#  Riemann Hypothesis.

Durmagambetov A.A111 [<EMAIL_ADDRESS>010000, Kazakhstan

###### Abstract

Results by J. Hadamard, E. Landau, A. Walvis, T. Estarmann, and P.R. Chernoff
concerning pseudo zeta functions are further developed and applied to the
Riemann hypothesis. Specifically, the Riemann hypothesis is reduced to a
Riemann–Hilbert boundary value problem, and the integral Hilbert transforms
arising in its solution allow the calculation of an exact lower bounds for the
zeta function.

###### keywords:

Euler product, Dirichlet, Riemann, Hilbert, Poincaré, Riemann hypothesis, zeta
function, Hadamard, Landau, Walvis, Estarmann, Chernoff

###### MSC:

[2010] 11M26

††journal: Journal of LaTeX Templates

url]

##  1 INTRODUCTION

аппррлол This study is concerned with the properties of modified zeta
functions. Riemann’s zeta function is defined by the Dirichlet series

$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^{s}},\,\ s=\sigma+it,$ (1)

which is absolutely and uniformly convergent in any finite region of the
complex $s$-plane for which $\sigma\geq 1+\epsilon,\epsilon>0.$ If $\sigma>1$,
then $\zeta$ is represented by the following Euler product formula

$\zeta(s)=\prod_{p}\left[1-\frac{1}{p^{s}}\right]^{-1},$ (2)

where $p$ runs over all prime numbers. $\zeta(s)$ was first introduced by
Euler in 1737 [1], who also obtained formula (2). Dirichlet and Chebyshev
considered this function in their study on the distribution of prime numbers
[2]. However, the most profound properties of $\zeta(z)$ were only discovered
later, when it was extended to the complex plane. In 1876, Riemann [3] proved
that $\zeta(s)$ allows analytical continuation to the entire $z$-plane as
follows:

$\pi^{-s/2}\Gamma(s/2)\zeta(s)=1/(s(s-1))+\int\limits_{1}^{+\infty}(x^{s/2-1}+x^{-(1+s)/2})\theta(x)dx,$
(3)

where $\Gamma(z)$ is the gamma function and

$\theta(x)=\sum_{n=1}^{\infty}\exp(-\pi n^{2}x).$

$\zeta(s)$ is a regular function for all values of $s$, except $s=1$, where it
has a simple pole with residue $1$; moreover, it satisfies the following
functional equation:

$\pi^{-s/2}\Gamma(s/2)\zeta(s)=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s)$ (4)

This equation is called Riemann’s functional equation.

Riemann’s zeta function is an important subject of study and has numerous
interesting generalizations. The role of the zeta function is highly
significant in number theory, where it is connected with various fundamental
functions, such as the Möbius function, the Liouville function, the number of
divisors, and the number of prime divisors. The detailed theory of zeta
functions is presented in [4]. The zeta function has found application in
various other fields, notably in quantum statistical mechanics and quantum
field theory [5–7]. Riemann’s zeta function is often introduced in quantum
statistics formulas. A well-known example is the Stefan-Boltzman law of a
black body’s radiation. Its ubiquitous use in seemingly unrelated areas
demonstrates the necessity for further investigation.

The present study is concerned with the analytical properties of the following
generalized zeta functions:

$P(s)=\sum_{j\geq 1}\frac{1}{p_{j}^{s}},\,Re(s)>1+\delta,\delta>0,$

where $\\{p_{j}:j\geq 1\\}$ is an increasing enumeration of all prime numbers.
The form of $P(s)$ suggests that it possesses the same properties as the zeta
function; however, this is not quite obvious and can be seen by considering.

$ln(\zeta(s))=\sum_{n=1}^{\infty}P(ns)/n,\,\,Re(s)>1+\delta,\delta>0.$

Hadamard was the first to apply $P(s)$ in the study of the zeta function [8].
Chernoff made significant progress in the Riemann hypothesis using $P(s)$ [9].
In the present study, modifications of Chernoff’s results are obtained.
Specifically, his study on the pseudo zeta function is completed. Chernoff
obtained an equivalent formulation of the Riemann hypothesis in terms of a
pseudo zeta function as follows.

THEOREM. (Chernoff) Let

$C(s)=\prod_{n>1}\left[1-\frac{1}{(nln(n))^{s}}\right]^{-1}$

. Then, $C(s)$ continues analytically into the critical strip and has no zeros
there.

The significance of this theorem is that if the primes were distributed more
regularly (i.e., if $p_{n}\equiv n\log n$), then the Riemann hypothesis would
be trivially true. In an effort to further develop the work of Chernoff and
Hadamard, the following question naturally arises: Does the pseudo zeta
function $P(s)$ continues analytically into the critical strip? It should be
noted that analytic extensions of $P(s)$ were first studied by E. Landau and
A. Walvis [10] and T. Estarmann [11], [12]; however, no satisfactory estimates
for $P(s)$ were obtained, and the present study is concerned with this
question.

THEOREM. ( E. Landau, A. Walvis, T. Estarmann)
Let $\mu(n)-functionM\ddot{o}bius\,\,\,$. Then,

$P(s)=\sum_{n\geq
1}\frac{\mu(n)ln\zeta(ns)}{n}\,\,as\,\,\,Re(s)>1+\delta,\,\,\delta>0,$
$P_{0}(s)=\sum_{n\geq
1}\frac{\mu(n)ln\zeta(ns)}{n}-meromorphic\,\,function\,\,as\,\,Re(s)>\delta,\,\,\delta>0.$

We introduce the following analogs of the function P(s):

$Q_{m}(s)=ln(\zeta(s))-\sum_{n=m}^{\infty}P(ns)/n,\,\,Re(s)>1/m+\delta,\\\ $
$Q_{m}(1-s)=ln(\zeta(1-s))-\sum_{n=m}^{\infty}P(n(1-s))/n,\,\,Re(s)<1-1/m-\delta\\\
$ (5)

The paper is organized as follows. Intermediate estimates are first obtained
for the $ln\zeta(s)$. Subsequently, the sets where the logarithm of the zeta
function is uniquely determined are defined. These sets are composed of
rectangles in which the zeta function has no roots, and they cover the entire
critical strip except for the rectangular regions in which the zeros of the
zeta functions are located. In the rectangles in which there are no zeros of
the zeta function, the real value of its logarithm can be defined, and in
these sets, the mirror-symmetric equation that arises by taking the logarithm
on both sides of the Riemann functional equation is investigated. Then, the
Fourier transform is applied to it, and it is multiplied by a regulating
factor. Thus, a Riemann-Hilbert boundary value problem is obtained for the
$Q_{m}(s)$ The properties of the solution to the Riemann–Hilbert boundary
value problem are expressed in terms of the Hilbert integral transform. In the
rectangles in which the zeta function has no roots, the Hilbert transform can
be used to obtain exact lower bounds for the zeta function in the critical
strip.

##  2 RESULTS

As mentioned in Introduction, certain simple intermediate estimates are first
obtained.

The rectangles in which the zeta function hasn’t zeros are first introduced as
follows:

$\,\,D(n,\epsilon)=(s|\epsilon<Re(s)<1-\epsilon,\,\,Im(s)\neq
Im(s_{n}),Im(s_{n})-d_{n}\leq Im(s)\leq Im(s_{n})+d_{n}$
$\overline{D(n,\epsilon)}=(s|\epsilon<Re(s)<1-\epsilon,\,Im(-s)\neq
Im(-s_{n}),-Im(s_{n})-d_{n}\leq Im(s)\leq-Im(s_{n})+d_{n},$

where

$\zeta(s_{n+1})=0,\,\zeta(s_{n})=0,\,\zeta(1-s_{n})=0,\zeta(1-s_{n+1})=0,\,\,\zeta(1-s_{n})=0,\\\
$ $d_{n}=(Im(s_{n+1})-Im(s_{n}))/2$

The sets of $D(n,\epsilon),$ is shown in the figure below.
$Re(s)$$Im(s)$$0$1/2$+\epsilon$1/2$-\epsilon$1 $-\epsilon$ $D(n,\epsilon)$
$\overline{D(n,\epsilon)}$

######  Theorem 1.

Let $s\in D(n,\epsilon)\cup\overline{D(n,\epsilon)}$ ,
$F(s)=\frac{s}{2}ln(\pi)-ln(\Gamma(s/2))-\frac{1-s}{2}ln(\pi)+ln(\Gamma(1-s)/2)).$
Then,

$\displaystyle\sup_{s\in
D(n,\epsilon)\cup\overline{D(n,\epsilon)}}|F(\tau+i\alpha)|+\sup_{s\in
D(n,\epsilon)\cup\overline{D(n,\epsilon)}}|\frac{dF(\tau+i\alpha)}{d\tau}|<CC_{n}$

###### Proof.

As $\epsilon<\tau<1-\epsilon$ implies that F - is holomorphic which completes
the proof. ∎

As mentioned in Introduction, a Riemann–Hilbert boundary value problem should
be obtained. To this end, an equation should be derived that determines the
difference between the boundary values of the analytic functions in the upper
plane and the lower plane.

######  Theorem 2.

Let $\epsilon>2/m\in D(n,\epsilon)$ or $s\in\overline{D(n,\epsilon)}$ and

$F_{m}(s)=F(s)-\sum_{n=m}^{\infty}P(ns)/n+\sum_{n=m}^{\infty}P(n(1-s))/n$

Then,

$\displaystyle Q_{m}(s)=Q_{m}(1-s)+F_{m}(s)$ $\displaystyle\sup_{s\in
D(n,\epsilon)\cup\overline{D(n,\epsilon)}}|F_{m}(s)|\leq C_{m}C_{n}$

###### Proof.

The equation (5) implies for $s\in D(n,\epsilon)$ or
$s\in\overline{D(n,\epsilon)}$, and an equation for $Q_{m}(s)$ may be obtained
. Theorem 1 and condition of Theorem 2 implies estimate for$F_{m}$ ∎

As the resulting equation is not suitable for the formulation of a
Riemann–Hilbert boundary value problem, a number of transformations should be
performed. Certain preliminary arguments suggest that the Fourier transform is
an appropriate choice. In Riemann–Hilbert boundary value problems, the
asymptotic behavior of the unknown functions is highly important. To ensure
this behavior, it is necessary to have estimates for the derivatives. The
evaluation of the derivatives is usually carried out by integration by parts.
Thus, the following functions are introduced:

$\left\\{\begin{array}[]{l}\nu_{\epsilon}(s)=0,\,\,\,Re(s)<1/3+2\epsilon;\\\
\nu_{\epsilon}(s)=1,\,\,\,\epsilon<Re(s)<1/2-2\epsilon;\\\
\nu_{\epsilon}(s)=0,\,\,\,1/2-2\epsilon<Re(s)<1/2+2\epsilon;\\\
\nu_{\epsilon}(s)=1,\,\,\,1/2+2\epsilon<Re(s)<1-2\epsilon,\\\
\nu_{\epsilon}(s)=0,\,\,\,Re(s)>1-2\epsilon;\\\ \psi(t)=\sigma
e^{\frac{1}{t^{2}-1}},\,\,\
t^{2}<1;\frac{1}{\sigma}=\int\limits_{-1}^{1}e^{\frac{1}{t^{2}-1}}dt\\\
\psi(t)=0,\,\,t^{2}\geq 1;\\\
\mu_{\epsilon}(x)=\int\psi(s/\epsilon)\nu_{\epsilon}(x-t)dt/\epsilon=\int\psi(x-t/\epsilon)\nu_{\epsilon}(t)dt/\epsilon;\end{array}\right.$

The graph of $\nu_{\epsilon}$ is shown in the figure below.

$s$$\nu_{\epsilon}$$0$1/2$+\epsilon$1/2$-\epsilon$1 $-\epsilon$

As the equation is multiplied by $\mu_{\epsilon}$, it should be verified that
the symmetry of the equation for $Q_{m}$ is not affected.

######  Lemma 1.

For $\nu_{\epsilon}$ and $\mu_{\epsilon}$, we have
$\nu_{\epsilon}(x)=\nu_{\epsilon}(1-x),\,\,\mu_{\epsilon}(x)=\mu_{\epsilon}(1-x)$.

###### Proof.

We have $\nu_{\epsilon}(x)=\nu_{\epsilon}(1-x)$ by definition. Moreover,

$\mu_{\epsilon}(x)=\int\psi(s/\epsilon)\nu_{\epsilon}(x-s)ds/\epsilon=\int\psi(s/\epsilon)\nu_{\epsilon}(1-x+s)ds/\epsilon=\mu_{\epsilon}(1-x).$

∎

######  Lemma 2.

Let $1/2+3\epsilon<x<1-3\epsilon$. Then, $\mu_{\epsilon}(x)=1$.

###### Proof.

$\displaystyle\mu_{\epsilon}(x)=\int\psi(s/\epsilon)\nu_{\epsilon}(x-s)ds/\epsilon=\int_{-\epsilon}^{\,\,\epsilon}\psi(s/\epsilon)\nu_{\epsilon}(x-s)ds/\epsilon=\int_{-\epsilon}^{\,\,\epsilon}\psi(s/\epsilon)ds/\epsilon=1.$

∎

To compute the Fourier transform of the equation for $Q_{m}$, the functions
$R(k)$, $Q_{\epsilon}(s)$, $Q_{\epsilon}(1-s,m)$, $(F_{\epsilon}(s)$, the
Fourier transform of $Q_{\epsilon}(s)$, the Fourier transform of
$Q_{\epsilon}(s)$, the Fourier transform of $F_{\epsilon}(s)$, are introduced
as follows:

$\displaystyle
Q_{\epsilon}(s)=Q_{m}(s)\mu_{\epsilon}(Re(s)),\,\,\,\,Q_{\epsilon}(1-s)=Q_{m}(1-s)\mu_{\epsilon}(1-Re(s))\,\,$
$\displaystyle
R(k)=\frac{e^{-ik}}{k-ia}+1\,\,,F_{\epsilon}(s)=F_{m}(s)\mu_{\epsilon}(Re(s));$
$\displaystyle\frac{1}{\sqrt{2\pi}}\int_{\epsilon}^{1-\epsilon}Q_{\epsilon}(1-\tau-i\alpha)e^{-ik\tau}d\tau=\frac{e^{-ik}}{\sqrt{2\pi}}\int_{\epsilon}^{1-\epsilon}Q_{\epsilon}(\tau-i\alpha)e^{ik\tau}d\tau;$
$\displaystyle\widetilde{Q_{\epsilon}}(k,-\alpha)=\frac{e^{-ik}}{\sqrt{2\pi}}\int_{\epsilon}^{1-\epsilon}Q_{\epsilon}(\tau-i\alpha)e^{-ik\tau}d\tau,\,$
$\displaystyle\widetilde{Q_{\epsilon}}(k,\alpha)=\frac{1}{\sqrt{2\pi}}\int_{\epsilon}^{1-\epsilon}Q_{\epsilon}(\tau+i\alpha)e^{-ik\tau}d\tau,\,$
$\displaystyle\
\widetilde{F_{\epsilon}}(k,\alpha)=\frac{1}{\sqrt{2\pi}}\int_{\epsilon}^{1-\epsilon}F_{\epsilon}(\tau+i\alpha)e^{-ik\tau}ds.$
$\displaystyle
J_{\epsilon}(k,\alpha)=\frac{1}{\sqrt{2\pi}}\int_{\epsilon}^{1-\epsilon}Q_{\epsilon}(\tau-i\alpha)e^{ik\tau}d\tau\,\,\,,I_{\epsilon}(k,\alpha)=\frac{1}{\sqrt{2\pi}}\int_{\epsilon}^{1-\epsilon}Q_{\epsilon}(\tau+i\alpha)e^{-ik\tau}d\tau.$
(6)

To obtain the Riemann-Hilbert boundary value problem , the following lemma is
required.

######  Lemma 3.

Let $a>2$ then $ind(R)=0$

###### Proof.

By definition

$ind(R)=\frac{1}{2\pi i}\int_{-\infty}^{+\infty}\frac{R(k)^{\prime}}{R(k)}dk$

As

$Im(k)<0,|e^{-ik}|<1\,\,and\,\,|k-ia|>2\,\,yield\,\,\,\frac{R(k)^{\prime}}{R(k)}$

have nothing pole. Latest statement and Lemma of Jordan yield $ind(R)=0$.

∎

To obtain the necessary asymptotics, the following lemma is required.

######  Lemma 4.

Let

$a>2$

then $ln(R(k))$ is single-valued analitical fuction in lower half plane.

###### Proof.

As

$Im(k)\leq 0\,\,\ \,$

yeild

$Re(R(k))=1+Re\left[\frac{e^{ik}}{k-ia}\right]>0$

which completes proof. ∎

Denote $\Omega_{n}(s)={\rm Re}(s-s_{n})$ and
$\omega(s)=\frac{(s-s_{n})^{\rho(s_{n})}(s-1+s_{n})^{\rho(s_{n})}.}{(1-s)}$

$\rho(s_{n})$ is multiplicity root of $\zeta(s)$ as $s=s_{n}$ The following
presents results of [14]
Theorem of Backlund R.
Let $\zeta(s_{n})=0$ then

$\rho(s_{n})<C_{0}ln|s_{n}|.$

All the arguments given below are based on the assumption of the error of the
Riemann hypothesis i.e $|\Omega_{n}(1/2)|=\epsilon_{n}>0$. At the end of our
work we will encounter a contradiction with our assumption from which the
truth of the Riemann hypothesis will follow

######  Lemma 5.

Let $\gamma_{n}=\frac{1}{4\rho(s_{n})}$, $|\Omega_{n}(1/2)|=\epsilon_{n}\geq
2/m$, and $d_{n}=({\rm Im}(s_{n+1})-{\rm Im}(s_{n}))/2.$

Then, we have the following estimate as
$\epsilon=0.01\epsilon_{n}(1-Re(s_{n}))$:

$\displaystyle\sup\limits_{Ims_{n}-d_{n}\leq\alpha\leq
Im{s_{n}+d_{n}}}\int_{\epsilon}^{1-\epsilon}\left|Q_{\epsilon}(\tau+i\alpha)\right|^{2}+\left|Q_{\epsilon}(\tau+i\alpha)\right|d\tau<C_{n}C_{\epsilon_{n}}C_{\gamma_{n}}.$
$\displaystyle\sup\limits_{-Ims_{n}-d_{n}\leq\alpha\leq-
Im{s_{n}+d_{n}}}\int_{\epsilon}^{1-\epsilon}\left|Q_{\epsilon}(\tau-i\alpha)\right|^{2}+\left|Q_{\epsilon}(\tau-i\alpha)\right|d\tau<C_{n}C_{\epsilon_{n}}C_{\gamma_{n}}.$

$\displaystyle\sup\limits_{Ims_{n}-d_{n}\leq\alpha\leq
Im{s_{n}+d_{n}}}\int_{\epsilon}^{1-\epsilon}\left|F_{\epsilon}(\tau+i\alpha)\right|^{2}+\left|F_{\epsilon}(\tau+i\alpha)\right|d\tau<C_{n}C_{\epsilon_{n}}C_{\gamma_{n}}.$
$\displaystyle\sup\limits_{-Ims_{n}-d_{n}\leq\alpha\leq-
Im{s_{n}+d_{n}}}\int_{\epsilon}^{1-\epsilon}\left|F_{\epsilon}(\tau-i\alpha)\right|^{2}+\left|F_{\epsilon}(\tau-i\alpha)\right|d\tau<C_{n}C_{\epsilon_{n}}C_{\gamma_{n}}.$

###### Proof.

by definion $Q_{\epsilon}(s)$,

$I_{Q}=\int_{\epsilon}^{1-\epsilon}\left|Q_{\epsilon}(\tau+i\alpha)\right|^{2}+\left|Q_{\epsilon}(\tau+i\alpha)\right|d\tau\leq$
$\int_{\epsilon}^{1-\epsilon}\left|ln|\zeta(\tau+i\alpha)|\right|^{2}+\left|ln\left|\zeta(\tau+i\alpha)\right|\right|d\tau+\mu_{\epsilon}(s)|\sum_{n=m}^{\infty}P(ns)/n|+\mu_{\epsilon}(s)|\sum_{n=m}^{\infty}P(n(1-s))/n|\leq$
$C_{\epsilon}+\int_{\epsilon}^{1-\epsilon}\left|ln\left|\frac{\zeta(\tau+i\alpha)}{\omega(\tau+i\alpha)}\right|\right|^{2}+\left|ln\left|\frac{1}{\omega(\tau+i\alpha)}\right|\right|^{2}+\left|ln\left|\frac{\zeta(\tau+i\alpha)}{\omega(\tau+i\alpha)}\right|\right|+\left|ln\left|\frac{1}{\omega(\tau+i\alpha)}\right|\right|d\tau$

Denote

$L_{max}=\max\limits_{s\in
D(n,\epsilon)\cup\overline{D(n,\epsilon)}}\left|\frac{\zeta(s)}{\omega(s)}\right|,\,\,\,L_{min}=\min\limits_{s\in
D(n,\epsilon)\cup\overline{D(n,\epsilon)}}\left|\frac{\zeta(s)}{\omega(s)}\right|$

$I_{Q}\leq
C_{\epsilon}+\left|ln\left|L_{max}+\frac{1}{L_{min}}\right|\right|+\left|ln\left|L_{max}+\frac{1}{L_{min}}\right|\right|^{2}+C_{\gamma_{n}}\int_{\epsilon}^{1-\epsilon}\left|\frac{1}{\omega(s)}\right|^{2\gamma_{n}}+\left|\frac{1}{\omega(s)}\right|^{\gamma_{n}}d\tau$

which completes the proof. ∎

The previous constructions allow the calculation of the asymptotics as
follows.

######  Lemma 6.

Let $(3/4+i\alpha)\in D(n,\epsilon)$,$\Omega_{n}(1/2)=\epsilon>2/m,$.Then

$\displaystyle\lim_{Im(k)\to-\infty}I_{\epsilon}(k,\alpha)=0,\,\,\lim_{Im(k)\to\infty}J_{\epsilon}(k,\alpha)=0.$
(7)

and as Im(k)=0

$\displaystyle\lim_{Re(k)\to\infty}I_{\epsilon}(k,\alpha)=0,\,\,\lim_{Re(k)\to\infty}J_{\epsilon}(k,\alpha)=0.$
(8)

###### Proof.

To study the asymptotics, by Lemma 5 and the finiteness of $\mu_{\epsilon}$
yield

$\displaystyle\left|I_{\epsilon}(k,\alpha)\right|=\left|\int_{\epsilon}^{1-\epsilon}Q_{\epsilon}(\tau+i\alpha)e^{-ik\tau}d\tau\right|\leq\int_{\epsilon}^{1-\epsilon}\left(\left|Q_{\epsilon}(\tau+i\alpha)\right|^{2}d\tau\right)^{1/2}\frac{1}{|Im(k)|^{1/2}}$

A similar argument is used for the function

$J_{\epsilon}(k,\alpha)=\frac{1}{\sqrt{2\pi}}\int_{\epsilon}^{1-\epsilon}Q_{\epsilon}(\tau-i\alpha)e^{ik\tau}d\tau.$

As $Im(k)>0\,\,,J_{\epsilon}(\tau,\alpha)$ can be estimated using the last
expression and Lemma 5 as follows:

$\left|J_{\epsilon}(k,\alpha)\right|<\int_{\epsilon}^{1-\epsilon}\left(\left|Q_{\epsilon}(\tau-i\alpha)\right|^{2}d\tau\right)^{1/2}\frac{1}{|Im(k)|^{1/2}}$

As Im(k)=0, by the Riemann-Lebesgue lemma yield

$\displaystyle\lim_{Re(k)\to\infty}I_{\epsilon}(k,\alpha)=0,\,\,\lim_{Re(k)\to\infty}J_{\epsilon}(k,\alpha)=0.$
(9)

which completes the proof. ∎

For $f\in W_{2}^{1}(R)=\\{f\in
L_{2}(R):(1+|\omega|^{2})^{{1/2}}{\hat{f}}(\omega)\in L_{2})\\}.$, the
operators $T_{\pm}$ and $T$ are defined as follows:

$\displaystyle T_{+}f=\frac{1}{2\pi i}\lim\limits_{Imz\rightarrow
0}\int\limits_{-\infty}^{\infty}\frac{f({s})}{s-z}ds,\leavevmode\nobreak\
Im\leavevmode\nobreak\ z>0,T_{-}f=\frac{1}{2\pi i}\lim\limits_{Imz\rightarrow
0}\int\limits_{-\infty}^{\infty}\frac{f({s})}{s-z}ds,\leavevmode\nobreak\
Im\leavevmode\nobreak\ z<0.$ $\displaystyle Tf=\frac{1}{2}(T_{+}+T_{-})f.$

These operators are closely related to the Hilbert transform, whose isometric
properties were studied by Poincaré. The following result is from [13].

######  Lemma 7.

$\displaystyle TT=\frac{1}{4}I,\leavevmode\nobreak\
TT_{+}=\frac{1}{2}T_{+},\leavevmode\nobreak\ TT_{-}=-\frac{1}{2}T_{-},\
T_{+}=T+\frac{1}{2}I,\leavevmode\nobreak\ T_{-}=T-\frac{1}{2}I,$

where $I$ is the identity operator $If=f$.

The reduction to a Riemann–Hilbert boundary value problem can now be
formulated as follows.

######  Theorem 3.

Let

$\displaystyle(3/4+i\alpha)\in D(n,\epsilon),a>2,\Omega_{n}(1/2)=\epsilon>2/m$
(10) $\displaystyle\Gamma_{+}(k)=-\frac{1}{2\pi
i}\int_{-\infty}^{\,\,\infty}\frac{ln(R(t))dt}{t-k-i0}$ (11)
$\displaystyle\Gamma_{-}(k)=-\frac{1}{2\pi
i}\int_{-\infty}^{\,\,\infty}\frac{ln(R(t))dt}{t-k+i0}$ (12) $\displaystyle
X_{+}(k)=e^{\Gamma_{+}(k)},\,\,\,X_{-}(k)=e^{\Gamma_{-}(k)},\,\,R(k)=\frac{X_{-}(k)}{X_{+}(k)},\,G_{\epsilon}(k,\alpha)=J_{\epsilon}(k,\alpha).$
(13)

Then,

$J_{\epsilon}(k,\alpha)=-\frac{X_{+}(k)}{2\pi
i}\int_{-\infty}^{\,\,\infty}\frac{G_{\epsilon}(t,\alpha)}{X_{-}(t)}\frac{dt}{t-k-i0}=X_{+}(k)T_{+}\frac{G_{\epsilon}}{X_{-}}$
$\frac{I_{\epsilon}(k,\alpha)}{k-ia}-\frac{\widetilde{F_{\epsilon}}(k,\alpha)}{k-ia}=-\frac{X_{-}(k)}{2\pi
i}\int_{-\infty}^{\,\,\infty}\frac{G_{\epsilon}(t,\alpha)}{X_{-}(t)}\frac{dt}{t-k+i0}dt=X_{-}(k)T_{-}\frac{G_{\epsilon}}{X_{-}}.$

###### Proof.

By Theorem 2 and Lemma 2 we have

$\displaystyle Q_{\epsilon}(s)=Q_{\epsilon}(1-s)+F_{\epsilon}(s).$ (14)

Using the Fourier transform, we obtain

$\displaystyle
I_{\epsilon}(k,\alpha)=e^{-ik}J_{\epsilon}(k,\alpha)+\widetilde{F_{\epsilon}}(k,\alpha).$
(15)

Multiplying this equation by $\frac{1}{k-ia}$ we get

$\displaystyle\frac{I_{\epsilon}(k,\alpha)}{k-ia}=\frac{e^{-ik}J_{\epsilon}(k,\alpha)}{k-ia}+\frac{\widetilde{F_{\epsilon}}(k,\alpha)}{k-ia}.$
(16)

Rewriting latest equation

$\displaystyle\frac{I_{\epsilon}(k,\alpha)}{k-ia}-\frac{\widetilde{F_{\epsilon}}(k,\alpha)}{k-ia}=R(k)J_{\epsilon}(k,\alpha)+J_{\epsilon}(k,\alpha).$
(17)

$\displaystyle\Psi_{-}(k,\alpha)=\frac{I_{\epsilon}(k,\alpha)}{k-ia}-\frac{\widetilde{F_{\epsilon}}(k,\alpha)}{k-ia}$
(18) $\displaystyle\Psi_{+}(k,\alpha)=J_{\epsilon}(k,\alpha)$ (19)
$\displaystyle G_{\epsilon}(k,\alpha)=J_{\epsilon}(k,\alpha).$ (20)

Using Lemma 5, the following Riemann-Hilbert boundary value problem is
obtained regarding the definition of an analytic function from its boundary
values on the real line:

Hilbert’s formula and Lemma 4-Lemma 6 gives the solution to the Riemann-
Hilbert boundary value problem (21),(22)

$\displaystyle\Psi_{+}(k,\alpha)=-\frac{X_{+}(k)}{2\pi
i}\int_{-\infty}^{\,\,\infty}\frac{G_{\epsilon}(t,\alpha)}{X_{-}(t)}\frac{dt}{t-k-i0}$
(23) $\displaystyle\Psi_{-}(k,\alpha)=-\frac{X_{-}(k)}{2\pi
i}\int_{-\infty}^{\,\,\infty}\frac{G_{\epsilon}(t,\alpha)}{X_{-}(t)}\frac{dt}{t-k+i0}$
(24)

Denote

$\displaystyle\Phi_{+}(k,\alpha)=\Psi_{+}(k,\alpha)-\frac{J_{\epsilon}(k,\alpha)R(k)}{X_{-}(k)}$
(25)
$\displaystyle\Phi_{-}(k,\alpha)=\frac{\Psi_{-}(k,\alpha)}{X_{-}(k)}-\frac{I_{\epsilon}(k,\alpha)-\widetilde{F_{\epsilon}}(k,\alpha)}{X_{-}(k)(k-ia)}$
(26)

For $\Phi_{+}(k,\alpha),\Phi_{-}(k,\alpha)$ we have new boundary-value problem
of Riemann-Hilbert :

$\displaystyle\Phi_{-}(k,\alpha)=\Phi_{+}(k,\alpha)$ (27)
$\displaystyle\lim_{Re(k)\to\infty}\Phi_{+}(k,\alpha)=0\,\,\,as\,\,\,Im(k)\geq
0,\,\,\lim_{Re(k)\to-\infty}\Phi_{-}(k,\alpha)=0\,\,\,as\,Im(k)\leq 0$ (28)

Liouville theorem implies that

$\displaystyle\Phi_{-}(k,\alpha)=0\,\,\,,\Phi_{+}(k,\alpha)=0.$ (29)

which completes the proof. ∎

##  3 DISCUSSION

_Our computations led to a new definition of the functions
$I_{\epsilon}(k),\,J_{\epsilon}(k)$, which we obtained from the Riemann-
Hilbert boundary-value problem. From the uniqueness of the solution of the
Riemann-Hilbert boundary value problem - functions
$I_{\epsilon}(k),\,J_{\epsilon}(k)$, defined earlier in (6) and obtained from
the Hilbert formula are equal! _
To obtain the final estimates for the zeta function, the isometric properties
of the integral Hilbert transform will be used.

######  Theorem 4.

Let $(3/4+i\alpha)\in D(n,\epsilon)$ and $a>2$,$\Omega_{n}(1/2)=\epsilon>2/m$.
Then,

$\displaystyle
C^{-1}<\left|X_{-}(t)\right|<C,\,\,\,C^{-1}<\left|X_{+}(t)\right|<C.$
$\displaystyle||\Psi_{+}||_{L_{2}}\leq
C_{\epsilon},\,\,\,\,||\Psi_{-}||_{L_{2}}\leq C_{\epsilon},\,\,\,\,\,\,\,\ $

###### Proof.

By Lemma 5 and Lemma 8 we get

$\displaystyle\Gamma_{-}(k)=\frac{1}{2\pi
i}\int_{-\infty}^{\,\,\infty}\frac{ln(R(t))dt}{t-k+i0}=T_{-}ln(R)=ln(R)$
$\displaystyle\Gamma_{+}(k)=\frac{1}{2\pi
i}\int_{-\infty}^{\,\,\infty}\frac{ln(R(t))dt}{t-k-i0}=T_{+}ln(R)=0$

$T_{-}ln(R)=ln(R)$, $T_{+}ln(R)=0$ implies

$\displaystyle X_{-}(t)=R(t),\,\,\ X_{+}(t)=1$ $\displaystyle
C^{-1}<\left|X_{-}(t)\right|<C,\,\,\,\left|X_{+}(t)\right|=1.$

Using Theorem 4 and Lemma 6, we obtain

$\displaystyle||\Psi_{-}||_{L_{2}}^{2}+||\Psi_{+}||_{L_{2}}^{2}=\int_{-\infty}^{+\infty}\left|\frac{I_{\epsilon}(k,\alpha)}{k-ia}-\frac{\widetilde{F_{\epsilon}}(k,\alpha)}{k-ia}\right|^{2}dk+\int_{-\infty}^{+\infty}\left|J_{\epsilon}(k,\alpha)\right|^{2}dk\leq
C_{n}C_{\epsilon}$

∎

######  Lemma 8.

Let $\beta_{n},\phi_{n}$ satisfies equations

$e^{\beta}=\sqrt{(2\pi n+\phi)^{2}+(\beta-a)^{2}},$
$\phi=-\arg\left(\frac{1}{2\pi n+\phi+i(-a+\beta)}\right)$

Then

$t_{n}=2\pi n+i\beta_{n}+\phi_{n}$

root of equation

$R(k)=0$

and

$\beta_{n}=ln(2n\pi)+o(1),\left|\frac{d\beta_{n}}{da}\right|\leq\frac{Cln(n)}{n^{2}}$

,

$\phi_{n}=\pi+O(ln(n)/n),\,\,\,\,\left|\frac{d\phi_{n}}{da}\right|\leq\frac{Cln(n)}{n^{2}}$

###### Proof.

$R(t_{n})=\frac{e^{-it_{n}}}{t_{n}-ia}+1=\frac{e^{-i2\pi
n+\beta_{n}+i\phi_{n}}}{2\pi n+\phi_{n}+i(\beta_{n}-a)}+1=$
$-\frac{e^{\beta_{n}}}{\sqrt{(2\pi
n+\phi_{n})^{2}+(\beta_{n}-a)^{2}}}+1=-1+1=0$

take $\beta_{n}=ln(n\pi)+\gamma_{n}$ then

$e^{\gamma_{n}}=\sqrt{1+\frac{(ln(n\pi)+\gamma_{n}-a)^{2}}{(2\pi
n+\phi_{n})^{2}}},$

for $\phi_{n}$ we have

$\phi_{n}=\pi-\arctan\left(\frac{(-a+\beta)}{2\pi n+\phi_{n}}\right)$

and we get

$\phi_{n}=\pi+O(ln(n)/n)$

Estimates for derivatives follow from the results obtained.

proof complete. ∎

The following presents results of [15]
Theorem of Existence and invertibility of the Fourier

If f is in L1 (i.e., f is absolutely integrable) and if it is of bounded
variation on every finite interval, then
$\int_{-\infty}^{\infty}f(t)e^{ikt}dt$ exists and f(t) can be recovered from
the inverse Fourier transform relationship at each point at which f is
continuous.

######  Lemma 9.

Let

$Q_{\epsilon}(s)=Q_{m}(s)\mu_{\epsilon}(Re(s)),\Omega_{n}(1/2)=\epsilon>2/m$
$I_{\epsilon}(k,\alpha)=\int_{\epsilon}^{1-\epsilon}Q_{\epsilon}(\tau+i\alpha)e^{-ik\tau}d\tau$

$s=\tau+i\alpha\in D$ , where $\alpha\in(Im(s_{n})<\alpha<Im(s_{n+1})$ is
fixed
then

$0<\left|\zeta(\tau_{min}+i\alpha)\right|\leq\left|\zeta(s)\right|\leq|\zeta(\tau_{max}+i\alpha)$
$\max_{\epsilon\leq\tau\leq
1-\epsilon}\left|Q_{\epsilon}(\tau+i\alpha)\right|\leq C(\alpha,n,\epsilon)$

$\max_{\epsilon\leq\tau\leq
1-\epsilon}\left|\frac{dQ_{\epsilon}(\tau+i\alpha)}{d\tau}\right|\leq
C(\alpha,n,\epsilon)$

$\lim\limits_{N\to\infty}\int_{N}^{-N}e^{itk}\frac{dI_{\epsilon}(k,\alpha)}{dk}dk=-itf_{\epsilon}(t)$

###### Proof.

From the holomorphic of the $\zeta(s)$ follows $|\zeta(s)|$ is harmonic
function. According to the Weierstrass theorem, the functions
$Q_{\epsilon}(s,n)$ its exact maximum and minimum on a compact set

$0<\left|\zeta(\tau_{min}+i\alpha)\right|\leq\left|\zeta(\tau+i\alpha)\right|\leq|\zeta(\tau_{max}+i\alpha),\epsilon\leq\tau\leq
1-\epsilon$
$\left|Q_{\epsilon}(q_{min}+i\alpha)\right|\leq\max_{\epsilon\leq\tau\leq
1-\epsilon}\left|Q_{\epsilon}(\tau+i\alpha)\right|\leq\left|Q_{\epsilon}(q_{max}+i\alpha)\right|$

From the holomorphic of the $\frac{d\zeta(s)}{ds}$ follows
$\left|\frac{d\zeta(s)}{ds}\right|$ is harmonic function and
$|\frac{dQ_{\epsilon}}{ds}|$ is continuous function. And we have for its the
same estimates

$\max_{\epsilon\leq\tau\leq
1-\epsilon}\left|\frac{d\zeta(\tau+i\alpha)}{d\tau}\right|\leq\left|\frac{d\zeta(\tau+i\alpha)}{d\tau}\big{|}_{\tau=\tau_{dmax}}\right|$

$\max_{\epsilon\leq\tau\leq
1-\epsilon}\left|\frac{dQ_{\epsilon}(\tau+i\alpha)}{d\tau}\right|\leq\left|\frac{dQ_{\epsilon}(\tau+i\alpha)}{d\tau}\big{|}_{\tau=\tau_{qmax}}\right|$

For last statement of Lemma 9, we have

$\lim\limits_{N\to\infty}\int_{N}^{-N}e^{itk}\frac{dI_{\epsilon}}{dk}dk=\lim\limits_{N\to\infty}e^{itk}I_{\epsilon}\bigg{|}_{N}^{-N}-it\lim\limits_{N\to\infty}\int_{N}^{-N}e^{itk}I_{\epsilon}dk=$
$\lim\limits_{N\to\infty}[A_{N}+B_{N}]$

$Q_{\epsilon}\in L_{1}(-\infty,\infty)$ and the Riemann-Lebesgue lemma yield

$\lim\limits_{N\to\infty}[A_{N}]=0.$

Last estimates
$\left|Q_{\epsilon}\right|,\left|\frac{dQ_{\epsilon}}{d\tau}\right|$ ,Theorem
of Existence and invertibility of the Fourier  implies final statement of
Lemma 9 ∎

##  4 DISCUSSION

_Since we calculate the inversion of the Fourier transform only on a line
separated from the line where the zeta function has a root, the growth of
these estimates when approaching zero does not affect the final result. After
calculating the inverse Fourier transform, we begin to use completely
different estimates, which are already uniform, although the line tends to a
straight line, where the zeta function has a root and the intermediate
estimates do not satisfy the final goal!_

##  5 DISCUSSION

_Pay particular attention to the example of Davenport and Heilbronn-Type of
Functions.See in [16]
Not applicable to the Davenport and Heilbronn-Type of Functions. This method
can be applied only under the conditions of the existence of the Euler
product. _

######  Lemma 10.

Next statements is true

$I_{\epsilon}(k,\alpha)=F_{\epsilon}(k,\alpha)+(k-ia)\sum_{0}^{\infty}\frac{G_{\epsilon}(t_{n},\alpha)}{X^{\prime}_{-}(t_{n})(t_{n}-k)}$
$-i\sum_{0}^{\infty}\frac{G_{\epsilon}(t_{n},\alpha)}{X^{\prime}_{-}(t_{n})(t_{n}-k)}=k\sum_{0}^{\infty}\frac{dt_{n}}{da}\frac{d}{dt_{n}}\left(\frac{G_{\epsilon}(t_{n},\alpha)}{X^{\prime}_{-}(t_{n})(t_{n}-k)}\right)+\frac{dt_{n}}{da}\frac{d}{dt_{n}}\left(\frac{I_{\epsilon}(k,\alpha)-F_{\epsilon}(k,\alpha)}{X_{-}(k)}\right)$

###### Proof.

By Theorem 3,

$\Psi_{-}(k,\alpha)=\frac{I_{\epsilon}(k,\alpha)}{k-ia}-\frac{\widetilde{F_{\epsilon}}(k,\alpha)}{k-ia}=-\frac{X_{-}(k)}{2\pi
i}\int_{-\infty}^{\,\,\infty}\frac{G_{\epsilon}(t,\alpha)}{X_{-}(t)}\frac{dt}{t-k+i0}.$

Denote

$I_{1}=\int_{-\infty}^{\infty}\frac{G_{\epsilon}(t,\alpha)}{X_{-}(t)(t-k+i0)}dt,\,\,I_{2}=(k-ia)I_{1}.$

Holomorphics of the function $\frac{G_{\epsilon}(t,\alpha)}{t-k+i\delta}$ as
$\delta>0$ and analyticity of the function $X_{-}(t)$ in upper plane and Lemma
of Jordan yield

$I_{1}=\lim\limits_{\delta\downarrow
0}\int_{-\infty}^{\infty}\frac{G_{\epsilon}(t,\alpha)}{X_{-}(t)(t-k+i\delta)}dt=\sum_{0}^{\infty}\frac{G_{\epsilon}(t_{n},\alpha)}{X^{\prime}_{-}(t_{n})(t_{n}-k)}$
$\displaystyle\frac{I_{\epsilon}(k,\alpha)}{X_{-}(k)}=\frac{F_{\epsilon}(k,\alpha)}{X_{-}(k)}+(k-ia)\sum_{0}^{\infty}\frac{G_{\epsilon}(t_{n},\alpha)}{X^{\prime}_{-}(t_{n})(t_{n}-k)}$
(30)

differentiating (30) by a

$-i\sum_{0}^{\infty}\frac{G_{\epsilon}(t_{n},\alpha)}{X^{\prime}_{-}(t_{n})(t_{n}-k)}=k\sum_{0}^{\infty}\frac{dt_{n}}{da}\frac{d}{dt_{n}}\left(\frac{G_{\epsilon}(t_{n},\alpha)}{X^{\prime}_{-}(t_{n})(t_{n}-k)}\right)+\frac{dt_{n}}{da}\frac{d}{dt_{n}}\left(\frac{I_{\epsilon}(k,\alpha)-F_{\epsilon}(k,\alpha)}{X_{-}(k)}\right)$

∎

######  Theorem 5.

Let $s\in D(l,\epsilon)\cup\overline{D(l,\epsilon)}\,and\,a>2$, with
$3\epsilon<Re(s)<1-3\epsilon$ and $\Omega_{n}(1/2)=\epsilon>2/m$. Then,

$\displaystyle|Q(s)|<C_{l}C_{\epsilon}.$

###### Proof.

As $k\in(-N,N)$ uniformly-convergent series yields

$\int_{-N}^{+N}e^{itk}\frac{d}{dk}\left(\frac{I_{\epsilon}}{X_{-}}\right)dk=\int_{-N}^{+N}e^{itk}\frac{d}{dk}\left(\frac{F_{\epsilon}}{X_{-}}\right)dk+\int_{-N}^{+N}e^{itk}\frac{d}{dk}\left((k-ia)I_{1}\right)dk$

By definition $I_{1}$:

$\int_{-N}^{+N}e^{itk}\frac{d}{dk}\left((k-ia)I_{1}\right)dk=\int_{-N}^{+N}\sum_{1}^{N}e^{itk}\frac{d}{dk}\left(\frac{(k-ia)G_{\epsilon}(t_{n},\alpha)}{X^{\prime}_{-}(t_{n})(t_{n}-k)}\right)dk$
$+\int_{-N}^{+N}\sum_{N}^{\infty}\frac{d}{dk}\left(\frac{(k-ia)G_{\epsilon}(t_{n},\alpha)}{X^{\prime}_{-}(t_{n})(t_{n}-k)}\right)e^{itk}dk=W_{1}+W_{2}$

Finaly we get

$\left|\int_{-N}^{+N}e^{itk}\frac{d}{dk}\left(\frac{I_{\epsilon}}{X_{-}}\right)dk\right|\leq
C_{l}C_{\epsilon}$
$\left|\int_{-N}^{+N}e^{itk}\frac{d}{dk}\left(I_{\epsilon}\right)dk\right|\leq\left|\int_{-N}^{+N}e^{itk}\frac{d}{dk}\left(\frac{I_{\epsilon}}{X_{-}}-I_{\epsilon}\right)dk\right|+C_{l}C_{\epsilon}$

Lemma 2,Lemma 9-10,Theorem 4 and the last estimates yields

$|Q(s)|<2C_{l}C_{\epsilon}\,\,as\,\,3\epsilon<Re(s)<1-3\epsilon$

,

which completes the proof, ∎

As mentioned in Introduction, the values of the zeta function in adjacent
rectangles should be compared. This will be carried out in the following
theorem.

######  Theorem 6.

The Riemann’s function has nontrivial zeros only on the line $Re(s)=1/2$.

###### Proof.

Let it be assumed that there is a root of the zeta function with
$s_{n}=1/2+\delta_{n}+i*\alpha_{n}$, where $\delta_{n}>0$ i.e
$\Omega_{n}(1/2)=\delta_{n}>0$. Let $s_{n+1}=1/2+\delta_{n+1}+i\alpha_{n+1}$,
where $\delta_{n+1}\geq 0$ be another root nearest to it. Then, the following
sets corresponding to $s_{n}$ are constructed:

$\,\,D(n,\epsilon)=(s|\epsilon<Re(s)<1-\epsilon,Im(s)\neq
Im(s_{n}),Im(s_{n})-d_{n}\leq Im(s)\leq Im(s_{n})+d_{n}$
$\overline{D(n,\epsilon)}=(s|\epsilon<Re(s)<1-\epsilon,Im(-s)\neq
Im(-s_{n}),-Im(s_{n})-d_{n}\leq Im(s)\leq-Im(s_{n})+d_{n},$

where

$\zeta(s_{n+1})=0,\,\zeta(s_{n})=0,\,\zeta(1-s_{n})=0,\zeta(1-s_{n+1})=0,\,\,\zeta(1-s_{n})=0,\\\
$ $d_{n}=(Im(s_{n+1})-Im(s_{n}))/2$

where $\epsilon=0.01\delta_{n}(1/2-\delta_{n})>2/m$. As $1/2<Re(s)<1$ and
$s\in D(n,\epsilon)\cup\overline{D(n,\epsilon)}$, thus, we have the equation
for $Q_{m}$. Theorem 5 now yields

$|ln(\left|\zeta(1/2+\delta_{n}+i\alpha_{n}-i\delta)\right|)\leq|Q_{m}(1/2+\delta_{n}-i\alpha_{n}-i\delta)|+$
$|\sum_{n=m}^{\infty}P(ns)/n+\sum_{n=m}^{\infty}P(n(1-s))/n|<2C_{n}C_{\epsilon}$

Furthermore,

$\lim\limits_{\delta\rightarrow
0}\,\,\,|ln(\left|\zeta(1/2+\delta_{n}+i\alpha_{n}-i\delta)|)\right|=\infty.$

These estimates for $\left|Q_{m}(s)\right|,$ imply that the function does not
have zeros on the half plane $Re(s)>1/2$. By the integral representation (3),
these results are extended to the half plane $Re(s)<1/2$ i.e
$\Omega_{n}(1/2)=0$ . Thus, Riemann’s hypothesis has been proved. ∎

##  6 CONCLUSION

In this study, estimates were obtained for the logarithm of Riemann’s zeta
function off the line $Re(s)=1/2$. Thus, the work of great mathematicians
culminated by applying their achievements in this field. Without their
efforts, a solution to the problem would not have even been attempted.

This study on the Riemann hypothesis was completed by reducing it to a
Riemann-Hilbert boundary-value problem for analytic functions. This was
started by Riemann himself and continued by Hadamard among others, and the
present study has drawn on ideas by Landau, Walvis, Estarmann, and Chernoff.
It was possible to complete the proof of the Riemann hypothesis using the
solution to the Riemann-Hilbert boundary-value problem by Riemann, Hilbert,
and Poincaré.

After finishing this study, the author came to the conclusion that the problem
was actually solved by the joint efforts of Riemann, Hilbert, Poincaré, and
Fourier.

##  7 ACKNOWLEDGEMENTS

The author thanks the National Engineering Academy of the Republic of
Kazakhstan, in particular, Academician NAS RK B.Zhumagulov for constant
attention and support.

Moreover, the author thanks the Mathematics seminar at the Kazakhstan branch
of the Moscow State University for attention and valuable comments, as well as
Professors B. Kanguzhin and M. Otelbaev, and the organizers of
Automorphicformsworkshop.org/AFW2018 for their detailed review and valuable
comments.The author is especially grateful to Professor Steven Miller for a
thorough analysis of the work and detailed recommendations that have
significantly improved the paper. The author is especially grateful to P
Plotnikov, and A Mednykh for their thorough analysis of the work and detailed
recommendations that have significantly improved the paper. The author is
especially grateful to Mathematics seminar Nurlan Temirgaliev at the
L.N.Gumilyov Eurasian National University for their thorough analysis of the
work and detailed recommendations that have significantly improved the paper.

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