Explicit bounds for the Riemann zeta-function on the 1-line

Ghaith A. Hiary, Nicol Leong, and Andrew Yang GH: Department of Mathematics, The Ohio State University, 231 West 18th Ave, Columbus, OH 43210, USA hiary.1@osu.edu NL: University of New South Wales (Canberra) at the Australian Defence Force Academy, ACT, Australia nicol.leong@unsw.edu.au AY: University of New South Wales (Canberra) at the Australian Defence Force Academy, ACT, Australia andrew.yang1@unsw.edu.au

Abstract.

Explicit estimates for the Riemann zeta-function on the $1$ -line are derived using various methods, in particular van der Corput lemmas of high order and a theorem of Borel and Carathéodory.

Key words and phrases:

Van der Corput estimate, exponential sums, Riemann zeta function.

2020 Mathematics Subject Classification:

Primary: 11M06, 11Y35, 11L07.

1. Introduction

Let $\zeta(s)$ denote the Riemann zeta-function. An open problem in analytic number theory is to determine the growth rate of $\zeta(s)$ on the line $s=1+it$ as $t\to+\infty$ . Assuming the Riemann hypothesis, Littlewood [Lit26, Lit28] showed that the value of $|\zeta(1+it)|$ is limited to the range

\frac{1}{\log\log t}\ll\zeta(1+it)\ll\log\log t.

In comparison, the sharpest estimates requiring no assumption furnish a wider range of the form

(1)

\frac{1}{(\log t)^{2/3}(\log\log t)^{1/3}}\ll\zeta(1+it)\ll\log^{2/3}t.

The upper bound in the unconditional estimate (1) was made explicit by Ford [For02]. He computed numerical constants $c$ and $t_{0}$ such that¹¹1Trudgian [Tru14] later lowered Ford’s value of $c$ from $76.2$ to $62.6$ .

|\zeta(1+it)|\leq c\log^{2/3}t,\qquad(t\geq t_{0}).

The constants $c$ and $t_{0}$ are often too large for standard applications, including explicit bounds on $S(t)$ [Tru14a, HSW22] and explicit zero-density estimates [KLN18]. One instead defaults to an asymptotically worse bound of the form $\zeta(1+it)\ll\log t$ that nevertheless is sharper in a finite range of $t$ of interest.

This motivates our Theorem 1. We establish a new explicit bound on $\zeta(1+it)$ that is asymptotically sharper than $O(\log t)$ but still good at small values of $t$ .

Theorem 1.

For $t\geq 3$ , we have

|\zeta(1+it)|\leq 1.731\frac{\log t}{\log\log t}.

Additionally, using a different method, we prove explicit upper bounds of the same form, on both $1/|\zeta(1+it)|$ and $\left|\zeta^{\prime}(1+it)/\zeta(1+it)\right|$ .

Theorem 2.

For $t\geq 3$ , we have

\frac{1}{|\zeta(1+it)|}\leq 431.7\frac{\log t}{\log\log t}.

Furthermore, for $t\geq 500$ , if

1\leq\sigma\leq 1+\frac{9}{250}\frac{\log\log t}{\log t},

then

\left|\frac{\zeta^{\prime}}{\zeta}(\sigma+it)\right|\leq 154.5\frac{\log t}{\log\log t}.

Theorem 1 makes explicit a result due to Weyl [Wey21] (see also [Tit86, Section 5.16]), and is sharper than the prior explicit bounds due to Patel [Pat22] on $|\zeta(1+it)|$ for $t$ in the range $\exp(3382)\leq t\leq\exp(3.61\cdot 10^{8})$ .²²2We also note that Theorem 1 is sharper for $t\geq 23256$ than the bound $|\zeta(1+it)|\leq\tfrac{3}{4}\log t$ given in [Tru14].

As far as we know, the bounds in Theorem 2 are the first unconditional explicit bounds of their kind. The first assertion of Theorem 2 supersedes [Car+22, Proposition A.2] when $t\geq\exp(\exp(10.07))$ , while the second assertion supersedes [Tru15, Table 2 with $(W,R_{1})=(12,40.14)$ ] for $t\geq\exp(\exp(3.85))$ .

Under the assumption of the Riemann Hypothesis, sharper explicit estimates for $\zeta^{\prime}/\zeta(1+it)$ and $1/\zeta(1+it)$ of the order $\log\log t$ are known [Sim23], [CHS24, Theorem 5], [Chi23]. Much work has also been done on bounding more general $L$ -functions [Lum18], [PS22].

Lastly, it is reasonable to try to adapt the remarkable approaches in [Sou09, p. 984] and [LLS15, Lemma 2.5] to our context. In these articles, very precise bounds, conditional on the Riemann hypothesis, are obtained. One can arrange for these approaches to yield unconditional bounds, and the outcome is similarly constrained by the width of available explicit zeros-free regions. See [Li10], for example, where the method from [Sou09] is used to establish bounds on $L$ -functions for $\sigma\geq 1$ .

2. Discussion

One can derive an explicit bound on $|\zeta(1+it)|$ of the form $\log t+\alpha_{1}$ using a version of the Euler–Maclaurin summation. For example, using the version in [Sim20, Corollary 2], applied with $t_{0}=30\pi$ and $c=1/4$ , together with the elementary harmonic sum bound³³3Here, $\gamma=0.577216\ldots$ is the Euler constant. (see Lemma 8 below),

(2)

\sum_{n=1}^{N}\frac{1}{n}\leq\log N+\gamma+\frac{1}{2N},

yields after a small numerical computation, $|\zeta(1+it)|\leq\log t-0.45$ for $t\geq 3$ .

Patel [Pat22] replaced the use of the Euler–Maclaurin summation by an explicit version of the Riemann–Siegel formula on the $1$ -line. Consequently, the size of the main term in the Euler–Maclaurin bound was cut by half. Patel thus obtained $|\zeta(1+it)|\leq\frac{1}{2}\log t+1.93$ for $t\geq 3$ , a substantial improvement at the expense of a larger constant term.

In general, we expect that an explicit van der Corput lemma of order $k$ , where $k\geq 2$ , will produce a bound of the form

(3)

|\zeta(1+it)|\leq\frac{1}{k}\log t+\alpha_{k},\qquad(t\geq 3).

One proceeds by approximating $\zeta(1+it)$ by an Euler–Maclaurin sum $\sum_{n\ll t}n^{-1-it}$ , or better yet by a Riemann–Siegel sum $\sum_{n\ll\sqrt{t}}n^{-1-it}$ . In either case, roughly speaking, the term $(1/k)\log t$ arises from bounding the subsum with $n\leq t^{1/k}$ via the triangle inequality and the harmonic sum estimate (2), while the constant $\alpha_{k}$ predominantly arises from bounding the subsum with $n>t^{1/k}$ via van der Corput lemmas of order $\leq k$ .

However, the $\alpha_{k}$ obtained using this approach appear to grow fast with $k$ . For example, in [Pat21], $\alpha_{5}$ is about $23$ -times larger than $\alpha_{2}$ , which already suggests exponential growth. Despite this, there is still gain in letting $k$ grow with $t$ , provided the growth is slow enough. We let $k$ grow like $\log\log t$ , which is permitted by a recent optimized van der Corput test, uniform in $k$ , derived in [Yan24]. Subsequently, we obtain a bound on $|\zeta(1+it)|$ of the form $c_{0}\log t/\log\log t$ with an explicit number $c_{0}$ .

While we could let $k$ increase faster than $\log\log t$ , leading to a smaller contribution of the $(1/k)\log t$ term, the $\alpha_{k}$ thus obtained would likely take over as the main term, asymptotically surpassing $\log t/\log\log t$ in size. This limits the order of the van der Corput lemmas we employ.

In proving Theorem 1, we encounter an unexpected “gap interval”, not covered by any of the other explicit estimates, which requires a special treatment. Namely, the interval $I=[\exp(16),\exp(662)]$ . To prove the inequality in Theorem 1 for all $t\in I$ , we use the following proposition.

Proposition 3.

We have

|\zeta(1+it)|\leq\begin{cases}\displaystyle\frac{1}{3}\log t+4.66,&t\geq\exp(16),\\ &\\ \displaystyle\frac{8}{33}\log t+12.45,&t\geq\exp(88).\end{cases}

To clarify the need for Proposition 3, consider that if one inputs the bounds in [Pat22], as below, one finds

\max_{t\in I}\frac{\displaystyle\min\left(\log t,\frac{1}{2}\log t+1.93,\frac{1}{5}\log t+44.02\right)}{\log t/\log\log t}\geq 2.539,

and the maximum of this ratio occurs around $t_{1}=\exp(140.3)$ . On the other hand, the savings we otherwise establish here may materialize only when $t$ is well past $t_{1}$ . Therefore, without the improvement enabled by Proposition 3, we cannot do any better than the bound $|\zeta(1+it)|\leq 2.539\log t/\log\log t$ over $t\in I$ . Proposition 3, which is based on a hybrid of van der Corput lemmas of orders $k=3$ , $4$ and $5$ , allows us to circumvent this barrier.

In retrospect, a gap interval like $I$ should be expected. This is because the estimates in [Pat21] rely, essentially, on van der Corput lemmas of orders $k=2$ and $k=5$ .⁴⁴4The bound obtained via the Riemann–Siegel formula in [Pat21] should correspond to a $k=2$ van der Corput lemma. In particular, bounds that would follow from using van der Corput lemmas of orders $k=3$ and $k=4$ are not utilized, which presumably causes the gap.

3. Proof of Theorem 1

Let $t_{0}$ be a positive number to be chosen later, subject to $t_{0}\geq\exp(990/7)$ . We suppose $t\geq t_{0}$ throughout this section. (We will use another method for $t<t_{0}$ .)

Theorem 9 supplies a Riemann–Siegel approximation of $\zeta(1+it)$ in terms of two sums, plus a remainder term $\mathcal{R}$ . We use the triangle inequality to bound the entire second sum in this theorem. This gives

(4)

\begin{split}|\zeta(1+it)|&\leq\left|\sum_{n=1}^{n_{1}}\frac{1}{n^{1+it}}\right|+\frac{g(t)}{\sqrt{2\pi}}+\mathcal{R}(t),\qquad n_{1}=\lfloor\sqrt{t/(2\pi)}\rfloor.\end{split}

Since $g(t)$ and $\mathcal{R}(t)$ are well-understood, the bulk of the work is in bounding the main sum over $n\leq n_{1}$ .

We introduce two sequences, $\{X_{m}\}_{0\leq m\leq M}$ and $\{Y_{k}\}_{3\leq k\leq r+1}$ . The former sequence $X_{m}$ will be used to partition the main sum in (4) in a dyadic manner. The latter sequence $Y_{k}$ will be used to easily bound the partition points $X_{m}$ from above and below solely in terms of $t$ (or, what is essentially the same, in terms of $n_{1}$ ).

To this end, let $h$ and $h_{2}$ be real positive numbers, also to be chosen later, subject to $1<h\leq 2$ and $h_{2}<1/\log 2$ . We use $h_{2}$ immediately in defining

\begin{split}&R:=2^{r-1},\qquad r:=\lfloor h_{2}\log\log t\rfloor.\end{split}

Furthermore we will assume that $h_{2}$ and $t_{0}$ are chosen so that

(5)

r_{0}:=\lfloor h_{2}\log\log t_{0}\rfloor\geq 6.

Since $t\geq t_{0}$ by supposition, $r\geq r_{0}$ . The integer $r$ will correspond to the highest order van der Corput lemma we use in bounding the tail of the main sum in (4).

Furthermore, let

\begin{split}&K:=2^{k-1},\qquad\theta_{k}:=\frac{2K}{(k-2)K+4},\qquad(k\geq 3).\end{split}

In the definition of $\theta_{k}$ , $K$ is determined by $k$ . Also, $\theta_{k}$ decreases monotonically to zero with $k\geq 3$ , and is bounded from above by $\theta_{3}=1$ .

We now define the previously mentioned sequences $\{X_{m}\}$ and $\{Y_{k}\}$ . Set

\begin{split}&X_{0}:=\frac{1}{\sqrt{2\pi}}t^{1/2},\qquad X_{m}:=h^{-m}X_{0},\qquad(m\geq 1).\end{split}

So, $\lfloor X_{0}\rfloor=n_{1}$ and since $h>1$ , $X_{m}$ decreases monotonically with $m$ . Also set

(6)

Y_{k}:=X_{0}^{\theta_{k}},\qquad(k\geq 3),

so that $Y_{k}$ decreases monotonically with $k\geq 3$ , starting at $Y_{3}=X_{0}$ . Note that if $k$ is large, then $\theta_{k}\approx 2/k$ , and so $Y_{k}\approx t^{1/k}$ for large $k$ . Lastly, define

M:=\left\lceil\frac{(1-\theta_{r+1})\log X_{0}}{\log h}\right\rceil.

The integer $M$ corresponds to the number of points we will use to partition the main sum. From the definition, we have

(7)

\frac{(1-\theta_{r+1})\log X_{0}}{\log h}\leq M<\frac{(1-\theta_{r+1})\log X_{0}}{\log h}+1.

Thus, our choice of $M$ ensures that

(8)

X_{M}\leq X_{0}^{\theta_{r+1}}=Y_{r+1}<X_{M-1}\leq hX_{0}^{\theta_{r+1}}.

Additionally, we have

(9)

X_{M}\leq Y_{r+1}<Y_{r}<\cdots<Y_{4}<Y_{3}=X_{0}.

The assumption (5) is required for the subsequent analysis to be valid. This condition implies, among other things, that $\theta_{r+1}<1$ and, hence, $M\geq 1$ . This in turn ensures that some quantities appearing later (e.g. $M-1$ ) are nonnegative. All contingent conditions we impose (including on $r_{0}$ , $h$ , $h_{2}$ and $t_{0}$ ) will be verified at the end, after choosing the values of our free parameters.

By the chains of inequalities (8) and (9), for all $m\in[1,M-1]$ , $Y_{r+1}<X_{m}<Y_{3}$ . Therefore, since the sequence $Y_{r+1},\ldots,Y_{3}$ partitions the interval $[Y_{r+1},Y_{3}]$ , for each integer $m\in[1,M-1]$ , there is a unique integer $k\in[3,r]$ such that

(10)

X_{m}\in(Y_{k+1},Y_{k}].

Note, though, that multiple $m$ ’s could correspond to the same $k$ .

With this in mind, let us denote

(11)

S_{m}:=\sum_{X_{m}+1<n\leq X_{m-1}}\frac{1}{n^{1+it}}=\sum_{\lfloor X_{m}\rfloor+1<n\leq\lfloor X_{m-1}\rfloor}\frac{1}{n^{1+it}},

and observe that

(12)

\left|\sum_{X_{M-1}<n\leq X_{0}}\frac{1}{n^{1+it}}\right|\leq\sum_{m=1}^{M-1}\left|\frac{1}{(\lfloor X_{m}\rfloor+1)^{1+it}}+S_{m}\right|\leq\sum_{m=1}^{M-1}\left(\frac{1}{X_{m}}+|S_{m}|\right).

We calculate

(13)

\frac{\lfloor X_{m-1}\rfloor}{\lfloor X_{m}\rfloor+1}<\frac{X_{m-1}}{X_{m}}=h.

Here, we used the elementary inequality $x-1<\lfloor x\rfloor\leq x$ , valid for any real number $x$ together with the definition $X_{m}=h^{-m}X_{0}$ .

We bound $S_{m}$ using Lemma 13 with $a=\lfloor X_{m}\rfloor+1$ and $b=\lfloor X_{m-1}\rfloor\leq ha$ . For each integer $m\in[1,M-1]$ such that $X_{m}\in(Y_{k+1},Y_{k}]$ , and with $K=2^{k-1}$ , we have, for any choice of the free parameter $\eta>0$ ,

(14)

\begin{split}\left|S_{m}\right|&\leq C_{k}(\eta,h)(\lfloor X_{m}\rfloor+1)^{-k/(2K-2)}t^{1/(2K-2)}\\ &\qquad+D_{k}(\eta,h)(\lfloor X_{m}\rfloor+1)^{k/(2K-2)-2/K}t^{-1/(2K-2)}.\end{split}

To bound the RHS, observe that

(15)

Y_{k+1}<\lfloor X_{m}\rfloor+1\leq X_{m}+1=X_{m}\left(1+\frac{1}{X_{m}}\right)\leq Y_{k}\left(1+\frac{1}{X_{m}}\right).

In addition, $X_{m}\geq X_{M-1}>X_{0}^{\theta_{r+1}}$ . Moreover, it follows by definition that

(16)

\theta_{r+1}=\frac{2}{r-1+2^{2-r}}.

Since $\theta_{r+1}$ decreases continuously with $r$ , and $r\leq h_{2}\log\log t$ , we have $X_{0}^{\theta_{r+1}}=\exp(\theta_{r+1}\log X_{0})\geq\kappa(h_{2},t)$ where

(17)

\kappa(h_{2},t):=\exp\left(\frac{\log(t/(2\pi))}{h_{2}\log\log t-1+2^{2-h_{2}\log\log t}}\right).

It is straightforward to verify that $\kappa(h_{2},t)$ is increasing in $t$ for $t\geq t_{0}$ , so $\kappa(h_{2},t)\geq\kappa(h_{2},t_{0})$ . Inserting (15) and (17) into (14),

(18)

\begin{split}\left|S_{m}\right|&\leq C_{k}(\eta,h)Y_{k+1}^{-k/(2K-2)}t^{1/(2K-2)}\\ &\qquad+\widehat{D}_{k}(\eta,h,h_{2},t_{0})Y_{k}^{k/(2K-2)-2/K}t^{-1/(2K-2)}\end{split}

for $t\geq t_{0}$ , where

(19)

\widehat{D}_{k}(\eta,h,h_{2},t_{0}):=D_{k}(\eta,h)\left(1+\frac{1}{\kappa(h_{2},t_{0})}\right)^{k/(2K-2)-2/K}.

This is permissible since the exponents of $\lfloor X_{m}\rfloor+1$ in the first and second terms of (14) respectively satisfy

(20)

-\frac{k}{2K-2}<0,\qquad\text{and}\qquad\frac{k}{2K-2}-\frac{2}{K}\geq 0,\qquad(k\geq 3).

It is worth clarifying that the right-hand side of (18) depends on $m$ via the requirement (10).

Next, we express the bound on $S_{m}$ in (18) solely in terms of $t$ . Starting with the first term, we have

(21)

\displaystyle Y_{k+1}^{-k/(2K-2)}t^{1/(2K-2)}=(2\pi)^{\theta_{k+1}k/(4K-4)}t^{-\rho(k)},

where after a quick calculation,

(22)

\rho(k):=\frac{K-2}{K-1}\cdot\frac{1}{2(k-1)K+4}.

We estimate the second term in (18) similarly, yielding

(23)

\begin{split}Y_{k}^{k/(2K-2)-2/K}t^{-1/(2K-2)}&<(2\pi)^{-\theta_{k}(k/(4K-4)-1/K)}t^{-\frac{K}{K-1}\cdot\frac{1}{2(k-2)K+4}}\\ &<(2\pi)^{-\theta_{k}(k/(4K-4)-1/K)}t^{-\rho(k)}.\end{split}

The exponent of $2\pi$ in the estimate (21) may be bounded for $k>5$ via $\theta_{k+1}\leq 32/81$ . Additionally, recalling the second inequality of (20), we may bound the contribution of the $2\pi$ -factor in the estimate (23) simply by $1$ . This motivates us to define

(24)

A(k):=\begin{cases}(2\pi)^{\theta_{k+1}\frac{k}{4(K-1)}}C_{k}+(2\pi)^{-\theta_{k}(\frac{k}{4(K-1)}-\frac{1}{K})}\widehat{D}_{k},&3\leq k\leq 5,\\ (2\pi)^{\frac{8k}{81(K-1)}}C_{k}+\widehat{D}_{k},&6\leq k.\end{cases}

Note that $A(k)$ additionally depends on $\eta$ , $h$ , $h_{2}$ and $t_{0}$ . Assembling (18), (21), (23) and (24), we obtain that if $1\leq m\leq M-1$ and $k$ is determined via the requirement (10), then

(25)

|S_{m}|<A(k)\,t^{-\rho(k)}.

We now divide the argument into two cases: $3\leq k\leq 5$ and $k>5$ . In the former case, let $N_{k}$ denote the number of sums $S_{m}$ corresponding to each integer $k\geq 3$ . In other words, $N_{k}$ is the number of $m$ such that $X_{m}\in(Y_{k+1},Y_{k}]$ . We observe that $X_{m}\in(Y_{k+1},Y_{k}]$ if and only if $M_{k}\leq m<M_{k+1}$ , where

(26)

M_{k}:=\frac{(1-\theta_{k})\log X_{0}}{\log h}.

Therefore, by the bound on $|S_{m}|$ in (25),

(27)

\sum_{1\leq m<M_{6}}|S_{m}|<\sum_{k=3}^{5}N_{k}\,A(k)\,t^{-\rho(k)}.

By counting the maximum number of integers in a continuous interval, for $k\geq 3$ ,

(28)

N_{k}\leq\frac{(\theta_{k}-\theta_{k+1})\log X_{0}}{\log h}+1\leq\phi_{k}\log t,

where

(29)

\phi_{k}=\phi_{k}(h,t_{0}):=\frac{\theta_{k}-\theta_{k+1}}{2\log h}+\frac{1}{\log t_{0}}.

Thus, in view of (27) we obtain

(30)

\sum_{1\leq m<M_{6}}|S_{m}|\leq b_{0},

where by (28) and on explicitly computing the values of $\rho(k)$ for $3\leq k\leq 5$ ,

(31)

\begin{split}b_{0}=b_{0}(\eta,h,h_{2},t_{0})&:=A(3)\phi_{3}\frac{\log t_{0}}{t_{0}^{1/30}}+A(4)\phi_{4}\frac{\log t_{0}}{t_{0}^{3/182}}+A(5)\phi_{5}\frac{\log t_{0}}{t_{0}^{7/990}}.\end{split}

Here, we used the supposition $t\geq t_{0}\geq\exp(990/7)$ which ensures that $\log t/t^{\rho(k)}$ is decreasing for $t\geq t_{0}$ , with $3\leq k\leq 5$ .

Next, for $k>5$ , we use a tail argument that removes the dependence of $A$ on $k$ in the bound (25). Define

(32)

A_{0}=A_{0}(\eta,h,h_{2},t_{0}):=\sup_{k\geq 6}A(k).

We will bound $A_{0}$ explicitly later (Section 4), after determining $\eta$ , $h$ , $h_{2}$ and $t_{0}$ . Importantly, under our assumptions, it will transpire that $A_{0}$ is finite.

In the meantime, we observe that $\rho(k)$ is continuously decreasing in $k\geq 3$ (towards zero), as can be seen from computing the derivative of $\rho(k)$ with respect to $k$ . So, by simply counting the number of terms, and noting that as $m$ ranges over $[M_{6},M-1]$ , $k\leq r$ , we obtain from (25),

(33)

\begin{split}\sum_{M_{6}\leq m\leq M-1}|S_{m}|\leq|M-M_{6}|A_{0}\,t^{-\rho(r)}\leq A_{1}\,\frac{|b_{1}-\theta_{r+1}|}{t^{\rho(r)}}\log t,\end{split}

where, from (7) and (26),

A_{1}=A_{1}(\eta,h,h_{2},t_{0}):=\frac{1}{2}\cdot\frac{A_{0}}{\log h},\qquad b_{1}=b_{1}(h,t_{0}):=\theta_{6}+\frac{2\log h}{\log t_{0}}.

We want to further simplify the bound on the right-side of (33). Via (16) and a routine calculation,

(34)

\frac{2}{r-1+2^{2-r_{0}}}\leq\theta_{r+1}<b_{1},\qquad(r\geq r_{0}\geq 6).

Hence, utilizing the inequality $r\leq h_{2}\log\log t$ as well as $\log(u)\leq u-1$ , valid for any real number $u>0$ , yields that if $r\geq 6$ , then

(35)

\begin{split}\frac{b_{1}-\theta_{r+1}}{t^{\rho(r)}}\leq\exp(b_{1}-\theta_{r+1}-1-\rho(r)\log t)\leq\frac{1}{A_{2}}\frac{1}{\log t},\end{split}

where (identifying $x$ with $\log t$ below)

(36)

A_{2}=A_{2}(h,h_{2},t_{0}):=\inf_{x\geq\log t_{0}}\frac{1}{x}\exp\left(1-b_{1}+\frac{2}{h_{2}\log x-1+2^{2-r_{0}}}+x\,\rho(h_{2}\log x)\right).

In indicating the dependencies of $A_{2}$ , we used that $r_{0}$ is determined by $h_{2}$ and $t_{0}$ , and $b_{1}$ further depends on $h$ . Now, from the definition of $\rho$ , and using

2^{h_{2}\log x}=x^{h_{2}\log 2},

we obtain⁵⁵5By the lower bound on $h_{2}$ in (5), we have $x^{h_{2}\log 2}\geq 64$ for $x\geq\log t_{0}$ . So, $\rho(h_{2}\log x)$ is well-defined for $x\geq\log t_{0}$ .

(37)

\rho(h_{2}\log x)=\frac{x^{h_{2}\log 2}-4}{x^{h_{2}\log 2}-2}\cdot\frac{1}{(h_{2}\log x-1)x^{h_{2}\log 2}+4}.

Since⁶⁶6This step is where a slow enough growth on $r$ is utilized, to ensure that $(\log t)^{h_{2}\log 2}=o(\log t)$ as $t\to\infty$ , so that the constant $A_{2}(h_{2},t_{0})$ defined in (36) is positive. by supposition $0<h_{2}<1/\log 2$ , the quantity defining $A_{2}$ tends to $+\infty$ as $x\to+\infty$ . As this quantity is continuous and positive for $x\geq\log t_{0}$ , $A_{2}$ is finite and positive for all admissible $h_{2}$ . From (33) and (35) we conclude that

(38)

\sum_{M_{6}\leq m\leq M-1}|S_{m}|\leq\frac{A_{1}}{A_{2}}.

Lastly, to bound the remaining terms in (12), in view of (7), we have the rough estimate

\sum_{m=1}^{M-1}\frac{1}{X_{m}}=\frac{1}{X_{0}}\sum_{m=1}^{M-1}h^{m}=\frac{1}{X_{0}}\frac{h^{M}-h}{h-1}<\frac{X_{0}^{1-\theta_{r+1}}-1}{X_{0}}\frac{h}{h-1}<\frac{h}{h-1}\frac{1}{X_{0}^{\theta_{r+1}}}.

Since $r\leq h_{2}\log\log t$ and $\theta_{r+1}>2/r$ , we have, for $t\geq t_{0}$ ,

(39)

\sum_{m=1}^{M-1}\frac{1}{X_{m}}<\frac{h}{h-1}\exp\left(-\frac{\log(t_{0}/(2\pi))}{h_{2}\log\log t_{0}}\right)=:E(h,h_{2},t_{0}),

say.

In summary, combining (12), (30), (38) and (39), the tail sum satisfies the inequality

(40)

\begin{split}\left|\sum_{X_{M-1}<n\leq X_{0}}\frac{1}{n^{1+it}}\right|&\leq\frac{A_{1}(\eta,h,h_{2},t_{0})}{A_{2}(h,h_{2},t_{0})}+b_{0}(\eta,h,h_{2},t_{0})+E(h,h_{2},t_{0}).\end{split}

So, for such $r$ , the tail sum is bounded by a constant independent of $t$ .

It remains to bound the initial sum over $1\leq n\leq X_{M-1}$ . We use the triangle inequality followed by the harmonic sum bound (2), which yields

(41)

\begin{split}\left|\sum_{1\leq n\leq X_{M-1}}\frac{1}{n^{1+it}}\right|&\leq\log X_{M-1}+\gamma+\frac{1}{2(X_{M-1}-1)}.\end{split}

We use the inequality (8) to bound $X_{M-1}$ from above and below, so that the right-side of (41) satisfies

(42)

\begin{split}\leq\frac{\theta_{r+1}}{2}\log\frac{t}{2\pi}+\log h+\gamma+\frac{1}{2(X_{0}^{\theta_{r+1}}-1)}.\end{split}

Since $\theta_{r+1}<2/(r-1)$ by (16), the first term in (42) is bounded by

(43)

\frac{\log(t/(2\pi))}{r-1}<\frac{\log t-\log 2\pi}{h_{2}\log\log t-2}\leq\frac{1-\frac{\log 2\pi}{\log t_{0}}}{h_{2}-\frac{2}{\log\log t_{0}}}\frac{\log t}{\log\log t},

where the last inequality is due to the fact that $(1-\frac{\log 2\pi}{x})/(h_{2}-\frac{2}{\log x})$ is decreasing for $x\geq 990/7$ if $h_{2}<1/\log 2$ .

In addition, we use (17) to bound the last term. Putting this together with (41), (42) and (43), we obtain

(44)

\left|\sum_{1\leq n\leq X_{M-1}}\frac{1}{n^{1+it}}\right|<A_{3}\frac{\log t}{\log\log t},

where

\begin{split}A_{3}=A_{3}(h,h_{2},t_{0})&:=\frac{1-\frac{\log 2\pi}{\log t_{0}}}{h_{2}-\frac{2}{\log\log t_{0}}}+\left(\log h+\gamma+\frac{1}{2(\kappa(h_{2},t_{0})-1)}\right)\frac{\log\log t_{0}}{\log t_{0}}.\end{split}

Lastly, the bound on $\zeta(1+it)$ in (4) carries an extra term $g(t)/\sqrt{2\pi}+\mathcal{R}(t)$ . Using the definitions in Theorem 9, this term is easily bounded by $g(t_{0})/\sqrt{2\pi}+\mathcal{R}(t_{0})$ . Therefore, combining this with (40) and (44), we see that

|\zeta(1+it)|\leq\left|\sum_{1\leq n\leq n_{1}}\frac{1}{n^{1+it}}\right|+\frac{g(t_{0})}{\sqrt{2\pi}}+\mathcal{R}(t_{0})\leq A_{4}\frac{\log t}{\log\log t},

for $t\geq t_{0}$ , where

A_{4}=A_{4}(\eta,h,h_{2},t_{0}):=A_{3}+\left(\frac{A_{1}}{A_{2}}+b_{0}+E+\frac{g(t_{0})}{\sqrt{2\pi}}+\mathcal{R}(t_{0})\right)\frac{\log\log t_{0}}{\log t_{0}}.

We choose the following values for our free parameters, which are suggested by numerical experimentation.

(45)

\eta=1.2364,\qquad h=1.0224,\qquad h_{2}=0.85532,\qquad t_{0}=\exp(3381).

With these choices, our preconditions on $t_{0}$ and $h_{2}$ hold, and the conditions $h>1$ and $r_{0}\geq 6$ also hold (in fact, $r_{0}=6$ ). Additionally, in Section 4, we show that

	$\displaystyle A_{0}$	$\displaystyle=0.0976483443\ldots,$
	$\displaystyle A_{2}$	$\displaystyle=0.0618341521\ldots.$

Therefore, we can explicitly compute

	$\displaystyle b_{0}$	$\displaystyle\leq 4.629\cdot 10^{-8},$
	$\displaystyle A_{1}$	$\displaystyle\leq 2.2039724975\ldots,$
	$\displaystyle A_{3}$	$\displaystyle\leq 1.6420608139\ldots,$
	$\displaystyle A_{4}$	$\displaystyle\leq 1.7301296329\ldots.$

In summary, the assertion of Theorem 1 holds when $t\geq t_{0}=\exp(3381)$ .

For smaller $t\leq t_{0}$ we use the bound in Patel [Pat22] and Proposition 3. Specifically, the bound in [Pat22, Theorem 1.1] yields

\begin{split}&|\zeta(1+it)|\leq\frac{1}{5}\log t+44.02\leq 1.731\frac{\log t}{\log\log t},\qquad\exp(662)\leq t\leq\exp(3381),\\ &|\zeta(1+it)|\leq\frac{1}{2}\log t+1.93\leq 1.731\frac{\log t}{\log\log t},\qquad 3\leq t\leq\exp(16).\end{split}

This leaves the gap interval $\exp(16)\leq t\leq\exp(662)$ for which we invoke Proposition 3. Since the gap interval falls in the range of $t$ where Proposition 3 is applicable, our target bound is verified there as well.

4. Bounding $A_{0}$ and $A_{2}$

We derive an upper bound on $A_{0}(\eta,h,h_{2},t_{0})$ , where the parameter values are given in (45). For the convenience of the reader, we recall the definition $A_{0}:=\sup_{k\geq 6}A(k)$ , where for $k\geq 6$ ,

A(k):=(2\pi)^{k/(81K-81)}C_{k}(\eta,h)+\widehat{D}_{k}(\eta,h,h_{2},t_{0}).

With our choice of free parameter values in (45), we numerically compute the following list of $A(k)$ .

	$\displaystyle A(6)$	$\displaystyle=0.0976483443\ldots,\qquad A(7)=0.0914388403\ldots,$
	$\displaystyle A(8)$	$\displaystyle=0.0882612019\ldots,\qquad A(9)=0.0865130096\ldots,$
	$\displaystyle A(10)$	$\displaystyle=0.0855306439\ldots.$

The values of $C_{k}$ and $D_{k}$ required for the above computation are obtained using the formulas in Lemma 13. In turn, evaluating $C_{k}(\eta,h)$ and $D_{k}(\eta,h)$ relies on evaluating $A_{k}(\eta,h^{k})$ and $B_{k}(\eta)$ . When $k=3$ , $A_{k}$ and $B_{k}$ are computed directly using their definitions in Lemma 11. When $k\geq 4$ , the recursive formulas in Lemma 12 are used to compute them. Let us examine these recursive formulas.

First, the following quantity, which appears in the formula for $A_{k+1}(\eta,\omega)$ in (92), satisfies

(46)

\max_{k\geq 10}\frac{2^{19/12}(K-1)}{\sqrt{(2K-1)(4K-3)}}=\frac{2^{19/12}\cdot 511}{\sqrt{2092035}}=:\mu_{10}.

Here, $K=2^{k-1}$ and the maximum occurs when $k=10$ . Another quanitity appearing in the formula for $A_{k+1}(\eta,\omega)$ is $\delta_{k}$ , which is clearly seen to decrease with $k$ . Noting that $h>1$ , hence $\omega=h^{k+1}>1$ for any $k\geq 0$ , we deduce that

A_{k+1}(\eta,\omega)\leq\delta_{10}\left(1+\mu_{10}A_{k}(\eta,\omega)^{1/2}\right).

So, we are led to consider the discrete map

x_{n+1}=\delta_{10}\left(1+\mu_{10}x_{n}^{1/2}\right).

By a routine calculation, this map has a single fixed point at

(47)

x^{*}=\left(\frac{\mu_{10}\delta_{10}}{2}+\sqrt{\frac{\mu_{10}^{2}\delta_{10}^{2}}{4}+\delta_{10}}\right)^{2}=2.7600429449\ldots,

which is a stable point. Since $A_{k+1}(\eta,\omega)\leq x_{k+1}$ for $k\geq 10$ and, by a direct numerical computation, $A_{10}(\eta,\omega)\leq x^{*}$ , it follows that

(48)

A_{k}(\eta,h^{k})\leq x^{*},\qquad(k\geq 10).

The analysis of $B_{k+1}$ for $k\geq 10$ proceeds similarly. The recursive formula for $B_{k+1}$ contains the following quantity satisfying

(49)

\max_{k\geq 10}\frac{2^{3/2}(K-1)}{\sqrt{(2K-3)(4K-5)}}\leq\frac{2^{3/2}\cdot 511}{3\sqrt{231767}}=:\nu_{10}.

So, $B_{k+1}\leq\delta_{10}\nu_{10}B_{k}^{1/2}$ for $k\geq 10$ . Consider the discrete map $y_{n+1}=\delta_{10}\nu_{10}y_{n}^{1/2}$ . This map has a single fixed point, namely,

y^{*}=(\delta_{10}\nu_{10})^{2}=1.0023463404\ldots,

which is also stable. Since $B_{k+1}\leq y_{k+1}$ for $k\geq 10$ and, by a direct numerical computation, $B_{10}\leq y^{*}$ , it follows that

(50)

B_{k}(\eta)\leq y^{*},\qquad(k\geq 10).

Furthermore, we numerically verify that for $k\geq 10$ ,

h^{2k/K-k/(2K-2)}(h-1)\left(\frac{(k-1)!}{2\pi}\right)^{\frac{1}{2K-2}}<0.023,

h^{k/(2K-2)}(h-1)^{1-2/K}\left(\frac{2\pi}{(k-1)!}\right)^{\frac{1}{2K-2}}<0.023.

Combining this with (48) and (50), and recalling the definitions in Lemma 13 and (19), yields the inequalities $C_{k}(\eta,h)<0.023x^{*}$ and $\widehat{D}_{k}(\eta,h)<0.024y^{*}$ for $k\geq 10$ . Consequently, $A(k)<0.09$ for $k\geq 10$ .

Furthermore, in view of the values of $A(k)$ listed at the beginning, we also see that $A(k)\leq A(6)=0.09764\ldots$ for $6\leq k\leq 10$ . We therefore conclude $A_{0}=A(6)$ .

For the computation related to $A_{2}$ , we shall use a derivative calculation to show that the function

(51)

a_{2}(u):=1-b_{1}+e^{u}\,\rho(h_{2}u)+\frac{2}{h_{2}u-1+2^{2-r_{0}}}-u

is increasing in $u\geq u_{0}:=\log\log t_{0}$ . So, on identifying $u$ with $\log x$ in the definition of $A_{2}$ in (36), the value of $A_{2}$ is $\exp(a_{2}(u_{0}))$ .

To this end, we first calculate

\frac{d}{du}e^{u}\,\rho(h_{2}u)=e^{u}\rho(h_{2}u)\left(1+\frac{d}{du}\log\rho(h_{2}u)\right),\qquad\frac{d}{du}\log\rho(h_{2}u)=-h_{2}\tilde{a}_{2}(h_{2}u),

where

\tilde{a}_{2}(v)=\frac{(v-1)\log 2+1-2^{-v}(8(v-1)\log 2+6)+4^{-v}(8(v-2)\log 2+8)}{(v-1+2^{2-v})(1-2^{1-v})(1-2^{2-v})}.

Note that $v$ is identified with $h_{2}u$ , so we need only consider $v\geq h_{2}u_{0}=:v_{0}$ . It is easy to see that

\sup_{v\geq v_{0}}\tilde{a}_{2}(v)<\sup_{v\geq v_{0}}\frac{(v-1)\log 2+1}{(v-1+2^{2-v_{0}})(1-2^{1-v_{0}})(1-2^{2-v_{0}})}\leq\log 2+0.2,

where the supremum occurs at $v=v_{0}$ . Hence, the derivative of $e^{u}\,\rho(h_{2}u)$ is at least $e^{u}\rho(h_{2}u)(1-(\log 2+0.2)h_{2})$ for $u\geq u_{0}$ . Importantly, $1-(\log 2+0.2)h_{2}$ is positive. In comparison, the derivative of the remaining terms in the formula for $a_{2}(u)$ is

\frac{d}{du}\left(\frac{2}{h_{2}u-1+2^{2-r_{0}}}-u\right)=-\frac{2h_{2}}{(h_{2}u-1+2^{2-r_{0}})^{2}}-1\geq-1.04731\ldots,

since $u\geq u_{0}$ and $r_{0}=6$ . Since

(52)

e^{u}\rho(h_{2}u)(1-(\log 2+0.2)h_{2})\geq 1.06112\ldots

for $u\geq u_{0}$ , the derivative of $a_{2}(u)$ is positive for $u\geq u_{0}$ , proving our claim.

5. Proof of Proposition 3

The calculation is similar to that in [Hia16] on the $1/2$ -line, so we shall be brief in some details. We divide the main sum in (4) into pieces of length $\approx t^{1/3}$ , then apply the van der Corput Lemma 10 to each piece other than an initial segment, which is bounded by the triangle inequality.

Let $t_{1}$ be a positive number to be chosen later, subject to $t_{1}\geq\exp(12)$ . Suppose that $t\geq t_{1}$ , unless otherwise stated, and let $n_{1}$ be as defined in (4). Let $P=\lceil t^{1/3}\rceil$ , $Q=\lfloor n_{1}/P\rfloor$ , and let $q_{0}$ a positive integer to be chosen later, subject to $q_{0}\leq Q$ . Thus, we have

(53)

\begin{split}\left|\sum_{n=1}^{n_{1}}\frac{1}{n^{1+it}}\right|&\leq\sum_{n=1}^{q_{0}P-1}\frac{1}{n}+\sum_{q=q_{0}}^{Q-1}\left|\sum_{n=qP}^{(q+1)P-1}\frac{1}{n^{1+it}}\right|+\left|\sum_{QP}^{n_{1}}\frac{1}{n^{1+it}}\right|.\end{split}

The harmonic sum bound (2) yields, subject to $q_{0}P>1$ ,

(54)

\sum_{n=1}^{q_{0}P-1}\frac{1}{n}\leq\log(q_{0}P-1)+\gamma+\frac{1}{2(q_{0}P-1)}.

For $q_{0}\leq q\leq Q$ , let $N=qP$ and let $N^{\prime}$ be an integer such that $N\leq N^{\prime}<N+P$ . Partial summation gives

(55)

\displaystyle\left|\sum_{n=N}^{N^{\prime}}\frac{1}{n^{1+it}}\right|

\displaystyle\leq\frac{1}{N}\max_{1\leq L\leq P}|S_{N}(L)|,

where

S_{N}(L):=\sum_{n=0}^{L-1}\exp(2\pi if(n))\qquad\text{and}\qquad f(x)=\frac{t}{2\pi}\log(N+x).

We estimate $S_{N}(L)$ using Lemma 10 with $r=q$ , $K=P$ and

K_{0}=t_{1}^{1/3}\leq P.

This gives, for any positive $\eta$ ,

\displaystyle\frac{1}{N}\max_{1\leq L\leq P}|S_{N}(L)|

\displaystyle\leq\frac{1}{q}\sqrt{\frac{\alpha}{W^{1/3}}+\frac{\beta W^{1/3}}{P}+\frac{\alpha\eta}{P}+\frac{\eta\beta W^{2/3}}{P^{2}}}.

where $W$ , $\lambda$ , $\alpha=\alpha(W,\lambda)$ , $\beta=\beta(W)$ are defined in Lemma 10. We estimate the (nonnegative) terms under the square-root as follows.

	$\displaystyle\frac{\alpha}{W^{1/3}}<\frac{\alpha}{\pi^{1/3}q},$	$\displaystyle\frac{\beta W^{1/3}}{P}$	$\displaystyle\leq\frac{\beta\pi^{1/3}(1+1/q_{0})q}{t^{1/3}},$
	$\displaystyle\frac{\alpha\eta}{P}\leq\frac{\alpha\eta}{t^{1/3}},$	$\displaystyle\frac{\eta\beta W^{2/3}}{P^{2}}$	$\displaystyle\leq\frac{\eta\beta\pi^{2/3}(1+1/q_{0})^{2}q^{2}}{t^{2/3}}.$

Isolating the first term using the inequality $\sqrt{x+y}\leq\sqrt{x}+\sqrt{y}$ , valid for nonnegative $x$ and $y$ , and using the inequality $q\leq Q\leq t^{1/6}/(2\pi)^{1/2}$ to bound $q$ in the remaining terms, therefore gives

(56)

\frac{1}{N}\max_{1\leq L\leq P}|S_{N}(L)|\leq\frac{\alpha(W,\lambda)^{1/2}}{\pi^{1/6}q^{3/2}}+\frac{B(W,\lambda)}{qt^{1/12}},

where, since $t\geq t_{1}$ ,

B(W,\lambda):=\sqrt{\frac{1+1/q_{0}}{2^{1/2}\pi^{1/6}}\beta(W)+\frac{\eta}{t_{1}^{1/6}}\alpha(W,\lambda)+\frac{\eta(1+1/q_{0})^{2}}{2\pi^{1/3}t_{1}^{1/6}}\beta(W)}.

We also note that for all $q\geq q_{0}$ we have $W\geq W_{0}$ and $\lambda\leq\lambda_{0}$ , where

\begin{split}W_{0}:=\pi(q_{0}+1)^{3},\qquad&\lambda_{0}:=\left(1+\frac{1}{q_{0}}\right)^{3}.\end{split}

Therefore, for all $q$ under consideration, $\alpha(W)\leq\alpha(W_{0},\lambda_{0})$ and $\beta(W)\leq\beta(W_{0})$ . It follows that for $q\geq q_{0}$ ,

(57)

\frac{1}{N}\max_{1\leq L\leq P}|S_{N}(L)|\leq\frac{\alpha_{0}^{1/2}}{\pi^{1/6}q^{3/2}}+\frac{B_{0}}{qt^{1/12}}.

where $\alpha_{0}:=\alpha(W_{0},\lambda_{0})$ and $B_{0}:=B(W_{0},\lambda_{0})$ .

Put together, we have

(58)

\begin{split}\left|\sum_{n=q_{0}P}^{n_{1}}\frac{1}{n^{1+it}}\right|\leq&\frac{\alpha_{0}^{1/2}}{\pi^{1/6}}\sum_{q=q_{0}}^{Q}\frac{1}{q^{3/2}}+\frac{B_{0}}{t^{1/12}}\sum_{q=q_{0}}^{Q}\frac{1}{q}.\end{split}

Moreover, from the argument in [HPY24, (3.1)], and using the bound $Q\leq t^{1/6}/(2\pi)^{1/2}$ once again,

(59)

\begin{split}&\sum_{q=q_{0}}^{Q}\frac{1}{q^{3/2}}\leq\int_{q_{0}-1/2}^{Q+1/2}\frac{\text{d}x}{x^{3/2}}<\frac{2}{\sqrt{q_{0}-1/2}},\\ \sum_{q=q_{0}}^{Q}\frac{1}{q}&\leq\int_{q_{0}-1/2}^{Q+1/2}\frac{\text{d}x}{x}<\frac{1}{6}\log t+\log\tau_{1}-\log\left(q_{0}-1/2\right),\end{split}

where

\tau_{1}:=\frac{1}{(2\pi)^{1/2}}+\frac{1}{2t_{1}^{1/6}}.

Therefore, observing that $\frac{1}{6}\frac{\log t}{t^{1/12}}$ is decreasing for $t\geq t_{1}\geq\exp(12)$ , and combining this with the initial sum bound (54), we obtain

	$\displaystyle\left\|\sum_{n=1}^{n_{1}}\frac{1}{n^{1+it}}\right\|$	$\displaystyle<\log\left(q_{0}P-1\right)+\gamma+\frac{1}{2(q_{0}P-1)}+\frac{\alpha_{0}^{1/2}}{\pi^{1/6}}\frac{2}{\sqrt{q_{0}-1/2}}$
		$\displaystyle\quad+\frac{B_{0}}{t_{1}^{1/12}}\left(\frac{1}{6}\log t_{1}+\log\tau_{1}-\log\left(q_{0}-1/2\right)\right).$

We now estimate

\log\left(q_{0}P-1\right)\leq\log\left(q_{0}t^{1/3}+q_{0}-1\right)\leq\frac{1}{3}\log t+\log\left(q_{0}+(q_{0}-1)t_{1}^{-1/3}\right),

which follows using $P<t^{1/3}+1$ . Furthermore,

\frac{1}{2(q_{0}P-1)}\leq\frac{1}{2\big{(}q_{0}t_{1}^{1/3}-1\big{)}}.

This yields

(60)

|\zeta(1+it)|\leq\frac{1}{3}\log t+B_{1}(q_{0})+\frac{g(t_{1})}{\sqrt{2\pi}}+\mathcal{R}(t_{1}),

where

	$\displaystyle B_{1}(q_{0}):=$	$\displaystyle\,\gamma+\log\left(q_{0}+(q_{0}-1)t_{1}^{-1/3}\right)+\frac{1}{2\big{(}q_{0}t_{1}^{1/3}-1\big{)}}$
		$\displaystyle+\frac{\alpha_{0}^{1/2}}{\pi^{1/6}}\frac{2}{\sqrt{q_{0}-1/2}}+\frac{B_{0}}{t_{1}^{1/12}}\left(\frac{1}{6}\log t_{1}+\log\tau_{1}-\log\left(q_{0}-1/2\right)\right).$

Choosing $t_{1}=\exp(16)$ , $\eta=1.267$ and $q_{0}=5$ , we obtain

(61)

|\zeta(1+it)|\leq\frac{1}{3}\log t+4.66,\qquad t\geq\exp(16).

This implies Theorem 1 in the range $\exp(16)\leq t\leq\exp(88)$ .

Consider now the second part of Proposition 3. In this critical region, we make use of a few additional tools to sharpen our estimates. In particular, we employ a different method of bounding the second sum appearing in Theorem 9, leading to an estimate of size $O(t^{-1/12})$ instead of $O(1)$ . Also, we employ the $k$ -th derivative tests used in the proof of Theorem 1, however this time we introduce additional scaling parameters to further optimise the switching points between derivative tests of different order.

To this end, let $t_{2}$ be a real number to be chosen later, subject to $t_{2}\geq\exp(182/3)$ . Assume $t\geq t_{2}$ . Recall from (58) and (59) we showed for $q_{0}\geq 1$ , $P=\lceil t^{1/3}\rceil$ and $t\geq t_{2}$ that the tail sum satisfies

(62)

\left|\sum_{n=q_{0}P}^{n_{1}}\frac{1}{n^{1+it}}\right|\leq\frac{\alpha_{0}^{1/2}}{\pi^{1/6}}\frac{2}{\sqrt{q_{0}-1/2}}+\frac{B_{0}}{t_{2}^{1/12}}\left(\frac{1}{6}\log t_{2}+\log\tau_{1}-\log\left(q_{0}-1/2\right)\right).

As for the initial sum over $1\leq n<q_{0}P$ , instead of bounding it using the triangle inequality throughout as in (54), we will refine our estimate by using $k$ -th order van der Corput lemmas with $k=4$ and $5$ in the subintervals $(Z_{5},Z_{4}]$ and $(Z_{6},Z_{5}]$ respectively, where

\displaystyle Z_{4}:=q_{0}P-1,\qquad Z_{5}:=\mu_{5}t^{\theta_{5}/2},\qquad Z_{6}:=\mu_{6}t^{\theta_{6}/2},

with $\theta_{5},\theta_{6}$ the same as in the proof of Theorem 1 and $\mu_{5},\mu_{6}$ real positive numbers to be chosen later, subject to

(63)

1<\mu_{6}t_{2}^{\theta_{6}/2}\leq\mu_{5}t_{2}^{\theta_{5}/2}\leq q_{0}\lceil t_{2}^{1/3}\rceil-1.

We remark that $Z_{5},Z_{6}$ will serve a similar purpose as $Y_{k}$ in the proof of Theorem 1, whereas the new parameters $\mu_{5},\mu_{6}$ will allow us to fine-tune our estimates in this critical region.

Let $h^{\prime}>1$ be a real number to be chosen later, and let $X_{m}^{\prime}=(h^{\prime})^{-m}Z_{4}$ for integer $m$ . Define

M^{\prime}:=\left\lceil\frac{\log(Z_{4}/Z_{6})}{\log h^{\prime}}\right\rceil.

The definition of $M^{\prime}$ guarantees $X^{\prime}_{M^{\prime}}\leq Z_{6}<X^{\prime}_{M^{\prime}-1}$ . Next, define the subsum

S_{m}^{\prime}:=\sum_{X_{m}^{\prime}+1<n\leq X_{m-1}^{\prime}}\frac{1}{n^{1+it}},

so that, similarly to (12),

\left|\sum_{X^{\prime}_{M^{\prime}-1}<n\leq Z_{4}}\frac{1}{n^{1+it}}\right|\leq\sum_{1\leq m\leq M^{\prime}-1}\left(\frac{1}{X_{m}^{\prime}}+|S_{m}^{\prime}|\right).

By a similar calculation as in (13), for each $1\leq m\leq M^{\prime}-1$ ,

\frac{\lfloor X_{m-1}^{\prime}\rfloor}{\lfloor X_{m}^{\prime}\rfloor+1}\leq h^{\prime}.

Let $k=4$ or $5$ . Following the argument of Theorem 1, if $Z_{k+1}<X_{m}^{\prime}\leq Z_{k}$ , then

	$\displaystyle\left\|S_{m}^{\prime}\right\|$	$\displaystyle\leq C_{k}(\eta,h^{\prime})(\lfloor X_{m}^{\prime}\rfloor+1)^{-k/(2K-2)}t^{1/(2K-2)}$
		$\displaystyle\qquad\qquad+D_{k}(\eta,h^{\prime})(\lfloor X_{m}^{\prime}\rfloor+1)^{k/(2K-2)-2/K}t^{-1/(2K-2)}$
		$\displaystyle\leq C_{k}(\eta,h^{\prime})Z_{k+1}^{-k/(2K-2)}t^{1/(2K-2)}+D_{k}(\eta,h^{\prime})(Z_{k}+1)^{k/(2K-2)-2/K}t^{-1/(2K-2)}.$

We write $\mu_{4}:=q_{0}+(q_{0}-1)t_{2}^{-1/3}$ so that $Z_{4}\leq\mu_{4}t^{1/3}$ . Since $k/(2K-2)-2/K\geq 0$ , for $t\geq t_{2}$ ,

|S_{m}^{\prime}|\leq A_{k}^{\prime}(t):=\begin{cases}\displaystyle\mu_{5}^{-2/7}C_{4}t^{-3/182}+(\mu_{4}+t_{2}^{-1/3})^{1/28}D_{4}t^{-5/84},&k=4,\\ \displaystyle\mu_{6}^{-1/6}C_{5}t^{-7/990}+(\mu_{5}+t_{2}^{-4/13})^{1/24}D_{5}t^{-4/195},&k=5.\end{cases}

Let $N_{k}^{\prime}$ be the number of $X_{m}^{\prime}$ satisfying $Z_{k+1}<X_{m}^{\prime}\leq Z_{k}$ . Then

\displaystyle N_{k}^{\prime}

\displaystyle\leq\frac{\log Z_{k}-\log Z_{k+1}}{\log h^{\prime}}+1,\qquad\text{therefore}\qquad N_{k}^{\prime}\leq B^{\prime}_{k}\frac{\log t}{\log h^{\prime}},

where

B^{\prime}_{k}:=\begin{cases}\displaystyle\frac{1}{39}+\frac{\max\{0,\log\mu_{4}-\log\mu_{5}+\log h^{\prime}\}}{\log t_{2}},&k=4,\\ \,&\\ \displaystyle\frac{28}{429}+\frac{\max\{0,\log\mu_{5}-\log\mu_{6}+\log h^{\prime}\}}{\log t_{2}},&k=5.\end{cases}

Put together, we have

\sum_{Z_{k+1}<X^{\prime}_{m}\leq Z_{k}}|S_{m}^{\prime}|\leq\frac{A^{\prime}_{k}(t)B^{\prime}_{k}}{\log h^{\prime}}\log t.

For $t\geq t_{2}$ , the function $A_{4}^{\prime}(t)\log t$ is decreasing, so $A_{4}^{\prime}(t)\log t\leq A_{4}^{\prime}(t_{2})\log t_{2}$ . For $k=5$ , we use

t^{-7/990}\log t\leq\frac{990}{7e}\qquad\text{and}\qquad t^{-4/195}\log t\leq t_{2}^{-4/195}\log t_{2},\qquad(t\geq t_{2}).

Therefore,

(64)

\begin{split}\sum_{1\leq m\leq M^{\prime}-1}|S^{\prime}_{m}|&\leq\frac{1}{\log h^{\prime}}\bigg{(}A^{\prime}_{4}(t_{2})B^{\prime}_{4}\log t_{2}+\frac{990}{7e}\frac{B^{\prime}_{5}C_{5}}{\mu_{6}^{1/6}}\\ &\qquad\qquad+\left(\mu_{5}+t_{2}^{-4/13}\right)^{1/24}B^{\prime}_{5}D_{5}t_{2}^{-4/195}\log t_{2}\bigg{)}.\end{split}

Next, similarly to (39), we obtain

(65)

\sum_{1\leq m\leq M^{\prime}-1}\frac{1}{X_{m}^{\prime}}<\frac{h^{\prime}}{h^{\prime}-1}\frac{1}{Z_{6}}\leq\frac{h^{\prime}}{(h^{\prime}-1)\mu_{6}t_{2}^{\theta_{6}/2}},\qquad(t\geq t_{2}).

In the remaining range $1\leq n\leq X^{\prime}_{M^{\prime}-1}$ , we use that $h^{\prime}Z_{6}\geq X^{\prime}_{M^{\prime}-1}$ . Combined with the triangle inequality and (2), this yields

(66)

\begin{split}\left|\sum_{1\leq n\leq\lfloor X_{M^{\prime}-1}\rfloor}\frac{1}{n^{1+it}}\right|&\leq\sum_{1\leq n\leq\lfloor h^{\prime}Z_{6}\rfloor}\frac{1}{n}\leq\log h^{\prime}Z_{6}+\gamma+\frac{1}{2(h^{\prime}Z_{6}-1)}\\ &\leq\frac{\theta_{6}}{2}\log t+\log\mu_{6}+\log h^{\prime}+\gamma+\frac{1}{2\big{(}h^{\prime}\mu_{6}t_{2}^{\theta_{6}/2}-1\big{)}}.\end{split}

Next, consider the second sum $\sum_{1\leq n\leq n_{1}}n^{it}$ in Theorem 9, which was just bounded by $n_{1}$ in our previous calculations. Let $q$ , $q_{0}$ , $P$ , $Q$ and $N$ be as defined in the proof of Proposition 3, and suppose as before that $t\geq t_{2}\geq\exp(182/3)$ . Using $P\leq t^{1/3}+1$ , the estimate (57) gives for $q_{0}\leq q\leq Q$ ,

\max_{1\leq L\leq P}|S_{N}(L)|\leq\left(t^{1/3}+1\right)\left(\frac{\alpha_{0}^{1/2}}{\pi^{1/6}q^{1/2}}+\frac{B_{0}}{t^{1/12}}\right).

We combine this with the inequality

\sum_{q=q_{0}}^{Q}\frac{1}{q^{1/2}}\leq\int_{q_{0}-1}^{Q}\frac{\text{d}x}{x^{1/2}}<2Q^{1/2},

as well as bound the sum over $1\leq n<q_{0}P$ by the number of terms in the sum, which gives

\displaystyle\left|\sum_{n=1}^{n_{1}}n^{it}\right|

\displaystyle<q_{0}P-1+\left(t^{1/3}+1\right)\left(\frac{2\alpha_{0}^{1/2}Q^{1/2}}{\pi^{1/6}}+\frac{B_{0}Q}{t^{1/12}}\right).

Since $Q\leq t^{1/6}/\sqrt{2\pi}$ , we obtain for $t\geq t_{2}$ ,

(67)

\left|\sum_{n=1}^{n_{1}}n^{it}\right|<C\,t^{5/12},

where

\displaystyle C(q_{0},\eta,t_{2})

\displaystyle:=\left(1+t_{2}^{-1/3}\right)\left(\frac{q_{0}}{t_{2}^{1/12}}+\frac{2^{3/4}\alpha_{0}^{1/2}}{\pi^{5/12}}+\frac{B_{0}}{\sqrt{2\pi}}\right).

Therefore, after multiplying by $g(t)t^{-1/2}$ , we obtain a sharper bound on the contribution of the second sum in Theorem 9. Namely, the bound $Cg(t_{2})\,t^{-1/12}$ , which replaces the original estimate $g(t_{2})/\sqrt{2\pi}$ .

Lastly, combining (62), (64), (65), (66) and (67), and noting that $\theta_{6}=16/33$ , we have for $t\geq t_{2}$

|\zeta(1+it)|\leq\frac{8}{33}\log t+B_{3},

where

	$\displaystyle B_{3}:=$	$\displaystyle\,\frac{1}{\log h^{\prime}}\left(A^{\prime}_{4}(t_{2})B^{\prime}_{4}\log t_{2}+\frac{990}{7e}\frac{B^{\prime}_{5}C_{5}}{\mu_{6}^{1/6}}+(\mu_{5}+t_{2}^{-4/13})^{1/24}B^{\prime}_{5}D_{5}t_{2}^{-4/195}\log t_{2}\right)$
		$\displaystyle\qquad+\frac{h^{\prime}}{(h^{\prime}-1)\mu_{6}t_{2}^{\theta_{6}/2}}+\frac{\alpha_{0}(W_{0},\lambda_{0})^{1/2}}{\pi^{1/6}}\frac{2}{\sqrt{q_{0}-1/2}}$
		$\displaystyle\qquad+\frac{B_{0}(W_{0},\lambda_{0})}{t_{2}^{1/12}}\left(\frac{1}{6}\log t_{2}+\log\tau_{1}-\log\left(q_{0}-1/2\right)\right)$
		$\displaystyle\qquad+\log\mu_{6}+\log h^{\prime}+\gamma+\frac{1}{2\big{(}h^{\prime}\mu_{6}t_{2}^{\theta_{6}/2}-1\big{)}}+C(q_{0},\eta,t_{2})\frac{g(t_{2})}{t_{2}^{1/12}}+\mathcal{R}(t_{2}).$

We choose

	$\displaystyle\eta$	$\displaystyle=1.3348,\qquad\qquad$	$\displaystyle h^{\prime}=1.056,\qquad\qquad$	$\displaystyle q_{0}=46,$
	$\displaystyle\mu_{5}$	$\displaystyle=48.575,\qquad\qquad$	$\displaystyle\mu_{6}=51.296,\qquad\qquad$	$\displaystyle t_{2}=\exp(90).$

Verifying that (63) holds, we calculate $B_{3}\leq 12.45$ , i.e. that

|\zeta(1+it)|\leq\frac{8}{33}\log t+12.45,\qquad t\geq\exp(88).

This implies Theorem 1 for $\exp(88)\leq t\leq\exp(662)$ .

6. Proof of Theorem 2

We begin by proving some bounds on $\zeta(s)$ . To help make our notation more intuitive, we will allow reuse of variable names from previous sections. The reader should keep in mind these variables mean different things in the confines of this section. Since this section is independent from Section 3, Section 4 and Section 5, this will hopefully cause no confusion.

Lemma 4.

Let $C>0$ and $t_{0}\geq e^{e}$ be constants. Suppose $t\geq t_{0}$ . Define

\sigma_{t}:=1-\frac{C(\log\log t)^{2}}{\log t}.

Additionally, suppose that $C$ satisfies

(68)

\frac{C(\log\log t_{e})^{2}}{\log t_{e}}\leq\frac{1}{2},\qquad\text{where}\qquad t_{e}:=e^{e^{2}}.

For each $t\geq t_{0}$ , if $\sigma_{t}\leq\sigma\leq 2$ , then

(69)

|\zeta(\sigma+it)|\leq A\log^{B}t\qquad\text{where}\qquad A=76.2,\qquad B=\frac{2}{3}+\frac{71.2\,C^{3/2}}{e^{2}}.

Proof.

From Ford [For02], if $1/2\leq\sigma\leq 1$ , then

(70)

|\zeta(\sigma+it)|\leq 76.2\,t^{4.45(1-\sigma)^{3/2}}\log^{2/3}t.

As $\sigma_{t}\geq 1/2$ , a consequence of the assumption (68) and considering that $(\log\log t)^{2}/\log t$ reaches its maximum at $t=t_{e}$ , we may use (70) for any $\sigma\in[\sigma_{t},1]$ .

In view of this, since $(1-\sigma)^{3/2}$ is monotonically decreasing with $\sigma$ , if $\sigma_{t}\leq\sigma\leq 1$ , then (70) gives

(71)

\displaystyle|\zeta(\sigma+it)|\leq 76.2\exp\left(4.45(1-\sigma_{t})^{3/2}\log t+\frac{2}{3}\log\log t\right).

Or, written differently,

(72)

|\zeta(\sigma+it)|\leq A\exp\left(\tilde{B}\log\log t\right),

where $A$ is defined as in (69) and

\tilde{B}:=\frac{2}{3}+4.45\,C^{3/2}\,\frac{(\log\log t)^{2}}{\sqrt{\log t}}.

As $(\log\log t)^{2}/\sqrt{\log t}$ reaches its maximum at $t=e^{e^{4}}$ , attaining the value $16/e^{2}$ , we obtain after a simple computation that $\tilde{B}\leq B$ , with $B$ defined as in (69). The desired result therefore follows for $\sigma_{t}\leq\sigma\leq 1$ on replacing $\tilde{B}$ with $B$ in (72).

To show that the result holds for $1\leq\sigma\leq 2$ as well, we use the Phragmén–Lindelöf Principle on the holomorphic function

f(s)=(s-1)\zeta(s).

To this end, on the $1$ -line we have

|f(1+it)|=|t|\,|\zeta(1+it)|\leq 62.6\,|2+it|\log^{2/3}|2+it|.

This inequality is verified numerically for $|t|<3$ and is a consequence of the bound $|\zeta(1+it)|\leq 62.6\log^{2/3}t$ for $|t|\geq 3$ , given in [Tru14]. Also, on the $2$ -line, plainly,

|f(2+it)|\leq|1+it|\,\zeta(2)<62.6|3+it|\log^{2/3}|3+it|,

for all real $t$ . By [Tru14a, Lemma 3], we thus obtain⁷⁷7We apply [Tru14a, Lemma 3] with the following parameter values (in the notation in the cited paper): $a=1$ , $b=2$ , $Q=1$ , $\alpha_{1}=\beta_{1}=1$ , $\alpha_{2}=\beta_{2}=2/3$ and $A=B=62.6$ . Note that the $A$ and $B$ in [Tru14a, Lemma 3] are different from the $A$ and $B$ in the statement of our lemma.

(73)

|f(s)|\leq 62.6|1+s|\log^{2/3}|1+s|,\qquad(1\leq\sigma\leq 2).

Moreover, for $1\leq\sigma\leq 2$ we have the easy estimates,

\begin{split}&\frac{|1+s|}{|s-1|}\leq\frac{\sqrt{9+t^{2}}}{t}\leq\sqrt{1+9t_{0}^{-2}},\\ &\log|1+s|\leq\log\sqrt{9+t^{2}}\leq\log t+\log\sqrt{1+9t_{0}^{-2}}.\end{split}

So that, for $1\leq\sigma\leq 2$ and $t\geq t_{0}$ , the inequality (73) implies that

\displaystyle|\zeta(\sigma+it)|

\displaystyle\leq 62.6\sqrt{1+9t_{0}^{-2}}\left(1+\frac{\log\sqrt{1+9t_{0}^{-2}}}{\log t_{0}}\right)\log^{2/3}t<A\log^{B}t,

where $A$ and $B$ are defined as in (69), as desired. ∎

Remark.

Following Titchmarsh’s argument [Tit86, Theorem 5.17], we could make use of [Yan24, Theorem 1.1]. For simplicity, however, we used the Richert-type bound (70) due to Ford [For02] instead.

Lemma 5.

Let $t_{0}\geq e^{e}$ and $d_{1}>0$ be constants. Suppose $t\geq t_{0}$ . We have

\left|\zeta\left(1+d_{1}\frac{\log\log t}{\log t}+it\right)\right|\leq C^{\prime}(d_{1},t_{0})\frac{\log t}{\log\log t},

where $\gamma$ is Euler’s constant, and

(74)

C^{\prime}(d_{1},t_{0})=\frac{1}{d_{1}}\exp\left(\gamma d_{1}\frac{\log\log t_{0}}{\log t_{0}}\right).

Proof.

We use the following uniform bound due to Ramaré [Ram16, Lemma 5.4]:

|\zeta(\sigma+it)|\leq\zeta(\sigma)\leq\frac{e^{\gamma(\sigma-1)}}{\sigma-1},\qquad(\sigma>1).

Substituting $\sigma=1+d_{1}\log\log t/\log t$ therefore gives for $t\geq t_{0}$ ,

\zeta(\sigma)\leq\frac{e^{\gamma d_{1}\log\log t/\log t}}{d_{1}\log\log t/\log t}\leq\frac{1}{d_{1}}\exp\left(\frac{\gamma d_{1}\log\log t_{0}}{\log t_{0}}\right)\frac{\log t}{\log\log t},

as claimed. In the last inequality, we used that $\log\log t/\log t$ reaches its maximum at $t=e^{e}$ and is monotonically decreasing after that. ∎

6.1. Bounding $|\zeta^{\prime}(1+it)/\zeta(1+it)|$

We follow the argument of [Tru15], which gives an explicit version of results in [Tit86, §3]. The method relies on Theorem 14.

We will therefore construct concentric disks, centred just to the right of the line $s=1+it$ , and extend their radius slightly to the left and into the critical strip. Ultimately, we want to apply Lemmas 15 and 16 with $f(s)=\zeta(s)$ , which will give us the desired results.

The process is as follows. Let $t_{0}\geq e^{e}$ and suppose $t^{\prime}\geq t_{0}$ . Let $d$ be a real positive constant, which will be chosen later. Denote the center of the concentric disks to be constructed by $s^{\prime}=\sigma^{\prime}+it^{\prime}$ , where

(75)

\sigma^{\prime}=1+\delta_{t^{\prime}},\qquad\text{with}\qquad\delta_{t^{\prime}}:=\frac{d\log\log t^{\prime}}{\log t^{\prime}}.

Notice that $\delta_{t^{\prime}}$ is decreasing with $t^{\prime}$ and is at most $d/e$ . Let $r>0$ denote the radius of our largest concentric disk. In the sequel, $r$ will be determined as function of $s^{\prime}$ .

In addition, suppose $\sigma^{\prime}+r\leq 1+\epsilon$ where $\epsilon\leq 1$ is another positive constant to be chosen later. This implies, for instance, that $r\leq\epsilon\leq 1$ . In order to enforce this inequality on $\sigma^{\prime}$ and $r$ , it is enough to demand that

(76)

\frac{d}{e}+r\leq\epsilon.

Let $s=\sigma+it$ be a complex number. Aiming to apply Lemma 15 in the disk $|s-s^{\prime}|\leq r$ , we first seek a valid $A_{1}$ in that disk. Similarly, for Lemma 16, we seek a valid $A_{2}$ at the disk center $s=s^{\prime}$ . In applying Lemma 16, we moreover need to ensure that the non-vanishing condition on $f(s)=\zeta(s)$ is fulfilled, which we do by way of a zero-free region of zeta.

Let us first determine a valid $A_{1}$ , with the aid of Lemma 4. To this end, let $C>0$ be a parameter restricted by the inequality in (68), so that (in the notation of Lemma 4) $\sigma_{t}\geq 1/2$ for $t\geq e$ . Note that $\sigma_{t}$ is monotonically decreasing for $e\leq t\leq t_{e}$ and monotonically increasing thereafter. So, the maximum of $\sigma_{t}$ over the interval $t^{\prime}-r\leq t\leq t^{\prime}+r$ occurs at one of the end-points. For example, if $t^{\prime}-r\geq t_{e}$ , then the maximum occurs at $t=t^{\prime}+r$ . And if $t^{\prime}+r\leq t_{e}$ , then the maximum occurs at $t=t^{\prime}-r$ . With this in mind, we calculate

(77)

\max_{|t-t^{\prime}|\leq r}\sigma_{t}=\max\{\sigma_{t^{\prime}-r},\sigma_{t^{\prime}+r}\}<1-\frac{C_{0}\,(\log\log t^{\prime})^{2}}{\log t^{\prime}},

where, since $r\leq\epsilon$ ,

C_{0}:=C\,\min\left\{\left(\frac{\log\log(t^{\prime}-\epsilon)}{\log\log t^{\prime}}\right)^{2},\frac{\log t^{\prime}}{\log(t^{\prime}+\epsilon)}\right\}.

Clearly, $C_{0}=C_{0}(t^{\prime})$ increases monotonically with $t^{\prime}$ (towards $C$ ). We therefore set

(78)

C_{1}:=C_{0}(t_{0})\qquad\text{and}\qquad\sigma_{1,t^{\prime}}:=1-\frac{C_{1}\,(\log\log t^{\prime})^{2}}{\log t^{\prime}}.

So that, by (77), no matter $t^{\prime}$ , $r$ and $\epsilon$ , so long as they are subject to our conditions,

\sigma_{1,t^{\prime}}\geq\max_{|t-t^{\prime}|\leq r}\sigma_{t}.

Consequently, Lemma 4 gives that for each $t\in[t^{\prime}-r,t^{\prime}+r]$ and any $\sigma\in[\sigma_{1,t^{\prime}},2]$ , and on recalling $r\leq\epsilon$ ,

(79)

|\zeta(s)|\leq A\log^{B}(t^{\prime}+r)\leq A_{3}\log^{B}t^{\prime},\qquad A_{3}=A\left(1+\frac{\log(1+\epsilon/t^{\prime})}{\log t^{\prime}}\right)^{B}.

We would like the inequality (79) to hold throughout the disk $|s-s^{\prime}|\leq r$ . This will be certainly the case if this disk lies entirely in the rectangle specified by $\sigma\in[\sigma_{1,t^{\prime}},2]$ and $t\in[t^{\prime}-r,t^{\prime}+r]$ . This follows, in turn, on requiring $\sigma^{\prime}-r\geq\sigma_{1,t^{\prime}}$ or, equivalently, that

r\leq\frac{C_{1}(\log\log t^{\prime})^{2}}{\log t^{\prime}}+\delta_{t^{\prime}}.

Therefore, subject to the constraint (76), the following choice of $r$ works:

(80)

r=r_{t^{\prime}}:=\left(C_{1}+\frac{d}{\log\log t^{\prime}}\right)\frac{(\log\log t^{\prime})^{2}}{\log t^{\prime}}.

To ensure the constraint (76) is met for all $t^{\prime}\geq t_{0}$ , it suffices that

(81)

\frac{d}{e}+\left(C_{1}+d\right)\frac{4}{e^{2}}\leq\epsilon\leq 1.

Here, we used that $(\log\log t^{\prime})^{2}/\log t^{\prime}$ reaches its maximum of $4/e^{2}$ at $t^{\prime}=t_{e}$ .

Overall, combining (79) and (80) with the trivial bound $|\zeta(s^{\prime})|\geq\zeta(2\sigma^{\prime})/\zeta(\sigma^{\prime})$ , easily seen on considering the Euler product of zeta, we obtain that throughout the disk $|s-s^{\prime}|\leq r$ ,

\displaystyle\left|\frac{\zeta(s)}{\zeta(s^{\prime})}\right|\leq

\displaystyle\,A_{3}\,\frac{\zeta(\sigma^{\prime})}{\zeta(2\sigma^{\prime})}\log^{B}t^{\prime}=A_{3}\,\frac{X(\sigma^{\prime})}{\sigma^{\prime}-1}\log^{B}t^{\prime},

where

X(\sigma^{\prime})=\frac{\zeta(\sigma^{\prime})(\sigma^{\prime}-1)}{\zeta(2\sigma^{\prime})}.

Since $\sigma^{\prime}=\sigma^{\prime}(t^{\prime})$ decreases to $1$ with $t^{\prime}\geq t_{0}$ , $X(\sigma^{\prime})$ decreases to $1/\zeta(2)$ with $t^{\prime}$ . Thus, recalling from (75) that $\sigma^{\prime}=1+\delta_{t^{\prime}}$ and observing $A_{3}$ is also decreasing in $t^{\prime}\geq t_{0}$ , we obtain that throughout the disk $|s-s^{\prime}|\leq r$ ,

(82)

\left|\frac{\zeta(s)}{\zeta(s^{\prime})}\right|\leq\frac{A_{\max}\,X_{\max}}{d}\frac{\log^{B+1}t^{\prime}}{\log\log t^{\prime}},

where $A_{\max}=A_{3}(t_{0})$ and $X_{\max}=X(\sigma^{\prime}(t_{0}))$ .

In summary, when applying Lemma 15 in the disk $|s-s^{\prime}|\leq r$ , with $r$ as in (80) and subject to (81), we may take

\log A_{1}=(B+1)\log\log t^{\prime}-\log\log\log t^{\prime}+\log\left(\frac{A_{\max}\,X_{\max}}{d}\right).

Next, in preparation for applying Lemma 16, we want to ensure that the intersection of the disk $|s-s^{\prime}|\leq r$ and the right half-plane $\sigma\geq\sigma^{\prime}-\alpha r$ lies entirely in a zero-free region for zeta. By [Yan24, Corollary 1.2], we have the following zero-free region, valid for $t\geq 3$ .

\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}.

Since $1-c_{0}\log\log t/\log t$ is monotonically increasing for $t\geq t_{0}$ and since $t^{\prime}\geq t_{0}$ and $r\leq\epsilon$ , it suffices for our purposes to require that

\sigma^{\prime}-\alpha r\geq 1-\frac{c_{0}\log\log(t^{\prime}+\epsilon)}{\log(t^{\prime}+\epsilon)}.

In other words, it suffices that

\alpha\leq\frac{1}{r}\left(\delta_{t^{\prime}}+\frac{c_{0}\log\log(t^{\prime}+\epsilon)}{\log(t^{\prime}+\epsilon)}\right).

Therefore, subject to $\alpha<1/2$ as demanded by Lemma 16, the following choice of $\alpha$ works.

(83)

\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}},

where (using monotonicity to deduce the inequality below)

(84)

c_{1}:=\frac{c_{0}\log t_{0}}{\log(t_{0}+\epsilon)},\qquad\text{so that}\qquad c_{1}\leq c_{0}.

In particular, the constraint $\alpha<1/2$ is fulfilled if

(85)

\frac{d+c_{1}}{d+C_{1}\log\log t_{0}}<\frac{1}{2}.

Now, to determine $A_{2}$ , we utilise [Del87], which asserts that if $\sigma>1$ , then

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

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

\left|\frac{\zeta^{\prime}(s^{\prime})}{\zeta(s^{\prime})}\right|<\frac{1}{\delta_{t^{\prime}}}=\frac{\log t^{\prime}}{d\log\log t^{\prime}}.

So, we may take $A_{2}$ in Lemma 16 to be

A_{2}=\frac{\log t^{\prime}}{d\log\log t^{\prime}}.

Lastly, in applying Lemma 16, we need to be able to reach the $1$ -line from our position at $s^{\prime}$ . So, we set

\beta=\frac{\delta_{t^{\prime}}}{\alpha r}=\frac{d}{c_{1}+d}.

Clearly, $\beta<1$ , as required by Lemma 16. The conclusion of the lemma therefore holds throughout the disk $|s-s^{\prime}|\leq\alpha\beta r=\delta_{t^{\prime}}$ . This includes, in particular, the horizontal line segment $s=\sigma+it^{\prime}$ , with $1\leq\sigma\leq 1+2\delta_{t^{\prime}}$ .

Putting it all together, and making the substitution $t^{\prime}\to t$ , Lemma 16 hence furnishes the bound

(86)

\left|\frac{\zeta^{\prime}(s)}{\zeta(s)}\right|\leq\frac{8\beta}{(1-\beta)(1-2\alpha)^{2}}\,\frac{\log A_{1}}{r}+\frac{1+\beta}{1-\beta}\,A_{2},

valid in the region

1\leq\sigma\leq 1+\frac{2d\log\log t}{\log t},\qquad t\geq t_{0}.

We now input the values we determined for $A_{1}$ , $A_{2}$ , $r$ , $\alpha$ and $\beta$ into (86). We also note

\frac{1+\beta}{1-\beta}=1+\frac{2d}{c_{1}},\qquad\text{so that}\qquad\frac{1+\beta}{1-\beta}\,A_{2}=\left(\frac{1}{d}+\frac{2}{c_{1}}\right)\frac{\log t}{\log\log t},

as well as

\frac{8\beta}{1-\beta}=\frac{8d}{c_{1}},\qquad\frac{1}{(1-2\alpha)^{2}}=\left(\frac{C_{1}\log\log t+d}{C_{1}\log\log t-d-2c_{1}}\right)^{2}.

Hence, on observing that $1/(1-2\alpha)^{2}$ is decreasing in $t$ , we obtain the following lemma.

Lemma 6.

Let $t_{0}\geq e^{e}$ and suppose that $t\geq t_{0}$ . Let $C$ , $d$ and $\epsilon$ be any positive constants satisfying the constraints (68), (81) and (85). If

1\leq\sigma\leq 1+\frac{2d\log\log t}{\log t},

then

\left|\frac{\zeta^{\prime}(\sigma+it)}{\zeta(\sigma+it)}\right|\leq Q_{1}(C,d,\epsilon,t_{0})\frac{\log t}{\log\log t},

where

\displaystyle Q_{1}(C,d,\epsilon,t_{0}):=\lambda_{1}\left(B+1+\frac{1}{\log\log t_{0}}\log\Bigg{(}\frac{A_{\max}X_{\max}}{d}\Bigg{)}\right)+\lambda_{2}.

Here,

\begin{split}\lambda_{1}:=\frac{1}{C_{1}}\cdot\frac{8d}{c_{1}}\cdot\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}{d}+\frac{2}{c_{1}}.\end{split}

The number $B$ is defined as in Lemma 4 and depends on $C$ . The numbers $A_{\max}$ and $X_{\max}$ are defined as in (82) and depend on $\epsilon$ and $t_{0}$ . The number $C_{1}$ is defined in (78), and depends on $C$ , $\epsilon$ and $t_{0}$ . The number $c_{1}$ is defined in (84) and depends on $\epsilon$ , $t_{0}$ and $c_{0}=1/21.233$ .

6.2. Bounding $|1/\zeta(1+it)|$

Moving from a bound on the logarithmic derivative of $\zeta$ to one on the reciprocal of $\zeta$ is done in the usual way (see [Tit86, Theorem 3.11]), with some improvements using the trigonometric polynomial (see [Car+22, Proposition A.2]).

Lemma 7.

Let $t_{0}\geq e^{e}$ and $d>0$ be constants. Suppose that for each $t\geq t_{0}$ ,

\text{if}\quad 1\leq\sigma\leq 1+\frac{2d\log\log t}{\log t},\quad\text{then}\quad\left|\frac{\zeta^{\prime}(\sigma+it)}{\zeta(\sigma+it)}\right|\leq R_{1}\frac{\log t}{\log\log t}.

Then for any $t\geq t_{0}$ and any real number $d_{1}$ such that $0<d_{1}\leq 2d$ , we have

\left|\frac{1}{\zeta(1+it)}\right|\leq Q_{2}(d_{1},t_{0})\frac{\log t}{\log\log t},

where

Q_{2}(d_{1},t_{0})=\exp(R_{1}d_{1})\left(\frac{\log\log t_{0}}{\log t_{0}}+\frac{1}{d_{1}}\right)^{3/4}\left(C^{\prime}(d_{1},2t_{0})\left(\frac{\log 2}{\log t_{0}}+1\right)\right)^{1/4}.

The number $C^{\prime}$ is defined as in (74).

Proof.

Suppose $t\geq t_{0}$ . Let $d_{1}$ be a real positive parameter such that $d_{1}\leq 2d$ . Let $\delta_{1}=d_{1}\log\log t/\log t$ . It is easy to see that

	$\displaystyle\log\left\|\frac{1}{\zeta(1+it)}\right\|$	$\displaystyle=-\textup{Re}\log\zeta(1+it)$
		$\displaystyle=-\textup{Re}\log\zeta\left(1+\delta_{1}+it\right)+\int_{1}^{1+\delta_{1}}\textup{Re}\left(\frac{\zeta^{\prime}}{\zeta}(\sigma+it)\right)\text{d}\sigma.$

So, using our suppositions, we have

(87)

\log\left|\frac{1}{\zeta(1+it)}\right|\leq-\log\left|\zeta\left(1+\delta_{1}+it\right)\right|+R_{1}d_{1}.

On the other hand, for $\sigma>1$ , and using the classical nonnegativity argument involving the trigonometric polynomial $3+4\cos\theta+\cos 2\theta$ [Tit86, Section 3.3], gives that

\zeta^{3}(\sigma)|\zeta^{4}(\sigma+it)\zeta(\sigma+2it)|\geq 1.

Combining this with the inequality $\zeta(\sigma)\leq\sigma/(\sigma-1)$ and Lemma 5, we therefore obtain, on taking $\sigma=1+\delta_{1}$ ,

(88)

\begin{split}\left|\frac{1}{\zeta(1+\delta_{1}+it)}\right|&\leq|\zeta(1+\delta_{1})|^{3/4}|\zeta(1+\delta_{1}+2it)|^{1/4}\\ &\leq\left(\frac{1+\delta_{1}}{\delta_{1}}\right)^{3/4}\left(C^{\prime}(d_{1},2t_{0})\frac{\log 2t}{\log\log 2t}\right)^{1/4}.\end{split}

We now observe

(89)

\begin{split}\frac{1+\delta_{1}}{\delta_{1}}&\leq\left(\frac{\log\log t_{0}}{\log t_{0}}+\frac{1}{d_{1}}\right)\frac{\log t}{\log\log t},\\ \frac{\log 2t}{\log\log 2t}&\leq\left(\frac{\log 2}{\log t_{0}}+1\right)\frac{\log t}{\log\log t}.\end{split}

So that, combining (89) and (88), and noting by (87) we have

\displaystyle\left|\frac{1}{\zeta(1+it)}\right|

\displaystyle\leq\frac{\exp(R_{1}d_{1})}{|\zeta(1+\delta_{1}+it)|},

the lemma follows. ∎

6.3. Numeric calculations

We define the following upper bounds:

Q_{1}(C,d,\epsilon,t_{0})\leq R_{1}\qquad\text{ and }\qquad Q_{2}(d_{1},t_{0})\leq R_{2}.

The numeric calculations for $R_{1}$ and $R_{2}$ proceed as follows. First fix $t_{0}\geq e^{e}$ and a positive $\epsilon\leq 1$ . For the purposes of obtaining an upper bound $R_{1}$ , we then optimise $Q_{1}(C,d,\epsilon,t_{0})$ in Lemma 6 over $C$ and $d$ , all the while ensuring $C$ , $d$ , and $\epsilon$ are subject to our constraints (68), (81) and (85)

Thereafter, with these values of $d$ and $R_{1}$ , the relationship $d_{1}\leq 2d$ allows us to obtain $R_{2}$ by optimising $Q_{2}(d_{1},t_{0})$ in Lemma 7 over $d_{1}$ .

Our choices of

\displaystyle t_{0}=500,\qquad\epsilon=0.517,

\displaystyle C=\frac{e^{2}}{8},\qquad d=0.018,\qquad d_{1}=0.0065,

give us

\displaystyle r

\displaystyle\leq 0.502,\qquad

\displaystyle\alpha\leq 0.0381,\qquad

\displaystyle\beta=0.274\ldots,

and the upper bounds

R_{1}=154.5,\qquad R_{2}=431.7.

In summary, we have for $t\geq 500$ ,

(90)

\frac{1}{|\zeta(1+it)|}\leq 431.7\frac{\log t}{\log\log t}.

Furthermore, in the region

1\leq\sigma\leq 1+\frac{9}{250}\frac{\log\log t}{\log t},

we have for $t\geq 500$ ,

(91)

\left|\frac{\zeta^{\prime}}{\zeta}(\sigma+it)\right|\leq 154.5\frac{\log t}{\log\log t}.

Finally, to cover the range from $3\leq t\leq 500$ in (90), we refer to the proof of Proposition $A.2$ in [Car+22], where via interval arithmetic, they computed that for $2\leq t\leq 500$ ,

\frac{1}{|\zeta(1+it)|}\leq 2.079\log t.

Comparing this with the estimate in (90), we see after a simple verification that the bound

\frac{1}{|\zeta(1+it)|}\leq 431.7\frac{\log t}{\log\log t}

holds for all $t\geq 3$ .

This concludes our proof of Theorem 2. These bounds can certainly be improved. Some possible avenues of attack would be to treat the sum over zeros in the proof of Lemma 16 more carefully, or to use a larger zero-free region.

Remark.

It is possible with our methods to directly get a bound like (91) for $t<500$ , especially if one were not concerned with the size of $R_{2}$ . For instance, choosing $t_{0}=16$ , $\epsilon=0.508$ , $C=e^{2}/8$ , $d=0.0147$ and $d_{1}=0.0056$ , would return for $t\geq 16$ that if

1\leq\sigma\leq 1+\frac{147}{5000}\frac{\log\log t}{\log t},

then

\left|\frac{\zeta^{\prime}}{\zeta}(\sigma+it)\right|\leq 177.5\frac{\log t}{\log\log t}.

While this would cause a modest jump in $R_{1}$ , the size of $R_{2}$ would increase significantly to $R_{2}=511.1$ . This is because $R_{2}$ increases exponentially with $R_{1}$ , as evident from Lemma 7. The dominant factor there is $C^{\prime}$ , and any improvement in that quantity would help greatly.

7. Background results

7.1. Background results related to Sections 3-5

Lemma 8.

[Fra21, Lemma 2.1] Let $\gamma$ denote Euler’s constant. For $N\geq 1$ ,

\sum_{n=1}^{N}\frac{1}{n}<\log N+\gamma+\frac{1}{2N}-\frac{1}{12N^{2}}+\frac{1}{64N^{4}}.

Theorem 9.

[Pat22, Theorem 3.9] For $t>0$ and $n_{1}=\lfloor\sqrt{t/(2\pi)}\rfloor$ we have

|\zeta(1+it)|\leq\left|\sum_{n=1}^{n_{1}}\frac{1}{n^{1+it}}\right|+\frac{g(t)}{t^{1/2}}\left|\sum_{n=1}^{n_{1}}\frac{1}{n^{-it}}\right|+\mathcal{R}

where

g(t)=\sqrt{2\pi}\exp\left(\frac{5}{3t^{2}}+\frac{\pi}{6t}\right),

and

\begin{split}\mathcal{R}=\mathcal{R}(t):=&\frac{1}{t^{1/2}}\left(\sqrt{\frac{\pi}{2}}+\frac{g(t)}{2}\right)+\frac{1}{t}\left(9\sqrt{\frac{\pi}{2}}+\frac{g(t)}{\sqrt{\pi(3-2\log 2)}}\right)\\ &+\frac{1}{t^{3/2}}\left(\frac{968\pi^{3/2}+g(t)242\pi}{700}\right).\end{split}

Next we review some explicit estimates of exponential sums from the literature. The following lemma is a specialised third-derivative test for a phase function commonly encountered when estimating the Riemann zeta-function. It is due to [HPY24, Lemma 2.6] which builds on [Hia16].

Lemma 10 (Explicit third derivative test).

[HPY24, Lemma 2.6] Let $r$ and $K$ be positive integers, and let $t$ and $K_{0}$ be positive numbers. Suppose $K\geq K_{0}>1$ . Let

\displaystyle f(x):=\frac{t}{2\pi}\log\left(rK+x\right),\qquad W:=\frac{\pi(r+1)^{3}K^{3}}{t},\qquad\lambda:=\frac{(1+r)^{3}}{r^{3}},

so that $1/W\leq|f^{\prime\prime\prime}(x)|\leq\lambda/W$ for $0\leq x\leq K$ . Then, for each positive integer $L\leq K$ and any $\eta>0$ ,

\left|\sum_{n=0}^{L-1}e^{2\pi if(n)}\right|^{2}\leq\left(\frac{K}{W^{1/3}}+\eta\right)(\alpha K+\beta W^{2/3})

where

	$\displaystyle\alpha$	$\displaystyle=\alpha(W,\lambda):=\frac{1}{\eta}+\frac{\eta\mu}{3W^{1/3}}+\frac{\mu}{3W^{2/3}}+\frac{32\mu}{15\sqrt{\pi}}\sqrt{\eta+W^{-1/3}},$
	$\displaystyle\beta$	$\displaystyle=\beta(W):=\frac{32}{3\sqrt{\pi\eta}}+\left(2-\frac{4}{\pi}\right)\frac{1}{W^{1/3}},$
	$\displaystyle\mu$	$\displaystyle=\mu(\lambda):=\frac{1}{2}\lambda^{2/3}\left(1+\frac{1}{(1-K_{0}^{-1})\lambda^{1/3}}\right).$

Remark.

If $W\leq 1$ , say, then $\alpha$ will be greater than $1$ , and so $\alpha L^{2}\geq L^{2}$ . Hence, for small $W$ , it is better to use the triangle inequality than Lemma 10.

Remark.

We mention that there is a typo in the statement of [HPY24, Lemma 2.6] where $N$ in the statement of the lemma should be $0$ . This lemma is only applied with $N=0$ in [HPY24] in any case.⁸⁸8In passing, let us also point out that on [HPY24, 211], the minimum in the definition of $y_{j}$ should be against $L-m$ rather than $L-m-1$ . This does not affect Equation 5.4 on p. 211 since, as stated in [HPY24, Lemma 2.5], the bound on $|g^{\prime\prime}(x)|$ holds up to $K-m$ , not only $K-m-1$ . None of the results in [HPY24] are affected.

The next two lemmas respectively represent another third derivative test and a $k$ th derivative test for $k\geq 4$ , due to [Yan24]. The reason we state this additional third derivative test is because $A_{3}$ and $B_{3}$ comprise the boundary cases for the recursive formulas in Lemma 12.

Lemma 11 (Another explicit third derivative test).

[Yan24, Lemma 2.4] Let $a$ and $N$ be integers such that $N\geq 1$ . Suppose $f(x)$ has three continuous derivatives and $f^{\prime\prime\prime}(x)$ is monotonic over the interval $(a,a+N]$ . Suppose further that there are numbers $\lambda_{3}>0$ and $\omega>1$ such that $\lambda_{3}\leq|f^{\prime\prime\prime}(x)|\leq\omega\lambda_{3}$ for all $x\in(a,a+N]$ . Then for any $\eta>0$ ,

\left|\sum_{n=a+1}^{a+N}e^{2\pi if(n)}\right|\leq A_{3}\omega^{1/2}N\lambda_{3}^{1/6}+B_{3}N^{1/2}\lambda_{3}^{-1/6}

where

\begin{split}&A_{3}=A_{3}(\eta,\omega):=\sqrt{\frac{1}{\eta\omega}+\frac{32}{15\sqrt{\pi}}\sqrt{\eta+\lambda_{0}^{1/3}}+\frac{1}{3}\left(\eta+\lambda_{0}^{1/3}\right)\lambda_{0}^{1/3}}\delta_{3},\end{split}

\begin{split}&B_{3}=B_{3}(\eta):=\frac{\sqrt{32}}{\sqrt{3}\pi^{1/4}\eta^{1/4}}\delta_{3},\end{split}

and $\lambda_{0}$ and $\delta_{3}$ are defined by

\begin{split}&\lambda_{0}=\lambda_{0}(\eta,\omega):=\left(\frac{1}{\eta}+\frac{32\eta^{1/2}\omega}{15\sqrt{\pi}}\right)^{-3},\end{split}

\begin{split}&\delta_{3}=\delta_{3}(\eta):=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{3}{8}\pi^{1/2}\eta^{3/2}}}.\end{split}

Lemma 12 (General explicit $k$ -th derivative test).

[Yan24, Lemma 2.5] Let $a$ , $N$ and $k$ be integers such that $N\geq 1$ and $k\geq 4$ . Suppose $f(x)$ is equipped with $k$ continuous derivatives and $f^{(k)}(x)$ is monotonic over the interval $(a,a+N]$ . Suppose further that there are numbers $\lambda_{k}>0$ and $\omega>1$ such that $\lambda_{k}\leq|f^{(k)}(x)|\leq\omega\lambda_{k}$ for all $x\in(a,a+N]$ . Then for any $\eta>0$ ,

\left|\sum_{n=a+1}^{a+N}e^{2\pi if(n)}\right|\leq A_{k}\omega^{2/K}N\lambda_{k}^{1/(2K-2)}+B_{k}N^{1-2/K}\lambda_{k}^{-1/(2K-2)},

where $K=2^{k-1}$ , $A_{k}$ and $B_{k}$ are defined recursively via the formulas

(92)

A_{k+1}(\eta,\omega):=\delta_{k}\left(\omega^{-1/K}+\frac{2^{19/12}(K-1)}{\sqrt{(2K-1)(4K-3)}}A_{k}(\eta,\omega)^{1/2}\right),

(93)

B_{k+1}(\eta):=\delta_{k}\frac{2^{3/2}(K-1)}{\sqrt{(2K-3)(4K-5)}}B_{k}(\eta)^{1/2},

with $A_{3}$ and $B_{3}$ as in Lemma 11, and $\delta_{k}$ defined by

(94)

\delta_{k}=\delta_{k}(\eta):=\sqrt{1+\frac{2}{2337^{1-2/K}}\left(\frac{9\pi}{1024}\eta\right)^{1/K}}.

Lemma 13.

Let $a$ , $b$ and $k$ be integers such that $a>0$ , $b>a$ and $k\geq 3$ . Let $t>0$ . Let $h>1$ be a number such that $b\leq ha$ . Then for any $\eta>0$ ,

\left|\sum_{a<n\leq b}\frac{1}{n^{1+it}}\right|\leq C_{k}a^{-k/(2K-2)}t^{1/(2K-2)}+D_{k}a^{k/(2K-2)-2/K}t^{-1/(2K-2)},

where

\begin{split}C_{k}=C_{k}(\eta,h)&:=A_{k}(\eta,h^{k})h^{2k/K-k/(2K-2)}(h-1)\left(\frac{(k-1)!}{2\pi}\right)^{\frac{1}{2K-2}},\\ D_{k}=D_{k}(\eta,h)&:=B_{k}(\eta)h^{k/(2K-2)}(h-1)^{1-2/K}\left(\frac{2\pi}{(k-1)!}\right)^{\frac{1}{2K-2}}.\end{split}

and $A_{k}(\eta,h^{k})$ and $B_{k}(\eta)$ defined in accordance with Lemma 12.

Proof.

This is a special case of Lemma 12 and Lemma 11 with the phase function

f(x)=-\frac{t}{2\pi}\log x,\qquad\text{and hence}\qquad f^{(k)}(x)=(-1)^{k}\frac{t}{2\pi}\frac{(k-1)!}{x^{k}}.

Since $a<b\leq ha$ by assumption, $\lambda_{k}\leq|f^{(k)}(x)|\leq\omega\lambda_{k}$ for $a<x\leq b$ , where

\lambda_{k}=\frac{t}{2\pi}\frac{(k-1)!}{h^{k}a^{k}},\qquad\omega=h^{k}.

Suppose that $k\geq 4$ . Then, we may apply Lemma 12 with these values of $\lambda_{k}$ and $\omega$ , and with $N=b-a$ . This furnishes a bound that increases with $N$ . Noting that $N\leq(h-1)a$ , we thus obtain

	$\displaystyle\left\|\sum_{a<n\leq b}\frac{1}{n^{it}}\right\|$	$\displaystyle\leq A_{k}(\eta,h^{k})(h^{k})^{2/K}(h-1)a\left(\frac{t}{2\pi}\frac{(k-1)!}{h^{k}a^{k}}\right)^{1/(2K-2)}$
		$\displaystyle\qquad\qquad+B_{k}(\eta)((h-1)a)^{1-2/K}\left(\frac{t}{2\pi}\frac{(k-1)!}{h^{k}a^{k}}\right)^{-1/(2K-2)}$
		$\displaystyle=C_{k}a^{1-k/(2K-2)}t^{1/(2K-2)}+D_{k}a^{1-2/K+k/(2K-2)}t^{-1/(2K-2)}.$

The desired result hence follows from partial summation in this case. If $k=3$ , then the result follows from Lemma 11 using a similar calculation. ∎

7.2. Background results related to Section 6

Here we give some preliminary lemmas for our proofs of estimates on $|\zeta^{\prime}(1+it)/\zeta(1+it)|$ and $|1/\zeta(1+it)|$ . We reiterate that Section 6 reuses variable names from previous sections. These variables mean different things in the confines of this section as this section is independent from Sections 3, 4, and 5.

The first two lemmas, which form the basis of our method, are built upon a theorem of Borel and Carathéodory. This theorem enables us to deduce an upper bound for the modulus of a function and its derivatives on a circle, from bounds for its real part on a larger concentric circle.

Theorem 14 (Borel–Carathéodory).

Let $s_{0}$ be a complex number. Let $R$ be a positive number, possibly depending on $s_{0}$ . Suppose that the function $f(s)$ is analytic in a region containing the disk $|s-s_{0}|\leq R$ . Let $M$ denote the maximum of $\textup{Re}\,f(s)$ on the boundary $|s-s_{0}|=R$ . Then, for any $r\in(0,R)$ and any $s$ such that $|s-s_{0}|\leq r$ ,

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

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

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

Proof.

Proofs for estimating both $|f(s)|$ and $|f^{\prime}(s)|$ are available in [Tit39, §5.5]. However, if we include the restriction $f(s_{0})=0$ , we can refer to [MV07, Lemma 6.2], which provides alternative proofs, resulting in a sharper estimate for $|f^{\prime}(s)|$ . ∎

Lemma 15.

Let $s_{0}$ be a complex number. Let $r$ and $\alpha$ be positive numbers (possibly depending on $s_{0}$ ) such that $\alpha<1/2$ . Suppose that the function $f(s)$ is analytic in a region containing the disk $|s-s_{0}|\leq r$ . Suppose further there is a number $A_{1}$ independent of $s$ such that,

\left|\frac{f(s)}{f(s_{0})}\right|\leq A_{1},\qquad|s-s_{0}|\leq r.

Then, for any $s$ in the disk $|s-s_{0}|\leq\alpha r$ we have

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

Proof.

We follow the proof in [Tit86, Section 3]. Let $g$ be the function

g(s)=f(s)\prod_{\rho}\frac{1}{s-\rho},

where $\rho$ in the (finite) product runs through the zeros of $f(s)$ , counted with multiplicity, that satisfy $|\rho-s_{0}|\leq\tfrac{1}{2}r$ . Since by construction the poles and zeros cancel, $g$ is analytic in $|s-s_{0}|\leq r$ , and does not vanish in the disk $|s-s_{0}|\leq\tfrac{1}{2}r$ . Therefore, the function

h(s)=\log\frac{g(s)}{g(s_{0})},

where the logarithm branch is determined by $h(s_{0})=0$ , is analytic in some region containing the disk $|s-s_{0}|\leq\tfrac{1}{2}r$ .

Now, on the circle $|s-s_{0}|=r$ , we have $|s-\rho|\geq\tfrac{1}{2}r\geq|s_{0}-\rho|$ . So, on this circle,

\left|\frac{g(s)}{g(s_{0})}\right|=\left|\frac{f(s)}{f(s_{0})}\prod_{\rho}\left(\frac{s_{0}-\rho}{s-\rho}\right)\right|\leq\left|\frac{f(s)}{f(s_{0})}\right|\leq A_{1}.

The inequality also holds in the interior of the circle, by the maximum modulus principle. Hence, $\textup{Re}\,h(s)\leq\log A_{1}$ throughout the disk $|s-s_{0}|\leq r$ .

Given this, we apply Theorem 14 to $h(s)$ with the concentric circles centered at $s_{0}$ of radii $\alpha r$ and $\frac{1}{2}r$ . This yields throughout the disk $|s-s_{0}|\leq\alpha r$ that

(95)

|h^{\prime}(s)|\leq\frac{4\log A_{1}}{r(1-2\alpha)^{2}}.

On the other hand, since $g(s_{0})$ is constant, we have

|h^{\prime}(s)|=\left|\frac{g^{\prime}(s)}{g(s)}\right|=\left|\frac{f^{\prime}(s)}{f(s)}-\sum_{\rho}\frac{1}{s-\rho}\right|.

Combining this formula with the inequality (95) completes the proof. ∎

Lemma 16.

Let $s$ and $s_{0}$ be complex numbers with real parts $\sigma$ and $\sigma_{0}$ , respectively. Let $r$ , $\alpha$ , $\beta$ , $A_{1}$ and $A_{2}$ be positive numbers, possibly depending on $s_{0}$ , such that $\alpha<1/2$ and $\beta<1$ . Suppose that the function $f(s)$ satisfies the conditions of Lemma 15 with $r$ , $\alpha$ and $A_{1}$ , and that

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

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

\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}.

Proof.

Using the inequality $-|z|\leq-\textup{Re}\,z\leq|z|$ , valid for any complex number $z$ , together with Lemma 15 in $|s-s_{0}|\leq\alpha r$ , we have throughout this disk,

(96)

\displaystyle-\textup{Re}\,\frac{f^{\prime}(s)}{f(s)}

\displaystyle\leq\frac{4\log A_{1}}{r(1-2\alpha)^{2}}-\sum_{\rho}\textup{Re}\,\frac{1}{s-\rho},

where $\rho$ runs through the zeros of $f$ (with multiplicity) satisfying $|\rho-s_{0}|\leq\tfrac{1}{2}r$ . By the assumption on the nonvanishing of $f$ , each $\rho$ in this sum satisfies $\textup{Re}\,\rho<\sigma_{0}-\alpha r$ . On the other hand, all $s$ in the disk $|s-s_{0}|\leq\alpha r$ clearly satisfy $\sigma\geq\sigma_{0}-\alpha r$ . Hence,

\sum_{\rho}\textup{Re}\,\frac{1}{s-\rho}=\sum_{\rho}\frac{\sigma-\textup{Re}\,\rho}{|s-\rho|^{2}}>0,

throughout the disk $|s-s_{0}|\leq\alpha r$ . We may thus drop the sum in (96) and still obtain a valid upper bound, call it $M_{1}$ .

Now, $-f^{\prime}(s)/f(s)$ is analytic in the disk $|s-s_{0}|\leq\alpha r$ since by assumption $f$ does not vanish there. We can therefore apply Theorem 14 to $-f^{\prime}(s)/f(s)$ using the two concentric circles $|s-s_{0}|\leq\alpha\beta r$ and $|s-s_{0}|\leq\alpha r$ , and the upper bound $M_{1}\geq M$ on the circle $|s-s_{0}|\leq\alpha r$ , which yields the result. ∎

Acknowledgements

We would like to thank Fatima Majeed for pointing out errors in an early version of this manuscript and giving us some helpful suggestions.

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Explicit bounds for the Riemann zeta-function on the 1-line

Abstract.

Key words and phrases:

2020 Mathematics Subject Classification:

1. Introduction

Theorem 1.

Theorem 2.

2. Discussion

Proposition 3.

3. Proof of Theorem 1

4. Bounding A0A_{0} and A2A_{2}

5. Proof of Proposition 3

6. Proof of Theorem 2

Lemma 4.

Proof.

Remark.

Lemma 5.

Proof.

6.1. Bounding |ζ′​(1+i​t)/ζ​(1+i​t)||\zeta^{\prime}(1+it)/\zeta(1+it)|

Lemma 6.

6.2. Bounding |1/ζ​(1+i​t)||1/\zeta(1+it)|

Lemma 7.

Proof.

6.3. Numeric calculations

Remark.

7. Background results

7.1. Background results related to Sections 3-5

Lemma 8.

Theorem 9.

Lemma 10 (Explicit third derivative test).

Remark.

Remark.

Lemma 11 (Another explicit third derivative test).

Lemma 12 (General explicit kk-th derivative test).

Lemma 13.

Proof.

7.2. Background results related to Section 6

Theorem 14 (Borel–Carathéodory).

Proof.

Lemma 15.

Proof.

Lemma 16.

Proof.

Acknowledgements

References

4. Bounding $A_{0}$ and $A_{2}$

6.1. Bounding $|\zeta^{\prime}(1+it)/\zeta(1+it)|$

6.2. Bounding $|1/\zeta(1+it)|$

Lemma 12 (General explicit $k$ -th derivative test).