Kids can be picky eaters, and getting them to try new things may be a constant struggle. We're going to explore food traditions and recipes from around the world, and encourage you and your family to start your own "Culture Night" right at home, where you create an entire experience around cooking and eating exotic food. At Chefs Plate, we're firm believers that everyone can benefit from a little bit of spice in their lives, and we want to help take you on a culinary adventure around the globe.
In the first instalment of our "Culture Night" series, we're going to start by taking a dive into Indian Cuisine! Food plays a big role in how they spend time and relax, and dinner time is family time – so gather your loved ones together, and let's get cooking!
Indian civilization is one of the oldest in the world, dating back around 4000 years, and during that time they have developed a rich history of culture and tradition. Food and family are at the at the centre of these traditions, from when dinner is scheduled to the way dishes are prepared and served – no one should eat alone!
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Indian cuisine is so unique because it combines elements and flavours from so many of its surrounding areas. Even across the country, the food varies greatly from region to region. One amazing meal that borrows elements from all corners of the country is biryani – a rice dish containing a variety of fragrant spices, meats, and vegetables.
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text:
Jaroslav Navrátil (tennisser), een voormalig Tsjechisch tennisser en huidig tenniscoach
Jaroslav Navrátil (voetballer), een Tsjechisch voetballer bij Heracles Almelo

redpajama_set_name: RedPajamaWikipedia


Show your Browns pride and team spirit with this single-sided sewn flag kit. It makes a great addition to your game room, man cave, or when flown outside. Officially licensed, made by Party Animal.
The included steel pole is not meant to withstand extreme wind conditions. If high winds are predicted, simply remove pole from plastic ground stake. For more permanent installation in areas where high winds are possible, replace pole with 10 foot length of inexpensive 1/2 inch PVC pipe.


Picking a theme for your child's birthday party is a fun thing for you todo together. Depending on their age, they can have a lot of input or just a little bit. Either way, the theme your child selects reflects just what he or she is enjoying at this time in his or her life. Typically, the theme has to do with a favorite character, a favorite movie or a favorite television program.
Already on children's radars is the upcoming spring release from Disney entitled Zootopia. With human like characters, a funny plot line and some great action and suspense, kids will be requesting this as their theme this year.
One easy way to make your party theme pop is to have Disney Zootopia party supplies. This table cover measures 54″ x 96″ and will protect your table from spills.


text:
module Data.Random.Sequence (randomElementT) where

import Data.Random
import Data.Sequence as SQ

randomElementT :: Seq a -> RVarT m a
randomElementT xs | SQ.null xs = error "randomElementT: empty seq!"
randomElementT xs = do
  n <- uniformT 0 (SQ.length xs - 1)
  return (xs `index` n)

redpajama_set_name: RedPajamaGithub


€249.00US$271.41£216.6372% di sconto In Stock.
2.Support for on-board diagnostics on all GM systems 1992 thru current year.
5.RS232 & RS485 ports provide the avenue to connect future devices to your Tech 2 Flash.
6.Ability to capture stored DTCs, Freeze Frame, and Failure Record for later review.
7.Ability to view the status of vehicle diagnostic tests as they run.
8.Bi-directional The authentic GM 1992-current year software will provide full bi-directional control when performing power train, chassis, and body diagnostics.
13.Communications: Integral RS232 communication port.
An industry change in protocol from OBD II to Controller Area Network (CAN) has begun. In order to perform diagnostics on a vehicle that uses the CAN interface; you will need the Tech 2 Flash and a CAN diagnostic interface module. The module connects between the Tech 2 Flashs diagnostic link cable and the OBD adapter. Once the module is connected, it can be used on all GM vehicle applications, even those vehicles using other platforms such as OBD I and OBD II.
RS232 cable and user`s manual.


text:
\section{Introduction}
\label{sec:1}

The theory of the Riemann zeta function $\zeta(s)$ has been studied for over 150 years. Among the topics of research, the study of its zeros has been one of the main focus of research. Lately, the study of the zeros of its derivatives has also been part of the research area.
In fact, in 1970, Berndt \cite[Theorem]{ber} proved that
\begin{equation} \label{eq:nkt}
N_k(T) = \frac{T}{2\pi}\log{\frac{T}{4\pi}}-\frac{T}{2\pi}+O(\log{T})
\end{equation}
where $N_k(T)$ denotes the number of zeros of the $k$-th derivative of the Riemann zeta function, we write as $\zeta^{(k)}(s)$, with $0<\text{Im}\,(s)\leq T$, counted with multiplicity, for any positive integer $k$.
And in 1974, Levinson and Montgomery \cite[Theorem 10]{lev} showed that for any positive integer $k$,
\begin{equation} \label{eq:kdist}
\begin{aligned}
\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ \zeta^{(k)}(\rho^{(k)})=0,\, 0<\gamma^{(k)}\leq T}} \left(\beta^{(k)}-\frac{1}{2}\right) = \frac{kT}{2\pi}\log{\log{\frac{T}{2\pi}}} &+ \frac{1}{2\pi}\left(\frac{1}{2}\log{2} - k\log{\log{2}}\right)T\\
&\quad\quad\quad\quad\quad\quad\quad- k\text{Li}\left(\frac{T}{2\pi}\right) + O(\log{T})
\end{aligned}
\end{equation}
where the sum is counted with multiplicity and
\begin{align*}
\text{Li}(x) := \int_2^x \frac{dt}{\log{t}}.
\end{align*}
In addition to the above result \eqref{eq:kdist}, Levinson and Montgomery \cite{lev} also studied the location of the zeros of $\zeta^{(k)}(s)$.
There are many other papers on the zeros of $\zeta^{(k)}(s)$; for example, Conrey and Ghosh \cite[Theorem 1]{con} in 1989, studied the zeros of $\zeta^{(k)}(s)$ near the critical line.

In 2012, Akatsuka \cite[Theorems 1 and 3]{aka} improved each of the error term of the results obtained by Berndt and by Levinson and Montgomery mentioned above (eq. \eqref{eq:nkt} and \eqref{eq:kdist}) for the case $k=1$ under the assumption of the truth of the Riemann hypothesis.
His results are
\begin{align*}
\sum_{\substack{\rho'=\beta'+i\gamma',\\ \zeta'(\rho')=0,\, 0<\gamma'\leq T}} \left(\beta'-\frac{1}{2}\right) = \frac{T}{2\pi}\log{\log{\frac{T}{2\pi}}} &+ \frac{1}{2\pi}\left(\frac{1}{2}\log{2} - \log{\log{2}}\right)T\\
&\quad\quad\quad\quad\quad\quad\quad- \text{Li}\left(\frac{T}{2\pi}\right) + O((\log{\log{T})^2})
\end{align*}
and
\begin{equation*}
N_1(T) = \frac{T}{2\pi}\log{\frac{T}{4\pi}}-\frac{T}{2\pi}+O\left(\frac{\log{T}}{(\log{\log{T}})^{1/2}}\right)
\end{equation*}
if the Riemann hypothesis is true.
In this paper, we generalize these two results of Akatsuka for any positive integer $k$.
Before we introduce our results, we define some notation.

\vspace{3mm}
In this paper we denote by $\mathbb{Z}$, $\mathbb{R}$, and $\mathbb{C}$
the set of all rational integers, the set of all real numbers, and the set of all complex numbers, respectively.
Throughout this paper, the letter $k$ is used as a fixed positive integer, that is $k\geq1$ ($k\in\mathbb{Z}$), unless otherwise specified.
Next, we let $\rho = \beta + i\gamma$ and $\rho^{(k)} = \beta^{(k)} + i\gamma^{(k)}$ represent the nontrivial zeros of the Riemann zeta function and the non-real zeros of the $k$-th derivative of the Riemann zeta function, respectively. Then we define $N(T)$ and $N_k(T)$ as follows:

\begin{definition}
For $T>0$, we define
\begin{equation*}
N(T) := \sharp'\{\rho=\beta+i\gamma \,|\, 0<\gamma\leq T\}
\end{equation*}
and
\begin{equation*}
N_k(T) := \sharp'\{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)} \,|\, 0<\gamma^{(k)}\leq T\}
\end{equation*}
where $\sharp'$ means the number of elements counted with multiplicity.
\end{definition}

\vspace{2mm}
Each of the following results in this paper is a generalization of Theorem 1, Corollary 2, and Theorem 3 of \cite{aka}, respectively. Note that each sum counts the non-real zeros of $\zeta^{(k)}(s)$ with multiplicity and that $O_k$ denotes the error terms which depend only on $k$.

\begin{thm} \label{cha1}
Assume that the Riemann hypothesis is true. Then for any $T>4\pi$, we have
\begin{equation*}
\begin{aligned}
\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ 0<\gamma^{(k)}\leq T}} \left(\beta^{(k)} -\frac{1}{2}\right) = \frac{kT}{2\pi}\log{\log{\frac{T}{2\pi}}} &+ \frac{1}{2\pi}\left(\frac{1}{2}\log{2} - k\log{\log{2}}\right)T\\
&\quad\quad\quad\quad\quad- k\emph{Li}\left(\frac{T}{2\pi}\right) + O_k((\log{\log{T}})^2).
\end{aligned}
\end{equation*}
\end{thm}

\begin{cor} \label{cha12} (Cf. \cite[Theorem 3]{lev}.)
Assume that the Riemann hypothesis is true. Then for $0<U<T$ (where $T$ is restricted to satisfy $T>2\pi$), we have
\begin{equation*}
\begin{aligned}
\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ T<\gamma^{(k)}\leq T+U}} \left(\beta^{(k)}-\frac{1}{2}\right) = \frac{kU}{2\pi}\log{\log{\frac{T}{2\pi}}} &+ \frac{1}{2\pi}\left(\frac{1}{2}\log{2} - k\log{\log{2}}\right)U\\
&\quad\quad\quad\quad+ O\left(\frac{U^2}{T\log{T}}\right) + O_k((\log{\log{T}})^2).
\end{aligned}
\end{equation*}
Here the error term $O\left(\frac{U^2}{T\log{T}}\right)$ holds uniformly, in other words, it does not depend on any parameter.
\end{cor}

\begin{thm} \label{cha2}
Assume that the Riemann hypothesis is true. Then for $T\geq2$, we have
\begin{equation*}
N_k(T) = \frac{T}{2\pi}\log{\frac{T}{4\pi}} - \frac{T}{2\pi} + O_k\left(\frac{\log{T}}{(\log{\log{T}})^{1/2}}\right).
\end{equation*}
\end{thm}

\vspace{2mm}
We write Re$(s)$ and Im$(s)$ (for any $s\in\mathbb{C}$) as $\sigma$ and $t$ respectively. We abbreviate the Riemann hypothesis as RH, and finally,
we define two functions $F(s)$ and $G_k(s)$, as follows:
\begin{definition}
\begin{align*}
F(s) := 2^s\pi^{s-1}\sin{\left(\frac{\pi s}{2}\right)}\Gamma(1-s),\quad G_k(s) := (-1)^k\frac{2^s}{(\log{2})^k}\zeta^{(k)}(s).
\end{align*}
\end{definition}
\noindent
By the above definition of $F(s)$, we can check easily that the functional equation for $\zeta(s)$ states
\begin{equation} \label{eq:fe}
\zeta(s) = F(s)\zeta(1-s).
\end{equation}

\begin{rmk}
The function $F(s)$ is the one appearing in \cite{aka} and \cite[Section 3]{lev} and the function $G_k(s)$ is the $\zeta^{(k)}$-version of the function $G(s)$ in \cite{aka}, meanwhile this function is denoted $Z_k(s)$ in \cite[Section 3]{lev}. Most of the symbols used in this paper follow those used in \cite{aka}.
\end{rmk}

Before we begin the next section, we intend to give a brief outline of the proofs.
Nevertheless, since the steps of our proofs basically follow those given in \cite{aka} with a few crucial modifications, instead of the outline of the proofs, we present the main modifications related to the proofs.

\vspace{2mm}
First of all, condition 2 of Lemma 2.1 of \cite{aka} is related to the functional equation for $\zeta'(s)$. In our case, we need to consider generally for $\zeta^{(k)}(s)$. As a result, we obtain a function which consists of terms that are not logarithmic derivatives of some functions so that this cannot be taken care as in the case of $\zeta'(s)$. In the present paper, we take care of these terms generally in a way that does not involve any calculation on logarithmic derivatives.

Secondly, similar to condition 2, in condition 3 of Lemma 2.1 of \cite{aka}, the factor to be estimated was $\frac{F'}{F}(s)$ which is just the logarithmic derivative of $F(s)$, whereas in the present paper, we need to take care of $\frac{F^{(k)}}{F}(s)$ which is not a logarithmic derivative of any function. Thus, as in condition 2, we estimate this term generally in a way which does not involve any calculation on logarithmic derivatives, and there we need to take a suitable logarithmic branch of the function $\log{\frac{F^{(k)}}{F}(s)}$.

The next is condition 4 of Lemma 2.1 of \cite{aka}. For $\zeta'(s)$, the term we need to estimate was $\frac{\zeta'}{\zeta}(s)$ which is just the logarithmic derivative of $\zeta(s)$. In \cite{aka}, the inequality $\text{Re}\,\left(\frac{\zeta'}{\zeta}(s)\right)<0$ was obtained, however for $\zeta^{(k)}(s)$, the sign of $\text{Re}\,\left(\frac{\zeta^{(k)}}{\zeta}(s)\right)$ does not seem to stay unchanged in any region defined by $x\leq\sigma<\frac{1}{2},\, t\geq y$ for some $x\leq-1$ and large $y>0$. Since it is sufficient to show that $\frac{\zeta^{(k)}}{\zeta}(s)$ is holomorphic and non-zero, and has bounded argument in some such region, we modify the condition in such a way.

Furthermore, with the modifications of these conditions of the first lemma, the choice of logarithmic branch of the function $\log{\left(\frac{1}{\frac{F^{(k)}}{F}(s)}\frac{\zeta^{(k)}}{\zeta}(s)\right)}$ in the proof of Proposition \ref{prop2} (which is the analogue of Proposition 2.2 in \cite{aka}) must be taken more carefully so that these conditions can be used in our calculations. In this present paper, in order to evaluate the function $\log{\left(\frac{1}{\frac{F^{(k)}}{F}(s)}\frac{\zeta^{(k)}}{\zeta}(s)\right)}$, we first define the functions $\log{\left(\frac{1}{\frac{F^{(k)}}{F}(s)}\frac{\zeta^{(k)}}{\zeta}(s)\right)}$, $\log{\frac{F^{(k)}}{F}(s)}$, and $\log{\frac{\zeta^{(k)}}{\zeta}(s)}$ independently. Then using the continuities of $\arg{\left(\frac{1}{\frac{F^{(k)}}{F}(s)}\frac{\zeta^{(k)}}{\zeta}(s)\right)}$, $\arg{\frac{F^{(k)}}{F}(s)}$, and $\arg{\frac{\zeta^{(k)}}{\zeta}(s)}$, we observe the difference $\arg{\left(\frac{1}{\frac{F^{(k)}}{F}(s_0)}\frac{\zeta^{(k)}}{\zeta}(s)\right)} - \left(-\arg{\frac{F^{(k)}}{F}(s)} + \arg{\frac{\zeta^{(k)}}{\zeta}(s)}\right)$ in the region under evaluation (see the evaluation of $I_{15}$ in Proposition \ref{prop2}).

Finally, the region $\frac{1}{2}<\sigma\leq a$ considered in Lemma 2.3 of \cite{aka} does not work well for $\frac{\zeta^{(k)}}{\zeta}(s)$. The reason is that the current best estimate of $\frac{\zeta^{(k)}}{\zeta}(s)$ depends on the usage of Cauchy's integral formula, hence we need to keep a certain distance between $\frac{1}{2}$ and the infimum of $\sigma$ in the region. Therefore, we put here a small distance $\epsilon_0>0$ (see the statement of our Lemma \ref{lem3}).

\section{Proof of Theorem \ref{cha1} and Corollary \ref{cha12}}
\label{sec:2}

In this section we give the proofs of Theorem 1 and Corollary 2. For that purpose, we need a few lemmas and a proposition which are analogues of those in \cite{aka}.

\vspace{3mm}
To begin with, we introduce the following lemma which is a generalization of Lemma 2.1 of \cite{aka} for the case of $\zeta^{(k)}(s)$.

\begin{lem} \label{lem1}
Assume RH. Then there exist $a_k\geq10$, $\sigma_k\leq-1$, and $t_k\geq\max{\{a_k^2,-\sigma_k\}}$ such that the following conditions are satisfied:
\begin{align*}
&1.\,\, \left|G_k(s)-1\right| \leq \frac{1}{2}\left(\frac{2}{3}\right)^{\sigma/2}, \quad \text{for any } \sigma\geq a_k;\\
&2.\,\, \left|\sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(s)}\frac{\zeta^{(j)}}{\zeta}(1-s)\right| \leq 2^\sigma, \quad \text{for } \sigma\leq\sigma_k \text{ and } t\geq2;\\
&3.\,\, \left|\frac{F^{(k)}}{F}(s)\right|\geq1 \text{ holds in the region } \sigma_k\leq\sigma\leq\frac{1}{2},\,  t\geq t_k-1. \text{ Furthermore, we can take}\\
&\quad \text{the logarithmic branch of } \log{\frac{F^{(k)}}{F}(s)} \text{ in that region such that it is holomorphic there}\quad\,\,\,\\
&\quad \text{and }\,\, \frac{\alpha_k\pi}{6} < \arg{\frac{F^{(k)}}{F}(s)} < \frac{\beta_k\pi}{6} \text{ holds, where } (\alpha_k,\beta_k)=
\begin{cases}
(5,7) \text{ if } k\text{ is odd},\\
(-1,1) \text{ if } k\text{ is even};
\end{cases}\\
&4.\,\, \frac{\zeta^{(k)}}{\zeta}(s)\neq0 \text{ holds in the region } \sigma_k\leq\sigma<\frac{1}{2},\, t\geq t_k-1. \text{ Furthermore, we can take}\\
&\quad \text{the logarithmic branch of } \log{\frac{\zeta^{(k)}}{\zeta}(s)} \text{ in that region such that it is holomorphic there}\\
&\quad \text{and }\,\, \frac{k\pi}{2} < \arg{\frac{\zeta^{(k)}}{\zeta}(s)} < \frac{3k\pi}{2} \text{ holds};\\
&5.\,\, \zeta(\sigma+it_k)\neq0,\, \zeta^{(k)}(\sigma+it_k)\neq0, \quad \text{for all}\,\, \sigma\in\mathbb{R}.\\
\end{align*}
\end{lem}

\begin{proof}
\raggedright
$\quad$\\

\vspace{2mm}
1. See \cite[inequality (3.2) (p. 54)]{lev}.

\vspace{3mm}
2. We begin by estimating $\frac{F^{(k)}}{F^{(k-j)}}(s)$ ($j=1,2,\cdots,k$) in the region $\sigma<1,\, t\geq2$.
We set $f(s):=\left(\frac{1}{2}-s\right)\left(\log{(1-s)} - \log{(2\pi)} + \frac{\pi i}{2} \right) + s + O(1)$, where $f(s)$ is an analytic function and
\begin{equation*}
f'(s) = -\log{(1-s)} + O(1),\quad\quad f^{(j)}(s) = O(1) \quad\quad (j\geq2).
\end{equation*}
As in \cite[pp. 54--55]{lev}, we can write
\begin{equation*}
F(s) = \exp(f(s)).
\end{equation*}
Using methods similar to \cite[Lemma 6 (p. 133)]{gon} and \cite[pp. 54--55]{lev},
we can show that
\begin{equation} \label{eq:Fj}
F^{(j)}(s) = F(s)(f'(s))^j\left(1+O\left(\frac{1}{|\log{s}|^2}\right)\right)
\end{equation}
holds for any positive integer $j$. In consequence, for $j=1,2,\cdots,k$, we have
\begin{equation*}
\left|\frac{F^{(k)}}{F^{(k-j)}}(s)\right| = \left|(f'(s))^j\left(1+O\left(\frac{1}{|\log{s}|^2}\right)\right)\right| \geq (\log{|1-s|})^j - \left|O\left((\log{|1-s|})^{j-1}\right)\right|.
\end{equation*}
Certainly, this also holds in the region $\sigma\leq-1,\, t\geq2$, so for any positive integer $k$, we can take $\sigma_{k_1}\leq-1$ sufficiently small (i.e. sufficiently large in the negative direction) so that for any $s$ with $\sigma\leq\sigma_{k_1}$ and $t\geq2$, we have
\begin{equation} \label{eq:Fkkj}
\left|\frac{F^{(k)}}{F^{(k-j)}}(s)\right| \geq \frac{1}{2k} (\log{|1-s|})^j \geq \frac{1}{2k}(\log{(1-\sigma)})^j.
\end{equation}

\vspace{2mm} \hspace{5mm}
Next we estimate $\frac{\zeta^{(j)}}{\zeta}(1-s)$ ($j=1,2,\cdots,k$).
In the region $\sigma\leq-1,\, t\geq2$, we have
\begin{align*}
\left|\zeta^{(j)}(1-s)\right| &\leq \left|\frac{(\log{2})^j}{2^{1-s}}\right| + \left|\sum_{n=3}^\infty \frac{(\log{n})^j}{n^{1-s}}\right|
\leq \frac{1}{2}(\log{2})^j2^\sigma + \int_2^\infty \frac{(\log{x})^j}{x^{1-\sigma}}dx\\
&= 2^\sigma\left(\frac{1}{2}(\log{2})^j + \sum_{l=0}^j\frac{(\log{2})^{j-l}\frac{j!}{(j-l)!}}{(-\sigma)^{l+1}}\right)
\end{align*}
and
\begin{equation*}
\left|\zeta(1-s)\right| \geq 1 - \left|\sum_{n=2}^\infty \frac{1}{n^{1-s}}\right|
\geq 1 - \sum_{n=2}^\infty \frac{1}{n^2} = 2 - \frac{\pi^2}{6}.
\end{equation*}
Thus,
\begin{equation} \label{eq:zetaj}
\left|\frac{\zeta^{(j)}}{\zeta}(1-s)\right| \leq \frac{2^\sigma}{2 - \frac{\pi^2}{6}}\left(\frac{1}{2}(\log{2})^j + \sum_{l=0}^j\frac{(\log{2})^{j-l}\frac{j!}{(j-l)!}}{(-\sigma)^{l+1}}\right).
\end{equation}

\hspace{5mm}
Now combining inequalities \eqref{eq:Fkkj} and \eqref{eq:zetaj}, for $\sigma\leq\sigma_{k_1}$ and $t\geq2$, we have
\begin{align*}
&\left|\sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(s)}\frac{\zeta^{(j)}}{\zeta}(1-s)\right| \leq \sum_{j=1}^k {{k}\choose{j}} \frac{1}{\left|\frac{F^{(k)}}{F^{(k-j)}}(s)\right|} \left|\frac{\zeta^{(j)}}{\zeta}(1-s)\right|\\
&\quad\quad\quad\quad\quad\quad\quad\quad
\leq 2^\sigma\frac{2k}{2 - \frac{\pi^2}{6}} \sum_{j=1}^k {{k}\choose{j}} \frac{1}{(\log{(1-\sigma)})^j} \left(\frac{1}{2}(\log{2})^j + \sum_{l=0}^j\frac{(\log{2})^{j-l}\frac{j!}{(j-l)!}}{(-\sigma)^{l+1}}\right).
\end{align*}

\vspace{2mm} \hspace{5mm}
Since for any positive integer $k$,
\begin{align*}
\lim_{\sigma\rightarrow-\infty} \frac{2k}{2 - \frac{\pi^2}{6}} \sum_{j=1}^k {{k}\choose{j}} \frac{1}{(\log{(1-\sigma)})^j} \left(\frac{1}{2}(\log{2})^j + \sum_{l=0}^j\frac{(\log{2})^{j-l}\frac{j!}{(j-l)!}}{(-\sigma)^{l+1}}\right) = 0,
\end{align*}
we can take $\sigma_k\leq\sigma_{k_1}$ ($\leq-1$) so that
\begin{align*}
\frac{2k}{2 - \frac{\pi^2}{6}} \sum_{j=1}^k {{k}\choose{j}} \frac{1}{(\log{(1-\sigma)})^j} \left(\frac{1}{2}(\log{2})^j + \sum_{l=0}^j\frac{(\log{2})^{j-l}\frac{j!}{(j-l)!}}{(-\sigma)^{l+1}}\right) \leq 1
\end{align*}
holds for any $\sigma\leq\sigma_k$.
This implies that
\begin{align*}
 \left|\sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(s)}\frac{\zeta^{(j)}}{\zeta}(1-s)\right| \leq 2^\sigma
\end{align*}
holds for $\sigma\leq\sigma_k$, $t\geq2$.

\vspace{2mm} \hspace{5mm}
Now with the above $\sigma_k$, we are going to look for $t_k\geq\max{\{a_k^2, -\sigma_k\}}$ for which conditions 3 to 5 hold.

\vspace{3mm}
3. We start by examining condition 3. We first consider the region $\sigma_k\leq\sigma\leq\frac{1}{2},\, t\geq99$.
It follows from equation \eqref{eq:Fj} that in this region,
\begin{equation} \label{eq:Fk}
F^{(k)}(s) = F(s)(-\log{(1-s)} + O(1))^k\left(1+O\left(\frac{1}{|\log{s}|^2}\right)\right)
\end{equation}
holds.
This gives us,
\begin{align*}
\left|\frac{F^{(k)}}{F}(s)\right| \geq |(\log{(1-s)})^k| - \left|O_{\sigma_k}((\log{t})^{k-1})\right| \geq (\log{t})^k - \left|O_{\sigma_k}((\log{t})^{k-1})\right|
\end{align*}
for $\sigma_k\leq\sigma\leq\frac{1}{2}$ and $t\geq99$.
Thus, for any integer $k\geq1$, we can take $t_{k_1}\geq100$ such that
\begin{equation} \label{eq:cond31}
\left|\frac{F^{(k)}}{F}(s)\right| \geq 1
\end{equation}
for $\sigma_k\leq\sigma\leq\frac{1}{2}$ and $t\geq t_{k_1}-1$.

\vspace{2mm} \hspace{5mm}
We note from equation \eqref{eq:Fk} that $\frac{F^{(k)}}{F}(s) = (-1)^k(\log{t})^k + O((\log{t})^{k-1})$ when $\sigma_k\leq\sigma\leq\frac{1}{2}$ and $t\geq99$.
Consequently, for odd integer $k\geq1$, we can find $t'_{k_2}\geq100$ sufficiently large such that
\begin{equation*}
\frac{5\pi}{6} < \arg{\frac{F^{(k)}}{F}(s)} < \frac{7\pi}{6}
\end{equation*}
holds for $\sigma_k\leq\sigma\leq\frac{1}{2}$ and $t\geq t'_{k_2}-1$.
Similarly, when $k$ is even, we can also find $t''_{k_2}\geq100$ large enough such that
\begin{equation*}
-\frac{\pi}{6} < \arg{\frac{F^{(k)}}{F}(s)} < \frac{\pi}{6}
\end{equation*}
holds for $\sigma_k\leq\sigma\leq\frac{1}{2}$ and $t\geq t''_{k_2}-1$.
Since all zeros and poles of $F(s)$ lie on $\mathbb{R}$, $\frac{F^{(k)}}{F}(s)$ has no poles in $t>0$. This along with inequality \eqref{eq:cond31} implies that $\log{\frac{F^{(k)}}{F}(s)}$ is holomorphic in the region with this branch.
Thus setting
\begin{align*}
(\alpha_k,\beta_k) :=
\begin{cases}
(5,7), &\quad \text{if}\,\, k \,\,\text{is odd},\\
(-1,1), &\quad \text{if}\,\, k \,\,\text{is even};
\end{cases}
\end{align*}
and
\begin{align*}
t_{k_2} :=
\begin{cases}
t'_{k_2}, &\quad \text{if}\,\, k \,\,\text{is odd},\\
t''_{k_2}, &\quad \text{if}\,\, k \,\,\text{is even};
\end{cases}
\end{align*}
we find that $\log{\frac{F^{(k)}}{F}(s)}$ is holomorphic and that
\begin{equation*}
\frac{\alpha_k\pi}{6} < \arg{\frac{F^{(k)}}{F}(s)} < \frac{\beta_k\pi}{6}
\end{equation*}
holds in the region $\sigma_k\leq\sigma\leq\frac{1}{2},\, t\geq t_{k_2}-1$.

\vspace{2mm} \hspace{5mm}
By the above calculations, we find that $\max{\{t_{k_1}, t_{k_2}, a_k^2, -\sigma_k\}}$ is a candidate for $t_k$. Thus we have proven that $t_k\geq\max{\{a_k^2, -\sigma_k\}}$ for which condition 3 holds exists. Since we want $t_k$ to also satisfy conditions 4 and 5, we need to examine those conditions to completely prove the existence of $t_k$.

\vspace{3mm}
4. Referring to \cite[Corollary of Theorem 7 (p. 51)]{lev}, we know that RH implies that for any positive integer $j$, $\zeta^{(j)}(s)$ has at most a finite number of non-real zeros in $\sigma<\frac{1}{2}$. Hence we can number all the non-real zeros of $\zeta^{(j)}(s)$ in $\sigma<\frac{1}{2}$ as $\rho_1^{(j)}, \rho_2^{(j)}, \rho_3^{(j)}, \cdots, \rho_{m_j}^{(j)}$ ($\rho_l^{(j)} = \beta_l^{(j)} + i\gamma_l^{(j)}$) for some integer $m_j\geq2$ (note that if $\zeta^{(j)}\left(\rho^{(j)}\right) = 0$, then $\zeta^{(j)}\left(\overline{\rho^{(j)}}\right) = 0$, so $m_j\geq2$) in the order such that $\gamma_l^{(j)}\leq\gamma_{l+1}^{(j)}$ for all $1\leq l\leq m_j-1$.
Therefore, $\zeta^{(j)}(s)\neq0$ when $\sigma<\frac{1}{2}$ and $t\geq\gamma_{m_j}^{(j)}+1$.
We set $t_{k_3} := \displaystyle{\max_{1\leq j\leq k}{(\gamma_{m_j}^{(j)}+2)}}$, then for all $j=1,2,\cdots,k$, we have
\begin{equation} \label{eq:4zenhan}
\zeta^{(j)}(s)\neq0
\end{equation}
in the region $\sigma<\frac{1}{2},\, t\geq t_{k_3}-1$.

\vspace{2mm} \hspace{5mm}
Next we show that we can take the logarithmic branch of $\log{\frac{\zeta^{(k)}}{\zeta}(s)}$ in the region $\sigma_k\leq\sigma<\frac{1}{2},\, t\geq t_{k_4}-1$ for some $t_{k_4}\geq100$, so that it is holomorphic there and
\begin{align*}
\frac{k\pi}{2} < \arg{\frac{\zeta^{(k)}}{\zeta}(s)} < \frac{3k\pi}{2}
\end{align*}
holds there by first claiming that we can find some $t_{k_4}\geq t_{k_3}$ for which
\begin{equation} \label{eq:claim1}
\text{Re}\left(\frac{\zeta^{(j)}}{\zeta^{(j-1)}}(s)\right) < 0 \quad\quad (\sigma_k\leq\sigma<\frac{1}{2},\, t\geq t_{k_4}-1)
\end{equation}
holds for all $j=1,2,\cdots,k$.
We first note that for any $j=1,2,\cdots,k$, $\frac{\zeta^{(j)}}{\zeta^{(j-1)}}(s)$ is holomorphic and has no zeros in the region defined by $\sigma<\frac{1}{2}$ and $t\geq t_{k_3}-1$.

\vspace{2mm} \hspace{5mm}
To show this, we refer to \cite[pp. 64--65]{lev} and we can show that for any $j=1,2,\cdots,k$,
\begin{align*}
\text{Re}\left(\frac{\zeta^{(j)}}{\zeta^{(j-1)}}(s)\right) \leq -\frac{2}{9}\log{|s|} + O_{\sigma_k}(1)
\end{align*}
holds when $\sigma_k\leq\sigma<\frac{1}{2}$, and $ t\geq t_{k_3}-1$.
Thus, we can take $t_{k_4}\geq t_{k_3}$ such that for all $j=1,2,\cdots,k$ inequalities \eqref{eq:claim1} holds.

\vspace{2mm} \hspace{5mm}
The above immediately implies that for each $j=1,2,\cdots,k$, there exists an integer $l_j$ such that
\begin{equation} \label{eq:forlj}
\frac{\pi}{2}+2l_j\pi<\arg{\frac{\zeta^{(j)}}{\zeta^{(j-1)}}(s)}<\frac{3\pi}{2}+2l_j\pi
\end{equation}
holds for $\sigma_k\leq\sigma<\frac{1}{2},\, t\geq t_{k_4}-1$.
We then choose the logarithmic branch of each of $\log{\frac{\zeta^{(j)}}{\zeta^{(j-1)}}(s)}$ such that each $l_j$ in inequalities \eqref{eq:forlj} is zero and take the logarithmic branch of $\log{\frac{\zeta^{(k)}}{\zeta}(s)}$ so that
\begin{equation*}
\arg{\frac{\zeta^{(k)}}{\zeta}(s)} = \sum_{j=1}^k \arg{\frac{\zeta^{(j)}}{\zeta^{(j-1)}}(s)}
\end{equation*}
holds in the region $\sigma_k\leq\sigma<\frac{1}{2},\, t\geq t_{k_4}-1$.
Note that from \eqref{eq:4zenhan} and the analyticity of $\zeta^{(k)}(s)$ in $\sigma<\frac{1}{2}$ (also note that we are assuming RH thus $\zeta(s)\neq0$ when $\sigma<\frac{1}{2}$ and $t\geq t_{k_4}-1$), $\log{\frac{\zeta^{(k)}}{\zeta}(s)}$ is holomorphic in this region with this branch.
We then obtain a holomorphic function $\log{\frac{\zeta^{(k)}}{\zeta}(s)}$ with inequalities
\begin{equation*}
\frac{k\pi}{2} < \arg{\frac{\zeta^{(k)}}{\zeta}(s)} < \frac{3k\pi}{2}
\end{equation*}
in the region $\sigma_k\leq\sigma<\frac{1}{2},\, t\geq t_{k_4}-1$.

\vspace{2mm} \hspace{5mm}
Combining the proof of condition 3 and the above calculations, we find that $\max{\{t_{k_1}, t_{k_2}, t_{k_4}, a_k^2, -\sigma_k\}}$ is a candidate for $t_k$. Therefore we have proven that $t_k\geq\max{\{a_k^2, -\sigma_k\}}$ for which conditions 3 and 4 hold exists.

\vspace{3mm}
5. Now we set $t_{k_5}:=\max{\{t_{k_1}, t_{k_2}, t_{k_4}, a_k^2, -\sigma_k\}}$.
\begin{itemize}
\item Since we are assuming RH, $\zeta(\sigma+it)\neq0$ for any $t>0$ if $\sigma\neq\frac{1}{2}$.
\item According to \cite[Table 1 (p. 678)]{spi2}, $\zeta'(\sigma+it)\neq0$ for any $t\in\mathbb{R}$ if $\sigma\geq3$ and $\zeta''(\sigma+it)\neq0$ for any $t\in\mathbb{R}$ if $\sigma\geq5$.
According to \cite[Theorem 1]{spi2}, for $k\geq3$, $\zeta^{(k)}(\sigma+it)\neq0$ for any $t\in\mathbb{R}$ if $\sigma\geq\frac{7}{4}k+2$.
Indeed, we can check that for $k=1$, $\frac{7}{4}+2>3$ and for $k=2$, $\frac{7}{2}+2>5$,
thus for any positive integer $k$,
\begin{align*}
\zeta^{(k)}(\sigma+it)\neq0 \quad\quad (\sigma\geq\frac{7}{4}k+2,\, t\in\mathbb{R}).
\end{align*}
\item Since $t_{k_5}\geq t_{k_3}$, from \eqref{eq:4zenhan}, we have
$\zeta^{(k)}(\sigma+it)\neq0$ for $\sigma<\frac{1}{2}$ and $t\geq t_{k_5}$.
\end{itemize}

Hence, for any positive integer $k$, we only need to find $t_k\in[t_{k_5}+1,t_{k_5}+2]$ for which
\begin{align*}
\zeta\left(\frac{1}{2}+it_k\right)\neq0 \quad\quad \text{and} \quad\quad \zeta^{(k)}(\sigma+it_k)\neq0\quad \text{for } \frac{1}{2} \leq \sigma \leq \frac{7}{4}k+2
\end{align*}
hold.
Note that this is possible by the identity theorem for complex analytic functions.
Thus, we have shown that $t_k$ defined above satisfies $t_k\geq\max{\{a_k^2,-\sigma_k\}}$ and also conditions 3 to 5.\\
\end{proof}

\begin{rmk}
For $k=1$ and $k=2$, more precise results are known. Refer to \cite{aka} and \cite{ade}, respectively.
These results are obtained based on the works of Speiser \cite{spe}, Spira \cite{spi3}, and Yildirim \cite{yil1} (also \cite{yil2}) on the zeros of $\zeta'(s)$ and $\zeta''(s)$.
\end{rmk}

\begin{prop} \label{prop2}
Assume RH. Take $a_k$ and $t_k$ which satisfy all conditions of Lemma \ref{lem1}.
Then for $T\geq t_k$ which satisfies $\zeta^{(k)}(\sigma+iT)\neq0$ and $\zeta(\sigma+iT)\neq0$ for any $\sigma\in\mathbb{R}$, we have
\begin{align*}
\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ 0<\gamma^{(k)}\leq T}} \left(\beta^{(k)}-\frac{1}{2}\right) &= \frac{kT}{2\pi}\log{\log{\frac{T}{2\pi}}} + \frac{1}{2\pi}\left(\frac{1}{2}\log{2} - k\log{\log{2}}\right)T - k\textnormal{Li}\left(\frac{T}{2\pi}\right)\\
&\quad\quad\quad+ \frac{1}{2\pi}\int_{1/2}^{a_k} \left(-\arg{\zeta(\sigma+iT)}+\arg{G_k(\sigma+iT)}\right)d\sigma + O_k(1),
\end{align*}
where the logarithmic branches are taken so that $\log{\zeta(s)}$ and $\log{G_k(s)}$ tend to 0 as $\sigma\rightarrow\infty$ and are holomorphic in $\mathbb{C}\backslash\{\rho+\lambda \,|\, \zeta(\rho)=0 \,\,\text{or}\,\, \infty, \, \lambda\leq0\}$ and $\mathbb{C}\backslash\{\rho^{(k)}+\lambda \,|\, \zeta^{(k)}(\rho^{(k)})=0 \,\,\text{or}\,\, \infty, \, \lambda\leq0\}$, respectively.
\end{prop}

\begin{proof}
The steps of the proof generally follow the proof of Proposition 2.2 of \cite{aka}.
We first take $a_k$, $\sigma_k$, and $t_k$ as in Lemma \ref{lem1} and fix them.
Then, we take $T\geq t_k$ such that $\zeta^{(k)}(\sigma+iT)\neq0$ and $\zeta(\sigma+iT)\neq0$ ($^\forall\sigma\in\mathbb{R}$).
We also let $\delta\in(0,1/2]$ and put $b:=\frac{1}{2}-\delta$.
We consider the rectangle with vertices $b+it_k$,  $a_k+it_k$, $a_k+iT$, and $b+iT$, and then we apply Littlewood's lemma (cf. \cite[pp. 132--133]{tit1}) to $G_k(s)$ there.
By taking the imaginary part, we obtain
\begin{equation} \label{eq:lit}
\begin{aligned}
2\pi\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ t_k<\gamma^{(k)}\leq T}} (\beta^{(k)}-b) &=
\int_{t_k}^T\log{|G_k(b+it)|}dt - \int_{t_k}^T\log{|G_k(a_k+it)|}dt\\
&\quad\quad\quad\quad\quad- \int_b^{a_k}\arg{G_k(\sigma+it_k)}d\sigma + \int_b^{a_k}\arg{G_k(\sigma+iT)}d\sigma\\
&=: I_1 + I_2 + I_3 + \int_b^{a_k}\arg{G_k(\sigma+iT)}d\sigma
\end{aligned}
\end{equation}
where the sum is counted with multiplicity.
By the same reasoning as in \cite[p. 2246]{aka}, we have
\begin{equation*}
I_2 = O_{a_k}(1),\quad I_3 = O_{a_k,t_k}(1).
\end{equation*}

\vspace{2mm}
Now we only need to estimate $I_1$.
From the functional equation for $\zeta(s)$ (eq. \eqref{eq:fe}), we can deduce that
\begin{align*}
\zeta^{(k)}(s) &= F^{(k)}(s)\zeta(1-s)\left(1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(s)}\frac{\zeta^{(j)}}{\zeta}(1-s)\right)\\
&= F(s)\frac{F^{(k)}}{F}(s)\zeta(1-s)\left(1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(s)}\frac{\zeta^{(j)}}{\zeta}(1-s)\right).
\end{align*}
Hence,
\begin{align*}
I_1 &= \int_{t_k}^T\log{|G_k(b+it)|}dt = \int_{t_k}^T\log{\frac{2^b}{(\log{2})^k}|\zeta^{(k)}(b+it)|}dt\\
&= \int_{t_k}^T\log{\frac{2^b}{(\log{2})^k}}dt +  \int_{t_k}^T\log{|\zeta^{(k)}(b+it)|}dt\\
&= (b\log{2}-k\log{\log{2}})(T-t_k) + \int_{t_k}^T\log{|F(b+it)|}dt + \int_{t_k}^T\log{\left|\frac{F^{(k)}}{F}(b+it)\right|}dt\quad\\
&\quad\quad\quad+ \int_{t_k}^T\log{|\zeta(1-b-it)|}dt\\
&\quad\quad\quad+ \int_{t_k}^T\log{\left|1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(b+it)}\frac{\zeta^{(j)}}{\zeta}(1-b-it)\right|}dt\\
&=: ((b\log{2}-k\log{\log{2}})T + O_{t_k}(1)) + I_{12} + I_{13} + I_{14} + I_{15}. \numberthis \label{eq:i1}
\end{align*}

\vspace{2mm}
As shown in \cite[pp. 2247--2249]{aka},
\begin{align*}
I_{12} = \left(\frac{1}{2}-b\right)\left(T\log{\frac{T}{2\pi}}-T\right) + O_{t_k}(1),
\end{align*}
\begin{equation*}
I_{14} = -\int_{1-b}^{a_k}\arg{\zeta(\sigma+iT)}d\sigma + O_{a_k,t_k}(1).
\end{equation*}

\vspace{2mm}
Below we estimate $I_{13}$ and $I_{15}$.

\vspace{2mm}
We begin with the estimation of $I_{13}$. We consider for $0<\sigma<\frac{1}{2}$ and $t\geq100$.
We first show that
\begin{equation} \label{eq:Fk2}
F^{(k)}(s) = F(s)(f'(s))^k\left(1 + O\left(\frac{e^{-t}}{|\log{s}|^2}\right) + O\left(\frac{1}{|s||\log{s}|^2}\right)\right)
\end{equation}
holds in the region $\sigma<1,\, t\geq100$.
It is obvious that the above error estimate is more precise than that in equation \eqref{eq:Fj}. The proof is similar to the proof of condition 2 of Lemma \ref{lem1}.
We begin by taking the logarithmic branch of $\log{\left(\sin{\frac{\pi s}{2}}\right)}$ as
\begin{equation} \label{eq:logsin}
\log{\left(\sin{\frac{\pi s}{2}}\right)} = -\frac{\pi is}{2}-\log{2}+\frac{\pi i}{2}-\sum_{n=1}^\infty\frac{e^{\pi ins}}{n}
\end{equation}
in the region $0<\sigma<1,\, t\geq2$ and analytically continue it to the region $\sigma<1,\,$ $t\geq2$.
Next, we apply Stirling's formula to $\Gamma(1-s)$ in the region $-\frac{\pi}{2}<\arg{(1-s)}<\frac{\pi}{2}$. Substituting these into $F(s)$, we obtain
\begin{align*}
F(s) = \exp\left(\frac{\pi i}{4} - 1 + \left(\frac{1}{2}-s\right)\log{\frac{(1-s)i}{2\pi}} + s + O(e^{-t}) + O\left(\frac{1}{|s|}\right)\right)
\end{align*}
for $\sigma<1$ and $t\geq100$, where the term $O(e^{-t})$ comes from the term $\sum_{n=1}^\infty\frac{e^{\pi ins}}{n}$ in equation \eqref{eq:logsin} and the term $O\left(\frac{1}{|s|}\right)$ originates from the Stirling's formula.

We now write $f(s):=\left(\frac{1}{2}-s\right)\log{\frac{(1-s)i}{2\pi}}+s+O(e^{-t})+O\left(\frac{1}{|s|}\right)$ and differentiate it with respect to $s$ to obtain
\begin{align*}
f'(s) = -\log{\frac{(1-s)i}{2\pi}} + \frac{1}{2(1-s)} + O(e^{-t}) + O\left(\frac{1}{|s|^2}\right)
\end{align*}
and
\begin{equation*}
f^{(j)}(s) = O(e^{-t}) + O\left(\frac{1}{|s|^{j-1}}\right)
\end{equation*}
for $j\geq2$.
Equation \eqref{eq:Fk2} immediately follows.
As a consequence to equation \eqref{eq:Fk2},
\begin{align*}
&\frac{F^{(k)}}{F}(b+it) = \left(-\log{\frac{t+(1-b)i}{2\pi}} + \frac{1}{2(1-b-it)} + O\left(\frac{1}{t^2}\right)\right)^k \left(1+ O\left(\frac{1}{t(\log{t})^2}\right)\right)\\
&\quad\quad\quad= \left(-\log{\frac{t}{2\pi}} + \frac{t^2-2(1-b)((1-b)^2+t^2)}{2((1-b)^2+t^2)t}i + O\left(\frac{1}{t^2}\right)\right)^k \left(1+ O\left(\frac{1}{t(\log{t})^2}\right)\right).
\end{align*}
This gives us
\begin{align*}
&\log{\frac{F^{(k)}}{F}(b+it)} = k\log{\log{\frac{t}{2\pi}}} + k\log{(-1)} + k\log\Bigg(1-\frac{t^2-2(1-b)((1-b)^2+t^2)}{2((1-b)^2+t^2)t\log{\frac{t}{2\pi}}}i\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+ O\left(\frac{1}{t^2\log{t}}\right)\Bigg) +  O\left(\frac{1}{t(\log{t})^2}\right)\\
&\quad\quad\quad\quad\quad= k\log{\log{\frac{t}{2\pi}}} + k\log{(-1)} - k\frac{t^2-2(1-b)((1-b)^2+t^2)}{2((1-b)^2+t^2)t\log{\frac{t}{2\pi}}}i +  O\left(\frac{1}{t(\log{t})^2}\right).
\end{align*}
Consequently we have
\begin{align*}
\text{Re}\left(\log{\frac{F^{(k)}}{F}(b+it)}\right) = k\log{\log{\frac{t}{2\pi}}} + O\left(\frac{1}{t(\log{t})^2}\right).
\end{align*}
Hence,
\begin{equation*}
\begin{aligned}
I_{13} &= \int_{t_k}^T\log{\left|\frac{F^{(k)}}{F}(b+it)\right|}dt =  \int_{t_k}^T\text{Re}\left(\log{\frac{F^{(k)}}{F}(b+it)}\right)dt\\
&= k\int_{t_k}^T \log{\log{\frac{t}{2\pi}}}dt + O\left(\int_{t_k}^T\frac{dt}{t(\log{t})^2}\right)\\
&= kT\log{\log{\frac{T}{2\pi}}} - 2\pi k\text{Li}\left(\frac{T}{2\pi}\right) + O_{t_k}(1).
\end{aligned}
\end{equation*}

\vspace{2mm}
Finally, we estimate $I_{15}$.
Again from the functional equation for $\zeta(s)$ (eq. \eqref{eq:fe}), we have
\begin{align*}
\zeta^{(k)}(s) &= F^{(k)}(s)\zeta(1-s)\left(1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(s)}\frac{\zeta^{(j)}}{\zeta}(1-s)\right)\\
&= \frac{F^{(k)}}{F}(s)\zeta(s)\left(1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(s)}\frac{\zeta^{(j)}}{\zeta}(1-s)\right)
\end{align*}
which gives us
\begin{equation} \label{eq:fe2}
\frac{1}{\frac{F^{(k)}}{F}(s)}\frac{\zeta^{(k)}}{\zeta}(s) = 1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(s)}\frac{\zeta^{(j)}}{\zeta}(1-s).
\end{equation}

It follows from condition 2 of Lemma \ref{lem1} that the right hand side of equation \eqref{eq:fe2} is holomorphic and has no zeros in the region defined by $\sigma\leq\sigma_k$ and $t\geq2$.
And from conditions 3 and 4 of Lemma \ref{lem1}, the left hand side of equation \eqref{eq:fe2} is holomorphic and has no zeros in the region defined by $\sigma_k\leq\sigma<\frac{1}{2}$ and $t\geq t_k-1$.
Thus, we can determine $\log{\left(1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(s)}\frac{\zeta^{(j)}}{\zeta}(1-s)\right)}$ so that it tends to $0$ as $\sigma\rightarrow-\infty$ which follows from condition 2 of Lemma \ref{lem1}, and is holomorphic in the region $\sigma<\frac{1}{2},\, t>t_k-1$.

Now we consider the trapezoid $C$ with vertices $b+it_k$, $b+iT$, $-T+iT$, and $-t_k+it_k$ (as in \cite[p. 2247]{aka}). Then by Cauchy's integral theorem,
\begin{equation} \label{eq:i15}
\int_C \log{\left(1 -  \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(s)}\frac{\zeta^{(j)}}{\zeta}(1-s)\right)}ds =0.
\end{equation}
By using condition 2 of Lemma \ref{lem1}, we can also show that (cf. \cite[p. 2248]{aka})
\begin{align*}
\left(\int_{\sigma_k+iT}^{-T+iT}+\int_{-T+iT}^{-t_k+it_k}+\int_{-t_k+it_k}^{\sigma_k+it_k}\right)\log{\left(1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(s)}\frac{\zeta^{(j)}}{\zeta}(1-s)\right)}ds = O(1).
\end{align*}
Next we estimate the integral from $\sigma_k+it_k$ to $b+it_k$ trivially and we obtain
\begin{align*}
\int_{\sigma_k+it_k}^{b+it_k} \log{\left(1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(s)}\frac{\zeta^{(j)}}{\zeta}(1-s)\right)} ds = O_{t_k}(1).
\end{align*}

Substituting the above two equations into equation \eqref{eq:i15} and taking the imaginary part, we obtain
\begin{align*}
&I_{15} =  \int_{t_k}^T \log{\left|1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(b+it)}\frac{\zeta^{(j)}}{\zeta}(1-b-it)\right|} dt\\
&= \int_{\sigma_k}^b \arg{\left(1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(\sigma+iT)}\frac{\zeta^{(j)}}{\zeta}(1-\sigma-iT)\right)} d\sigma + O_{t_k}(1)\\
&\overset{\eqref{eq:fe2}}= \int_{\sigma_k}^b \arg{\left(\frac{1}{\frac{F^{(k)}}{F}(\sigma+iT)}\frac{\zeta^{(k)}}{\zeta}(\sigma+iT)\right)}d\sigma + O_{t_k}(1).
\end{align*}

Now we determine the logarithmic branch of $\log{\frac{F^{(k)}}{F}(s)}$ and $\log{\frac{\zeta^{(k)}}{\zeta}(s)}$ in the region $\sigma_k\leq\sigma<\frac{1}{2},\, t\geq t_k-1$ as in conditions 3 and 4, respectively, of Lemma \ref{lem1}.
Note that
\begin{align*}
\log{\left|\frac{1}{\frac{F^{(k)}}{F}(s)}\frac{\zeta^{(k)}}{\zeta}(s)\right|} = -\log{\left|\frac{F^{(k)}}{F}(s)\right|} + \log{\left|\frac{\zeta^{(k)}}{\zeta}(s)\right|}
\end{align*}
holds in the region $\sigma_k\leq\sigma<\frac{1}{2},\, t\geq t_k-1$.
Furthermore, since $\log{\left(\frac{1}{\frac{F^{(k)}}{F}(s)}\frac{\zeta^{(k)}}{\zeta}(s)\right)}$ ($= \log{\left(1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(s)}\frac{\zeta^{(j)}}{\zeta}(1-s)\right)}$), $\log{\frac{F^{(k)}}{F}(s)}$, and $\log{\frac{\zeta^{(k)}}{\zeta}(s)}$ are holomorphic in this region, we know that $\arg{\left(\frac{1}{\frac{F^{(k)}}{F}(s)}\frac{\zeta^{(k)}}{\zeta}(s)\right)}$, $\arg{\frac{F^{(k)}}{F}(s)}$, and $\arg{\frac{\zeta^{(k)}}{\zeta}(s)}$ are continuous there. Since the region $\sigma_k\leq\sigma<\frac{1}{2},\, t\geq t_k-1$ is connected, there exists a constant $n\in\mathbb{Z}$ such that
\begin{equation*}
\arg{\left(\frac{1}{\frac{F^{(k)}}{F}(s)}\frac{\zeta^{(k)}}{\zeta}(s)\right)} = -\arg{\frac{F^{(k)}}{F}(s)} + \arg{\frac{\zeta^{(k)}}{\zeta}(s)} + 2n\pi
\end{equation*}
holds in $\sigma_k\leq\sigma<\frac{1}{2},\, t\geq t_k-1$.

From this choice of logarithmic branch, we have
\begin{equation} \label{eq:arg}
\frac{(3k-\beta_k)}{6}\pi+2n\pi < \arg{\left(\frac{1}{\frac{F^{(k)}}{F}(\sigma+iT)}\frac{\zeta^{(k)}}{\zeta}(\sigma+iT)\right)} < \frac{(9k-\alpha_k)}{6}\pi + 2n\pi
\end{equation}
for $\sigma_k\leq\sigma<\frac{1}{2}$.
Here, $\alpha_k$ and $\beta_k$  are the constants given in Lemma \ref{lem1}, that is,
\begin{align*}
(\alpha_k,\beta_k) =
\begin{cases}
(5,7), &\quad \text{if}\,\, k \,\,\text{is odd},\\
(-1,1), &\quad \text{if}\,\, k \,\,\text{is even}.
\end{cases}
\end{align*}
Since $n$ does not depend on $s$, $n=O_k(1)$. Therefore
\begin{align*}
\arg{\left(\frac{1}{\frac{F^{(k)}}{F}(\sigma+iT)}\frac{\zeta^{(k)}}{\zeta}(\sigma+iT)\right)} = O_k(1).
\end{align*}
From this, we can easily show that
\begin{equation*}
I_{15} = O_k(1),
\end{equation*}
for $\sigma_k$ and $t_k$ are fixed constants that depend only on $k$.

\vspace{2mm}
Inserting the estimates of $I_{12}$, $I_{13}$, $I_{14}$, and $I_{15}$ into equation \eqref{eq:i1}, we obtain
\begin{align*}
I_1 = (b\log{2}-k\log{\log{2}})T + \left(\frac{1}{2}-b\right)\left(T\log{\frac{T}{2\pi}}-T\right) &+ kT\log{\log{\frac{T}{2\pi}}} - 2k\pi\text{Li}\left(\frac{T}{2\pi}\right)\\
&- \int_{1-b}^{a_k}\arg{\zeta(\sigma+iT)}d\sigma + O_k(1),
\end{align*}
since $a_k$ and $t_k$ are fixed constants that depend only on $k$.

\vspace{2mm}
To finalize the proof of Proposition \ref{prop2}, we insert the estimates of $I_1$, $I_2$, and $I_3$ into equation \eqref{eq:lit} to obtain
\begin{align*}
2&\pi\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ 0<\gamma^{(k)}\leq T}} (\beta^{(k)}-b) =
kT\log{\log{\frac{T}{2\pi}}} + (b\log{2}-k\log{\log{2}})T - 2k\pi\text{Li}\left(\frac{T}{2\pi}\right)\\
&\quad+\left(\frac{1}{2}-b\right)\left(T\log{\frac{T}{2\pi}}-T\right) - \int_{1-b}^{a_k}\arg{\zeta(\sigma+iT)}d\sigma + \int_b^{a_k}\arg{G_k(\sigma+iT)}d\sigma + O_k(1).
\end{align*}
Taking the limit $\delta\rightarrow0$, we have $b\rightarrow\frac{1}{2}$, thus
\begin{align*}
\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ 0<\gamma^{(k)}\leq T}} \left(\beta^{(k)}-\frac{1}{2}\right) &= \frac{kT}{2\pi}\log{\log{\frac{T}{2\pi}}} + \frac{1}{2\pi}\left(\frac{1}{2}\log{2}-k\log{\log{2}}\right)T - k\text{Li}\left(\frac{T}{2\pi}\right)\\
&\quad\quad\quad+ \frac{1}{2\pi}\int_{\frac{1}{2}}^{a_k}(-\arg{\zeta(\sigma+iT)}+\arg{G_k(\sigma+iT)})d\sigma + O_k(1).
\end{align*}
\end{proof}

\begin{rmk}
The proof of Proposition \ref{prop2} (and thus of Proposition 2.2 of \cite{aka}) actually more or less follows the proof of Theorem 10 given in \cite[Section 3]{lev}.
One obvious difference is that we did not estimate the fourth integral in equation \eqref{eq:lit} while Levinson and Montgomery estimated the corresponding integral (the fourth integral in equation (3.1) of \cite[Section 3]{lev}) as $O(\log{T})$. As what Akatsuka \cite{aka} did, that term contributes to the integral appearing in Proposition \ref{prop2} which will be estimated in the following few lemmas. This integral will contribute to the error term to Theorem \ref{cha1} and in the proofs of the following lemmas, we shall use the assumption of RH to reduce the upper bound of this integral.
\end{rmk}

\begin{rmk}
In contrast to the proof of Theorem 10 of \cite{lev}, in this paper (and in \cite{aka} as well), we describe some important estimates, such as those on $G_k(s)$, $\frac{F^{(k)}}{F}(s)$, and $\frac{\zeta^{(k)}}{\zeta}(s)$, which are related to the existence of fixed constants $a_k$, $\sigma_k$, and $t_k$ in Lemma \ref{lem1} for the sake of clarity. Furthermore, we also explicitly state Proposition \ref{prop2} since it clearly points out the main terms of Theorem \ref{cha1} and thus this gives the readers clear information of the term that in the current research contributes to the error term which is to be possibly improved in future research.
\end{rmk}

To complete the proof of Theorem \ref{cha1}, we need to estimate
\begin{align*}
\int_{\frac{1}{2}}^{a_k}(-\arg{\zeta(\sigma+iT)}+\arg{G_k(\sigma+iT)})d\sigma
\end{align*}
in Proposition \ref{prop2}.
For that purpose, similar to the method taken in \cite{aka}, below we give two bounds for $-\arg{\zeta(\sigma+iT)}+\arg{G_k(\sigma+iT)}$.
We write
\begin{align*}
-\arg{\zeta(\sigma+iT)}+\arg{G_k(\sigma+iT)} = \arg{\frac{G_k}{\zeta}(\sigma+iT)}
\end{align*}
where the argument on the right hand side is taken so that
$\log{\frac{G_k}{\zeta}(s)}$ tends to $0$ as $\sigma\rightarrow\infty$ and is holomorphic in $\mathbb{C}\backslash\{z+\lambda \,|\, (\zeta^{(k)}/\zeta)(z)=0 \,\,\text{or}\,\, \infty, \, \lambda\leq0\}$.

\vspace{3mm}
\begin{lem} \label{lem3}
Assume RH and let $T\geq t_k$.
Then for any $\epsilon_0>0$ satisfying $\epsilon_0<\frac{1}{2\log{T}}$
(since $T\geq t_k\geq100$, $\epsilon_0<\frac{1}{8}$),
we have for $\frac{1}{2}+\epsilon_0<\sigma\leq a_k$,
\begin{align*}
\arg{\frac{G_k}{\zeta}(\sigma+iT)} = O_{a_k,t_k} \left(\frac{\log{\frac{\log{T}}{\epsilon_0}}}{\sigma-\frac{1}{2}-\epsilon_0}\right).
\end{align*}
\end{lem}

\begin{proof}
To begin with, we note that $\frac{G_k}{\zeta}(s)$ is uniformly convergent to $1$ as $\sigma\rightarrow\infty$ for $t\in\mathbb{R}$, so we can take a number $c_k\in\mathbb{R}$ satisfying $a_k+1\leq c_k\leq\frac{t_k}{2}$ and $\frac{1}{2}\leq\text{Re}\left(\frac{G_k}{\zeta}(s)\right)\leq\frac{3}{2}\,\, (\text{when}\,\, \sigma\geq c_k)$. In fact, we can check that taking $c_k=10+k^2$ is enough.

\vspace{2mm}
The proof also proceeds similarly to the proof of Lemma 2.3 of \cite{aka}.
We let $\sigma\in(1/2+\epsilon_0,a_k]$ and let
$q_{G_k/\zeta}=q_{G_k/\zeta}(\sigma,T)$ denote the number of times $\text{Re}\left(\frac{G_k}{\zeta}(u+iT)\right)$ vanishes in $u\in[\sigma,c_k]$.
Then, $\left|\arg{\frac{G_k}{\zeta}(\sigma+iT)}\right|\leq\left(q_{G_k/\zeta}+1\right)\pi$.

\vspace{2mm}
Now we estimate $q_{G_k/\zeta}$.
For that purpose, we set
\begin{align*}
H_k(z)=H_{k_T}(z):=\frac{\frac{G_k}{\zeta}(z+iT)+\frac{G_k}{\zeta}(z-iT)}{2} \quad (z\in\mathbb{C})
\end{align*}
and $n_{H_k}(r):=\sharp\{z\in\mathbb{C} \,|\, H_k(z)=0, |z-c_k|\leq r\}$.
Then, we have $q_{G_k/\zeta}\leq n_{H_k}(c_k-\sigma)$ for $\frac{1}{2}+\epsilon_0<\sigma\leq a_k$.
For each $\sigma\in(1/2+\epsilon_0,a_k]$, we take $\epsilon=\epsilon_{\sigma,T}$ satisfying $0<\epsilon<\sigma-\frac{1}{2}-\epsilon_0$, then $H_k(z)$ is holomorphic in the region $\{z\in\mathbb{C} \,|\, |z-c_k| \leq c_k-\sigma+\epsilon\}$.
As in \cite[p. 2250]{aka}, by using Jensen's theorem (cf. \cite[pp. 125--126]{tit1}), we can show that
\begin{align*}
n_{H_k}(c_k-\sigma) &\leq \frac{1}{C_1\epsilon}\int_0^{c_k-\sigma+\epsilon}\frac{n_{H_k}(r)}{r}dr\\
&= \frac{1}{C_1\epsilon}\frac{1}{2\pi}\int_0^{2\pi}\log{|H_k(c_k+(c_k-\sigma+\epsilon)e^{i\theta})|}d\theta - \frac{1}{C_1\epsilon}\log{|H_k(c_k)|}
\end{align*}
for some constant $C_1>0$,
which by our choice of $c_k$ gives us
\begin{equation} \label{eq:nhk}
n_{H_k}(c_k-\sigma) \leq \frac{1}{C_1\epsilon}\frac{1}{2\pi}\int_0^{2\pi}\log{|H_k(c_k+(c_k-\sigma+\epsilon)e^{i\theta})|}d\theta + \frac{1}{\epsilon}O_{a_k,t_k}(1).
\end{equation}

\vspace{2mm}
Finally we estimate $\frac{1}{2\pi}\int_0^{2\pi}\log{|H_k(c_k+(c_k-\sigma+\epsilon)e^{i\theta})|}d\theta$.
From \cite[Theorems 9.2 and 9.6(A)]{tit2} (similar to what stated in \cite[p. 2250]{aka}),
\begin{align*}
\frac{\zeta'}{\zeta}(\sigma\pm it) = O\left(\frac{\log{T}}{\sigma-\frac{1}{2}}\right)
\end{align*}
holds for $\frac{1}{2}<\sigma\leq 2c_k$ and $\frac{T}{2}\leq t\leq 2T$.
Thus, for $\frac{1}{2}+\epsilon_0<\sigma\leq 2c_k$ and $\frac{T}{2}\leq t\leq 2T$, we have
\begin{equation} \label{eq:k1}
\frac{\zeta'}{\zeta}(\sigma\pm it) = O\left(\frac{\log{T}}{\epsilon_0}\right).
\end{equation}

\vspace{2mm}
With this estimate, we show that
\begin{align*}
\frac{\zeta^{(k)}}{\zeta}(s) = O\left(\frac{(\log{T})^k}{\epsilon_0^k}\right)
\end{align*}
holds for $\frac{1}{2}+\epsilon_0<\sigma<2c_k$ and $\frac{T}{2}\leq |t|\leq 2T$.
We use induction on $k$ in the equation.
For $k=1$, $\frac{\zeta'}{\zeta}(\sigma\pm it) = O\left(\frac{\log{T}}{\epsilon_0}\right)$ follows from equation \eqref{eq:k1}.
Suppose that $\frac{\zeta^{(n)}}{\zeta}(s) = O\left(\frac{(\log{T})^n}{\epsilon_0^n}\right)$ hold in the region $\frac{1}{2}+\epsilon_0<\sigma<2c_k,\, \frac{T}{2}\leq |t|\leq 2T$ for a positive integer $n$, then
\begin{equation} \label{eq:prodr}
\left(\frac{\zeta^{(n)}}{\zeta}(s)\right)' =  \frac{1}{2\pi i}\int_{|z-s|=\epsilon_0} \frac{\frac{\zeta^{(n)}}{\zeta}(z)}{(z-s)^2}dz = O\left(\frac{(\log{T})^n}{\epsilon_0^{n+1}}\right).
\end{equation}
Meanwhile,
\begin{align*}
\left(\frac{\zeta^{(n)}}{\zeta}(s)\right)' = \frac{\zeta^{(n+1)}}{\zeta}(s) - \frac{\zeta^{(n)}}{\zeta}(s)\frac{\zeta'}{\zeta}(s)
\end{align*}
holds in the region.

Therefore, by equation \eqref{eq:prodr} and by the induction hypothesis,
\begin{align*}
\frac{\zeta^{(n+1)}}{\zeta}(s) &= \left(\frac{\zeta^{(n)}}{\zeta}(s)\right)' + \frac{\zeta^{(n)}}{\zeta}(s)\frac{\zeta'}{\zeta}(s) = O\left(\frac{(\log{T})^n}{\epsilon_0^{n+1}}\right) + O\left(\frac{(\log{T})^{n+1}}{\epsilon_0^{n+1}}\right)\\
&= O\left(\frac{(\log{T})^{n+1}}{\epsilon_0^{n+1}}\right)
\end{align*}
holds for $\frac{1}{2}<\sigma\leq 2c_k$ and $\frac{T}{2}\leq |t|\leq 2T$.
Hence, by induction, we find that
\begin{align*}
\frac{\zeta^{(k)}}{\zeta}(s) = O\left(\frac{(\log{T})^k}{\epsilon_0^k}\right)
\end{align*}
holds in the region defined by $\frac{1}{2}+\epsilon_0<\sigma<2c_k$ and $\frac{T}{2}\leq |t|\leq 2T$.
This immediately gives us
\begin{align*}
|H_k(c_k+(c_k-\sigma+\epsilon)e^{i\theta})| \ll_{a_k,t_k} \frac{(\log{T})^k}{\epsilon_0^k},
\end{align*}
and so
\begin{equation*}
|H_k(c_k+(c_k-\sigma+\epsilon)e^{i\theta})| \leq C_2(a_k,t_k) \frac{(\log{T})^k}{\epsilon_0^k}
\end{equation*}
for some constant $C_2>0$ which depends only on $a_k$ and $t_k$.
Thus,
\begin{align*}
\frac{1}{2\pi}\int_0^{2\pi}\log{|H_k(c_k+(c_k-\sigma+\epsilon)e^{i\theta})|}d\theta \leq \log{C_2(a_k,t_k)} + k\log{\frac{\log{T}}{\epsilon_0}} \ll_{a_k,t_k} \log{\frac{\log{T}}{\epsilon_0}}.
\end{align*}
Applying this to inequality \eqref{eq:nhk}, we obtain
\begin{align*}
n_{H_k}(c_k-\sigma) = \frac{1}{\epsilon}O_{a_k,t_k}\left(\log{\frac{\log{T}}{\epsilon_0}}\right)
\end{align*}
which implies
\begin{equation*}
\arg{\frac{G_k}{\zeta}(\sigma+iT)} = \frac{1}{\epsilon}O_{a_k,t_k}\left(\log{\frac{\log{T}}{\epsilon_0}}\right).
\end{equation*}
Taking $\epsilon=\frac{1}{2}\left(\sigma-\frac{1}{2}-\epsilon_0\right) \quad \left(<\sigma-\frac{1}{2}-\epsilon_0\right)$, we obtain
\begin{align*}
\arg{\frac{G_k}{\zeta}(\sigma+iT)} = O_{a_k,t_k}\left(\frac{\log{\frac{\log{T}}{\epsilon_0}}}{\sigma-\frac{1}{2}-\epsilon_0}\right).
\end{align*}
\end{proof}

\begin{lem} \label{lem4}
Assume RH and let $A\geq2$ be fixed. Then there exists a constant $C_0>0$ such that
\begin{equation*}
\left|\zeta^{(k)}(\sigma+it)\right|\leq\exp{\left( C_0\left(\frac{(\log T)^{2(1-\sigma)}}{\log\log{T}}+(\log{T})^{1/10}\right)\right)}
\end{equation*}
holds for $T\geq t_k$, $\frac{T}{2}\leq t\leq 2T$, $\frac{1}{2}-\frac{1}{\log{\log{T}}}\leq\sigma\leq A$.
\end{lem}

\begin{proof}
Referring to \cite[(14.14.2), (14.14.5) and the first equation on p. 384]{tit2} (cf. \cite[pp. 2251--2252]{aka}), we know that
\begin{equation} \label{eq:lem4zen}
|\zeta(\sigma+it)| \leq \exp{\left(C_3\left(\frac{(\log{T})^{2(1-\sigma)}}{\log{\log{T}}}\right)+(\log{T})^{1/10}\right)}
\end{equation}
holds for  $\frac{1}{2}-\frac{2}{\log{\log{T}}}\leq\sigma\leq A+1$, $\frac{T}{3}\leq t\leq 3T$ for some constant $C_3>0$.

Applying Cauchy's integral formula, we see that
\begin{align*}
\zeta^{(k)}(s) = \frac{k!}{2\pi i} \int_{|z-s|=\epsilon} \frac{\zeta(z)}{(z-s)^{k+1}}dz \quad\quad \text{for} \quad 0<\epsilon<\frac{1}{2}
\end{align*}
holds in the region defined by  $\frac{1}{2}-\frac{1}{\log{\log{T}}}\leq\sigma\leq A$ and $\frac{T}{2}\leq t\leq 2T$.
Applying inequality \eqref{eq:lem4zen} and by taking $\epsilon=\frac{1}{2(\log{\log{T}})^{1/k}} \,\, (<\frac{1}{2})$, we obtain Lemma \ref{lem4}.\\
\end{proof}

\begin{lem} \label{lem5}
Assume RH and let $T\geq t_k$.
Then for any $\frac{1}{2}\leq\sigma\leq\frac{3}{4}$, we have
\begin{align*}
\arg{G_k(\sigma+iT)} = O_{a_k}\left(\frac{(\log T)^{2(1-\sigma)}}{(\log{\log{T}})^{1/2}}\right).
\end{align*}
\end{lem}

\begin{proof}
The proof proceeds in the same way as the proof of Lemma 2.4 of \cite{aka}.
Refer to \cite[pp. 2252--2253]{aka} for the detailed proof and use Lemma \ref{lem4} above in place of Lemma 2.6 of \cite{aka}.\\
\end{proof}

\begin{rmk}
The restrictions of the lower bound of $T$ we gave in Lemmas \ref{lem3}, \ref{lem4}, and \ref{lem5} are not essential, but they are sufficient for our needs.
We may let $T$ be any positive number in Lemmas \ref{lem3}, \ref{lem4}, and \ref{lem5}, however in that case, we need to modify some calculations in the proofs. Thus we used these restrictions for our convenience.\\
\end{rmk}

\vspace{3mm}
\noindent
\underline{\textbf{Proof of Theorem \ref{cha1}:}}\\

First of all, we consider for $T\geq t_k$ which satisfies $\zeta^{(k)}(\sigma+iT)\neq0$ and $\zeta(\sigma+iT)\neq0$ for any $\sigma\in\mathbb{R}$. By Lemma \ref{lem3}, we have
\begin{align*}
\int_{\frac{1}{2}+2\epsilon_0}^{a_k} \arg{\frac{G_k}{\zeta}(\sigma+iT)}d\sigma \ll_{a_k,t_k} \int_{\frac{1}{2}+2\epsilon_0}^{a_k} \frac{\log{\frac{\log{T}}{\epsilon_0}}}{\sigma-\frac{1}{2}-\epsilon_0}d\sigma
\ll_{a_k} \log{\frac{\log{T}}{\epsilon_0}}\log{\frac{1}{\epsilon_0}}.
\end{align*}

\vspace{2mm}
Next, by Lemma \ref{lem5},
\begin{align*}
\arg{G_k(\sigma+iT)} = O_{a_k}\left(\frac{(\log T)^{2(1-\sigma)}}{(\log{\log{T}})^{1/2}}\right) \quad\quad \text{for} \quad \frac{1}{2}\leq\sigma\leq\frac{3}{4}
\end{align*}
and from equation (2.23) of \cite[p. 2251]{aka} (cf. \cite[equations (14.14.3) and (14.14.5)]{tit2}),
RH implies that
\begin{equation*}
\arg{\zeta(\sigma+iT)} = O\left(\frac{(\log T)^{2(1-\sigma)}}{\log{\log{T}}}\right)
\end{equation*}
holds uniformly for $\frac{1}{2}\leq\sigma\leq\frac{3}{4}$.
Thus,
\begin{align*}
\int_{\frac{1}{2}}^{\frac{1}{2}+2\epsilon_0} \arg{\frac{G_k}{\zeta}(\sigma+iT)}d\sigma \ll_{a_k} \frac{\log T}{(\log{\log{T}})^{1/2}}\epsilon_0.
\end{align*}

\vspace{2mm}
Now we take $\epsilon_0 = \frac{1}{4\log{T}} \,\, \left(<\frac{1}{2}\frac{1}{\log{T}}\right)$, then we have
\begin{align*}
\int_{\frac{1}{2}}^{a_k} \arg{\frac{G_k}{\zeta}(\sigma+iT)}d\sigma \ll_{a_k,t_k} (\log{\log{T}})^2.
\end{align*}
Applying this to Proposition \ref{prop2} and noting that $a_k$ and $t_k$ are fixed constants that depend only on $k$, we have
\begin{equation} \label{eq:kari1}
\begin{aligned}
\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ 0<\gamma^{(k)}\leq T}} \left(\beta^{(k)}-\frac{1}{2}\right) = \,\frac{kT}{2\pi}\log{\log{\frac{T}{2\pi}}} &+ \frac{1}{2\pi}\left(\frac{1}{2}\log{2} - k\log{\log{2}}\right)T - k\textnormal{Li}\left(\frac{T}{2\pi}\right)\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad+ O_k((\log{\log{T}})^2).
\end{aligned}
\end{equation}

\vspace{3mm}
Secondly, for $4\pi<T<t_k$, we are adding some finite number of terms which depend on $t_k$, and thus depend only on $k$ so this can be included in the error term.\\

Thirdly, for $T\geq t_k$ such that $\zeta^{(k)}(\sigma+iT)=0$ or $\zeta(\sigma+iT)=0$ for some $\sigma\in\mathbb{R}$, there is some increment in the value of $\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ 0<\gamma^{(k)}\leq T}} \left(\beta^{(k)}-\frac{1}{2}\right)$ as much as $\displaystyle{\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ \gamma^{(k)} = T}} \left(\beta^{(k)}-\frac{1}{2}\right)}$. Now we estimate this and we show that this can be included in the error term of equation \eqref{eq:kari1}.
We start by taking a small $0<\epsilon<1$ such that $\zeta^{(k)}(\sigma+i(T\pm\epsilon))\neq0$ and $\zeta(\sigma+i(T\pm\epsilon))\neq0$ for any $\sigma\in\mathbb{R}$.
According to equation \eqref{eq:kari1},
\begin{align*}
\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ 0<\gamma^{(k)}\leq T+\epsilon}} \left(\beta^{(k)}-\frac{1}{2}\right) = \frac{k(T+\epsilon)}{2\pi}\log{\log{\frac{T+\epsilon}{2\pi}}} &+ \frac{1}{2\pi}\left(\frac{1}{2}\log{2} - k\log{\log{2}}\right)(T+\epsilon)\\
&- k\textnormal{Li}\left(\frac{T+\epsilon}{2\pi}\right) + O_k((\log{\log{T}})^2),
\end{align*}
\begin{align*}
\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ 0<\gamma^{(k)}\leq T-\epsilon}} \left(\beta^{(k)}-\frac{1}{2}\right) = \frac{k(T-\epsilon)}{2\pi}\log{\log{\frac{T-\epsilon}{2\pi}}} &+ \frac{1}{2\pi}\left(\frac{1}{2}\log{2} - k\log{\log{2}}\right)(T-\epsilon)\\
&- k\textnormal{Li}\left(\frac{T-\epsilon}{2\pi}\right) + O_k((\log{\log{T}})^2).
\end{align*}
Thus,
\begin{align*}
\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ T-\epsilon<\gamma^{(k)}\leq T+\epsilon}} \left(\beta^{(k)}-\frac{1}{2}\right) &= \frac{k(T+\epsilon)}{2\pi}\log{\log{\frac{T+\epsilon}{2\pi}}} - \frac{k(T-\epsilon)}{2\pi}\log{\log{\frac{T-\epsilon}{2\pi}}}\\
&\quad+ \frac{\epsilon}{\pi}\left(\frac{1}{2}\log{2} - k\log{\log{2}}\right) - k\left(\textnormal{Li}\left(\frac{T+\epsilon}{2\pi}\right) - \textnormal{Li}\left(\frac{T-\epsilon}{2\pi}\right)\right)\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+ O_k((\log{\log{T}})^2)\\
&= \frac{k\epsilon}{\pi}\log{\log{\frac{T}{2\pi}}} + \frac{k\epsilon}{\pi\log{\frac{T}{2\pi}}} + \frac{\epsilon}{\pi}\left(\frac{1}{2}\log{2} - k\log{\log{2}}\right) - \frac{k\epsilon}{\pi\log{\frac{T}{2\pi}}}\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+ O\left(\frac{\epsilon^2}{T\log{T}}\right) + O_k((\log{\log{T}})^2)\\
&= \frac{k\epsilon}{\pi}\log{\log{\frac{T}{2\pi}}} + \frac{\epsilon}{\pi}\left(\frac{1}{2}\log{2} - k\log{\log{2}}\right) + O_k((\log{\log{T}})^2).
\end{align*}
This gives us
\begin{equation*}
\begin{aligned}
\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ T-\epsilon<\gamma^{(k)}\leq T+\epsilon}} \left(\beta^{(k)}-\frac{1}{2}\right) = O_k((\log{\log{T}})^2)
\end{aligned}
\end{equation*}
which implies
\begin{equation*}
\begin{aligned}
\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ \gamma^{(k)} = T}} \left(\beta^{(k)}-\frac{1}{2}\right) = O_k((\log{\log{T}})^2).
\end{aligned}
\end{equation*}
Therefore, this increment can also be included in the error term.\\

Hence,
\begin{align*}
\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ 0<\gamma^{(k)}\leq T}} \left(\beta^{(k)}-\frac{1}{2}\right) = \,\frac{kT}{2\pi}\log{\log{\frac{T}{2\pi}}} + \frac{1}{2\pi}\left(\frac{1}{2}\log{2} - k\log{\log{2}}\right)T &- k\textnormal{Li}\left(\frac{T}{2\pi}\right)\\
&+ O_k((\log{\log{T}})^2)
\end{align*}
holds for any $T>4\pi$.\\
\qed

\vspace{3mm}
\noindent
\underline{\textbf{Proof of Corollary \ref{cha12}:}}\\

This is an immediate consequence of Theorem \ref{cha1}.
For the proof, refer to \cite[p. 58 (the ending part of Section 3)]{lev}.\\
\qed

\section{Proof of Theorem \ref{cha2}}
\label{sec:3}

In this section we give the proof of Theorem \ref{cha2}.
We first show the following proposition.

\begin{prop} \label{prop6}
Assume RH. Take $t_k$ which satisfies all conditions of Lemma \ref{lem1}. Then for $T\geq2$ which satisfies $\zeta(\sigma+iT)\neq0$ and $\zeta^{(k)}(\sigma+iT)\neq0$ for all $\sigma\in\mathbb{R}$, we have
\begin{align*}
N_k(T) = \frac{T}{2\pi}\log{\frac{T}{4\pi}}-\frac{T}{2\pi}+\frac{1}{2\pi}\arg{G_k\left(\frac{1}{2}+iT\right)}+\frac{1}{2\pi}\arg{\zeta\left(\frac{1}{2}+iT\right)}+O_k(1)
\end{align*}
where the arguments are taken as in Proposition \ref{prop2}.
\end{prop}

\begin{proof}
The steps of the proof also follow those of the proof of Proposition 3.1 of \cite{aka}.
We take $a_k$, $\sigma_k$, $t_k$, $T$, $\delta$, and $b$ as in the beginning of the proof of Proposition \ref{prop2}. We let $b' := \frac{1}{2}-\frac{\delta}{2}$. Replacing $b$ by $b'$ in equation \eqref{eq:lit}, we have
\begin{align*}
2\pi\sum_{\substack{\rho^{(k)}=\beta^{(k)}+i\gamma^{(k)},\\ 0<\gamma^{(k)}\leq T}} (\beta^{(k)}-b') &=
\int_{t_k}^T\log{|G_k(b'+it)|}dt - \int_{t_k}^T\log{|G_k(a_k+it)|}dt\\
&\quad\quad\quad\quad\quad\quad- \int_{b'}^{a_k}\arg{G_k(\sigma+it_k)}d\sigma + \int_{b'}^{a_k}\arg{G_k(\sigma+iT)}d\sigma.
\end{align*}

Subtracting this from equation \eqref{eq:lit}, we have
\begin{equation} \label{eq:lit'}
\begin{aligned}
\pi\delta (N_k(T) - N_k(t_k)) &=
\int_{t_k}^T\log{|G_k(b+it)|}dt - \int_{t_k}^T\log{|G_k(b'+it)|}dt\\
&\quad\quad\quad\quad\quad\quad\quad- \int_b^{b'}\arg{G_k(\sigma+it_k)}d\sigma + \int_b^{b'}\arg{G_k(\sigma+iT)}d\sigma\\
&=: J_1 + J_2 + J_3 + \int_b^{b'}\arg{G_k(\sigma+iT)}d\sigma.
\end{aligned}
\end{equation}

\vspace{2mm}
Referring to the estimate of $I_3$ in the proof of Proposition \ref{prop2} (cf. \cite[p. 2246]{aka}), we can easily show that
\begin{equation*}
J_3 = O_{t_k}(\delta).
\end{equation*}

\vspace{2mm}
Now we estimate $J_1 + J_2$.
From equation \eqref{eq:i1}, we have
\begin{align*}
J_1+J_2 &= \int_{t_k}^T\log{|G_k(b+it)|}dt - \int_{t_k}^T\log{|G_k(b'+it)|}dt\\
&= ((b-b')\log{2})(T-t_k) + \int_{t_k}^T (\log{|F(b+it)|}-\log{|F(b'+it)|})dt\\
&\quad\quad+ \int_{t_k}^T \left(\log{\left|\frac{F^{(k)}}{F}(b+it)\right|}-\log{\left|\frac{F^{(k)}}{F}(b'+it)\right|}\right)dt\\
&\quad\quad+ \int_{t_k}^T (\log{|\zeta(1-b-it)|}-\log{|\zeta(1-b'-it)|})dt\\
&\quad\quad+ \int_{t_k}^T \Bigg(\log{\left|1 -  \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(b+it)}\frac{\zeta^{(j)}}{\zeta}(1-b-it)\right|}\\
&\quad\quad\quad\quad\quad\quad\quad- \log{\left|1 -  \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(b'+it)}\frac{\zeta^{(j)}}{\zeta}(1-b'-it)\right|}\Bigg)dt\\
&=: \left(\left(-\frac{\delta}{2}\log{2}\right)T + O_{t_k}(\delta)\right) + J_{12} + J_{13} + J_{14} + J_{15}.
\end{align*}

\vspace{2mm}
Referring to \cite[pp. 2255--2256]{aka}, we have
\begin{align*}
J_{12} = \frac{\delta}{2}\left(T\log{\frac{T}{2\pi}}-T\right) + O_{t_k}(\delta),
\end{align*}
\begin{equation*}
J_{14} = \int_{1-b'}^{1-b}\arg{\zeta(\sigma+iT)}d\sigma + O_{t_k}(\delta).
\end{equation*}

\vspace{2mm}
We only need to estimate $J_{13}$ and $J_{15}$.
We begin with the estimation of $J_{13}$. We determine the logarithmic branch of $\log{\frac{F^{(k)}}{F}(s)}$ for $0<\sigma<\frac{1}{2}$ and $t>t_k-1$ as in condition 3 of Lemma \ref{lem1}. We then have $\arg{\frac{F^{(k)}}{F}(s)}\in(\alpha_k\pi/6,\beta_k\pi/6)$, where the pair $(\alpha_k,\beta_k)$ is defined as in Lemma \ref{lem1}.

As in \cite[p. 2255]{aka}, we apply Cauchy's integral theorem to $\log{\frac{F^{(k)}}{F}(s)}$ on the rectangle with vertices $b+it_k$, $b'+it_k$, $b'+iT$, and $b+iT$ and take the imaginary part, then we obtain
\begin{align*}
J_{13} =  \int_b^{b'} \arg{\frac{F^{(k)}}{F}(\sigma+it_k)}d\sigma - \int_b^{b'} \arg{\frac{F^{(k)}}{F}(\sigma+iT)}d\sigma = O_k(\delta).
\end{align*}

\vspace{2mm}
Finally, we estimate $J_{15}$.
We determine the logarithmic branch of
\begin{align*}
\log{\left(1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(s)}\frac{\zeta^{(j)}}{\zeta}(1-s)\right)}
\end{align*}
in the same manner as that in the estimation of $I_{15}$ in the proof of Proposition \ref{prop2}, then it is holomorphic in the region $0<\sigma<\frac{1}{2},\, t>t_k-1$.
Applying Cauchy's integral theorem to it on the path taken for estimating $J_{13}$, we have
\begin{align*}
J_{15} = &\int_b^{b'} \arg{\left(1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(\sigma+it_k)}\frac{\zeta^{(j)}}{\zeta}(1-\sigma-it_k)\right)}\\
&\quad\quad\quad- \int_b^{b'} \arg{\left(1 - \sum_{j=1}^k {{k}\choose{j}} (-1)^{j-1} \frac{1}{\frac{F^{(k)}}{F^{(k-j)}}(\sigma+iT)}\frac{\zeta^{(j)}}{\zeta}(1-\sigma-iT)\right)}d\sigma.
\end{align*}
Again using equation \eqref{eq:fe2},
\begin{align*}
J_{15} = \int_b^{b'} \arg{\left(\frac{1}{\frac{F^{(k)}}{F}(\sigma+it_k)}\frac{\zeta^{(k)}}{\zeta}(\sigma+it_k)\right)}d\sigma - \int_b^{b'} \arg{\left(\frac{1}{\frac{F^{(k)}}{F}(\sigma+iT)}\frac{\zeta^{(k)}}{\zeta}(\sigma+iT)\right)}d\sigma.
\end{align*}
Applying inequalities \eqref{eq:arg}, we obtain
\begin{align*}
J_{15} = O_k(\delta).
\end{align*}
Hence, since $t_k$ is a fixed constant that depends only on $k$,
\begin{align*}
J_1+J_2 = \frac{\delta}{2}\left(T\log{\frac{T}{4\pi}}-T\right) + \int_{1-b'}^{1-b} \arg{\zeta(\sigma+iT)}d\sigma + O_k(\delta).
\end{align*}

\vspace{3mm}
Inserting the estimates of $J_1+J_2$ and $J_3$ into equation \eqref{eq:lit'}, we have
\begin{equation} \label{eq:n2t}
N_k(T) = \frac{T}{2\pi}\log{\frac{T}{4\pi}} - \frac{T}{2\pi} + \frac{1}{\pi\delta}\left(\int_{1-b'}^{1-b} \arg{\zeta(\sigma+iT)}d\sigma + \int_b^{b'} \arg{G_k(\sigma+iT)}d\sigma \right)+ O_k(1).
\end{equation}

\vspace{3mm}
Taking the limit $\delta\rightarrow0$ and applying the mean value theorem,
\begin{align*}
\lim_{\delta\rightarrow0} \frac{1}{\pi\delta}\int_{1-b'}^{1-b} \arg{\zeta(\sigma+iT)}d\sigma = \frac{1}{2\pi}\arg{\zeta\left(\frac{1}{2}+iT\right)}
\end{align*}
by noting that $b=\frac{1}{2}-\delta$ and $b'=\frac{1}{2}-\frac{\delta}{2}$.
And similarly,
\begin{align*}
\lim_{\delta\rightarrow0} \frac{1}{\pi\delta}\int_b^{b'} \arg{G_k(\sigma+iT)}d\sigma = \frac{1}{2\pi}\arg{G_k\left(\frac{1}{2}+iT\right)}.
\end{align*}
Substituting these into equation \eqref{eq:n2t}, we have
\begin{align*}
N_k(T) = \frac{T}{2\pi}\log{\frac{T}{4\pi}} - \frac{T}{2\pi} + \frac{1}{2\pi}\arg{G_k\left(\frac{1}{2}+iT\right)} + \frac{1}{2\pi}\arg{\zeta\left(\frac{1}{2}+iT\right)} + O_k(1).
\end{align*}

\vspace{3mm}
If $2\leq T<t_k$, then $N_k(T)\leq N_k(t_k) = O_{t_k}(1) = O_k(1)$.
Hence the above equation holds for any $T\geq2$ which satisfies the conditions of Proposition \ref{prop6}.\\
\end{proof}

\vspace{3mm}
\noindent
\underline{\textbf{Proof of Theorem \ref{cha2}:}}\\

Firstly we consider for $T\geq2$ which satisfies $\zeta^{(k)}(\sigma+iT)\neq0$ and $\zeta(\sigma+iT)\neq0$ for any $\sigma\in\mathbb{R}$.
By Lemma \ref{lem5},
\begin{align*}
\arg{G_k\left(\frac{1}{2}+iT\right)} = O_{a_k}\left(\frac{\log T}{(\log{\log{T}})^{1/2}}\right)
\end{align*}
and again from equation (2.23) of \cite[p. 2251]{aka}, we have
\begin{align*}
\arg{\zeta\left(\frac{1}{2}+iT\right)} = O\left(\frac{\log T}{\log{\log{T}}}\right).
\end{align*}
Substituting these into Proposition \ref{prop6}, we obtain
\begin{align*}
N_k(T) = \frac{T}{2\pi}\log{\frac{T}{4\pi}} - \frac{T}{2\pi} + O_k\left(\frac{\log{T}}{(\log{\log{T}})^{1/2}}\right).
\end{align*}

Next, if $\zeta(\sigma+iT)=0$ or $\zeta^{(k)}(\sigma+iT)=0$ for some $\sigma\in\mathbb{R} \quad (T\geq2)$, then we again take a small $0<\epsilon<1$ such that $\zeta^{(k)}(\sigma+i(T\pm\epsilon))\neq0$ and $\zeta(\sigma+i(T\pm\epsilon))\neq0$ for any $\sigma\in\mathbb{R}$ as in the proof of Theorem \ref{cha1}.
Then similarly, we can show that the increment of the value of $N_k(T)$ can be included in the error term of the above equation.

Therefore
\begin{equation*}
N_k(T) = \frac{T}{2\pi}\log{\frac{T}{4\pi}} - \frac{T}{2\pi} + O_k\left(\frac{\log{T}}{(\log{\log{T}})^{1/2}}\right)
\end{equation*}
holds for any $T\geq2$.\\
\qed

\begin{rmk}
It is well-known that in the case of the Riemann zeta function $\zeta(s)$ itself, the number $N(T)$ of zeros of $\zeta(s)$ is estimated as
\begin{align*}
N(T) = \frac{T}{2\pi}\log{\frac{T}{2\pi}} - \frac{T}{2\pi} + S(T) + O\left(\frac{1}{T}\right),
\end{align*}
where $S(T) = \frac{1}{\pi}\arg{\zeta\left(\frac{1}{2}+iT\right)}$ with a standard branch (cf. \cite[Theorem 9.3]{tit2}).
Thus, the function $S(T)$ determines the error term in the estimate of $N(T)$.
Under RH, we have
\begin{align} \label{eq:st}
S(T) = O\left(\frac{\log{T}}{\log{\log{T}}}\right)
\end{align}
(cf. \cite[Theorem 14.13, equation (14.13.1)]{tit2}).
In comparison to the above estimate, the term that determines the error term of $N_k(T)$ is
\begin{align*}
\frac{1}{2\pi}\arg{G_k\left(\frac{1}{2}+iT\right)}+\frac{1}{2\pi}\arg{\zeta\left(\frac{1}{2}+iT\right)}
\end{align*}
by Proposition \ref{prop6} and under RH, they are currently estimated as follows:
\begin{align} \label{eq:skt}
\arg{G_k\left(\frac{1}{2}+iT\right)} =  O_k\left(\frac{\log{T}}{(\log{\log{T}})^{1/2}}\right),\quad \arg{\zeta\left(\frac{1}{2}+iT\right)} = O\left(\frac{\log{T}}{\log{\log{T}}}\right).
\end{align}
This estimate of $\arg{G_k\left(\frac{1}{2}+iT\right)}$ determines the error term of $N_k(T)$
and it results in $N_k(T)$ having error term slightly greater in magnitude than that of $N(T)$.
However, this is the best known estimate on $N_k(T)$ under RH at present.
\end{rmk}

Furthermore, the size of the implied $O$-constant in equation \eqref{eq:st} has been studied in many papers, such as \cite{car} and \cite{far}. In contrast to this, we currently have no information about the implied $O$-constant in the first equation of \eqref{eq:skt}.

\section*{Acknowledgements}

The author would like to express her gratitude to Prof. Kohji Matsumoto, Prof. Hirotaka Akatsuka, senior Ryo Tanaka, and also the referee for their valuable advice.

redpajama_set_name: RedPajamaArXiv


Your Maileg friends only fit into the clothes and furniture that corresponds to their size (medium bunnies fit into medium clothes and medium furniture).
Big Brother and Big Sister Mice can fit into Micro clothing and furniture.
Baby Boy and Baby Girl Mouse can fit into the My Collection high chairs and bouncy chairs.
Medium Mice can wear the same size clothes as Best Friends.
Best friends clothing options can be worn by Best Friends and Medium Mice.
Best Friends also fit into many Medium Furniture options.


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text:
'CALL YOU IN'
Watch: Ashley Judd Confronted by Sex-Work Activist at The Wing: 'It's Not OK to Not Listen'
As sex workers protested Judd's talk outside, a sex-work activist questioned the actress's anti-prostitution stance inside. Watch exclusive video of the exchange here.
Photo Illustration by The Daily Beast
Last month, actor and activist Ashley Judd sparked backlash on social media when she posted her controversial positions on sex work and consent.
After facing criticism for an Instagram post about "#survivors of #prostitution who are saying #MeToo" in which she wrote that "prostitution is paid rape," Judd stood by her stance. She tweeted, "I believe body invasion is indeed inherently harmful, and cash is the proof of coercion. Buying sexual access commodifies something that is beyond the realm of capitalism and entrepreneurship: girls and women's orfices [sic]."
As The Daily Beast previously reported, Judd has consistently conflated sex work and sex trafficking, and is on the advisory council of Demand Abolition, an organization that seeks to eradicate "the illegal commercial sex industry in the US."
Unsurprisingly, sex workers and activists did not take kindly to Judd's offensive takes. Kate D'Adamo, a sex-worker rights advocate and partner with Reframe Health and Justice, was one of the first to call her out, tweeting, "Congrats, @AshleyJudd, on your hard work trying to make #MeToo a space where those most likely to face and harm are unwelcome and unsafe. #sexworkerlivesmatter."
Another social media user who responded to Judd on Twitter, @BrookeBrou, told The Daily Beast that, "At best it sends a confusing, conflicting message when a celebrity uses the #MeToo movement as a way to support their anti-sex work crusade. At worst it gives everyday people wrong and harmful information about [sex work], while also missing the mark on what consent truly is."
Despite her track record of reducing folks in the sex trade to trafficked "orifices," Judd is still welcomed and celebrated in feminist spaces. On Thursday night, a group of activists congregated outside of The Wing Flatiron in Manhattan to protest a conversation with Ashley Judd. The Wing is a members-only women's co-working space that's "dedicated to the advancement of women in our wider community."
When @SXNOIR, who asked to be identified by the name of the platform that she founded, saw that Judd would be coming to The Wing, she saw it as an opportunity to engage with the actress about her recent anti-sex work statements. She worked with activists Brooke Kelty and Clara Cline to organize a two-part action: inside, @SXNOIR would ask Judd a question during the Q&A, while outside, a silent protest would gather to catch Judd and The Wing members on their way out.
Just minutes before the event, @SXNOIR walked The Daily Beast through their strategy and objectives, emphasizing a "call-in" as opposed to a "call-out": "We want to be loving, we want to be caring, but we want to hold people accountable."
"I think it's very important to be mindful and also educate the people who could be misinformed about how [Judd's remarks] put women in potential danger," she explained. "Everyone on Twitter who has responded to her cannot physically be here right now, but I'm here right now, and so we want to elevate their voices the best way we can. We're physically here, you cannot shadowban us."
She made it clear that, "We're not coming for The Wing," but did express disappointment in their handling of the Judd event. "One thing that I will criticize The Wing on is, when I joined, I was so impressed with how diverse their programming was. Like when they had the women of color and wellness event, they passed the mic to the people who could actually talk about it. And with this event with Ashley Judd, they're not passing the mic at all out of fear of making someone uncomfortable. And it's like, imagine how we feel sitting in the audience listening to someone who is actively working against us? That makes me very uncomfortable."
@SXNOIR mentioned that there were initially hopes of having a sex worker or an educated ally engage with Judd onstage in an official capacity. "I can't say it was ignored, but it wasn't taken seriously," she explained. "It was somewhat dismissed. And so I think that speaks volumes, and I think The Wing could use their resources and use their members better, because I know for a fact they have members who are sex workers. So are you inclusive of them, or are you not?"
Judd's conversation with The New York Times' Cara Buckley lasted about 40 minutes and covered a number of topics, from going on the record about Harvey Weinstein to Judd's humanitarian work; at one point, the actress even performed parts of the "Nasty Woman" poem she recited at the Women's March. Near the end of the talk, Buckley did ask about the recent backlash against Judd, and her statements essentially erasing consensual sex work. In response, Judd reiterated that, "There's no such thing as sex work." She continued, "It's paid sexual access, it's not sex, which is something that's consensual and sacred."
When @SXNOIR got the chance to pose her question, she began by asking Judd about her work with Demand Abolition. Judd called for the decriminalization of women and girls in the sex trade and the criminalization of people who purchase sex, a "solution" that many sex workers and activists take issue with. As Kate D'Adamo previously told The Daily Beast, "We have the data—in places where they increased policing of only clients/johns, violence increased, the level of violence people were willing to endure before they reported it to police increased, and people end up seeing more clients for less money. It's not a question mark anymore. This is willfully valuing an opinion over someone's life."
@SXNOIR continued, "I'm here today to call you in and to say that we're ready to talk. We want to sit down with you, we want you to hear our stories. And I understand that you have your opinion and we have our opinion, but it's not OK to not listen to all of the women who are affected by sex work, and there are a lot of women who do consensual sex work." After a back and forth between @SXNOIR and another member of the audience, Judd finally addressed her, saying, "You are beautiful and eloquent."
"I'm on your side," Judd concluded. "I'm not on the side of the pimps, the brothel keepers, and the pornographers."
"She was conflating the idea of sex trafficking, which is real and is terrible and needs to be fixed, and the idea of a woman who is consenting to using her body in a way that she wants to."
Clara Cline, a co-organizer of the protest who also attended the talk, told The Daily Beast that she found Judd's response to be lacking. "I thought it was very cute how she was like, well I appreciate you, clap clap clap. As in, this is over. She didn't have a real thoughtful response, as if she had listened to anything, any of the talking points [@SXNOIR] mentioned."
When asked if she thought Judd might seriously reconsider her positions, Cline laughed, "Hell no! She's probably going to go home, get on Twitter and write something like, wow, good for these people exercising their freedom of speech! She's not going to change her stance, but we're here to show her, and to show the people who are following her on social media that we are here, and she can't talk over us." Cline added, "We're not out every day and so it's hard for us to have a voice, especially when we are shadowbanned."
As the crowd at The Wing dispersed, multiple attendees came up to @SXNOIR to thank her for her contribution. Outside, a small band of activists lined the streets with signs reading "Don't be a SWERF [sex worker exclusionary radical feminist]" and "Hey Ashley Judd, please don't reduce me to my orifices."
While most of the decamping Wing members walked by quickly, two attendees from the talk who were not affiliated with the protesters picked up signs and joined in. One of them, a Wing member named Brittany, told The Daily Beast that Judd's statements during the event "rubbed [her] the wrong way," particularly "that she was conflating the idea of sex trafficking, which is real and is terrible and needs to be fixed, and the idea of a woman who is consenting to using her body in a way that she wants to." She said that @SXNOIR's comments articulated the discomfort that she was feeling, "and so as soon as she told me inside that there was a potential to stand with women in a way that I felt more authentically connected to outside, that felt like the obvious next step." Brittany concluded that while she found some of Judd's points during the talk interesting, the conversation ultimately "wasn't encompassing of all women. And I want to be a part of the conversation that is."
Eventually Judd exited the building into the protest. She greeted @SXNOIR and went up to many of the activists individually, shaking their hands and introducing herself. Brooke Kelty, a sex worker and sex worker rights activist who co-organized the protest, told The Daily Beast that she was pleased with the outcome, saying, "I didn't think [Judd] would actually talk with us, so that was nice." Kelty continued, "I was hoping for a little bit more people, but I was super happy some members of The Wing actually joined us."
Moving forward, Kelty offered, "The best case scenario is that [Judd] would be in contact with a group like SWOP and have conversations with other sex workers, because sex work is huge and it's very diverse. There are lots of different voices. There are survivors who are unhappy and don't agree with me and the people who are protesting here, but like I said, it's a very big community, so you need to have a more diverse group that you're listening to."
Kelty concluded, "I understand in some part where [Judd's] coming from—she is a rape survivor, and so she wants to protect women. She wants to protect people. At the same time, I just don't agree with how she's going about it."
@ahzimm

redpajama_set_name: RedPajamaCommonCrawl


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simply thirteen km (8 miles) north of Kuala Lumpur a towering limestone outcrop is domestic to these outstanding caves. Inside the 19th century a small Hindu shrine changed into built in the largest cave, later called Temple Cave. An huge golden statue of Lord Murugan, to whom the caves are dedicated, stands at the foot of a flight of 272 steps leading up to Temple Cave. At the pinnacle of the stairs, traffic will discover three massive limestone caves plus smaller ones filled with Hindu artwork and statues visitors may additionally need to carry peanuts or bananas for the monkeys in an effort to greet them upon arrival.
Melaka (or Malacca) is an vintage metropolis that was one the capital of a powerful Malay nation before the colonial technology, in addition to a town below next Portuguese, Dutch and British rule. Many buildings here have a beautiful aggregate of colonial architecture and vibrant Asian color. Huge, white pillared manses with red and red roofs take a seat above golden gardens. The crumbling remains of A Famosa, the Portuguese fort based by using Afonso de Albequerque, sits most of the modern metropolis skyscrapers of Melaka city even as the Dutch district features some of the oldest Dutch architecture inside the East. Of special word here is the large spires of the Church of St Francis Xavier, that are expansive and easy to see at the skyline.


text:
Gardening, Landscaping, & Building
General and Misc.
Left back to General and Misc.
The Art and Flair of Mary Blair by John Canemaker 2003
See more: General and Misc.
The Art and Flair of Mary Blair: An Appreciation. By John Canemaker, foreword by Roy E. Disney. 2003, hardcover with dustjacket. 10x10", 110 pages. Loaded with gorgeous color and b&w photos.
"The style and vibrant color of Disney films in the early 1940s through mid-1950s came primarily from the artist Mary Blair. In her prime, she was an amazingly prolific American artist who enlivened and influenced the not-so-small worlds of film, print, theme parks, architectural decor, and advertising.
"She dominated design at the Walt Disney Studios with a joyful creativity and exuberant color palette that stamped to look of many classic Disney animated features including Cinderella and Peter Pan. Favorite theme park attractions, most notably "It's a Small World" ride, originally created for the 1964 New York World's Fair."
THINGS TO MAKE FOR CHILDREN A SUNSET BOOK (1961)
OPTICAL ART THEORY AND PRACTICE BY RENE PAROLA (1969)
CRAFTS OF THE MODERN WORLD BY ROSE SLIVKA 1968
BEAUTY BOOK OF KNOWLEDGE (1950's)
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1. supportedFeatures in serviceDescription: should we prefix these with wsrp: thereby allowing non-standard features to referenced [aka extensions].
2. Lifetime in both RegistrationState and RegistrationContext. This sticks out like a sore thumb. It seems to be there so ModifyRegistration can continue to return a RegistrationState. I suggest instead it return a ModifyRegistrationResult that contains both a RegistrationState and a Lifetime. If we do this should each be optional and signify that that omission means no changes? Or does the producer have to explicitly return items because omission means null?
3. ClientData Type section. First sentence is: "The MarkupParams structure carries information concerning the user agent using this type." This seems out of date. Suggest a better introduction to what the clientdata structure is about.
4. HandleEventsFailed is inconsistent with our CopyPortlets/Import/ExportsFailed structures. Should we reconsider and have the structure return the QNames of the events that failed [and multiple for any given reason]. Do we really care to handle the odd situation that the array contains the same events twice? We could always tell the Consumer that if it is bothered by this to send them in different requests. Another possibility is that the Event type get a new field to carry a consumer supplied ID that can be used to reference in the subsequent failed array.
5. Section 6.1.23 EventParams Type -- description of portletStateChange: reference that this impacts processing of an event needs to be updated to reflect that handleEvents is now a bulk operation. Second sentence seems to have a misplaced semicolon.
6. Can someone clarify the use case for resource form submission/file upload? Is this merely because we didn't prohibit it in 1.0? Should we be more restrictive in our API? Part of my objection is this gives the portlet a loophole to change state during render but without access to navigational state updates.
<rt> I think there were several; 1) Certain technologies (including ASP, I think) carry their state using hidden form fields and therefore form submission is the expected means for receiving this back with activated URLs, 2) Consider a search result that says it has 342 pages (i.e. a large number) and allows the user to enter a page number to jump to ... rather than refreshing the entire page, such a portlet could consider the search results as a resource and fetch a different one for placement within the page. Again, browser technology makes this easier if the entry field is encapsulated within a form, 3) any other use cases people would want to raise?
7. Section 6.3 Getresource: sentence [line 2253] "This operation is used whenever the URL activated has ..." would be better as "This operation is used whenever the activated URL has ..."


Quality of life - When summer hits its wings and everything blooms again - even humans. Spending time with loved ones, raising his glass and spread a lot of love around him. Special events and moments you will remember - or not.
We have party lighting for all occasions; Midsummer party, wedding or evening barbecue with friends. In late summer evenings in the cottage, or the party inside the barn of our dark months.
All our party loops are equipped with LED lighting of the highest quality, which not only provides a good light, but is also energy efficient and keep year after year.
We at lampgallerian might not be invited to you on these occasions - but we can at least help you with festive lighting that will set the mood. We have something for everyone, from vibrant colors to the romantic white lights with a pleasant light.
Do you have any questions you can always contact us. Otherwise it is just to choose, choose and dream away to those glorious moments - moments of life.