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\begin{document}

\keywords{Riemann zeta--function, entire Riemann zeta--function, completed Riemann zeta--function, Riemann $\Xi$--function, Riemann hypothesis}

\subjclass[2020]{Primary: 11-04; Secondary: 11M06, 11M26}

\title[Riemann hypothesis]{On the Riemann hypothesis for the zeta function}

\author{Xiao-Jun Yang}

\email{dyangxiaojun@163.com; xjyang@cumt.edu.cn}

\address{School of Mathematics, China University of Mining and Technology, Xuzhou 221116, China}


\begin{abstract}
In this paper we address some variants for the products of Hadamard and Patterson.
We prove that all zeros of the Riemann $\Xi$--function are real. We also prove that the
Riemann hypothesis is true. The equivalence theorems associated with the Riemann zeta--function
are obtained in detail.
\end{abstract}

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

In 1774, Euler~\cite{1} introduced the zeta--function by the real series
\begin{equation}
\label{eq1}
\zeta \left( s \right)=\sum\limits_{n=1}^\infty {m^{-s}} ,
\end{equation}
where $s\in \mathds{R}$, $n\in \mathds{N}$ and $s>1$, and proposed his product by
\begin{equation}
\label{eq2}
\zeta \left( s \right)=\prod\limits_p {\left( {1-p^{-s}} \right)^{-1}} ,
\end{equation}
where $p$ run through all primes and $s>1$, if we denote the sets of real
and integer numbers by $\mathds{R}$ and $\mathds{N}$, respectively.

\subsection{Functions}

In his remarkable paper published in 1859, Riemann~\cite{2} extended (\ref{eq1}) to be
meromorphic continuation to the whole complex plane $s\in \mathds{C}$
except for the simple pole $s=1$ with residue $Res_{s=1} \left( {\zeta
\left( s \right)} \right)=1$ and proposed that (\ref{eq1}) and (\ref{eq2}) are valid
for $Re\left( s \right)>1$ if $\mathds{C}$ is denoted as the set of complex
numbers, and $Re\left( s \right)$ and $Im\left( s \right)$
are denoted as the real and imaginary part of the complex variable $s\in \mathds{C}$.
It has been proved that for $s\in \mathds{C}\backslash \{1\}$, Eq. (\ref{eq1}),
which is well-known Riemann zeta--function, has the
trivial zeros $s=-2m$ with $m\in \mathds{N}$~\cite{3}.

Riemann suggested the completed Riemann zeta--function by the expression~\cite{2}
\begin{equation}
\label{eq3}
\varsigma \left( s \right)=\pi ^{-s/2}\Gamma \left( {s/2} \right)\zeta
\left( s \right),
\end{equation}
where $s\in \mathds{C}$ and $\Gamma $ is the gamma function~\cite{4}.
In his book~\cite{5}, Neukirch discovered that (\ref{eq3}) is meromorphic continuation to the
whole complex plane $s\in \mathds{C}$ except for two poles $s=1$ and $s=0$ with
residues $Res_{s=1} \left( {\varsigma \left( s \right)} \right)=1$ and
$Res_{s=0} \left( {\varsigma \left( s \right)} \right)=-1$, respectively.

To remove the poles and trivial zeros of them, Riemann discovered
the entire Riemann zeta--function, defined by~\cite{2}
\begin{equation}
\label{eq4}
\xi \left( s \right)=\frac{1}{2}s\left( {s-1} \right)\pi ^{-s/2}\Gamma
\left( {s/2} \right)\zeta \left( s \right)
\end{equation}
for $s\in \mathds{C}$.

In order to study the zeros of (\ref{eq4}), Riemann defined the entire Riemann
zeta--function at the critical line $Re\left( s \right)=1/2$ by~\cite{2}
\begin{equation}
\label{eq5}
\Xi \left( t \right)=\xi \left( {1/2+it} \right),
\end{equation}
which is denoted by Landau~\cite{6}, and called the Riemann $\Xi$ function for $t\in \mathds{C}$ and $i=\sqrt {-1}$ by P\'{o}lya~\cite{7}, and discovered by Riemann~\cite{2}
\begin{equation}
\label{eq6}
\log \Xi \left( t \right)=\sum\limits_{n=1}^\infty {\left(
{1-\frac{t^2}{\tau _n^2 }} \right)} +\log \Xi \left( 0 \right),
\end{equation}
where $\tau _n $ run the positive roots of $\Xi \left( t \right)=0$.

\subsection{Analytic continuations}

By (\ref{eq4}), Riemann discovered that~\cite{2}
\begin{equation}
\label{eq7}
\xi \left( s \right)=4\int\limits_1^\infty {\frac{d\left(
{x^{\frac{3}{2}}\psi ^{\left( 1 \right)}\left( t \right)}
\right)}{dt}t^{-\frac{1}{4}}\cosh \left( {\frac{\left( {s-\frac{1}{2}}
\right)\log t}{2}} \right)dt}
\end{equation}
for $s\in \mathds{C}$ and
\begin{equation}
\label{eq8}
\psi \left( x \right)=\sum\limits_{n=1}^\infty {e^{-n^2\pi x}} ,
\end{equation}
with $x>0$.

To further study the zeros of (\ref{eq1}) for $s\in \mathds{C}$, Riemann showed~\cite{2}
\begin{equation}
\label{eq9}
\Xi \left( t \right)=\frac{1}{2}-\left( {t^2+\frac{1}{4}}
\right)\int\limits_1^\infty {\psi \left( x \right)x^{-\frac{3}{4}}\cos
\left( {\frac{t}{2}\log x} \right)dx} ,
\end{equation}
or, alternatively,
\begin{equation}
\label{eq10}
\Xi \left( t \right)=4\int\limits_1^\infty {\frac{d\left(
{x^{\frac{3}{2}}\psi ^{\left( 1 \right)}\left( x \right)}
\right)}{dx}x^{-\frac{1}{4}}\cos \left( {\frac{t}{2}\log x} \right)dx} .
\end{equation}
By (\ref{eq9}), Hadamard~\cite{8} in 1893 set up the theory of the entire function to
deal with the series of (\ref{eq5}), and suggested that
\begin{equation}
\label{eq11}
\Xi \left( t \right)=\frac{1}{2}-\left( {t^2+\frac{1}{4}}
\right)\int\limits_1^\infty {\psi \left( x \right)x^{-\frac{3}{4}}\cos
\left( {\frac{t}{2}\log x} \right)dx} =\sum\limits_{k=0}^\infty {\left( {-1}
\right)^k\varphi _k t^{2m}} ,
\end{equation}
where
\begin{equation}
\label{eq12}
\varphi _k =\left( {\frac{D_{2k} }{4}-D_{2k-2} } \right)
\end{equation}
and
\begin{equation}
\label{eq13}
D_k =\frac{1}{2^{2k}\Gamma \left( {2k+1} \right)}\int\limits_1^\infty {\psi
\left( x \right)x^{-\frac{3}{4}}\left( {\log x} \right)^kdx}.
\end{equation}
In this case, (\ref{eq5}) is an even entire function.

By (\ref{eq10}), P\'{o}lya~\cite{7} in 1927 showed that
\begin{equation}
\label{eq14}
\varphi _k =\frac{{(-1)^k}\Xi ^{\left( {2k} \right)}\left( 0 \right)}{\Gamma \left(
{2k+1} \right)}.
\end{equation}
To find that the zeros of (\ref{eq5}) are real, Hardy in 1914~\cite{9} proposed
the well-known result is that (\ref{eq5}) has infinitely many
zeros, proved by Hardy and Littlewood in 1921~\cite{10}. On the other
hand, the necessary condition for the real zeros of (\ref{eq5}) is that the
coefficients $\varphi _k $ satisfy the Tur\'{a}n inequalities, which was
suggested in~\cite{11,12}. P\'{o}lya and Schur~\cite{13} considered that the product and
series for (\ref{eq5}) are equivalent to each others. They have played the important roles in
the analysis of the zeta function because it is connected with the Laguerre-P\'{o}lya
class in theory of the entire function~\cite{13a}.

In 1893, Hadamard developed~\cite{8} a well-known product for the entire
Riemann--zeta function by the expression~\cite{3}
\begin{equation}
\label{eq15}
\xi \left( s \right)=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{s_n }} \right)} ,
\end{equation}
where $s_n $ run the roots of $\xi \left( s \right)=0$ and $s\in \mathds{C}$.

After one year, Cahen~\cite{14} discovered that
\begin{equation}
\label{eq16}
\Xi \left( t \right)=\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{t^2}{\tau _n^2 }} \right)} ,
\end{equation}
which can be deduced by (\ref{eq6}) and was rediscovered by Titchmarsh~\cite{15}, where
$\tau _n $ run the positive roots of $\Xi \left( t \right)=0$.

In 1974, Edwards proved that~\cite{3}
\begin{equation}
\label{eq17}
\xi \left( s \right)=A_1 \prod\limits_{n=1}^\infty {\left(
{1-\frac{s-\frac{1}{2}}{s_n -\frac{1}{2}}} \right)}
\end{equation}
and
\begin{equation}
\label{eq18}
\Xi \left( t \right)=A_2 \prod\limits_{n=1}^\infty {\left( {1-\frac{t}{t_n
}} \right)} ,
\end{equation}
where $t_n $ run the roots of $\Xi \left( t \right)=0$, and both $A_1 $ and
$A_2 $ are the constants.

In 1988, Patterson~(\cite{16}, p.34) discovered an alternative product by
\begin{equation}
\label{eq19}
\xi \left( s \right)=\frac{1}{2}\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{\widehat{s_n }}} \right)} \left( {1-\frac{s}{1-\widehat{s_n }}}
\right),
\end{equation}
where $s\in \mathds{C}$ and
\begin{equation}
\label{eq20}
\widehat{s_n }=Re\left( {s_n } \right)+i\tau _n .
\end{equation}

\subsection{Functional equations}

In 1859, Riemann showed that~\cite{2}
\begin{equation}
\label{eq21}
\zeta \left( s \right)=\Gamma \left( {1-s} \right)\left( {2\pi }
\right)^{s-1}2\sin \left( {s\pi /2} \right)\zeta \left( {1-s} \right),
\end{equation}
where $s\in \mathds{C}$.

For $s\in \mathds{C}$ Landau in 1918~\cite{17} proposed an alternative form of (\ref{eq21}),
given by
\begin{equation}
\label{eq22}
\zeta \left( {1-s} \right)=2\left( {2\pi } \right)^{-s}\cos \left( {\pi s/2}
\right)\Gamma \left( s \right)\zeta \left( s \right),
\end{equation}
and discovered that~\cite{17}
\begin{equation}
\label{eq23}
\varsigma \left( s \right)=\varsigma \left( {1-s} \right),
\end{equation}
\begin{equation}
\label{eq24}
\xi \left( s \right)=\xi \left( {1-s} \right)
\end{equation}
and
\begin{equation}
\label{eq25}
\Xi \left( t \right)=\Xi \left( {1-t} \right),
\end{equation}
where $s\in \mathds{C}$.

It was reported in 1768 that Euler~\cite{18} discovered (\ref{eq21}) and (\ref{eq21}) in the real case.
In his book (for the details, see~\cite{6}, p.288), Landau
also suggested that
\begin{equation}
\label{eq26}
\xi \left( s \right)=\xi \left( {1-s} \right)=\xi \left( {\frac{1}{2}+it}
\right)=\xi \left( {\frac{1}{2}-it} \right)=\Xi \left( t \right)=\Xi \left(
{-t} \right),
\end{equation}
where $t\in \mathds{R}$ and $s\in \mathds{C}$.

\subsection{Critical trip}

The critical trip for the Riemann zeta--function is
\begin{equation}
\label{eq27}
0<Re\left( s \right)<1,
\end{equation}
which was proposed in 1859 by Riemann~\cite{2} and proved in 1896 by Hadamard
~\cite{19} and Vallee-Poussin~\cite{20}. This was connected with the number of the
nontrivial zeros of the Riemann zeta--function in critical trip $0<Re\left( s
\right)<1$ and in the trip $0<t<T$, which is defined by~\cite{20a,20b}
\begin{equation}
\label{eq28}
\widetilde{N}\left( T \right)=\# \left\{ {s=\sigma +it:0\le t\le T,0<\sigma <1,\zeta
\left( s \right)=0} \right\}.
\end{equation}
The completed and entire Riemann zeta--functions also have the same critical trip
$0<Re\left( s \right)<1$~\cite{3}.

\subsection{Critical line}

Riemann proposed that the critical line for the Riemann zeta--function is
that all nontrivial zeros of the Riemann zeta--function lie on the line
$Re\left( s \right)=1/2$, which is also the critical line of the completed
and entire Riemann zeta--functions~\cite{21,21a}.

\subsection{Riemann-von Mangoldt formula}

The number of the nontrivial zeros for the Riemann zeta--function at the
critical line and in the trip $0<t<T$ is equal to the number of the zeros
for $\Xi \left( t \right)$ in the trip $0<t<T$, and we denote it by~\cite{2,20a,20b}
\begin{equation}
\label{eq29}
\widetilde{N}_0 \left( T \right)=\# \left\{ {s=1/2+it:0\le t\le T, \zeta \left( s
\right)=0} \right\}.
\end{equation}
Riemann reported that~\cite{2,20a}
\begin{equation}
\label{eq30}
\widetilde{N}_0 \left( T \right)\approx \frac{T}{2\pi }\log \frac{T}{2\pi e},
\end{equation}
and von Mangoldt proved in 1905 that~\cite{22,20b}
\begin{equation}
\label{eq31}
\widetilde{N}_0 \left( T \right)\mbox{=}\frac{T}{2\pi }\log \frac{T}{2\pi
e}\mbox{+}O\left( {\log T} \right).
\end{equation}
Eq. (\ref{eq31}) is the Riemann-von Mangoldt formula due to them.

\subsection{Riemann-Siegel formula}

In the Riemann's Nachlass, Riemann proposed the formula of the method to
find the imaginary part of the complex variable $s_n $, where $s_n $ run the
roots of $\xi \left( s \right)=0$ and $s\in \mathds{C}$. In fact, Riemann
obtained few values for the imaginary part of the complex variable $s_n $
~\cite{2}. Based on the work of Hardy and Littlewood~\cite{10}, Siegel in 1932 revised
the Riemann result and obtained~\cite{23,24}
\begin{equation}
\label{eq32}
e^{i\mathds{A}\left( t \right)}\zeta \left( {\frac{1}{2}+it}
\right)=2\sum\limits_{j=1}^{g\left( t \right)} {\frac{1}{\sqrt j }\cos
\left( {\mathds{A}\left( t \right)-t\log j} \right)} +\mathds{S}\left( t \right),
\end{equation}
where $t>0$,
\begin{equation}
\label{eq33}
\mbox{ }g\left( t \right)=\left[ {\sqrt {\frac{t}{2\pi }} } \right],
\end{equation}
\begin{equation}
\label{eq34}
\mathds{A}\left( t \right)=-\frac{t}{2}\log \pi +\arg \Gamma \left(
{\frac{1}{4}+\frac{it}{2}} \right)
\end{equation}
and~\cite{10}
\begin{equation}
\label{eq35}
\mathds{S}\left( t \right)=\mbox{O}\left( {1/\sqrt[4]{t}} \right).
\end{equation}
It is clear that Riemann and Siegel~\cite{23,24} showed that
\begin{equation}
\label{eq36}
Im\left( {s_n } \right)=\tau _n \mbox{ }\left( {\tau _n >0} \right)
\end{equation}
subject to the condition, which implies that either~\cite{20a}
\begin{equation}
\label{eq37}
\widetilde{N}\left( T \right)=\widetilde{N}_0 \left( T \right),
\end{equation}
or ~\cite{23}
\begin{equation}
\label{eq38}
Im\left( {s_n } \right)=t_n \mbox{ }\left( {t_n >0} \right),
\end{equation}
where $\Xi \left( t \right)=0$.
Different from the techniques~\cite{24a}, a fast algorithm for finding the nontrivial zeros for the Riemann zeta--function
was developed by Odlyzko and Sch\"{o}nhage~\cite{24b}.

It is well known that (\ref{eq38}) might be obtained by the theorem of Hardy~\cite{9} and the
Tur\'{a}n inequalities~\cite{24c}, and also deduced by the product for the Riemann
$\Xi$ function~\cite{3}, which is an analogue of the Euler product
of the cosine~\cite{25}. Here, $t_n $ are called the Riemann-Siegel zeros, which
are well-known constants in the present paper.

\subsection{Riemann hypothesis}

In order to present the distribution of the prime, Riemann in 1895~\cite{2}
conjectured well-known Riemann hypothesis, which gives the assert as
follows~\cite{2,20a,20b}:

\begin{conjecture}
\label{C1}
The nontrivial zeros of $\zeta \left( s \right)$ have real part equal to
$1/2$, i.e., $Re\left( s \right)=1/2$.
\end{conjecture}

This famous hypothesis remains the most important unsolved problem in theory
of the analytic number theory for more than 160 years~\cite{25a,25b,25c}. As a matter of fact,
Riemann~\cite{2} showed that the zeros of the completed~\cite{5} and entire~\cite{25d} Riemann zeta--function
are the same as those of the Riemann zeta--function. This implies that
the real part of the zeros of the completed~\cite{5} and entire~\cite{25d} Riemann zeta--functions
functions is the same as that of the Riemann zeta--function~\cite{2}. For the details, see~\cite{2,20a,20b}. The
consequence of its truth implies that Lindel\"{o}f conjecture~(\cite{16}, p.34),
Montgomery pair correlation conjecture~\cite{26a,26b}
and the Berry--Keating conjecture in quantum chaos
~\cite{27,27a} are true and is also related to the Yang-Lee theorem~\cite{28,28a}.
For the more equivalences of this conjecture, see~\cite{29}.

\subsection{The target of this paper}

To prove \textit{Conjecture \ref{C1}}, our main aim is to find a best way to provide new
products for the Riemann $\Xi$--function with the nontrivial
zeros $s_n $ of the Riemann zeta--function by the variants of the Hadamard~\cite{8} and Patterson~(\cite{16}, p.34) products at the
critical line because by the result coming from the theorem of Hardy~\cite{9},
it is known that it may has infinitely many real zeros, which is the same as the fact that
the Jensen polynomials for the Riemann zeta--function have real zeros~\cite{29a}.
With the
assistance of the Riemann-Siegel zeros $t_n $, we obtain the real part of
the (nontrivial) zeros for the entire Riemann zeta--function (\ref{eq4}), which is
derived from the fact, which, so--called conjecture of Jensen~\cite{7,25c,29a} proved
in our paper, that all zeros for the Riemann $\Xi$--function (\ref{eq5}) are real.
This real part
implies the truth of \textit{Conjecture \ref{C1}}.

The structure of the paper is designed as follows. In Section \ref{sec:2} we
investigate the integral, series and product for the entire Riemann zeta--
function. In Section \ref{sec:3} we study integral, series and product for the
Riemann $\Xi$ function and give the existence theorem of
the zeros for the Riemann $\Xi$--function, and prove the
Riemann hypothesis. In Section \ref{sec:4} we mainly give the equivalence theorems.

\subsection{Acknowledgements}
I work on the plan of the Riemann conjecture from autumn 2018. At first, I would like to express thanks for Professor Nouzha El Yacoubi (President of African Mathematical Union), Professor Mahmoud Abdel-Aty (Vice-President of African Academy of Sciences and Sohag University), Professor Jeffrey Hoffstein (Brown University),  Professor Mircea Merca (University of Craiova and Academy of Romanian Scientists), Professor J. A. Tenreiro Machado (Instituto Superior de Engenharia do Porto), Professor Mustafa Bayram (Biruni  University), Professor Sohail Nadeem (Quaid-i-Azam University), Professor Miguel A. F. Sanjuan (King Juan Carlos University), Professor Minvydas Ragulskis (Kaunas University of Technology), Professor Hossein Jafari (University of South Africa), and Professor Dumitru Baleanu (Cankaya University), for supporting the plan of the Riemann conjecture from spring 2019.
I would also like to thank Professor Mahouton Norbert Hounkonnou (University of Abomey-Calavi) and Professor Carlo Cattani (Tuscia University) for giving the help to improve the manuscript and discuss the Riemann-zeta function. I would like to remember Professor He-Ping Xie (Shenzhen University) and Professor Feng Gao (China University of Mining and Technology) who support me to study
mathematics.
This work is supported by the Yue-Qi Scholar of the China University of Mining and
Technology (No. 102504180004).

\section{Theory of the entire Riemann zeta--function} {\label{sec:2}}

In this section we investigate the theory of the entire Riemann zeta--function
involving the integral, series and product formulas.

\subsection{The integral representation}

The integral representation of the entire Riemann zeta--function was first
given by Riemann~\cite{2}.

\begin{lemma}
(Riemann integral representation theorem)
\label{L1}

Let $s\in \mathds{C}$. If $\psi \left( x \right)$ is defined in (\ref{eq8}), then
\begin{equation}
\label{eq39}
\xi \left( s \right)=4\int\limits_1^\infty {\frac{d\left(
{x^{\frac{3}{2}}\psi ^{\left( 1 \right)}\left( x \right)}
\right)}{dx}x^{-\frac{1}{4}}\cosh \left( {\frac{\left( {s-1} \right)\log
x}{2}} \right)dx} .
\end{equation}
Moreover,
\begin{equation}
\label{eq40}
\xi \left( s \right)=\xi \left( {1-s} \right).
\end{equation}
\end{lemma}

\begin{proof}
See, for (\ref{eq39}), Riemann~\cite{2} and, for (\ref{eq40}), Landau~\cite{17}.
\end{proof}

\subsection{The series representation}

The series representation for the entire Riemann zeta--function was first
suggested by Edwards~\cite{3}.

\begin{lemma}
(Edwards series theorem)
\label{L2}

Let $s\in \mathds{C}$ and $k\in \mathds{N}\cup \left\{ 0 \right\}$. If $\psi
\left( x \right)$ is defined in (\ref{eq8}), then
\begin{equation}
\label{eq41}
\xi \left( s \right)=\sum\limits_{k=0}^\infty {\varphi _k \left(
{s-\frac{1}{2}} \right)^{2k}}
\end{equation}
where
\begin{equation}
\label{eq42}
\varphi _k =\frac{4}{\Gamma \left( {2k+1} \right)}\int\limits_1^\infty
{\frac{d\left( {x^{\frac{3}{2}}\psi ^{\left( 1 \right)}\left( x \right)}
\right)}{dx}x^{-\frac{1}{4}}\left( {\frac{\log x}{2}} \right)^{2k}dx} .
\end{equation}
\end{lemma}

\begin{proof}
For the detailed proof, see~\cite{3}.
\end{proof}

\subsection{The product representations}

We now introduce the products of Hadamard and Patterson and propose the
variants of them.

To begin with, we investigate the special values for the entire Riemann zeta--
function.

\begin{lemma}
(Non-vanishing theorem)
\label{L3}

There exist
\begin{equation}
\label{eq43}
\xi \left( 0 \right)=\xi \left( 1 \right)\ne 0
\end{equation}
and
\begin{equation}
\label{eq44}
\xi \left( {1/2} \right)\ne 0.
\end{equation}
\end{lemma}

\begin{proof}
See~\cite{17} and related references~(\cite{6}, p.288;~\cite{30}, p.49).
\end{proof}

\begin{remark}
Here, $\xi \left( 0 \right)=\xi \left( 1 \right)=1/2$
and $\xi \left( {1/2} \right)\simeq0.994$~(\cite{30}, p.49).
\end{remark}


\begin{lemma}
(Product of Hadamard I)
\label{L4}

Let $s\in \mathds{C}$ and $n\in \mathds{N}$. Then
\begin{equation}
\label{eq45}
\xi \left( s \right)=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{s_n }} \right)} ,
\end{equation}
where the product runs over zeroes $s_n $ of (\ref{eq4}).
\end{lemma}

\begin{proof}
See Hadamard~\cite{8}, Edwards~(\cite{3}, p.47) and Broughan~(\cite{30}, p.47).
\end{proof}

\begin{lemma}
(Product of Hadamard II)
\label{L5}

Let $\gamma $ be the Euler's constant such that
\begin{equation}
\label{eq46}
\mu =\log 2+\frac{1}{2}\log \pi -1-\frac{1}{2}\gamma .
\end{equation}
Then for $s\in \mathds{C}$ and $n\in \mathds{N}$,
\begin{equation}
\label{eq47}
\xi \left( s \right)=\xi \left( 0 \right)e^{s\mu }\prod\limits_{n=1}^\infty
{\left( {1-\frac{s}{s_n }} \right)e^{s/s_n }} ,
\end{equation}
where the product runs over zeroes $s_n $ of (\ref{eq4}).
\end{lemma}

\begin{proof}
See the work of Hadamard~\cite{8}.
\end{proof}

\begin{remark}
It is known that (\ref{eq45}) and (\ref{eq47}) have been proved by Hadamard~\cite{8}, (\ref{eq45}) has
been reported in~\cite{3}, and (\ref{eq47}) has been commented in~(\cite{6}, p.32), discussed
by Valiron and Titchmarsh (\cite{32}, p.3), and proved by Ingham (\cite{31}, p.57-58). For
more detailed work, see~\cite{30}.
\end{remark}

Now, we start with the work from Hadamard to Patterson.

\begin{lemma}
(Products of Patterson)
\label{L6}

Let $s\in \mathds{C}$ and $n\in \mathds{N}$. Suppose that
\begin{equation}
\label{eq48}
\widehat{s_n }=Re\left( {s_n } \right)+i\tau _n ,
\end{equation}
which are defined in (\ref{eq20}). Then
\begin{equation}
\label{eq49}
2\xi \left( s \right)=\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{\widehat{s_n }}} \right)} \left( {1-\frac{s}{1-\widehat{s_n }}}
\right),
\end{equation}
where the product runs over zeroes $\widehat{s_n }$.
\end{lemma}

\begin{proof}
See the work of Patterson~(\cite{16}, p.34).
\end{proof}

We now propose some variants from the products of Hadamard~\cite{8} by
\textit{Lemma \ref{L3}}.

\begin{theorem}
\label{T1}
Let $s\in \mathds{C}$ and $n\in \mathds{N}$. Then
\begin{equation}
\label{eq50}
\xi \left( s \right)=\xi \left( {\frac{1}{2}}
\right)\prod\limits_{n=1}^\infty {\left( {1-\frac{s-\frac{1}{2}}{s_n
-\frac{1}{2}}} \right)} ,
\end{equation}
where the product runs over zeroes $s_n $ of (\ref{eq4}).
\end{theorem}

\begin{proof}
By \textit{Lemmas \ref{L3}} and \textit{\ref{L4}}, we show
\begin{equation}
\label{eq51}
\begin{array}{l}
 \xi \left( s \right)\\
 =\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{s_n }} \right)} \\
=\xi \left( 0 \right)\prod\limits_{n=1}^\infty
{\left( {\frac{s_n -s}{s_n }} \right)} \\
=\xi \left( 0\right)\prod\limits_{n=1}^\infty {\left( {\frac{s_n -\frac{1}{2}}{s_n
-\frac{1}{2}}\cdot \frac{s_n -s}{s_n }} \right)} \\
=\xi \left( 0 \right)\left( {\prod\limits_{n=1}^\infty {\frac{s_n
-\frac{1}{2}}{s_n }} } \right)\cdot \left( {\prod\limits_{n=1}^\infty
{\frac{s_n -s}{s_n -\frac{1}{2}}} } \right) \\
 =\xi \left( 0 \right)\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{1}{2s_n }} \right)} } \right]\cdot \left(
{\prod\limits_{n=1}^\infty {\frac{s_n -s}{s_n -\frac{1}{2}}} } \right)\\
=\xi
\left( 0 \right)\left[ {\prod\limits_{n=1}^\infty {\left( {1-\frac{1}{2s_n
}} \right)} } \right]\cdot \left[ {\prod\limits_{n=1}^\infty {\frac{s_n
-\frac{1}{2}-\left( {s-\frac{1}{2}} \right)}{s_n -\frac{1}{2}}} }
\right]\\
=\xi \left( {\frac{1}{2}} \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s-\frac{1}{2}}{s_n -\frac{1}{2}}} \right)} , \\
 \end{array}
\end{equation}
which is the required result.
\end{proof}

\begin{theorem}
\label{T2}
If $\mu $ is defined in (\ref{eq46}), then for $s\in \mathds{C}$ and $n\in \mathds{N}$,
\begin{equation}
\label{eq52}
\xi \left( s \right)=\xi \left( {\frac{1}{2}} \right)e^{s\mu
}\prod\limits_{n=1}^\infty {\left( {1-\frac{s-\frac{1}{2}}{s_n
-\frac{1}{2}}} \right)e^{s/s_n }} ,
\end{equation}
where the product runs over zeroes $s_n $ of (\ref{eq4}).
\end{theorem}

\begin{proof}
In view of \textit{Lemmas \ref{L3}} and \textit{\ref{L5}}, we have
\begin{equation}
\label{eq53}
\begin{array}{l}
 \xi \left( s \right)\\
 =\xi \left( 0 \right)e^{s\mu }\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{s_n }} \right)e^{s/s_n }} \\
 =\left( {e^{s\mu }\prod\limits_{n=1}^\infty {e^{s/s_n }} } \right)\cdot
\left[ {\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{s_n }} \right)} } \right]. \\
 \end{array}
\end{equation}
By \textit{Lemma \ref{L3}} and \textit{Theorem \ref{T1}}, Eq. (\ref{eq53}) can be rewritten in
the form
\begin{equation}
\label{eq54}
\begin{array}{l}
 \xi \left( s \right)\\
 =\left( {e^{s\mu }\prod\limits_{n=1}^\infty {e^{s/s_n }} } \right)\cdot
\left[ {\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{s_n }} \right)} } \right] \\
 =\left( {e^{s\mu }\prod\limits_{n=1}^\infty {e^{s/s_n }} } \right)\cdot
\left[ {\xi \left( {\frac{1}{2}} \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s-\frac{1}{2}}{s_n -\frac{1}{2}}} \right)} } \right] \\
 =\xi \left( {\frac{1}{2}} \right)e^{s\mu }\prod\limits_{n=1}^\infty {\left(
{1-\frac{s-\frac{1}{2}}{s_n -\frac{1}{2}}} \right)e^{s/s_n }}, \\
 \end{array}
\end{equation}
and the result follows.
\end{proof}

We now suggest a variant from the product of Patterson~(\cite{16}, p.34) by
\textit{Lemma \ref{L3}}.

\begin{theorem}
\label{T3}
Let $s\in \mathds{C}$ and $n\in \mathds{N}$. Then
\begin{equation}
\label{eq55}
\xi \left( s \right)=\xi \left( {\frac{1}{2}}
\right)\prod\limits_{n=1}^\infty {\left( {1-\frac{\left( {s-\frac{1}{2}}
\right)^2}{\left( {\widehat{s_n }-\frac{1}{2}} \right)^2}} \right)} ,
\end{equation}
where the product runs over $\widehat{s_n }$, which are defined in (\ref{eq19}).
\end{theorem}

\begin{proof}
With (\ref{eq49}) we arrive at
\begin{equation}
\label{eq56}
\begin{array}{l}
 \xi \left( s \right)\\
 =\frac{1}{2}\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{\widehat{s_n }}} \right)} \prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{1-\widehat{s_n }}} \right)} \\
 =\frac{1}{2}\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{1}{2\widehat{s_n }}} \right)} \cdot \prod\limits_{n=1}^\infty
{\left( {1-\frac{s-\frac{1}{2}}{\widehat{s_n }-\frac{1}{2}}} \right)} }
\right] \cdot \left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}}{1-\widehat{s_n }}} \right)} \cdot
\prod\limits_{n=1}^\infty {\left( {1+\frac{s-\frac{1}{2}}{\widehat{s_n
}-\frac{1}{2}}} \right)} } \right] \\
 =\frac{1}{2}\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{1}{2\widehat{s_n }}} \right)} \cdot \prod\limits_{n=1}^\infty
{\left( {1-\frac{\frac{1}{2}}{1-\widehat{s_n }}} \right)} } \right] \cdot \left[
{\prod\limits_{n=1}^\infty {\left( {1-\frac{s-\frac{1}{2}}{\widehat{s_n
}-\frac{1}{2}}} \right)} \cdot \prod\limits_{n=1}^\infty {\left(
{1+\frac{s-\frac{1}{2}}{\widehat{s_n }-\frac{1}{2}}} \right)} } \right] \\
 =\frac{1}{2}\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{1}{2\widehat{s_n }}} \right)} \cdot \prod\limits_{n=1}^\infty
{\left( {1-\frac{\frac{1}{2}}{1-\widehat{s_n }}} \right)} }
\right] \cdot \left[\prod\limits_{n=1}^\infty { {\left(
{1-\frac{s-\frac{1}{2}}{\widehat{s_n }-\frac{1}{2}}} \right)\left(
{1+\frac{s-\frac{1}{2}}{\widehat{s_n }-\frac{1}{2}}} \right)} } \right]\\
 =\frac{1}{2}\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{1}{2\widehat{s_n }}} \right)} \cdot \prod\limits_{n=1}^\infty
{\left( {1-\frac{\frac{1}{2}}{1-\widehat{s_n }}} \right)} } \right] \cdot \left[
\prod\limits_{n=1}^\infty {\left( {1-\frac{\left( {s-\frac{1}{2}}
\right)^2}{\left( {\widehat{s_n }-\frac{1}{2}} \right)^2}} \right)}\right] \\
 =\xi \left( {\frac{1}{2}} \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{\left( {s-\frac{1}{2}} \right)^2}{\left( {\widehat{s_n
}-\frac{1}{2}} \right)^2}} \right)} , \\
 \end{array}
\end{equation}
where
\begin{equation}
\label{eq57}
\begin{array}{l}
 \prod\limits_{n=1}^\infty {\left( {1-\frac{s}{\widehat{s_n }}} \right)}\\
=\prod\limits_{n=1}^\infty {\left( {\frac{\widehat{s_n }-s}{\widehat{s_n }}}
\right)} \\
=\prod\limits_{n=1}^\infty {\left( {\frac{\widehat{s_n
}-\frac{1}{2}}{\widehat{s_n }-\frac{1}{2}}\cdot \frac{\widehat{s_n
}-s}{\widehat{s_n }}} \right)} \\
=\left( {\prod\limits_{n=1}^\infty
{\frac{\widehat{s_n }-\frac{1}{2}}{\widehat{s_n }}} } \right)\cdot \left(
{\prod\limits_{n=1}^\infty {\frac{\widehat{s_n }-s}{\widehat{s_n
}-\frac{1}{2}}} } \right)\\
=\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{1}{2\widehat{s_n }}} \right)} } \right]\cdot \left(
{\prod\limits_{n=1}^\infty {\frac{\widehat{s_n }-s}{\widehat{s_n
}-\frac{1}{2}}} } \right) \\
 =\left[ {\prod\limits_{n=1}^\infty {\left( {1-\frac{1}{2\widehat{s_n }}}
\right)} } \right]\cdot \left[ {\prod\limits_{n=1}^\infty
{\frac{\widehat{s_n }-\frac{1}{2}-\left( {s-\frac{1}{2}}
\right)}{\widehat{s_n }-\frac{1}{2}}} } \right]\\
 =\left[ {\prod\limits_{n=1}^\infty {\left( {1-\frac{1}{2\widehat{s_n }}}
\right)} } \right]\cdot \left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{s-\frac{1}{2}}{\widehat{s_n }-\frac{1}{2}}} \right)} } \right] \\
 \end{array}
\end{equation}
and
\begin{equation}
\label{eq58}
\begin{array}{l}
 \prod\limits_{n=1}^\infty {\left( {1-\frac{s}{1-\widehat{s_n }}} \right)}\\
=\prod\limits_{n=1}^\infty {\left( {\frac{1-\widehat{s_n }-s}{1-\widehat{s_n
}}} \right)} \\
=\prod\limits_{n=1}^\infty {\left( {\frac{1-\widehat{s_n
}-\frac{1}{2}}{1-\widehat{s_n }-\frac{1}{2}}\cdot \frac{1-\widehat{s_n
}-s}{1-\widehat{s_n }}} \right)} \\
 =\left( {\prod\limits_{n=1}^\infty {\frac{1-\widehat{s_n
}-\frac{1}{2}}{1-\widehat{s_n }}} } \right)\cdot \left(
{\prod\limits_{n=1}^\infty {\frac{1-\widehat{s_n }-s}{1-\widehat{s_n
}-\frac{1}{2}}} } \right)\\
=\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}}{1-\widehat{s_n }}} \right)} } \right]\cdot \left(
{\prod\limits_{n=1}^\infty {\frac{1-\widehat{s_n }-s}{1-\widehat{s_n
}-\frac{1}{2}}} } \right) \\
 =\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}}{1-\widehat{s_n }}} \right)} } \right]\cdot \left[
{\prod\limits_{n=1}^\infty {\frac{1-\widehat{s_n }-\frac{1}{2}-\left(
{s-\frac{1}{2}} \right)}{1-\widehat{s_n }-\frac{1}{2}}} } \right] \\
 =\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}}{1-\widehat{s_n }}} \right)} } \right]\cdot \left[
{\prod\limits_{n=1}^\infty {\left( {1-\frac{s-\frac{1}{2}}{1-\widehat{s_n
}-\frac{1}{2}}} \right)} } \right]\\
=\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}}{1-\widehat{s_n }}} \right)} } \right]\cdot \left[
{\prod\limits_{n=1}^\infty {\left( {1+\frac{s-\frac{1}{2}}{\widehat{s_n
}-\frac{1}{2}}} \right)} } \right]. \\
 \end{array}
\end{equation}
\end{proof}


\begin{theorem}
\label{T4}
If $\mu $ is defined in (\ref{eq46}), then for $s\in \mathds{C}$ and $n\in \mathds{N}$,
\begin{equation}
\label{eq59}
e^{s\mu }\prod\limits_{n=1}^\infty {e^{s/s_n }} =1,
\end{equation}
where the product runs over zeroes $s_n $ of (\ref{eq4}).
\end{theorem}

\begin{proof}
By (\ref{eq45}), we have
\begin{equation}
\label{eq60}
\xi \left( s \right)=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{s_n }} \right)} ,
\end{equation}
and by (\ref{eq47}), we show
\begin{equation}
\label{eq61}
\xi \left( s \right)=\xi \left( 0 \right)e^{s\mu }\prod\limits_{n=1}^\infty
{\left( {1-\frac{s}{s_n }} \right)e^{s/s_n }} .
\end{equation}
Thus, we deduce from (\ref{eq60}) and (\ref{eq61}) that
\begin{equation}
\label{eq62}
\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left( {1-\frac{s}{s_n }}
\right)} =\xi \left( 0 \right)e^{s\mu }\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{s_n }} \right)e^{s/s_n }} ,
\end{equation}
which implies that
\begin{equation}
\label{eq63}
e^{s\mu }\prod\limits_{n=1}^\infty {e^{s/s_n }} =1.
\end{equation}
So, the proof is completed.
\end{proof}

\subsection{Order, exponent of convergence, and genus}

To show the sufficient condition for the existence of the zeros of the
entire Riemann zeta--function, we now introduce the theory of the entire
functions.

Let
\begin{equation}
\label{eq64}
\mathds{M}\left( r \right)=\mathop {\max }\limits_{r=\left| s \right|} \left|
{\xi \left( s \right)} \right|
\end{equation}
be the maximum modulus of $\xi \left( s \right)$ on a disk of radius $r$.

We denote the order $\vartheta $ of $\xi \left( s \right)$ by
\begin{equation}
\label{eq65}
\vartheta =\mathop {\lim }\limits_{r\to \infty } \sup \frac{\log \log \mathds{
M}\left( r \right)}{\log r}.
\end{equation}
The genus of the canonical product, $\delta $, is defined by a canonical
product
\begin{equation}
\label{eq66}
\prod\limits_{n=1}^\infty {\mathds{E}\left( {s/s_n ,\delta } \right)}
=\prod\limits_{n=1}^\infty {\left( {1-\frac{s}{s_n }} \right)} =\frac{\xi
\left( s \right)}{\xi \left( 0 \right)}.
\end{equation}
By \textit{Lemma \ref{L4}}, we find that $\xi \left( s \right)$ is an entire
function having infinite number of zeros $s_n $. Let
\begin{equation}
\label{eq67}
\mathds{M}=\left\{ {s_n } \right\}_{n=1}^\infty ,\mbox{ }\left| {s_n }
\right|\le \left| {s_{n+1} } \right|\mbox{ }\left( {n\in \mathds{N}} \right)
\end{equation}
be a sequence of complex numbers with $\mathop {\lim }\limits_{n\to \infty }
\left| {s_n } \right|=+\infty $. The exponent of convergence $\nu $ of the
sequence $\mathds{M}$ is defined by
\begin{equation}
\label{eq68}
\nu =\inf \left\{ {\alpha \left| {\left| {s_n } \right|^{-\alpha }<\infty }
\right.,\xi \left( {s_n } \right)=0} \right\}.
\end{equation}

\begin{theorem}
\label{T5}
$\xi \left( s \right)$ is an integral function of order $\vartheta =1$ with
the convergence exponent of their zeros $\nu =1$ and genus $\delta
=0$.
\end{theorem}

\begin{proof}
Applying \textit{Lemmas \ref{L3}} and \textit{\ref{L4}}, we have
\begin{equation}
\label{eq69}
\xi \left( s \right)=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{s_n }} \right)} ,
\end{equation}
where $\xi \left( 0 \right)\ne 0$.

Hadamard proved~\cite{8} that $\xi \left( s \right)$ is an integral function of
finite order with genus $\delta =0$. Titchmarsh~(\cite{32}, p.3) proved that $\xi \left(
s \right)$ is an integral function of order $\vartheta =1$. By applying the
theorem of Borel~(\cite{34}, p.30;~\cite{35}, p.19), we find that an entire function of
order $\vartheta =1$ is equal to the convergence exponent of its zeros,
and we show $\vartheta =\nu =1$.

Thus, the result follows.
\end{proof}

\begin{remark}
By \textit{Theorem \ref{T4}}, we deduce that $\xi \left( s \right)$ is of genus $\delta
=0$ since $\xi \left( s \right)$ is an integral function of order $\vartheta =1$.
\end{remark}

\subsection{Some equivalent classes}

Here, we now present some equivalent classes, which are connected with the
series and product representations of them.

\textit{Class I}

If (\ref{eq40}) is valid, by \textit{Theorem \ref{T1}} and \textit{Lemma \ref{L2}}, there exist
\begin{equation}
\label{eq70}
\sum\limits_{k=0}^\infty {\varphi _k \left( {s-\frac{1}{2}} \right)^{2k}}
=\xi \left( {1/2} \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s-\frac{1}{2}}{s_n -\frac{1}{2}}} \right)} ,
\end{equation}
or, alternatively, by \textit{Lemma \ref{L2}} and \textit{Theorem \ref{T2}},
\begin{equation}
\label{eq71}
\sum\limits_{k=0}^\infty {\varphi _k \left( {s-\frac{1}{2}} \right)^{2k}}
=\xi \left( {\frac{1}{2}} \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{\left( {s-\frac{1}{2}} \right)^2}{\left( {\widehat{s_n
}-\frac{1}{2}} \right)^2}} \right)} ,
\end{equation}
where $\xi \left( {1/2} \right)\ne 0$, $s\in \mathds{C}$ , $k\in \mathds{N}\cup
\left\{ 0 \right\}$, $n\in \mathds{N}$, $\varphi _k $ is defined in (\ref{eq42}),
$\widehat{s_n }$ are defined in (\ref{eq20}), and the product runs over zeroes $s_n
$ of (\ref{eq4}).

\textit{Class II}

If (\ref{eq40}) is valid, by \textit{Lemmas \ref{L2}} and \textit{\ref{L4}}, there is
\begin{equation}
\label{eq72}
\sum\limits_{k=0}^\infty {\varphi _k \left( {s-\frac{1}{2}} \right)^{2k}}
=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left( {1-\frac{s}{s_n }}
\right)} ,
\end{equation}
where $\xi \left( 0 \right)\ne 0$, $s\in \mathds{C}$ , $k\in \mathds{N}\cup
\left\{ 0 \right\}$, $n\in \mathds{N}$, $\varphi _k $ is defined in (\ref{eq42}), and
the product runs over zeroes $s_n $ of (\ref{eq4}).

\textit{Class III}

If (\ref{eq40}) is valid, by \textit{Lemma \ref{L4}} and \textit{Theorems \ref{T1}},
this implies that there exists
\begin{equation}
\label{eq73}
\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left( {1-\frac{s}{s_n }}
\right)} =\xi \left( {1/2} \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s-\frac{1}{2}}{s_n -\frac{1}{2}}} \right)} ,
\end{equation}
where $s\in \mathds{C}$, $n\in \mathds{N}$, $\xi \left( 0 \right)\ne 0$ ($\xi
\left( {1/2} \right)\ne 0)$ , and the products run over
zeroes $s_n $ of (\ref{eq4}).

\section{Theory of the Riemann $\Xi$ function} {\label{sec:3}}

In this section we study the theory of the Riemann $\Xi$ function.
We consider its order, convergence exponent of its zeros,
and genus, theorems of Hardy and Ingham,
Riemann-von Mangoldt formula and Riemann-Siegel formula. Then, we consider
its integral, series and product formulas.
Finally, we study its real zeros
based on the theorem of Hardy and give the proof of the Riemann hypothesis.

Let us recall the Riemann $\Xi$ function, defined by~\cite{2}
\begin{equation}
\label{eq74}
\Xi \left( t \right)=\xi \left( {\frac{1}{2}+it} \right).
\end{equation}

\subsection{Order, convergence exponent, and genus}

Now, we start with its order, convergence exponent, and genus
to study the existence of its zeros.

\begin{theorem}
\label{T6}
Let $t\in \mathds{C}$. Then, $\Xi \left( t \right)$ is an integral function of order
$\vartheta =1$ with the convergence exponent of their zeros $\nu =1_{
}$ and genus $\delta =0$.
\end{theorem}

\begin{proof}
Titchmarsh~(\cite{33}, p.29) proved that $\Xi \left( t \right)$ is an integral function
of order $\vartheta =1$.

By (\ref{eq66}), we know
\begin{equation}
\label{eq75}
\prod\limits_{n=1}^\infty {\mathds{E}\left( {it/\left( {s_n -1/2}
\right),\delta } \right)} =\prod\limits_{n=1}^\infty {\left(
{1-\frac{it}{s_n -\frac{1}{2}}} \right)} =\frac{\Xi \left( t \right)}{\Xi
\left( 0 \right)},
\end{equation}
and we find $\delta =0$.

By applying the theorem of Borel~(\cite{34}, p.30; ~\cite{35}, p.19), we have that an
entire function of order $\vartheta =1$ is equal to the convergence
exponent of its zeros, and we get $\vartheta =\nu =1$~(\cite{33}, p.30)).

Thus, we finish the proof.
\end{proof}

\subsection{Integral representations}

We now report the Riemann, Jensen and Hardy
integral representations of it.

\begin{lemma}
(Riemann integral theorem)
\label{L7}

Let $t\in \mathds{C}$. If $\psi \left( x \right)$ is defined in (\ref{eq8}), then
\begin{equation}
\label{eq76}
\Xi \left( t \right)=4\int\limits_1^\infty {\frac{d\left(
{x^{\frac{3}{2}}\psi ^{\left( 1 \right)}\left( x \right)}
\right)}{dx}x^{-\frac{1}{4}}\cos \left( {\frac{t\log x}{2}} \right)dx}
\end{equation}
or, alternatively,
\begin{equation}
\label{eq77}
\Xi \left( t \right)=\frac{1}{2}-\left( {t^2+\frac{1}{4}}
\right)\int\limits_1^\infty {\psi \left( x \right)x^{-\frac{3}{4}}\cos
\left( {\frac{t}{2}\log x} \right)dx} .
\end{equation}
Moreover,
\begin{equation}
\label{eq78}
\Xi \left( {-t} \right)=\Xi \left( t \right).
\end{equation}
\end{lemma}

\begin{proof}
See~\cite{2}.
\end{proof}

\begin{lemma}
(Jensen integral theorem)
\label{L8}

Let $t\in \mathds{R}$. Then,
\begin{equation}
\label{eq79}
\Xi \left( t \right)=\int\limits_0^\infty {\mathds{k}\left( x \right)\cos
\left( {tx} \right)dx} ,
\end{equation}
where
\begin{equation}
\label{eq80}
\mathds{k}\left( x \right)=4\sum\limits_{n=1}^\infty {\left( {2n^4\pi
^2e^{9x/2}-3n^2\pi e^{5x/2}} \right)e^{-n^2\pi e^{2x}}} .
\end{equation}
\end{lemma}

\begin{proof}
See the work of Jensen~\cite{36,36a}.
\end{proof}

\begin{remark}
Wintner~\cite{37} proved that
\begin{equation}
\label{eq81}
\mathds{k}\left( x \right)>0
\end{equation}
and
\begin{equation}
\label{eq82}
\mathds{k}\left( x \right)=\mathds{k}\left( {-x} \right).
\end{equation}
\end{remark}

\begin{lemma}
(Hardy integral theorem)
\label{L9}

Let $t\in \mathds{R}$. Then,
\begin{equation}
\label{eq83}
\mathds{S}\left( x \right)=\int\limits_0^\infty {\frac{\Xi \left( t
\right)}{t^2+\frac{1}{4}}\cos \left( {tx} \right)dx} ,
\end{equation}
where
\begin{equation}
\label{eq84}
\mathds{S}\left( x \right)=\frac{\pi }{2}\left( {e^{x/2}-2e^{-x/2}\psi \left(
{e^{-2x}} \right)} \right),
\end{equation}
and $\psi \left( x \right)$ is defined in (\ref{eq8}). Moreover,
\begin{equation}
\label{eq85}
\mathds{S}\left( {-x} \right)=\mathds{S}\left( x \right).
\end{equation}
\end{lemma}

\begin{proof}
See~\cite{9,38} and the Titchmarsh's book~(\cite{33}, p.260).
\end{proof}

\begin{remark}
The Riemann integral theorem~\cite{2} was used to obtain the existence of the real
zeros~\cite{3}. The Jensen integral theorem~\cite{36,36a} was presented to get the P\'{o}lya
series theorem~\cite{13}, which is related to the Jensen polynomials for the Riemann
zeta--function~\cite{29a}. The Hardy integral theorem~\cite{9} was considered to obtain the
theorem of Hardy~\cite{9,38}.
\end{remark}

\subsection{Series representations}

We now investigate the series representations for the Riemann $\Xi$ function.

\begin{lemma}
(Hadamard series theorem)
\label{L11}

Let $t\in \mathds{R}$. Then,
\begin{equation}
\label{eq86}
\Xi \left( t \right)=\sum\limits_{k=0}^\infty {\left( {-1} \right)^k\varphi
_k t^{2k}}
\end{equation}
and
\begin{equation}
\label{eq87}
\varphi _k =\frac{D_{2k} }{4}-D_{2k-2} ,
\end{equation}
where
\begin{equation}
\label{eq88}
D_k =\frac{1}{2^{2k}\Gamma \left( {2k+1} \right)}\int\limits_1^\infty {\psi
\left( x \right)x^{-\frac{3}{4}}\left( {\log x} \right)^kdx} .
\end{equation}
\end{lemma}

\begin{proof}
See~\cite{8}.
\end{proof}

\begin{lemma}
(P\'{o}lya series theorem)
\label{L12}

Let $t\in \mathds{R}$. Then,
\begin{equation}
\label{eq89}
\Xi \left( t \right)=\sum\limits_{k=0}^\infty {\left( {-1} \right)^k\varphi
_k t^{2k}}
\end{equation}
and
\begin{equation}
\label{eq90}
\varphi _k =\frac{2}{\pi \Gamma \left( {2k+1} \right)}\int\limits_0^\infty
{\mathds{k}\left( x \right)x^{2k}dx} =\frac{\Xi ^{\left( {2m} \right)}\left( 0
\right)}{\Gamma \left( {2k+1} \right)}.
\end{equation}
\end{lemma}

\begin{proof}
See~\cite{13}.
\end{proof}


\begin{theorem}
\label{T8}
Let $t\in \mathds{R}$. Then,
\begin{equation}
\label{eq91}
\Xi \left( t \right)=\sum\limits_{k=0}^\infty {\left( {-1} \right)^k\varphi
_k t^{2k}}
\end{equation}
and
\begin{equation}
\label{eq92}
\varphi _k =\frac{4}{\Gamma \left( {2k+1} \right)}\int\limits_1^\infty
{\frac{d\left( {x^{\frac{3}{2}}\psi ^{\left( 1 \right)}\left( x \right)}
\right)}{dx}x^{-\frac{1}{4}}\left( {\frac{\log x}{2}} \right)^{2k}dx} .
\end{equation}
\end{theorem}

\begin{proof}
Substituting $s=1/2+it$ into \textit{Lemma \ref{L2}}, the result follows.
\end{proof}

\begin{remark}
The coefficients of the series read
\begin{equation}
\label{eq93}
\begin{array}{l}
 \varphi _k =\frac{D_{2k} }{4}-D_{2k-2} \\
 =\frac{1}{2^{2k}\Gamma \left( {2k+1}
\right)}\int\limits_1^\infty {\psi \left( x \right)x^{-\frac{3}{4}}\left(
{\log x} \right)^kdx} \\
 =\frac{2}{\pi \Gamma \left( {2k+1} \right)}\int\limits_0^\infty {\mathds{
k}\left( x \right)x^{2k}dx} \\
=\frac{\Xi ^{\left( {2k} \right)}\left( 0
\right)}{\Gamma \left( {2k+1} \right)} \\
 =\frac{4}{\Gamma \left( {2k+1} \right)}\int\limits_1^\infty {\frac{d\left(
{x^{\frac{3}{2}}\psi ^{\left( 1 \right)}\left( x \right)}
\right)}{dx}x^{-\frac{1}{4}}\left( {\frac{\log x}{2}} \right)^{2k}dx} \\
 \end{array}
\end{equation}
with $\varphi _k >0$ , where (\ref{eq88}) is valid.

By \textit{Theorem \ref{T6}}, the coefficients of the series, obtained by Hadamard series theorem,
P\'{o}lya series theorem and \textit{Theorem \ref{T8}}, is related to
order $\vartheta$ of the entire function $\Xi \left( t \right)$ by~(\cite{35}, Theorem 2.2.2, p.9)
\[
\vartheta =\mathop {\lim }\limits_{k\to \infty } \sup \frac{\log \log {
k}}{\log{\left(1/{|\varphi _k|}\right)}}=1,
\]
and the convergence exponent of its zeros $\nu$ by
\[
\nu=\mathop {\lim }\limits_{k\to \infty } \sup \frac{\log \log {
k}}{\log{\left(1/{|\varphi _k|}\right)}}=1
\]
if one uses the theorem of Borel~(\cite{34}, p.30; ~\cite{35}, p.19).
Moreover, it is related to the formula of Jensen, which is connected with
the number of zeros of $\Xi \left( t \right)$~(\cite{20b}, p.14).
\end{remark}

\subsection{Product formulas}

We now consider the products for the Riemann $\Xi$ function.

\begin{lemma}
\label{L13}
Let $t\in \mathds{R}$ and $n\in \mathds{N}$. Then
\begin{equation}
\label{eq94}
\Xi \left( t \right)=\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{t^2}{\tau _n^2 }} \right)} ,
\end{equation}
where $\tau _n $ run the positive roots of $\Xi \left( t \right)=0$.
Moreover,
\begin{equation}
\label{eq95}
\log \Xi \left( t \right)=\log \Xi \left( 0 \right)+\log
\sum\limits_{n=1}^\infty {\left( {1-\frac{t^2}{\tau _n^2 }} \right)} .
\end{equation}
\end{lemma}

\begin{proof}
(\ref{eq94}) was reported by Cahen~\cite{14} and Titchmarsh~\cite{15}. (\ref{eq95}) was proposed
by Riemann~\cite{2}.
\end{proof}

\begin{lemma}
\label{L14}
Let $t\in \mathds{R}$ and $n\in \mathds{ N}$. Then
\begin{equation}
\label{eq96}
\Xi \left( t \right)=\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{t}{t_n }} \right)} ,
\end{equation}
where $t_n $ run the roots of $\Xi \left( t \right)=0$.
\end{lemma}

\begin{proof}
(\ref{eq96}) was proposed by Hadamard~\cite{8} and discussed by Landau (\cite{6}, p.199).
\end{proof}

\begin{theorem}
\label{T9}
Let $t\in \mathds{R}$ and $n\in \mathds{N}$. Then
\begin{equation}
\label{eq97}
\Xi \left( t \right)=\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{it}{s_n -\frac{1}{2}}} \right)},
\end{equation}
where $s_n $ run the roots of $\xi \left( s \right)=0$.
\end{theorem}

\begin{proof}
Taking $s=1/2+it$ into \textit{Lemma \ref{L4}}, we present
\begin{equation}
\label{eq98}
\Xi \left( t \right)=\xi \left( {1/2+it} \right)=\xi \left( 0
\right)\prod\limits_{n=1}^\infty {\left( {1-\frac{\frac{1}{2}+it}{s_n }}
\right)} .
\end{equation}
Hence,
\begin{equation}
\label{eq99}
\begin{array}{l}
 \Xi \left( t \right)
 =\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}+it}{s_n }} \right)} \\
=\xi \left( 0
\right)\prod\limits_{n=1}^\infty {\left( {\frac{s_n -\frac{1}{2}-it}{s_n }}
\right)} \\
=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left( {\frac{s_n
-\frac{1}{2}}{s_n -\frac{1}{2}}\cdot \frac{s_n -\frac{1}{2}-it}{s_n }}
\right)} \\
 =\xi \left( 0 \right)\left( {\prod\limits_{n=1}^\infty {\frac{s_n
-\frac{1}{2}}{s_n }} } \right)\cdot \left( {\prod\limits_{n=1}^\infty
{\frac{s_n -\frac{1}{2}-it}{s_n -\frac{1}{2}}} } \right)\\
 =\xi \left( 0 \right)\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{1}{2s_n }} \right)} } \right]\cdot \left[
{\prod\limits_{n=1}^\infty {\frac{s_n -\frac{1}{2}-it}{s_n -\frac{1}{2}}} }
\right]\\
=\xi \left( {\frac{1}{2}} \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{it}{s_n -\frac{1}{2}}} \right)} \\
=\Xi \left( 0
\right)\prod\limits_{n=1}^\infty {\left( {1-\frac{it}{s_n -\frac{1}{2}}}
\right)} , \\
 \end{array}
\end{equation}
and the required result follows.
\end{proof}

\begin{theorem}
\label{T10}
If $\mu $ is defined in (\ref{eq46}), then for $t\in \mathds{R}$ and $n\in \mathds{N}$,
\begin{equation}
\label{eq100}
\Xi \left( s \right)=\Xi \left( 0 \right)e^{\left( {1/2+it} \right)\mu
}\prod\limits_{n=1}^\infty {\left( {1-\frac{it}{s_n -\frac{1}{2}}}
\right)e^{\left( {1/2+it} \right)/s_n }} ,
\end{equation}
where the product runs over zeroes $s_n $ of (\ref{eq4}).
\end{theorem}

\begin{proof}
Taking $s=1/2+it$ into \textit{Lemma \ref{L5}}, we give
\begin{equation}
\label{eq101}
\Xi \left( t \right)=\xi \left( {1/2+it} \right)=\xi \left( 0 \right)e^{\mu
\left( {1/2+it} \right)}\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}+it}{s_n }} \right)e^{\left( {1/2+it} \right)/s_n }} .
\end{equation}
Hence, by \textit{Theorem \ref{T9}}, we show
\begin{equation}
\label{eq102}
\begin{array}{l}
 \Xi \left( t \right)\\
 =\xi \left( 0 \right)e^{\mu \left( {1/2+it}
\right)}\prod\limits_{n=1}^\infty {\left( {1-\frac{\frac{1}{2}+it}{s_n }}
\right)e^{\left( {1/2+it} \right)/s_n }} \\
=\left( {e^{\mu \left( {1/2+it}
\right)}\prod\limits_{n=1}^\infty {e^{\left( {1/2+it} \right)/s_n }} }
\right)\cdot \left[ {\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}+it}{s_n }} \right)} } \right] \\
 =\left( {e^{\mu \left( {1/2+it} \right)}\prod\limits_{n=1}^\infty
{e^{\left( {1/2+it} \right)/s_n }} } \right)\cdot \left[ {\Xi \left( 0
\right)\prod\limits_{n=1}^\infty {\left( {1-\frac{it}{s_n -\frac{1}{2}}}
\right)} } \right]\\
=\Xi \left( 0 \right)e^{\left( {1/2+it} \right)\mu
}\prod\limits_{n=1}^\infty {\left( {1-\frac{it}{s_n -\frac{1}{2}}}
\right)e^{\left( {1/2+it} \right)/s_n }} . \\
 \end{array}
\end{equation}
Thus, we finish the proof.
\end{proof}

\begin{theorem}
\label{T11}
Let $t\in \mathds{R}$ and $n\in \mathds{N}$. Then
\begin{equation}
\label{eq103}
\Xi \left( t \right)=\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left[
{1+\frac{t^2}{\left( {\widehat{s_n }-\frac{1}{2}} \right)^2}} \right]} ,
\end{equation}
where the product runs over $\widehat{s_n }$, which are defined in (\ref{eq20}).
\end{theorem}

\begin{proof}
Taking $s=1/2+it$ into (\ref{eq49}), we arrive at
\begin{equation}
\label{eq104}
\Xi \left( t \right)=\xi \left( {1/2+it}
\right)=\frac{1}{2}\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}+it}{\widehat{s_n }}} \right)} \prod\limits_{n=1}^\infty
{\left( {1-\frac{\frac{1}{2}+it}{1-\widehat{s_n }}} \right)} .
\end{equation}
Therefore,
\begin{equation}
\label{eq105}
\begin{array}{l}
 \xi \left( s \right)\\
 =\frac{1}{2}\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}+it}{\widehat{s_n }}} \right)} \prod\limits_{n=1}^\infty
{\left( {1-\frac{\frac{1}{2}+it}{1-\widehat{s_n }}} \right)} \\
 =\frac{1}{2}\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{1}{2\widehat{s_n }}} \right)} \cdot \prod\limits_{n=1}^\infty
{\left( {1-\frac{it}{\widehat{s_n }-\frac{1}{2}}} \right)} } \right]\cdot \left[
{\prod\limits_{n=1}^\infty {\left( {1-\frac{\frac{1}{2}}{1-\widehat{s_n }}}
\right)} \cdot \prod\limits_{n=1}^\infty {\left( {1+\frac{it}{\widehat{s_n
}-\frac{1}{2}}} \right)} } \right] \\
 =\frac{1}{2}\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{1}{2\widehat{s_n }}} \right)} \cdot \prod\limits_{n=1}^\infty
{\left( {1-\frac{\frac{1}{2}}{1-\widehat{s_n }}} \right)} } \right]\cdot \left[
{\prod\limits_{n=1}^\infty {\left( {1-\frac{it}{\widehat{s_n }-\frac{1}{2}}}
\right)} \cdot \prod\limits_{n=1}^\infty {\left( {1+\frac{it}{\widehat{s_n
}-\frac{1}{2}}} \right)} } \right] \\
 =\frac{1}{2}\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{1}{2\widehat{s_n }}} \right)} \cdot \prod\limits_{n=1}^\infty
{\left( {1-\frac{\frac{1}{2}}{1-\widehat{s_n }}} \right)} }
\right] \cdot \left[ \prod\limits_{n=1}^\infty {{\left( {1-\frac{it}{\widehat{s_n
}-\frac{1}{2}}} \right)\left( {1+\frac{it}{\widehat{s_n }-\frac{1}{2}}}
\right)}} \right]\\
 =\frac{1}{2}\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{1}{2\widehat{s_n }}} \right)} \cdot \prod\limits_{n=1}^\infty
{\left( {1-\frac{\frac{1}{2}}{1-\widehat{s_n }}} \right)} } \right]\cdot \left[
\prod\limits_{n=1}^\infty {\left( {1+\frac{t^2}{\left( {\widehat{s_n
}-\frac{1}{2}} \right)^2}} \right)} \right]\\
 =\xi \left( {\frac{1}{2}} \right)\prod\limits_{n=1}^\infty {\left(
{1+\frac{t^2}{\left( {\widehat{s_n }-\frac{1}{2}} \right)^2}} \right)} \\
=\Xi
\left( 0 \right)\prod\limits_{n=1}^\infty {\left[ {1+\frac{t^2}{\left(
{\widehat{s_n }-\frac{1}{2}} \right)^2}} \right]} , \\
 \end{array}
\end{equation}
where
\begin{equation}
\label{eq106}
\begin{array}{l}
 \prod\limits_{n=1}^\infty {\left( {1-\frac{\frac{1}{2}+it}{\widehat{s_n }}}
\right)}\\
=\prod\limits_{n=1}^\infty {\left( {\frac{\widehat{s_n
}-\frac{1}{2}-it}{\widehat{s_n }}} \right)} \\
=\prod\limits_{n=1}^\infty
{\left( {\frac{\widehat{s_n }-\frac{1}{2}}{\widehat{s_n }-\frac{1}{2}}\cdot
\frac{\widehat{s_n }-\frac{1}{2}-it}{\widehat{s_n }}} \right)}\\
=\left(
{\prod\limits_{n=1}^\infty {\frac{\widehat{s_n }-\frac{1}{2}}{\widehat{s_n
}}} } \right)\cdot \left( {\prod\limits_{n=1}^\infty {\frac{\widehat{s_n
}-\frac{1}{2}-it}{\widehat{s_n }-\frac{1}{2}}} } \right)\\
=\left[
{\prod\limits_{n=1}^\infty {\left( {1-\frac{1}{2\widehat{s_n }}} \right)} }
\right]\cdot \left( {\prod\limits_{n=1}^\infty {\frac{\widehat{s_n
}-\frac{1}{2}-it}{\widehat{s_n }-\frac{1}{2}}} } \right) \\
 =\left[ {\prod\limits_{n=1}^\infty {\left( {1-\frac{1}{2\widehat{s_n }}}
\right)} } \right]\cdot \left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{it}{\widehat{s_n }-\frac{1}{2}}} \right)} } \right] \\
 \end{array}
\end{equation}
and
\begin{equation}
\label{eq107}
\begin{array}{l}
 \prod\limits_{n=1}^\infty {\left( {1-\frac{\frac{1}{2}+it}{1-\widehat{s_n
}}} \right)}\\
=\prod\limits_{n=1}^\infty {\left( {\frac{1-\widehat{s_n
}-\frac{1}{2}-it}{1-\widehat{s_n }}} \right)} \\
=\prod\limits_{n=1}^\infty
{\left( {\frac{1-\widehat{s_n }-\frac{1}{2}}{1-\widehat{s_n
}-\frac{1}{2}}\cdot \frac{1-\widehat{s_n }-\frac{1}{2}-it}{1-\widehat{s_n
}}} \right)} \\
 =\left( {\prod\limits_{n=1}^\infty {\frac{1-\widehat{s_n
}-\frac{1}{2}}{1-\widehat{s_n }}} } \right)\cdot \left(
{\prod\limits_{n=1}^\infty {\frac{1-\widehat{s_n
}-\frac{1}{2}-it}{1-\widehat{s_n }-\frac{1}{2}}} } \right)\\
=\left[
{\prod\limits_{n=1}^\infty {\left( {1-\frac{\frac{1}{2}}{1-\widehat{s_n }}}
\right)} } \right]\cdot \left[ {\prod\limits_{n=1}^\infty
{\frac{1-\widehat{s_n }-\frac{1}{2}-it}{1-\widehat{s_n }-\frac{1}{2}}} }
\right] \\
 =\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}}{1-\widehat{s_n }}} \right)} } \right]\cdot \left[
{\prod\limits_{n=1}^\infty {\left( {1-\frac{it}{1-\widehat{s_n
}-\frac{1}{2}}} \right)} } \right]\\
=\left[ {\prod\limits_{n=1}^\infty {\left(
{1-\frac{\frac{1}{2}}{1-\widehat{s_n }}} \right)} } \right]\cdot \left[
{\prod\limits_{n=1}^\infty {\left( {1+\frac{it}{\widehat{s_n }-\frac{1}{2}}}
\right)} } \right]. \\
 \end{array}
\end{equation}
Thus, the proof is completed.
\end{proof}

\subsection{Well-known results}

By \textit{Theorem \ref{T6}}, it is seen that $\Xi \left( t \right)$ has infinity
many zeros. We need to know $\Xi \left( t \right)$ has an infinity of real
zeros. We now investigate the Riemann-von Mangoldt formula and
Riemann-Siegel formula and theorems of Hardy and Ingham.

\begin{lemma}
(Theorem of Hardy)
\label{L15}

There are infinitely many zeros of $\zeta \left( s \right)$ on the critical
line $Re\left( s \right)=1/2$.
\end{lemma}

\begin{proof}
See~\cite{9} and~(\cite{33}, p.254-282).
\end{proof}

\begin{lemma}
(Theorem of Ingham)
\label{L16}

There are infinitely many zeros of $\xi \left( s \right)$ on the critical
line $Re\left( s \right)=1/2$.
\end{lemma}

\begin{proof}
See the proof~(\cite{31}, p.48).
\end{proof}

\begin{lemma}
(Riemann-von Mangoldt formula)
\label{L17}

Let
\begin{equation}
\label{eq108}
N_0 \left( T \right)=\# \left\{ {s=1/2+it:0\le t\le T,\xi \left( s
\right)=0} \right\}.
\end{equation}
Then,
\begin{equation}
\label{eq109}
N_0 \left( T \right)\mbox{=}\frac{T}{2\pi }\log \frac{T}{2\pi
e}\mbox{+}O\left( {\log T} \right).
\end{equation}
\end{lemma}

\begin{proof}
See the work of von Mangoldt~\cite{22}.
\end{proof}

\begin{lemma}
(Riemann-Siegel formula)
\label{L18}

Let
\begin{equation}
\label{eq110}
e^{i\mathds{A}\left( t \right)}\zeta \left( {\frac{1}{2}+it}
\right)=2\sum\limits_{j=1}^{g\left( t \right)} {\frac{1}{\sqrt j }\cos
\left( {\mathds{A}\left( t \right)-t\log j} \right)} +\mathds{S}\left( t \right),
\end{equation}
where $t>0$,
\begin{equation}
\label{eq111}
\mbox{ }g\left( t \right)=\left[ {\sqrt {\frac{t}{2\pi }} } \right],
\end{equation}
\begin{equation}
\label{eq112}
\mathds{A}\left( t \right)=-\frac{t}{2}\log \pi +\arg \Gamma \left(
{\frac{1}{4}+\frac{it}{2}} \right)
\end{equation}
and
\begin{equation}
\label{eq113}
\mathds{S}\left( t \right)=\mbox{O}\left( {1/\sqrt[4]{t}} \right).
\end{equation}
Then, the Riemann-Siegel zeros $\tau _n $ for $n\in \mathds{N}$ are determined
by (\ref{eq110}).
\end{lemma}

\begin{proof}
See the work of Siegel~\cite{23}.
\end{proof}

\subsection{All zeros are real}

\textit{Lemmas \ref{L13}} and \textit{\ref{L14}} showed $\Xi \left( t \right)$ has an
infinity of real zeros, which was not proved by Riemann~\cite{2}, Cahen~\cite{14}, and
Titchmarsh~\cite{15}. This problem was reported by Titchmarsh~(\cite{32}, p.44; \cite{33}, p.255).
\textit{Theorem \ref{T6}} gives us to show that $\Xi \left( t
\right)$ has an infinity of zeros. The theorems of Hardy and Ingham imply that
infinitely many zeros lie on the critical line $Re\left( s \right)=1/2$ but
do not confirm all zeros lie on the critical line $Re\left( s \right)=1/2$.
Riemann and Siegel~\cite{23} discovered the Riemann-Siegel zeros $t_n $. Riemann
and von Mangoldt~\cite{22} presented that the number of the zeros of $\Xi \left(
t \right)$. The Jensen polynomials for the Riemann zeta--function have real zeros~\cite{29a} but
the case of all zeros of $\Xi \left( t \right)$ has not been known.
However, this statement of the conjecture of Jensen~\cite{7,25c,29a}, that all the zeros of $\Xi \left( t \right)$ are real,
remains unproved to this day.
Now, our aim is to prove it.

\begin{theorem}
(Conjecture of Jensen)
\label{T12}

Let $t\in \mathds{R}$. Then, all zeros of $\Xi \left( t \right)$ are real.
\end{theorem}

\begin{proof}
Now, by \textit{Theorem \ref{T9}}, we structure
\begin{equation}
\label{eq114}
\Xi \left( t \right)=\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{it}{s_n -\frac{1}{2}}} \right)}
\end{equation}
where $t\in \mathds{R}$ and $n\in \mathds{N}$.

To prove it, our aim is to consider two cases are that all zeros are complex number
$\widehat{s_n }-\frac{1}{2}=\sigma _n +ig_n$ for $\sigma _n \in \mathds{R}\backslash \left\{ 0 \right\}$
and $g_n \in \mathds{R}\backslash \left\{ 0 \right\}$,
and that $\widehat{s_n }-\frac{1}{2}=f_n$ for $f_n \in \mathds{R}\backslash \left\{ 0 \right\}$.
If they are false, all zeros of $\Xi \left( t \right)$ are real.

Now, we begin with the followings:

\textit{Case I.}
We consider the hypothesis that $\Xi \left( t \right)$ has the complex zeros, given by
\begin{equation}
\label{eq115}
\widehat{s_n }-\frac{1}{2}=\sigma _n +ig_n,
\end{equation}
where $\sigma _n \in \mathds{R}\backslash \left\{ 0 \right\}$ and $g_n \in \mathds{R}\backslash \left\{ 0 \right\}$.

Putting (\ref{eq115}) into (\ref{eq114}), we get
\begin{equation}
\label{eq116}
\Xi \left( t \right)=\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{it}{\sigma _n +ig_n }} \right)} .
\end{equation}
Substituting $t=\left( {s-1/2} \right)/i$ into (\ref{eq116}), we have
\begin{equation}
\label{eq117}
\xi \left( s \right)=\Xi \left( {\left( {s-1/2} \right)/i} \right)=\Xi
\left( 0 \right)\prod\limits_{n=1}^\infty {\left( {1-\frac{s-1/2}{\sigma _n
+ig_n }} \right)} =\xi \left( {\frac{1}{2}} \right)\prod\limits_{n=1}^\infty
{\left( {1-\frac{s-1/2}{\sigma _n +ig_n }} \right)} .
\end{equation}
Since $s_n =Re\left( {s_n } \right)+it_n $ with $\xi \left( {s_n }
\right)=0$, where $Re\left( {s_n } \right)\in \mathds{R}\backslash \left\{ 0
\right\}$ and $t_n \in \mathds{R}\backslash \left\{ 0 \right\}$, (\ref{eq117}) can be
rewritten as
\begin{equation}
\label{eq118}
\xi \left( {\frac{1}{2}} \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s_n -1/2}{\sigma _n +ig_n }} \right)} =0,
\end{equation}
which leads to
\begin{equation}
\label{eq119}
\sigma _n +ig_n -\left( {s_n -1/2} \right)=0.
\end{equation}
For $\sigma _n \in \mathds{R}\backslash \left\{ 0 \right\}$ and $g_n \in \mathds{R}\backslash \left\{ 0 \right\}$ we get
\begin{equation}
\label{eq120}
s_n =Re\left( {s_n } \right)+it_n =\sigma _n +1/2+ig_n.
\end{equation}
By $\sigma _n \in \mathds{R}\backslash \left\{ 0 \right\}$ and (\ref{eq120}),
we show that the critical line is $Re\left( {s_n } \right)=\sigma _n
+1/2$, which is in contradiction with the fact that the theorems of Hardy and Ingham
imply that infinitely many zeros lie on the critical line $Re\left( s
\right)=1/2$.

\textit{Case II.}
Now, we give the hypothesis that $\Xi \left( t \right)$ has the zeros
\begin{equation}
\label{eq121}
\widehat{s_n }-\frac{1}{2}=f_n,
\end{equation}
where $f_n \in \mathds{R}\backslash \left\{ 0 \right\}$.

In a similar way, substituting (\ref{eq121}) into (\ref{eq114}), we suggest that
\begin{equation}
\label{eq122}
\Xi \left( t \right)=\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{it}{f_n }} \right)} .
\end{equation}
Combining $t=\left( {s-1/2} \right)/i$ and (\ref{eq122}), we arrive at
\begin{equation}
\label{eq123}
\xi \left( s \right)=\Xi \left( {\left( {s-1/2} \right)/i} \right)=\Xi
\left( 0 \right)\prod\limits_{n=1}^\infty {\left( {1-\frac{s-1/2}{f_n }}
\right)} =\xi \left( {\frac{1}{2}} \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s-1/2}{f_n }} \right)} .
\end{equation}
Let $s_n =Re\left( {s_n } \right)+it_n $ with $Re\left( {s_n } \right)\in
\mathds{R}\backslash \left\{ 0 \right\}$ and $t_n \in \mathds{R}\backslash \left\{
0 \right\}$ such that $\xi \left( {s_n } \right)=0$.

From (\ref{eq123}) we have
\begin{equation}
\label{eq124}
\xi \left( {\frac{1}{2}} \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s_n -1/2}{f_n }} \right)} =0,
\end{equation}
which yields that
\begin{equation}
\label{eq125}
f_n -\left( {s_n -1/2} \right)=0.
\end{equation}
Since $f_n \in \mathds{R}\backslash \left\{ 0 \right\}$, by (\ref{eq125}), we show
\begin{equation}
\label{eq126}
s_n =f_n +1/2\in \mathds{R},
\end{equation}
which is in contradiction with the fact $s_n =Re\left( {s_n } \right)+it_n $
with $Re\left( {s_n } \right)\in \mathds{R}\backslash \left\{ 0 \right\}$ and
$t_n \in \mathds{R}\backslash \left\{ 0 \right\}$ since (\ref{eq126}) do not belong to the nontrivial zeros of $\zeta \left( s \right)$. Moreover,
(\ref{eq126}) are also in contradiction with
the theorems of Hardy and Ingham.

Because two cases are false, all zeros of $\Xi \left( t \right)$ are real.

Thus, the result follows.
\end{proof}

\subsection{A detailed proof of Conjecture \ref{C1}}

Here, our target is to prove that $Re\left( {\widehat{s_n }} \right)=1/2$ and
$Re\left( {s_n } \right)=1/2$ since all zeros of $\Xi \left( t \right)$ are
real.

By \textit{Theorem \ref{T10}}, we structure for $t\in \mathds{R}$ and $n\in \mathds{N}$,
\begin{equation}
\label{eq127}
\Xi \left( t \right)=\Xi \left( 0 \right)e^{\left( {1/2+it} \right)\mu
}\prod\limits_{n=1}^\infty {\left( {1-\frac{it}{s_n -\frac{1}{2}}}
\right)e^{\left( {1/2+it} \right)/s_n }} ,
\end{equation}
where $\mu $ is defined in (\ref{eq46})
and the product runs over zeroes $s_n $ of (\ref{eq4}).

Combining (\ref{eq127}) and \textit{Theorem \ref{T12}}, we find that there exist real
zeros $\widetilde{\tau _n }\in \mathds{ R}$ of $\Xi \left( t \right)$ such that
\begin{equation}
\label{eq128}
\Xi \left( {\widetilde{\tau _n }} \right)
=\Xi \left( 0 \right)e^{\left( {1/2+i\widetilde{\tau _n}} \right)\mu
}\prod\limits_{n=1}^\infty {\left( {1-\frac{i\widetilde{\tau _n
}}{s_n -\frac{1}{2}}}
\right)e^{\left( {1/2+i\widetilde{\tau _n
}} \right)/s_n }} =0,
\end{equation}
where $\mu $ is defined in (\ref{eq46})
and the product runs over zeroes $s_n $ of (\ref{eq4}).

Now, by (\ref{eq128}),
\begin{equation}
\label{eq129}
1-\frac{i\widetilde{\tau _n }}{ {{s_n }-\frac{1}{2}}}=0,
\end{equation}
which implies that
\begin{equation}
\label{eq130}
{s_n }=\frac{1}{2}+i\widetilde{\tau _n }.
\end{equation}
With (\ref{eq130}), we have
\begin{equation}
\label{eq131}
Re\left( {{s_n }} \right)=\frac{1}{2},
\end{equation}
and
\begin{equation}
\label{eq132}
Im\left( {{s_n }} \right)=t_n=\widetilde{\tau _n }.
\end{equation}
Thus,
\begin{equation}
\label{eq133}
\tau _n=\mid{t_n}\mid=\mid{\widetilde{\tau _n }}\mid
\end{equation}
and
\begin{equation}
\label{eq134}
s_n =\frac{1}{2}+it_n=\frac{1}{2}+i\widetilde{\tau _n }=\frac{1}{2}\pm \tau _n .
\end{equation}
By \textit{Lemma \ref{L18}}, the Riemann-Siegel formula gives all values of (\ref{eq132})
and the number of all real zeros is confirmed by the Riemann-von Mangoldt
formula.

Thus, we finish the proof of \textit{Conjecture \ref{C1}}.

\begin{remark}
By \textit{Theorem \ref{T9}}, we also obtain the same result.
\end{remark}

\subsection{Tur\'{a}n inequalities}

The Tur\'{a}n inequalities was proved by many researchers~\cite{11,12,13}.
Here, we give a new proof of the Tur\'{a}n inequalities from the point of
view of theory of the entire functions.

\begin{theorem}
(Tur\'{a}n inequalities)
\label{T13}

Let $k\in \mathds{ N}\cup \left\{ 0 \right\}$. Then,
\begin{equation}
\label{eq135}
\varphi _k ^2-\left( {\frac{2k-1}{2k+1}} \right)\varphi _{k-1} \varphi
_{k+1} >0.
\end{equation}
\end{theorem}

\begin{proof}
By \textit{Theorem \ref{T6}}, $\Xi \left( t \right)$ is an even integral function
of order $\vartheta =1$ and real for real $t$. With \textit{Theorem \ref{T12}},
$\Xi \left( t \right)$ has only real zeros. From \textit{Lemma \ref{L11}},
\textit{Theorem \ref{T8}} and \textit{Lemma \ref{L12}}, we show
\begin{equation}
\label{eq136}
\Xi \left( t \right)=\sum\limits_{k=0}^\infty {\left( {-1} \right)^k\varphi
_k t^{2k}} .
\end{equation}
By theory of the entire functions~(\cite{35}, Theorem 2.8.2., p.24), we arrive at

\[
\varphi _k ^2-\left( {\frac{2k-1}{2k+1}} \right)\varphi _{k-1} \varphi
_{k+1} >0.
\]

Thus, the proof is completed.
\end{proof}

\subsection{Proofs of Lemmas \ref{T13} and \ref{L14}}

There is the exact significance of this remark, reported by Titchmarsh~(\cite{32}, p.44; \cite{33}, p.255),
that Riemann~\cite{2} considered that $\tau_n$ are real. Riemann~\cite{2} and Cahen~\cite{14} did not prove it.
Hadamard~\cite{8} and Landau (\cite{6}, p.199) did not report the proof of it. Here, we give their proofs.

At first, we prove \textit{Lemma \ref{L13}}.

Putting (\ref{eq133}) into \textit{Theorem \ref{T11}}, we show

\[
\begin{array}{l}
\Xi \left( t \right)
=\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left[
{1+\frac{t^2}{\left( {\widehat{s_n }-\frac{1}{2}} \right)^2}} \right]} \\
=\Xi
\left( 0 \right)\prod\limits_{n=1}^\infty {\left\{ {1+\frac{t^2}{\left[
{\left( {\frac{1}{2}+i\tau _n } \right)-\frac{1}{2}} \right]^2}} \right\}}\\
=\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left( {1-\frac{t^2}{\tau
_n^2 }} \right)}
\end{array}
\]
and
\[
\begin{array}{l}
 \log \Xi \left( t \right)
 =\log \Xi \left( 0 \right)+\log
\sum\limits_{n=1}^\infty {\left[ {1+\frac{t^2}{\left( {\widehat{s_n
}-\frac{1}{2}} \right)^2}} \right]} \\
 =\log \Xi \left( 0 \right)+\log \sum\limits_{n=1}^\infty {\left\{
{1+\frac{t^2}{\left[ {\left( {\frac{1}{2}+i\tau _n } \right)-\frac{1}{2}}
\right]^2}} \right\}} \\
=\log \Xi \left( 0 \right)+\log
\sum\limits_{n=1}^\infty {\left( {1-\frac{t^2}{\tau _n^2 }} \right)} , \\
\end{array}
\]
which are the required result.

Now, we give the proof of \textit{Lemma \ref{L14}}.

Combining (\ref{eq134}) and \textit{Theorem \ref{T9}}, we give

\[
\begin{array}{l}
\Xi \left( t \right)\\
=\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{it}{s_n -\frac{1}{2}}} \right)} \\
=\Xi \left( 0
\right)\prod\limits_{n=1}^\infty {\left[ {1-\frac{it}{\left(
{\frac{1}{2}+it_n } \right)-\frac{1}{2}}} \right]} \\
=\Xi \left( 0
\right)\prod\limits_{n=1}^\infty {\left( {1-\frac{it}{it_n }} \right)} \\
=\Xi
\left( 0 \right)\prod\limits_{n=1}^\infty {\left( {1-\frac{t}{t_n }}
\right)},
\end{array}
\]

and the required result follows.

\subsection{A equivalent class}

We now discuss a equivalent class, which is connected with the series and
product representations.

By \textit{Lemmas \ref{L11}} and \textit{\ref{L14}}, there exist

\[
\sum\limits_{k=0}^\infty {\left( {-1} \right)^k\varphi _k t^{2k}} =\Xi
\left( {0} \right)\prod\limits_{n=1}^\infty {\left( {1-\frac{t}{t_n }}
\right)} ,
\]
or, alternatively, by \textit{Lemmas \ref{L11}} and \textit{\ref{L13}},
\[
\sum\limits_{k=0}^\infty {\left( {-1} \right)^k\varphi _k t^{2k}} =\Xi
\left( 0 \right)\prod\limits_{n=1}^\infty {\left( {1-\frac{t^2}{\tau _n^2 }}
\right)} ,
\]
where $\Xi \left( 0 \right)\ne 0$, $s\in \mathds{C}$ , $k\in \mathds{N}\cup
\left\{ 0 \right\}$, $n\in \mathds{N}$, $\varphi _k $ is defined in (\ref{eq93}),
$t_n $ run the roots of $\Xi \left( t \right)=0$ and $\tau _n $ run the
positive roots of $\Xi \left( t \right)=0$.

\section{Comments} {\label{sec:4}}

\subsection{Nontrivial zeros}

By applying \textit{Lemma \ref{L13}} and \textit{Theorem \ref{T11}}, we also obtain
\begin{equation}
\label{eq137}
\Xi \left( t \right)=\Xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left[
{1+\frac{t^2}{\left( {\widehat{s_n }-\frac{1}{2}} \right)^2}} \right]} =\Xi
\left( 0 \right)\prod\limits_{n=1}^\infty {\left( {1-\frac{t^2}{\tau _n^2 }}
\right)} ,
\end{equation}
which yields that
\begin{equation}
\label{eq138}
\widehat{s_n }=\frac{1}{2}\pm \tau _n .
\end{equation}
It is observed that $s_n =1/2+it_n $, $1-s_n =1/2-it_n $, $\overline {s_n }
=1/2-it_n $ and $1-\overline {s_n } =1/2+it_n $ are the zeros (nontrivial zeros) for the entire
Riemann zeta--function, where $t_n =\pm \tau _n $ since all zeros are nontrivial zeros.

\subsection{Some equivalence theorems }
We now present some equivalence theorems associated with the Riemann zeta--function.
\begin{theorem}
(Equivalence theorem I)
\label{T14}

Let $s\in \mathds{C}$ and $n\in\mathds{N}$. Then the following representations
are equivalent:
\begin{description}
\item[A1]
\begin{equation}
\label{eq139}
\xi \left( s \right)=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{s_n }} \right)} ,
\end{equation}
where the product runs over zeroes $s_n $ of (\ref{eq4});
 \item[A2]
\begin{equation}
\label{eq140}
\xi \left( s \right)=\xi \left( {1/2} \right)\prod\limits_{n=1}^\infty
{\left( {1+\frac{i\left( {s-1/2} \right)}{t_n }} \right)} ,
\end{equation}
where $t_n $ run the roots of $\Xi \left( t \right)=0$;
 \item[A3]
\begin{equation}
\label{eq141}
\xi \left( s \right)=\xi \left( 0 \right)e^{s\mu }\prod\limits_{n=1}^\infty
{\left( {1-\frac{s}{s_n }} \right)e^{s/s_n }} ,
\end{equation}
where $\mu $ is defined in (\ref{eq46}) and the product runs over zeroes $s_n
$ of (\ref{eq4});
 \item[A4]
\begin{equation}
\label{eq142}
\xi \left( s \right)=\xi \left( {\frac{1}{2}} \right)e^{s\mu
}\prod\limits_{n=1}^\infty {\left( {1-\frac{s-\frac{1}{2}}{it_n }}
\right)e^{s/s_n }} ,
\end{equation}
where $\mu $ is defined in (\ref{eq46}), $t_n $ run the roots of $\Xi \left( t
\right)=0$, and the product runs over zeroes $s_n $ of (\ref{eq4});
 \item[A5]
\begin{equation}
\label{eq143}
\xi \left( s \right)=\xi \left( 0 \right)\prod\limits_{n=1}^\infty {\left(
{1-\frac{s}{\widehat{s_n }}} \right)} \left( {1-\frac{s}{1-\widehat{s_n }}}
\right),
\end{equation}
where the product runs over zeroes $s_n $ of (\ref{eq4});
 \item[A6]
\begin{equation}
\label{eq144}
\xi \left( s \right)=\xi \left( {1/2} \right)\prod\limits_{n=1}^\infty
{\left( {1+\frac{\left( {s-1/2} \right)^2}{\tau _n^2 }} \right)} ,
\end{equation}
where $\tau _n $ run the positive roots of $\Xi \left( t \right)=0$;
 \item[A7]
\begin{equation}
\label{eq145}
\xi \left( s \right)=4\int\limits_1^\infty {\frac{d\left(
{x^{\frac{3}{2}}\psi ^{\left( 1 \right)}\left( x \right)}
\right)}{dx}x^{-\frac{1}{4}}\cosh \left( {\frac{\left( {s-1} \right)\log
x}{2}} \right)dx} .
\end{equation}
where$_{ }\psi \left( x \right)$ is defined in (\ref{eq8});
 \item[A8]
\begin{equation}
\label{eq146}
\xi \left( s \right)=\sum\limits_{k=0}^\infty {\varphi _k \left(
{s-\frac{1}{2}} \right)^{2k}} ,
\end{equation}
where $\varphi _k $ are expressed in (\ref{eq93}).
\end{description}
\end{theorem}

\begin{proof}
$A1$ is the product of Hadamard~\cite{8}. $A2$ is the consequence
of \textit{Theorem \ref{T1}} and \textit{Conjecture \ref{C1}}. $A3$ is the product of
Hadamard, which was proved in~\cite{8} and the book of Ingham~(\cite{31}, p.57-58). $A4$ is derived from
\textit{Theorem \ref{T2}} and \textit{Conjecture \ref{C1}}. $A5$ is the product of
Patterson~(\cite{16}, p.34). $A6$ is deduced from \textit{Theorem \ref{T3}} and
\textit{Conjecture \ref{C1}}. $A7$ is the Riemann integral representation theorem
~\cite{2}. $A8$ is the Edwards series theorem, where $\varphi _k$ are
expressed in (\ref{eq93}).
\end{proof}

\begin{theorem}
\label{T15}
Let
\begin{equation}
\label{eq147}
N\left( T \right)=\# \left\{ {s=\sigma +it:0\le t\le T,0<\sigma <1,\xi
\left( s \right)=0} \right\}
\end{equation}
and
\begin{equation}
\label{eq148}
N_0 \left( T \right)=\# \left\{ {s=1/2+it:0\le t\le T,\xi \left( s
\right)=0} \right\}.
\end{equation}
Then
\begin{equation}
\label{eq149}
N\left( T \right)=N_0 \left( T \right).
\end{equation}
\end{theorem}

\begin{proof}
With \textit{Theorem \ref{T14}}, we give
\begin{equation}
\label{eq150}
\xi \left( {s_n } \right)=\xi \left( {1/2} \right)\prod\limits_{n=1}^\infty
{\left( {1+\frac{\left( {s_n -1/2} \right)^2}{\tau _n^2 }} \right)} =0,
\end{equation}
and we show
\begin{equation}
\label{eq151}
s_n =1/2\pm i\tau _n ,
\end{equation}
where $n\in \mathds{N}$ and $\tau _n $ run the positive roots of $\Xi \left( t
\right)=0$.

Thus, the required result follows.
\end{proof}

The zeros (nontrivial zeros) for the entire Riemann zeta--function $\xi \left( {s} \right)$ are showed in Fig. \ref{fig1}.




\begin{figure}
  \centering
  \includegraphics[width=8cm]{fig1.eps}
  \caption{
  The distribution of the zeros (nontrivial zeros) for the entire Riemann zeta--function $\xi \left( {s} \right)$.
  The red points are represented as
  the nontrivial zeros for $\xi \left( {s} \right)$.
  All nontrivial zeros
for $\xi \left( {s} \right)$ lie on the critical line $Re\left(s \right)=1/2$ and in the critical
strip $0<Re\left( s \right)<1$.}\label{fig1}
\end{figure}

\begin{theorem}
(Equivalence theorem II)
\label{T16}

Let $s\in \mathds{C}\backslash \left\{ {0,1} \right\}$ and $n\in \mathds{N}$. Then
the following representations are equivalent:
\begin{description}
\item[B1]
\begin{equation}
\label{eq152}
\varsigma \left( s \right)=\frac{1}{s\left( {s-1}
\right)}\prod\limits_{n=1}^\infty {\left( {1-\frac{s}{s_n }} \right)} ,
\end{equation}
where the product runs over zeroes $s_n $ of (\ref{eq4});
\item[B2]
\begin{equation}
\label{eq153}
\varsigma \left( s \right)=\frac{2\xi \left( {1/2} \right)}{s\left( {s-1}
\right)}\prod\limits_{n=1}^\infty {\left( {1+\frac{i\left( {s-1/2}
\right)}{t_n }} \right)} ,
\end{equation}
where $t_n $ run the roots of $\Xi \left( t \right)=0$;
\item[B3]
\begin{equation}
\label{eq154}
\varsigma \left( s \right)=\frac{1}{s\left( {s-1} \right)}e^{s\mu
}\prod\limits_{n=1}^\infty {\left( {1-\frac{s}{s_n }} \right)e^{s/s_n }} ,
\end{equation}
where$_{ }\mu $ is defined in (\ref{eq46}) and the product runs over zeroes $s_n
$ of (\ref{eq4});
\item[B4]
\begin{equation}
\label{eq155}
\varsigma \left( s \right)=\frac{2\xi \left( {1/2} \right)}{s\left( {s-1}
\right)}e^{s\mu }\prod\limits_{n=1}^\infty {\left(
{1-\frac{s-\frac{1}{2}}{it_n }} \right)e^{s/s_n }} ,
\end{equation}
where$\mu $ is defined in (\ref{eq46}), $t_n $ run the roots of $\Xi \left( t
\right)=0$, and the product runs over zeroes $s_n $ of (\ref{eq4});
\item[B5]
\begin{equation}
\label{eq156}
\varsigma \left( s \right)=\frac{1}{s\left( {s-1}
\right)}\prod\limits_{n=1}^\infty {\left( {1-\frac{s}{\widehat{s_n }}}
\right)} \left( {1-\frac{s}{1-\widehat{s_n }}} \right),
\end{equation}
where the product runs over zeroes $s_n $ of (\ref{eq4});
\item[B6]
\begin{equation}
\label{eq157}
\varsigma \left( s \right)=\frac{2\xi \left( {1/2} \right)}{s\left( {s-1}
\right)}\prod\limits_{n=1}^\infty {\left( {1+\frac{\left( {s-1/2}
\right)^2}{\tau _n^2 }} \right)} ,
\end{equation}
where $\tau _n $ run the positive roots of $\Xi \left( t \right)=0$;
\item[B7]
\begin{equation}
\label{eq158}
\varsigma \left( s \right)=\frac{8}{s\left( {s-1}
\right)}\int\limits_1^\infty {\frac{d\left( {x^{\frac{3}{2}}\psi ^{\left( 1
\right)}\left( x \right)} \right)}{dx}x^{-\frac{1}{4}}\cosh \left(
{\frac{\left( {s-1} \right)\log x}{2}} \right)dx} ,
\end{equation}
where$_{ }\psi \left( x \right)$ is defined in (\ref{eq8});
\item[B8]
\begin{equation}
\label{eq159}
\varsigma \left( s \right)=\frac{1}{s\left( {s-1}
\right)}\sum\limits_{k=0}^\infty {\varphi _k \left( {s-\frac{1}{2}}
\right)^{2k}} ,
\end{equation}
where $\varphi _k $ are expressed in (\ref{eq93}).
\end{description}
\end{theorem}

\begin{proof}
By (\ref{eq3}) and (\ref{eq4}), we arrive at
\begin{equation}
\label{eq160}
\xi \left( s \right)=\frac{1}{2}s\left( {s-1} \right)\varsigma \left( s
\right),
\end{equation}
which implies that
\begin{equation}
\label{eq161}
\varsigma \left( s \right)=\frac{2\xi \left( s \right)}{s\left( {s-1}
\right)}.
\end{equation}
Thus, combining (\ref{eq161}) and \textit{Theorem \ref{T14}}, we obtain the required
results since $\xi \left( 0 \right)=1/2$~(\cite{30}, p.49).
\end{proof}

\begin{theorem}
\label{T17}
Let
\begin{equation}
\label{eq162}
\widehat{N}\left( T \right)=\# \left\{ {s=\sigma +it:0\le t\le T,0<\sigma <1,\varsigma
\left( s \right)=0} \right\}
\end{equation}
and
\begin{equation}
\label{eq163}
\widehat{N}_0 \left( T \right)=\# \left\{ {s=1/2+it:0\le t\le T,\varsigma \left( s
\right)=0} \right\}.
\end{equation}
Then
\begin{equation}
\label{eq164}
\widetilde{N}\left( T \right)=\widetilde{N}_0 \left( T \right).
\end{equation}
\end{theorem}

\begin{proof}
By \textit{Theorem \ref{T16}}, we show
\begin{equation}
\label{eq165}
\varsigma \left( {s_n } \right)=\frac{2\xi \left( {1/2} \right)}{s_n \left( {s_n
-1} \right)}\prod\limits_{n=1}^\infty {\left( {1+\frac{i\left( {s_n -1/2}
\right)}{t_n }} \right)} =0,
\end{equation}
since $s_n \left( {s_n -1} \right)\ne 0$, and we obtain
\begin{equation}
\label{eq166}
s_n =1/2+it_n ,
\end{equation}
where $n\in \mathds{N}$ and $t_n $ run the roots of $\Xi \left( t \right)=0$.

Thus, we finish the proof.
\end{proof}

The nontrivial zeros for the completed Riemann zeta--function $\varsigma \left( {s} \right)$ and its poles are showed in Fig. \ref{fig2}.

\begin{figure}
  \centering
  \includegraphics[width=8cm]{fig2.eps}
  \caption{
  The distribution of the nontrivial zeros and poles for the completed Riemann zeta--function $\varsigma \left( {s} \right)$.
  The red points are represented as
  the nontrivial zeros for $\varsigma \left( {s} \right)$.
  The yellow points are represented as
  simple poles for $\varsigma \left( {s} \right)$.
  All nontrivial zeros
for $\varsigma \left( {s} \right)$ lie on the critical line $Re\left(s \right)=1/2$ and in the critical
strip $0<Re\left( s \right)<1$.}\label{fig2}
\end{figure}

\begin{theorem}
(Equivalence theorem III)
\label{T18}

Let $s\in \mathds{C}\backslash \left\{ 1 \right\}$ and $n\in \mathds{N}$. If
$\Lambda \left( s \right)=2\pi ^{s/2}/\left( {s-1} \right)\Gamma \left(
{s/2+1} \right)$, then the following representations are equivalent:
\begin{description}
\item[C1]
\begin{equation}
\label{eq167}
\zeta \left( s \right)=\xi \left( 0 \right)\Lambda \left( s
\right)\prod\limits_{n=1}^\infty {\left( {1-\frac{s}{s_n }} \right)} ,
\end{equation}
where the product runs over zeroes $s_n $ of (\ref{eq4});
\item[C2]
\begin{equation}
\label{eq168}
\zeta \left( s \right)=\xi \left( {1/2} \right)\Lambda \left( s
\right)\prod\limits_{n=1}^\infty {\left( {1+\frac{i\left( {s-1/2}
\right)}{t_n }} \right)} ,
\end{equation}
where $t_n $ run the roots of $\Xi \left( t \right)=0$;
\item[C3]
\begin{equation}
\label{eq169}
\zeta \left( s \right)=\xi \left( 0 \right)\Lambda \left( s \right)e^{s\mu
}\prod\limits_{n=1}^\infty {\left( {1-\frac{s}{s_n }} \right)e^{s/s_n }} ,
\end{equation}
where $\mu $ is defined in (\ref{eq46}) and the product runs over zeroes $s_n
$ of (\ref{eq4});
\item[C4]
\begin{equation}
\label{eq170}
\zeta \left( s \right)=\xi \left( {\frac{1}{2}} \right)\Lambda \left( s
\right)e^{s\mu }\prod\limits_{n=1}^\infty {\left(
{1-\frac{s-\frac{1}{2}}{it_n }} \right)e^{s/s_n }} ,
\end{equation}
where $\mu $ is defined in (\ref{eq46}), $t_n $ run the roots of $\Xi \left( t
\right)=0$, and the product runs over zeroes $s_n $ of (\ref{eq4});
\item[C5]
\begin{equation}
\label{eq171}
\zeta \left( s \right)=\xi \left( 0 \right)\Lambda \left( s
\right)\prod\limits_{n=1}^\infty {\left( {1-\frac{s}{\widehat{s_n }}}
\right)} \left( {1-\frac{s}{1-\widehat{s_n }}} \right),
\end{equation}
where the product runs over zeroes $s_n $ of (\ref{eq4});
\item[C6]
\begin{equation}
\label{eq172}
\zeta \left( s \right)=\xi \left( {1/2} \right)\Lambda \left( s
\right)\prod\limits_{n=1}^\infty {\left( {1+\frac{\left( {s-1/2}
\right)^2}{\tau _n^2 }} \right)} ,
\end{equation}
where $\tau _n $ run the positive roots of $\Xi \left( t \right)=0$;
\item[C7]
\begin{equation}
\label{eq173}
\zeta \left( s \right)=4\Lambda \left( s \right)\int\limits_1^\infty
{\frac{d\left( {x^{\frac{3}{2}}\psi ^{\left( 1 \right)}\left( x \right)}
\right)}{dx}x^{-\frac{1}{4}}\cosh \left( {\frac{\left( {s-1} \right)\log
x}{2}} \right)dx} .
\end{equation}
where $\psi \left( x \right)$ is defined in (\ref{eq8});
\item[C8]
\begin{equation}
\label{eq174}
\zeta \left( s \right)=\Lambda \left( s \right)\sum\limits_{k=0}^\infty
{\varphi _k \left( {s-\frac{1}{2}} \right)^{2k}} ,
\end{equation}
where $\varphi _k $ are expressed in (\ref{eq93}).
\end{description}
\end{theorem}

\begin{proof}
With (\ref{eq4}) we present
\begin{equation}
\label{eq175}
\xi \left( s \right)=\frac{1}{2}s\left( {s-1} \right)\pi ^{-s/2}\Gamma
\left( {s/2} \right)\zeta \left( s \right),
\end{equation}
which leads to
\begin{equation}
\label{eq176}
\zeta \left( s \right)=\Lambda \left( s \right)\xi \left( s
\right),
\end{equation}
where
\begin{equation}
\label{eq177}
\Lambda \left( s \right)=2\pi ^{s/2}/\left( {s-1} \right)\Gamma \left(
{s/2+1} \right).
\end{equation}
Thus, by (\ref{eq176}) and \textit{Theorem \ref{T14}}, we get the desired results.
\end{proof}

\begin{remark}
In the Landau's book~(\cite{6}, p.313), he suggested that
\begin{equation}
\label{eq178}
\zeta \left( s \right)=\frac{e^{s\ell }}{2\left( {s-1} \right)\Gamma \left(
{s/2+1} \right)}\prod\limits_{n=1}^\infty {\left( {1-\frac{s}{s_n }}
\right)e^{s/s_n }} ,
\end{equation}
where $s_n $ run the nontrivial zeros of $\zeta \left( s \right)=0$ and
$\ell $ is a constant.

In his paper~(\cite{6}, p.199), Landau proved that
\begin{equation}
\label{eq179}
\zeta \left( s \right)=\frac{\pi ^{s/2}\Xi \left( 0 \right)}{\left( {s-1}
\right)\Gamma \left( {1+s/2} \right)}\prod\limits_{n=1}^\infty {\left(
{1+\frac{\left( {s-\frac{1}{2}} \right)^2}{\tau _n^2 }} \right)} ,
\end{equation}
where $\tau _n $ run the positive roots of $\Xi \left( t \right)=0$.

Here, we have to point out that $C3$ and $C6$ are equivalent to (\ref{eq178})
and (\ref{eq179}), respectively.
\end{remark}

\begin{theorem}
\label{T19}
Let
\begin{equation}
\label{eq180}
\widetilde{N}\left( T \right)=\# \left\{ {s=\sigma +it:0\le t\le T,0<\sigma
<1,\zeta \left( s \right)=0} \right\}
\end{equation}
and
\begin{equation}
\label{eq181}
\widetilde{N}_0 \left( T \right)=\# \left\{ {s=1/2+it:0\le t\le T,\zeta
\left( s \right)=0} \right\}.
\end{equation}
Then
\begin{equation}
\label{eq182}
\widetilde{N}\left( T \right)=\widetilde{N}_0 \left( T \right).
\end{equation}
\end{theorem}

\begin{proof}

Making use of \textit{Theorem \ref{T18}}, we give
\begin{equation}
\label{eq183}
\begin{array}{l}
 \zeta \left( {s_n } \right)\\
 =\xi \left( {1/2} \right)\Lambda \left( {s_n }
\right)e^{\mu s_n }\prod\limits_{n=1}^\infty {\left( {1-\frac{s_n
-\frac{1}{2}}{it_n }} \right)e^{s_n /s_n }} \\
 =\xi \left( {1/2} \right)\Lambda \left( {s_n } \right)e^{\mu s_n
}\prod\limits_{n=1}^\infty {\left( {1-\frac{s_n -\frac{1}{2}}{it_n }}
\right)}\\
=0 \\
 \end{array}
\end{equation}
since
\[
\Lambda \left( {s_n } \right)=2\pi ^{s_n /2}/\left( {s_n -1} \right)\Gamma
\left( {s_n /2+1} \right)\ne 0,
\]
and we have
\begin{equation}
\label{eq184}
s_n =1/2+it_n ,
\end{equation}
where $n\in \mathds{N}$ and $t_n $ run the roots of $\Xi \left( t \right)=0$.

Thus, the proof is completed.
\end{proof}

The trivial, pole, and nontrivial zeros for the Riemann zeta--function $\zeta \left( {s} \right)$ are showed in Fig. \ref{fig3}.
\begin{figure}
  \centering
  \includegraphics[width=10cm]{fig3.eps}
  \caption{
  The distribution of the trivial, pole, and nontrivial zeros for the Riemann zeta--function $\zeta \left( {s} \right)$.
  The red points are represented as
  the nontrivial zeros for $\zeta \left( {s} \right)$.
  The blue points are represented as
  the trivial zeros for $\zeta \left( {s} \right)$.
  The yellow point is represented as a
  simple pole for $\zeta \left( {s} \right)$.
  All nontrivial zeros
for $\zeta \left( {s} \right)$ lie on the critical line $Re\left(s \right)=1/2$ and in the critical
strip $0<Re\left( s \right)<1$.}\label{fig3}
\end{figure}

\begin{remark}
For the history of \textit{Theorems \ref{T15}}, \textit{\ref{T17}} and \textit{\ref{T19}},
see~\cite{20a,20b}.

By applying the theory of the entire function~(\cite{35}, p.29), we show that the canonical product
\begin{equation}
\label{eq185}
\prod\limits_{n=1}^\infty {\mathds{E}\left( {t/t_n,0 } \right)}=\Xi
\left( 0 \right)\prod\limits_{n=1}^\infty {\left( {1-\frac{t^2}{\tau _n^2 }}
\right)}
=\prod\limits_{n=1}^\infty {\left( {1-\frac{t}{t_n }} \right)} =\frac{\Xi
\left( t \right)}{\Xi \left( 0 \right)},
\end{equation}
is of growth $(1,0)$ since ${\sum\limits_{n=1}^\infty {t_n^{-2}}}$ converges.

In a similar manner, by theory of the entire function~(\cite{35}, p.29),
the canonical product
\begin{equation}
\label{eq186}
\prod\limits_{n=1}^\infty {\mathds{E}\left( {s/s_n,0 } \right)}
=\prod\limits_{n=1}^\infty {\left( {1-\frac{s}{s_n }} \right)} =\frac{\xi
\left( s \right)}{\xi \left( 0 \right)},
\end{equation}
is also of growth $(1,0)$ since ${\sum\limits_{n=1}^\infty {s_n^{-2}}}$ converges.

Heath--Brown in 2005~\cite{25d} pointed out that all zeros for ${\xi \left( s \right)}$ are simple.
We now deduce it is true since by (\ref{eq140}), we find that all zeros for ${\xi \left( s \right)}$
are different and that ${\xi \left( s \right)}$ is an entire function.
\end{remark}

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\end{document}
