# An Explicit Vinogradov–Korobov Zero-Free Region for Dirichlet $L$-functions

Tanmay Khale

###### Abstract.

We establish the first explicit form of the Vinogradov–Korobov zero-free
region for Dirichlet $L$-functions.

##  1\. Introduction and statement of main results

Vinogradov [vinogradov] and Korobov [korobov] proved that there exists an
absolute and effectively computable constant $c>0$ such that the Riemann zeta
function $\zeta(\sigma+it)$ does not vanish in the region

$\sigma>1-\frac{c}{(\log|t|)^{\frac{2}{3}}(\log\log|t|)^{\frac{1}{3}}},\qquad\absolutevalue{t}\geq
10.$ (1.1)

Ford [fordzfr0]111The results of this paper depend on the calculations in
[fordzfr0]. An updated version with minor numerical corrections can be found
in [fordzfr]. proved that one may take $c=1/57.54$. If $|t|$ is sufficiently
large, then this improves to $1/49.13$. In this paper, we prove the first
explicit version of (1.1) for Dirichlet $L$-functions.

######  Theorem 1.1.

Let $q\geq 3$, and let $\chi\pmod{q}$ be a Dirichlet character. The Dirichlet
$L$-function $L(\sigma+it,\chi)$ does not vanish in the region

$\sigma\geq 1-\frac{1}{10.5\log
q+61.5(\log|t|)^{2/3}(\log\log|t|)^{1/3}},\quad|t|\geq 10.$ (1.2)

Also, there exists an absolute and effectively computable constant $Y>0$ such
that $L(\sigma+it,\chi)$ does not vanish in the region

$\sigma\geq 1-\frac{1}{10.1\log
q+49.13(\log|t|)^{\frac{2}{3}}(\log\log|t|)^{\frac{1}{3}}},\qquad\absolutevalue{t}\geq
Y.$ (1.3)

As part of our proof of Theorem 1.1, we first require that there exists some
absolute and effectively computable constant $c^{\prime}>0$ such that
$L(\sigma+it,\chi)\neq 0$ in the region

$\sigma\geq 1-\frac{c^{\prime}}{\log
q+(\log|t|)^{\frac{2}{3}}(\log\log|t|)^{\frac{1}{3}}},\qquad\absolutevalue{t}\geq
10.$ (1.4)

Montgomery [montgomery, p. 176] states this result without reference or proof.
A zero-free region of the form (1.4) with weaker $q$-dependence follows from
Mitsui [Mitsui, Lemma 11]. Several authors (e.g., [Bartz, coleman, Hinz]) have
proved an analogue of (1.4) for $L$-functions of Hecke characters over a
number field $K$, and the result for Dirichlet characters follows by taking
$K=\mathbb{Q}$. For completeness, we give a short proof of (1.4) in in
Appendix B that relies only on well-known bounds for the sum in (2.5) along
with a convenient version of Jensen’s formula from complex analysis.

### Acknowledgements

The author thanks Jesse Thorner both for suggesting this project and for his
extensive feedback which improved the quality of the presentation in this
paper. The author also thanks Kevin Ford for numerous enlightening
conversations. All numerical computations were carried out using Mathematica
12.

##  2\. Notation and Preliminaries

###  2.1. Preliminaries on Dirichlet $L$-functions

The results in this subsection can be found in [davenport]. Let $\chi\pmod{q}$
be a Dirichlet character. The Dirichlet $L$-function $L(s,\chi)$ is defined by
the absolutely convergent Dirichlet series

$L(s,\chi)=\sum_{n=1}^{\infty}\frac{\chi(n)}{n^{s}}$

in the half-plane $\mathrm{Re}(s)>1$. If $\chi$ is nonprincipal, then
$L(s,\chi)$ extends to an entire function, while if $\chi$ is principal, then
$L(s,\chi)$ extends to a meromorphic function on the complex plane with a
single simple pole at $s=1$.

We say that $\chi$ is odd (resp. even) if $\chi(-1)=-1$ (resp. $\chi(-1)=1$).
We write

$\mathfrak{a}{(\chi)}=\begin{cases}0&\text{ when }\chi\text{ is even,}\\\
1&\text{ when }\chi\text{ is odd.}\end{cases}$

When there is no ambiguity concerning the Dirichlet character under
consideration, we write $\mathfrak{a}$ instead of $\mathfrak{a}(\chi)$. We
also write

$\delta(\chi)=\begin{cases}0&\text{ when }\chi\text{ is nonprincipal,}\\\
1&\text{ when }\chi\text{ is principal.}\end{cases}$

Given $L(s,\chi)$ we define the completed $L$-function $\xi(s,\chi)$ by

$\xi(s,\chi)=(s(s-1))^{\delta(\chi)}\Big{(}\frac{q}{\pi}\Big{)}^{\frac{s+\mathfrak{a}}{2}}\Gamma\Big{(}\frac{s+\mathfrak{a}}{2}\Big{)}L(s,\chi).$
(2.1)

If $\chi$ is primitive, there exists a constant $W(\chi)$ of modulus one such
that the completed $L$-function satisfies the functional equation

$\xi(1-s,\overline{\chi})=W(\chi)\xi(s,\chi).$ (2.2)

Furthermore, $\xi(s,\chi)$ is an entire function of order one, and hence
possesses a Hadamard factorization

$\xi(s,\chi)=\prod_{\xi(\rho,\chi)=0}e^{\frac{s}{\rho}}\Big{(}1-\frac{s}{\rho}\Big{)}.$

It follows from the analytic continuation and functional equation for
$\xi(s,\chi)$ that if $\chi$ is primitive, then $L(s,\chi)$ has trivial zeros
at $(-2n-2\delta(\chi)-\mathfrak{a}(\chi))_{n=0}^{\infty}$. If $\chi\pmod{q}$
is induced by the primitive character $\chi^{\prime}\pmod{q^{\prime}}$, then

$L(s,\chi)=L(s,\chi^{\prime})\prod_{\begin{subarray}{c}p\mid q\\\ p\nmid
q^{\prime}\end{subarray}}\Big{(}1-\frac{1}{p^{s}}\Big{)},$

and $L(s,\chi)$ has zeros on the line $\mathrm{Re}(s)=0$ corresponding to the
zeros of the factors $1-p^{-s}$. We define

$\mathscr{I}(\chi)=\begin{cases}0&\text{if }\chi\text{ is primitive,}\\\
1&\text{otherwise.}\end{cases}$

Observe that since $L(s,\overline{\chi})=\overline{L(\overline{s},\chi)}$, the
zeros of $L(s,\chi)L(s,\overline{\chi})$ come in conjugate pairs.
Consequently, if the region $R\subset\mathbb{C}$ is free of zeros of
$L(s,\chi)L(s,\overline{\chi})$, the reflection $\overline{R}$ is also zero-
free. It therefore suffices to demonstrate a zero-free region holding for
arbitrary $\chi$ and $\mathrm{Im}(s)\geq 0$ to conclude a zero-free region
holding for arbitrary $\chi$ and $\mathrm{Im}(s)$ positive or negative.
Throughout the remainder of the paper we will impose the restriction
$\mathrm{Im}(s)\geq 0$ without loss of generality. By McCurley’s [mccurley,
Theorem 1.1] explicit version of a classical zero-free region of de la Vallée
Poussin, $L(\sigma+it,\chi)$ is non-vanishing for $t\geq 0$ and

$\sigma\geq 1-\frac{1}{9.64590880\log(\max\\{q,qt,10\\})}.$ (2.3)

apart from a possible real zero which is necessarily simple.

###  2.2. Estimates for exponential sums

Let $N\in\mathbb{N}$ and $u,t\in\mathbb{R}$, satisfying

$0<u\leq 1,\quad 1\leq N\leq t,\quad\lambda=\frac{\log t}{\log N}.$

Throughout this paper, we will assume that there exist constants $A,B>0$ such
that the Hurwitz zeta function, defined for $\sigma>1$ by

$\zeta(\sigma+it,u)=\sum_{n=0}^{\infty}(n+u)^{-(\sigma+it)},$

and extended to a meromorphic function on the complex plane by analytic
continuation, satisfies the bound

$\displaystyle|\zeta(\sigma+it,u)-u^{-(\sigma+it)}|\leq
At^{B(1-\sigma)^{3/2}}(\log t)^{2/3},\qquad t\geq 3,~{}1/2\leq\sigma\leq 1.$
(2.4)

Ford [ford, Theorem 1] shows that one may take $A=76.2$ and $B=4.45$ in (2.4).
More generally, in [ford, Lemma 7.3], he establishes how the values of $A,B$
above explicitly depend on the constants in an upper bound for the exponential
sums

$S(N,t)=\max_{0<u\leq 1}\max_{N<R\leq 2N}\Big{|}\sum_{N<n\leq
R}(n+u)^{-it}\Big{|}.$ (2.5)

Namely, suppose that there exist constants $C,D>0$ such that $S(N,t)\leq
CN^{1-1/(D\lambda^{2})}$. Let $B=\frac{2}{9}\sqrt{3D}$. Assume further that
$t\geq 10^{100}$ and $\frac{15}{16}\leq\sigma\leq 1$. By [ford, Lemma 7.3], we
then have that

$\begin{gathered}|\zeta(\sigma+it,u)-u^{-(\sigma+it)}|\leq\Big{(}\frac{C+1+10^{-80}}{(\log
t)^{2/3}}+1.569CD^{1/3}\Big{)}t^{B(1-\sigma)^{3/2}}(\log
t)^{2/3}.\end{gathered}$

###  2.3. Notation

We let

$\gamma_{\mathbb{\mathbb{Q}}}=\lim_{N\to\infty}\Big{(}\sum_{n=1}^{N}n^{-1}-\log
N\Big{)}=0.5772156649\dots$

denote the Euler–Mascheroni constant $\gamma_{\mathbb{\mathbb{Q}}}$. We write
$\log_{3}(x)=\log\log\log(x)$ to denote the third-iterate of the logarithm
function. Finally, any sum written in the form

$\sum_{L(\rho,\chi)=0}f(\rho)$

in this paper is always a sum over zeros $\rho$ satisfying
$0<\mathrm{Re}(\rho)<1$, each counted with multiplicity.

##  3\. General Results

Theorem 1.1 is a consequence of the following more general result.

######  Theorem 3.1.

Assume that there exist constants $A>0$ and $0<B\leq 4.45$ such that (2.4)
holds. Let $T_{0}\geq e^{1938}$ satisfy $\frac{\log T_{0}}{\log\log
T_{0}}\geq\frac{1139}{B}$. For $t\geq T_{0}-1$, define

$X(t)=1.16\log(A+1)+0.845593+0.23531\log\Big(\frac{B}{\log\log\tau}\Big{missing})+\Big{(}\frac{1.694}{B^{4/3}}-\frac{2.087}{B^{1/3}}\Big{)}\Big{(}\frac{\log\log\tau}{\log\tau}\Big{)}^{\frac{1}{3}},$

where $\tau=4t+1$, and

$M_{1}=\min\Big{(}0.055071,\frac{0.16521-\frac{0.184833}{\log
T_{0}}}{2.9997+\max_{t\geq T_{0}}X(t)/\log\log t}\Big{)}.$

If for all characters $\chi\pmod{q}$ we have that $L(\sigma+it,\chi)$ does not
vanish in the region $t\in[T_{0}-1,T_{0}]$ and

$\sigma\geq 1-\frac{1}{(10.082+\frac{1.607}{\log\log T_{0}})\log
q+9.791\log\log q+B^{2/3}M_{1}^{-1}(\log t)^{2/3}(\log\log t)^{1/3}},$ (3.1)

then for all $t\geq T_{0}$ and all $\chi\pmod{q}$, $L(\sigma+it,\chi)$ is
nonvanishing in the region (3.1).

Theorem 3.1 has the following corollary for large $t$.

######  Corollary 3.2.

For all $0<B\leq 4.45$, there exists an effectively computable constant
$T_{0}(B)$ depending at most on $B$ such that $L(\sigma+it,\chi)$ is
nonvanishing in the region

$\sigma\geq 1-\frac{1}{10.1\log
q+(0.05507)^{-1}B^{2/3}(\log|t|)^{2/3}(\log\log|t|)^{1/3}},\qquad t\geq
T_{0}(B).$ (3.2)

###### Proof.

We first show that if $T_{0}=T_{0}(B)$ is sufficiently large and if (3.1)
holds for all $t\geq T_{0}$, then (3.2) holds in this range. Observe that
since $0.055071\leq\frac{0.16520}{2.9997}$ and $X(t)$ is bounded,
$M_{1}=0.055071$ for $t\geq T_{0}$ and $T_{0}$ sufficiently large. Next, in
order to bound the $9.791\log\log q$ term in the denominator of (3.1) from
above, we split into the two cases $q\geq Q_{0}$ and $q<Q_{0}$, where $Q_{0}$
is a sufficiently large absolute constant. If $T_{0}$ and $Q_{0}$ are
sufficiently large (where the thresholds for both $T_{0}$ and $Q_{0}$ are
absolute), then for all $q\geq Q_{0}$ we have that
$10.082+\frac{1.607}{\log\log T_{0}}+9.791\frac{\log\log q}{\log q}\leq 10.1$.
On the other hand, if $q<Q_{0}$ and $T_{0}$ is sufficiently large (in terms of
$Q_{0}$ and $B$), then

$\log\log q<\log\log
Q_{0}<(0.05507^{-1}-0.055071^{-1})B^{2/3}(\log|t|)^{2/3}(\log\log|t|)^{1/3}.$

These observations together imply that if $T_{0}$ is sufficiently large and
$t\geq T_{0}$, then the denominator in (3.1) is bounded above by $10.1\log
q+(0.05507)^{-1}B^{2/3}(\log|t|)^{2/3}(\log\log|t|)^{1/3}$. Consequently,
Theorem 3.1 implies Corollary 3.2 as long as for some positive $T_{0}$, the
condition that (3.1) holds for all zeros $\beta+i\gamma$ with
$T_{0}-1\leq\gamma\leq T_{0}$ is realized. We will show that this is indeed
the case using a log-free zero-density estimate for Dirichlet $L$-functions.

Define $N(\sigma,T,\chi)=\\#\\{\rho=\beta+i\gamma\colon
L(\rho,\chi)=0,~{}\sigma\leq\beta\leq 1,|\gamma|\leq T\\}$, where each zero
$\beta+i\gamma$ is counted with multiplicity. Write $c=(0.055071)^{-1}$. Let
$\varepsilon>0$ be arbitrary. By [jutila, Theorem 1], there exists an
effectively computable constant $c_{\varepsilon}>0$ such that in the range
$4/5\leq\sigma\leq 1$ and $T\geq 1$, we have

$N(\sigma,T,\chi)\leq c_{\varepsilon}(qT)^{(2+\varepsilon)(1-\sigma)}.$

Taking $\sigma=1-(10.2\log q+(c/2)B^{2/3}(\log T)^{2/3}(\log\log
T)^{1/3})^{-1}$, we find that if $T$ is sufficiently large, then

$N(\sigma,T,\chi)\leq
c_{\varepsilon}e^{\frac{2+\varepsilon}{10.2}}\exp\Big(\frac{(2+\varepsilon)}{(c/2)B^{2/3}}\Big{(}\frac{\log
T}{\log\log T}\Big{)}^{1/3}\Big{missing})\leq\frac{T}{2}.$

As such, if $T$ is sufficiently large, then there exists
$T_{0}\in[\max\\{e^{1938},\sqrt{T}\\},T]$ such that all zeros $\beta+i\gamma$
of $L(s,\chi)$ with $T_{0}-1\leq\gamma\leq T_{0}$ satisfy

$\sigma<1-(10.2\log q+(c/2)B^{2/3}(\log T)^{2/3}(\log\log T)^{1/3})^{-1}\\\
\leq 1-(10.2\log q+cB^{2/3}(\log\gamma)^{2/3}(\log\log\gamma)^{1/3})^{-1},$

as desired. ∎

We now demonstrate how Theorem 1.1 may be deduced from Theorem 3.1.

###### Proof of Theorem 1.1, assuming Theorem 3.1.

Note that (1.3) is immediate from Corollary 3.2 with $B=4.45$. It remains to
demonstrate that (1.2) follows from Theorem 3.1. Note that (1.3) is immediate
from Corollary 3.2 with $B=4.45$. It remains to demonstrate that (1.2) follows
from Theorem 3.1. We will show the slightly stronger statement that
$L(s,\chi)$ is nonvanishing in the region

$\sigma\geq 1-\frac{1}{10.3\log q+9.791\log\log q+61.306(\log
t)^{2/3}(\log\log t)^{1/3}},\quad|t|\geq 10$ (3.3)

To see that Theorem 1.1 follows from (3.3), suppose that $t\geq e^{1944}$,
since Theorem 1.1 is a consequence of (2.3) when $t\leq e^{1944}$. If
$q<e^{428}$, then $9.791\log\log q<9.791\log(428)<0.194(\log t)^{2/3}(\log\log
t)^{1/3}$. On the other hand, if $q\geq e^{428}$, then $9.791\log\log
q<0.13861\log q$. In either case, (1.2) follows from (3.3).

Observe that if $t\geq T_{0}\geq e^{1944}$, $A=76.2$, $B=4.45$ and
$\tau=4t+1$, then

$\displaystyle
1.16\log(A+1)+0.845593+0.23531\log\Big(\frac{B}{\log\log\tau}\Big{missing})+\Big{(}\frac{1.694}{B^{4/3}}-\frac{2.087}{B^{1/3}}\Big{)}\Big{(}\frac{\log\log\tau}{\log\tau}\Big{)}^{\frac{1}{3}}$
$\displaystyle\leq
6.238712-0.23531\log\log\log\tau-1.037389\Big{(}\frac{\log\log\tau}{\log\tau}\Big{)}^{1/3}\leq
5.61718,$

since the first derivative test shows that the middle expression above is
maximized at $\tau\approx e^{24\,012.6}$ on the domain $\tau\geq e^{1944}$.
Theorem 3.1 implies that for $T_{0}\geq e^{1944}$,

$M_{1}\geq\min\Big{(}0.055071,\frac{0.16511492}{2.9997+\frac{5.61718}{\log\log
T_{0}}}\Big{)}.$ (3.4)

Moreover, if $T_{0}\geq e^{1944}$, then

$10.082+\frac{1.607}{\log\log T_{0}}\leq 10.3.$ (3.5)

For $t\leq e^{1944}$, (3.3) follows from (2.3). On the other hand, for $t\geq
e^{1944}$, we apply Theorem 3.1 with $A=76.2$, $B=4.45$, and $T_{0}=e^{1944}$,
utilizing (3.4) and (3.5) to bound the constants in front of $(\log
t)^{2/3}(\log\log t)^{1/3}$ and $\log q$ respectively. ∎

We note that if the unpublished result [kadiri, (17)] is used in place of
[mccurley, Theorem 1.1], the constant $61.306$ in (3.3) may be lowered to
$59.521$.

##  4\. Basic Estimates for $|\zeta(s)|$ and $L(s,\chi)$

First, we will require the following elementary estimates for $L(s,\chi)$.

######  Lemma 4.1.

If $1<\sigma\leq 1.06$ and $t$ is real, then for any Dirichlet character
$\chi\pmod{q}$,

$\frac{1}{\zeta(\sigma)}\leq|L(\sigma+it,\chi)|\leq\zeta(\sigma)\leq
0.6+\frac{1}{\sigma-1}.$

For all $\sigma>1$ and real $t$, we have

$\Big{|}-\frac{L^{\prime}}{L}(\sigma+it,\chi)\Big{|}<\frac{1}{\sigma-1}.$

###### Proof.

This follows immediately from [fordzfr, Lemma 3.1]. ∎

######  Lemma 4.2.

Let $\chi\pmod{q}$ be a primitive Dirichlet character, with $q\geq 3$, and let
$u$ be real. Then, we have

$\displaystyle\Big{|}\frac{L^{\prime}}{L}(-\textstyle{\frac{1}{2}}+iu,\chi)\Big{|}\leq
8.21+\log q+\frac{1}{2}\log(1+u^{2}/4).$

###### Proof.

The functional equation for Dirichlet $L$-functions (see (2.1), (2.2)) implies
that

$\displaystyle\frac{1}{2}\log\Big(\frac{q}{\pi}\Big{missing})+\frac{1}{2}\frac{\Gamma^{\prime}}{\Gamma}\Big{(}\frac{s+\mathfrak{a}}{2}\Big{)}+\frac{L^{\prime}}{L}(s,\chi)=-\frac{1}{2}\log\Big(\frac{q}{\pi}\Big{missing})-\frac{1}{2}\frac{\Gamma^{\prime}}{\Gamma}\Big{(}\frac{1-s+\mathfrak{a}}{2}\Big{)}-\frac{L^{\prime}}{L}(1-s,\overline{\chi}),$

or, by rearranging,

$\displaystyle-\frac{L^{\prime}}{L}(s,\chi)=\frac{L^{\prime}}{L}(1-s,\overline{\chi})+\log\Big(\frac{q}{\pi}\Big{missing})+\frac{1}{2}\Big{(}\frac{\Gamma^{\prime}}{\Gamma}\Big{(}\frac{s+\mathfrak{a}}{2}\Big{)}+\frac{\Gamma^{\prime}}{\Gamma}\Big{(}\frac{1-s+\mathfrak{a}}{2}\Big{)}\Big{)}.$

If $s=-\frac{1}{2}+iu$, then it follows from the functional equation
$\Gamma(z+1)=z\Gamma(z)$ and the identity
$\overline{\Gamma(z)}=\Gamma(\overline{z})$ twice that

$\displaystyle\frac{1}{2}\Big{(}\frac{\Gamma^{\prime}}{\Gamma}\Big{(}\frac{s+\mathfrak{a}}{2}\Big{)}+\frac{\Gamma^{\prime}}{\Gamma}\Big{(}\frac{1-s+\mathfrak{a}}{2}\Big{)}\Big{)}$
$\displaystyle=\mathrm{Re}\Big{(}\frac{\Gamma^{\prime}}{\Gamma}\Big{(}\frac{3}{4}+\frac{\mathfrak{a}}{2}+\frac{iu}{2}\Big{)}\Big{)}+\frac{1}{\frac{1}{2}-\mathfrak{a}-iu}$
(4.1)
$\displaystyle=\mathrm{Re}\Big{(}\frac{\Gamma^{\prime}}{\Gamma}\Big{(}\frac{7}{4}+\frac{\mathfrak{a}}{2}+\frac{iu}{2}\Big{)}-\frac{1}{\frac{3}{4}+\frac{\mathfrak{a}}{2}+\frac{iu}{2}}\Big{)}+\frac{1}{\frac{1}{2}-\mathfrak{a}-iu}.$
(4.2)

Ono and Soundararajan show in [sound, Lemma 4] that if $\mathrm{Re}(z)\geq 1$,
then

$\Big{|}\frac{\Gamma^{\prime}}{\Gamma}(z)\Big{|}\leq\frac{11}{3}+\frac{\log(1+x^{2})}{2}+\frac{\log(1+y^{2})}{2}.$

We use the inequality $|\log(q/\pi)|\leq\log(q)-1$ (which holds for $q\geq
3$), combined with (4.1) if $\mathfrak{a}=1$ and (4.2) if $\mathfrak{a}=0$ to
find that

$\displaystyle\Big{|}\frac{L^{\prime}}{L}(-\tfrac{1}{2}+iu,\chi)\Big{|}$
$\displaystyle\leq\log(q)-1+\Big{|}\frac{L^{\prime}}{L}\Big{(}\frac{3}{2}+iu,\overline{\chi}\Big{)}\Big{|}+\sqrt{\frac{4}{1+4u^{2}}}+\sqrt{\frac{16}{9+4u^{2}}}$
$\displaystyle\qquad+\frac{11}{3}+\frac{\log(1+49/16)}{2}+\frac{\log(1+u^{2}/4)}{2}.$

Since
$|\frac{L^{\prime}}{L}(\frac{3}{2}+iu,\overline{\chi})|\leq-\frac{\zeta^{\prime}}{\zeta}(\frac{3}{2})\leq
1.505236$, the desired result follows. ∎

In standard treatments, the zeros of Dirichlet $L$-functions are counted
“symmetrically,” i.e. in the region $\absolutevalue{t}\leq T$. However, we
require an estimate for zeros of $L(s,\chi)$ in the upper half plane, for
which we could find no explicit estimates in the literature. Let

$N^{\pm}(T,\chi)=\\#\\{\rho=\beta+i\gamma:0<\pm\gamma<T\\}+\frac{\\#\\{\rho=\beta+i\gamma:\gamma\in\\{0,\pm
T\\}\\}}{2}.$

######  Lemma 4.3.

Let $\chi$ be a primitive Dirichlet character modulo $q\geq 3$. For all $T>1$
and $0<\eta\leq 1/2$,

$\Big{|}N^{\pm}(T,\chi)-\frac{T}{2\pi}\log\frac{qT}{2\pi
e}-C_{0}(\chi,\eta)\Big{|}\leq C_{1}\log(qT)+C_{2},$

where

$C_{1}=\frac{1+2\eta}{2\pi\log(2)},\qquad
C_{2}=0.1529-0.134\eta+\frac{2\log\zeta(1+\eta)}{\log
2}-\frac{\log\zeta(2+2\eta)}{\log
2}+\frac{2}{\pi}\log\zeta\Big{(}\frac{3}{2}+2\eta\Big{)}.$

###### Proof.

By interchanging $\chi$ and $\overline{\chi}$, it suffices to prove the
desired estimate for $N^{+}(T,\chi)$. If we define

$S(T,\chi):=\frac{1}{2\pi}\Big{[}\lim_{t\to
T^{+}}\int_{\frac{1}{2}}^{\infty}\mathrm{Im}\Big{(}-\frac{L^{\prime}}{L}(\sigma+it,\chi)\Big{)}d\sigma+\lim_{t\to
T^{-}}\int_{\frac{1}{2}}^{\infty}\mathrm{Im}\Big{(}-\frac{L^{\prime}}{L}(\sigma+it,\chi)\Big{)}d\sigma\Big{]},$

then by [mv, Theorem 14.5], we have that

$N(T,\chi)=\frac{1}{\pi}\arg\Big{(}\Gamma\Big{(}\frac{1}{4}+\frac{\mathfrak{a}}{2}+i\frac{T}{2}\Big{)}\Big{)}+\frac{T}{2\pi}\log\frac{q}{\pi}+S(T,\chi)-S(0,\chi).$
(4.3)

Let $\sigma_{1}=\frac{3}{2}+2\eta>1$, and let
$C_{0}(\chi,\eta)=\frac{1}{\pi}\arg(L(\sigma_{1},\chi))-S(0,\chi)$.

By [mccurley, (2.8)] and the subsequent unlabeled display, we have that for
$T\geq 1$,

$\displaystyle\Big{|}\arg\Big{(}\Gamma\Big{(}\frac{1}{4}+\frac{\mathfrak{a}}{2}+i\frac{T}{2}\Big{)}\Big{)}-\frac{T}{2}\log\frac{T}{2e}\Big{|}\leq
0.6909.$ (4.4)

Next, we bound $|S(T,\chi)-\frac{1}{\pi}\arg(L(\sigma_{1},\chi))|$ by
considering the variation of $\arg(L(s,\chi))$ along the contour
$\mathcal{C}:\sigma_{1}\to\sigma_{1}+iT\to 1/2+iT$. By [mccurley, (2.15)], we
have for all $T\geq 1$ that

$\Big{|}\Delta_{\sigma_{1}+iT\to\frac{1}{2}+iT}~{}\arg(L(s,\chi))\Big{|}\leq\frac{1+2\eta}{2\log
2}\log(0.74685qT)+\frac{2\pi\log\zeta(1+\eta)}{\log
2}-\frac{\pi\log\zeta(2+2\eta)}{\log 2}.$ (4.5)

Moreover, since $|\arg L(s,\chi)|\leqslant|\log
L(s,\chi)\mid\leqslant\log\zeta(\mathrm{Re}(s))$ and
$\sigma_{1}=\frac{3}{2}+2\eta$, we have

$|\Delta_{\sigma_{1}+it_{0}\to\sigma_{1}+iT}~{}{\arg(L(s,\chi))}|\leq
2\log\Big(\zeta\Big{(}\frac{3}{2}+2\eta\Big{)}\Big{missing}).$ (4.6)

Combining (4.4), (4.5), and (4.6) yields the desired result. ∎

We obtain the following corollary.

######  Corollary 4.4.

Let $\chi$ be a primitive Dirichlet character modulo $q\geq 3$. For all $T>1$,

$\absolutevalue{N^{\pm}(T,\chi)-\frac{T}{2\pi}\log\frac{qT}{2\pi
e}-C_{0}(\chi)}\leq 0.2297\log(qT)+24.77,$

where $C_{0}$ is as given in Lemma 4.3.

###### Proof.

Take $\eta=0.00019$, in Lemma 4.3. ∎

##  5\.  Additional Bounds for $|\zeta(s)|$ and $|L(s,\chi)|$

Unlike the results of the previous section, the estimates in this section will
depend on the explicit exponential sum estimates derived in Ford [ford]. We
first extract an upper bound for $L(s,\chi)$ from (2.4).

######  Lemma 5.1.

Assume that there exist positive constants $A$ and $B$ such that (2.4) holds.
Let $q\geq 3$, and let $\chi\pmod{q}$ be a Dirichlet character. If $|t|\geq 3$
and $1/2\leq\sigma\leq 1$, then

$\displaystyle|L(\sigma+it,\chi)|\leq
Aq^{1-\sigma}|t|^{B(1-\sigma)^{3/2}}(\log|t|)^{2/3}+1.92\frac{q^{1-\sigma}-1}{1-\sigma}.$

###### Proof.

Note by the periodicity of the values of $\chi(n)$ that

$\displaystyle L(\sigma+it,\chi)$
$\displaystyle=q^{-(\sigma+it)}\sum_{a=1}^{q}\chi(a)\zeta\Big{(}\sigma+it,\frac{a}{q}\Big{)}.$

Hence, it follows from (2.4) that

$\displaystyle|L(\sigma+it,\chi)|$
$\displaystyle=\Big{|}q^{-(\sigma+it)}\sum_{a=1}^{q}\chi(a)\Big{(}\zeta\Big{(}\sigma+it,\frac{a}{q}\Big{)}-\Big{(}\frac{a}{q}\Big{)}^{-(\sigma+it)}\Big{)}+q^{-(\sigma+it)}\sum_{a=1}^{q}\Big{(}\frac{a}{q}\Big{)}^{-(\sigma+it)}\Big{|}$
$\displaystyle\leq
Aq^{1-\sigma}|t|^{B(1-\sigma)^{3/2}}(\log|t|)^{2/3}+\sum_{a=1}^{q}a^{-\sigma}.$

To bound the sum above, observe that

$\displaystyle\sum_{a=1}^{q}a^{-\sigma}\leq
1+\int_{1}^{q}u^{-\sigma}du=1+\frac{u^{1-\sigma}}{1-\sigma}\Big{\rvert}_{u=1}^{q}=1+\frac{q^{1-\sigma}-1}{1-\sigma}.$

Furthermore, note that $\sigma\leq 1$ implies that
$\frac{q^{1-\sigma}-1}{1-\sigma}=\int_{1}^{q}u^{-\sigma}du\geq\int_{1}^{q}\frac{du}{u}=\log
q$. Since $q\geq 3$, it follows that $1+\frac{q^{1-\sigma}-1}{1-\sigma}\leq
1.92\frac{q^{1-\sigma}-1}{1-\sigma}$. ∎

This following lemma is a variant of [fordzfr, Lemma 3.4].

######  Lemma 5.2.

Let $q\geq 3$, and let $\chi\pmod{q}$ be a Dirichlet character. Fix
$\sigma\in[1/2,1)$, and $a\in(0,1/2]$. Assume that there exist positive
constants $A$ and $B$ such that (2.4) holds. Suppose that $t,q$ satisfy

$\displaystyle 1-\sigma\geq 1.92(\log(t/100))^{-2/3},\qquad
t\geq\max\\{q^{\frac{1}{100\,000}},e^{1938}\\}.$ (5.1)

Then, it follows that

$\int_{-\infty}^{\infty}\frac{\log|L(\sigma+it+iau,\chi)|}{(\cosh
u)^{2}}du\leq 2\Big{(}\log(A+1)+(1-\sigma)\log q\\\ +B(1-\sigma)^{3/2}\log
t+\frac{2}{3}\log\log t\Big{)}.$ (5.2)

###### Proof.

Note that even if $L(\sigma+it+iau,\chi)=0$ on the path of integration, the
integral in (5.2) converges, since all zeros have finite order. Let

$S=\Big{\\{}u\in\mathbb{R}:\frac{1.92}{1-\sigma}\geq|t+au|^{B(1-\sigma)^{3/2}}(\log|t+au|)^{2/3}\Big{\\}}.$

Note that if $u\notin S$ and $|t+au|>3$, then

$\displaystyle|L(\sigma+it+iau,\chi)|\leq(A+1)q^{1-\sigma}|t+au|^{B(1-\sigma)^{3/2}}(\log|t+au|)^{2/3}.$
(5.3)

To see this, observe that if $u\notin S$, then

$\displaystyle 1.92\frac{q^{1-\sigma}-1}{1-\sigma}\leq
q^{1-\sigma}|t+au|^{B(1-\sigma)^{3/2}}(\log|t+au|)^{2/3}.$

Consequently, (5.3) follows from Lemma 5.1 (which is applicable since
$|t+au|\geq 3$). We now split up the the integral in (5.2) into six disjoint
pieces.

Case 1: $|t+au|\leq 3$. Note that in this range,
$|\sigma+it+iau|\leq\sqrt{10}$. For $\chi$ nonprincipal, we use

$\displaystyle|L(s,\chi)|=|s|\Big{|}\int_{1}^{\infty}\sum_{n\leq
x}\chi(n)x^{-s-1}dx\Big{|}\leq|s|q\int_{1}^{\infty}x^{-3/2}dx=2q|s|$

with $s=\sigma+it+iau$, which implies that
$\log|L(\sigma+it+iau,\chi)|\leq\log(2q|\sigma+it+iau|)\leq\log(2\sqrt{10}q)$.
By the assumptions $t\geq q^{\frac{1}{100\,000}}$ and $t\geq e^{1938}$, we
have that

$|L(\sigma+it+iau,\chi)|\leq 100\,000\log t+\log(2\sqrt{10})\leq 100\,001\log
t.$ (5.4)

On the other hand, if $\chi$ is principal, then the same conclusion follows
from the identity (see [tit, (2.1.4)])

$\zeta(s)=\frac{1}{s-1}+\frac{1}{2}+s\int_{1}^{\infty}\frac{\lfloor
x\rfloor-x+1/2}{x^{s+1}}dx,$

combined with (5.1) and the fact that

$|\log|L(s,\chi_{0})|-\log|\zeta(s)||\leq\sum_{p\mid
q}\log\frac{1}{|1-p^{-\sigma}|}\leq-\log\big((1-2^{-\frac{1}{2}})(1-3^{-\frac{1}{2}})\big{missing})-\log\big(1-5^{-\frac{1}{2}}\big{missing})\frac{\log
q}{\log 5}.$

Case 2: $u\in S$ and $|t+au|>3$. Observe that

$\frac{q^{1-\sigma}-1}{1-\sigma}\leq\begin{cases}(e-1)\log q&\mbox{if
$\sigma\geq 1-(\log q)^{-1}$,}\\\ q^{1-\sigma}\log q&\mbox{if $\sigma<1-(\log
q)^{-1}$.}\end{cases}$

Thus, in either case, we have that
$\frac{q^{1-\sigma}-1}{1-\sigma}\leq\sqrt{q}\log q$. Consequently, by the
assumptions that $q\leq t^{100\,000}$ and $t\geq e^{1938}$, we have that

$\displaystyle\log\Big(1.92\frac{q^{1-\sigma}-1}{1-\sigma}\Big{missing})\leq\log(1.92)+\frac{1}{2}\log
q+\log\log q\leq 50\,001\log t.$ (5.5)

Moreover, by (5.1), the assumption $t\geq e^{1938}$, and the fact that $u\in
S$, we have that

$\displaystyle\log(Aq^{1-\sigma}|t+au|^{B(1-\sigma)^{3/2}}(\log|t+au|)^{2/3})$
$\displaystyle\leq\log(A)+\frac{1}{2}\log q+\log((\log t)^{2/3})$ (5.6)
$\displaystyle\leq\log(A)+50\,000\log(t)+\frac{2}{3}\log\log t$
$\displaystyle\leq\log(A)+50\,001\log(t).$

It follows from Corollary 5.1, (5.5) and (5.6) that for $u\in S$ such that
$|t+au|>3$, we have

$\displaystyle\log|L(\sigma+it+iau,\chi)|$
$\displaystyle\leq\log\Big(Aq^{1-\sigma}|t+au|^{B(1-\sigma)^{3/2}}(\log|t+au|)^{2/3}+1.92\frac{q^{1-\sigma}-1}{1-\sigma}\Big{missing})$
(5.7) $\displaystyle\leq\log(2)+\log(A+1)+50\,001\log(t)$
$\displaystyle\leq\log(A+1)+50\,002\log(t).$

Case 3: $\frac{-2t}{a}\leq u<\frac{-t-3}{a}$ and $u\notin S$. In this case,
(5.3) gives the inequality

$\displaystyle|L(\sigma+it+iu,\chi)|\leq(A+1)q^{1-\sigma}t^{B(1-\sigma)^{3/2}}(\log
t)^{2/3}.$ (5.8)

Case 4: $u>\frac{3-t}{a}$ and $u\notin S$. We use the inequalities
$\log(1+x)\leq x$ and $\log(1+x)\leq x-\frac{1}{2}x^{2}+\frac{1}{3}x^{3}$,
both valid for $x>-1$, alongside (5.3) to obtain that

$\displaystyle\log|L(\sigma+it+iau,\chi)|$
$\displaystyle\leq\log(A+1)+(1-\sigma)\log q$ (5.9)
$\displaystyle\qquad+B(1-\sigma)^{3/2}\log(t+au)+\frac{2}{3}\log\log(t+au)$
$\displaystyle\leq\log\Big((A+1)q^{1-\sigma}t^{B(1-\sigma)^{3/2}}(\log
t)^{2/3}\Big{missing})$
$\displaystyle\qquad+\Big{(}B(1-\sigma)^{3/2}+\frac{2}{3\log
t}\Big{)}\Big{(}\frac{au}{t}-\frac{(au)^{2}}{2t^{2}}+\frac{(au)^{3}}{3t^{3}}\Big{)}.$

Case 5: $u<-\frac{2t}{a}$ and $u\notin S$. Using $\log(1+x)\leq x$ and (5.3),
we find that

$\displaystyle\log|L(\sigma+it+iau,\chi)|$
$\displaystyle\leq\log\Big((A+1)q^{1-\sigma}t^{B(1-\sigma)^{3/2}}(\log
t)^{2/3}\Big{missing})$ (5.10)
$\displaystyle\qquad+\Big{(}B(1-\sigma)^{3/2}+\frac{2}{3\log
t}\Big{)}\Big{(}\frac{-au-2t}{t}\Big{)}.$

The estimates (5.4) and (5.7)-(5.10), along with the identity
$\int_{-\infty}^{\infty}(\sech(x))^{2}dx=2$, yield that

$\displaystyle\int_{-\infty}^{\infty}\frac{\log|L(\sigma+it+iau,\chi)|}{(\cosh
u)^{2}}du$ $\displaystyle\qquad\leq 2\Big{(}\log(A+1)+(1-\sigma)\log
q+B(1-\sigma)^{3/2}\log t+\frac{2}{3}\log\log t\Big{)}+E,$

where

$\displaystyle E$ $\displaystyle=50\,002\log(t)\int_{\begin{subarray}{c}u\in
S\\\ |t+au|>3\end{subarray}}\frac{1}{(\cosh
u)^{2}}du+\frac{600\,006\log(t)}{a(\cosh(\frac{3-t}{a}))^{2}}$
$\displaystyle+\Big{(}B(1-\sigma)^{3/2}+\frac{2}{3\log
t}\Big{)}\Big{(}\int_{-\infty}^{-\frac{2t}{a}}\frac{-au-2t}{t(\cosh
u)^{2}}du+\int_{\frac{3-t}{a}}^{\infty}\frac{\frac{au}{t}-\frac{(au)^{2}}{2t^{2}}+\frac{(au)^{3}}{3t^{3}}}{(\cosh
u)^{2}}du\Big{)}.$

By (5.1), $S$ is a subset of $\\{u:u\leq-0.99t/a\\}$. Since (for $M\geq 0$) we
have that $\int_{u\leq-M}\frac{du}{(\cosh u)^{2}}\leq 2e^{-2M}$, and since
$a\leq 1/2$, it follows that

$\displaystyle 50\,002\log(t)\int_{\begin{subarray}{c}u\in S\\\
|t+au|>3\end{subarray}}\frac{1}{(\cosh u)^{2}}du$ $\displaystyle\leq
100\,004\log(t)e^{-1.98t/a}$ $\displaystyle\leq e^{-t/a}\cdot e^{-1.96t}\cdot
100\,004\log(t)\leq 0.001e^{-t/a}.$

Since $a\leq 1/2$ and $t\geq e^{1938}$, we have that $ae^{(t-6)/a}\geq
2e^{2t-12}$. Moreover, since $(\cosh u)^{2}\geq\frac{1}{4}e^{2|u|}$, we have
that
$a(\cosh(\frac{3-t}{a}))^{2}\geq\frac{1}{4}ae^{(t-6)/a}e^{t/a}\geq\frac{1}{2}e^{2t-12}e^{t/a}$.
Consequently, we find that

$\displaystyle
0.001e^{-t/a}+\frac{600\,006\log(t)}{a(\cosh(\frac{3-t}{a}))^{2}}$
$\displaystyle\leq 0.001e^{-t/a}+\frac{1\,200\,012\log(t)}{e^{t/a+2t-12}}\leq
e^{-t/a}.$

The final lines of the proof of [fordzfr, Lemma 3.4] show, mutatis mutandis,
that

$\displaystyle\Big{(}B(1-\sigma)^{3/2}+\frac{2}{3\log
t}\Big{)}\Big{(}\int_{-\infty}^{-\frac{2t}{a}}\frac{-au-2t}{t(\cosh
u)^{2}}du+\int_{\frac{3-t}{a}}^{\infty}\frac{\frac{au}{t}-\frac{(au)^{2}}{2t^{2}}+\frac{(au)^{3}}{3t^{3}}}{(\cosh
u)^{2}}du\Big{)}\leq-\frac{a}{20t^{2}\log t}.$

It follows that $E\leq 0$. ∎

##  6\. Detecting zeros of $L(s,\chi)$

We first recall the form of Jensen’s formula shown for functions in a vertical
strip proven in [fordzfr, Lemma 2.2].

######  Lemma 6.1 (Lemma 2.2 of [ford]).

Suppose $f$ is the quotient of two entire functions of finite order, and does
not have a zero or a pole at $z=z_{0}$ nor at $z=0$. Then, for all $\eta>0$
except for a set of Lebesgue measure 0 (the exceptional set may depend on $f$
and $z_{0}$), we have

$\displaystyle-\mathrm{Re}\Big{(}\frac{f^{\prime}(z_{0})}{f(z_{0})}\Big{)}$
$\displaystyle=\frac{\pi}{2\eta}\sum_{|\mathrm{Re}(z_{0}-\omega)|\leq\eta}m_{\omega}\mathrm{Re}\Big{(}\cot\Big(\frac{\pi(\omega-
z_{0})}{2\eta}\Big{missing})\Big{)}$
$\displaystyle+\frac{1}{4\eta}\int_{-\infty}^{\infty}\frac{\log|f(z_{0}-\eta+\frac{2\eta
iu}{\pi})|-\log|f(z_{0}+\eta+\frac{2\eta iu}{\pi})|}{(\cosh u)^{2}}du.$

where $\omega$ runs over the zeros and poles of $f$, and where the
multiplicity $m_{\omega}$ is positive if $\omega$ is a zero or negative if
$\omega$ is a pole.

The following lemma is our main method for detecting zeros of $L(s,\chi)$.

######  Lemma 6.2.

Assume that there exist positive constants $A$ and $B$ such that (2.4) holds.
Let $q\geq 3$, and let $\chi\pmod{q}$ be a Dirichlet character. Let
$t\geq\max\\{e^{1938},q^{\frac{1}{100\,000}}\\}$, $0<\eta<10$,
$\sigma-\eta>1/2$, $1\leq\sigma\leq 1+\eta-1.92(\log(t/100))^{-2/3}$, and
$s=\sigma+it$. If
$S\subseteq\\{z\in\mathbb{C}\colon\sigma-\eta\leq\mathrm{Re}(z)\leq 1\\}$,
then

$\displaystyle-\mathrm{Re}\Big{(}\frac{L^{\prime}(s,\chi)}{L(s,\chi)}\Big{)}$
$\displaystyle\leq-\sum_{\rho\in
S,L(\rho,\chi)=0}\mathrm{Re}\Big{(}\frac{\pi}{2\eta}\cot\Big(\frac{\pi(s-\rho)}{2\eta}\Big{missing})\Big{)}$
$\displaystyle\quad+\frac{1}{2\eta}\Big{(}(1-\sigma+\eta)\log
q+\frac{2}{3}\log\log t+B(1-\sigma+\eta)^{3/2}\log t+\log(A+1)\Big{)}$
$\displaystyle\quad-\frac{1}{4\eta}\int_{-\infty}^{\infty}\frac{\log|L(s+\eta+2\eta
iu/\pi,\chi)|}{(\cosh u)^{2}}du+(1-\mathfrak{a})(1-\delta(\chi))e^{-1937}.$

###### Proof.

Apply Lemma 6.1 with $f=s^{(1-\delta(\chi))(\mathfrak{a}-1)}L(s,\chi)$ and
$z_{0}=s$. The conclusion follows precisely as in the proof of [fordzfr, Lemma
4.1] when $\mathfrak{a}=1$ or $\delta(\chi)=1$, with Lemma 5.2 in place of
[fordzfr, Lemma 3.4]. When $\delta(\chi)=0$ and $\mathfrak{a}=0$, we have that

$\displaystyle\frac{1}{4\eta}\int_{-\infty}^{\infty}\frac{\log|(s-\eta+\frac{2\eta
iu}{\pi})^{-1}L(s-\eta+\frac{2\eta iu}{\pi},\chi)|-\log|(s+\eta+\frac{2\eta
iu}{\pi})^{-1}L(s+\eta+\frac{2\eta iu}{\pi},\chi)|}{(\cosh u)^{2}}du$
$\displaystyle=\frac{1}{4\eta}\int_{-\infty}^{\infty}\frac{\log|L(s-\eta+\frac{2\eta
iu}{\pi},\chi)|-\log|L(s+\eta+\frac{2\eta iu}{\pi},\chi)|}{(\cosh u)^{2}}du$
$\displaystyle\qquad+\frac{1}{4\eta}\int_{-\infty}^{\infty}\log|\frac{s+\eta+\frac{2\eta
iu}{\pi}}{s-\eta+\frac{2\eta iu}{\pi}}\Big{|}\frac{du}{(\cosh u)^{2}}.$

Moreover, we have that

$\displaystyle\frac{1}{4\eta}\int_{-\infty}^{\infty}\log|\frac{s+\eta+\frac{2\eta
iu}{\pi}}{s-\eta+\frac{2\eta iu}{\pi}}\Big{|}\frac{du}{(\cosh u)^{2}}$
$\displaystyle=\frac{1}{4\eta}\int_{-\infty}^{\infty}\log|1+\frac{2\eta}{s-\eta+\frac{2\eta
iu}{\pi}}\Big{|}\frac{du}{(\cosh u)^{2}}$
$\displaystyle\leq\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{\absolutevalue{s-\eta+\frac{2\eta
iu}{\pi}}}\frac{du}{(\cosh u)^{2}}.$

Finally, we crudely estimate the latter integral. Utilizing the facts that
$t\geq e^{1938}$, $\eta\leq 10$, $\sigma-\eta>1/2$, and that for $M\geq 0$,
$\int_{u\leq-M}\frac{du}{(\cosh u)^{2}}\leq 2e^{-2M}$, we obtain

$\displaystyle\int_{-\infty}^{\infty}\frac{1}{\absolutevalue{s-\eta+\frac{2\eta
iu}{\pi}}}\frac{du}{(\cosh u)^{2}}$
$\displaystyle=\int_{\absolutevalue{t+\frac{2\eta
u}{\pi}}<0.5t}\frac{1}{\absolutevalue{s-\eta+\frac{2\eta
iu}{\pi}}}\frac{du}{(\cosh u)^{2}}+\int_{\absolutevalue{t+\frac{2\eta
u}{\pi}}\geq 0.5t}\frac{1}{\absolutevalue{s-\eta+\frac{2\eta
iu}{\pi}}}\frac{du}{(\cosh u)^{2}}$ $\displaystyle<4e^{-\frac{\pi
t}{40}}+4/t\leq 2e^{-1937},$

which completes the proof. ∎

In our proof, we require strong upper bounds for the number of zeros of
$L(s,\chi)$ near the point $1+it$. Let $N_{\chi}(t,R)$ denote the number of
zeros $\rho$ of $L(s,\chi)$ satisfying $|1+it-\rho|\leq R$.

######  Lemma 6.3.

Assume that there exist positive constants $A$ and $B$ such that (2.4) holds.
Let $q\geq 3$, and let $\chi\pmod{q}$ be a Dirichlet character. If
$1.04(\log(t/100))^{-2/3}\leq R\leq 1/4$ and
$t\geq\max\\{e^{1938},q^{\frac{1}{100\,000}}\\}$, then we have

$\displaystyle N_{\chi}(t,R)\leq 1.3478R^{3/2}B\log
t+0.49+\frac{\log(A+1)-\log R+1.8579R\log q+\frac{2}{3}\log\log t}{1.879}.$

###### Proof.

Apply Lemma 6.2 with $s=1+0.6421R+it$, $\eta=2.5R$, and

$S=\\{z\in\mathbb{C}:|1+it-z|\leq R,~{}\mathrm{Re}(z)\leq 1\\}.$

(Note that with these choices, $\sigma-\eta\geq 1/2$.) By Lemma 4.1,

$\displaystyle\Big{|}\frac{L^{\prime}}{L}(s+\eta+iv,\chi)\Big{|}$
$\displaystyle\leq\frac{1}{3.1421R},$ (6.1)
$\displaystyle|L(s+\eta+iv,\chi)|^{-1}$ $\displaystyle\leq\zeta(1+3.1421R)\leq
0.6+\frac{1}{3.1421R}.$

By [ford, (4.2)], in the region $U=\\{z:\mathrm{Re}(z)\geq
0.6421,|z-0.6421|\leq 1\\}$, we have that

$\mathrm{Re}\Big{(}\frac{\pi}{5}\cot\Big(\frac{\pi
z}{5}\Big{missing})\Big{)}\geq 0.3758.$ (6.2)

Thus, it follows from (6.1), (6.2), and Lemma 6.2 that

$\displaystyle-\frac{1}{3.1421R}$
$\displaystyle\leq-0.3758\frac{N_{\chi}(t,R)}{R}+\frac{1}{5R}\Big{(}1.8579R\log
q+\frac{2}{3}\log\log t+(1.8579R)^{3/2}B\log t$
$\displaystyle\qquad+\log(A+1)+\log\Big(0.6+\frac{1}{3.1421R}\Big{missing})\Big{)}+(1-\mathfrak{a})(1-\delta(\chi))e^{-1937}.$

By the argument concluding the proof of [ford, Lemma 4.2], we have that
$\log(0.6+\frac{1}{3.1421}R)\leq-\log R-0.6735$, which completes the proof. ∎

######  Lemma 6.4.

Let $q\geq 3$, and let $\chi\pmod{q}$ be a Dirichlet character. If
$t\geq\max\\{e^{1938},q^{\frac{1}{100\,000}}\\}$ and
$1.04(\log(t/100))^{-2/3}<v\leq 1/4$, then

$\displaystyle\sum_{L(\rho,\chi)=0\atop|1+it-\rho|>v}\frac{1}{|1+it-\rho|^{2}}$
$\displaystyle\leq(8.14467+5.3912B(v^{-1/2}-2))\log t-8.5\log(A+1)$
$\displaystyle+518.7+\frac{1}{v^{2}}\Big{(}\frac{\log(A+1)-\log
v+\frac{2}{3}\log\log t}{1.879}+0.224\Big{)}-\frac{N_{\chi}(t,v)}{v^{2}}$
$\displaystyle+(\log q)(1.978v^{-1}+0.23267).$

###### Proof.

Assume without loss of generality that $\chi$ is primitive. We also assume
without loss of generality that $t\pm 1$ is not the ordinate of a zero. (If
$\chi$ is imprimitive and $\chi^{\prime}$ induces $\chi$, then $\chi^{\prime}$
has the same nontrivial zeros as $\chi$ and smaller modulus. If $t\pm 1$ is
the ordinate of a zero, apply the result to a sequence with $t^{\prime}$
tending to $t$ and $v^{\prime}$ tending to $v$ from above.) By [fordzfr, Lemma
4.3], we may also assume that $\chi$ is nonprincipal. We partition the zeros
with $|1+it-\rho|>v$ into the sets

$\displaystyle Z_{0}$ $\displaystyle=\\{\rho:|\mathrm{Im}(\rho)|\leq 1\\},$
$\displaystyle Z_{1}$ $\displaystyle=\\{\rho\notin
Z_{0}:|\mathrm{Im}(\rho)-t|\geq 1\\},$ $\displaystyle Z_{2}$
$\displaystyle=\Big{\\{}\rho\notin Z_{0}\cup
Z_{1}:|1+it-\rho|\geq\frac{1}{4}\text{ and
}|it-\rho|\geq\frac{1}{4}\Big{\\}},$ $\displaystyle Z_{3}$
$\displaystyle=\\{\rho:\rho\notin Z_{0}\cup Z_{1}\cup Z_{2}\text{ and
}|1+it-\rho|>v\\}.$

For $k\in\\{0,1,2,3\\}$, we define

$S_{k}=\sum_{\rho\in Z_{k}}\frac{1}{|1+it-\rho|^{2}}.$

First, note that by [mccurley, Theorem 2.1] with $\eta=0.5$, the number of
zeros of $L(s,\chi)$ with $t\leq 1$ is at most $1.919\log q+2.54$. Since
$q\leq t^{100\,000}$ and $t\geq e^{1938}$, it follows that $|S_{0}|\leq
e^{-1800}$.

Let $Q^{\pm}(T)=N^{\pm}(T,\chi)-\frac{T}{2\pi}\log\frac{qT}{2\pi
e}-C_{0}(\chi)$. By Corollary 4.4, we have that

$\displaystyle|Q^{\pm}(T)|\leq 0.2297\log(qT)+24.77.$

Write $C_{1}=0.2297$ and $C_{2}=24.77$. Observe that

$\displaystyle
S_{1}\leq\int_{t+1}^{\infty}\frac{dN^{+}(u)}{(u-t)^{2}}+\int_{1}^{t-1}\frac{dN^{+}(u)}{(t-u)^{2}}+\int_{1}^{\infty}\frac{dN^{-}(u)}{(u+t)^{2}}=I_{1}+I_{2}+I_{3}.$

Since $dN^{\pm}(u)=\frac{1}{2\pi}\log
q+\frac{1}{2\pi}\log(\frac{u}{2\pi})+dQ^{\pm}(u)$ and
$\log(t+x)\leq\log(t)+\frac{x}{t}$, we find that

$\displaystyle I_{1}$
$\displaystyle\leq\frac{1}{2\pi}\int_{1}^{\infty}\frac{\log(t+x)+\log q-\log
2\pi}{x^{2}}dx+|Q^{+}(t+1)|+2\int_{1}^{\infty}\frac{|Q^{+}(t+x)|}{x^{3}}dx$
$\displaystyle=\frac{\big{(}1+\frac{1}{t}\big{)}\log(1+t)+\log
q-\log(2\pi)}{2\pi}+|Q^{+}(t+1)|+2\int_{1}^{\infty}\frac{|Q^{+}(t+x)|}{x^{3}}dx$
$\displaystyle\leq(2C_{1}+0.15916)\log(qt)+(2C_{2}-0.292507)+2\int_{1}^{\infty}\frac{C_{2}x/t}{x^{3}}dx$
$\displaystyle\leq(2C_{1}+0.15916)\log(qt)+(2C_{2}-0.2925).$

Similarly, we find

$\displaystyle I_{2}$
$\displaystyle\leq\frac{1}{2\pi}\log\Big(\frac{qt}{2\pi}\Big{missing})+\frac{|Q^{+}(1)|}{(t-1)^{2}}+2\max_{1\leq
u\leq t-1}|Q^{+}(u)|$
$\displaystyle\leq(2C_{1}+0.159155)\log(qt)+(2C_{2}-0.292507),$

and since $q\leq t^{100\,000}$ and $t\geq e^{1938}$,

$I_{3}\leq\frac{1}{2\pi}\int_{1}^{\infty}\frac{\log\big(\frac{u+t}{2\pi}\big{missing})+\log
q}{(u+t)^{2}}du+\frac{\absolutevalue{Q^{-}(1)}}{(1+t)^{2}}+2\int_{1}^{\infty}\frac{|Q^{-}(u)|}{(u+t)^{3}}du\leq
0.000001.$

Consequently, $S_{1}$ is upper bounded by

$\displaystyle|S_{1}|$
$\displaystyle\leq(4C_{1}+0.31831)\log(qt)+(4C_{2}-0.58)$ (6.3)
$\displaystyle=1.23711\log(qt)+98.5.$

It follows from the proof of [fordzfr, Lemma 4.3], mutatis mutandis, that

$\displaystyle S_{2}\leq\frac{80N_{2}}{9},\qquad
S_{3}\leq\frac{80N_{3}}{9}+\Big{(}\frac{80}{9}-\frac{1}{v^{2}}\Big{)}N_{\chi}(t,v)+2\int_{v}^{1/4}\frac{N_{\chi}(t,u)}{u^{3}}du.$

Furthermore, since $t\geq e^{1938}$, we have

$\displaystyle N_{2}+N_{3}$
$\displaystyle=N^{+}(t+1,\chi)-N^{+}(t-1,\chi)-N_{\chi}(t,v)$ (6.4)
$\displaystyle\leq(2C_{1}+0.31831)\log(qt)+(2C_{2}-0.585)-N_{\chi}(t,v)$
$\displaystyle\leq 0.7771\log(qt)+49-N_{\chi}(t,v).$

Lemma 6.3 implies that

$\displaystyle
2\int_{v}^{1/4}\frac{N_{\chi}(t,u)}{u^{3}}du\leq\Big{(}\frac{\log(A+1)+\frac{2}{3}\log\log
t}{1.879}+0.49\Big{)}(v^{-2}-16)$ $\displaystyle\quad+5.3912B(v^{-1/2}-2)\log
t+\frac{1}{1.879}\Big{(}8-16\log 4-\frac{1+2\log v}{2v^{2}}\Big{)}+1.978(\log
q)(v^{-1}-4).$

By (6.4), it therefore follows that

$\displaystyle S_{2}+S_{3}\leq$
$\displaystyle(6.907556+5.3912B(v^{-1/2}-2))\log t-8.5\log(A+1)$
$\displaystyle+420.2+\frac{1}{v^{2}}\Big{(}\frac{\log(A+1)-\log
v+\frac{2}{3}\log\log t}{1.879}+0.224\Big{)}-\frac{N_{\chi}(t,v)}{v^{2}}$
$\displaystyle+(\log q)(1.978v^{-1}-1.00444).$

Combining this with the upper bound for $S_{1}$ in (6.3) completes the proof.
∎

The next two lemmata correspond to [ford, Lemmata 4.5 and 4.6], which in turn
are based on [hb, Lemmata 5.1 and 5.2].

######  Lemma 6.5.

Let $0<\eta\leq\frac{3}{2}$. Suppose that $f$ is a non-negative real function
which has continuous derivative on $(0,\infty)$. Suppose that the Laplace
transform $F(z)=\int_{0}^{\infty}f(y)e^{-zy}dy$ is absolutely convergent for
$\mathrm{Re}(z)>0$. Let $F_{0}(z)=F(z)-f(0)/z$, and suppose that there exists
a positive constant $D$ possibly depending on $f$ such

$\displaystyle\mathrm{Re}(z)\geq
0,\qquad|z|\geq\eta,\qquad|F_{0}(z)|\leq\frac{D}{|z|^{2}}.$ (6.5)

For $\mathrm{Re}(s)>1$, $\mathrm{Im}(s)\geq 0$, and $\chi\pmod{q}$ a Dirichlet
character with $q\geq 3$, we define

$K_{\chi}(s):=\sum_{n=1}^{\infty}\Lambda(n)\chi(n)n^{-s}f(\log n),$

and $Q(q)=\log\log q+0.66$. We have that

$\displaystyle\Big{|}K_{\chi}(s)-\Big{(}-f(0)\frac{L^{\prime}(s,\chi)}{L(s,\chi)}-\sum_{L(\rho,\chi)=0}F_{0}(s-\rho)-(1-\mathfrak{a})(1-\delta(\chi))F_{0}(s)+\delta(\chi)F_{0}(s-1)\Big{)}\Big{|}$
$\displaystyle\leq D\Big{(}3.44+\frac{1}{3}\log
q+\frac{1}{3}\log(1+\mathrm{Im}(s))+\mathscr{I}(\chi)(f(0)+\max_{x\in(0,\infty)}f(x))Q(q)\Big{)}.$

###### Proof.

We follow the proof of [fordzfr, Lemma 4.5]. Suppose that $\chi$ is primitive
and that $q\geq 3$. Suppose $s=\sigma+it$ and $1<\alpha<\sigma$. Define

$\displaystyle I=\frac{1}{2\pi
i}\int_{\alpha-i\infty}^{\alpha+i\infty}-\frac{L^{\prime}(w,\chi)}{L(w,\chi)}F_{0}(s-w)dw.$

Since $-L^{\prime}(w,\chi)/L(w,\chi)=\sum_{n}\chi(n)\Lambda(n)n^{-s}$, the sum
converging uniformly on $\mathrm{Re}(w)=\alpha$, we may integrate term by
term. Thus, with

$\displaystyle J_{n}=\frac{1}{2\pi
i}\int_{\alpha-i\infty}^{\alpha+i\infty}n^{-w}F_{0}(s-w)dw=\frac{n^{-s}}{2\pi
i}\int_{\sigma-\alpha-i\infty}^{\sigma-\alpha+i\infty}n^{u}F_{0}(u)du,$

we have $I=\sum_{n}\Lambda(n)\chi(n)J_{n}$ where $J_{n}=n^{-s}(f(\log
n)-f(0))$ via the proof of [fordzfr, Lemma 4.5]. Consequently, we have

$I=K_{\chi}(s)+f(0)\frac{L^{\prime}(s,\chi)}{L(s,\chi)}.$

Moving the line of integration to $\mathrm{Re}(w)=-1/2$, we have

$I=\frac{1}{2\pi
i}\int_{-1/2-i\infty}^{-1/2+i\infty}-\frac{L^{\prime}(w,\chi)}{L(w,\chi)}F_{0}(s-w)dw-\sum_{\rho}F_{0}(s-\rho)-(1-\mathfrak{a})(1-\delta(\chi))F_{0}(s)+\delta(\chi)F_{0}(s-1).$
(6.6)

By (6.5), (6.6) and Lemma 4.2, the integral in (6.6) is bounded above by
$\frac{D}{2\pi}I^{\prime}$, where

$\displaystyle I^{\prime}$ $\displaystyle\leq\int_{-\infty}^{\infty}\frac{\log
q+8.21+\frac{1}{2}\log(1+(u/2)^{2})}{9/4+(u-t)^{2}}du$
$\displaystyle=\frac{2\pi}{3}\log
q+5.48\pi+\frac{1}{3}\int_{-\infty}^{\infty}\frac{\log(1+(t/2+3v/4)^{2})}{1+v^{2}}dv$
$\displaystyle\leq\frac{2\pi}{3}\log
q+5.48\pi+\frac{1}{3}\int_{-\infty}^{\infty}\frac{\log(1+t^{2})+\log(1+v^{2})}{1+v^{2}}dv$
$\displaystyle\leq 2\pi\Big{(}\frac{1}{3}\log
q+2.74+\log(2)+\frac{\log(1+t)}{3}\Big{)}.$

This proves that when $\chi$ is primitive and $q\geq 3$,

$\displaystyle K_{\chi}(s)$
$\displaystyle=-f(0)\frac{L^{\prime}(s,\chi)}{L(s,\chi)}-\sum_{L(\rho,\chi)=0}F_{0}(s-\rho)-(1-\mathfrak{a})(1-\delta(\chi))F_{0}(s)+\delta(\chi)F_{0}(s-1)$
$\displaystyle\quad+D\Big{(}3.44+\frac{1}{3}\log
q+\frac{1}{3}\log(1+\mathrm{Im}(s))\Big{)}.$

The same conclusion follows when $\chi$ is primitive and $q=1$ by [fordzfr,
Lemma 4.5].

Now, let $\chi$ be possibly imprimitive with $q\geq 3$, and let
$\chi^{\prime}$ be the character inducing $\chi$. By Corollary A.2, we have
that

$\displaystyle|K_{\chi}(s)-K_{\chi^{\prime}}(s)|$
$\displaystyle\leq\sum_{n\in\mathbb{N}\atop(n,q)>1}f(\log
n)\frac{\Lambda(n)}{n}\leq Q(q)\max_{x\in(0,\infty)}f(x),$
$\displaystyle\Big{|}f(0)\frac{L^{\prime}}{L}(s,\chi)-f(0)\frac{L^{\prime}}{L}(s,\chi^{\prime})\Big{|}$
$\displaystyle\leq
f(0)\sum_{n\in\mathbb{N}\atop(n,q)>1}\frac{\Lambda(n)}{n}\leq Q(q)f(0).$

Consequently, the desired result holds for all Dirichlet characters with
$q\geq 3$. ∎

######  Corollary 6.6.

Assume that there exist positive constants $A$ and $B$ such that (2.4) holds.
Let $\chi\pmod{q}$ be a Dirichlet character with $q\geq 3$. Suppose that
$1.92(\log(t/100))^{-2/3}<\eta<\frac{1}{2}$. Let $f$ be a compactly supported
function satisfying (6.5) for some positive constant $D$. Suppose $s=1+it$
with $t\geq\max\\{q^{\frac{1}{100\,000}},e^{1938}\\}$. Then, we have

$\displaystyle\mathrm{Re}(K_{\chi}(s))$
$\displaystyle\leq-\sum_{L(\rho,\chi)=0\atop|1+it-\rho|\leq\eta}\mathrm{Re}\Big{\\{}F(s-\rho)+f(0)\Big{(}\frac{\pi}{2\eta}\cot\Big(\frac{\pi(s-\rho)}{2\eta}\Big{missing})-\frac{1}{s-\rho}\Big{)}\Big{\\}}$
$\displaystyle\quad+\frac{f(0)}{2\eta}\Big{[}\frac{2\log\log
t}{3}+B\eta^{3/2}\log
t+\log(A+1)-\frac{1}{2}\int_{-\infty}^{\infty}\frac{\log|L\big{(}s+\eta+\frac{2\eta
ui}{\pi},\chi\big{)}\big{|}}{(\cosh u)^{2}}du\Big{]}$
$\displaystyle\quad+D\Big{(}3.5+\frac{\log q}{3}+\frac{\log
t}{3}+\sum_{L(\rho,\chi)=0\atop|1+it-\rho|>\eta}\frac{1}{|1+it-\rho|^{2}}\Big{)}$
$\displaystyle\quad+\mathscr{I}(\chi)(f(0)+\max_{x\in(0,\infty)}f(x))Q(q)+\frac{f(0)}{2}\log
q+f(0)(1-\delta(\chi))(1-\mathfrak{a})e^{-1937}.$

Moreover, $K_{\chi_{0}}(1)\leq F(0)+1.8D$.

###### Proof.

The final assertion of the lemma follows immediately from [fordzfr, Lemma
4.6]. Assume $\sigma>1$. Since $t\geq e^{1938}$,
$\mathrm{Re}(F_{0}(s-1))\leq\frac{D}{t}\leq 0.0001D$, and similarly
$-\mathrm{Re}(F_{0}(s))\leq 0.0001D$ by (6.5). Furthermore, by (6.5),

$\displaystyle\sum_{L(\rho,\chi)=0\atop|1+it-\rho|>\eta}|F_{0}(s-\rho)|\leq
D\sum_{L(\rho,\chi)=0\atop|1+it-\rho|>\eta}\frac{1}{|1+it-\rho|^{2}}.$

By combining Lemma 6.2 (with $S=\\{z:\mathrm{Re}(z)\leq
1,~{}|1+it-z|\leq\eta\\}$) and Lemma 6.5, we obtain the inequality

$\displaystyle\mathrm{Re}(K_{\chi}(s))$
$\displaystyle\leq-\sum_{L(\rho,\chi)=0\atop|1+it-\rho|\leq\eta}\mathrm{Re}\Big{\\{}F(s-\rho)+f(0)\Big{(}\frac{\pi}{2\eta}\cot\Big(\frac{\pi(s-\rho)}{2\eta}\Big{missing})-\frac{1}{s-\rho}\Big{)}\Big{\\}}$
$\displaystyle\quad+\frac{f(0)}{2\eta}\Big{[}\frac{2\log\log
t}{3}+B\eta^{3/2}\log
t+\log(A+1)-\frac{1}{2}\int_{-\infty}^{\infty}\frac{\log|L\big{(}s+\eta+\frac{2\eta
ui}{\pi},\chi\big{)}\big{|}}{(\cosh u)^{2}}du\Big{]}$
$\displaystyle\quad+D\Big{(}3.5+\frac{\log q}{3}+\frac{\log
t}{3}+\sum_{L(\rho,\chi)=0\atop|1+it-\rho|>\eta}\frac{1}{|1+it-\rho|^{2}}\Big{)}$
$\displaystyle\quad+\mathscr{I}(\chi)(f(0)+\max_{x\in(0,\infty)}f(x))Q(q)+\frac{f(0)}{2}\log
q+f(0)(1-\delta(\chi))(1-\mathfrak{a})e^{-1937}.$

As in [ford, (4.10)], we may take $\sigma\to 1^{+}$, which proves the lemma. ∎

##  7\. A Trigonometric Inequality

We use an inequality that plays the same role as the inequality
$3+4\cos\theta+\cos(2\theta)\geq 0$ in Mertens’ treatment of the proof of the
prime number theorem due to Hadamard and de la Vallée Poussin. For any real
numbers $a_{1},a_{2}$, there exist real numbers $b_{0},\dots,b_{4}$ such that

$\sum_{j=0}^{4}b_{j}\cos(j\theta)=8(a_{1}+\cos\theta)^{2}(a_{2}+\cos\theta)^{2}\geq
0\quad(\theta\in\mathbb{R}).$ (7.1)

In particular, we have that

$\displaystyle b_{4}=1,\quad b_{3}=4(a_{1}+a_{2}),\quad
b_{2}=4(1+a_{1}^{2}+a_{2}^{2}+4a_{1}a_{2})$ (7.2) $\displaystyle
b_{1}=(a_{1}+a_{2})(12+16a_{1}a_{2}),\quad b_{0}=b_{2}-1+8(a_{1}a_{2})^{2}.$

######  Lemma 7.1.

Suppose that $a_{1},a_{2}\in\mathbb{R}$, and that $b_{1},b_{2},\dots,b_{4}$
are defined by (7.2). We have

$\int_{-\infty}^{\infty}\frac{1}{(\cosh
u)^{2}}\sum_{j=1}^{4}b_{j}\log|L(\sigma+\eta+ijt_{1}+iut_{2},\chi^{j})|du\geq-2b_{0}\log\zeta(1+\eta).$

###### Proof.

Denote by $I$ the integral in the lemma. By [fordzfr, (5.4)],

$U(y):=\int_{-\infty}^{\infty}\frac{e^{iyu}}{\cosh^{2}u}du=\frac{\pi
y}{\sinh(\pi y/2)}\geq 0.$

By [ford, (5.2)-(5.4)], alongside

$\log|L(s,\chi)|=\mathrm{Re}\Big{(}\sum_{\begin{subarray}{c}p\\\ m\geq
1\end{subarray}}\frac{1}{m}p^{-ms}(\chi(p))^{m}\Big{)},$

we find that

$\displaystyle I$
$\displaystyle=\sum_{p,m}\frac{1}{m}p^{-m(\sigma+\eta)}\mathrm{Re}\Big{(}\sum_{j=1}^{4}b_{j}p^{-ijmt_{1}}e^{imj\arg(\chi(p))}\int_{-\infty}^{\infty}\frac{p^{-imut_{2}}}{(\cosh
u)^{2}}du\Big{)}$
$\displaystyle=\sum_{p,m}\frac{1}{m}p^{-m(\sigma+\eta)}U(mt_{2}\log
p)\sum_{j=1}^{4}b_{j}\cos(j(mt_{1}\log p-m\arg(\chi(p))))$ $\displaystyle\geq-
b_{0}\sum_{p,m}\frac{1}{m}p^{-m(\sigma+\eta)}U(mt_{2}\log p).$

Finally, since $U(y)\leq 2$, we obtain that $I\geq-2b_{0}\zeta(1+\eta)$. ∎

##  8\. The functions $f$, $F$, and $K$

Let $f$ be a function with compact support, define $F$ and $K_{\chi}$ as in
Lemma 6.5 and assume that (6.5) holds for some positive constant $D$. Let
$a_{1},a_{2}$ be real numbers and define $b_{0},\dots,b_{4}$ by (7.2). Let
$b_{5}=b_{1}+\dots+b_{4}$, and write $L_{1}=\log(4t+1)$,
$L_{2}=\log\log(4t+1)$. By (7.1),

$\mathrm{Re}\sum_{j=0}^{4}b_{j}K_{\chi^{j}}(1+ijt)=\sum_{n=1}^{\infty}\Lambda(n)n^{-1}f(\log
n)\sum_{j=0}^{4}b_{j}\cos(j(t\log n-\arg(\chi(n))))\geq 0.$ (8.1)

Assume that $\chi\pmod{q}$ is a primitive Dirichlet character with $q\geq 3$.
Applying Lemma 6.6 with $s=1+ijt$ ($1\leq j\leq 4$), (7.1) and Lemma 7.1 with
$t_{2}=\frac{2\eta}{\pi}$,

$\displaystyle 0\leq$ $\displaystyle-\mathrm{Re}\sum_{\begin{subarray}{c}1\leq
j\leq 4\\\ L(\rho,\chi^{j})=0\\\
|1+ijt-\rho|\leq\eta\end{subarray}}b_{j}\Big{(}F(1+ijt-\rho)+f(0)\Big{(}\frac{\pi}{2\eta}\cot\Big(\frac{\pi(1+ijt-\rho)}{2\eta}\Big{missing})-\frac{1}{1+ijt-\rho}\Big{)}\Big{)}$
(8.2)
$\displaystyle\quad+\frac{f(0)}{2\eta}\Big{[}b_{5}\Big{(}\frac{2L_{2}}{3}+B\eta^{3/2}L_{1}+\log(A+1)\Big{)}+b_{0}\log\zeta(1+\eta)\Big{]}+b_{0}F(0)$
$\displaystyle\quad+D\Big{(}b_{5}\Big{(}3.5+\frac{\log
q}{3}+\frac{L_{1}}{3}\Big{)}+1.8b_{0}+\sum_{1\leq j\leq
4}b_{j}\sum_{\begin{subarray}{c}L(\rho,\chi^{j})=0\\\
|1+ijt-\rho|>\eta\end{subarray}}\frac{1}{|1+ijt-\rho|^{2}}\Big{)}$
$\displaystyle\quad+(b_{2}+b_{3}+b_{4})(f(0)+\max_{x\in(0,\infty)}f(x))Q(q)+b_{5}\frac{f(0)}{2}\log
q+(b_{2}+b_{4})e^{-1937}.$

Note that since $\chi$ is primitive, $\mathscr{I}(\chi)=0$ and hence
$(f(0)+\max_{x\in(0,\infty)}f(x))Q(q)$ is multiplied by a factor of
$b_{2}+b_{3}+b_{4}$ (rather than $b_{5}=b_{1}+b_{2}+b_{3}+b_{4}$) above.
Similarly, $1-\mathfrak{a}(\chi^{j})$ can only be nonzero when $j$ is odd, and
hence $e^{-1937}$ is multiplied by a factor of $b_{2}+b_{4}$.

We now explicitly define the functions $f$, $F$ and $K$. Suppose $\gamma\geq
e^{1938}$, $L(\beta+i\gamma,\chi)=0$, and that there exists $0<\lambda\leq
1-\beta$ such that

$L(\sigma+it,\chi^{j})\neq 0\quad(1-\lambda<\sigma\leq 1,\gamma-1\leq t\leq
4\gamma+1,0\leq j\leq 4).$ (8.3)

Let $\theta$ be the unique solution to

$(\sin\theta)^{2}=\frac{b_{1}}{b_{0}}(1-\theta\cot\theta),\quad
0<\theta<\pi/2,$ (8.4)

and define the real function

$g(u)=\begin{cases}(\cos(u\tan\theta)-\cos\theta)\sec^{2}\theta&|u|\leq\frac{\theta}{\tan\theta},\\\
0&\text{ else. }\end{cases}$ (8.5)

For $u\geq 0$, let

$w(u)=(g*g)(u)=\int_{-\infty}^{\infty}g(u-t)g(t)dt,\qquad
W(z)=\int_{0}^{\infty}e^{-zu}w(u)du.$

From the above definitions it follows (see [fordzfr, (6.6)], [hb, Lemma 6.1])
that

$\displaystyle W(0)$ $\displaystyle=2\sec^{2}\theta(1-\theta\cot\theta)^{2},$
(8.6) $\displaystyle W(-1)$
$\displaystyle=2\tan^{2}\theta+3-3\theta(\tan\theta+\cot\theta),$
$\displaystyle w(0)$
$\displaystyle=\sec^{2}\theta(\theta\tan\theta+3\theta\cot\theta-3).$

Then, set

$f(u)=\lambda e^{\lambda u}w(\lambda u)\quad(u\geq 0),$ (8.7)

and

$F(z)=\int_{0}^{\infty}e^{-zu}f(u)du=W\Big{(}\frac{z}{\lambda}-1\Big{)}.$
(8.8)

##  9\. An Inequality for the Real Part of a Zero

######  Lemma 9.1.

Assume that there exist positive constants $A,B$ such that (2.4) holds.
Suppose $\chi\pmod{q}$ is a primitive Dirichlet character with $q\geq 3$.
Suppose that $L(\beta+i\gamma)=0$, with
$\gamma\geq\max\\{q^{\frac{1}{100\,000}},e^{1938}\\}$ and
$1.92(\log(\gamma/100))^{-2/3}<\eta\leq 1/4$. Suppose that $\lambda>0$
satisfies (8.3). Finally, suppose that

$1-\beta\leq\eta/2,\quad
0<\lambda\leq\min\Big{(}1-\beta,\frac{1}{198}\eta\Big{)}.$ (9.1)

Write $L_{1}=\log(4\gamma+1)$ and $L_{2}=\log\log(4\gamma+1)$. Then, we have
that

$\frac{1}{\lambda}\Big{[}0.16521-0.1876\Big{(}\frac{1-\beta}{\lambda}-1\Big{)}\Big{]}\\\
\leq
1.471\frac{1-\beta}{\eta^{2}}+\frac{1}{2\eta}\Big{[}\frac{666550}{200211}\Big{(}\frac{2}{3}L_{2}+B\eta^{3/2}L_{1}+\log(A+1)\Big{)}+\log\zeta(1+\eta)\Big{]}\\\
+3.5146\lambda\Big{[}(8.47801+5.392B(\eta^{-\frac{1}{2}}-2))L_{1}+\frac{\frac{\log((A+1)/\eta)+\frac{2}{3}L_{2}}{1.879}+0.224}{\eta^{2}}\\\
+(1.978\eta^{-1}+0.56601)\log q+522.75\Big{]}\\\ +\frac{b_{5}}{2b_{0}}\log
q+\frac{b_{2}+b_{3}+b_{4}}{b_{0}}Q(q)+\frac{b_{2}+b_{4}}{b_{0}}e^{-1937}.$

###### Proof.

We take $a_{1}=0.225$ and $a_{2}=0.9$. Then, by (7.2), we have that

$\begin{array}[]{ll}b_{0}=10.01055&b_{3}=4.5\\\ b_{1}=17.14500&b_{4}=1.0\\\
b_{2}=10.68250&b_{5}=33.3275,\end{array}$

and by (8.4) and (8.6),

$\theta=1.152214629976363048877\ldots,\quad
w(0)=6.82602968445295450905\ldots.$

Let

$\displaystyle
c_{0}=\frac{1}{\sin\theta(\cos\theta)^{3}}=16.2983216223932350562\ldots$
$\displaystyle
c_{1}=(\theta-\sin\theta\cos\theta)(\tan\theta)^{4}=19.9352005926244107856\ldots$
$\displaystyle
c_{2}=(\tan\theta)^{3}(\sin\theta)^{2}=9.4813169452950521682\ldots$
$\displaystyle
c_{3}=(\theta-\sin\theta\cos\theta)(\tan\theta)^{2}=3.945405755634895592\ldots$

Let $R=197$, let

$H(R)=\frac{c_{0}(c_{2}\frac{(R+1)^{2}}{R^{3}}(e^{2\theta/\tan\theta}+1)+\frac{c_{1}}{R^{2}}+c_{3})}{(1-\frac{\tan^{2}\theta}{R^{2}})^{2}},$
$c_{4}=\frac{H(R)(R+1)^{2}}{R^{3}w(0)}+1+1/R,$

By [fordzfr, (7.6)] and the assumption $\eta\geq 198\lambda$, (6.5) holds with

$D=c_{4}\lambda f(0).$ (9.2)

Furthermore, a Mathematica calculation shows that $e^{x}w(x)\leq 7.23$, and
thus that

$f(0)+\max_{x\in(0,\infty)}f(x)\leq
f(0)+7.23\lambda=f(0)+f(0)\frac{7.23}{w(0)}\leq 2.06f(0)$ (9.3)

Let

$z=\frac{\pi}{2\eta}(1+ijt-\rho),\qquad c=\frac{\pi\lambda}{2\eta}.$ (9.4)

Define

$V_{c}(z)=cw(0)\Big{(}\cot z-\frac{1}{z}\Big{)}+W\Big{(}\frac{z}{c}-1\Big{)}.$
(9.5)

By (8.7) and (8.8),

$F(1+ijt-\rho)+f(0)\Big{(}\frac{\pi}{2\eta}\cot\Big(\frac{\pi(1+ijt-\rho)}{2\eta}\Big{missing})-\frac{1}{1+ijt-\rho}\Big{)}=V_{c}(z).$

We aim to bound the contribution to the double sum appearing in (8.2) from all
zeros besides $\beta+i\gamma$. By [fordzfr, (7.7)], we have the lower bound

$\mathrm{Re}(V_{c}(z))\geq-c_{5}c^{2}w(0)\quad\Big{(}\mathrm{Re}(z)\geq
c,|z|\leq\frac{\pi}{2}\Big{)},$ (9.6)

where

$c_{5}=\frac{4}{\pi^{2}}\Big{(}1+\frac{(R+1)^{2}H(R)}{w(0)R^{3}}\Big{)}=\frac{4}{\pi^{2}}\Big{(}c_{4}-1/R\Big{)}.$

This condition is satisfied for all zeros in the double sum appearing in (8.2)
(with $z,c$ as defined in (9.4)) by the assumption on $\lambda$ in (8.3).
Thus, by (9.4) (with $t=\gamma$), (9.5), and (9.6), we have that

$\displaystyle-\mathrm{Re}\sum_{\begin{subarray}{c}1\leq j\leq 4\\\
L(\rho,\chi^{j})=0\\\
|1+ij\gamma-\rho|\leq\eta\end{subarray}}b_{j}\Big{(}F(1+ij\gamma-\rho)+f(0)\Big{(}\frac{\pi}{2\eta}\cot\Big(\frac{\pi(1+ij\gamma-\rho)}{2\eta}\Big{missing})-\frac{1}{1+ij\gamma-\rho}\Big{)}\Big{)}$
(9.7) $\displaystyle\quad\leq-
b_{1}V_{c}\Big{(}\frac{\pi}{2\eta}(1-\beta)\Big{)}+c_{5}c^{2}w(0)\sum_{j=1}^{4}b_{j}N_{\chi^{j}}(j\gamma,\eta).$

Write $L_{1}=\log(4\gamma+1)$ and $L_{2}=\log\log(4\gamma+1)$. Combining Lemma
6.4 (with $t=\gamma$), (8.2), (8.7), (9.2), (9.3), and (9.7), we find that

$\displaystyle
b_{0}F(0)-b_{1}V_{c}\Big{(}\frac{\pi}{2\eta}(1-\beta)\Big{)}+\frac{\lambda
f(0)}{\eta^{2}}\Big{(}\frac{\pi^{2}c_{5}}{4}-c_{4}\Big{)}\sum_{j=1}^{4}b_{j}N(j\gamma,\eta)$
$\displaystyle+\frac{f(0)}{2\eta}\Big{[}b_{5}\Big{(}\frac{2L_{2}}{3}+B\eta^{3/2}L_{1}+\log(A+1)\Big{)}+b_{0}\log\zeta(1+\eta)\Big{]}+b_{0}F(0)$
$\displaystyle+c_{4}\lambda f(0)b_{5}\Big{[}\Big{(}3.5+\frac{\log
q}{3}+\frac{L_{1}}{3}\Big{)}+1.8\frac{b_{0}}{b_{5}}+\Big{(}8.14467+5.3912B(v^{-1/2}-2)\Big{)}\log
t$ (9.8)
$\displaystyle+518.7-8.5\log(A+1)+\frac{1}{\eta^{2}}\Big{(}\frac{\log(A+1)-\log\eta+\frac{2}{3}L_{2}}{1.879}+0.224\Big{)}+(\log
q)(1.978v^{-1}+0.23267)\Big{]}$
$\displaystyle+(b_{2}+b_{3}+b_{4})f(0)Q(q)+b_{5}\frac{f(0)}{2}\log
q+(b_{2}+b_{4})e^{-1937}\geq 0.$

Since $c_{5}<\frac{4}{\pi^{2}}c_{4}$, we may drop the summation term in the
first line above. Furthermore, the proof of [ford, Lemma 7.1] shows that

$\displaystyle
F(0)-\frac{b_{1}}{b_{0}}V_{c}\left(\frac{\pi}{2\eta}(1-\beta)\right)$
$\displaystyle\leq
0.348f(0)\frac{\pi^{2}}{4\eta^{2}}\frac{b_{1}}{b_{0}}(1-\beta)$ (9.9)
$\displaystyle+\frac{f(0)}{\lambda}\left(-\cos^{2}\theta+\frac{0.7475b_{1}}{b_{0}w(0)}\left(\frac{1-\beta}{\lambda}-1\right)\right).$

Dividing both sides of the equation by $b_{0}f(0)$, employing the numerical
values of $b_{0}$, $b_{1}$, $b_{5}$ and $\theta$, the inequalities $H(R)\leq
66.3307$ and $c_{4}\leq 1.055656$ (for $R=197$), and utilizing (9.9) in (9)
gives the desired result. ∎

##  10\. Proof of Theorem 3.1

Recall the notation and hypotheses of Lemma 9.1. In this section, we choose
the free parameters $\lambda$ and $\eta$ appearing in Lemma 9.1 in order to
complete the proof of Theorem 3.1. The optimal choices are given in (10.3) and
(10.4). Throughout this section, we assume without loss of generality that
$\chi$ is primitive.

###  10.1. Final choices of $\lambda$ and $\eta$

Suppose that $T_{0}$ satisfies the hypotheses of Theorem 3.1. If
$1-\beta\geq((10.082+1.607/\log\log T_{0})\log q+9.791\log\log q)^{-1}$ for
all zeros $\beta+i\gamma$ of $L(s,\chi)$ satisfying $\gamma\geq T_{0}$, then
Theorem 3.1 is immediate. We therefore assume that
$1-\beta<((10.082+1.607/\log\log T_{0})\log q+9.791\log\log q)^{-1}$ and
$\gamma\geq T_{0}$ for some zeros of $L(s,\chi)$. Define

$\displaystyle
Z(\beta,\gamma)=\frac{(1-\beta)B^{2/3}(\log\gamma)^{2/3}(\log\log\gamma)^{1/3}}{1-(1-\beta)\big{(}\big{(}10.082+\frac{1.607}{\log\log
T_{0}}\big{)}\log q+\log\log q\big{)}}$ (10.1) $\displaystyle
S=\Big{\\{}\beta+i\gamma:\exists\chi\pmod{q},~{}L(\beta+i\gamma,\chi)=0,~{}\gamma\geq
T_{0},$ $\displaystyle\qquad\qquad
1-\beta\leq\Big{(}\Big{(}10.082+\frac{1.607}{\log\log T_{0}}\Big{)}\log
q+9.791\log\log q\Big{)}^{-1}\Big{\\}},$ $\displaystyle
M=\inf_{\beta+i\gamma\in S}Z(\beta,\gamma).$

By Corollary B.2, we have that $M>0$. To prove Theorem 3.1, it suffices to
prove that $M\geq M_{1}$. Suppose to the contrary that

$M<M_{1}\leq 0.055071.$ (10.2)

Under the hypothesis of (10.2), there exists a zero $\beta+i\gamma$ of
$L(s,\chi)$ with $\gamma\geq T_{0}$, $1-\beta\leq(10.082+\frac{1.607}{\log\log
T_{0}})\log q+9.791\log\log q$, and

$Z(\beta,t)\in[M,M(1+\delta)],\quad\delta=\min\Big{(}\frac{10^{-100}}{\log
T_{0}},\frac{M_{1}-M}{2M}\Big{)}.$

By (2.3), we necessarily must have $\gamma\geq q^{\frac{1}{100\,000}}$, since

$\frac{1}{9.64590880\log(q\gamma)}\leq 1-\beta\leq\frac{1}{10.082\log q}.$

Let $L_{1}=\log(4\gamma+1)$, and $L_{2}=\log\log(4\gamma+1)$. By (10.1), (8.3)
holds with

$\lambda=\frac{1}{(10.082+\frac{1.607}{\log\log T_{0}})\log q+9.791\log\log
q+\frac{1}{M}B^{2/3}L_{1}^{2/3}L_{2}^{1/3}}.$ (10.3)

Define

$E=\Big{(}\frac{4(1+b_{0}/b_{5})}{3}\Big{)}^{\frac{2}{3}}=\Big{(}\frac{1733522}{999825}\Big{)}^{\frac{2}{3}},\quad\eta=EB^{-\frac{2}{3}}\Big{(}\frac{L_{2}}{L_{1}}\Big{)}^{\frac{2}{3}}.$
(10.4)

We now verify that our choices of $\lambda$ and $\eta$ satisfy the hypotheses
of Lemma 9.1. The lower bound $\frac{\log T_{0}}{\log\log
T_{0}}\geq\frac{1139}{B}$ guarantees that $\eta\leq 0.014$, and the bound
$T_{0}\geq\exp(1938)$ ensures that $\lambda\leq
0.055071(BL_{1})^{-2/3}L_{2}^{-1/3}\leq\frac{\eta}{198}$. Furthermore, the
hypotheses $T_{0}\geq e^{1938}$ and $B\leq 4.45$ imply that
$\eta>1.92(\log(\gamma/100))^{-2/3}$. Finally, the inequalities $T_{0}\geq
e^{1938}$ and $M_{1}\leq 0.055071$ ensure that the remaining hypotheses of
Lemma 9.1 are met.

###  10.2. Proof of Theorem 3.1

In order to contradict (10.2), we apply Lemma 9.1 with $\lambda$ and $\eta$ as
defined in (10.3) and (10.4), respectively. With these choices, we will prove
that

$\displaystyle\frac{1}{2\eta}\Big{[}\frac{666550}{200211}\Big{(}\frac{2}{3}L_{2}+B\eta^{3/2}L_{1}+\log(A+1)\Big{)}+\log\zeta(1+\eta)\Big{]},$
(10.5) $\displaystyle\leq
2.99968(BL_{1})^{\frac{2}{3}}L_{2}^{\frac{1}{3}}+\Big{(}\frac{BL_{1}}{L_{2}}\Big{)}^{\frac{2}{3}}\Big{(}1.1534\log(A+1)-0.124193+0.23096\log(B/L_{2})\Big{)},$

that

$3.5146\lambda\Big{[}(8.47801+5.392B(\eta^{-\frac{1}{2}}-2))L_{1}+\frac{\frac{\log(A/\eta)+\frac{2}{3}L_{2}}{1.879}+0.224}{\eta^{2}}\\\
+(1.978\eta^{-1}+0.56601)\log q+522.75\Big{]}\\\
\leq\Big{(}\frac{BL_{1}}{L_{2}}\Big{)}^{\frac{2}{3}}\Big{[}\frac{1.694-2.087B}{B^{4/3}}\Big{(}\frac{L_{2}}{L_{1}}\Big{)}^{\frac{1}{3}}+0.006534\log(A+1)+0.0043556\log\Big(\frac{B}{L_{2}}\Big{missing})+0.954863\Big{]}\\\
+\frac{0.265271\log q}{\log\log T_{0}},$ (10.6)

and that

$\displaystyle\frac{b_{5}}{2b_{0}}\log
q+\frac{b_{2}+b_{3}+b_{4}}{b_{0}}Q(q)+\frac{b_{2}+b_{4}}{b_{0}}e^{-1937}$
(10.7) $\displaystyle\leq 1.66462\log q+1.6166\log\log
q+0.009783B^{2/3}\Big{(}\frac{L_{1}}{L_{2}}\Big{)}^{2/3}.$

in subsections 10.3, 10.4, and 10.5 respectively. Assuming these inequalities
for now, we proceed with our proof. Observe that since $M\leq M_{1}\leq
0.055071$ and $T_{0}\geq e^{1938}$,

$1.471\frac{1-\beta}{\eta^{2}}\leq 0.0389B^{2/3}L_{1}^{2/3}L_{2}^{-5/3}\leq
0.00514B^{2/3}\Big{(}\frac{L_{1}}{L_{2}}\Big{)}^{2/3}.$ (10.8)

Next, note that by the lower bound on $T_{0}$,

$\displaystyle\frac{1-\beta}{\lambda}-1$
$\displaystyle\leq\frac{(10.082+\frac{1.607}{\log\log T_{0}})\log
q+9.791\log\log
q+\frac{1}{M}B^{2/3}L_{1}^{2/3}L_{2}^{1/3}}{(10.082+\frac{1.607}{\log\log
T_{0}})\log q+9.791\log\log
q+\frac{1}{M(1+\delta)}B^{2/3}(\log\gamma)^{2/3}(\log\log\gamma)^{1/3}}$
(10.9) $\displaystyle\leq\frac{0.98525}{\log T_{0}}.$

Applying (10.5)-(10.8) in Lemma 9.1 and gathering terms, we obtain the
inequality

$\displaystyle\frac{1}{\lambda}\Big{(}0.16521-\frac{0.184833}{\log
T_{0}}\Big{)}$
$\displaystyle\leq(BL_{1})^{\frac{2}{3}}L_{2}^{\frac{1}{3}}\Big{(}2.99968+\frac{X(\gamma)}{L_{2}}\Big{)}$
$\displaystyle\qquad+1.66462\log q+1.6166\log\log q+\frac{0.265271\log
q}{\log\log T_{0}}.$

Rearranging and using the definition of $\lambda$ from (10.3), we find that

$\displaystyle\lambda$
$\displaystyle=\frac{1}{\frac{1}{M}(BL_{1})^{\frac{2}{3}}L_{2}^{1/3}+(10.082+\frac{1.607}{\log\log
T_{0}})\log q+9.791\log\log q}$ (10.10)
$\displaystyle\geq\frac{0.16521-\frac{0.184833}{\log
T_{0}}}{(BL_{1})^{\frac{2}{3}}L_{2}^{\frac{1}{3}}(2.99968+\frac{X(\gamma)}{L_{2}})+\big{(}1.66462+\frac{0.265271}{\log\log
T_{0}}\big{)}\log q+1.6166\log\log q}.$

Note that since $0.16521-\frac{0.184833}{\log T_{0}}\geq 0.1651146$ for
$T_{0}\geq e^{1938}$, we have that

$\displaystyle\Big{(}(10.082+\frac{1.607}{\log\log T_{0}})(\log
q)+9.791\log\log q\Big{)}\Big{(}0.16521-\frac{0.184833}{\log T_{0}}\Big{)}$
(10.11) $\displaystyle>\Big{(}1.66462+\frac{0.265271}{\log\log
T_{0}}\Big{)}\log q+1.6166\log\log q.$

By utilizing (10.10) and (10.11), we obtain the inequality

$\displaystyle\frac{0.16521-\frac{0.184833}{\log
T_{0}}}{M}(BL_{1})^{\frac{2}{3}}L_{2}^{1/3}+\Big{(}1.66462+\frac{0.265271}{\log\log
T_{0}}\Big{)}\log q+1.6166\log\log q$
$\displaystyle\qquad<(BL_{1})^{\frac{2}{3}}L_{2}^{\frac{1}{3}}\Big{(}2.99968+\frac{X(\gamma)}{L_{2}}\Big{)}+\Big{(}1.66462+\frac{0.265271}{\log\log
T_{0}}\Big{)}\log q+1.6166\log\log q,$

which implies that

$M>\frac{0.16521-\frac{0.184833}{\log
T_{0}}}{2.99968+\frac{X(\gamma)}{\log\log\gamma}}\geq M_{1}.$

This contradicts (10.2), as desired, and our proof of Theorem 3.1 is complete.

###  10.3. Proof of (10.5)

By Lemma 4.1, we have that

$\log\zeta(1+\eta)\leq\log(1/\eta+0.6)\leq\log(1/\eta)+0.0084.$ (10.12)

Since $\log(1/\eta)\sim\frac{2}{3}L_{2}$, it follows that the main term
involving $t$ in (10.5) (which in fact gives the main term in the final
result) is bounded above by

$\frac{1}{2\eta}\Big{[}\frac{b_{5}}{b_{0}}\Big{(}\frac{2L_{2}}{3}+B\eta^{\frac{3}{2}}L_{1}\Big{)}+\frac{2L_{2}}{3}\Big{]}=\frac{b_{5}}{b_{0}}\Big{(}1+\frac{b_{0}}{b_{5}}\Big{)}^{\frac{1}{3}}\Big{(}\frac{3B}{4}\Big{)}^{\frac{2}{3}}L_{1}^{\frac{2}{3}}L_{2}^{\frac{1}{3}}\leq
2.99968(BL_{1})^{\frac{2}{3}}L_{2}^{\frac{1}{3}}.$ (10.13)

Note that $\log\log T_{0}\geq 7.569>4.45\geq B$, meaning that
$\log(B/L_{2})<0$. Hence, the remaining terms in (10.5) give

$\displaystyle\leq\frac{1}{2\eta}\Big{[}\frac{b_{5}}{b_{0}}\log(A+1)-\log
E+\frac{2}{3}\log(B/L_{2})+0.0084\Big{]}$
$\displaystyle\leq\frac{B^{\frac{2}{3}}}{2E}\Big{(}\frac{L_{1}}{L_{2}}\Big{)}^{\frac{2}{3}}\Big{[}\frac{b_{5}}{b_{0}}\log(A+1)-0.35848+\frac{2}{3}\log(B/L_{2})\Big{]}$
$\displaystyle\leq\Big{(}\frac{BL_{1}}{L_{2}}\Big{)}^{\frac{2}{3}}\Big{(}1.1534\log(A+1)-0.124193+0.23096\log(B/L_{2})\Big{)}.$

###  10.4. Proof of (10.6)

We next consider the terms in (10.6) not involving $q$. Since $522.75\leq
0.26974\log\gamma$ and $T_{0}\geq e^{1938}$, these terms contribute at most

$\displaystyle
0.193553L_{1}^{-\frac{2}{3}}L_{2}^{-\frac{1}{3}}B^{-\frac{2}{3}}\Big{[}\Big{(}8.74775+\frac{5.392B^{\frac{4}{3}}}{\sqrt{E}}\Big{(}\frac{L_{1}}{L_{2}}\Big{)}^{\frac{1}{3}}-10.784B\Big{)}L_{1}\Big{.}$
$\displaystyle\Big{.}\quad+\frac{B^{\frac{4}{3}}}{E^{2}}\Big{(}\frac{L_{1}}{L_{2}}\Big{)}^{\frac{4}{3}}\Big{(}\frac{\log(A+1)+\frac{4}{3}L_{2}+\frac{2}{3}\log(B/L_{2})-\log
E}{1.879}+0.224\Big{)}\Big{]}$
$\displaystyle\leq\Big{(}\frac{BL_{1}}{L_{2}}\Big{)}^{\frac{2}{3}}\Big{[}0.9347+\frac{1.694-2.087B}{B^{\frac{4}{3}}}\Big{(}\frac{L_{2}}{L_{1}}\Big{)}^{\frac{1}{3}}+\frac{0.049454}{L_{2}}\Big{(}\log(A+1)+\frac{2}{3}\log(B/L_{2})+0.05401\Big{)}\Big{]}$
$\displaystyle\leq\Big{(}\frac{BL_{1}}{L_{2}}\Big{)}^{\frac{2}{3}}\Big{[}\frac{1.694-2.087B}{B^{4/3}}\Big{(}\frac{L_{2}}{L_{1}}\Big{)}^{\frac{1}{3}}+0.006534\log(A+1)+0.0065334\cdot\frac{2}{3}\log\Big(\frac{B}{L_{2}}\Big{missing})+0.935053\Big{]}.$

Utilizing the assumption that $\frac{\log T_{0}}{\log\log
T_{0}}\geq\frac{1139}{B}$, we find that the terms in (10.6) involving $q$ are
bounded by

$\displaystyle 3.5146\lambda\Big{(}\frac{1.978}{\eta}+0.56601\Big{)}\log q$
(10.14) $\displaystyle\leq\frac{3.5146}{10.082\log
q+\frac{1}{0.055071}B^{2/3}L_{1}^{2/3}L_{2}^{1/3}}\cdot\log
q\cdot\Big{(}1.978\frac{B^{2/3}}{E}\cdot\frac{L_{1}^{2/3}}{L_{2}^{2/3}}+0.56601\Big{)}$
$\displaystyle\leq\frac{6.9519}{E}\frac{B^{2/3}L_{1}^{2/3}}{\frac{1}{0.055071}B^{2/3}L_{1}^{2/3}L_{2}}\log
q+0.197312\cdot 1139^{-2/3}\cdot
B^{2/3}\Big{(}\frac{L_{1}}{L_{2}}\Big{)}^{2/3}$
$\displaystyle\leq\frac{1}{L_{2}}0.055071\frac{6.9519}{E}\log
q+0.00181B^{2/3}\Big{(}\frac{L_{1}}{L_{2}}\Big{)}^{2/3}$
$\displaystyle\leq\frac{0.265271\log q}{\log\log
T_{0}}+0.00181B^{2/3}\Big{(}\frac{L_{1}}{L_{2}}\Big{)}^{2/3}.$

###  10.5. Proof of (10.7)

Utilizing the numerical values of $b_{0}$, $b_{2}$, $b_{3}$ and $b_{4}$, along
with the lower bound $\frac{\log T_{0}}{\log\log T_{0}}\geq\frac{1139}{B}$ ,
we find that

$\displaystyle\frac{b_{2}+b_{3}+b_{4}}{b_{0}}Q(q)$ $\displaystyle\leq
1.6166(\log\log q+0.66)\leq 1.6166\log\log
q+0.009783B^{2/3}\Big{(}\frac{L_{1}}{L_{2}}\Big{)}^{2/3}.$

Finally, we utilize the numerical values of $b_{0},\dots,b_{5}$ to bound
$\frac{b_{5}}{2b_{0}}\log q+\frac{b_{2}+b_{4}}{b_{0}}e^{-1937}$ above by
$1.66462\log q$.

##  Appendix A Explicit estimate for $\sum_{p\mid q}\frac{1}{p}.$

######  Lemma A.1.

Let

$E=-\gamma_{\mathbb{Q}}-\sum_{n=2}^{\infty}\sum_{p}\frac{\log p}{p^{n}}.$

If $q\geq 2310$, then $\sum_{p\mid q}\frac{\log p}{p}<\log\log q+E+0.1313$.

###### Proof.

Let $P_{x}=\prod_{p\leq x}p$ (a primorial). Choose $x$ so that $P_{x}$ is the
greatest primorial less than or equal to $q$. Note that
$\log(3)/3>\log(2)/2>\log(5)/5$, and that for $n\geq 3$ the function
$\log(n)/n$ is decreasing. Since $q\geq 2310$, this observation implies that

$\sum_{p\mid q}\frac{\log p}{p}\leq\sum_{p\leq x}\frac{\log p}{p}=\sum_{p\mid
P_{x}}\frac{\log p}{p}.$

It therefore follows that any upper bound for $\sum_{p\mid q}\frac{\log p}{p}$
which is increasing in $q$ and applies when $q\geq 2310$ is primorial in fact
holds for all $q\geq 2310$. We thus assume without loss of generality that
there exists an $x$ such that $q=P_{x}$. Since $P_{x}=e^{\theta(x)}$, and it
follows from [rs, (3.14)] that $e^{\theta(x)}>e^{x(1-\frac{1}{2\log(2310)})}$,
we have that $x<\frac{\log q}{1-\frac{1}{2\log(2310)}}$. By [rs, (2.8),
(3.22)], we have that

$\displaystyle\sum_{p\mid q}\frac{\log p}{p}$ $\displaystyle=\sum_{p\leq
x}\frac{\log p}{p}<\log x+E+0.06456.$

The desired result follows in our range of $x$. ∎

From the preceding lemma, we immediately obtain the following corollary.

######  Corollary A.2.

If $q\geq 3$, then $\sum_{(n,q)>1}\frac{\Lambda(n)}{n}<\log\log q+0.66$.

###### Proof.

For $3\leq q\leq 2310$, the statement can be verified by computer. For $q\geq
2310$, observe that by Proposition A.1,

$\sum_{(n,q)>1}\frac{\Lambda(n)}{n}\leq\sum_{p\mid q}\frac{\log
p}{p}+\sum_{n=2}^{\infty}\sum_{p}\frac{\log p}{p^{n}}=\sum_{p\mid q}\frac{\log
p}{p}-\gamma_{\mathbb{Q}}-E<\log\log q+0.1313-\gamma_{\mathbb{Q}}.$

∎

##  Appendix B Preliminary Vinogradov–Korobov for Dirichlet $L$-functions

######  Theorem B.1.

Let $T_{0}\geq e^{10\,650}$. Assume that there exist positive constants $A$
and $B\leq 4.45$ such that (2.4) holds, that $\frac{\log T_{0}}{\log\log
T_{0}}\geq\frac{5110.6}{B}$, and that $\log\log T_{0}\geq\frac{183}{B^{2}}$.
For any Dirichlet character $\chi\pmod{q}$ with $q\geq 3$,
$L(\sigma+it,\chi)\neq 0$ when $t\geq T_{0}$ and

$\sigma>1-\frac{1}{18\log q+cB^{2/3}(\log t)^{2/3}(\log\log t)^{1/3}},$ (B.1)

where

$c=31.76+\max\Big{\\{}\max_{t\geq
T_{0}}\frac{-2.89\log_{3}t+14.44\log(A+1)+3.59}{\log\log t},0\Big{\\}}.$

In particular, if $\log_{3}t\geq 5\log(A+1)+1.25$, then $c=31.76$.

###### Proof.

Suppose for sake of contradiction that there existed a zero $\beta+i\gamma$ of
$L(s,\chi)$ with

$\beta\geq 1-\frac{1}{18\log
q+cB^{2/3}(\log\gamma)^{2/3}(\log\log\gamma)^{1/3}},\qquad t\geq 3,$ (B.2)

and $\gamma\geq T_{0}$. By the zero-free region in (2.3), we must have
$\gamma\geq q^{\frac{1}{100\,000}}$, since

$\frac{1}{9.64590880\log(q\gamma)}\leq 1-\beta\leq\frac{1}{18\log q}.$

Set $a_{1}=0.225$ and $a_{0}=9$ in (7.2). Let $C=4/3$ and

$\eta=\Big{(}\frac{C}{B}\frac{\log\log\gamma}{\log\gamma}\Big{)}^{2/3}.$ (B.3)

By the assumption $\frac{\log\log T_{0}}{\log T_{0}}\geq\frac{5110.6}{B}$, we
have that $(\frac{4\log\log\gamma}{3B\log\gamma})^{2/3}\leq(\frac{4\log\log
T_{0}}{3B\log T_{0}})^{2/3}<1/2$, i.e., that $\eta<1/2$. We apply Lemma 6.2
with $s=\sigma+ij\gamma$ for $1\leq j\leq 4$, with $S=\\{\beta+i\gamma\\}$ for
$j=1$ and $S=\emptyset$ for $j\neq 1$, and with some $\sigma$ (to be specified
below) such that $1\leq\sigma\leq 1+\eta-(\log(\gamma/100))^{-2/3}$. We also
apply (7.1) and Lemma 7.1 with $t=\frac{2\eta}{\pi}$, to obtain that

$\displaystyle 0$ $\displaystyle\leq-
b_{1}\mathrm{Re}\Big{(}\frac{\pi}{2\eta}\cot\Big(\frac{\pi(\sigma-\beta)}{2\eta}\Big{missing})\Big{)}$
(B.4) $\displaystyle\quad+\frac{b_{5}}{2\eta}\Big{(}(1-\sigma+\eta)\log
q+\frac{2}{3}\log\log\gamma+B(1-\sigma+\eta)^{3/2}\log\gamma+\log(A+1)\Big{)}+\frac{b_{0}}{\sigma-1}$
$\displaystyle\quad-\frac{1}{4\eta}\sum_{j=1}^{4}b_{j}\int_{-\infty}^{\infty}\frac{\log|L(\sigma+\eta+it+2\eta
iu/\pi,\chi^{j})|}{(\cosh u)^{2}}du+(b_{2}+b_{4})e^{-1937}$
$\displaystyle\leq-
b_{1}\mathrm{Re}\Big{(}\frac{\pi}{2\eta}\cot\Big(\frac{\pi(\sigma-\beta)}{2\eta}\Big{missing})\Big{)}+\frac{b_{5}}{2\eta}\Big{(}\frac{2}{3}\log\log\gamma+B\eta^{3/2}\log\gamma+\log(A+1)\Big{)}$
$\displaystyle\quad+\frac{b_{0}}{\sigma-1}+\frac{b_{0}}{2\eta}\log(\zeta(1+\eta))+\frac{b_{5}}{2}\log
q+(b_{2}+b_{4})e^{-1937}.$

We now set $\sigma=1+F(1-\beta)$ with $F=3.238$. Observe that since $\log\log
T_{0}\geq\frac{183}{B^{2}}$ and $\gamma\geq T_{0}\geq e^{10\,650}$,

$\displaystyle 1+\eta-\sigma=\eta-F(1-\beta)$
$\displaystyle\geq\Big{(}\frac{C}{B}\frac{\log\log\gamma}{\log\gamma}\Big{)}^{2/3}-\frac{F}{31.76B^{2/3}(\log\gamma)^{2/3}(\log\log\gamma)^{1/3}}$
$\displaystyle\geq 1.92(\log(\gamma/100))^{-2/3},$

so indeed we have $1\leq\sigma\leq 1+\eta-1.92(\log(\gamma/100))^{-2/3}$.
Furthermore, (B.2) and (B.3) imply that
$\frac{\sigma-\beta}{\eta}=(F+1)\frac{1-\beta}{\eta}\leq 0.012$. Since
$\cot(x)\geq 0.99988x^{-1}$ when $0\leq x\leq 0.012\frac{\pi}{2}$, we find
that

$\displaystyle-
b_{1}\mathrm{Re}\Big{(}\frac{\pi}{2\eta}\cot\Big(\frac{\pi(\sigma-\beta)}{2\eta}\Big{missing})\Big{)}\leq-0.99988\frac{b_{1}}{\sigma-\beta}.$

Hence, by the numerical values of $b_{0}$, $b_{5}$, and $F$, we have that

$\displaystyle-
b_{1}\mathrm{Re}\Big{(}\frac{\pi}{2\eta}\cot\Big(\frac{\pi(\sigma-\beta)}{2\eta}\Big{missing})\Big{)}+\frac{b_{0}}{\sigma-1}\leq-0.99988\frac{b_{1}}{\sigma-\beta}+\frac{b_{0}}{F(1-\beta)}$
(B.5)
$\displaystyle\qquad\leq\Big{(}\frac{10.01055}{F}-0.99988\frac{17.14500}{F+1}\Big{)}\frac{1}{1-\beta}\leq-0.953\frac{1}{1-\beta}.$

Combining (B.5) with (B.4), we obtain the inequality

$\displaystyle 0.953\frac{1}{1-\beta}$
$\displaystyle\leq\frac{b_{5}}{2\eta}\Big{(}\frac{2}{3}\log\log\gamma+B\eta^{3/2}\log\gamma+\log(A+1)\Big{)}$
(B.6)
$\displaystyle\quad+\frac{b_{0}}{2\eta}\log(\zeta(1+\eta))+\frac{b_{5}}{2}\log
q+(b_{2}+b_{4})e^{-1937}.$

Note that by Lemma 4.1 and the inequality $\log(1+x)\leq x$, we have that

$\frac{b_{0}}{2\eta}\log(\zeta(1+\eta))\leq\frac{b_{0}}{2\eta}\log\Big(\frac{1}{\eta}\Big{missing})+0.3b_{0}.$
(B.7)

Since $\log(1/\eta)\sim\frac{2}{3}\log\log\gamma$, it follows that the main
term on the right-hand side of (B.6) is

$\displaystyle\frac{1}{\eta}\Big{(}\frac{b_{5}}{2}\Big{(}\frac{2}{3}\log\log\gamma+B\eta^{3/2}\log\gamma\Big{)}+\frac{b_{0}}{3}\log\log\gamma\Big{)}=\frac{\log\log\gamma}{\eta}\Big{(}b_{5}\Big{(}\frac{1}{3}+\frac{C}{2}\Big{)}+\frac{b_{0}}{3}\Big{)}$
(B.8)
$\displaystyle\qquad=C^{-2/3}\Big{(}b_{5}\Big{(}\frac{1}{3}+\frac{C}{2}\Big{)}+\frac{b_{0}}{3}\Big{)}B^{2/3}(\log\gamma)^{2/3}(\log\log\gamma)^{1/3}$
$\displaystyle\qquad\leq
30.26576B^{2/3}(\log\gamma)^{2/3}(\log\log\gamma)^{1/3}$

via (B.7). Note that since $\frac{\log T_{0}}{\log\log
T_{0}}\geq\frac{5110.6}{B}$ and $\gamma\geq T_{0}\geq e^{10\,650}$,

$0.3b_{0}+(b_{2}+b_{4})e^{-1937}\leq
0.010122B^{2/3}\Big{(}\frac{\log\gamma}{\log\log\gamma}\Big{)}^{2/3}.$ (B.9)

Consequently, by (B.9), (B.7), and the assumption $B\leq 4.45$, all of the
remaining terms in (B.6) excluding $\frac{b_{5}}{2}\log q$ contribute at most

$\displaystyle\frac{b_{5}}{2\eta}\log(A+1)+\frac{b_{0}}{3\eta}(\log(B/C)-\log_{3}\gamma)+0.010122B^{2/3}\Big{(}\frac{\log\gamma}{\log\log\gamma}\Big{)}^{2/3}$
$\displaystyle\leq\Big{(}C^{-2/3}\Big{(}\frac{b_{5}}{2}\log(A+1)+\frac{b_{0}}{3}(\log(B/C)-\log_{3}\gamma)\Big{)}+0.010122\Big{)}B^{2/3}\Big{(}\frac{\log\gamma}{\log\log\gamma}\Big{)}^{2/3}$
$\displaystyle\leq(-2.7545\log_{3}\gamma+13.75563\log(A+1)+3.33)B^{2/3}\Big{(}\frac{\log\gamma}{\log\log\gamma}\Big{)}^{2/3}.$

Consequently, we have that

$\displaystyle 0.953\frac{1}{1-\beta}$ $\displaystyle\leq 16.66375\log
q+30.26576B^{2/3}(\log\gamma)^{2/3}(\log\log\gamma)^{1/3}$
$\displaystyle\quad+(-2.7545\log_{3}\gamma+13.75563\log(A+1)+3.33)B^{2/3}\Big{(}\frac{\log\gamma}{\log\log\gamma}\Big{)}^{2/3},$

meaning that

$\frac{1}{1-\beta}\leq\Big{(}31.76+\frac{-2.89\log_{3}\gamma+14.44\log(A+1)+3.495}{\log\log\gamma}\Big{)}B^{2/3}(\log\gamma)^{2/3}(\log\log\gamma)^{1/3}+17.49\log
q,$

a contradiction. ∎

######  Corollary B.2.

Let $q\geq 3$, and let $\chi\pmod{q}$ be a Dirichlet character. Then,
$L(\sigma+it,\chi)$ is nonvanishing in the regions

$\displaystyle\sigma$ $\displaystyle\geq 1-\frac{1}{18\log
q+86(\log|t|)^{\frac{2}{3}}(\log\log|t|)^{\frac{1}{3}}},\qquad\absolutevalue{t}\geq\exp(\exp(\exp(23))),$
$\displaystyle\sigma$ $\displaystyle\geq 1-\frac{1}{18\log
q+104(\log|t|)^{\frac{2}{3}}(\log\log|t|)^{\frac{1}{3}}},\qquad\absolutevalue{t}\geq
3.$

###### Proof.

The first part follows from Theorem B.1 with $B=4.45$. For the second part, we
take $A=76.2$ and $B=4.45$. If $T_{0}\geq e^{11\,450}$, then
$-2.89\log_{3}t+14.435\log(A+1)+3.495\leq 59.8$. Thus, by Theorem B.1, we
obtain $c\leq 31.76+\frac{59.8}{\log\log T_{0}}$. The desired result follows
once we take $T_{0}=e^{11\,450}$ and apply (2.3) for $t\leq T_{0}$. ∎

## References