1\\ |t|\leq 3\end{subarray}}\Big{(}|(s-1)\zeta(s)|-1.546|s+1|\log|s+1|\Big{)}<-0.21,

the maximum occurring at t=0t=0 and σ=1/2\sigma=1/2. Second, it can be seen that

|s1|t1/(2k2)logt\displaystyle|s-1|t^{1/(2^{k}-2)}\log t |s+1|1+1/(2k2)log|s+1|\displaystyle\leq|s+1|^{1+1/(2^{k}-2)}\log|s+1|

for σ=σk\sigma=\sigma_{k} and |t|3|t|\geq 3, and lastly

|(s1)ζ(s)|ζ(1+κ)|s+1||(s-1)\zeta(s)|\leq\zeta(1+\kappa)|s+1|

for σ=1+κ\sigma=1+\kappa and |t|3|t|\geq 3. Then, by [17, Lemma 3] we have

|ζ(s)|\displaystyle|\zeta(s)| 1.546ζ(1+κ)|s+1|1+1/(2k2)log|s+1||s1|\displaystyle\leq 1.546\zeta(1+\kappa)\frac{|s+1|^{1+1/(2^{k}-2)}\log|s+1|}{|s-1|}
Aκt1/(2k2)logt\displaystyle\leq A_{\kappa}t^{1/(2^{k}-2)}\log t (3)

for σkσ1+κ\sigma_{k}\leq\sigma\leq 1+\kappa and tt03t\geq t_{0}\geq 3, where AκA_{\kappa} is defined in (2).

From here we can use the proof of Titchmarsh’s Theorem 5.17 [15]. Let

k=1log2log(ω1logtloglogt),k=\left\lfloor\frac{1}{\log 2}\log\left(\frac{\omega_{1}\log t}{\log\log t}\right)\right\rfloor,

where ω1\omega_{1} is chosen such that we have k3k\geq 3 for all tt0t\geq t_{0}. This kk implies

k2k2k2k1ω1log2log(ω1logtloglogt)loglogtlogtω2(loglogt)2logt\displaystyle\frac{k}{2^{k}-2}\geq\frac{k}{2^{k}}\geq\frac{1}{\omega_{1}\log 2}\log\left(\frac{\omega_{1}\log t}{\log\log t}\right)\frac{\log\log t}{\log t}\geq\frac{\omega_{2}(\log\log t)^{2}}{\log t} (4)

with ω2\omega_{2} as defined in the statement of the lemma.

The bound in (4) indicates that since (3) holds for σσk\sigma\geq\sigma_{k}, it also holds for

σ1ω2(loglogt)2logt.\sigma\geq 1-\frac{\omega_{2}(\log\log t)^{2}}{\log t}.

The expression for kk also implies

12k2432k8loglogt3ω1logt,\displaystyle\frac{1}{2^{k}-2}\leq\frac{4}{3\cdot 2^{k}}\leq\frac{8\log\log t}{3\omega_{1}\log t},

so (3) becomes

|ζ(s)|Aκt1/(2k2)logtAκt8loglogt/(3ω1logt)logt=Aκ(logt)1+8/3ω1,\displaystyle|\zeta(s)|\leq A_{\kappa}t^{1/(2^{k}-2)}\log t\leq A_{\kappa}t^{8\log\log t/(3\omega_{1}\log t)}\log t=A_{\kappa}(\log t)^{1+8/3\omega_{1}},

where AκA_{\kappa} is in (2). For specific t0t_{0}, we can find a κ\kappa that minimises AκA_{\kappa}, then find the smallest ω1\omega_{1} that permits our choice of kk for all tt0t\geq t_{0}. It is then possible to compute BB and ω2\omega_{2}. This allows us to compute the values in Table 1 mentioned in the statement of the lemma. ∎

We can compare this result to that of Hiary, Leong and Yang [6, Lemma 7], derived using a Richert-type bound for ζ(s)\zeta(s) by Ford in [4]. For instance, in the range teet\geq e^{e} and 10.309(loglogt)2logtσ21-\frac{0.309(\log\log t)^{2}}{\log t}\leq\sigma\leq 2, we have

|ζ(s)|2.5(logt)1.91,|\zeta(s)|\leq 2.5(\log t)^{1.91},

whereas [6, Lemma 7] has

|ζ(s)|76.2(logt)3.29.|\zeta(s)|\leq 76.2(\log t)^{3.29}.
      t0t_{0}       κ\kappa       AκA_{\kappa}       ω1\omega_{1}       BB       ω2\omega_{2}
      33       1.51.5       1010       8/e8/e       1.911.91       0.3090.309
      eee^{e}       3.23.2       2.52.5       8/e8/e       1.911.91       0.3090.309
      500500       8.18.1       1.61.6       2.362.36       2.132.13       0.3860.386
      HH       41.541.5       1.61.6       0.940.94       3.843.84       0.9410.941
Table 1: Admissible values of parameters in Lemma 3.
Lemma 4.

[20, Corollary 1.2] There are no zeroes of ζ(σ+it)\zeta(\sigma+it) in the region

σ>1loglogt21.233logt,t3.\sigma>1-\frac{\log\log t}{21.233\log t},\qquad t\geq 3.
Lemma 5.

[14, Lemma 5.4] Let γ\gamma denote Euler’s constant. For σ>1\sigma>1 we have

|ζ(σ+it)|ζ(σ)eγ(σ1)σ1.|\zeta(\sigma+it)|\leq\zeta(\sigma)\leq\frac{e^{\gamma(\sigma-1)}}{\sigma-1}.
Theorem 3.

[6, Theorem 1] For t3t\geq 3 we have

|ζ(1+it)|\displaystyle|\zeta(1+it)| 1.731logtloglogt.\displaystyle\leq 1.731\frac{\log t}{\log\log t}.
Lemma 6.

Let a,b,Q,a,b,Q, and kk be real numbers such that Q+a>eQ+a>e, and let f(s)f(s) be a holomorphic function in the strip aRe(s)ba\leq\textup{Re}(s)\leq b such that it satisfies the growth condition

|f(s)|<Cexp(ek|t|)|f(s)|<C\exp\big{(}e^{k|t|}\big{)}

for some C>0C>0 and 0<k<π/(ba)0<k<\pi/(b-a). Suppose further that

|f(s)|{A|Q+s|α1(log|Q+s|loglog|Q+s|)α2for Re(s)=aB|Q+s|β1(log|Q+s|loglog|Q+s|)β2for Re(s)=b,\displaystyle|f(s)|\leq\begin{cases}A|Q+s|^{\alpha_{1}}\left(\frac{\log|Q+s|}{\log\log|Q+s|}\right)^{\alpha_{2}}\quad\text{for ${\textup{Re}}(s)=a$}\\ B|Q+s|^{\beta_{1}}\left(\frac{\log|Q+s|}{\log\log|Q+s|}\right)^{\beta_{2}}\quad\text{for ${\textup{Re}}(s)=b$},\end{cases} (5)

where α1β1\alpha_{1}\geq\beta_{1} and where A,B,α1,α2,β1,β20A,B,\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}\geq 0. Then throughout the strip aRe(s)ba\leq\textup{Re}(s)\leq b, one has

|f(s)|\displaystyle|f(s)|\leq (A|Q+s|α1(log|Q+s|loglog|Q+s|)α2)bRe(s)ba\displaystyle\left(A|Q+s|^{\alpha_{1}}\left(\frac{\log|Q+s|}{\log\log|Q+s|}\right)^{\alpha_{2}}\right)^{\tfrac{b-\textup{Re}(s)}{b-a}}
(B|Q+s|β1(log|Q+s|loglog|Q+s|)β2)Re(s)aba.\displaystyle\qquad\qquad\bm{\cdot}\left(B|Q+s|^{\beta_{1}}\left(\frac{\log|Q+s|}{\log\log|Q+s|}\right)^{\beta_{2}}\right)^{\tfrac{\textup{Re}(s)-a}{b-a}}.
Proof.

This is derived in a similar way to Lemma 3 of [17]. Trudgian’s lemma is an adaption of a theorem from Rademacher on a generalisation of the Phragmén–Lindelöf theorem [13, §33]. The Phragmén–Lindelöf theorem [13, §29] is applied to a function F(s)F(s), to which Trudgian adds a factor of log(Q+s)\log(Q+s), to use the theorem for bounds on f(s)f(s) that have a factor of log|Q+s|\log|Q+s|.

We will make a similar change to Rademacher’s theorem. However, instead of a factor of log(Q+s)\log(Q+s), we will include a factor of log(Q+s)/loglog(Q+s)\log(Q+s)/\log\log(Q+s) to form the function

F(s)=f(s)ϕ(s;Q)E1evs(log(Q+s)loglog(Q+s))α2(σb)+β2(aσ)ba,F(s)=f(s)\phi(s;Q)E^{-1}e^{-vs}\left(\frac{\log(Q+s)}{\log\log(Q+s)}\right)^{\frac{\alpha_{2}(\sigma-b)+\beta_{2}(a-\sigma)}{b-a}},

where EE is defined in [17, Lem. 2]. We also include the restriction Q+a>eQ+a>e to ensure that F(s)F(s) is holomorphic in the strip aσba\leq\sigma\leq b. The proof then proceeds as in [17, Lemma 3] and [13, §32]. ∎

Lemma 7.

For 1σσ11\leq\sigma\leq\sigma_{1} with σ1>1\sigma_{1}>1, and tt03t\geq t_{0}\geq 3, we have

|ζ(σ+it)|Z(σ1,t0)logtloglogt,|\zeta(\sigma+it)|\leq Z(\sigma_{1},t_{0})\frac{\log t}{\log\log t},

where

Z(σ1,t0):=1.731ζ(σ1)(1+3t0)log(t0+σ1+2)logt0.Z(\sigma_{1},t_{0}):=1.731\zeta(\sigma_{1})\left(1+\frac{3}{t_{0}}\right)\frac{\log(t_{0}+\sigma_{1}+2)}{\log t_{0}}.

Admissible values of ZZ are given in Table 2 for specific t0t_{0} and optimized σ1\sigma_{1}.

      t0t_{0}       σ1\sigma_{1}       ZZ
      33       4.324.32       7.4797.479
      eee^{e}       5.885.88       2.4392.439
      2ee2e^{e}       6.776.77       2.0652.065
      500500       11.1411.14       1.7501.750
      10001000       11.9711.97       1.7411.741
      HH       11.9611.96       1.7321.732
Table 2: Values for ZZ in Lemma 7 for specific t0t_{0} after optimizing over σ1\sigma_{1}.
Proof.

This is similar to the proof of Lemma 3. We take f(s):=(s1)ζ(s)f(s):=(s-1)\zeta(s) in Lemma 6 and bound ζ(s)\zeta(s) in 1σσ11\leq\sigma\leq\sigma_{1} for σ1>1\sigma_{1}>1. Note that f(s)f(s) is entire in this strip, and satisfies the growth condition given by Rademacher or [17, Lemma 3].

We will need two bounds on |ζ(s)||\zeta(s)| to verify the condition in (5): one for σ=1\sigma=1 and one for σ=σ1\sigma=\sigma_{1}. The latter is relatively simple to derive, as we can use |ζ(σ1+it)|ζ(σ1)|\zeta(\sigma_{1}+it)|\leq\zeta(\sigma_{1}), which is true for all tt and computable. For the other bound we use Theorem 3.

With a=1a=1, b=2b=2, α1=1\alpha_{1}=1, α2=1\alpha_{2}=1, β1=1\beta_{1}=1, β2=0\beta_{2}=0, A=1.731A=1.731 and B=ζ(σ1)B=\zeta(\sigma_{1}), we will take Q=2Q=2 to check that (5) holds for both |t|3|t|\leq 3 and |t|3|t|\geq 3. First, it can be numerically verified that for σ=1\sigma=1,

|(s1)ζ(s)|1.731|s+2|log|s+2|loglog|s+2|\displaystyle|(s-1)\zeta(s)|\leq 1.731|s+2|\frac{\log|s+2|}{\log\log|s+2|}

for |t|3|t|\leq 3. Note that the choice of QQ comes from this particular condition. Second, Theorem 3 implies that

|(s1)ζ(s)|1.731tlogtloglogt\displaystyle|(s-1)\zeta(s)|\leq 1.731\frac{t\log t}{\log\log t} 1.731|s+2|log|s+2|loglog|s+2|\displaystyle\leq 1.731|s+2|\frac{\log|s+2|}{\log\log|s+2|}

for σ=1\sigma=1 and |t|3|t|\geq 3. Lastly,

|(s1)ζ(s)|ζ(σ1)|s+2||(s-1)\zeta(s)|\leq\zeta(\sigma_{1})|s+2|

holds for σ=σ1\sigma=\sigma_{1} and all tt.

Then, following a similar conclusion to [17, Lemma 3], we have

|ζ(s)|\displaystyle|\zeta(s)| 1.731ζ(σ1)|s+2||s1|log|s+2|loglog|s+2|Z(σ1,t0)logtloglogt,\displaystyle\leq 1.731\zeta(\sigma_{1})\frac{|s+2|}{|s-1|}\frac{\log|s+2|}{\log\log|s+2|}\leq Z(\sigma_{1},t_{0})\frac{\log t}{\log\log t},

for 1σσ11\leq\sigma\leq\sigma_{1} and tt03t\geq t_{0}\geq 3. Finally, we optimize over σ1\sigma_{1} with the aim of minimizing ZZ, to obtain the values in Table 2, as mentioned in the statement of the lemma. ∎

Lemma 8.

Let d,σ1d,\,\sigma_{1} be real positive parameters, σ=1+dloglogt/logt\sigma=1+d\log\log t/\log t, γ\gamma be Euler’s constant, and Z(σ1,t0)Z(\sigma_{1},t_{0}) be defined as in Lemma 7. Then for tt03t\geq t_{0}\geq 3 and all σ>1\sigma>1, we have

|1ζ(σ+it)|V1(d,σ1,t0)logtloglogt,\left|\frac{1}{\zeta(\sigma+it)}\right|\leq V_{1}(d,\sigma_{1},t_{0})\frac{\log t}{\log\log t},

where

V1(d,σ1,t0):=(1d+loglogt0logt0)3/4(Z(σ1,2t0)(1+log2logt0))1/4.V_{1}(d,\sigma_{1},t_{0}):=\left(\frac{1}{d}+\frac{\log\log t_{0}}{\log t_{0}}\right)^{3/4}\left(Z(\sigma_{1},2t_{0})\left(1+\frac{\log 2}{\log t_{0}}\right)\right)^{1/4}. (6)

In addition, for σ>1\sigma>1 such that σσ1\sigma\leq\sigma_{1} and σ(1+γ)/γ\sigma\leq(1+\gamma)/\gamma, we have for tt03t\geq t_{0}\geq 3 that

|1ζ(σ+it)|V2(d,σ1,t0)logtloglogt,\left|\frac{1}{\zeta(\sigma+it)}\right|\leq V_{2}(d,\sigma_{1},t_{0})\frac{\log t}{\log\log t},

where

V2(d,σ1,t0):=(1dexp(γdloglogt0logt0))3/4(Z(σ1,2t0)(1+log2logt0))1/4.V_{2}(d,\sigma_{1},t_{0}):=\left(\frac{1}{d}\exp\left(\frac{\gamma d\log\log t_{0}}{\log t_{0}}\right)\right)^{3/4}\left(Z(\sigma_{1},2t_{0})\left(1+\frac{\log 2}{\log t_{0}}\right)\right)^{1/4}. (7)
Proof.

Since σ>1\sigma>1, we use the classical non-negativity argument involving the trigonometric polynomial 2(1+cosθ)2=3+4cosθ+cos2θ2(1+\cos\theta)^{2}=3+4\cos\theta+\cos 2\theta (see [16, § 3.3]). That is, we use ζ3(σ)|ζ4(σ+it)ζ(σ+2it)|1\zeta^{3}(\sigma)|\zeta^{4}(\sigma+it)\zeta(\sigma+2it)|\geq 1, which implies

|1ζ(σ+it)||ζ(σ)|3/4|ζ(σ+2it)|1/4.\left|\frac{1}{\zeta(\sigma+it)}\right|\leq|\zeta(\sigma)|^{3/4}|\zeta(\sigma+2it)|^{1/4}. (8)

Taking σ=1+δ:=1+dloglogt/logt\sigma=1+\delta:=1+d\log\log t/\log t in the above gives

|1ζ(1+δ+it)||ζ(1+δ)|3/4|ζ(1+δ+2it)|1/4.\left|\frac{1}{\zeta(1+\delta+it)}\right|\leq|\zeta(1+\delta)|^{3/4}|\zeta(1+\delta+2it)|^{1/4}. (9)

These two factors can be bounded with Lemmas 5 and 7. First consider 1<σmin{σ1,(1+γ)/γ}1<\sigma\leq\min\{\sigma_{1},(1+\gamma)/\gamma\}. In this range, eγδ/δe^{\gamma\delta}/\delta is decreasing. Hence, we have

eγδδ\displaystyle\frac{e^{\gamma\delta}}{\delta} 1dexp(γdloglogt0logt0)logtloglogt,\displaystyle\leq\frac{1}{d}\exp\left(\frac{\gamma d\log\log t_{0}}{\log t_{0}}\right)\frac{\log t}{\log\log t},
log2tloglog2t\displaystyle\frac{\log 2t}{\log\log 2t} (1+log2logt0)logtloglogt.\displaystyle\leq\left(1+\frac{\log 2}{\log t_{0}}\right)\frac{\log t}{\log\log t}.

Applying these bounds in Lemmas 5 and 7, and then (9), we find

|1ζ(1+δ+it)|V2(d,σ1,t0)logtloglogt.\left|\frac{1}{\zeta(1+\delta+it)}\right|\leq V_{2}(d,\sigma_{1},t_{0})\frac{\log t}{\log\log t}.

Second, to obtain a bound over all σ>1\sigma>1, we simply repeat the process with the trivial bound ζ(σ)σ/(σ1)\zeta(\sigma)\leq\sigma/(\sigma-1) in place of Lemma 5, as it is decreasing for all σ>1\sigma>1. However, it is worth noting that this bound is not as sharp as Lemma 5 for σ\sigma close to 11. ∎

4 Bounding |ζ(s)/ζ(s)||\zeta^{\prime}(s)/\zeta(s)|

In this section, we use the lemmas in the previous section to bound |ζ(s)/ζ(s)||\zeta^{\prime}(s)/\zeta(s)|. We follow the argument of [18], which gives an explicit version of results in [16, §3]. This has been done in [6] to obtain bounds for |ζ(s)/ζ(s)||\zeta^{\prime}(s)/\zeta(s)| and |1/ζ(s)||1/\zeta(s)| on the 11-line, and we now extend these results to hold within a zero-free region. The method is identical to that of [6], but uses sharper bounds on ζ\zeta, and a more careful handling of constants to obtain some savings. Given this, we will only give an outline of how to obtain such bounds, defining the functions with the same notation, but omitting details. The reader is referred to [6] for a full elucidation on the proof, especially on justification of certain parameter choices.

As before, we construct concentric discs, centred just to the right of the line σ=1+it\sigma=1+it, and extend them slightly to the left and into the critical strip. We aim to apply Lemmas 1 and 2 with f(s)=ζ(s)f(s)=\zeta(s).

Let t0eet_{0}\geq e^{e} and tt0t^{\prime}\geq t_{0} be constants. The centre of the concentric discs will be denoted by

s=σ+it=1+δt+it=1+dloglogtlogt+it,s^{\prime}=\sigma^{\prime}+it^{\prime}=1+\delta_{t^{\prime}}+it^{\prime}=1+\frac{d\log\log t^{\prime}}{\log t^{\prime}}+it^{\prime},

where dd is a real positive constant to be chosen later. Notice that δt\delta_{t^{\prime}} is decreasing for t>eet^{\prime}>e^{e}, and is at most d/ed/e. Also, let rr and ϵ1\epsilon\leq 1 be positive parameters chosen such that σ+r1+ϵ\sigma^{\prime}+r\leq 1+\epsilon, which will be satisfied if

de+rϵ.\frac{d}{e}+r\leq\epsilon. (10)

Let s=σ+its=\sigma+it be a complex number. Aiming to apply Lemma 1 in the disc |ss|r|s-s^{\prime}|\leq r, we first seek a valid A1A_{1} in that disc. Similarly, for Lemma 2, we seek a valid A2A_{2} at the disc centre s=ss=s^{\prime}. In applying Lemma 2, we also need to ensure that the non-vanishing condition on f(s)=ζ(s)f(s)=\zeta(s) is met, which we do with a zero-free region.

First, let

σ1,t:=1C1(loglogt)2logt,C1:=ω2min{(loglog(tϵ)loglogt)2,logtlog(t+ϵ)},\sigma_{1,t^{\prime}}:=1-\frac{C_{1}\,(\log\log t^{\prime})^{2}}{\log t^{\prime}},\,\,C_{1}:=\omega_{2}\cdot\min\left\{\left(\frac{\log\log(t^{\prime}-\epsilon)}{\log\log t^{\prime}}\right)^{2},\frac{\log t^{\prime}}{\log(t^{\prime}+\epsilon)}\right\}, (11)

where ω2\omega_{2} is defined in Lemma 3. Then, by the same lemma, for each t[tr,t+r]t\in[t^{\prime}-r,t^{\prime}+r] and any σ[σ1,t,1+κ]\sigma\in[\sigma_{1,t^{\prime}},1+\kappa], we have

|ζ(s)|AκlogB(t+r)A3logBt,A3=A3(t):=Aκ(1+log(1+ϵ/t)logt)B,|\zeta(s)|\leq A_{\kappa}\log^{B}(t^{\prime}+r)\leq A_{3}\log^{B}t^{\prime},\quad A_{3}=A_{3}(t^{\prime}):=A_{\kappa}\left(1+\frac{\log(1+\epsilon/t^{\prime})}{\log t^{\prime}}\right)^{B}, (12)

with AκA_{\kappa} and BB given in Table 1.

For our purposes, we choose a radius of

r=rt:=(C1+dloglogt)(loglogt)2logtC1(loglogt)2logt+δt.r=r_{t^{\prime}}:=\left(C_{1}+\frac{d}{\log\log t^{\prime}}\right)\frac{(\log\log t^{\prime})^{2}}{\log t^{\prime}}\leq\frac{C_{1}(\log\log t^{\prime})^{2}}{\log t^{\prime}}+\delta_{t^{\prime}}. (13)

To ensure the constraint (10) is met for all tt0t^{\prime}\geq t_{0}, we need

de+(C1+d)4e2ϵ1,\frac{d}{e}+\left(C_{1}+d\right)\frac{4}{e^{2}}\leq\epsilon\leq 1, (14)

as (loglogt)2/logt(\log\log t^{\prime})^{2}/\log t^{\prime} reaches a maximum of 4/e24/e^{2} at t=te:=ee2t^{\prime}=t_{e}:=e^{e^{2}}. This choice of rr also requires Lemma 3 to be applicable for σ\sigma up to 1+δt+r1+\delta_{t^{\prime}}+r, which corresponds to

κδt+r=(C1+2dloglogt)(loglogt)2logt.\kappa\geq\delta_{t^{\prime}}+r=\left(C_{1}+\frac{2d}{\log\log t^{\prime}}\right)\frac{(\log\log t^{\prime})^{2}}{\log t^{\prime}}.

This will be satisfied for all t0eet_{0}\geq e^{e} for

κ4e2(ω2+2dloglogt0),\kappa\geq\frac{4}{e^{2}}\left(\omega_{2}+\frac{2d}{\log\log t_{0}}\right),

and the values of κ\kappa in Table 1 satisfy this condition.

To fulfil the bounding condition of Lemma 1, we can take either lower bound for |ζ(s)||\zeta(s^{\prime})| from Lemma 8, and use a constant

V{V1(d,σ1,t0),V2(d,σ1,t0)},V\in\left\{V_{1}(d,\sigma_{1},t_{0}),\,V_{2}(d,\sigma_{1},t_{0})\right\}, (15)

where dd and tt^{\prime} are always taken such that 1+δt(1+γ)/γ1+\delta_{t^{\prime}}\leq(1+\gamma)/\gamma. With this, by (12) and (13), throughout the disc |ss|r|s-s^{\prime}|\leq r we have

|ζ(s)ζ(s)|\displaystyle\left|\frac{\zeta(s)}{\zeta(s^{\prime})}\right|\leq A3VlogB+1tloglogt.\displaystyle\,A_{3}\,V\frac{\log^{B+1}t^{\prime}}{\log\log t^{\prime}}.

Noting that A3A_{3} is decreasing in tt0t^{\prime}\geq t_{0}, we obtain that throughout |ss|r|s-s^{\prime}|\leq r,

log|ζ(s)ζ(s)|logA1:=(B+1)loglogtlogloglogt+log(AmaxV),\log\left|\frac{\zeta(s)}{\zeta(s^{\prime})}\right|\leq\log A_{1}:=(B+1)\log\log t^{\prime}-\log\log\log t^{\prime}+\log\left(A_{\max}\,V\right), (16)

where Amax=A3(t0)A_{\max}=A_{3}(t_{0}).

Next, we fulfill the zero-free condition of Lemma 2 by way of the zero-free region in Lemma 4, where for t3t\geq 3,

ζ(σ+it)0forσ>1c0loglogtlogt,wherec0:=121.233.\zeta(\sigma+it)\neq 0\quad\text{for}\quad\sigma>1-\frac{c_{0}\log\log t}{\log t},\quad\text{where}\quad c_{0}:=\frac{1}{21.233}. (17)

This leads to the following choice of

α=1r(d+c1)loglogtlogt=d+c1d+C1loglogt,\alpha=\frac{1}{r}\cdot\frac{(d+c_{1})\log\log t^{\prime}}{\log t^{\prime}}=\frac{d+c_{1}}{d+C_{1}\log\log t^{\prime}}, (18)

where

c1:=c0logt0log(t0+ϵ),c_{1}:=\frac{c_{0}\log t_{0}}{\log(t_{0}+\epsilon)}, (19)

to keep c1c0c_{1}\leq c_{0}, and we require

d+c1d+C1loglogt0<12\frac{d+c_{1}}{d+C_{1}\log\log t_{0}}<\frac{1}{2} (20)

to satisfy the constraint α<1/2\alpha<1/2 in Lemma 1.

To determine A2A_{2}, we utilise [5, Lemma 70.1], which states that for σ>1\sigma>1,

|ζ(s)ζ(s)|ζ(σ)ζ(σ)<1σ1.\left|\frac{\zeta^{\prime}(s)}{\zeta(s)}\right|\leq-\frac{\zeta^{\prime}(\sigma)}{\zeta(\sigma)}<\frac{1}{\sigma-1}. (21)

Taking s=ss=s^{\prime} in this inequality leads to

|ζ(s)ζ(s)|<1δt=A2:=logtdloglogt.\left|\frac{\zeta^{\prime}(s^{\prime})}{\zeta(s^{\prime})}\right|<\frac{1}{\delta_{t^{\prime}}}=A_{2}:=\frac{\log t^{\prime}}{d\log\log t^{\prime}}.

Finally, we turn to the choice of β\beta in Lemma 2. From here on, we differ from [6] as we want the circles to be able to extend into the zero-free region. For any results from Lemma 2 to be meaningful to us, we want the left-most point of |ss|αβr|s-s^{\prime}|\leq\alpha\beta r to lie between the zero-free region and not exceed the 11-line. In other words, we want σαβr1\sigma^{\prime}-\alpha\beta r\leq 1, which means β\beta must satisfy

1>βδtαr=dc1+d.1>\beta\geq\frac{\delta_{t^{\prime}}}{\alpha r}=\frac{d}{c_{1}+d}. (22)

Up to this point, Lemma 2 holds for

1(βc1d(1β))loglogtlogtσ1+(βc1+d(1+β))loglogtlogt.1-(\beta c_{1}-d(1-\beta))\frac{\log\log t^{\prime}}{\log t^{\prime}}\leq\sigma^{\prime}\leq 1+(\beta c_{1}+d(1+\beta))\frac{\log\log t^{\prime}}{\log t^{\prime}}.

For larger σ\sigma^{\prime}, we will just use (21). To finish, observe that 1/(12α)21/(1-2\alpha)^{2} is decreasing in tt^{\prime}, so we can substitute ttt^{\prime}\mapsto t to arrive at the following lemma.

Lemma 9.

Let t0eet_{0}\geq e^{e}, σ1σ\sigma_{1}\geq\sigma, d>0d>0, and ϵ1\epsilon\leq 1 be constants satisfying the constraints (14) and (20). Let β>0\beta>0 (subject to (22)) and c1=c1(ϵ,t0)c_{1}=c_{1}(\epsilon,t_{0}) (defined in (19)) be chosen such that W:=1/(βc1d(1β))>1/c0W:=1/(\beta c_{1}-d(1-\beta))>1/c_{0}, where c0:=1/21.233c_{0}:=1/21.233. Then, for tt0t\geq t_{0} and

σ1loglogtWlogt\sigma\geq 1-\frac{\log\log t}{W\log t}

we have

|ζ(σ+it)ζ(σ+it)|Q1(d,ϵ,σ1,t0)logtloglogt,\left|\frac{\zeta^{\prime}(\sigma+it)}{\zeta(\sigma+it)}\right|\leq Q_{1}(d,\epsilon,\sigma_{1},t_{0})\frac{\log t}{\log\log t},

where

Q1(d,ϵ,σ1,t0):=minV{max\displaystyle Q_{1}(d,\epsilon,\sigma_{1},t_{0}):=\min_{V}\Bigg{\{}\max {1βc1+d(1+β),\displaystyle\Bigg{\{}\frac{1}{\beta c_{1}+d(1+\beta)},\,
λ1(B+1+log(AmaxV)loglogt0)+λ2}},\displaystyle\lambda_{1}\left(B+1+\frac{\log\left(A_{\max}\,V\right)}{\log\log t_{0}}\right)+\lambda_{2}\Bigg{\}}\Bigg{\}},

with

λ1:=8βC1(1β)(C1loglogt0+dC1loglogt0d2c1)2,λ2:=1+βd(1β).\begin{split}\lambda_{1}:=\frac{8\beta}{C_{1}(1-\beta)}\left(\frac{C_{1}\log\log t_{0}+d}{C_{1}\log\log t_{0}-d-2c_{1}}\right)^{2},\qquad\lambda_{2}:=\frac{1+\beta}{d(1-\beta)}.\end{split}

The constant BB is defined in Lemma 3, AmaxA_{\max} is defined in (16) and depends on t0t_{0}, C1C_{1} is defined in (11) and depends on ϵ\epsilon and t0t_{0}, and VV is defined in (15) and depends on d,σ1d,\,\sigma_{1} and t0t_{0}.

4.1 Computations

We can compute admissible R1R_{1} in Corollary 1 using Lemma 9.

Proof of Corollary 1.

Although we can use Lemma 9 with t0=eet_{0}=e^{e}, we can reduce the size of Q1Q_{1} by computing it for t0=Ht_{0}=H, the height to which the Riemann Hypothesis has been verified in [12], and combining it with a result for eet0t<He^{e}\leq t_{0}\leq t<H. Over the latter range, we know there are no zeros of ζ(s)\zeta(s) with σ>1/2\sigma>1/2. We can therefore use a wider zero-free region than in Lemma 4 when verifying the condition in Lemma 2. Since

12<1(e/2)loglogtlogt\frac{1}{2}<1-\frac{(e/2)\log\log t}{\log t}

for all t>0t>0, we can replace c0c_{0} with any cRH<e/2c_{RH}<e/2 in (17) as long as it fulfils (20). Given (19), we see that (20) is satisfied by

cRH<C1loglogt0d2,c_{RH}<\frac{C_{1}\log\log t_{0}-d}{2}, (23)

since log(t0+ϵ)/logt0>1\log(t_{0}+\epsilon)/\log t_{0}>1 for all t0>0t_{0}>0. When optimising over dd to calculate Q1Q_{1} for tHt\geq H, we consistently found an optimal d<0.05d<0.05. Hence, we will assume d<0.05d<0.05, which, along with ϵ1\epsilon\leq 1 and t0=eet_{0}=e^{e}, gives us that the right-hand side of (23) is

12(0.309min{(loglog(ee1))2,elog(ee+1)}0.05)0.121,\frac{1}{2}\left(0.309\cdot\min\left\{(\log\log(e^{e}-1))^{2},\frac{e}{\log(e^{e}+1)}\right\}-0.05\right)\geq 0.121\ldots,

by Table 1. Therefore, we can choose any cRH1/8.22<0.121<e/2c_{RH}\leq 1/8.22<0.121\ldots<e/2 for the range of values eetHe^{e}\leq t\leq H.

Although a larger cRHc_{RH} would mean we have a larger zero-free region, taking it too large can have an adverse effect on the parameter α\alpha. Smaller cRHc_{RH} reduces α\alpha, which means a better bound in Lemma 2, and hence in Lemma 9. However, larger cRHc_{RH} means we can take smaller β\beta to satisfy the expression for WW, which leads to a smaller value for λ1\lambda_{1} in Lemma 9.

Starting by fixing a value for WW, we compute Q1Q_{1} in the range tHt\geq H by optimising over ϵ\epsilon, β\beta, and dd, subject to the constraints of Lemma 9, and using c0c_{0} as in (17). We similarly obtain Q1Q_{1} in the range t0tHt_{0}\leq t\leq H, but instead optimise over ϵ\epsilon, β\beta, d<0.05d<0.05, and cRH1/8.22c_{RH}\leq 1/8.22. We also require the result to hold for

σαβr1loglogtWlogt,\sigma^{\prime}-\alpha\beta r\leq 1-\frac{\log\log t}{W\log t},

which leads to replacing the condition in (22) with

βd+1/Wd+c1.\beta\geq\frac{d+1/W}{d+c_{1}}.

Finally we take the maximum of the two results, which is an admissible Q1Q_{1} over all tt0t\geq t_{0}. This value is R1R_{1} in Corollary 1. In Table 3, we opt to use the smallest t0t_{0} that makes the Q1Q_{1} over t0tHt_{0}\leq t\leq H the smallest of the two Q1Q_{1}. For cRHc_{RH} we use 1212 for all entries up to and including W=31W=31; for W=35W=35 we use cRH=10.5c_{RH}=10.5, and for all larger WW we use cRH=9c_{RH}=9. ∎

WW t0t_{0} R1R_{1} α1\alpha_{1} ϵ\epsilon dd β\beta σ1\sigma_{1}
21.2421.24 eee^{e} 586798586798 0.027470.02747 0.9840.984 0.0408330.040833 0.9998234790.999823479 9.599.59
21.321.3 eee^{e} 6141161411 0.027360.02736 0.6690.669 0.0404690.040469 0.9983081980.998308198 13.5313.53
21.421.4 eee^{e} 2479324793 0.028990.02899 0.5710.571 0.0458260.045826 0.9960447810.996044781 13.0213.02
21.521.5 eee^{e} 1554715547 0.026150.02615 0.7940.794 0.0365030.036503 0.9930038440.993003844 12.7112.71
21.621.6 eee^{e} 1134811348 0.029020.02902 0.750.75 0.0459350.045935 0.9913985490.991398549 9.029.02
21.721.7 eee^{e} 89188918 0.028220.02822 0.6030.603 0.0433090.043309 0.9887888890.988788889 9.779.77
21.821.8 eee^{e} 73677367 0.026610.02661 0.7920.792 0.0379980.037998 0.9856049870.985604987 13.3613.36
21.921.9 eee^{e} 62726272 0.027420.02742 0.6020.602 0.0406760.040676 0.9836577020.983657702 7.527.52
2222 eee^{e} 54715471 0.027060.02706 0.6230.623 0.0394790.039479 0.9810343720.981034372 6.456.45
22.522.5 eee^{e} 33573357 0.027720.02772 0.7670.767 0.0416520.041652 0.9701172190.970117219 6.976.97
2323 eee^{e} 24392439 0.027040.02704 0.5980.598 0.0394110.039411 0.9581743130.958174313 9.939.93
2424 1818 15991599 0.027310.02731 0.8370.837 0.0402950.040295 0.937867550.93786755 9.839.83
2525 2424 12051205 0.02730.0273 0.9080.908 0.0402740.040274 0.9187773270.918777327 13.4113.41
2626 3333 976976 0.027230.02723 0.7350.735 0.0400440.040044 0.900907440.90090744 13.3813.38
2727 4545 826826 0.026980.02698 0.5830.583 0.0391990.039199 0.8834298740.883429874 8.348.34
2929 8585 643643 0.026450.02645 0.5620.562 0.0374560.037456 0.8508172940.850817294 7.117.11
3131 163163 534534 0.026590.02659 0.8490.849 0.0379450.037945 0.8255151090.825515109 8.048.04
3535 300300 412412 0.02620.0262 0.5830.583 0.0366610.036661 0.7788255080.778825508 10.6510.65
4040 490490 332332 0.026090.02609 0.9560.956 0.0363010.036301 0.735047140.73504714 10.0410.04
5050 500500 256256 0.02570.0257 0.6070.607 0.0350050.035005 0.6699628630.669962863 14.2714.27
6060 500500 219219 0.025430.02543 0.8380.838 0.0341130.034113 0.6252943360.625294336 11.6611.66
7070 500500 197197 0.025190.02519 0.810.81 0.0333520.033352 0.5921506870.592150687 14.7714.77
Table 3: Values for W,R1W,R_{1} in Corollary 1, based on (α,β,d,ϵ,σ1)(\alpha,\beta,d,\epsilon,\sigma_{1}) from Lemma 9, where α1\alpha_{1} is an upper bound on α\alpha, and each entry is valid for tt0t\geq t_{0}.

In addition, we have Corollary 2, especially relevant for σ=1\sigma=1.

Proof of Corollary 2.

The process of obtaining the bound for σ1\sigma\geq 1 is the same as that of [6, Theorem 2], but with the corresponding changes in AmaxA_{\max}, BB, and C1C_{1} due to Lemma 3. Note that the method is essentially the same as our Corollary 1, but we take β=d/(c1+d)\beta=d/(c_{1}+d). The rest of the proof follows mutatis mutandis.

As per Corollary 1, the computations are split into cases t0tHt_{0}\leq t\leq H and tHt\geq H, and we take the larger of the two resulting constants. An optimised value of cRHc_{RH} is used for each t0<Ht_{0}<H, and we note that both entries in Table 4 are determined by the value of Q1Q_{1} for t0tHt_{0}\leq t\leq H. ∎

t0t_{0} K1K_{1} 1/cRH1/c_{RH} α1\alpha_{1} dd s1s_{1} ϵ\epsilon
eee^{e} 238.4238.4 16.716.7 0.21880.2188 0.009460.00946 7.897.89 0.17450.1745
500500 113.3113.3 8.38.3 0.19380.1938 0.02010.0201 11.2611.26 0.40980.4098
HH 110.6110.6 21.23321.233 0.02400.0240 0.02950.0295 8.878.87 0.88150.8815
Table 4: Values for K1K_{1} in Corollary 2, based on (α,d,ϵ)(\alpha,d,\epsilon) from Lemma 9, where α1\alpha_{1} is an upper bound on α\alpha, and each entry is valid for tt0t\geq t_{0}.

5 Bounding |1/ζ(s)||1/\zeta(s)|

Moving from a bound on the logarithmic derivative of ζ(s)\zeta(s) to one on the reciprocal of ζ(s)\zeta(s) is done in the same way as [6, Lemma 10]. In essence, a bound on the latter depends on a bound for the former in some range of Re(s){\textup{Re}}(s). We are able to make some savings by using multiple bounds for |ζ(s)/ζ(s)||\zeta^{\prime}(s)/\zeta(s)| across the desired range, using corresponding pairs of (W,R1)(W,R_{1}) from Tables 3 and 4.

Lemma 10.

Let t0eet_{0}\geq e^{e} and d1>0d_{1}>0 be constants. Let {Wj:1jJ}\{W_{j}:1\leq j\leq J\} be a sequence of increasing real numbers where (Wj,R1,j)(W_{j},R_{1,j}) is a pair for which tt0t\geq t_{0} and

σ1loglogtWjlogtimplies|ζ(σ+it)ζ(σ+it)|R1,jlogtloglogt.\sigma\geq 1-\frac{\log\log t}{W_{j}\log t}\quad\text{implies}\quad\left|\frac{\zeta^{\prime}(\sigma+it)}{\zeta(\sigma+it)}\right|\leq R_{1,j}\frac{\log t}{\log\log t}.

Then, for

σ1loglogtW1logt\sigma\geq 1-\frac{\log\log t}{W_{1}\log t}

we have

|1ζ(σ+it)|Q2(d1,σ1,t0)logtloglogt,\left|\frac{1}{\zeta(\sigma+it)}\right|\leq Q_{2}(d_{1},\sigma_{1},t_{0})\frac{\log t}{\log\log t},

where

Q2(d1,σ1,t0)\displaystyle Q_{2}(d_{1},\sigma_{1},t_{0}) =minV{max{σ1σ11loglogt0logt0,V1(d1,σ1,t0),\displaystyle=\min_{V}\Biggr{\{}\max\Biggl{\{}\frac{\sigma_{1}}{\sigma_{1}-1}\frac{\log\log t_{0}}{\log t_{0}},\,V_{1}(d_{1},\sigma_{1},t_{0}), (24)
V(d1,σ1,t0)exp(j=1J1R1,j(1Wj1Wj+1)+R1,JWJ+d1K1)}},\displaystyle V(d_{1},\sigma_{1},t_{0})\cdot\exp\Bigg{(}\sum_{j=1}^{J-1}R_{1,j}\left(\frac{1}{W_{j}}-\frac{1}{W_{j+1}}\right)+\frac{R_{1,J}}{W_{J}}+d_{1}K_{1}\Bigg{)}\Biggr{\}}\Biggr{\}},

with any σ11+d1loglogt0/logt0\sigma_{1}\geq 1+d_{1}\log\log t_{0}/\log t_{0}, γ\gamma denoting Euler’s constant, K1K_{1} defined in Corollary 2, ZZ defined in Lemma 7, and V,V1V,\,V_{1} defined in (15) and Lemma 8, respectively.

Proof.

Let δ1=δ1(t):=d1loglogt/logt\delta_{1}=\delta_{1}(t):=d_{1}\log\log t/\log t. We will split the proof into three cases: σ1+δ1\sigma\leq 1+\delta_{1}, 1+δ1σσ11+\delta_{1}\leq\sigma\leq\sigma_{1}, and σσ1\sigma\geq\sigma_{1}. First see that in the range 1loglogt/(W1logt)σ1+δ11-\log\log t/(W_{1}\log t)\leq\sigma\leq 1+\delta_{1} we have

log|1ζ(σ+it)|\displaystyle\log\left|\frac{1}{\zeta(\sigma+it)}\right| =Relogζ(σ+it)\displaystyle=-\textup{Re}\log\zeta(\sigma+it)
=Relogζ(1+δ1+it)+σ1+δ1Re(ζζ(x+it))dx.\displaystyle=-\textup{Re}\log\zeta\left(1+\delta_{1}+it\right)+\int_{\sigma}^{1+\delta_{1}}\textup{Re}\left(\frac{\zeta^{\prime}}{\zeta}(x+it)\right)\text{d}x. (25)

Writing Δ=loglogt/logt\Delta=\log\log t/\log t, we can split the integral and rewrite it as

(1ΔW11ΔW2++1ΔWj1ΔWj+1++1ΔWJ1+11+δ1)Re(ζζ(x+it))dx.\displaystyle\left(\int_{1-\tfrac{\Delta}{W_{1}}}^{1-\tfrac{\Delta}{W_{2}}}+\cdots+\int_{1-\tfrac{\Delta}{W_{j}}}^{1-\tfrac{\Delta}{W_{j+1}}}+\cdots+\int_{1-\tfrac{\Delta}{W_{J}}}^{1}+\int_{1}^{1+\delta_{1}}\right)\textup{Re}\left(\frac{\zeta^{\prime}}{\zeta}(x+it)\right)\text{d}x.

By Corollary 2 and the assumption in the lemma we have

σ1+δ1Re(ζζ(x+it))dx\displaystyle\int_{\sigma}^{1+\delta_{1}}\textup{Re}\left(\frac{\zeta^{\prime}}{\zeta}(x+it)\right)\text{d}x j=1J1R1,j(1Wj1Wj+1)+R1,JWJ+d1K1.\displaystyle\leq\sum_{j=1}^{J-1}R_{1,j}\left(\frac{1}{W_{j}}-\frac{1}{W_{j+1}}\right)+\frac{R_{1,J}}{W_{J}}+d_{1}K_{1}.

We therefore have, from (5),

log|1ζ(σ+it)|<\displaystyle\log\left|\frac{1}{\zeta(\sigma+it)}\right|< log|ζ(1+d1loglogtlogt+it)|\displaystyle-\log\left|\zeta\left(1+\frac{d_{1}\log\log t}{\log t}+it\right)\right| (26)
+j=1J1R1,j(1Wj1Wj+1)+R1,JWJ+d1K1.\displaystyle+\sum_{j=1}^{J-1}R_{1,j}\left(\frac{1}{W_{j}}-\frac{1}{W_{j+1}}\right)+\frac{R_{1,J}}{W_{J}}+d_{1}K_{1}.

To estimate the first term on the right-hand side of (26) we take σ=1+δ1\sigma=1+\delta_{1} and apply (15) and Lemma 8, keeping in mind that we set σ11+δ1\sigma_{1}\geq 1+\delta_{1}. This gives us

|1ζ(1+δ1+it)|V(d1,σ1,t0)logtloglogt.\left|\frac{1}{\zeta(1+\delta_{1}+it)}\right|\leq V(d_{1},\sigma_{1},t_{0})\frac{\log t}{\log\log t}.

We combine this with (26) and exponentiate to obtain Q2Q_{2} for σ1+δ1\sigma\leq 1+\delta_{1}.

For 1+δ1σσ11+\delta_{1}\leq\sigma\leq\sigma_{1} we might not have the condition σ(1+γ)/γ\sigma\leq(1+\gamma)/\gamma, so we apply (6) from Lemma 8 to see that

|1ζ(1+δ1+it)|V1(d1,σ1,t0)logtloglogt.\left|\frac{1}{\zeta(1+\delta_{1}+it)}\right|\leq V_{1}(d_{1},\sigma_{1},t_{0})\frac{\log t}{\log\log t}.

Finally, for σσ1>1\sigma\geq\sigma_{1}>1 we have from (8) and the bound |ζ(s)|ζ(σ)|\zeta(s)|\leq\zeta(\sigma) that

|1ζ(σ+it)|ζ(σ)(σ1σ11loglogt0logt0)logtloglogt.\left|\frac{1}{\zeta(\sigma+it)}\right|\leq\zeta(\sigma)\leq\left(\frac{\sigma_{1}}{\sigma_{1}-1}\frac{\log\log t_{0}}{\log t_{0}}\right)\frac{\log t}{\log\log t}.

Taking the maximum Q2Q_{2} from all three cases completes the proof. ∎

Corollary 3 follows from Lemma 10 by computing Q2Q_{2} for specific choices of t0t_{0} and WjW_{j}, and optimizing over d1d_{1} and σ1\sigma_{1}. These computed values of Q2Q_{2} are labelled R2R_{2}.

Proof of Corollary 3.

After choosing t0t_{0}, we aim to minimise Q2Q_{2} by optimising over d1>0d_{1}>0 and σ1\sigma_{1}. This requires choosing JJ, which corresponds to the number of WjW_{j} values. Using more WjW_{j} will give a better result, but the improvements eventually become negligible, as Q2Q_{2} is largely determined by the initial few WjW_{j}.

For computations, we only used values for WjW_{j} from Table 3 (Corollary 1). This meant that using W=70W=70, for instance, would only be valid for t500t\geq 500. On the plus side, larger t0t_{0} meant we could also use smaller K1K_{1} from Corollary 2. With these factors in mind, we chose to compute two results for a selection of WW, one for the largest possible range, and the other for the smallest achievable constant. This meant that computing Q2Q_{2} for t0=eet_{0}=e^{e} and W23W\leq 23, for instance, only used the entries of Table 3 for t0=eet_{0}=e^{e}. Computing Q2Q_{2} for t0=500t_{0}=500, however, allowed us to use all available WW greater than the one chosen.

For W=21.24W=21.24, we can take R2=Q2(0.0031,6.52,ee)=44910R_{2}=Q_{2}(0.0031,6.52,e^{e})=44910. Other values of R2R_{2} are listed in Table 5 for the specified t0t_{0}, and we note that all results for t0<500t_{0}<500 had the same approximate optimal d1=0.0031d_{1}=0.0031 alongside the stated σ1\sigma_{1}, and the results for t0=500t_{0}=500 all had d1=0.0067d_{1}=0.0067 and σ1=12.35\sigma_{1}=12.35. ∎

      WW       t0t_{0}       R2R_{2}       σ1\sigma_{1}       t0t_{0}       R2R_{2}
      21.2421.24       eee^{e}       4491044910       6.526.52       500500       1497814978
      2222       eee^{e}       2322723227       6.526.52       500500       34383438
      2323       eee^{e}       2145321453       6.526.52       500500       24942494
      2424       1818       1334913349       7.147.14       500500       20182018
      2525       2424       95269526       6.946.94       500500       17311731
      2626       3333       73327332       8.138.13       500500       15321532
      2727       4545       59225922       8.398.39       500500       13821382
Table 5: Values of R2R_{2} for specific WW in Corollary 3, for tt0t\geq t_{0}.

If we only consider the case σ1\sigma\geq 1 it is possible to reduce R2R_{2}. This is done in Corollary 4. This result is especially useful for bounds on the line s=1+its=1+it.

Proof of Corollary 4.

The proof follows the same method as Lemma 10, but in the case σ1+δ1\sigma\leq 1+\delta_{1}, we only need the following bound in (5),

σ1+δ1Re(ζζ(x+it))dxd1K1.\int_{\sigma}^{1+\delta_{1}}\textup{Re}\left(\frac{\zeta^{\prime}}{\zeta}(x+it)\right)\text{d}x\leq d_{1}K_{1}.

This simplifies Q2(d1,σ1,t0)Q_{2}(d_{1},\sigma_{1},t_{0}), replacing the third case in (24) with

V2(d1,σ1,t0)exp(d1K1).V_{2}(d_{1},\sigma_{1},t_{0})\cdot\exp\left(d_{1}K_{1}\right).

We can now fix t0t_{0} and optimise over d1d_{1} using the values of K1K_{1} in Table 4.

For the second assertion of the corollary, Carneiro, Chirre, Helfgott, and Mejía-Cordero verify, using interval arithmetic in the proof of Proposition A.2 [1], that

|1ζ(1+it)|2.079logt\left|\frac{1}{\zeta(1+it)}\right|\leq 2.079\log t

for 2t5002\leq t\leq 500. This can then be used in conjunction with the result for t0=500t_{0}=500. All that remains is to check that

107.7logtloglogt2.079logt107.7\frac{\log t}{\log\log t}\geq 2.079\log t

holds for 3t5003\leq t\leq 500. ∎

      t0t_{0}       K2K_{2}       d1d_{1}       σ1\sigma_{1}
      eee^{e}       202.3202.3       0.00320.0032       6.646.64
      500500       107.7107.7       0.00670.0067       9.809.80
      HH       103.5103.5       0.00680.0068       9.779.77
Table 6: Values of K2K_{2} for specific t0t_{0} in Corollary 4.

6 Discussion

Using the methods of this paper, the main obstruction to obtaining better bounds on |ζ(s)/ζ(s)||\zeta^{\prime}(s)/\zeta(s)| and |1/ζ(s)||1/\zeta(s)| is the width of the zero-free region. Lemmas 1 and 2 rely on the Borel–Carathéodory theorem (Theorem 1), which, in the setting of a general function ff, already has the best possible constant (see [9, §5] for a discussion). Thus, to get improvements, one would have to input additional information on the specific function used. For example, in Lemmas 1 and 2 we use information on the zero-free region of ζ(s)\zeta(s), so one could utilise more information to do better.

Because we sought bounds of the form logt/loglogt\log t/\log\log t rather than logt\log t, we were restricted by the bound on |ζ(s)||\zeta(s)| in the region σ1ω2(loglogt)2/logt\sigma\geq 1-\omega_{2}(\log\log t)^{2}/\log t. The width of this region hampers us in a similar fashion to the width of the zero-free region, so improving ω2\omega_{2} is another possibility.

We conclude with a discussion on the trigonometric inequality (8) used in the proof of Lemma 8. When bounding |1/ζ(s)||1/\zeta(s)| for σ\sigma close to and above 1, (8) is superior to the trivial bound |ζ(s)|ζ(2σ)/ζ(σ)|\zeta(s)|\geq\zeta(2\sigma)/\zeta(\sigma) if

ζ(σ)ζ(2σ)4|ζ(σ+2it)|.\frac{\zeta(\sigma)}{\zeta(2\sigma)^{4}}\geq|\zeta(\sigma+2it)|. (27)

The validity of this inequality is not obvious, as the left-hand side of (27) decreases to (6/π2)4ζ(σ)0.14ζ(σ)(6/\pi^{2})^{4}\zeta(\sigma)\geq 0.14\zeta(\sigma) as σ1\sigma\to 1, but the right-hand side can be trivially bounded: |ζ(σ+2it)|ζ(σ)|\zeta(\sigma+2it)|\leq\zeta(\sigma). However, if this trivial bound is replaced by a bound depending on tt, then having sufficiently large tt on the right-hand side of (27) will prevent the bound from approaching infinity as σ1\sigma\rightarrow 1. Indeed, when using Lemma 7 to bound (8) over 1<σ6.771<\sigma\leq 6.77, we have

|ζ(σ+2it)|2.065(1+log2e)logtloglogt2.6logtloglogt,|\zeta(\sigma+2it)|\leq 2.065\left(1+\frac{\log 2}{e}\right)\frac{\log t}{\log\log t}\leq 2.6\frac{\log t}{\log\log t},

for t2eet\geq 2e^{e}. In contrast, for σ=1+δ1\sigma=1+\delta_{1} (with δ1\delta_{1} defined as in the proof of Lemma 10) we would have

ζ(σ)ζ(2σ)4=ζ(σ)(σ1)ζ(2σ)4d1logtloglogt(6/π2)4d1logtloglogt.\frac{\zeta(\sigma)}{\zeta(2\sigma)^{4}}=\frac{\zeta(\sigma)(\sigma-1)}{\zeta(2\sigma)^{4}d_{1}}\frac{\log t}{\log\log t}\geq\frac{(6/\pi^{2})^{4}}{d_{1}}\frac{\log t}{\log\log t}.

Thus (27) is true for d10.0525d_{1}\leq 0.0525, and our optimised d1d_{1} always fell in this range.

Notice that ZZ in Lemma 7 depends on, and also tends to, the constant in Theorem 3 (as t0t_{0} gets large). Unfortunately, an increase or decrease in this constant has a minimal impact on our main results, since our main constants are quite large. Thus, even though Lemma 8 offers a bound better than trivial, one would be better served seeking improvements from other avenues like the ones mentioned at the beginning of this section, for more substantial savings.

Another interesting point is that the conditions for trigonometric inequalities like (8) to give good bounds on |1/ζ(s)||1/\zeta(s)| for σ>1\sigma>1 differ from the conditions needed for good zero-free regions. For instance, when dealing with the latter, one requires a non-negative trigonometric polynomial

n=0Nancos(nθ)=a0+a1cos(θ)++aNcos(Nθ),\sum_{n=0}^{N}a_{n}\cos(n\theta)=a_{0}+a_{1}\cos(\theta)+\ldots+a_{N}\cos(N\theta), (28)

where a1>a0a_{1}>a_{0}, with each coefficient ana_{n} non-negative, to state a few conditions (see [8], [11], for more information). However, for |1/ζ(s)||1/\zeta(s)| the criteria is to have the sum of all coefficients in the polynomial n|an|2a1\sum_{n}|a_{n}|\leq 2a_{1}, with a1>0a_{1}>0. This condition is to ensure the overall bound is of the desired order. To illustrate this, if we had used a polynomial in the proof of Lemma 8 with n|an|=2a1+ϵ\sum_{n}|a_{n}|=2a_{1}+\epsilon, for any ϵ>0\epsilon>0, then we would have ended up with a factor of (logt/loglogt)1+(ϵ/a1)(\log t/\log\log t)^{1+(\epsilon/a_{1})} instead of logt/loglogt\log t/\log\log t.

The natural question to ask is if one could do better with a different choice of inequality than (8) when bounding |1/ζ(s)||1/\zeta(s)|. This is equivalent to asking for (28) with 0<a0/a1<3/40<a_{0}/a_{1}<3/4, while at the same time satisfying our new criteria. We briefly did a search for such polynomials of higher degree, to no avail. Although there have been many trigonometric polynomials found in the arena of refining zero-free regions, none of these fit our stated new criteria while also improving on 3/43/4. We hypothesise that no better polynomials exist. Certainly for a degree-22 polynomial of the form

(x+ycosθ)2=x2+y22+2xycosθ+y22cos2θ(x+y\cos\theta)^{2}=x^{2}+\frac{y^{2}}{2}+2xy\cos\theta+\frac{y^{2}}{2}\cos 2\theta

(which is the prequel to (8)), it is easily seen that to fulfil our new criteria, one needs (xy)20(x-y)^{2}\leq 0. Thus the only option is x=yx=y, and so (8) is the best possible in this situation.

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