On small values of the Riemann zeta-function at Gram points

The questions considered in the present paper belong essentially to the discrete theory of the Riemann zeta-function $\zeta(s)$. This is the name of a whole class of problems where the behaviour of $\zeta(s)$ and related functions on different discrete sets of points $\{s_{n}\}$ is studied.

Let $\vartheta(t)$ be the increment of a continuous branch of the argument of the function $\pi^{-s/2}\Gamma\bigl(s/2\bigr)$ along the line segment joining the points $s = 0.5$ and $s = 0.5+it$ $(t>0)$. Then Hardy's $Z$-function, defined by the formula $Z(t) = e^{i\vartheta(t)}\zeta(0.5+it)$, is real for real $t$, and its real zeros coincides with the ordinates of zeros of $\zeta(s)$ lying on the critical line $\Re s = 0.5$ (see, for example, [1], Ch. III, § 4).

Let $\varrho_{n} = \beta_{n} + i\gamma_{n}, n = 1,2,\dots$, be complex zeros of $\zeta(s)$ lying in the upper half-plane and indexed in ascending order of their ordinates (and arbitrarily in the case of the coincidence of the latter): $0< \gamma_{1}< \gamma_{2}< \dots\leqslant \gamma_{n}\leqslant\gamma_{n+1}\leqslant \dotsb$. Further, for any $n\geqslant 0$, we define $t_{n}$ as the unique solution of the equation

$$\begin{equation*} \vartheta(t_{n})=(n-1)\pi \end{equation*}$$

that satisfies the condition $\vartheta'(t_{n}) > 0$. The values $t_{n}$ are called Gram points. Some characteristics of the sequences $\{\gamma_{n}\}$ and $\{t_{n}\}$ appear to be very close to one another, which is the reason for the interest of researchers in Gram points. In particular, it is known that the asymptotic relations

$$\begin{equation*} \gamma_{n}\sim t_{n}\sim \frac{2\pi n}{\ln{n}} \end{equation*}$$

hold when $n$ increases unboundedly, and that the number of ordinates $\gamma_{n}$ lying on the sufficiently long segment $(0,T]$, and expressed by the Riemann-von Mangoldt formula as

$$\begin{equation*} \frac{T}{2\pi}\ln\frac{T}{2\pi}-\frac{T}{2\pi}+O(\ln{T}), \end{equation*}$$

coincides with the number of Gram points $t_{n}$ lying on the same segment to within the term $O(\ln T)$ (see [2], Ch. X, § 6 and [3], § 7).

The equations

$$\begin{equation*} \zeta(0.5+it)=e^{-i\vartheta(t)}Z(t)=Z(t)\cos{\vartheta(t)}-iZ(t)\sin{\vartheta(t)} \end{equation*}$$

and the definition of $t_{n}$ imply that the values $\zeta(0.5+it_{n})$ are real and coincide with the $x$-coordinates of the points of intersection of the curve

$$\begin{equation*} \mathscr{C}\colon t\mapsto (\operatorname{Re}\zeta(0.5+it),\operatorname{Im}\zeta(0.5+it)) \end{equation*}$$

and the real axis. The zeros of Hardy's function $Z(t)$ also correspond to the intersections of $\mathscr{C}$ and the real axis, but such points coincide with the origin.

It is natural to ask how these points of intersection are distributed. For example, with increasing $n$, can they run arbitrarily far to the left or to the right of the origin, or, conversely, approach it arbitrarily closely?

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A positive answer to the first question was given in the papers [4] and [5]. It appears that in the case of sufficiently large $N$ there exist indices $m$ and $n$ such that $1\leqslant m,n\leqslant N$ and

$$\begin{equation*} \zeta(0.5+it_{n}) > c(\ln{N})^{5/4},\qquad \zeta(0.5+it_{m}) < -c(\ln{N})^{5/4}, \end{equation*}$$

for some absolute constant $c > 0$. Moreover, for any $0 < \varepsilon < 0.5$ and $N\geqslant N_{0}(\varepsilon)$, there exists a number $k$ lying on the same segment such that

$$\begin{equation*} |\zeta(0.5+it_{k})|>c(\varepsilon)\exp{\biggl((0.5-\varepsilon) \sqrt{\frac{\ln{N}}{\ln\ln{N}}}\biggr)}. \end{equation*}$$

The aim of this paper is to answer the second question.

On the one hand, the coincidence of a Gram point $t_{n}$ with any ordinate $\gamma_{m}$ of a zero of $\zeta(s)$ seems absolutely improbable. This allows one to suppose that

$$\begin{equation} \zeta(0.5+it_{n})\ne 0 \end{equation} \tag{ 1.1 }$$

for any ¹ $n\geqslant 0$. At the same time, the numerical data provides examples of Gram points with values $\zeta(0.5+it_{n})$ that are 'anomalously' close to zero. The table below contains all the indices $n$ that do not exceed $10^{8}$ and are such that these values do not exceed $10^{-7}$ in modulus.

no.	$n$	$\zeta(0.5+it_{n})$	no.	$n$	$\zeta(0.5+it_{n})$
1	368 383	$8.908459\cdot 10^{-8}$	7	55 785 549	$3.751899\cdot 10^{-8}$
2	12 984 109	${-}2.052298\cdot 10^{-8}$	8	61 769 885	$8.026449\cdot 10^{-8}$
3	21 567 185	${-}6.709404\cdot 10^{-8}$	9	65 463 721	$1.083783\cdot 10^{-8}$
4	45 898 152	$8.302024\cdot 10^{-8}$	10	69 612 841	$9.033652\cdot 10^{-8}$
5	50 550 325	${-}6.397503\cdot 10^{-8}$	11	74 201 244	$2.875945\cdot 10^{-8}$
6	52 220 649	${-}6.529425\cdot 10^{-8}$	–	–	–

The same segment contains 60 values of $n$ with the condition $|\zeta(0.5+it_{n})| < 10^{-6}$.

In what follows, the theoretical explanation of this phenomenon is given. Namely, the following assertion holds.

Theorem 1. Suppose that $0< \varepsilon < 10^{-3}$ is fixed, $N\geqslant N_{0}(\varepsilon)>0,$ $M=[N^{\alpha+\varepsilon}]$, and $\alpha=27/82$. Further, let $\varphi(x)$ be any unbounded and monotonically increasing function that satisfies the conditions

$$\begin{equation*} a\varepsilon^{-1}\leqslant \varphi(x)\leqslant \sqrt{\ln\ln{x}}, \end{equation*}$$

where $a$ is a sufficiently large positive constant. Then the inequality

$$\begin{equation*} |\zeta(0.5+it_{n})| < \exp{\biggl(-\frac{1}{\varphi}\ln\ln{N}\biggr)}, \end{equation*}$$

holds with $\varphi=\varphi(N)$ for at least

$$\begin{equation*} \frac{M\varphi}{2\pi\sqrt{\ln\ln{N}}}\, \exp{\biggl(-\frac{3}{\varphi^{2}}\ln\ln{N}\biggr)} \end{equation*}$$

Gram points $t_{n},$ $N < n\leqslant N+M$.

The lower bound for the negative 'discrete' moment of $\zeta(s)$ is an immediate consequence of the above assertion.

Theorem 2. Suppose that $\varepsilon$ and $\delta$ are any fixed numbers that satisfy the conditions $0 < \varepsilon < 10^{-3}$ and $\delta>0$. Then the estimate

$$\begin{equation*} \sum_{N < n\leqslant N+M}|\zeta(0.5+it_{n})|^{-1} >M\exp{\biggl(\frac{\ln\ln{N}}{(\ln\ln\ln{N})^{\delta}}\biggr)} \end{equation*}$$

holds for any $N\geqslant N_{0}(\varepsilon,\delta)$ and $M=[N^{\alpha+\varepsilon}],$ $\alpha=27/82$.

In proving these theorems, we essentially use the ideas and methods of the classical paper [7] of Selberg and the recent paper [8] of Radziwiłł. In particular, this relates to the notation of the number of solutions of the equation (3.10) proposed in [8] by means of the multiplicative function $f(n)$ (see Lemma 7). This also relates to the trick of excluding from consideration the 'bad' set where the function under study is 'anomalously' large (the set $\mathscr{A}_{3}$ in Lemma 7).

In what follows, we use the following notations: $|\mathscr{A}|$ denotes the number of elements of a finite set $\mathscr{A}$; $G(x)$ denotes the Gaussian distribution, that is,

$$\begin{equation*} \begin{gathered} G(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{-u^{2}/2}\,du;\\ J_{0}(z)=\sum_{m=0}^{+\infty}\frac{(-1)^{m}}{(m!)^{2}} \biggl(\frac{z}{2}\biggr)^{2m} \end{gathered} \end{equation*}$$

is the Bessel function of the first kind; the symbols $\sum_{n}$ and $\sum_{n\in \mathscr{A}}$ stand for the sum over all $n$ from the interval $(N,N+M]$ and for the sum over all $n$ under the condition $n\in \mathscr{A}$ lying in the same interval, correspondingly; $\mathbf{E}(\xi)$ denotes the mean value of the random variable $\xi$ and $\mathbf{P}\{\xi\leqslant x\}$ means the probability of the event 'the random variable $\xi$ does not exceed $x$'; finally, $\theta, \theta_{1}, \theta_{2},\dots$ are complex numbers (generally speaking, different in different formulae) whose absolute value does not exceed 1.

This section contains auxiliary lemmas of a 'technical' nature.

Lemma 1. Let $\{\xi_{N}\}$ be a sequence of random variables. Suppose that there exist some fixed $r>0$ and some positive parameters $\sigma(N)$ and $\kappa(N)$ that increase unboundedly as $N\to+\infty$ and such that the generating function $K_{N}(s) =\mathbf{E}(e^{s\xi_{N}})$ for the moments of the random variable $\xi_{N}$ satisfies the relation

$$\begin{equation*} K_{N}(s)=\exp{\biggl(\frac{s^{2}}{4}\sigma(N)\biggr)}\Phi(s)(1+O(\kappa^{-1}(N))) \end{equation*}$$

in the disc $|s|\leqslant r$, where the implied constant in the $O$-symbol is absolute and the function $\Phi(s),$ $\Phi(0) = 1$, is analytic in the disc $|s|\leqslant r$ and independent of $N$. Then the distribution

$$\begin{equation*} F_{N}(u)=\mathbf{P}\Biggl\{\frac{\xi_{N}\sqrt{2}}{\sqrt{\sigma(N)}}\leqslant u\Biggr\} \end{equation*}$$

of the normalized random variable $\xi_{N}$ satisfies the following relations for any $u=o(\min{(\kappa(N),\sqrt{\sigma(N)})}),$ $u>1$:

$$\begin{equation*} \frac{1-F_{N}(u)}{1-G(u)}=1+O(r(u;N)),\qquad \frac{F_{N}(-u)}{G(-u)}=1+O(r(u;N)), \end{equation*}$$

where

$$\begin{equation*} r(u;N)=u\biggl(\frac{1}{\kappa(N)}+\frac{1}{\sqrt{\sigma(N)}}\biggr), \end{equation*}$$

and the constants in $O$ are absolute.

This is a particular case of Hwang's theorem in [9], where $u(s)=s^{2},$ $v(s)=\ln\Phi(s),$ $\phi(N)=\sigma(N)/4$.

Lemma 2. Let $k, \mu, \nu$ be integers, $\mu,\nu\geqslant 0,$ $k\geqslant 1,$ $\mu+\nu =k,$ $y_{0}\leqslant y\leqslant N^{1/(2k)}$. Further, suppose that $p_{1},\dots, p_{\mu}, q_{1},\dots, q_{\nu}$ run through the prime numbers from the interval $(1,y]$ that satisfy the condition $p_{1}\dotsb p_{\mu}\ne q_{1}\dotsb q_{\nu}$. Finally, suppose that all the terms of a sequence $a(p)$ for $p\leqslant y$ satisfy the condition $|a(p)|\leqslant \delta$. Then the sum

$$\begin{equation*} S=\sum_{n}\sum_{\substack{p_{1},\dots, p_{\mu}\\ q_{1},\dots,q_{\nu}}}\frac{a(p_{1})\dotsb \overline{a}(q_{\nu})}{\sqrt{p_{1} \dotsb q_{\nu}}}\biggl(\frac{q_{1}\dotsb q_{\nu}}{p_{1}\dotsb p_{\mu}}\biggr)^{it_{n}} \end{equation*}$$

satisfies the bound $|S|< (\delta x)^{k}\ln{N}$.

The proof repeats almost verbatim that of Lemma 6 in [10].

First we introduce some additional notation. Suppose that $2\leqslant x\leqslant t^{2}$. Following Selberg (see [7]), we define

$$\begin{equation*} \sigma_{x,t}=0.5+2\max{(|\beta-0.5|,(\ln x)^{-1})}, \end{equation*}$$

where the maximum is taken over all the zeros $\varrho = \beta + i\gamma$ of the Riemann zeta-function that satisfy the condition

$$\begin{equation*} |t-\gamma|\leqslant \frac{x^{\scriptstyle 3|\beta-0.5|}}{\ln x}. \end{equation*}$$

Further, let $n\geqslant 2$ be an integer. We set

$$\begin{equation*} \Lambda_{x}(n)=\begin{cases} \Lambda(n) &\text{if } 1\leqslant n\leqslant x,\\ \Lambda(n)\dfrac{\ln^{2}{(x^{3}/n)-2\ln^{2}{(x^{2}/n)}}}{2\ln^{2}x} &\text{if } x< n\leqslant x^{2},\\ \Lambda(n)\dfrac{\ln^{2}{(x^{3}/n)}}{2\ln^{2}x} &\text{if } x^{2}< n\leqslant x^{3}. \end{cases} \end{equation*}$$

The proof, going back to Selberg's theorem, stating that the function

$$\begin{equation*} \xi(t)=\frac{\sqrt{2}\ln|\zeta(0.5+it)|}{\sqrt{\ln\ln T}},\qquad 0< t\leqslant T, \end{equation*}$$

has a distribution tending to the Gaussian as $T\to +\infty$, is based on the following 'explicit' formula for the real part of the logarithm of the Riemann zeta-function:

$$\begin{equation} \begin{gathered} \begin{split} &\ln{|\zeta(0.5+it)|} =\sum_{m\leqslant x^{3}}\frac{\Lambda_{x}(m)}{m^{\sigma_{x,t}}} \,\frac{\cos{(t\ln{m})}}{\ln{m}}\\ &\qquad +O\biggl((\sigma_{x,t}-0.5)\bigl((\sigma_{x,t}-0.5)\ln{x}+\ln{(\chi(t))}\bigr) \biggl(\biggl|\sum_{m\leqslant x^{3}}\frac{\Lambda_{x}(m)}{m^{\sigma_{x,t}+it}}\biggr| +\ln{t}\biggr)\biggl), \end{split}\\ \chi(t)=1+\frac{1}{\eta(t)\ln{x}}; \end{gathered} \end{equation} \tag{ 3.1 }$$

here $\eta(t)=\min_{\varrho=\beta+i\gamma}|t-\gamma|$ is the distance between $t$ and the closest ordinate of a zero of $\zeta(s)$, so $\chi(\gamma_{m})=+\infty$ (see [11], Theorem 5.2).

However, this formula is unsuitable for handling the random variable with the values

$$\begin{equation*} \xi(t_{n})=\frac{\sqrt{2}\ln|\zeta(0.5+it_{n})|}{\sqrt{\ln\ln N}}, \qquad 0< n\leqslant N. \end{equation*}$$

The reason is that the proof of Selberg's theorem is based on estimates for the integrals of the positive powers of the $O$-term in the right-hand side of (3.1) and, in particular, on estimates for integrals of the following type:

$$\begin{equation} \int_{0}^{T}(\ln{\chi(t)})^{2m}\,dt,\qquad m=1,2,\dotsc. \end{equation} \tag{ 3.2 }$$

All these integrals are finite because, for example, the integrals

$$\begin{equation*} \int_{0}^{1}\biggl(\ln{\frac 1t}\biggr)^{2m}\,dt \end{equation*}$$

are finite. In the discrete case, one should handle the sums

$$\begin{equation} \sum_{n\leqslant N}(\ln{\chi(t_{n})})^{2m},\qquad m=1,2,\dots, \end{equation} \tag{ 3.3 }$$

instead of the integrals (3.2). The absence of precise information about the relative location of Gram points and ordinates of zeros of $\zeta(s)$ does not enable us even to assert the finiteness of the quantities (3.3).

The next lemma partially eliminates this inconvenience and enables us to estimate the logarithm of $|\zeta(0.5+it)|$ above by some function $C(t)$ for any $t$. Roughly speaking, this function can be obtained by omitting the factor $\ln{\chi(t)}$ from the right-hand side of (3.1).

Lemma 3. Suppose that $10\leqslant x \leqslant t^{\delta},$ $800\delta=1$. Then the inequality

$$\begin{equation*} \ln{|\zeta(0.5+it)|}\leqslant C(t) \end{equation*}$$

holds with

$$\begin{equation*} C(t)=\sum_{m\leqslant x^{3}}\frac{\Lambda_{x}(m)}{m^{\sigma_{x,t}}}\, \frac{\cos{(t\ln{m})}}{\ln{m}} + (\sigma_{x,t}-0.5)^{2}(\ln{x})\biggl(a_{1}\ln{t}+a_{2}\biggl|\sum_{p\leqslant x^{3}}\frac{\Lambda_{x}(p)}{p^{\sigma_{x,t}+it}}\biggr|\biggr), \end{equation*}$$

$a_{1}=4.01,$ $a_{2}=7.52$.

Proof. If $0.5+it$ is a zero of $\zeta(s)$ then the inequality of the lemma holds. So in what follows we assume that $\zeta(0.5+it)\ne 0$. In this case, we have

$$\begin{equation*} \ln{|\zeta(0.5+it)|} =-\operatorname{Re}\biggl(\operatorname{v.p.} \biggl(\int_{0.5}^{\sigma_{x,t}} +\int_{\sigma_{x,t}}^{+\infty}\biggr)\frac{\zeta'(\sigma + it)}{\zeta(\sigma+it)}\,d\sigma\biggr)=-I_{1}-I_{2}+I_{3}, \end{equation*}$$

where

$$\begin{equation*} \begin{aligned} I_{1}&=(\sigma_{x,t}-0.5)\operatorname{Re} \frac{\zeta'(\sigma_{x,t}+it)}{\zeta(\sigma_{x,t}+it)},\\ I_{2}&=\operatorname{Re}\biggl(\operatorname{v.p.}\int_{\sigma_{x,t}}^{+\infty} \frac{\zeta'(\sigma+it)}{\zeta(\sigma+it)}\,d\sigma\biggr),\\ I_{3}&=\operatorname{Re}\biggl(\operatorname{v.p.}\int_{0.5}^{\sigma_{x,t}} \biggl(\frac{\zeta'(\sigma_{x,t}+it)}{\zeta(\sigma_{x,t}+it)} -\frac{\zeta'(\sigma+it)}{\zeta(\sigma+it)}\biggr)\,d\sigma\biggr). \end{aligned} \end{equation*}$$

To estimate $I_{1}$ we use the inequality

$$\begin{equation} \biggl|\frac{\zeta'(\sigma_{x,t}+it)}{\zeta(\sigma_{x,t}+it)} \biggr|<0.8\ln{t}+2.6|r(x,t)|+4.82, \qquad r(x,t)=\sum_{m\leqslant x^{3}}\frac{\Lambda_{x}(m)}{m^{\sigma_{x,t}+it}}, \end{equation} \tag{ 3.4 }$$

proved in [7] (the calculation of the constants is contained in Lemma 7 of the survey [12]). Thus we get

$$\begin{equation} |I_{1}| \leqslant (\sigma_{x,t}-0.5)(0.8\ln{t}+2.6|r(x,t)|+4.82). \end{equation} \tag{ 3.5 }$$

Using the estimate $0.5(\sigma_{x,t}-0.5)\ln{x}\geqslant 1$ we rewrite (3.5) as follows:

$$\begin{equation*} |I_{1}|\leqslant (\sigma_{x,t}-0.5)^{2}(\ln{x})\{a_{1}^{(1)}\ln{t} +a_{2}^{(1)}|r(x,t)|+a_{3}^{(1)}\}, \end{equation*}$$

where $a_{1}^{(1)}=0.4,$ $a_{2}^{(1)}=1.3,$ $a_{3}^{(1)}=2.41$.

To calculate $I_{2}$ we apply the formula

$$\begin{equation*} \frac{\zeta'(s)}{\zeta(s)}=-\sum_{m\leqslant x^{3}} \frac{\Lambda_{x}(m)}{m^{s}}+\frac{13\theta}{3}x^{1/4-\sigma/2}(0.5\ln{t}+|r(x,t)|+4), \end{equation*}$$

where $s=\sigma+it,$ $\sigma\geqslant\sigma_{x,t}$ (see [7] and [12], Lemma 3)². Thus we find

$$\begin{equation*} \begin{aligned} I_{2}&=-\operatorname{Re}\int_{\sigma_{x,t}}^{+\infty} \biggl(\sum_{m\leqslant x^{3}}\frac{\Lambda_{x}(m)}{m^{\sigma+it}} +\frac{13\theta}{3}x^{1/4-\sigma/2} (0.5\ln{t} +|r(x,t)|+ 4)\biggr)\,d\sigma\\ &=-\operatorname{Re}\biggl(\sum_{m\leqslant x^{3}} \frac{\Lambda_{x}(m)}{\ln{m}}\frac{1}{m^{s}}\biggr) +\theta(\sigma_{x,t}-0.5)^{2}(\ln{x})\{a_{1}^{(2)}\ln{t} + a_{2}^{(2)}|r(x,t)|+a_{3}^{(2)}\}, \end{aligned} \end{equation*}$$

where $a_{1}^{(2)}=13/(12e),$ $a_{2}^{(2)}=13/(6e),$ $a_{3}^{(2)}=26/(3e)$.

Finally, using the equality

$$\begin{equation*} \frac{\zeta'(s)}{\zeta(s)}=\sum_{\varrho}\frac{1}{s-\varrho}+\theta(0.5\ln{|t|}+3) \end{equation*}$$

(see [13], §§ 76, 77 and also [12], Lemma 2), for $\zeta(\sigma+it)\ne 0$ we obtain

$$\begin{equation*} \operatorname{Re}\biggl(\frac{\zeta'(\sigma_{x,t}+it)}{\zeta(\sigma_{x,t}+it)} -\frac{\zeta'(\sigma+it)}{\zeta(\sigma+it)}\biggr)=\sum_{\varrho}\kappa(\varrho) + \theta_{1}(\ln{t}+6), \end{equation*}$$

where

$$\begin{equation*} \begin{aligned} \kappa(\varrho) &=\frac{\sigma_{x,t}-\beta}{(\sigma_{x,t}-\beta)^{2}+(t-\gamma)^{2}} -\frac{\sigma-\beta}{(\sigma-\beta)^{2}+(t-\gamma)^{2}}\\ &=\frac{(\sigma_{x,t}-\sigma)(t-\gamma)^{2}+(\sigma_{x,t}-\sigma)(\sigma_{x,t}-\beta)(\beta-\sigma)} {\{(\sigma_{x,t}-\beta)^{2}+(t-\gamma)^{2}\}\{(\sigma-\beta)^{2}+(t-\gamma)^{2}\}}\\ &=\kappa_{1}(\varrho)+\kappa_{2}(\varrho). \end{aligned} \end{equation*}$$

We also put

$$\begin{equation*} j_{r}(\varrho)=\int_{0.5}^{\sigma_{x,t}}\kappa_{r} (\varrho)\,d\sigma,\qquad j(\varrho)=j_{1}(\varrho)+j_{2}(\varrho). \end{equation*}$$

Then

$$\begin{equation*} I_{3}=\sum_{\varrho}j(\varrho)+\theta(\sigma_{x,t}-0.5)(\ln{t}+6). \end{equation*}$$

Since the estimates

$$\begin{equation*} \begin{aligned} 0\leqslant \kappa_{1}&=\frac{\sigma_{x,t}-\sigma}{(\sigma_{x,t}-\beta)^{2} +(t-\gamma)^{2}}\,\frac{(t-\gamma)^{2}}{(\sigma-\beta)^{2}+(t-\gamma)^{2}}\\ &\leqslant \frac{\sigma_{x,t}-\sigma}{(\sigma_{x,t}-\beta)^{2}+(t-\gamma)^{2}} \end{aligned} \end{equation*}$$

are valid on the whole segment $0.5\leqslant \sigma\leqslant \sigma_{x,t}$, for any zero $\varrho = \beta+i\gamma$ we have

$$\begin{equation*} \begin{aligned} j_{1} &\leqslant \int_{0.5}^{\sigma_{x,t}}\frac{(\sigma_{x,t}-\sigma) \,d\sigma}{(\sigma_{x,t}-\beta)^{2}+(t-\gamma)^{2}} =\frac{1}{2}\,\frac{(\sigma_{x,t}-0.5)^{2}}{(\sigma_{x,t}-\beta)^{2}+(t-\gamma)^{2}}\\ &\leqslant \frac{1}{4}(\sigma_{x,t}-0.5)(\ln{x}) \frac{(\sigma_{x,t}-0.5)^{2}}{(\sigma_{x,t}-\beta)^{2}+(t-\gamma)^{2}}. \end{aligned} \end{equation*}$$

To estimate $\kappa_{2}$ and $j_{2}$, we note first that the numerator of $\kappa_{2}$ is nonpositive on the whole segment of integration in both the cases $\beta\leqslant 0.5$ and $\beta\geqslant \sigma_{x,t}$. Hence, for all such zeros, we have $j_{2}\leqslant 0$. So, in what follows we assume that $0.5< \beta< \sigma_{x,t}$. Let us put $\tau = |t-\gamma|$ and consider two cases.

• Case 1:  
  $\tau > (\ln{x})^{-1}$. Transforming $\kappa_{2}$, we find

  $$\begin{equation*} \begin{aligned} \kappa_{2} &=\frac{(\sigma_{x,t}-\sigma)(\sigma_{x,t}-\sigma+\sigma-\beta)(\beta -\sigma)}{\{(\sigma_{x,t}-\beta)^{2}+\tau^{2}\}\{(\sigma-\beta)^{2}+\tau^{2}\}}\\ &=\frac{(\sigma_{x,t}-\sigma)^{2}(\beta-\sigma)-(\sigma_{x,t}-\sigma)(\beta-\sigma)^{2}} {\{(\sigma_{x,t}-\beta)^{2}+\tau^{2}\}\{(\sigma-\beta)^{2}+\tau^{2}\}}\\ &\leqslant \frac{(\sigma_{x,t}-\sigma)^{2}(\beta-\sigma)}{\{(\sigma_{x,t}-\beta)^{2} +\tau^{2}\}\{(\sigma-\beta)^{2}+\tau^{2}\}}. \end{aligned} \end{equation*}$$

  Further, noting that the last fraction is negative for $\sigma > \beta$ we get

  $$\begin{equation} \begin{aligned} j_{2} &\leqslant \int_{0.5}^{\beta}\frac{(\sigma_{x,t}-\sigma)^{2} (\beta-\sigma)\,d\sigma}{\{(\sigma_{x,t}-\beta)^{2}+\tau^{2}\} \{(\sigma-\beta)^{2}+\tau^{2}\}}\\ &\leqslant \frac{(\sigma_{x,t}-0.5)^{2}}{(\sigma_{x,t}-\beta)^{2} +\tau^{2}}\int_{0.5}^{\beta} \frac{(\beta-\sigma)\,d\sigma}{(\beta-\sigma)^{2}+\tau^{2}}\\ &=\frac{1}{2}\,\frac{(\sigma_{x,t}-0.5)^{2}}{(\sigma_{x,t}-\beta)^{2}+\tau^{2}} \ln\frac{(\beta-0.5)^{2}+\tau^{2}}{\tau^{2}}\\ &\leqslant \frac{1}{2}\,\frac{(\sigma_{x,t}-0.5)^{2}}{(\sigma_{x,t}-\beta)^{2} +\tau^{2}}\ln\frac{(\sigma_{x,t}-0.5)^{2}+\tau^{2}}{\tau^{2}}\\ &\leqslant \frac{(\sigma_{x,t}-0.5)^{2}}{(\sigma_{x,t}-\beta)^{2}+\tau^{2}} \ln{\biggl(1\,{+}\,\frac{\sigma_{x,t}-0.5}{\tau}\biggr)}\\ &\leqslant \frac{(\sigma_{x,t}-0.5)^{2}}{(\sigma_{x,t}-\beta)^{2}+\tau^{2}} \,\frac{\sigma_{x,t}-0.5}{\tau}\\ & < \frac{(\sigma_{x,t}-0.5)^{2}}{(\sigma_{x,t}-\beta)^{2}+\tau^{2}} (\sigma_{x,t}-0.5)\ln{x}. \end{aligned} \end{equation} \tag{ 3.6 }$$
• Case 2:  
  $0 \leqslant \tau\leqslant (\ln{x})^{-1}$. First let us assume that $\sigma_{x,t}=0.5+2(\beta_{0}-0.5)$, where $\beta_{0}>0.5$ is the abscissa of some zero $\varrho_{0}=\beta_{0}+i\gamma_{0}$ that satisfies the condition

  $$\begin{equation*} |t-\gamma_{0}|\leqslant \frac{x^{3|\beta_{0}-0.5|}}{\ln{x}}. \end{equation*}$$

Then the definition of $\sigma_{x,t}$ implies that $\beta_{0}$ is the largest among the abscissas of the zeros $\varrho^{*} =\beta^{*}+i\gamma^{*}$ satisfying the condition

$$\begin{equation} |t-\gamma^{*}|\leqslant \frac{x^{3|\beta^{*}-0.5|}}{\ln{x}}. \end{equation} \tag{ 3.7 }$$

But the estimate

$$\begin{equation*} \frac{x^{3|\beta^{*}-0.5|}}{\ln{x}}\geqslant\frac{1}{\ln{x}} \end{equation*}$$

holds for any $\beta^{*}$. Since the zero $\varrho = \beta + i\gamma$ under consideration satisfies the inequality $\tau =|t-\gamma|\leqslant (\ln{x})^{-1}$, this zero $\varrho$ is certainly contained in the set of zeros $\varrho^{*}$ satisfying (3.7). Hence its abscissa $\beta$ cannot exceed $\beta_{0}$:

$$\begin{equation*} \beta\leqslant \beta_{0}=0.5(\sigma_{x,t}+0.5). \end{equation*}$$

The last inequality implies that

$$\begin{equation} 0\leqslant \beta-0.5\leqslant 0.5(\sigma_{x,t}-0.5)\leqslant \sigma_{x,t}-\beta. \end{equation} \tag{ 3.8 }$$

Now let us assume that $\sigma_{x,t}=0.5+2(\ln{x})^{-1}$. Then for all the zeros under condition (3.7) (and, in particular, for the zero $\varrho = \beta + i\gamma$) we have

$$\begin{equation*} |\beta^{*}-0.5|\leqslant (\ln{x})^{-1}=0.5(\sigma_{x,t}-0.5) \leqslant \sigma_{x,t}-\beta^{*}. \end{equation*}$$

Thus, in this case the abscissa $\beta$ satisfies (3.8), too. Passing to the estimate of $j_{2}(\varrho)$, we find

$$\begin{equation*} \begin{aligned} j_{2} &=\frac{\sigma_{x,t}-\beta}{(\sigma_{x,t}-\beta)^{2} +\tau^{2}}\biggl(\int_{0.5}^{\beta}+\int_{\beta}^{\sigma_{x,t}}\biggr) \frac{(\sigma_{x,t}-\sigma)(\beta-\sigma)}{(\sigma-\beta)^{2}+\tau^{2}}\,d\sigma\\ &=\frac{\sigma_{x,t}-\beta}{(\sigma_{x,t}-\beta)^{2}+\tau^{2}}(j_{3}-j_{4}), \end{aligned} \end{equation*}$$

where

$$\begin{equation*} j_{3}=\int_{0.5}^{\beta}\frac{(\sigma_{x,t}-\sigma) (\beta-\sigma)\,d\sigma}{(\sigma- \beta)^{2}+\tau^{2}}, \qquad j_{4}=\int_{\beta}^{\sigma_{x,t}}\frac{(\sigma_{x,t}-\sigma) (\sigma-\beta)\,d\sigma}{(\sigma- \beta)^{2}+\tau^{2}}. \end{equation*}$$

Setting $u=\beta-\sigma$ and $v=\sigma-\beta$ in $j_{3}$ and $j_{4}$ respectively, we obtain

$$\begin{equation*} \begin{aligned} j_{3}&=(\sigma_{x,t}-\beta)\int_{0}^{\beta-0.5}\frac{u\,du}{u^{2}+\tau^{2}} +\int_{0}^{\beta-0.5}\frac{u^{2}\,du}{u^{2}+\tau^{2}}\\ &\leqslant (\sigma_{x,t}-\beta)\frac{1}{2}\ln {\frac{(\beta-0.5)^{2}+\tau^{2}}{\tau^{2}}}+(\beta-0.5),\\ j_{4}&=(\sigma_{x,t}-\beta)\int_{0}^{\sigma_{x,t}-\beta}\frac{v\,dv}{v^{2}+\tau^{2}} -\int_{0}^{\sigma_{x,t}-\beta}\frac{v^{2}\,dv}{v^{2}+\tau^{2}}\\ &\geqslant(\sigma_{x,t}-\beta)\frac{1}{2} \ln{\frac{(\sigma_{x,t}-\beta)^{2}+\tau^{2}}{\tau^{2}}}-(\sigma_{x,t}-\beta), \end{aligned} \end{equation*}$$

and hence

$$\begin{equation*} j_{3}-j_{4}\leqslant (\sigma_{x,t}-\beta)\frac{1}{2} \ln{\frac{(\beta-0.5)^{2}+\tau^{2}}{(\sigma_{x,t}-\beta)^{2} +\tau^{2}}}+(\sigma_{x,t}-0.5). \end{equation*}$$

By (3.8), the fraction under the logarithm sign does not exceed 1. Therefore,

$$\begin{equation} j_{2}\leqslant \frac{\sigma_{x,t}-\beta}{(\sigma_{x,t}-\beta)^{2}+\tau^{2}} (\sigma_{x,t}-0.5)\leqslant \frac{1}{2}\, \frac{(\sigma_{x,t}-0.5)^{2}}{(\sigma_{x,t}-\beta)^{2}+\tau^{2}} (\sigma_{x,t}-0.5)\ln{x}. \end{equation} \tag{ 3.9 }$$

Comparing (3.6) and (3.9) we conclude that inequality (3.6) holds for any $\beta$. Hence,

$$\begin{equation*} j(\varrho)=j_{1}(\varrho)+j_{2}(\varrho)\leqslant \frac{5}{4}\, \frac{(\sigma_{x,t}-0.5)^{2}}{(\sigma_{x,t}-\beta)^{2}+\tau^{2}} (\sigma_{x,t}-0.5)\ln{x}. \end{equation*}$$

Now let us use the inequality

$$\begin{equation*} \sum_{\varrho}\frac{\sigma_{x,t}-0.5}{(\sigma_{x,t}-\beta)^{2}+(t-\gamma)^{2}} \leqslant \frac{5}{3}\biggl(\operatorname{Re}\frac{\zeta'(\sigma_{x,t}+it)} {\zeta(\sigma_{x,t}+it)}+0.5\ln{t}+3\biggr) \end{equation*}$$

from [7] (see also [12], Lemma 6), together with the estimate (3.4). Thus we get

$$\begin{equation*} \begin{aligned} \sum_{\varrho}j(\varrho) &\leqslant \frac{5}{4}(\ln{x})(\sigma_{x,t}-0.5)^{2}\sum_{\varrho} \frac{\sigma_{x,t}-0.5}{(\sigma_{x,t}-\beta)^{2}+(t-\gamma)^{2}}\\ &\leqslant \frac{5}{4}(\ln{x})(\sigma_{x,t}-0.5)^{2}\biggl(\frac{5}{3} \operatorname{Re}\frac{\zeta'(\sigma_{x,t}+it)}{\zeta(\sigma_{x,t}+it)} +\frac{5}{6}\ln{t}+5\biggr)\\ &\leqslant (\ln{x})(\sigma_{x,t}-0.5)^{2}\biggl(\frac{65}{24} \ln{t}+\frac{13}{4}|r(x,t)|+16.3\biggr),\\ I_{3}&\leqslant \sum_{\varrho}j(\varrho)+(\ln{x})(\sigma_{x,t}-0.5)^{2}(0.5\ln{t}+3)\\ &\leqslant (\ln{x})(\sigma_{x,t}-0.5)^{2}\{a_{1}^{(3)}\ln{t} + a_{2}^{(3)}|r(x,t)|+a_{3}^{(3)}\}, \end{aligned} \end{equation*}$$

where $a_{1}^{(3)}=77/24,$ $a_{2}^{(3)}=13/4,$ $a_{3}^{(3)}=19.3$. Therefore,

$$\begin{equation*} \begin{gathered} \ln{|\zeta(0.5+it)|}\leqslant \sum_{m\leqslant x^{3}}\frac{\Lambda_{x}(m)}{\ln{m}} \,\frac{\cos{(t\ln{m})}}{m^{\sigma_{x,t}}}+R,\\ R=(\ln{x})(\sigma_{x,t}-0.5)^{2}\{a_{1}^{(4)}\ln{t}+ a_{2}^{(4)}|r(x,t)|+a_{3}^{(4)}\}, \qquad a_{j}^{(4)}=\sum_{k=1}^{3}a_{j}^{(k)}, \end{gathered} \end{equation*}$$

where $a_{1}^{(4)} < 4.00687,$ $a_{2}^{(4)} < 7.51374,$ $a_{3}^{(4)} < 24.89829$. Further, setting

$$\begin{equation*} q(x,t)=\sum_{p\leqslant x^{3}}\frac{\Lambda_{x}(p)}{p^{\sigma_{x,t}+it}} \end{equation*}$$

and noting that

$$\begin{equation*} \begin{aligned} |r(x,t)| &\leqslant |q(x,t)|+\sum_{k\geqslant 2} \sum_{p\leqslant x^{3/k}}\frac{\ln{p}}{p^{k/2}}\\ &\leqslant |q(x,t)|+\sum_{p\leqslant x^{3/2}} \frac{\ln{p}}{p}+\sum_{p}\frac{\ln{p}}{p(\sqrt{p}-1)}\\ & < |q(x,t)|+1.5\ln{x}+2.48, \end{aligned} \end{equation*}$$

for $10\leqslant x\leqslant t^{\delta},$ $\delta=800^{-1}$, we obtain

$$\begin{equation*} \begin{aligned} & a_{1}^{(4)}\ln{t} + a_{2}^{(4)}|r(x,t)|+a_{3}^{(4)}\\ &\qquad < a_{1}^{(4)}\ln{t}+a_{2}^{(4)}(|q(x,t)|+1.5\delta\ln{t}+2.48)+a_{3}^{(4)}\\ &\qquad \leqslant (\ln{t})\biggl(a_{1}^{(4)}+1.5\delta a_{2}^{(4)}+\frac{\delta}{\ln{2}} (2.48a_{2}^{(4)}+a_{3}^{(4)})\biggr)+a_{2}^{(4)}|q(x,t)|\\ &\qquad < 4.01\ln{t}+7.52|q(x,t)|. \end{aligned} \end{equation*}$$

The lemma is proved.

For what follows, we set $X=N^{0.1\varepsilon}$ and $L=\ln\ln N$.

Lemma 4. Let $k$ be an integer, $1\leqslant k\leqslant c\ln{X}$, where $c>0$ is a sufficiently small absolute constant, and $x=X^{1/(8k+3)}$. Then the following inequality holds:

$$\begin{equation*} \sum_{n}\biggl|C(t_{n})-\sum_{p\leqslant x} \frac{\cos{(t_{n}\ln{p})}}{\sqrt{p}}\biggr|^{2k} < \frac{c_{0}}{280}(c_{0}k)^{2k}M, \end{equation*}$$

where $c_{0}=a_{0}\varepsilon^{-1},$ $a_{0}=5132.8$.

The proof of this assertion is similar to the proof of Theorem 4 in [7] and to the proof of Lemma 2 in [10].

Let $\varphi(x)$ be an unbounded and monotonically increasing function for $x\to +\infty$ and suppose that the inequalities $1\leqslant \varphi(x)\leqslant c_{0}L(3\ln{L})^{-1}$ hold for $N\leqslant x\leqslant N+M$, where the constant $c_{0}$ is defined in Lemma 4. We set

$$\begin{equation*} \varphi=\varphi(N), \qquad y=\exp{\biggl(\frac{\ln{N}}{6^{3}L}\biggr)}, \qquad V_{z}(t)=\sum_{p\leqslant z}\frac{\cos{(t\ln{p})}}{\sqrt{p}}, \qquad \mathfrak{S}=\sum_{p\leqslant y}\frac{1}{p}. \end{equation*}$$

Lemma 5. The inequality

$$\begin{equation*} C(t_{n})\leqslant V_{y}(t_{n})+\frac{2L}{3\varphi} \end{equation*}$$

holds for any $n$ under the condition $N < n\leqslant N+M$, except for a set of at most $M\Delta_{1}$ indices, where

$$\begin{equation*} \Delta_{1}=4e\sqrt{\frac{L}{c_{0}\varphi}} \exp{\biggl(-\frac{2L}{3ec_{0}\varphi}\biggr)}. \end{equation*}$$

Proof. We put $k_{1}=[(3ec_{0}\varphi)^{-1}L],$ $x=X^{1/(8k_{1}+3)}$. Let $\mathscr{A}_{1}$ be the set of $n, N< n\leqslant N+M$, such that $|C(t_{n})-V_{x}(t_{n})| > (3\varphi)^{-1}L$. Then by Lemma 4 we find

$$\begin{equation*} \begin{aligned} |\mathscr{A}_{1}|\biggl(\frac{L}{3\varphi}\biggr)^{2k_{1}}&\leqslant \sum_{n\in \mathscr{A}_{1}}(C(t_{n})-V_{x}(t_{n}))^{2k_{1}}\\ &\leqslant \sum_{n}(C(t_{n})-V_{x}(t_{n}))^{2k_{1}} < \frac{c_{0}}{280} (c_{0}k_{1})^{2k_{1}}M, \end{aligned} \end{equation*}$$

and hence

$$\begin{equation*} |\mathscr{A}_{1}|<\frac{c_{0}}{280}\biggl(\frac{3c_{0} \varphi k_{1}}{L}\biggr)^{2k_{1}}M < \frac{c_{0}}{280}e^{-2k_{1}}M. \end{equation*}$$

Further, it is easy to note that

$$\begin{equation*} \ln{x}=\frac{0.1\varepsilon\ln{N}}{8k_{1}+3} > \frac{0.1\varepsilon\ln{N}}{3L}ec_{0} \varphi > \frac{a_{0}e}{30}\,\frac{\ln{N}}{L} > 10^{5}\ln{y}, \end{equation*}$$

so $x > y^{10^{5}}$. Denoting the set of $n$ under the condition $|V_{x}(t_{n})-V_{y}(t_{n})| > L (3\varphi)^{-1}$ by $\mathscr{A}_{2}$ and applying Lemma 2, we find

$$\begin{equation*} \begin{aligned} &|\mathscr{A}_{2}|\biggl(\frac{L}{3\varphi}\biggr)^{2k_{1}} \leqslant \sum_{n}(V_{x}(t_{n})-V_{y}(t_{n}))^{2k_{1}} \leqslant \sum_{n}\biggl|\sum_{y<p \leqslant x} \frac{p^{it_{n}}}{\sqrt{p}}\biggr|^{2k_{1}}\\ &\qquad=\sum_{n}\sum_{\substack{y < p_{1},\dots,p_{k_{1}}\leqslant x\\ y < q_{1},\dots,q_{k_{1}}\leqslant x}}(p_{1}\dotsb q_{k_{1}})^{-0.5} \biggl(\frac{p_{1}\dotsb p_{k_{1}}}{q_{1}\dotsb q_{k_{1}}}\biggr)^{it_{n}}\\ &\qquad\leqslant k_{1}!\,M\sum_{y<p_{1},\dots, p_{k_{1}}\leqslant x}(p_{1} \dotsb p_{k_{1}})^{-1}+x^{2k_{1}}\ln{N} < k_{1}!\,M\biggl(\sum_{y < p\leqslant x} \frac{1}{p}\biggr)^{k_{1}}+X^{0.25}\\ &\qquad\leqslant k_{1}!\,M\biggl(\ln\frac{\ln{x}}{\ln{y}} +O(\ln^{-2}y)\biggr)^{k_{1}}+X^{0.25}\ < k_{1}!\,M(\ln L)^{k_{1}}+X^{0.25}\\ &\qquad < 3\sqrt{k_{1}}\biggl(\frac{k_{1}\ln L}{e}\biggr)^{k_{1}}M, \end{aligned} \end{equation*}$$

and therefore

$$\begin{equation*} \begin{aligned} |\mathscr{A}_{2}| & < 3M\sqrt{k_{1}}\biggl(\frac{9\varphi^{2}}{L^{2}} \frac{\ln L}{e}\frac{L}{3ec_{0}\varphi}\biggr)^{k_{1}} =3M\sqrt{k_{1}}\biggl(\frac{3\varphi}{e^{2}c_{0}}\frac{\ln L}{L}\biggr)^{k_{1}}\\ &\leqslant 3M\sqrt{k_{1}}\biggl(\frac{3}{e^{2}c_{0}}\frac{c_{0}}{3} \frac{L}{\ln L}\frac{\ln L}{L}\biggr)^{k_{1}}=3Me^{-2k_{1}}\sqrt{k_{1}}. \end{aligned} \end{equation*}$$

Thus, excluding from consideration at most

$$\begin{equation*} \begin{aligned} |\mathscr{A}_{1}|+|\mathscr{A}_{2}|&\leqslant \biggl(\frac{c_{0}}{280}e^{-2k_{1}}+3 \sqrt{k_{1}}e^{-2k_{1}}\biggr)M < 4\sqrt{k_{1}}e^{-2k_{1}}M\\ &\leqslant 4e\sqrt{\frac{L}{c_{0}\varphi}} \exp{\biggl(-\frac{2L}{3ec_{0}\varphi}\biggr)}M \end{aligned} \end{equation*}$$

Gram points $t_{n}$, for the remaining points we get

$$\begin{equation*} C(t_{n})=V_{y}(t_{n})+(V_{x}(t_{n})-V_{y}(t_{n}))+(C(t_{n})-V_{x}(t_{n})) \leqslant V_{y}(t_{n})+\frac{2L}{3\varphi}. \end{equation*}$$

The lemma is proved.

Lemma 6. The inequality

$$\begin{equation*} |V_{y}(t_{n})|\leqslant 3\mathfrak{S} \end{equation*}$$

holds for any $n$ under the condition $N< n\leqslant N+M$, except for a set of at most $M\Delta_{2}$ indices, where $\Delta_{2}=9e\sqrt{\mathfrak{S}}e^{-9\mathfrak{S}}$.

Proof. Let $\mathscr{A}_{3}$ be the set of $n$ satisfying the condition $|V_{y}(t_{n})|> 3 \mathfrak{S}$. Setting $k_{2}=[9\mathfrak{S}]$ we therefore get $y^{2k_{2}}\ln{N}< M^{1/3}$. Using Lemma 4, we obtain the inequalities

$$\begin{equation*} \begin{aligned} |\mathscr{A}_{3}|\cdot(3\mathfrak{S})^{2k_{2}} &\leqslant \sum_{n}V_{y}^{2k_{2}}(t_{n})\leqslant \sum_{n} \biggl|\sum_{p\leqslant y}\frac{p^{it_{n}}}{\sqrt{p}}\biggr|^{2k_{2}} \leqslant k_{2}!\,M\mathfrak{S}^{k_{2}}+y^{2k_{2}}\ln{N}\\ & < 3\sqrt{k_{2}}\,M\biggl(\frac{k_{2}\mathfrak{S}}{e}\biggr)^{k_{2}}, \end{aligned} \end{equation*}$$

which lead us to the desired bound.

Denote by $\mathscr{A}$ the complement of $\mathscr{A}_{3}$ in the set of all integers in the segment $N< n\leqslant N+M$.

Lemma 7. Let $K(s)=\sum_{n\in\mathscr{A}}\exp{(sV_{y}(t_{n}))}$. Then the formula

$$\begin{equation*} K(s)=MF_{y}(s)(1+\theta(\ln{N})^{-3}) \end{equation*}$$

holds with $F_{y}(s)=\prod_{p\leqslant y}J_{0}(is/\sqrt{p})$ for any complex $s$ satisfying the condition $|s|\leqslant 1$.

Proof. We put $K=[3e^{2}\mathfrak{S}]+1$. Then

$$\begin{equation*} K(s)=\sum_{n\in\mathscr{A}}\biggl(\sum_{k=0}^{K}+\sum_{k>K}\biggr) \frac{s^{k}}{k!}\,V_{y}^{k}(t_{n})=K_{1}(s)+R_{1}(s) \end{equation*}$$

and, moreover,

$$\begin{equation*} \begin{aligned} |R_{1}(s)|&\leqslant M\sum_{k >K}\frac{(3\mathfrak{S})^{k}}{k!}< M\sum_{k > K} \biggl(\frac{3e\mathfrak{S}}{k}\biggr)^{k}< M\sum_{k > K} \biggl(\frac{3e\mathfrak{S}}{K}\biggr)^{k}\\ & < M\sum_{k>K}e^{-k}< Me^{-K}< Me^{-22\mathfrak{S}}. \end{aligned} \end{equation*}$$

Further, using the notation of Lemma 6 we represent $K_{1}(s)$ as follows:

$$\begin{equation*} \sum_{k=0}^{K}\frac{s^{k}}{k!}\sum_{n\in\mathscr{A}}V_{y}^{k}(t_{n}) =\sum_{k=0}^{K}\frac{s^{k}}{k!}\biggl(\sum_{n}V_{y}^{k}(t_{n}) -\sum_{n\in\mathscr{A}_{3}}V_{y}^{k}(t_{n})\biggr)=K_{2}(s)-R_{2}(s) \end{equation*}$$

and note that

$$\begin{equation*} |R_{2}(s)|\leqslant |\mathscr{A}_{3}|+\sum_{k=1}^{K}\frac{1}{k!} \sum_{n\in\mathscr{A}_{3}}|V_{y}(t_{n})|^{k}. \end{equation*}$$

Since $y^{2K}\ln{N}< M^{2/3}$, the Cauchy inequality and Lemma 6 imply

$$\begin{equation*} \begin{aligned} \frac{1}{k!}\sum_{n\in \mathscr{A}_{3}}|V_{y}(t_{n})|^{k} &\leqslant \frac{1}{k!}\biggl(|\mathscr{A}_{3}|\sum_{n}V_{y}^{2k} (t_{n})\biggr)^{1/2}\leqslant \frac{1}{k!} \biggl(M\Delta_{2}(k!\,M\mathfrak{S}^{k}+y^{2k}\ln{N})\biggr)^{1/2}\\ & < \frac{M}{k!}(2\Delta_{2}k!\,\mathfrak{S}^{k})^{1/2} < \frac{M\sqrt{\Delta_{2}}}{\sqrt[4]{k}}\biggl(\frac{\mathfrak{S}e}{k}\biggr)^{k/2} < \frac{M\sqrt{\Delta_{2}}}{\sqrt[4]{k}}e^{0.5\mathfrak{S}}. \end{aligned} \end{equation*}$$

Hence

$$\begin{equation*} \begin{aligned} |R_{2}(s)| &\leqslant M\Delta_{2}+M\sqrt{\Delta_{2}}e^{0.5\mathfrak{S}} \sum_{k=1}^{K}\frac{1}{\sqrt[4]{k}} \\ & < M\biggl(\Delta_{2}+\frac{4}{3}K^{3/4} \sqrt{\Delta_{2}}e^{0.5\mathfrak{S}}\biggr) < 61M\mathfrak{S}e^{-4\mathfrak{S}}. \end{aligned} \end{equation*}$$

To transform $K_{2}(s)$, we set

$$\begin{equation*} v_{k}=\sum_{n}V_{y}^{k}(t_{n}),\qquad U(t)=\sum_{p\leqslant y}\frac{p^{it}}{\sqrt{p}}. \end{equation*}$$

If $k\leqslant K$ is odd, then Lemma 2 implies

$$\begin{equation*} \begin{aligned} v_{k} & =2^{-k}\sum_{n}(U(t_{n})+\overline{U}(t_{n}))^{k} =2^{-k}\sum_{n}\sum_{\nu=0}^{k}\binom{k}{\nu}U^{\nu}(t_{n})\overline{U}^{k-\nu}(t_{n})\\ &=2^{-k}\sum_{\nu=0}^{k}\binom{k}{\nu}\sum_{n} \sum_{\substack{p_{1},\dots,p_{\nu}\leqslant y \\ q_{1},\dots,q_{k-\nu}\leqslant y}} (p_{1}\dotsb q_{k-\nu})^{-0.5} \biggl(\frac{p_{1}\dotsb p_{\nu}}{q_{1}\dotsb q_{k-\nu}}\biggr)^{it_{n}}\\ &=2^{-k}\theta\sum_{\nu=0}^{k}\binom{k}{\nu}y^{k}\ln{N}=y^{k} \theta\ln{N}=\theta_{1}M^{1/3}. \end{aligned} \end{equation*}$$

The contribution to the sum $v_{k}$ of the terms satisfying the condition $p_{1}\dotsb p_{\nu} = q_{1}\dotsb q_{k-\nu}$ for $k = 2m,$ $\nu = m$ is estimated in the same way. Therefore, in this case

$$\begin{equation*} v_{k}=2^{-2m}\binom{2m}{m}M\mathfrak{S}_{m}+\theta_{1}M^{1/3}, \end{equation*}$$

where

$$\begin{equation*} \mathfrak{S}_{m}=\sum_{\substack{p_{1}\dotsb p_{m}=q_{1}\dotsb q_{m} \\ p_{1},\dots,q_{m}\leqslant y}} (p_{1}\dotsb q_{m})^{-0.5} =\sum_{\substack{p_{1}\dotsb p_{m}=q_{1}\dotsb q_{m} \\ p_{1},\dots,q_{m}\leqslant y}} (p_{1}\dotsb p_{m})^{-1}. \end{equation*}$$

Thus,

$$\begin{equation*} \begin{aligned} K_{2}(s) &=\sum_{0\leqslant 2m\leqslant K}\frac{s^{2m}}{(2m)!}\,2^{-2m} \binom{2m}{m}M\mathfrak{S}_{m}+\theta M^{1/3}\sum_{k=1}^{+\infty}\frac{1}{k!}\\ &=M\biggl(\sum_{m=0}^{+\infty}-\sum_{m>0.5K}\biggr) \frac{\mathfrak{S}_{m}}{(m!)^{2}}\biggl(\frac{s}{2}\biggr)^{2m}+\theta eM^{1/3} =K_{3}(s)-R_{3}(s)+\theta eM^{1/3}. \end{aligned} \end{equation*}$$

Noting that

$$\begin{equation*} \mathfrak{S}_{m}\leqslant m!\sum_{p_{1},\dots,p_{m}\leqslant y} (p_{1}\dotsb p_{m})^{-1}\leqslant m!\,\mathfrak{S}^{m}, \end{equation*}$$

we find

$$\begin{equation*} \begin{aligned} |R_{3}(s)| &\leqslant \sum_{m>0.5K}\frac{1}{m!}\biggl(\frac{\mathfrak{S}}{4}\biggr)^{m} < M\sum_{m > 0.5K}\biggl(\frac{e\mathfrak{S}}{4m}\biggr)^{m}< M\sum_{m>0.5K} \biggl(\frac{e\mathfrak{S}}{2K}\biggr)^{m}\\ & < M\sum_{m > 0.5K}(6e)^{-m} < Me^{-30\mathfrak{S}}. \end{aligned} \end{equation*}$$

Now let $\varpi_{1}^{\alpha_{1}}\dotsb\varpi_{r}^{\alpha_{r}}$ be the canonical factorization of $n$ satisfying the condition $\Omega(n)= \alpha_{1}+\dots+\alpha_{r}=m$. Then the number of solutions of the equation

$$\begin{equation} n=p_{1}\dotsb p_{m}=q_{1}\dotsb q_{m} \end{equation} \tag{ 3.10 }$$

in primes $p_{1},\dots, q_{m}$ not exceeding $y$ is equal to

$$\begin{equation*} \biggl(\frac{m!}{\alpha_{1}!\dotsb \alpha_{r}!}\biggr)^{2} \end{equation*}$$

or 0 according to whether or not all the prime divisors $\varpi_{1},\dots, \varpi_{r}$ of $n$ are less than or equal to $y$. Defining the multiplicative function $f(n)$ for the powers of primes by the relations

$$\begin{equation*} f(\varpi^{\alpha})=\begin{cases} (\alpha!)^{-2} & \text{if } \varpi\leqslant y, \\ 0 & \text{otherwise}, \end{cases} \end{equation*}$$

we conclude that the number of solutions of (3.10) is represented as $(m!)^{2}f(n)$. Hence,

$$\begin{equation*} \begin{aligned} K_{3}(s) &=M\sum_{m=0}^{+\infty}\biggl(\frac{s}{2}\biggr)^{2m} \sum_{n:\Omega(n)=m}\frac{f(n)}{n}=M\sum_{n=1}^{+\infty}\frac{f(n)}{n} \biggl(\frac{s}{2}\biggr)^{2\Omega(n)}\\ &=M\prod_{p}\biggl(\sum_{m=0}^{+\infty}\frac{f(p^{m})}{p^{m}} \biggl(\frac{s}{2}\biggr)^{2m}\biggr)=M\prod_{p\leqslant y} \biggl(\sum_{m=0}^{+\infty}\frac{(-1)^{m}}{(m!)^{2}} \biggl(\frac{is}{2\sqrt{p}}\biggr)^{2m}\biggr)\\ &=M\prod_{p\leqslant y}J_{0}\biggl(\frac{is}{\sqrt{p}}\biggr)=MF_{y}(s). \end{aligned} \end{equation*}$$

Summing the bounds obtained above for the quantities $R_{j}(s)$, we find

$$\begin{equation*} \begin{aligned} K(s) &=MF_{y}(s)+\theta M(e^{-30\mathfrak{S}}+M^{-2/3} +61\mathfrak{S}e^{-4\mathfrak{S}}+e^{-22\mathfrak{S}})\\ &=M(F_{y}(s)+62\theta \mathfrak{S}e^{-4\mathfrak{S}}). \end{aligned} \end{equation*}$$

It remains to estimate the quantity $\min_{|s|\leqslant 1}|F_{y}(s)|$ below. We represent $F_{y}(s)$ in the form

$$\begin{equation*} \prod_{p\leqslant y}\exp\biggl(\frac{s^{2}}{4p}\biggr) \exp\biggl(-\frac{s^{2}}{4p}\biggr)J_{0}\biggl(\frac{is}{\sqrt{p}}\biggr) =\exp\biggl(\frac{s^{2}}{4}\mathfrak{S}\biggr)\Psi_{y}(s), \end{equation*}$$

where

$$\begin{equation*} \Psi_{y}(s)=\prod_{p\leqslant y}\psi\biggl(\frac{s^{2}}{4p}\biggr), \qquad \psi(v)=e^{-v}\sum_{\alpha=0}^{+\infty}\frac{v^{\alpha}}{(\alpha!)^{2}}. \end{equation*}$$

Expanding $\psi(v)$ in a series we find

$$\begin{equation*} \psi(v)=\sum_{m=0}^{+\infty}q_{m}v^{m}, \qquad q_{m}=\frac{(-1)^{m}}{m!}\sum_{\alpha=0}^{m} \frac{(-1)^{\alpha}}{\alpha!}\binom{m}{\alpha}. \end{equation*}$$

In particular, $q_{0}=1,$ $q_{1}=0,$ $q_{2}=-1/4$, and $q_{3}=1/9$. These identities, together with the obvious estimate

$$\begin{equation} |q_{m}|\leqslant \frac{1}{m!}\sum_{\alpha=0}^{m}\binom{m}{\alpha}=\frac{2^{m}}{m!}, \end{equation} \tag{ 3.11 }$$

imply that $|q_{m}|\leqslant 1$ for any $m\geqslant 1$. Therefore,

$$\begin{equation*} \begin{aligned} \Psi_{y}(s) & =\prod_{p\leqslant y}\biggl(\sum_{m=0}^{+\infty}q_{m} \biggl(\frac{s^{2}}{4p}\biggr)^{m}\biggr)\\ & =\sum_{m_{1},\dots,m_{r}\geqslant 0}q_{m_{1}}\dotsb q_{m_{r}} \sum_{p_{1}<\dots < p_{r}\leqslant y} \biggl(\frac{s^{2}}{4p_{1}}\biggr)^{m_{1}}\dotsb \biggl(\frac{s^{2}}{4p_{r}}\biggr)^{m_{r}}=\sum_{n=0}^{+\infty}\Phi_{n}s^{n}, \end{aligned} \end{equation*}$$

where $\Phi_{0}=1$ and

$$\begin{equation*} \Phi_{n}=\sum_{r\geqslant 1}\sum_{m_{1}+\dots+m_{r}=n}q_{m_{1}}\dotsb q_{m_{r}} \sum_{p_{1}<\dots < p_{r}\leqslant y}(p_{1}^{m_{1}}\dotsb p_{r}^{m_{r}})^{-1} \end{equation*}$$

for $n\geqslant 1$. Since $q_{1}=0$, a nontrivial contribution to $\Phi_{n}$ for $n\geqslant 2$ comes only from the tuples $(m_{1},\dots, m_{r})$ with $m_{1},\dots, m_{r}\geqslant 2$ (which is possible only in the case when $1\leqslant r\leqslant n/2$); moreover, the number of such tuples is equal to

$$\begin{equation*} \binom{n-1-r}{r-1} \end{equation*}$$

for given $n$ and $r$. Therefore, for any $n\geqslant 2$, we have

$$\begin{equation*} \begin{aligned} |\Phi_{n}| &\leqslant \sum_{1\leqslant r\leqslant n/2}\, \sum_{\substack{m_{1}+\dots+m_{r}=n \\ m_{1},\dots,m_{r}\geqslant 2}}\, \sum_{p_{1}<\dots<p_{r}\leqslant y} (p_{1}\dotsb p_{r})^{-2}\\ &\leqslant \sum_{1\leqslant r\leqslant n/2}\,\sum_{\substack{m_{1}+\dots+m_{r}=n \\ m_{1},\dots,m_{r}\geqslant 2}} \biggl(\sum_{k=1}^{+\infty}\frac{1}{k^{2}}-1\biggr) < \sum_{1\leqslant r\leqslant n/2}\binom{n-r-1}{r-1}. \end{aligned} \end{equation*}$$

It is easy to check that the last sum coincides with the coefficient of $x^{[n/2]-1}$ in the expansion of the fraction

$$\begin{equation*} \frac{(1+x)^{[n/2]-1+j}}{1-x-x^{2}} \end{equation*}$$

in a power series; here $j$ denotes the remainder occurring when $n$ is divided by 2. Expanding this expression into partial fractions, we get the estimate $|\Phi_{n}|< \phi^{n}$ where $\phi = (1+\sqrt{5})/2$. Thus, for $|s|\leqslant 1$ we have

$$\begin{equation*} |\Psi_{y}(s)-1|=\biggl|\sum_{n=2}^{+\infty}\Phi_{n} \biggl(\frac{s}{2}\biggr)^{2n}\biggr|\leqslant \sum_{n=2}^{+\infty}\biggl(\frac{\phi}{4}\biggr)^{n} =\frac{\phi^{2}}{16-4\phi} < \frac{1}{3}, \end{equation*}$$

and hence

$$\begin{equation*} \begin{gathered} \frac{2}{3} < |\Psi_{y}(s)|< \frac{4}{3}, \qquad |F_{y}(s)| > \frac{2}{3}\biggl|\exp{\biggl(\frac{s^{2}}{4}\mathfrak{S}\biggr)}\biggr| > \frac{2}{3}e^{-0.25\mathfrak{S}},\\ K(s)=MF_{y}(s)(1+62\theta_{1}F_{y}^{-1}(s)\mathfrak{S}e^{-4\mathfrak{S}}) =MF_{y}(s)(1+62\theta_{2}\mathfrak{S}e^{-3.75\mathfrak{S}}). \end{gathered} \end{equation*}$$

Finally, noting that $62\mathfrak{S}e^{-3.75\mathfrak{S}}< (\ln{N})^{-3}$, we obtain the result of the lemma.

Let $K_{N}(s)$ be the generating function for the moments of the discrete random quantity $\xi_{N}$ with the values $V_{y}(t_{n}),$ $N< n\leqslant N+M,$ $n\in \mathcal{A}$. Then, using the notations and the results of the Lemmas 6 and 7, we easily conclude that the relations

$$\begin{equation*} \begin{aligned} K_{N}(s) &=|\mathscr{A}|^{-1}\sum_{n\in \mathscr{A}}\exp{\{sV_{y}(t_{n})\}} =|\mathscr{A}|^{-1}MF_{y}(s)(1+\theta_{1}(\ln{N})^{-3})\\ &=F_{y}(s)(1-3e\theta_{2}\sqrt{\mathfrak{S}}e^{-9\mathfrak{S}})^{-1} (1+\theta_{1}(\ln{N})^{-3}) \end{aligned} \end{equation*}$$

hold in the disc $|s|\leqslant 1$. Denoting the products

$$\begin{equation*} \prod_{p}\psi\biggl(\frac{s^{2}}{4p}\biggr)\quad\text{and} \quad\prod_{p > y}\psi\biggl(\frac{s^{2}}{4p}\biggr) \end{equation*}$$

by $\Phi(s)$ and $\Phi_{y}(s)$, respectively, we obtain

$$\begin{equation*} K_{N}(s)=\exp{\biggl(\frac{s^{2}}{4}\mathfrak{S}\biggr)} \Phi(s)\Phi_{y}^{-1}(s)(1+2\theta_{3}(\ln{N})^{-3}). \end{equation*}$$

Suppose that $p > y$. Using inequality (3.11), we get

$$\begin{equation*} \begin{gathered} \biggl|\psi\biggl(\frac{s^{2}}{4p}\biggr)-1\biggr| \leqslant \sum_{m=2}^{+\infty}\frac{|q_{m}|}{(4p)^{m}} \leqslant \sum_{m=2}^{+\infty}\frac{(2p)^{-m}}{m!}< \frac{1}{7p^{2}},\\ \ln{\varphi\biggl(\frac{s^{2}}{4p}\biggr)} =\ln\biggl(1+\frac{\theta_{4}}{7p^{2}}\biggr)=\frac{\theta_{5}}{6p^{2}}, \end{gathered} \end{equation*}$$

and hence

$$\begin{equation*} \begin{gathered} \Phi_{y}(s)=\exp{\biggl(\theta_{6}\sum_{p>y}\frac{1}{6p^{2}}\biggr)} =\exp{\biggl(\frac{\theta_{7}}{6y}\biggr)}=1+\frac{\theta_{8}}{5y},\\ K_{N}(s)=\exp{\biggl(\frac{s^{2}}{4}\mathfrak{S}\biggr)} \Phi(s)(1+ 3\theta_{9}(\ln{N})^{-3}). \end{gathered} \end{equation*}$$

Taking

$$\begin{equation*} \sigma(N)=\mathfrak{S}=\sum_{p\leqslant y}\frac{1}{p}, \qquad \kappa(N)=(\ln{N})^{3},\qquad u=\frac{\sqrt{6L}}{\varphi} \end{equation*}$$

in Lemma 1, we conclude that the inequality $V(t_{n})\leqslant -u\sqrt{\mathfrak{S}/2}$ holds for at least

$$\begin{equation*} \frac{|\mathscr{A}|}{\sqrt{2\pi}}\biggl(\int_{u}^{+\infty}e^{-v^{2}/2}\,dv\biggr) (1+O(\varphi^{-1}))> 1.01\frac{M\varphi}{2\pi\sqrt{L}} \exp{\biggl(-\frac{3L}{\varphi^{2}}\biggr)} \end{equation*}$$

Gram points. Using Lemma 5 and excluding from the above quantity at most

$$\begin{equation*} M\Delta_{1} < 4eM\sqrt{\frac{L}{\varphi c_{0}}} \exp{\biggl(-\frac{2L}{3ec_{0}\varphi}\biggr)} \end{equation*}$$

points $t_{n}$, in the case $\varphi\geqslant 1.3\cdot 10^{5}\varepsilon^{-1}>9ec_{0}$, for at least

$$\begin{equation*} \begin{aligned} & 1.01\frac{M\varphi}{2\pi\sqrt{L}}\exp{\biggl(-\frac{3L}{\varphi^{2}}\biggr)} -4eM\sqrt{\frac{L}{\varphi c_{0}}}\exp{\biggl(-\frac{2L}{3ec_{0}\varphi}\biggr)}\\ &\qquad =\frac{M\varphi}{2\pi\sqrt{L}}\exp{\biggl(-\frac{3L}{\varphi^{2}}\biggr)} \biggl(1.01-\frac{8\pi e}{\sqrt{c_{0}}}\frac{L}{\varphi^{3/2}} \exp{\biggl(-\frac{2L}{3ec_{0}\varphi} \biggl(1-\frac{9ec_{0}}{2\varphi}\biggr)\biggr)}\biggr)\\ &\qquad > \frac{M\varphi}{2\pi\sqrt{L}}\exp{\biggl(-\frac{3L}{\varphi^{2}}\biggr)} \end{aligned} \end{equation*}$$

remaining points we have the inequalities

$$\begin{equation*} \ln{|\zeta(0.5+it_{n})|}\leqslant C(t_{n})\leqslant V_{y}(t_{n})+\frac{2L}{3\varphi} \leqslant -\frac{1}{\varphi}\sqrt{3\mathfrak{S}L} +\frac{2L}{3\varphi} < -\frac{L}{\varphi}. \end{equation*}$$

Theorem 1 is proved.

Now taking $\varphi(x)=0.5(\ln\ln\ln{x})^{\delta}$, we get $\varphi=0.5(\ln{L})^{\delta}$ and hence

$$\begin{equation*} \sum_{n}|\zeta(0.5+it_{n})|^{-1} > \frac{M\varphi}{\pi} \frac{e^{-3L/\varphi^{2}}e^{L/\varphi}}{\sqrt{L}} > M \exp{\biggl(\frac{L}{(\ln{L})^{\delta}}\biggr)}. \end{equation*}$$

Theorem 2 is proved.