# Explicit zero density for the Riemann zeta function

Habiba Kadiri  Department of Mathematics and Computer Science
University of Lethbridge
4401 University Drive
Lethbridge, Alberta
T1K 3M4 Canada<EMAIL_ADDRESS>,  Allysa Lumley  Department of
Mathematics and Statistics
York University
4700 Keele St
Toronto, Ontario
M3J 1P3 Canada<EMAIL_ADDRESS>and  Nathan Ng  Department of Mathematics and
Computer Science
University of Lethbridge
4401 University Drive
Lethbridge, Alberta
T1K 3M4 Canada<EMAIL_ADDRESS>
###### Abstract.

Let $N(\sigma,T)$ denote the number of nontrivial zeros of the Riemann zeta
function with real part greater than $\sigma$ and imaginary part between $0$
and $T$. We provide explicit upper bounds for $N(\sigma,T)$ commonly referred
to as a zero density result. In 1937, Ingham showed the following asymptotic
result $N(\sigma,T)=\mathcal{O}(T^{\frac{8}{3}(1-\sigma)}(\log T)^{5})$.
Ramaré recently proved an explicit version of this estimate. We discuss a
generalization of the method used in these two results which yields an
explicit bound of a similar shape while also improving the constants.

###### Key words and phrases:

Riemann zeta function, zero density, explicit results

###### 2010 Mathematics Subject Classification:

Primary 11M06, 11M26; Secondary 11Y35

Research for this article is partially supported by the NSERC Discovery grants
of H.K. (RGPIN-2015-06799) and N.N. (RGPIN-2015-05972). The calculations were
executed on the University of Lethbridge Number Theory Group Eudoxus machine,
supported by an NSERC RTI grant.

##  1\. Introduction

Throughout this article $\zeta(s)$ denotes the Riemann zeta function and
$\varrho$ denotes a non-trivial zero of $\zeta(s)$ lying in the critical
strip, $0<{\mathfrak{Re}}(s)<1$. Let $\frac{1}{2}<\sigma<1,T>0$, and define

(1.1) $N(\sigma,T)=\\#\\{\varrho=\beta+i\gamma:\
\zeta(\varrho)=0,0<\gamma<T\text{ and }\sigma<\beta<1\\}.$

We shall prove a non-trivial, explicit upper bound for $N(\sigma,T)$. Such a
bound is commonly referred to as a zero-density estimate. We denote RH the
Riemann Hypothesis and $\text{RH}(H_{0})$ the statement:

(1.2) $\text{RH}(H_{0}):\text{ all non-trivial zeros }\varrho\text{ of
}\zeta(s)\ \text{ with }\ |{\mathfrak{Im}}(\varrho)|\leq H_{0}\text{ satisfy
}{\mathfrak{Re}}(\varrho)=\frac{1}{2}.$

Currently, the best published value of $H_{0}$ for which (1.2) is true is due
to David Platt [19]:

$H_{0}=3.0610046\cdot 10^{10}$

with $N(H_{0})=103\,800\,788\,359$. Other strong evidence towards the RH is
the large body of zero-density estimates for $\zeta(s)$. Namely, very good
bounds for $N(\sigma,T)$ in various ranges of $\sigma$.
Let $\sigma>\frac{1}{2}$. In 1913 Bohr and Landau [2] showed that

(1.3) $N(\sigma,T)=\mathcal{O}\left(\frac{T}{\sigma-\frac{1}{2}}\right)$

for $T$ asymptotically large. This result implies that for any fixed
$\varepsilon>0$, almost all zeros of $\zeta(s)$ lie in the band
$|\frac{1}{2}-{\mathfrak{Re}}(s)|<\varepsilon$. This was improved in 1937 by
Ingham [12], who showed

(1.4) $N(\sigma,T)=\mathcal{O}\left(T^{(2+4c)(1-\sigma)}(\log T)^{5}\right)$

assuming that $\zeta(\frac{1}{2}+it)=\mathcal{O}\left(t^{c+\epsilon}\right)$.
In particular, the Lindelöf Hypothesis
$\zeta(\frac{1}{2}+it)=\mathcal{O}\left(t^{\epsilon}\right)$ implies that
$N(\sigma,T)=\mathcal{O}\left(T^{2(1-\sigma)+\epsilon}\right)$, also known as
the Density Hypothesis. There is a prolific literature on the bounds for
$\zeta(s)$, starting with the convexity bound of $c=\frac{1}{4}=0.25$
(Lindelöf), the first subconvexity bound of Hardy & Littlewood [8]
$c=\frac{1}{6}=0.1666\ldots$, to some more recent results of Huxley [10]
(2005) $c=\frac{32}{205}=0.1560\ldots$ and of Bourgain [3] (2017)
$c=\frac{13}{84}=0.1547\ldots$. In addition, there are also many articles on
estimates for $N(\sigma,T)$. A selection of some notable results may be found
in [10], [11], [13], and [3]. On the other hand, there are few explicit bounds
for $N(\sigma,T)$. We refer the reader to a result of the first author [14]
for an explicit version of Bohr and Landau’s bound. The method provides two
kind of results: for $T$ asymptotically large, as in $N(0.90,T)\leq
0.4421T+0.6443\log T-363\,301,$ and for $T$ taking a specific value, as in
$N(0.90,H_{0})<96.20$. These bounds are useful to improve estimates of prime
counting functions, as in [5], [4], [20], [26] and in [15] to find primes in
short intervals. Ramaré had earlier proven a version of (1.4) in his D.E.A.
memoire, which remained unpublished until recently. Let $\sigma\geq 0.52$ be
fixed. In [24] he proves 111Equation (1.1) [24, p. 326 ] gives the bound
$N(\sigma,T)\leq 4.9(3T)^{\frac{8(1-\sigma)}{3}}(\log T)^{5-2\sigma}+51.5(\log
T)^{2}$. However, there is a mistake in [24]. The authors have been in
communication with Professor Ramaré and he has sent us a proof of the revised
inequality (1.5).  that for any $T\geq 2000$

(1.5) $N(\sigma,T)\leq 965(3T)^{\frac{8(1-\sigma)}{3}}(\log
T)^{5-2\sigma}+51.5(\log T)^{2},$

which gives $N(0.90,T)<1293.48(\log
T)^{\frac{16}{5}}T^{\frac{4}{15}}+51.50(\log T)^{2},$ which gives the bound
for $T=H_{0}$: $N(0.90,H_{0})<2.1529\cdot 10^{10}.$ The purpose of this
article is to bound $N(\sigma,T)$ by applying Ingham’s argument with a general
weight and to improve both [14] and [24].

######  Theorem 1.1.

Let $\frac{10^{9}}{H_{0}}\leq k\leq 1,d>0,H\in[1002,H_{0})$, $\alpha>0$,
$\delta\geq 1$, $\eta_{0}=0.23622\ldots$, $1+\eta_{0}\leq\mu\leq 1+\eta$, and
$\eta\in(\eta_{0},\tfrac{1}{2})$ be fixed. Let
$\sigma>\frac{1}{2}+\frac{d}{\log H_{0}}$.
Then there exist $\mathcal{C}_{1},\mathcal{C}_{2}>0$ such that, for any $T\geq
H_{0}$,

(1.6) $N(\sigma,T)\leq\frac{(T-H)(\log T)}{2\pi
d}\log\Big{(}1+\frac{\mathcal{C}_{1}(\log(kT))^{2\sigma}(\log
T)^{4(1-\sigma)}T^{\frac{8}{3}(1-\sigma)}}{T-H}\Big{)}+\frac{\mathcal{C}_{2}}{2\pi
d}(\log T)^{2},$

where $\mathcal{C}_{1}=\mathcal{C}_{1}(\alpha,d,\delta,k,H,\sigma)$ and
$\mathcal{C}_{2}=\mathcal{C}_{2}(d,\eta,k,H,\mu,\sigma)$ are defined in (4.72)
and (4.73). Since $\log(1+x)\leq x$ for $x\geq 0$, (1.6) implies

(1.7) $N(\sigma,T)\leq\frac{\mathcal{C}_{1}}{2\pi d}(\log(kT))^{2\sigma}(\log
T)^{5-4\sigma}T^{\frac{8}{3}(1-\sigma)}+\frac{\mathcal{C}_{2}}{2\pi d}(\log
T)^{2}.$

In addition, numerical results are displayed in tables in Section 5.

For instance (1.7) gives $N(0.90,T)<11.499(\log
T)^{\frac{16}{5}}T^{\frac{4}{15}}+3.186(\log T)^{2},$ and (1.6) gives
$N(0.90,H_{0})<130.07.$ This improves previous results both numerically and
methodologically (one of the key ingredients is the choice of a more efficient
weight function in Ingham’s method). Note that choosing $k<1$ and optimizing
in $H$ can provide extra improvements to (1.5). In addition, we prove a
stronger bound for the argument of a holomorphic function. We now explain the
main ideas to prove Theorem 1.1.

##  2\. Setting up the proof

###  2.1. Littlewood’s classical method to count the zeros

Let $h(s)=\zeta(s)M(s)$ where $M(s)$ is entire and

(2.1)
$N_{h}(\sigma,T)=\\#\Big{\\{}\varrho^{\prime}=\beta^{\prime}+i\gamma^{\prime}\in\mathbb{C}\
:\ h(\varrho^{\prime})=0,\sigma<\beta^{\prime}<1,\text{ and
}0<\gamma^{\prime}<T\Big{\\}}.$

Then for a parameter $H\in(0,H_{0})$, we have by (1.2) that

$N(\sigma,T)=N(\sigma,T)-N(\sigma,H)\leq N_{h}(\sigma,T)-N_{h}(\sigma,H)$

for $T\geq H_{0}$. We compare the above number of zeros for $h$ to its
average:

$N_{h}(\sigma,T)-N_{h}(\sigma,H)\leq\frac{1}{\sigma-\sigma^{\prime}}\int_{\sigma^{\prime}}^{\mu}(N_{h}(\tau,T)-N_{h}(\tau,H))\,d\tau$

where $\mu>1$ and $\sigma^{\prime}$ is a parameter satisfying
$\frac{1}{2}<\sigma^{\prime}<\sigma$. Let $\mathcal{R}$ be the rectangle with
vertices $\sigma^{\prime}+iH$, $\mu+iH$, $\mu+iT$, and $\sigma^{\prime}+iT$.
We apply the classical lemma of Littlewood as stated in [25, (9.9.1)]:

(2.2)
$\int_{\sigma^{\prime}}^{\mu}\Big{(}N_{h}(\tau,T)-N_{h}(\tau,H)\Big{)}d\tau=-\frac{1}{2\pi
i}\int_{\mathcal{R}}\log h(s)ds.$

Thus

(2.3)
$N(\sigma,T)\leq\frac{1}{2\pi(\sigma-\sigma^{\prime})}\Big{(}\int_{H}^{T}\log|h(\sigma^{\prime}+it)|dt\Big{.}\\\
\Big{.}+\int_{\sigma^{\prime}}^{\mu}\arg
h(\tau+iT)d\tau-\int_{\sigma^{\prime}}^{\mu}\arg
h(\tau+iH)d\tau-\int_{H}^{T}\log|h(\mu+it)|dt\Big{)}.$

As $T$ grows larger, the main contribution arises from the first integral. The
second and third integrals can be treated by using a general result for
bounding $\arg f(s)$ for $f$ a holomorphic function. To do this we give an
improvement of a lemma of Titchmarsh [25, p. 213] (see Proposition 4.10 and
Corollary 4.11 below). The fourth integral can be estimated with a standard
mean value theorem for Dirichlet polynomials (see Lemma 3.6). A key goal is to
minimize the above expression over admissible functions $h$. We now give an
idea of how to estimate the first integral in (2.3).

###  2.2. How the second mollified moment of $\zeta(s)$ occurs

Let $X\geq 1$ be a parameter and define the mollifier to be

(2.4) $M_{X}(s)=\sum_{n\leq X}\frac{\mu(n)}{n^{s}}$

where $\mu(n)$ is the Möbius function. Note that this is a truncation of the
Dirichlet series for $\zeta(s)^{-1}$. These mollifiers were invented by Bohr
and Landau [2] to help control the size of $\zeta(s)$ in the critical strip.
Futhermore, let

(2.5) $f_{X}(s)=\zeta(s)M_{X}(s)-1.$

Note that the series expansion for $f_{X}$ is given by

(2.6) $\displaystyle f_{X}(s)=\sum_{n>X}\Big{(}\sum_{\begin{subarray}{c}d\mid
n\\\ d\leq X\end{subarray}}\mu(d)\Big{)}n^{-s}=\sum_{n\geq
1}\frac{\lambda_{X}(n)}{n^{s}},$ (2.7) with $\displaystyle\lambda_{X}(n)=0\
\text{ if }n\leq X,\ \ \lambda_{X}(n)=\sum_{\begin{subarray}{c}d\mid n\\\
d\leq X\end{subarray}}\mu(d)\ \text{ if }n>X.$

We shall choose $h=h_{X}$ with

(2.8) $h_{X}(s)=1-f_{X}(s)^{2}=\zeta(s)M_{X}(s)(2-\zeta(s)M_{X}(s)).$

Since we have

$\frac{1}{b-a}\int_{a}^{b}\log
f(t)dt\leq\log\left(\frac{1}{b-a}\int_{a}^{b}f(t)dt\right),$

for any $f$ non-negative and continuous, and $|h_{X}(s)|\leq
1+|f_{X}(s)|^{2}$, we deduce that

(2.9)
$\int_{H}^{T}\log\left(|h_{X}(\sigma^{\prime}+it)|\right)dt\leq(T-H)\log\left(1+\frac{1}{T-H}\int_{H}^{T}|f_{X}(\sigma^{\prime}+it)|^{2}dt\right).$

We denote

(2.10) $F_{X}(\sigma,T)=\int_{0}^{T}|f_{X}(\sigma+it)|^{2}dt\text{ where
}\sigma\geq\frac{1}{2}.$

To resume, the key point for getting a good bound on $N(\sigma,T)-N(\sigma,H)$
is to obtain a good bound for $F_{X}(\sigma,T)$. Following a classical method
due to Ingham we compare it to a smoothed version of itself.

###  2.3. Ingham’s smoothing method

Let $\sigma_{1}$ and $\sigma_{2}$ be such that $\sigma_{1}<\sigma<\sigma_{2}$.
Let $T>0$ and $g=g_{T}$ be a non-negative, real valued function, depending on
the parameter $T$, and holomorphic in
$\sigma_{1}\leq{\mathfrak{Re}}(s)\leq\sigma_{2}$. We define

(2.11)
$\mathcal{M}_{g,T}(X,\sigma)=\int_{-\infty}^{+\infty}|g(\sigma+it)|^{2}|f_{X}(\sigma+it)|^{2}dt.$

We shall consider $g$ of a special shape. For $\alpha,\beta>0$, assume that
there exist positive functions $\omega_{1},\omega_{2}$ such that $g$
satisfies, for all $\sigma\in[\sigma_{1},\sigma_{2}]$,

(2.12)
$\displaystyle|g(\sigma+it)|\leq\omega_{1}(\sigma,T,\alpha)e^{-\alpha\left(\frac{|t|}{T}\right)^{\beta}}\
\text{ for all}\ t,$ (2.13)
$\displaystyle\omega_{2}(\sigma,T,\alpha)\leq|g(\sigma+it)|\ \text{ for all}\
t\in[H,T].$

In addition, we assume that $|g|$ is even in $t$:

(2.14) $|g(\sigma-it)|=|g(\sigma+it)|\text{ for
}\sigma\in(\sigma_{1},\sigma_{2})\text{ and }t\in\mathbb{R}.$

Thus $F_{X}(\sigma,T)\ll_{g}\mathcal{M}_{g,T}(X,\sigma)$, and more precisely

(2.15)
$F_{X}(\sigma,T)\leq\frac{\mathcal{M}_{g,T}(X,\sigma)}{2(\omega_{2}(\sigma,T,\alpha))^{2}}.$

In this article, we shall choose a family of weights of the form

(2.16) $g(s)=g_{T}(s)=\frac{s-1}{s}e^{\alpha(\frac{s}{T})^{2}},\text{ where
}\alpha>0.$

These weights will satisfy the above conditions with $\beta=2$. We remark that
Ingham [12] made use of the weight $g(s)=\frac{s-1}{s\cos(\frac{1}{2T})}$ and
Ramaré [24] used $g(s)=\frac{s-1}{s(\cos s)^{\frac{1}{2T}}}$. These weights
satisfy (2.12) with $\beta=1$. We also studied the weights
$g(s)=\frac{s-1}{s(\cos s)^{\frac{\alpha}{T}}}$ and
$g(s)=\frac{s-1}{s(\cos\frac{\alpha}{T})}$. However, we obtained the best
results with $g$ given by (2.16). The functions $g$ are chosen so that for
fixed $\sigma$, $g(\sigma+it)$ behave likes the indicator function,
$\mathds{1}_{[0,T]}(t)$, and for $t$ large, $g(\sigma+it)$ has rapid decay.
Nevertheless, it is an open problem to determine the best weights $g$ to use
in this problem.

###  2.4. Final bound

Finally, to bound the integral $\mathcal{M}_{g,T}$, we appeal to a convexity
estimate for integrals (see [7]). For $\sigma_{2}>1$ (and $\sigma_{2}$ close
to $1$), if $\frac{1}{2}\leq\sigma\leq\sigma_{2}$, then

(2.17)
$\mathcal{M}_{g,T}(X,\sigma)\leq\mathcal{M}_{g,T}(X,\tfrac{1}{2})^{\frac{\sigma_{2}-\sigma}{\sigma_{2}-\frac{1}{2}}}\mathcal{M}_{g,T}(X,\sigma_{2})^{\frac{\sigma-\frac{1}{2}}{\sigma_{2}-\frac{1}{2}}}.$

The largest contribution arises from $\mathcal{M}_{g,T}(X,\frac{1}{2})$. To
bound this we make use of:

  * •

bounds (2.12), (2.13) for $g$ (see Lemma 3.7),

  * •

a version of Montgomery and Vaughan’s Mean Value Theorem for Dirichlet
polynomials (see Lemma 3.6),

  * •

bounds for arithmetic sums to bound the second moment of the mollifier $M_{X}$
(we use Ramaré’s bounds, see Lemma 3.3 and 3.4),

  * •

the most recent explicit subconvexity bound for the Riemann zeta function (due
to Hiary [9], see Lemma 3.2).

##  3\. Preliminary lemmas

###  3.1. Bounds for the Riemann zeta function

In this section we record a number of bounds for the zeta function. Rademacher
[22, Theorem 4] established the following explicit convexity bound.

######  Lemma 3.1.

For $-\frac{1}{2}\leq-\eta\leq\sigma\leq 1+\eta\leq\frac{3}{2}$, we have

(3.1) $|\zeta(s)|\leq
3\frac{|1+s|}{|1-s|}\left(\frac{|1+s|}{2\pi}\right)^{\frac{1}{2}(1-\sigma+\eta)}\zeta(1+\eta).$

The next lemma is an explicit version of van der Corput’s subconvexity bound
for $\zeta$ on the critical line, recently proven by Hiary. [9].

######  Lemma 3.2.

We have

(3.2) $\displaystyle\ |\zeta(\tfrac{1}{2}+it)|\leq a_{1}t^{\frac{1}{6}}\log t$
$\displaystyle\ \text{for all }\ t\geq 3,$ (3.3) $\displaystyle\max_{|t|\leq
T}|\zeta(\tfrac{1}{2}+it)|\leq a_{1}T^{\frac{1}{6}}\log T+a_{2}$
$\displaystyle\ \text{for all }\ T>0,$

with

(3.4) $a_{1}=0.63\text{ and }a_{2}=2.851.$

###### Proof of Lemma 3.2.

Statement (3.2) is [9, Theorem 1.1]. For $T\in[0,3]$, [9, Theorem 1.1]
provides that $|\zeta(\tfrac{1}{2}+it)|\leq 1.461$. We find that the minimum
of the function $t^{\frac{1}{6}}\log(t)$ occurs when $t=e^{-6}$. We require
the polynomial $a_{1}t^{\frac{1}{6}}\log(t)+a_{2}\geq 1.461$, choosing $a_{2}$
as in the statement of the lemma achieves this. ∎

###  3.2. Bounds for arithmetic sums

We list here some preliminary lemmas from [24] providing estimates for finite
arithmetic sums. Let

(3.5) $\begin{split}b_{1}=0.62,\ b_{2}=1.048,\ b_{3}=0.605,\ \text{and}\
b_{4}=0.529.\end{split}$

######  Lemma 3.3.

We have

(3.6) $\displaystyle\sum_{n\leq X}\mu^{2}(n)\leq b_{1}X\ \text{ for all }\
X\geq 1700,$ (3.7) $\displaystyle\sum_{n\leq
X}\frac{\mu^{2}(n)}{n}-\frac{6}{\pi^{2}}\log X\leq b_{2}\ \text{ for all }\
X\geq 1002.$

(3.6) is [24, Lemma 3.1] and (3.7) is [24, Lemma 3.4].

######  Lemma 3.4.

Let $\tau>1,\delta>0$, $X\geq 10^{9}$, and $\gamma$ denotes Euler’s constant.
Then

(3.8)
$\displaystyle\sum_{X<n<5X}\frac{\lambda_{X}(n)^{2}}{n^{2}}\leq\frac{b_{3}}{X},$
(3.9) $\displaystyle\sum_{n\geq
1}\frac{\lambda_{X}(n)^{2}}{n^{\tau}}\leq\frac{b_{4}\tau^{2}}{\tau-1}e^{\gamma(\tau-1)}\log
X,$ (3.10) $\displaystyle\sum_{n\geq
1}\frac{\lambda_{X}(n)^{2}}{n^{1+\frac{\delta}{\log
X}}}\leq\frac{b_{4}}{\delta}\Big{(}1+\frac{\delta}{\log
X}\Big{)}^{2}e^{\frac{\delta\gamma}{\log X}}(\log X)^{2},$ (3.11)
$\displaystyle\sum_{n\geq 1}\frac{\lambda_{X}(n)^{2}}{n^{2+\frac{2\delta}{\log
X}}}\leq\frac{b_{4}}{5\delta e^{\delta}}\Big{(}1+\frac{{\delta}}{\log
X}\Big{)}^{2}e^{\frac{{\delta}(\gamma-\log 5)}{\log X}}\frac{(\log
X)^{2}}{X}+\frac{b_{3}e^{-2{\delta}}}{X}.$

###### Proof.

(3.8) is [24, Lemma 5.6] and (3.9) is [24, Lemma 5.5]. (3.10) is a direct
consequence of (3.9), taking $\tau=1+\frac{{\delta}}{\log X}$.
For (3.11) we set $\tau=2+\frac{2{\delta}}{\log X}$. Since
$\lambda_{X}(n)^{2}=0$ when $1\leq n\leq X$, then

$\sum_{n\geq
1}\frac{\lambda_{X}(n)^{2}}{n^{\tau}}=\sum_{X<n<5X}\frac{\lambda_{X}(n)^{2}}{n^{\tau}}+\sum_{n\geq
5X}\frac{\lambda_{X}(n)^{2}}{n^{\tau}}.$

Since $\tau\geq 2$, we use (3.8) and find that the first sum is

$\leq\frac{1}{X^{\tau-2}}\sum_{X<n<5X}\frac{\lambda_{X}(n)^{2}}{n^{2}}\leq\frac{1}{X^{\tau-2}}\frac{b_{3}}{X}=\frac{b_{3}e^{-2{\delta}}}{X}.$

We bound the second sum using $n^{\tau}\geq(5X)^{1+\frac{{\delta}}{\log
X}}n^{1+\frac{{\delta}}{\log X}}$ and (3.10). We find that it is

$\leq\frac{1}{(5X)^{1+\frac{{\delta}}{\log
X}}}\frac{b_{4}}{{\delta}}\Big{(}1+\frac{{\delta}}{\log
X}\Big{)}^{2}e^{\frac{{\delta}\gamma}{\log X}}(\log X)^{2}.$

Combining bounds

$\sum_{n\geq
1}\frac{\lambda_{X}(n)^{2}}{n^{\tau}}\leq\frac{b_{4}}{5{\delta}e^{\delta}}\Big{(}1+\frac{{\delta}}{\log
X}\Big{)}^{2}e^{\frac{{\delta}(\gamma-\log 5)}{\log X}}\frac{(\log
X)^{2}}{X}+\frac{b_{3}e^{-2{\delta}}}{X}.$

∎

######  Lemma 3.5.

Let $\tau>1$ and $\gamma$ is Euler’s constant. Then for $X\geq 1$,

(3.12) $\sum_{n\geq
X}\frac{d(n)}{n^{\tau}}\leq\frac{\tau}{X^{\tau-1}}\left(\frac{\log
X}{\tau-1}+\frac{1}{(\tau-1)^{2}}+\frac{\gamma}{\tau-1}+\frac{7}{12\tau
X}\right)\\\ $

and for $X\geq 47$,

(3.13) $\sum_{n\geq
X}\frac{d(n)^{2}}{n^{\tau}}\leq\frac{2\tau}{X^{\tau-1}}\Big{(}\frac{(\log
X)^{3}}{\tau-1}+\frac{3\log^{2}X}{(\tau-1)^{2}}+\frac{6\log
X}{(\tau-1)^{3}}+\frac{6}{(\tau-1)^{4}}\Big{)}.$

###### Proof.

By partial summation, we have

$\displaystyle\sum_{n\geq
X}\frac{d(n)}{n^{\tau}}\leq\tau\int_{X}^{\infty}\frac{\sum_{n\leq
t}d(n)}{t^{\tau+1}}dt.$

Using $\sum_{n\leq t}d(n)\leq t(\log t+\gamma+\frac{7}{12t})$, for $t\geq 1$,
which follows from [23, Equation 3.1], we have

$\displaystyle\sum_{n\geq
X}\frac{d(n)}{n^{\tau}}\leq\tau\left(\int_{X}^{\infty}\frac{\log
t}{t^{\tau}}dt+\gamma\int_{X}^{\infty}\frac{dt}{t^{\tau}}+\frac{7}{12}\int_{X}^{\infty}\frac{dt}{t^{\tau+1}}\right).$

By applying the integrals

$\int_{X}^{\infty}\frac{\log t}{t^{c}}dt=\frac{\log
X}{(c-1)X^{c-1}}+\frac{1}{(c-1)^{2}X^{c-1}}\text{ and
}\int_{X}^{\infty}\frac{dt}{t^{c}}=\frac{1}{(c-1)X^{c-1}},\text{ where }c>1,$

we obtain (3.12). The second estimate is similar. We have

$\sum_{n\geq
X}\frac{d(n)^{2}}{n^{\tau}}\leq\tau\int_{X}^{\infty}\frac{\sum_{n\leq
t}d(n)^{2}}{t^{\tau+1}}\,dt.$

It suffices to use the elementary bound $\sum_{n\leq t}d(n)^{2}\leq t(\log
t+1)^{3}\leq 2t\log^{3}t$ for $t\geq 47$, derived by Gowers [6]. Thus

$\sum_{n\geq X}\frac{d(n)^{2}}{n^{\tau}}\leq
2\tau\int_{X}^{\infty}\frac{\log^{3}t}{t^{\tau}}\,dt=2\tau\left(\frac{\frac{(\log
X)^{3}}{\tau-1}+\frac{3\log^{2}X}{(\tau-1)^{2}}+\frac{6\log
X}{(\tau-1)^{3}}+\frac{6}{(\tau-1)^{4}}}{X^{\tau-1}}\right).$

∎

###  3.3. Mean value theorem for Dirichlet polynomials

We require Montgomery and Vaughan’s mean value theorem for Dirichlet
polynomials in the form derived by Ramaré [24].

######  Lemma 3.6.

Let $(u_{n})$ be a real-valued sequence. For every $T\geq 0$ we have

(3.14)
$\int_{0}^{T}\Big{|}\sum_{n=1}^{\infty}u_{n}n^{it}\Big{|}^{2}dt\leq\sum_{n\geq
1}|u_{n}|^{2}(T+\pi m_{0}(n+1)),$

with

(3.15) $m_{0}=\sqrt{1+\frac{2}{3}\sqrt{\frac{6}{5}}}.$

Let $0<T_{1}<T_{2}$. Then

(3.16)
$\int_{T_{1}}^{T_{2}}|\sum_{n=1}^{\infty}u_{n}n^{it}|^{2}dt\leq\sum_{n\geq
1}|u_{n}|^{2}(T_{2}-T_{1}+2\pi m_{0}(n+1)).$

###### Proof.

The inequality (3.14) is [24, Lemma 6.5], and (3.16) follows by the same
proof. This argument is an explicit version of Corollary 3 of [18] which makes
use of the main theorem of [21]. Note that (3.16) follows from two
applications of (3.14). ∎

###  3.4. Choice for the smooth weight $g$

######  Lemma 3.7.

Let $\alpha>0$ and $\beta=2$. Let $s=\sigma+it$ and let $g$ be as defined in
(2.16):

(3.17) $g(s)=\frac{s-1}{s}e^{\alpha\left(\frac{s}{T}\right)^{2}}.$

Let $\sigma_{1}=\frac{1}{2},\sigma_{2}>1$, and $H<T$. Define

(3.18) $\displaystyle\omega_{1}(\sigma,T,\alpha)$
$\displaystyle=e^{\alpha\left(\frac{\sigma}{T}\right)^{2}},$ (3.19)
$\displaystyle\omega_{2}(\sigma,T,\alpha)$
$\displaystyle=\left(1-\frac{1}{H}\right)e^{\alpha\left(\frac{\sigma}{T}\right)^{2}-\alpha}.$

Then for $\frac{1}{2}\leq\sigma\leq\sigma_{2}$, $g$ satisfies (2.12) and
(2.13):

(3.20)
$\displaystyle|g(\sigma+it)|\leq\omega_{1}(\sigma,T,\alpha)e^{-\alpha\left(\frac{|t|}{T}\right)^{2}}$
$\displaystyle\text{ for all }\ t,$ (3.21)
$\displaystyle\omega_{2}(\sigma,T,\alpha)\leq$ $\displaystyle|g(\sigma+it)|$
$\displaystyle\text{ for }\ H\leq t\leq T.$

###### Proof.

Since $\sigma\geq\frac{1}{2}$, we have
$\left|\frac{s-1}{s}\right|^{2}=1-\frac{2\sigma-1}{\sigma^{2}+t^{2}}\leq 1.$
Thus
$|g(s)|\leq|e^{\alpha(\frac{s}{T})^{2}}|=e^{\frac{\alpha\sigma^{2}}{T^{2}}}e^{\frac{-\alpha
t^{2}}{T^{2}}}$ and we have the expression for $\omega_{1}(\sigma,T,\alpha)$.
In addition, $|\frac{s-1}{s}|=|1-\frac{1}{s}|\geq 1-\frac{1}{|s|}\geq
1-\frac{1}{|t|}$, so for all $t\in[H,T]$, we have

$|g(s)|\geq\left(1-|t|^{-1}\right)e^{\frac{\alpha\sigma^{2}}{T^{2}}}e^{\frac{-\alpha
t^{2}}{T^{2}}}\geq(1-H^{-1})e^{\frac{\alpha\sigma^{2}}{T^{2}}}e^{-\alpha},$

which gives $\omega_{2}(\sigma,T,\alpha)$. ∎

##  4\. Proof of the Main Theorem

Unless specified in the rest of the article, we set $H_{0}=3.0610046\cdot
10^{10}$ and we have the following conditions on the parameters
$k,\sigma_{1},\delta$, and $\sigma_{2}$:

(4.1) $\displaystyle k\geq\frac{10^{9}}{H_{0}},\ \sigma_{1}=\frac{1}{2},\
\delta>0,\ \text{and}\ \sigma_{2}=1+\frac{{\delta}}{\log X}.$

###  4.1. Bounding $F_{X}(\sigma,T)$

We establish here some preliminary lemmas to estimate $F_{X}(\sigma,T)$ at
$\frac{1}{2}$ and at $1+\frac{{\delta}}{\log X}$.

####  4.1.1. Bounding $F_{X}(\frac{1}{2},T)$

We first need to bound the second moment of $M_{X}(\frac{1}{2}+it)$, where
$M_{X}$ is defined in (2.4).

######  Lemma 4.1.

Let $T>0$, $X\geq kH_{0}$, and $k$ satisfies (4.1). Then

(4.2)
$\int_{0}^{T}\left|M_{X}(\tfrac{1}{2}+it)\right|^{2}dt\leq(C_{1}T+C_{2}X)(\log
X),$

where

(4.3) $\displaystyle C_{1}$
$\displaystyle=C_{1}(k)=\frac{6}{\pi^{2}}+\frac{b_{2}}{\log(kH_{0})},$ (4.4)
$\displaystyle C_{2}$ $\displaystyle=C_{2}(k)=\frac{\pi
m_{0}b_{1}}{\log(kH_{0})}+\frac{6m_{0}}{\pi kH_{0}}+\frac{\pi
m_{0}b_{2}}{kH_{0}\log(kH_{0})},$

and the $b_{i}$’s are defined in (3.5) and $m_{0}$ in (3.15).

###### Proof.

We apply (3.14) to $u_{n}=\frac{\mu(n)}{n^{\frac{1}{2}}}$:

$\int_{0}^{T}\left|M_{X}\left(\tfrac{1}{2}+it\right)\right|^{2}dt\leq\sum_{n\leq
X}\frac{\mu^{2}(n)}{n}(T+\pi m_{0}(n+1)).$

Since $X\geq 1700$, we apply (3.6) to $(T+\pi m_{0})\sum_{n\leq
X}\frac{\mu^{2}(n)}{n}$ and (3.7) to $(\pi m_{0})\sum_{n\leq X}\mu^{2}(n)$
respectively. We factor $\log X$ to give

$\displaystyle\int_{0}^{T}\left|M_{X}\left(\tfrac{1}{2}+it\right)\right|^{2}dt$
$\displaystyle=(T+\pi m_{0})\left(\frac{6}{\pi^{2}}\log X+b_{2}\right)+\pi
m_{0}b_{1}X$ $\displaystyle=\left(\left(\frac{6}{\pi^{2}}+\frac{b_{2}}{\log
X}\right)T+\left(\frac{6m_{0}}{\pi X}+\frac{\pi m_{0}b_{2}}{X\log X}+\frac{\pi
m_{0}b_{1}}{\log X}\right)X\right)(\log X),$

and use the fact that $X\geq kH_{0}$ to obtain the announced bound. ∎

######  Lemma 4.2.

Let $T>0$, $X\geq kH_{0}$, and $k$ satisfies (4.1). Then

(4.5) $F_{X}(\tfrac{1}{2},T)\leq C_{4}\left(T^{\frac{1}{6}}\log
T+\frac{a_{2}}{a_{1}}\right)^{2}\left(T+\frac{C_{2}}{C_{1}}X\right)(\log X),$

where $a_{1},a_{2}$ are defined in (3.4), $C_{1}$ in (4.3), $C_{2}$ in (4.4),
and

(4.6) $\displaystyle a_{3}=-\frac{6a_{1}}{e}+a_{2},$ (4.7) $\displaystyle
C_{3}=C_{3}(k)=a_{3}^{2}C_{1}(k)\log(kH_{0}),$ (4.8) $\displaystyle
C_{4}=C_{4}(k)=C_{1}(k)a_{1}^{2}\left(1+\frac{1}{\sqrt{C_{3}(k)}}\right)^{2}.$

###### Proof.

We have from the definition of $F_{X}(\sigma,T)$ given as (2.10) and
Minkowski’s inequality that

$\sqrt{|F_{X}(\tfrac{1}{2},T)|}\leq\sqrt{\int_{0}^{T}|\zeta(\tfrac{1}{2}+it)M_{X}(\tfrac{1}{2}+it)|^{2}dt}+\sqrt{T}.$

To the last integral we apply Hiary’s subconvexity bound (3.3) to bound zeta
and (4.2) to bound the mean square of $M_{X}$. We let $I_{0}$ denote the
resulting bound so that

$I_{0}=(a_{1}T^{\frac{1}{6}}\log T+a_{2})^{2}(C_{1}T+C_{2}X)(\log X),$

and thus

$|F_{X}(\tfrac{1}{2},T)|\leq\left(\sqrt{I_{0}}+\sqrt{T}\right)^{2}=I_{0}\left(1+\sqrt{\frac{T}{I_{0}}}\right)^{2}.$

We note that $a_{1}T^{\frac{1}{6}}\log T+a_{2}$ is minimized at $T=e^{-6}$ and
we let $a_{3}$ represent this minimum. Then

$I_{0}\geq a_{3}^{2}(C_{1}T+C_{2}X)\log X\geq a_{3}^{2}C_{1}T\log X.$

We conclude with the lower bound $\frac{I_{0}}{T}\geq
a_{3}^{2}C_{1}\log(kH_{0})$, which is labeled $C_{3}$, and

$I_{0}=C_{1}a_{1}^{2}\left(T^{\frac{1}{6}}\log
T+\frac{a_{2}}{a_{1}}\right)^{2}\left(T+\frac{C_{2}}{C_{1}}X\right)(\log X),$

which completes the proof. ∎

####  4.1.2. Bounding $F_{X}(\sigma_{2},T)$ at
$\sigma_{2}=1+\frac{\delta}{\log X}$

######  Lemma 4.3.

Let $T>0$, $X\geq kH_{0}$ and $k,\delta,\sigma_{2}$ satisfy (4.1). Then

(4.9)
$F_{X}(\sigma_{2},T)\leq\left(C_{5}(k,{\delta})+\frac{C_{6}(k,{\delta})(T+\pi
m_{0})}{X}\right)(\log X)^{2},$

where

(4.10) $\displaystyle C_{5}(k,{\delta})=\frac{\pi
m_{0}b_{4}}{2{\delta}}\Big{(}1+\frac{2{\delta}}{\log(kH_{0})}\Big{)}^{2}e^{\frac{2{\delta}\gamma}{\log(kH_{0})}},$
(4.11) $\displaystyle
C_{6}(k,{\delta})=\frac{b_{4}}{5{\delta}e^{\delta}}\Big{(}1+\frac{{\delta}}{\log(kH_{0})}\Big{)}^{2}+\frac{b_{3}e^{-2{\delta}}}{(\log(kH_{0}))^{2}},$

the $b_{i}$’s are defined in (3.5), $m_{0}$ in (3.15), and $\gamma$ is Euler’s
constant.

###### Proof.

Recall that $F_{X}$ is defined by (2.10) and by (2.6) we have

$F_{X}(\sigma_{2},T)=\int_{0}^{T}|f_{X}(\sigma_{2}+it)|^{2}dt=\int_{0}^{T}\Big{|}\sum_{n\geq
1}\frac{\lambda_{X}(n)}{n^{\sigma_{2}+it}}\Big{|}^{2}dt.$

Inequality (3.14) implies the bound

$F_{X}(\sigma_{2},T)\leq\pi m_{0}\sum_{n\geq
1}\frac{\lambda_{X}(n)^{2}}{n^{2\sigma_{2}-1}}+(T+\pi m_{0})\sum_{n\geq
1}\frac{\lambda_{X}(n)^{2}}{n^{2\sigma_{2}}}.$

For $2\sigma_{2}-1=1+\frac{2{\delta}}{\log X}$ and
$2\sigma_{2}=2+\frac{2{\delta}}{\log X}$, we apply the bounds for arithmetic
sums (3.10) and (3.11) to respectively bound the two above sums. Thus

$\displaystyle\sum_{n\geq
1}\frac{\lambda_{X}(n)^{2}}{n^{1+2\frac{{\delta}}{\log
X}}}\leq\frac{b_{4}}{2{\delta}}\Big{(}1+\frac{2{\delta}}{\log
X}\Big{)}^{2}e^{\frac{2{\delta}\gamma}{\log X}}(\log X)^{2},$ and
$\displaystyle\sum_{n\geq
1}\frac{\lambda_{X}(n)^{2}}{n^{2+\frac{2{\delta}}{\log
X}}}\leq\frac{b_{4}}{5{\delta}e^{\delta}}\Big{(}1+\frac{{\delta}}{\log
X}\Big{)}^{2}\frac{(\log X)^{2}}{X}+\frac{b_{3}e^{-2{\delta}}}{X}.$

We combine these results and use the fact that $X\geq kH_{0}$ to complete the
proof. ∎

From here we may derive a bound for $\mathcal{M}_{g,T}(X,\sigma)$.

###  4.2. Explicit upper bounds for the mollifier
$\mathcal{M}_{g,T}(X,\sigma)$

The results in this section are proven for a general weight $g$ satisfying the
conditions described in Section 2.3. In [7, Theorem 7], Hardy et al. proved
the following convexity estimate:

######  Lemma 4.4.

Let $\frac{1}{2}\leq\sigma_{1}<1<\sigma_{2}$, let $T>0$, and $X>1$. Then

(4.12)
$\mathcal{M}_{g,T}(X,\sigma)\leq\mathcal{M}_{g,T}(X,\sigma_{1})^{\frac{\sigma_{2}-\sigma}{\sigma_{2}-\sigma_{1}}}\mathcal{M}_{g,T}(X,\sigma_{2})^{\frac{\sigma-\sigma_{1}}{\sigma_{2}-\sigma_{1}}}.$

In order to obtain a bound for the mollifier $\mathcal{M}_{g,T}(X,\sigma)$
inside the strip $\frac{1}{2}\leq\sigma\leq 1+\frac{{\delta}}{\log X}$, we
need explicit bounds at the extremities $\frac{1}{2}$ and
$1+\frac{{\delta}}{\log X}$.

######  Lemma 4.5.

Let $T>0$, $X>0,\sigma\geq\frac{1}{2}$, and let $g$ satisfy conditions (2.12)
and (2.14). Then

(4.13) $\mathcal{M}_{g,T}(X,\sigma)\leq
4\omega_{1}(\sigma,T,\alpha)^{2}\alpha\beta\int_{0}^{\infty}x^{\beta-1}e^{-2\alpha
x^{\beta}}F_{X}(\sigma,xT)dx.$

###### Proof.

By (2.14) and $|g(\sigma+it)|=|g(\sigma-it)|$ for $t\in\mathbb{R}$ and by an
application of (2.12) to the weight $g$ in the definition (2.11) of
$\mathcal{M}_{g,T}(X,\sigma)$, we have

(4.14) $\mathcal{M}_{g,T}(X,\sigma)\leq
2\omega_{1}(\sigma,T,\alpha)^{2}\int_{0}^{\infty}e^{-2\alpha\left(\frac{t}{T}\right)^{\beta}}|f_{X}(\sigma+it)|^{2}dt.$

Note that $\int_{0}^{U}|f_{X}(\sigma+it)|^{2}dt=F_{X}(\sigma,U)$ with
$F_{X}(\sigma,0)=0$ and
$\displaystyle{\lim_{U\to\infty}\Big{(}F_{X}(\sigma,U)e^{-2\alpha\left(\frac{U}{T}\right)^{\beta}}\Big{)}=0}$.
Integrating by parts gives

$\displaystyle\int_{0}^{\infty}e^{-2\alpha\left(\frac{t}{T}\right)^{\beta}}|f_{X}(\sigma+it)|^{2}dt$
$\displaystyle=2\alpha\beta\int_{0}^{\infty}\Big{(}\frac{t}{T}\Big{)}^{\beta}e^{-2\alpha\left(\frac{t}{T}\right)^{\beta}}F_{X}(\sigma,t)\frac{dt}{t}$
$\displaystyle=2\alpha\beta\int_{0}^{\infty}x^{\beta}e^{-2\alpha
x^{\beta}}F_{X}(\sigma,xT)\frac{dx}{x},$

by the variable change $x=\frac{t}{T}$. This combined with (4.14) yields the
announced (4.13). ∎

####  4.2.1. Bounding $\mathcal{M}_{g,T}(X,\frac{1}{2})$

Let $\alpha,\beta,A>0$ and let $n$ be a non-negative integer. We define

(4.15) $I(A,n)=\int_{0}^{\infty}x^{A}e^{-2\alpha x^{\beta}}(\log x)^{n}dx.$

In our context, $I(A,n)$ is a constant depending on parameters $A$ and $n$ and
is $\mathcal{O}(1)$ in comparison with $T$. The change of variable $y=2\alpha
x^{B}$ leads to the identity

(4.16)
$I(A,n)=(2\alpha)^{-\frac{A+1}{\beta}}\beta^{-(n+1)}\sum_{j=0}^{n}\binom{n}{j}(-\log(2\alpha))^{j}\Gamma^{(n-j)}\left(\frac{A+1}{\beta}\right),$

where $\Gamma^{(j)}(z)$ denotes the $j$-th derivative of Euler’s gamma
function. We also define

(4.17)
$\begin{split}\mathcal{J}(k,T)=&I(\beta+\tfrac{1}{3},0)+\frac{C_{2}}{C_{1}}kI(\beta-\tfrac{2}{3},0)+\frac{2I(\beta+\frac{1}{3},1)+2\frac{C_{2}}{C_{1}}kI(\beta-\frac{2}{3},1)}{(\log
T)}\\\
&+\frac{I(\beta+\frac{1}{3},2)+\frac{C_{2}}{C_{1}}kI(\beta-\frac{2}{3},2)}{(\log
T)^{2}}+\frac{2a_{2}\left(I(\beta+\frac{1}{6},0)+\frac{C_{2}k}{C_{1}}I(\beta-\frac{5}{6},0)\right)}{a_{1}T^{\frac{1}{6}}(\log
T)}\\\
&+\frac{2a_{2}\left(I(\beta+\frac{1}{6},1)+\frac{C_{2}k}{C_{1}}I(\beta-\frac{5}{6},1)\right)}{a_{1}T^{\frac{1}{6}}(\log
T)^{2}}+\frac{a_{2}^{2}\left(I(\beta,0)+\frac{C_{2}k}{C_{1}}I(\beta-1,0)\right)}{a_{1}^{2}T^{\frac{1}{3}}(\log
T)^{2}},\end{split}$ (4.18) $\mathcal{U}(\alpha,k,T)=4\alpha\beta
C_{4}\omega_{1}(\tfrac{1}{2},T,\alpha)^{2}\mathcal{J}(k,T),$

where $\omega_{1}$ and $C_{4}$ are respectively defined in (3.18) and (4.8).
We remark that in the case of our weight $g$, we have $\beta=2$. Thus in our
calculations of $\mathcal{J}(k,T)$ we specialize to $\beta=2$.

######  Lemma 4.6.

Let $\alpha,\beta>0$ and $g$ be a function satisfying (2.12) and (2.14). Let
$T\geq H_{0}$, $X=kT$, and $k$ satisfies (4.1). Then

$\mathcal{M}_{g,T}(X,\tfrac{1}{2})\leq\mathcal{U}(\alpha,k,T)(\log(kT))(\log
T)^{2}T^{\frac{4}{3}}.$

###### Proof.

We combine the bound (4.13) for $\mathcal{M}_{g,T}$ with the bound (4.5) for
$F_{X}(\frac{1}{2},xT)$:

$\mathcal{M}_{g,T}(X,\tfrac{1}{2})\leq 4\alpha\beta
C_{4}\omega_{1}(\tfrac{1}{2},T,\alpha)^{2}(\log
X)\left\\{T^{\frac{4}{3}}\int_{0}^{\infty}x^{\beta+\frac{1}{3}}(\log(xT))^{2}e^{-2\alpha
x^{\beta}}dx\right.\\\
\left.+\frac{2a_{2}}{a_{1}}T^{\frac{7}{6}}\int_{0}^{\infty}x^{\beta+\frac{1}{6}}\log(xT)e^{-2\alpha
x^{\beta}}dx+\frac{a_{2}^{2}}{a_{1}^{2}}T\int_{0}^{\infty}x^{\beta}e^{-2\alpha
x^{\beta}}dx\right.\\\
\left.+\frac{C_{2}}{C_{1}}XT^{\frac{1}{3}}\int_{0}^{\infty}x^{\beta-\frac{2}{3}}(\log(xT))^{2}e^{-2\alpha
x^{\beta}}dx+\frac{2a_{2}}{a_{1}}\frac{C_{2}}{C_{1}}XT^{\frac{1}{6}}\int_{0}^{\infty}x^{\beta-\frac{5}{6}}\log(xT)e^{-2\alpha
x^{\beta}}dx\right.\\\
\left.+\frac{a_{2}^{2}}{a_{1}^{2}}\frac{C_{2}}{C_{1}}X\int_{0}^{\infty}x^{\beta-1}e^{-2\alpha
x^{\beta}}dx\right\\}.$

We also use the fact that $(\log(xT))^{2}=(\log x)^{2}+2(\log x)(\log T)+(\log
T)^{2}$ and obtain

$\mathcal{M}_{g,T}(X,\tfrac{1}{2})\leq 4\alpha\beta
C_{4}\omega_{1}(\tfrac{1}{2},T,\alpha)^{2}(\log
X)\left\\{T^{\frac{4}{3}}\left(I(\beta+\tfrac{1}{3},2)+2(\log
T)I(\beta+\tfrac{1}{3},1)\right.\right.\\\ \left.\left.+(\log
T)^{2}I(\beta+\tfrac{1}{3},0)\right)+\frac{2a_{2}}{a_{1}}T^{\frac{7}{6}}\left(I(\beta+\tfrac{1}{6},1)+(\log
T)I(\beta+\tfrac{1}{6},0)\right)+\frac{a_{2}^{2}}{a_{1}^{2}}TI(\beta,0)\right.\\\
\left.+\frac{C_{2}}{C_{1}}XT^{\frac{1}{3}}\left(I(\beta-\tfrac{2}{3},2)+2(\log
T)I(\beta-\tfrac{2}{3},1)+(\log T)^{2}I(\beta-\tfrac{2}{3},0)\right)\right.\\\
\left.+\frac{2a_{2}}{a_{1}}\frac{C_{2}}{C_{1}}XT^{\frac{1}{6}}\left(I(\beta-\tfrac{5}{6},1)+(\log
T)I(\beta-\tfrac{5}{6},0)\right)+\frac{a_{2}^{2}}{a_{1}^{2}}\frac{C_{2}}{C_{1}}XI(\beta-1,0)\right\\},$

where $I$ is the integral defined in (4.15). At this point we choose $X=kT$ so
as to optimize the above bound, and we factor out the main term
$T^{\frac{4}{3}}(\log T)^{2}$:

$\mathcal{M}_{g,T}(X,\tfrac{1}{2})\leq 4\alpha\beta
C_{4}\omega_{1}(\tfrac{1}{2},T,\alpha)^{2}(\log(kT))(\log
T)^{2}T^{\frac{4}{3}}\left\\{I(\beta+\tfrac{1}{3},0)+\frac{kC_{2}}{C_{1}}I(\beta-\tfrac{2}{3},0)\right.\\\
\left.+2\frac{I(\beta+\tfrac{1}{3},1)+\frac{kC_{2}}{C_{1}}I(\beta-\tfrac{2}{3},1)}{(\log
T)}+\frac{I(\beta+\frac{1}{3},2)+\frac{kC_{2}}{C_{1}}I(\beta-\frac{2}{3},2)}{(\log
T)^{2}}+\frac{2a_{2}}{a_{1}}\frac{I(\beta+\frac{1}{6},0)+\frac{kC_{2}}{C_{1}}I(\beta-\frac{5}{6},0)}{(\log
T)T^{\frac{1}{6}}}\right.\\\
\left.+\frac{2a_{2}}{a_{1}}\frac{I(\beta+\frac{1}{6},1)+\frac{kC_{2}}{C_{1}}I(\beta-\frac{5}{6},1)}{(\log
T)^{2}T^{\frac{1}{6}}}+\frac{a_{2}^{2}}{a_{1}^{2}}\frac{I(\beta,0)+\frac{kC_{2}}{C_{1}}I(\beta-1,0)}{(\log
T)^{2}T^{\frac{1}{3}}}\right\\}.$

We recognize in the above term between brackets $\mathcal{J}(k,T)$ as
introduced in (4.17). ∎

####  4.2.2. Bounding $\mathcal{M}_{g,T}(X,\sigma_{2})$ at
$\sigma_{2}=1+\frac{\delta}{\log X}$

######  Lemma 4.7.

Let $g$ be as defined in Lemma 3.7. Let $T\geq H_{0}$, $X=kT$, and
$k,\delta,\sigma_{2}$ satisfy (4.1). Then

$\mathcal{M}_{g,T}(X,\sigma_{2})\leq\mathcal{V}(\alpha,k,{\delta},T)(\log(kT))^{2},$

where

(4.19)
$\displaystyle\mathcal{V}(\alpha,k,{\delta},T)=8\alpha\omega_{1}(\sigma_{2},T,\alpha)^{2}\mathcal{K}(k,{\delta},T),$
(4.20)
$\displaystyle\mathcal{K}(k,{\delta},T)=\left(C_{5}(k,{\delta})+\frac{C_{6}(k,{\delta})\pi
m_{0}}{kT}\right)I(1,0)+\frac{C_{6}(k,{\delta})}{k}I(2,0),$

and $m_{0},\omega_{1},C_{5},C_{6},$ and $I$ are respectively defined in
(3.15), (3.18), (4.10), (4.11), and (4.15).

###### Proof.

We combine the bound (4.13) for $\mathcal{M}_{g,T}$ with the bound (4.9) for
$F_{X}(\sigma_{2},xT)$ (since $X\geq kH_{0}$) to obtain

(4.21) $\mathcal{M}_{g,T}(X,\sigma_{2})\leq
4\alpha\beta\omega_{1}(\sigma_{2},T,\alpha)^{2}\Bigg{(}\int_{0}^{\infty}x^{\beta-1}e^{-2\alpha
x^{\beta}}\Big{(}C_{5}(k,\delta)+\frac{C_{6}(k,\delta)(xT+\pi
m_{0})}{X}\Big{)}(\log X)^{2}dx\Bigg{)}.$

Rearranging this and recalling the definition for $I$ in (4.15) we obtain

$\mathcal{M}_{g,T}(X,\sigma_{2})\leq
4\alpha\beta\omega_{1}(\sigma_{2},T,\alpha)^{2}(\log
X)^{2}\left(\left(C_{5}(k,{\delta})+\frac{C_{6}(k,{\delta})\pi
m_{0}}{X}\right)I(\beta-1,0)\right.\\\
\left.+\frac{C_{6}(k,{\delta})T}{X}I(\beta,0)\right).$

We conclude by noting that $X=kT$ and for our $g$, $\beta=2$. ∎

####  4.2.3. Conclusion

Finally, we provide bounds for $\mathcal{M}_{g,T}$.

######  Lemma 4.8.

Let $g$ be as defined in Lemma 3.7. Let $T\geq H_{0}$, $X=kT$, and $k$
satisfies (4.1). Assume $\frac{1}{2}\leq\sigma\leq 1+\frac{\delta}{\log X}$.
Then

(4.22) $\begin{split}\mathcal{M}_{g,T}(X,\sigma)&\leq
e^{\frac{8}{3}\delta(2\sigma-1)M(k,\sigma)+\frac{4\delta(2\sigma-1)\log\log
H_{0}}{\log(kH_{0})+2\delta}}\mathcal{U}(\alpha,k,T)^{2(1-\sigma)+\frac{2\delta(2\sigma-1)}{\log(kT)+2\delta}}\times\\\
&\mathcal{V}(\alpha,k,{\delta},T)^{2\sigma-1-\frac{2\delta(2\sigma-1)}{\log(kT)+2\delta}}(\log(kT))^{2\sigma}(\log
T)^{4(1-\sigma)}T^{\frac{8}{3}(1-\sigma)},\end{split}$

where $\mathcal{U}$ and $\mathcal{V}$ are respectively defined in (4.18) and
(4.19) and

(4.23) $M(k,\delta)=\max\Big{(}\frac{\log
H_{0}}{\log(kH_{0})+2\delta},1\Big{)}.$

###### Proof.

Let $\sigma_{1}=\frac{1}{2}$ andf $\sigma_{2}=1+\frac{\delta}{\log X}$ and
$\sigma\in[\sigma_{1},\sigma_{2}]$. We apply the convexity inequality (4.12)
with exponents

(4.24)
$a=\frac{\sigma_{2}-\sigma}{\sigma_{2}-\sigma_{1}}=\frac{1+\frac{\delta}{\log
X}-\sigma}{(1+\frac{\delta}{\log X})-\frac{1}{2}}\text{ and
}b=1-a=\frac{\sigma-\sigma_{1}}{\sigma_{2}-\sigma_{1}}=\frac{\sigma-\frac{1}{2}}{(1+\frac{\delta}{\log
X})-\frac{1}{2}}$

in combination with Lemmas 4.6, Lemma 4.7 to obtain

(4.25)
$\mathcal{M}_{g,T}(X,\sigma)\leq\mathcal{U}(\alpha,k,T)^{a}\mathcal{V}(\alpha,k,{\delta},T)^{b}(\log(kT))^{a+2b}(\log
T)^{2a}T^{\frac{4}{3}a}.$

Next, from the definitions of (4.24) it may be checked that

(4.26) $a=2(1-\sigma)+\frac{2\delta(2\sigma-1)}{\log X+2\delta},\text{ and }\
b=2\sigma-1-\frac{2\delta(2\sigma-1)}{\log X+2\delta}.$

From these equalities it follows that $a+2b\leq 2\sigma$. Using (4.26) and the
bound for $a+2b$ (since $\log(kT)\geq\log(kH_{0})\geq\log(10^{9})>1$), we have

(4.27) $\begin{split}\mathcal{M}_{g,T}(X,\sigma)&\leq
e^{\frac{4}{3}\times\frac{2\delta(2\sigma-1)\log
T}{\log(kT)+2\delta}+2\times\frac{2\delta(2\sigma-1)\log\log
T}{\log(kT)+2\delta}}\mathcal{U}(\alpha,k,T)^{2(1-\sigma)+\frac{2\delta(2\sigma-1)}{\log(kT)+2\delta}}\times\\\
&\mathcal{V}(\alpha,k,{\delta},T)^{2\sigma-1-\frac{2\delta(2\sigma-1)}{\log(kT)+2\delta}}(\log(kT))^{2\sigma}(\log
T)^{4(1-\sigma)}T^{\frac{8}{3}(1-\sigma)}.\end{split}$

Next we observe that the function $\frac{\log T}{\log(kT)+2\delta}$ decreases
if $\log k+2\delta<0$ and increases if $\log k+2\delta>0$ and thus

(4.28) $\frac{\log T}{\log(kT)+2\delta}\leq
M(k,\delta):=\begin{cases}\frac{\log H_{0}}{\log(kH_{0})+2\delta}&\text{ if
}\log k+2\delta<0,\\\ 1&\text{ if }\log k+2\delta\geq 0\end{cases}$

where $M(k,\delta)$ was defined in (4.23). Furthermore, it may be checked by
the conditions on $k$, that $\frac{\log\log T}{\log(kT)+2\delta}$ decreases as
long as $0<\delta<\frac{\log(H_{0})(\log\log H_{0}-1)}{2}$. Using these
observations in (4.27) we deduce (4.22). ∎

###  4.3. Bounding $F_{X}(\sigma,T)-F_{X}(\sigma,H)$

######  Lemma 4.9.

Let $g$ be as defined in Lemma 3.7. Let $\sigma\in[\frac{1}{2},1]$ and
$\alpha>0$. Let $T\geq H_{0}\geq H>0$, $X=kT$, $k$ satisfies (4.1), and
$0<\delta<\frac{\log(H_{0})(\log\log H_{0}-1)}{2}=26.36\ldots$. Then

(4.29)
$\begin{split}F_{X}(\sigma,T)-F_{X}(\sigma,H)&\leq\frac{e^{\frac{8}{3}\delta(2\sigma-1)M(k,\delta)+\frac{4\delta(2\sigma-1)\log\log
H_{0}}{\log(kH_{0})+2\delta}}\mathcal{U}(\alpha,k,T)^{2(1-\sigma)+\frac{2\delta(2\sigma-1)}{\log(kT)+2\delta}}\mathcal{V}(\alpha,k,{\delta},T)^{2\sigma-1-\frac{2\delta(2\sigma-1)}{\log(kT)+2\delta}}}{2(\omega_{2}(\sigma,T,\alpha))^{2}}\\\
&\times(\log(kT))^{2\sigma}(\log
T)^{4(1-\sigma)}T^{\frac{8}{3}(1-\sigma)},\end{split}$

where $\omega_{2},\mathcal{U},\mathcal{V}$ are respectively defined in (3.19),
(4.18), (4.19).

###### Proof.

By the assumed lower bound on $g$, (2.13), we have

$F_{X}(\sigma,T)-F_{X}(\sigma,H)=\int_{H}^{T}|f_{X}(\sigma+it)|^{2}dt\leq\frac{1}{(\omega_{2}(\sigma,T,\alpha))^{2}}\int_{H}^{T}|g(\sigma+it)|^{2}|f_{X}(\sigma+it)|^{2}dt.$

Since $t\to|g(\sigma+it)f_{X}(\sigma+it)|$ is even, it follows that

$F_{X}(\sigma,T)-F_{X}(\sigma,H)\leq\frac{\mathcal{M}_{g,T}(X,\sigma)}{2(\omega_{2}(\sigma,T,\alpha))^{2}}$

and we conclude by inserting the bound (4.22) for
$\mathcal{M}_{g,T}(X,\sigma)$. ∎

###  4.4. Explicit upper bounds for $\int_{\sigma^{\prime}}^{\mu}\arg
h_{X}(\tau+iT)d\tau-\int_{\sigma^{\prime}}^{\mu}\arg h_{X}(\tau+iH)d\tau$

The following Proposition and Corollary are a variant of Titchmarsh [25,
Lemma, p. 213]. This proposition gives a bounds for $\arg f(\sigma+iT)$ where
$f$ is a holomorphic function. The argument we use here is due to Backlund [1]
in the case that $f(s)=\zeta(s)$. The cases of Dirichlet $L$-functions and
Dedekind zeta functions have been worked out by McCurley [17] and by the first
and third authors [16] respectively.

######  Proposition 4.10.

Let $\eta>0$. Let $f(s)$ be a holomorphic function, for
${\mathfrak{Re}}(s)\geq-\eta$, real for real $s$. Assume there exist positive
constants $M$ and $m$ such that

(4.30) $\displaystyle|f(s)|\leq M\text{ for }{\mathfrak{Re}}(s)\geq 1+\eta,$
(4.31) $\displaystyle|{\mathfrak{Re}}f(1+\eta+it)|\geq m>0\text{ for all
}t\in\mathbb{R}.$

Let $\sigma\in(0,1+\eta]$ and assume that $U$ is not the ordinate of a zero of
$f(s)$. Then there exists an increasing sequence of natural numbers
$\\{N_{k}\\}_{k=1}^{\infty}$ such that

(4.32) $\left|\arg f(\sigma+iU)\right|\leq\frac{\pi}{\log
2}\mathscr{L}_{k}+\frac{\pi\log M}{2\log 2}-\frac{\pi\log m}{\log
2}+\frac{\pi}{2}+o_{k}(1)$

where

(4.33) $\mathscr{L}_{k}=\frac{1}{2\pi
N_{k}}\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\log\Big{(}\frac{1}{2}\sum_{j=0}^{1}|f(1+\eta+(1+2\eta)e^{i\theta}+(-1)^{j}iU)|^{N_{k}}\Big{)}\,d\theta$

and $o_{k}(1)$ is a term that approaches $0$ as $k\to\infty$.

###### Proof of Proposition 4.10.

Let $\eta>0$. We define $\arg f(1+\eta)=0$, and $\arg
f(s)=\arctan\frac{{\mathfrak{Im}}f(s)}{{\mathfrak{Re}}f(s)}$ for
${\mathfrak{Re}}(s)=1+\eta$, since, by (4.31), ${\mathfrak{Re}}(f(s))$ does
not vanish on ${\mathfrak{Re}}(s)=1+\eta$. It follows that

(4.34) $|\arg f(1+\eta+iU)|<\frac{\pi}{2}.$

Recall that $\arg f(\sigma+iU)$ is defined by continuous variation, moving
along the line $\mathcal{C}$ from $1+\eta+iU$ to $\sigma+iU$. It follows that

(4.35) $|\arg f(\sigma+iU)|\leq\left|\Delta_{\mathcal{C}}\arg
f(s)\right|+\frac{\pi}{2}.$

We now bound the argument change on $\mathcal{C}$. Let $N\in\mathbb{N}$ and
let

(4.36) $F_{N}(w)=\frac{1}{2}(f(w+iU)^{N}+f(w-iU)^{N}).$

Since $f(s)$ is real when $s$ is real, the reflection principle gives
$F_{N}(\sigma)={\mathfrak{Re}}\,f(\sigma+iU)^{N}$ for all $\sigma$ real.
Suppose $F_{N}(\sigma)$ has $n$ real zeros in the interval $[\sigma,1+\eta]$.
These zeros partition the interval into $n+1$ subintervals. On each of these
subintervals $\arg f(\sigma+iU)^{N}$ can change by at most $\pi$, since
${\mathfrak{Re}}\,f(\sigma+iU)^{N}$ is nonzero on the interior of each
subinterval. It follows that

(4.37) $|\Delta_{\mathcal{C}}\arg f(s)|=\frac{1}{N}|\Delta_{\mathcal{C}}\arg
f(s)^{N}|\leq\frac{(n+1)\pi}{N}.$

We now provide an upper bound for $n$. Jensen’s theorem asserts that

$\log|F_{N}(1+\eta)|+\int_{0}^{1+2\eta}\frac{n(u)du}{u}=\frac{1}{2\pi}\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}\log|F_{N}(1+\eta+(1+2\eta)e^{i\theta}|d\theta,$

where $n(u)$ denotes the number of zeros of $F_{N}(z)$ in the circle centered
at $1+\eta$ of radius $u$. Observe that $n(u)\geq n$ for
$u\geq\frac{1}{2}+\eta$ and thus

(4.38) $n\log
2\leq\frac{1}{2\pi}\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}\log|F_{N}(1+\eta+(1+2\eta)e^{i\theta}|d\theta-\log|F_{N}(1+\eta)|.$

Trivially from (4.36),

$|F_{N}(1+\eta+(1+2\eta)e^{i\theta})|\leq\frac{1}{2}\sum_{j=0}^{1}|f(1+\eta+(1+2\eta)e^{i\theta}+(-1)^{j}iU)|^{N},$

so for the left part of the contour in (4.38),

(4.39)
$\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\log|F_{N}(1+\eta+(1+2\eta)e^{i\theta}|d\theta\leq\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\log\Big{(}\frac{1}{2}\sum_{j=0}^{1}|f(1+\eta+(1+2\eta)e^{i\theta}+(-1)^{j}iU)|^{N}\Big{)}\,d\theta.$

For the right part of the contour in (4.38), we have
$-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$, so
${\mathfrak{Re}}(1+\eta+(1+2\eta)e^{i\theta})\geq 1+\eta$. We apply (4.30) and
obtain

(4.40)
$\frac{1}{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\log|F_{N}(1+\eta+(1+2\eta)e^{i\theta}|d\theta\leq\frac{N}{2}\log
M.$

To complete our bound for $n$, we require a lower bound for
$\log|F_{N_{k}}(1+\eta)|$.
We write $\displaystyle{f(1+\eta+iU)=re^{i\phi}}$ and then choose (by
Dirichlet’s approximation theorem) an increasing sequence of positive integers
$N_{k}$ tending to infinity such that $N_{k}\phi$ tends to $0$ modulo $2\pi$.
Since
$\displaystyle{\frac{F_{N_{k}}(1+\eta)}{|f(1+\eta+iU)|^{N_{k}}}=\frac{r^{N_{k}}\cos(N_{k}\phi)}{r^{N_{k}}}}$,
it follows that
$\displaystyle{\lim_{k\to\infty}\frac{F_{N_{k}}(1+\eta)}{|f(1+\eta+iU)|^{N_{k}}}=1}$.
Thus we derive

$\log|F_{N_{k}}(1+\eta)|\geq N_{k}\log|f(1+\eta+iU)|+o_{k}(1),$

where the term $o_{k}(1)\rightarrow 0$ as $k\rightarrow\infty$. Together with
(4.31), we obtain

(4.41) $\log|F_{N_{k}}(1+\eta)|\geq N_{k}\log m+o_{k}(1).$

Then (4.38), (4.39), (4.40), and (4.41) give

(4.42) $n\log
2\leq\frac{1}{2\pi}\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\log\Big{(}\frac{1}{2}\sum_{j=0}^{1}|f(1+\eta+(1+2\eta)e^{i\theta}+(-1)^{j}iU)|^{N_{k}}\Big{)}\,d\theta\\\
+\frac{N_{k}\log M}{2}-N_{k}\log m+o_{k}(1).$

By (4.37) it follows that

$\left|\Delta_{\mathcal{C}}\arg f(s)\right|\leq\frac{\pi}{\log
2}\mathscr{L}_{k}+\frac{\pi\log M}{2\log 2}-\frac{\pi\log m}{\log
2}+o_{k}(1),$

where $\mathscr{L}_{k}$ is defined by (4.33) . We conclude by combining this
with (4.35). ∎

We derive the following Corollary for $\arg h_{X}(s)$ from Proposition 4.10.

######  Corollary 4.11.

Let $\eta_{0}=0.23622\ldots$, $\eta\in[\eta_{0},\frac{1}{2})$, and $X\geq
10^{9}$. Assume that $U\geq H\geq 1002$ and that $U$ is not the ordinate of a
zero of $h_{X}(s)$. Then for all $\tau\in(0,1+\eta]$,

$|\arg h_{X}(\tau+iU)|\leq\frac{(1+2\eta)}{\log
2}\log\Big{(}\frac{b_{8}(\eta,H)}{2\pi}U\Big{)}+\frac{\pi(1+\eta)}{\log
2}(\log X)+\frac{\pi\log b_{7}(k,\eta,H_{0})}{2\log 2}+\frac{\pi\log
b_{5}(\eta)}{2\log 2}\\\ -\frac{\pi\log(1-b_{6}(10^{9},\eta)^{2})}{\log
2}+\frac{\pi}{2},$

where $b_{5},b_{6},b_{7},b_{8}$ are defined in (4.44), (4.45), (4.50), and
(4.51).

###### Proof of Corollary.

We apply Proposition 4.10 to $f=h_{X}$ as defined in (2.8):

$h_{X}(s)=1-f_{X}(s)^{2}=\zeta(s)M_{X}(s)(2-\zeta(s)M_{X}(s)).$

Let $\sigma\geq\eta+1$ and $t\in{\mathbb{R}}$. We establish an upper bound for
$|h_{X}(\sigma+it)|$. The triangle inequality in conjunction with
$\displaystyle{|\zeta(s)|\leq\zeta(1+\eta)}$ and with
$\displaystyle{|M_{X}(s)|\leq\sum_{n=1}^{\infty}\frac{|\mu(n)|}{n^{1+\eta}}=\frac{\zeta(1+\eta)}{\zeta(2+2\eta)}}$
give

(4.43) $|h_{X}(s)|\leq b_{5}(\eta)$

with

(4.44)
$b_{5}(\eta)=\frac{\zeta(1+\eta)^{4}}{\zeta(2+2\eta)^{2}}+\frac{2\zeta(1+\eta)^{2}}{\zeta(2+2\eta)}.$

We now give a lower bound for $|{\mathfrak{Re}}h_{X}(1+\eta+it)|$. We use the
reverse triangle inequality $|h_{X}(s)|\geq 1-|f_{X}(s)|^{2}$. It remains to
provide an upper bound for $|f_{X}(s)|$. Trivially from (2.5),

$|f_{X}(s)|\leq\sum_{n>X}\frac{|\lambda_{X}(n)|}{n^{1+\eta}}\leq\sum_{n>X}\frac{d(n)}{n^{1+\eta}},$

and by Lemma 3.5, we obtain

(4.45) $|f_{X}(1+\eta+it)|\leq b_{6}(X,\eta)=\frac{(1+\eta)(\log X)}{\eta
X^{\eta}}\Big{(}1+\frac{1}{\eta\log X}+\frac{\gamma}{\log
X}+\frac{7\eta}{12(1+\eta)X(\log X)}\Big{)}.$

Note that $\frac{(\log X)}{X^{\eta}}$ decreases when $\eta>\frac{1}{\log X}$,
which is the case since we assumed
$\eta>\frac{1}{\log(10^{9})}=0.048254\ldots$ and $X\geq 10^{9}$. Thus
$|f_{X}(s)|\leq b_{6}(10^{9},\eta)$ and

(4.46)
$|{\mathfrak{Re}}(h_{X}(s))|=|1-{\mathfrak{Re}}(f_{X}(s))^{2}|\geq|1-|f_{X}(s)|^{2}|\geq
1-|f_{X}(s)|^{2}\geq 1-b_{6}(10^{9},\eta)^{2}.$

Note our assumption $\eta\geq\eta_{0}=0.23622\ldots$ ensures
$1-b_{6}(10^{9},\eta)^{2}>0$.
Finally, we must bound $\mathscr{L}_{k}$ as defined in (4.33) in the case
$f=h_{X}$. We assume $w$ is a complex number such that
$-\eta\leq{\mathfrak{Re}}w\leq 1+\eta$ and $|{\mathfrak{Im}}w|\geq
U-(1+2\eta)$. Recall that by Lemma 3.1

$|\zeta(w)|\leq
3\frac{|1+w|}{|1-w|}\Big{(}\frac{|w+1|}{2\pi}\Big{)}^{\frac{1+\eta-{\mathfrak{Re}}w}{2}}\zeta(1+\eta).$

Since $\frac{|1+w|}{|1-w|}=\Big{|}1+\frac{2}{w-1}\Big{|}\leq
1+\frac{2}{|{\mathfrak{Im}}(w)|}\leq 1.002$ when $|{\mathfrak{Im}}(w)|\geq
1000$, then

(4.47) $|\zeta(w)|\leq
3.006\zeta(1+\eta)\Big{(}\frac{|w+1|}{2\pi}\Big{)}^{\frac{1+\eta-u}{2}}\text{
for }|{\mathfrak{Im}}(w)|\geq 1000.$

From the definition (2.4), we have the trivial bound

(4.48) $|M_{X}(w)|\leq X^{1+\eta}.$

It follows from

$|h_{X}(w)|\leq|\zeta(w)M_{X}(w)|^{2}+2|\zeta(w)||M_{X}(w)|,$

the bounds (4.47), (4.48),
$\frac{|w+1|}{2\pi}>1,-\frac{1+\eta-{\mathfrak{Re}}w}{2}<0$, and $X\geq
kH_{0}$, that

(4.49) $|h_{X}(w)|\leq
b_{7}(k,\eta,H_{0})\Big{(}\frac{|w+1|}{2\pi}\Big{)}^{1+\eta-u}X^{2(1+\eta)}\text{
for }|{\mathfrak{Im}}(w)|\geq 1000,$

with

(4.50)
$b_{7}(k,\eta,H_{0})=\left(1+\frac{2}{3.006\zeta(1+\eta)(kH_{0})^{1+\eta}}\right)\left(3.006\zeta(1+\eta)\right)^{2}.$

We apply this with $w=1+\eta+(1+2\eta)e^{i\theta}\pm iU$. Since
$\cos\theta\leq 0$, a little calculation gives

$|w+1|=|2+\eta+(1+2\eta)e^{i\theta}\pm
iU|\leq\sqrt{(2+\eta)^{2}+(1+2\eta+U)^{2}}\leq b_{8}(\eta,H)U,$

with

(4.51)
$b_{8}(\eta,H)=\sqrt{\frac{(2+\eta)^{2}}{H^{2}}+\Big{(}\frac{1+2\eta}{H}+1\Big{)}^{2}}.$

In addition
$1+\eta-u=1+\eta-(1+\eta+(1+2\eta)\cos\theta)=-(1+2\eta)(\cos\theta)$, and
(4.49) gives

(4.52) $|h_{X}(1+\eta+(1+2\eta)e^{i\theta}\pm iU)|\leq
b_{7}(k,\eta,H_{0})\Big{(}\frac{b_{8}(\eta,H)}{2\pi}U\Big{)}^{-(1+2\eta)(\cos\theta)}X^{2(1+\eta)},$

since $|{\mathfrak{Im}}(1+\eta+(1+2\eta)e^{i\theta}\pm iU)|\geq
U-(1-2\eta)\geq H-2\geq 1000$. We use this to bound $\mathscr{L}_{k}$ as
defined in (4.33):

$\mathscr{L}_{k}\leq\frac{1}{2\pi}\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\left(\log
b_{7}(k,\eta,H_{0})-(1+2\eta)(\cos\theta)\log\Big{(}\frac{b_{8}(\eta,H)}{2\pi}U\Big{)}+2(1+\eta)(\log
X)\right)d\theta.$

Calculating the integrals give

(4.53) $\mathscr{L}_{k}\leq\frac{\log
b_{7}(k,\eta,H_{0})}{2}+\frac{(1+2\eta)}{\pi}\log\Big{(}\frac{b_{8}(\eta,H)}{2\pi}U\Big{)}+(1+\eta)(\log
X).$

By (4.43) and (4.46) we may take $M=b_{5}(\eta)$ and
$m=1-b_{6}(10^{9},\eta)^{2}$ in (4.30) and (4.31) in the case of
$f(s)=h_{X}(s)$. Therefore by Proposition 4.10

(4.54) $\left|\arg h_{X}(\sigma+iU)\right|\leq\frac{\pi}{\log
2}\mathscr{L}_{k}+\frac{\pi\log b_{5}(\eta)}{2\log
2}-\frac{\pi\log(1-b_{6}(10^{9},\eta)^{2})}{\log 2}+\frac{\pi}{2}+o_{k}(1).$

Inserting the upper bound for $\mathscr{L}_{k}$ from (4.53) and letting
$k\to\infty$ we complete the proof as the $o_{k}(1)$ terms goes to zero. ∎

We are now in a position to bound the arguments.

######  Lemma 4.12.

Let $0<H\leq H_{0}\leq T$ and $X\leq T$. Let $\eta\in(\eta_{0},\frac{1}{2})$
with $\eta_{0}=0.23622\ldots$, $\sigma^{\prime}$ and $\mu$ satisfying
$\frac{1}{2}\leq\sigma^{\prime}<1<\mu\leq 1+\eta$. Then

(4.55) $\Big{|}\int_{\sigma^{\prime}}^{\mu}\arg
h_{X}(\tau+iT)d\tau-\int_{\sigma^{\prime}}^{\mu}\arg
h_{X}(\tau+iH)d\tau\Big{|}\leq C_{7}(\eta,H)\,(\mu-\sigma^{\prime})(\log T),$

where

(4.56) $C_{7}(\eta,H)=\frac{2(1+2\eta)+2\pi(1+\eta)}{\log
2}+\frac{b_{9}(\eta,H)}{\log H_{0}}.$

with $b_{9}(\eta,H)$ defined in (4.59).

###### Proof.

Note that

(4.57) $\Big{|}\int_{\sigma^{\prime}}^{\mu}\arg
h_{X}(\tau+iT)d\tau-\int_{\sigma^{\prime}}^{\mu}\arg
h_{X}(\tau+iH)d\tau\Big{|}\leq(\mu-\sigma^{\prime})\max_{\tau\in(\sigma^{\prime},\mu)}\Big{(}|\arg
h_{X}(\tau+iT)|+|\arg h_{X}(\tau+iH)|\Big{)}.$

By Corollary 4.11 we have

(4.58) $|\arg h_{X}(\tau+iH)|+|\arg h_{X}(\tau+iT)|\\\ \leq
b_{9}(\eta,H)+\frac{(1+2\eta)}{\log 2}(\log(HT))+\frac{2\pi(1+\eta)}{\log
2}(\log X)$

with

(4.59) $b_{9}(\eta,H)=\frac{\pi\log b_{7}(k,\eta,H_{0})}{\log 2}+\frac{\pi\log
b_{5}(\eta)}{\log 2}-\frac{2\pi\log(1-b_{6}(10^{9},\eta)^{2})}{\log
2}+\pi+\frac{2(1+2\eta)}{\log 2}\log\Big{(}\frac{b_{8}(\eta,H)}{2\pi}\Big{)}$

where $b_{7},b_{5},b_{6},b_{8}$ are defined in (4.50), (4.44), (4.45), (4.51).
Factoring $\log T$ in the right hand side of (4.58), using $H\leq T$, $X\leq
T$, and $H_{0}\leq T$ yields

(4.60) $|\arg h_{X}(\tau+iH)|+|\arg h_{X}(\tau+iT)|\leq(\log
T)\left(\frac{2(1+2\eta)+2\pi(1+\eta)}{\log 2}+\frac{b_{9}(\eta,H)}{\log
H_{0}}\right).$

Combining (4.57) and (4.60) leads to (4.55). ∎

###  4.5. Explicit lower bounds for $\int_{H}^{T}\log|h_{X}(\mu+it)|dt$

First, observe that (4.45) implies for

(4.61) $\mu\geq 1+\eta_{0}=1.23622\ldots,\ |f_{X}(\mu+it)|<1.$

This fact is used in the next lemma.

######  Lemma 4.13.

Assume $\mu\geq 1+\eta_{0}$ where $\eta_{0}=0.23622\ldots$. Let $X=kT$ where
$T\geq H_{0}$, $k$ satisfies (4.1), $k\leq 1$, and $2\pi m_{0}\leq H<T$. Then

(4.62) $-\int_{H}^{T}\log|h_{X}(\mu+it)|dt\leq C_{8}(k,\mu)(\log T).$

with

(4.63)
$C_{8}(k,\mu)=b_{10}(k,\mu)\frac{(\log(kH_{0}))^{2}}{(kH_{0})^{2\mu-2}}\left(\frac{4\mu
b_{11}(kH_{0},2\mu)}{k(2\mu-1)}+\frac{2\pi
m_{0}(2\mu-1)b_{11}(kH_{0},2\mu-1)}{(\mu-1)}\right),$

$b_{10}$ is defined in (4.66), $b_{11}$ in (4.68), and $m_{0}$ in (3.15).

###### Proof.

We begin by remarking that (4.45) implies $|f_{X}(\mu+it)|\leq
b_{6}(kH_{0},\mu-1)<1$ since $X\geq kH_{0}\geq 10^{9}$ and $\mu\geq
1+\eta_{0}$. Next, observe that $|h_{X}(\mu+it)|\geq|1-f_{X}(\mu+it)^{2}|\geq
1-|f_{X}(\mu+it)|^{2}$ and thus

(4.64) $-\log|h_{X}(\mu+it)|\leq-\log(1-|f_{X}(\mu+it)|^{2}).$

Since $-\frac{\log(1-u^{2})}{u^{2}}$ increases with $u\in(0,1)$, we have

(4.65) $-\log(1-|f_{X}(\mu+it)|^{2})\leq b_{10}(k,\mu)|f_{X}(\mu+it)|^{2},$

with

(4.66)
$b_{10}(k,\mu)=-\frac{\log\left(1-b_{6}(kH_{0},\mu-1)^{2}\right)}{b_{6}(kH_{0},\mu-1)^{2}}$

where $b_{6}$ is defined in (4.45). It follows from (4.64) and (4.65) that

(4.67) $-\int_{H}^{T}\log|h_{X}(\mu+it)|dt\leq
b_{10}(k,\mu)\int_{H}^{T}|f_{X}(\mu+it)|^{2}dt.$

We apply Lemma 3.6 and the bound $|\lambda_{X}(n)|\leq d(n)$ with
$\lambda_{X}(n)=0$ if $n\leq X$. We obtain

$\displaystyle\int_{H}^{T}|f_{X}(\mu+it)|^{2}dt$
$\displaystyle\leq\sum_{n=1}^{\infty}\frac{|\lambda_{X}(n)|^{2}}{n^{2\mu}}(T-H+2\pi
m_{0}(n+1))$ $\displaystyle\leq(T-H+2\pi
m_{0})\sum_{n>X}\frac{d(n)^{2}}{n^{2\mu}}+2\pi
m_{0}\sum_{n>X}\frac{d(n)^{2}}{n^{2\mu-1}}.$

We appeal to (3.13) to bound the above sums:

$\sum_{n\geq X}\frac{d(n)^{2}}{n^{\tau}}\leq\frac{(\log
X)^{3}}{X^{\tau-1}}\frac{2\tau b_{11}(kH_{0},\tau)}{(\tau-1)},$

since $X\geq kH_{0}$ where

(4.68) $b_{11}(X,\tau)=1+\frac{3}{(\tau-1)(\log X)}+\frac{6}{(\tau-1)^{2}(\log
X)^{2}}+\frac{6}{(\tau-1)^{3}(\log X)^{3}}.$

Since $X=kT$ we deduce that

$\int_{H}^{T}|f_{X}(\mu+it)|^{2}dt\leq\frac{(\log(kT))^{3}}{(kT)^{2\mu-2}}\left(\frac{4\mu
b_{11}(kH_{0},2\mu)}{k(2\mu-1)}+\frac{2\pi
m_{0}(2\mu-1)b_{11}(kH_{0},2\mu-1)}{(\mu-1)}\right).$

Note that $\frac{(\log(kT))^{2}}{(kT)^{2\mu-2}}$ decreases with $T$ as long as
$10^{9}>e^{\frac{1}{\mu-1}}$ (i.e. $\mu>\mu_{2}=1.072382\ldots$). Using this
and $\log(kT)\leq\log T$ (since $k\leq 1$) implies

(4.69)
$\int_{H}^{T}|f_{X}(\mu+it)|^{2}dt\leq\frac{(\log(kH_{0}))^{2}}{(kH_{0})^{2\mu-2}}\left(\frac{4\mu
b_{11}(kH_{0},2\mu)}{k(2\mu-1)}+\frac{2\pi
m_{0}(2\mu-1)b_{11}(kH_{0},2\mu-1)}{(\mu-1)}\right)(\log T).$

We conclude by combining this with (4.67). ∎

###  4.6. Proof of Zero Density Result

Finally, we are able to compile our bounds to obtain an upper bound for
$N(\sigma,T)$.

######  Lemma 4.14.

Assume
$\alpha>0,d>0,\delta>0,\eta_{0}=0.23622\ldots,\eta\in[\eta_{0},\frac{1}{2}),$
and $\mu\in[1+\eta_{0},1+\eta]$. Let $H_{0}=3.0610046\cdot 10^{10},\ 1002\leq
H\leq H_{0},\ \frac{10^{9}}{H_{0}}\leq k\leq 1$, $T\geq H_{0}$, and $X=kT$
Assume $\sigma>\frac{1}{2}+\frac{d}{\log H_{0}}$,
$\mathcal{U}(\alpha,k,H_{0})>1$, and $\mathcal{U}(\alpha,k,T)$ decreases in
$T$. Thus

(4.70) $N(\sigma,T)\leq\frac{(T-H)(\log T)}{2\pi
d}\log\left(1+\mathcal{C}_{1}\frac{(\log(kT))^{2\sigma}(\log
T)^{4(1-\sigma)}T^{\frac{8}{3}(1-\sigma)}}{T-H}\right)+\frac{\mathcal{C}_{2}}{2\pi
d}(\log T)^{2},$ (4.71) $N(\sigma,T)\leq\frac{\mathcal{C}_{1}}{2\pi
d}(\log(kT))^{2\sigma}(\log
T)^{5-4\sigma}T^{\frac{8}{3}(1-\sigma)}+\frac{\mathcal{C}_{2}}{2\pi d}(\log
T)^{2},$

with

(4.72)
$\displaystyle\mathcal{C}_{1}=\mathcal{C}_{1}(\alpha,d,\delta,k,H,\sigma)$
$\displaystyle=b_{12}(H)e^{\frac{8}{3}\delta(2\sigma-1)M(k,\delta)+\frac{4\delta(2\sigma-1)\log\log
H_{0}}{\log(kH_{0})+2\delta}}\mathcal{U}(\alpha,k,H_{0})^{2(1-\sigma)+\frac{2d}{\log
H_{0}}+\frac{2\delta(2\sigma-1)}{\log(kH_{0})+2\delta}}\times$
$\displaystyle\mathcal{V}(\alpha,k,{\delta},H_{0})^{2\sigma-1}e^{\frac{2d(2\log\log
H_{0}-\log\log(kH_{0}))}{\log H_{0}}+\frac{8d}{3}+2\alpha},$ (4.73)
$\displaystyle\mathcal{C}_{2}=\mathcal{C}_{2}(d,\eta,k,H,\mu,\sigma)$
$\displaystyle=C_{7}(\eta,H)\Big{(}\mu-\sigma+\frac{d}{\log
H_{0}}\Big{)}+C_{8}(k,\mu),$

and $\mathcal{U},\mathcal{V},M(k,\delta),C_{7},C_{8}$ and $b_{12}$ are
respectively defined in (4.18), (4.19), (4.23), (4.56), (4.63), and (4.75).

Remark. 1. The assumptions that $U(\alpha,k,H_{0})>1$ and $U(\alpha,k,T)$ are
decreasing can be removed from the theorem. However, this would overly
complicate the statement of the theorem. In all instances that we apply this
theorem (for various values of $\alpha$ and $k$) these conditions hold.

###### Proof.

We begin by assuming that $T$ is not the ordinate of a zero of $\zeta(s)$.
From (2.3), (2.9), and the definition (2.10) of $F_{X}$, we have for
$\sigma\in[\sigma^{\prime},1]$ where $\sigma^{\prime}\geq\frac{1}{2}$ and
$\mu\in[1+\eta_{0},1+\eta]$

$N(\sigma,T)\leq\frac{1}{2\pi(\sigma-\sigma^{\prime})}\Big{(}(T-H)\log\left(1+\frac{F_{X}(\sigma^{\prime},T)-F_{X}(\sigma^{\prime},H)}{(T-H)}\right)\Big{.}\\\
\Big{.}+\int_{\sigma^{\prime}}^{\mu}\arg
h_{X}(\tau+iT)d\tau-\int_{\sigma^{\prime}}^{\mu}\arg
h_{X}(\tau+iH)d\tau-\int_{H}^{T}\log|h_{X}(\mu+it)|dt\Big{)}.$

We apply Lemma 4.9, Lemma 4.12, and Lemma 4.13 to achieve

(4.74)
$\begin{split}&N(\sigma,T)\leq\frac{(T-H)}{2\pi(\sigma-\sigma^{\prime})}\times\\\
&\log\Bigg{(}1+\frac{e^{\frac{8}{3}\delta(2\sigma^{\prime}-1)M(k,\delta)+\frac{4\delta(2\sigma^{\prime}-1)\log\log
H_{0}}{\log(kH_{0})+2\delta}}\mathcal{U}(\alpha,k,T)^{2(1-\sigma^{\prime})+\frac{2\delta(2\sigma^{\prime}-1)}{\log(kT)+2\delta}}\mathcal{V}(\alpha,k,{\delta},T)^{2\sigma^{\prime}-1-\frac{2\delta(2\sigma^{\prime}-1)}{\log(kT)+2\delta}}}{2(\omega_{2}(\sigma^{\prime},T,\alpha))^{2}}\times\\\
&\frac{(\log(kT))^{2\sigma^{\prime}}(\log
T)^{4(1-\sigma^{\prime})}T^{\frac{8}{3}(1-\sigma^{\prime})}}{(T-H)}\Bigg{)}+\frac{\left(C_{7}(\eta,H)\,(\mu-\sigma^{\prime})+C_{8}(k,\mu)\right)(\log
T)}{2\pi(\sigma-\sigma^{\prime})}.\end{split}$

We make the choice $\sigma^{\prime}=\sigma-\frac{d}{\log T}$, for some $d>0$.
From the definition (4.19), we note that $\mathcal{V}(\alpha,k,{\delta},T)$
decreases with $T$. Since by assumption $\mathcal{U}(\alpha,k,H_{0})>1$ and
$T\to U(k,\alpha,T)$ decreases, it follows that
$\mathcal{U}(\alpha,k,H_{0})^{\frac{2d}{\log
T}+\frac{2\delta(2\sigma^{\prime}-1)}{\log(kT)+2\delta}}$ decreases with $T$
and thus

$\mathcal{U}(\alpha,k,T)^{2(1-\sigma^{\prime})+\frac{2\delta(2\sigma^{\prime}-1)}{\log(kT)+2\delta}}\leq\mathcal{U}(\alpha,k,H_{0})^{2(1-\sigma)+\frac{2d}{\log
H_{0}}+\frac{2\delta(2\sigma^{\prime}-1)}{\log(kH_{0})+2\delta}}.$

It may be shown that for our choice of parameters $\alpha,k,\delta$ that
$\mathcal{V}(\alpha,k,{\delta},T)>1$ for all $T\geq H_{0}$ and thus

$\mathcal{V}(\alpha,k,{\delta},T)^{2\sigma^{\prime}-1-\frac{2\delta(2\sigma^{\prime}-1)}{\log(kT)+2\delta}}\leq\mathcal{V}(\alpha,k,{\delta},T)^{2\sigma^{\prime}-1}.$

In addition,

$\displaystyle(\log(kT))^{2\sigma^{\prime}}(\log
T)^{4(1-\sigma^{\prime})}T^{\frac{8}{3}(1-\sigma^{\prime})}$
$\displaystyle=e^{\frac{2d}{\log T}(2\log\log
T-\log\log(kT))+\frac{8d}{3}}(\log(kT))^{2\sigma}(\log
T)^{4(1-\sigma)}T^{\frac{8}{3}(1-\sigma)}$ $\displaystyle\leq
e^{\frac{2d(2\log\log H_{0}-\log\log(kH_{0}))}{\log
H_{0}}+\frac{8d}{3}}(\log(kT))^{2\sigma}(\log
T)^{4(1-\sigma)}T^{\frac{8}{3}(1-\sigma)},$

since $T\geq H_{0}$ and $\frac{10^{9}}{H_{0}}\leq k\leq 1$ imply
$\frac{2\log\log T-\log\log(kT)}{\log T}$ decreases in $T$. Since
$\omega_{2}(\sigma^{\prime},T,\alpha)$ as defined in (3.19) increases with
$\sigma^{\prime}\geq\sigma-\frac{d}{\log H_{0}}$ and decreases with $T$, then

(4.75) $\frac{1}{2(\omega_{2}(\sigma^{\prime},T,\alpha))^{2}}\leq
b_{12}(H)e^{2\alpha}\ \text{with}\ b_{12}(H)=\frac{1}{2(1-\frac{1}{H})^{2}}.$

Combining the above inequalities establishes (4.70), and thus (4.71) (applying
$\log(1+y)\leq y$).

(4.76) $\begin{split}N(\sigma,T)&\leq\frac{(T-H)(\log T)}{2\pi
d}\log\left(1+b_{12}(H)e^{\frac{8}{3}\delta(2\sigma^{\prime}-1)M(k,\delta)+\frac{4\delta(2\sigma^{\prime}-1)\log\log
H_{0}}{\log(kH_{0})+2\delta}}\right.\\\
&\times\mathcal{U}(\alpha,k,H_{0})^{2(1-\sigma)+\frac{2d}{\log
H_{0}}+\frac{2\delta(2\sigma^{\prime}-1)}{\log(kH_{0})+2\delta}}\mathcal{V}(\alpha,k,{\delta},H_{0})^{2\sigma-1}e^{\frac{2d(2\log\log
H_{0}-\log\log(kH_{0}))}{\log H_{0}}+\frac{8d}{3}+2\alpha}\\\
&\times\left.\frac{(\log(kT))^{2\sigma}(\log
T)^{4(1-\sigma)}T^{\frac{8}{3}(1-\sigma)}}{(T-H)}\right)\\\
&+\frac{\left(C_{7}(\eta,H)\,(\mu-\sigma+\frac{d}{\log
H_{0}})+C_{8}(k,\mu)\right)(\log T)^{2}}{2\pi d}.\end{split}$

Since $\sigma^{\prime}\leq\sigma$, each remaining occurrence of
$\sigma^{\prime}$ may be replaced by $\sigma$. Finally, by a continuity
argument these inequalities extend to the case where $T$ is the ordinate of a
zero of the zeta function. ∎

##  5\. Tables of Computation

For fixed values of $\sigma$, Table 1 provides bounds for $N(\sigma,T)$ of the
shape (4.71). We fix values for $k$ in $[\frac{10^{9}}{H_{0}},1]$. The
parameters $\alpha,d,\delta,\eta$ and $H$ are chosen to make
$\frac{\mathcal{C}_{1}}{2\pi d}$ as small as possible with
$\mathcal{C}_{1}(\alpha,d,\delta,k,H,\sigma)$ as defined in (4.72) . The
program returns $H=H_{0}-1$ for all lines in the table. With this $H$ we
minimize of $C_{7}(\eta,H)$ which chooses $\eta=0.25618\ldots$. Then $\mu$ is
chosen to minimize $\mu C_{7}(\eta,H)+C_{8}(k,\mu)$ (as in the definition
(4.73) of $\mathcal{C}_{2}=\mathcal{C}_{2}(d,\eta,k,H,\mu,\sigma)$). We remark
that there is a small bit of subtlety when considering
$\mathcal{U}(\alpha,k,T)$, it is necessary to ensure all the coefficients in
$\mathcal{J}(k,T)$ are positive and this is checked with each set of
parameters used. This is to guarantee that $\mathcal{U}(\alpha,k,T)$ decreases
with $T$.

Table 1. The bound $N(\sigma,T)\leq A(\log(kT))^{2\sigma}(\log T)^{5-4\sigma}T^{\frac{8}{3}(1-\sigma)}+B(\log T)^{2}$ (4.71) for $\sigma=\sigma_{0}$ with $\frac{10^{9}}{H_{0}}\leq k\leq 1$. $\sigma_{0}$ | $k$ | $\mu$ | $\alpha$ | ${\delta}$ | $d$ | $A=\frac{\mathcal{C}_{1}}{2\pi d}$ | $B=\frac{\mathcal{C}_{2}}{2\pi d}$
---|---|---|---|---|---|---|---
$0.60$ | $0.5$ | $1.251$ | $0.288$ | $0.3140$ | $0.341$ | $2.177$ | $5.663$
$0.65$ | $0.6$ | $1.249$ | $0.256$ | $0.3070$ | $0.340$ | $2.963$ | $5.249$
$0.70$ | $0.8$ | $1.247$ | $0.222$ | $0.3040$ | $0.339$ | $3.983$ | $4.824$
$0.75$ | $1.0$ | $1.245$ | $0.189$ | $0.3030$ | $0.338$ | $5.277$ | $4.403$
$0.80$ | $1.0$ | $1.245$ | $0.160$ | $0.3030$ | $0.337$ | $6.918$ | $3.997$
$0.85$ | $1.0$ | $1.245$ | $0.133$ | $0.3030$ | $0.336$ | $8.975$ | $3.588$
$0.86$ | $1.0$ | $1.245$ | $0.127$ | $0.3030$ | $0.335$ | $9.441$ | $3.514$
$0.87$ | $1.0$ | $1.245$ | $0.122$ | $0.3030$ | $0.335$ | $9.926$ | $3.430$
$0.88$ | $1.0$ | $1.245$ | $0.116$ | $0.3030$ | $0.335$ | $10.431$ | $3.346$
$0.89$ | $1.0$ | $1.245$ | $0.111$ | $0.3030$ | $0.335$ | $10.955$ | $3.262$
$0.90$ | $1.0$ | $1.245$ | $0.105$ | $0.3030$ | $0.334$ | $11.499$ | $3.186$
$0.91$ | $1.0$ | $1.245$ | $0.100$ | $0.3030$ | $0.334$ | $12.063$ | $3.102$
$0.92$ | $1.0$ | $1.245$ | $0.095$ | $0.3030$ | $0.334$ | $12.646$ | $3.017$
$0.93$ | $1.0$ | $1.245$ | $0.089$ | $0.3030$ | $0.333$ | $13.250$ | $2.941$
$0.94$ | $1.0$ | $1.245$ | $0.084$ | $0.3030$ | $0.333$ | $13.872$ | $2.856$
$0.95$ | $1.0$ | $1.245$ | $0.079$ | $0.3030$ | $0.333$ | $14.513$ | $2.772$
$0.96$ | $1.0$ | $1.245$ | $0.074$ | $0.3030$ | $0.332$ | $15.173$ | $2.694$
$0.97$ | $1.0$ | $1.245$ | $0.069$ | $0.3030$ | $0.332$ | $15.850$ | $2.609$
$0.98$ | $1.0$ | $1.245$ | $0.064$ | $0.3030$ | $0.331$ | $16.544$ | $2.532$
$0.99$ | $1.0$ | $1.245$ | $0.060$ | $0.3030$ | $0.331$ | $17.253$ | $2.446$

For fixed values of $\sigma$, Table 2 provide bounds for $N(\sigma,H_{0})$ of
the shape (4.70). In this case, the choice of $H$ is essential and we choose
$H=H_{0}-10^{-6}$. As a consequence the “main term” is $\frac{10^{-6}}{2\pi
d}(\log H_{0})\log\Big{(}1+10^{6}\mathcal{C}_{1}(\log(kH_{0}))^{2\sigma}(\log
H_{0})^{4(1-\sigma)}H_{0}^{\frac{8}{3}(1-\sigma)}\Big{)}$ which becomes
insignificant in comparison to
$\frac{\mathcal{C}_{2}(d,\eta,k,H,\mu,\sigma)}{2\pi d}(\log H_{0})^{2}$, the
term arising from the argument. We take $\alpha=0.324$, $\delta=0.3000$, and
$k=1$ (as we did not find any other values giving better bounds). The
parameter $\eta$ is chosen to minimize $C_{7}(\eta,H)$, and then $\mu$ to
minimize $\mu C_{7}(\eta,H)+C_{8}(k,\mu)$: $\eta=0.2561\ldots$ and
$\mu=1.2453\ldots$.

Table 2. Bound (4.70) with $k=1$ $\sigma$ | $d$ | $\frac{1}{2\pi d}$ | $\mathcal{C}_{1}$ | $\frac{\mathcal{C}_{2}}{2\pi d}$ | $N(\sigma,H_{0})\leq$
---|---|---|---|---|---
$0.60$ | $2.414$ | $0.066$ | $2094.73$ | $0.893$ | $520.28$
$0.65$ | $3.621$ | $0.044$ | $97986.60$ | $0.595$ | $346.85$
$0.70$ | $4.828$ | $0.033$ | $4583580.34$ | $0.447$ | $260.14$
$0.75$ | $6.036$ | $0.027$ | $214409007.32$ | $0.357$ | $208.11$
$0.80$ | $7.243$ | $0.022$ | $10029544375.44$ | $0.298$ | $173.42$
$0.85$ | $8.450$ | $0.019$ | $469158276689.92$ | $0.255$ | $148.65$
$0.86$ | $8.691$ | $0.019$ | $1012341447042.27$ | $0.248$ | $144.52$
$0.87$ | $8.933$ | $0.018$ | $2184412502812.95$ | $0.242$ | $140.61$
$0.88$ | $9.174$ | $0.018$ | $4713486735514.76$ | $0.235$ | $136.91$
$0.89$ | $9.416$ | $0.017$ | $10170678467214.40$ | $0.229$ | $133.40$
$0.90$ | $9.657$ | $0.017$ | $21946110446020.33$ | $0.224$ | $130.07$
$0.91$ | $9.899$ | $0.017$ | $47354929689448.17$ | $0.218$ | $126.90$
$0.92$ | $10.140$ | $0.016$ | $102181631292174.11$ | $0.213$ | $123.88$
$0.93$ | $10.382$ | $0.016$ | $220485720114084.42$ | $0.208$ | $120.99$
$0.94$ | $10.623$ | $0.015$ | $475760194464125.94$ | $0.203$ | $118.24$
$0.95$ | $10.864$ | $0.015$ | $1026586948666903.92$ | $0.199$ | $115.62$
$0.96$ | $11.106$ | $0.015$ | $2215151194732183.30$ | $0.195$ | $113.10$
$0.97$ | $11.347$ | $0.015$ | $4779814142285142.58$ | $0.190$ | $110.70$
$0.98$ | $11.589$ | $0.014$ | $10313798574616601.14$ | $0.186$ | $108.39$
$0.99$ | $11.830$ | $0.014$ | $22254932487167323.15$ | $0.183$ | $106.18$

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