| Sbornik: Mathematics |
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On square of the Riemann zeta function modulus in the critical strip and estimates of means A. V. Zobnin Joint Institute for High Temperatures of the Russian Academy of Sciences, Moscow, Russia
Bibliographic databases:
    § 1. Introduction and tasksOne of the important problems in algebraic number theory is the behaviour of the Riemann zeta function in the critical strip. Because neither the Riemann nor even Lindelöf hypothesis has yet been proved, the investigations of mean values of the square of the zeta function modulus in the critical strip are relevant, in particular, estimates of the residual term (see [1] and [2])
$$
\begin{equation}
E_\sigma(T)=I_\sigma(T)-\zeta(2\sigma)T -\frac{\zeta(2\sigma-1)\Gamma(2\sigma-1)}{1-\sigma}\sin(\pi\sigma)T^{2-2\sigma},
\end{equation}
\tag{1.1}
$$
where as well as estimates for other intervals of integration [3] and estimates of other integrals containing the square of the zeta function [4]. In this paper a new integral representation of $|\zeta(s)|^2$ is presented (Theorem 1), which is used for the asymptotic evaluation of the Laplace transform of the residual term $E_\sigma(T)$ (Theorem 2) and for estimates of the integral
$$
\begin{equation*}
\int_{-\infty}^{\infty} |\zeta(\sigma+{i}(T+t))|^2\exp\left(\frac{-t^2}\Delta\right)\,dt
\end{equation*}
\notag
$$
for sufficiently large $\Delta$ (Theorem 3).
§ 2. Representation for the square of the modulus of the zeta functionTheorem 1. For $\operatorname{Re}(s)=\sigma>1/2$, $\sigma\ne1$, where (here $\{x\}=x-[x]$ is the fractional part of $x$). Proof. According to the Abel transform,
$$
\begin{equation*}
\sum_{n=1}^{N}\biggl(\frac{1}{n^{\overline{s}}}\sum_{k=1}^{n}\frac{1}{k^s}\biggr) =\sum_{n=1}^{N} \frac{1}{n^{\overline{s}}}\sum_{n=1}^{N}\frac{1}{n^s} -\sum_{n=1}^{N}\biggl(\frac{1}{n^{s}}\sum_{k=1}^{n}\frac{1}{k^{\overline{s}}}\biggr) +\sum_{n=1}^{N}\frac{1}{n^{2\sigma}},
\end{equation*}
\notag
$$
or
$$
\begin{equation*}
2\operatorname{Re}\biggl(\sum_{n=1}^{N}\biggl(\frac{1}{n^{\overline{s}}} \sum_{k=1}^{n}\frac{1}{k^s}\biggr)\biggr) =\sum_{n=1}^{N} \frac{1}{n^{\overline{s}}}\sum_{n=1}^{N}\frac{1}{n^s}+\sum_{n=1}^{N}\frac{1}{n^{2\sigma}}.
\end{equation*}
\notag
$$
From a well-known equality for a finite sum (see (2.1.3), (2.1.5) and (3.5.3) in [5])
$$
\begin{equation*}
\sum_{k=1}^{n} \frac{1}{k^{s}}=\zeta(s)+\frac{n^{1-s}}{1-s}+R_s(n),
\end{equation*}
\notag
$$
where
$$
\begin{equation}
R_s(n)=s\int_n^\infty\frac{\{y\}}{y^{s+1}}\,dy=O(n^{-\sigma}), \qquad n\to\infty,
\end{equation}
\tag{2.3}
$$
it follows that
$$
\begin{equation*}
\begin{aligned} \, &2\operatorname{Re}\biggl(\sum_{n=1}^{N}\frac{1}{n^{\overline{s}}} \biggl(\zeta(s)+\frac{n^{1-s}}{1-s}+R_s(n)\biggr)\biggr) = |\zeta(s)|^2+2\operatorname{Re} \biggl(\frac{\zeta(s)N^{1-\overline{s}}}{1-\overline{s}}\biggr) \\ &\qquad\qquad +\frac{N^{2-2\sigma}}{|1-s|^2}+\frac{N^{1-s} R_{\overline{s}}(N)}{1-s}+\frac{N^{1-\overline{s}} R_{s}(N)}{1-\overline{s}}+\zeta(s)R_{\overline{s}}(N)+\zeta(\overline{s})R_s(N) \\ &\qquad\qquad +R_s(N)R_{\overline{s}}(N)+\zeta(2\sigma)+\frac{N^{1-2\sigma}}{1-2\sigma}+R_{2\sigma}(N), \end{aligned}
\end{equation*}
\notag
$$
or
$$
\begin{equation}
\begin{aligned} \, \notag &2\operatorname{Re}\biggl(|\zeta(s)|^2 +\frac{\zeta(s)N^{1-\overline{s}}}{1-\overline{s}}+O(N^{-\sigma}) +\frac{1}{1-s}\sum_{n=1}^{N}\frac{1}{n^{2\sigma-1}}+\sum_{n=1}^{N}\frac{R_s(n)}{n^{\overline{s}}} \biggr) \\ &\qquad = |\zeta(s)|^2+2\operatorname{Re}\biggl(\frac{\zeta(s)N^{1-\overline{s}}}{1-\overline{s}}\biggr) +\frac{N^{2-2\sigma}}{|1-s|^2} \notag \\ &\qquad\qquad+\zeta(2\sigma)+O(N^{-\sigma})+O(N^{1-2\sigma}), \qquad N\to\infty. \end{aligned}
\end{equation}
\tag{2.4}
$$
(The $O$-estimates hold (not uniformly) for each fixed $s\ne1$.) Taking the relation
$$
\begin{equation*}
\operatorname{Re}\biggl(\frac{1}{1-s}\biggr)\sum_{n=1}^{N} \frac{1}{n^{2\sigma-1}}=\frac{1-\sigma}{|1-s|^2}\biggl(\zeta (2\sigma-1)+\frac{N^{2-2\sigma}}{2(1-\sigma)}+O(N^{1-2\sigma})\biggr), \qquad N\to\infty,
\end{equation*}
\notag
$$
into account we obtain
$$
\begin{equation}
\begin{aligned} \, |\zeta(s)|^2 &=\zeta(2\sigma)-2\frac{1-\sigma}{|1-s|^2}\zeta(2\sigma-1) -2\operatorname{Re}\sum_{n=1}^{N} \biggl( \frac{R_s(n)}{n^{\overline{s}}}\biggr) \nonumber \\ &\qquad +O(N^{1-2\sigma})+O(N^{-\sigma}),\qquad N\to\infty. \end{aligned}
\end{equation}
\tag{2.5}
$$
Remark 1. The terms in (2.4) increasing with $N$ (for $1/2<\sigma<1$) cancel one another! Substituting (2.3) into (2.5) and making the change of the variable $y=nx$, we derive the equation
$$
\begin{equation*}
\begin{aligned} \, |\zeta(s)|^2 &=\zeta(2\sigma)-2\frac{1-\sigma}{|1-s|^2}\zeta(2\sigma-1) -2\sum_{n=1}^{N}\int_1^\infty\frac{\{nx\}}{n^{2\sigma}} \operatorname{Re}\biggl(\frac{s}{x^{s+1}}\biggr)dx \\ &\qquad +O(N^{1-2\sigma})+O(N^{-\sigma}), \qquad N\to\infty. \end{aligned}
\end{equation*}
\notag
$$
Because the integral $\displaystyle \int_1^\infty\dfrac{\{nx\}}{x^{s+1}}\,dx$ converges absolutely, we have
$$
\begin{equation*}
\sum_{n=1}^{N}\int_1^\infty\frac{\{nx\}}{n^{2\sigma}} \operatorname{Re}\biggl(\frac{s}{x^{s+1}}\biggr)dx =\int_1^\infty \sum_{n=1}^{N}\frac{\{nx\}}{n^{2\sigma}}\operatorname{Re}\biggl(\frac{s}{x^{s+1}}\biggr)dx.
\end{equation*}
\notag
$$
Letting $N$ tend to infinity, we obtain the claim (2.1) of the theorem. Remark 2. The function $\varphi(s,x)=\sum_{n=1}^{\infty}\frac{\{nx\}-1/2}{n^s}$ was investigated by Hecke [6] as a function of $s$ for fixed $x$ (equal to quadratic irrationals). Let us formulate the properties of the function $\phi(\nu,x)$ that follow directly from the definition. Proposition 1. For fixed $\nu>1$ the function $\phi(\nu,x)$, as a function of $x$, is nonnegative, bounded, periodic with period $1$ and integrable on any finite interval. Now we prove one more property of $\phi(\nu,x)$, which will be used below. Proof. By the periodicity of $\phi(\nu,x)$ it is sufficient to perform the proof for $a=0$. For any integer $N$ we have
$$
\begin{equation*}
\begin{gathered} \, \sum_{n=1}^{N}\frac{\{nx\}}{n^{\nu}}=x\sum_{n=1}^{N}n^{1-\nu}-\sum_{n=1}^{N} n^{-\nu}\sum_{m\leqslant nx}1, \\ \begin{split} &\sum_{n=1}^{N} n^{-\nu}\sum_{m\leqslant nx}1 =\sum_{m\leqslant Nx}\sum_{m/x\leqslant n\leqslant N}n^{-\nu}=\sum_{m\leqslant Nx} \biggl( \frac{N^{1-\nu}-m^{1-\nu}x^{\nu-1}}{1-\nu}+O(x^\nu m^{-\nu}) \!\biggr) \\ &=\frac{xN^{2-\nu}}{1-\nu}-\frac{x^{\nu-1}\zeta(\nu-1)}{1-\nu}-\frac{x N^{2-\nu}}{(1-\nu)(2-\nu)}+O(N^{1-\nu})+O(xN^{1-\nu})+O(x^\nu) \\ &=\frac{xN^{2-\nu}}{2-\nu}\,{-}\,\frac{x^{\nu-1}\zeta(\nu\,{-}\,1)}{1-\nu} \,{+}\,O(N^{1-\nu})\,{+}\,O(xN^{1-\nu})\,{+}\,O(x^\nu),\qquad N\,{\to}\, \infty,\quad\! x\to 0. \end{split} \end{gathered}
\end{equation*}
\notag
$$
On the other hand so The lemma follows by letting $N \to \infty$. § 3. Integral estimatesNow we proceed to integral estimates of the Riemann zeta function in the critical strip which use Theorem 1 and Lemma 1. Proof. The definition (1.1) of the residual term implies that
$$
\begin{equation}
\begin{aligned} \, \notag \int_0^\infty E_\sigma(T)\exp(-\delta T)\,dT &=\int_0^\infty I_\sigma(T)\exp(-\delta T)\,dT-\frac{\zeta(2\sigma)}{\delta^2} \\ &\qquad -\frac{\zeta(2\sigma-1)\Gamma(2\sigma-1)\Gamma(3-2\sigma)\sin(\pi\sigma)}{1-\sigma}\, \delta^{2\sigma-3}. \end{aligned}
\end{equation}
\tag{3.2}
$$
Since the growth of the zeta-function is slower than exponential, we have
$$
\begin{equation*}
\int_0^\infty I_\sigma(T)\exp(-\delta T)\,dT =\frac{1}{\delta}\int_0^\infty|\zeta(\sigma+{i}t)|^2\exp(-\delta t)\,dt.
\end{equation*}
\notag
$$
It follows directly from (2.1) that
$$
\begin{equation}
\begin{aligned} \, \notag &\int_0^\infty|\zeta(\sigma+{i}t)|^2\exp(-\delta t)\,dt =\frac{\zeta(2\sigma)}{\delta}-\pi\zeta(2\sigma-1)+O(\delta) \\ &\qquad\qquad -2\operatorname{Re}\biggl(\int_0^\infty dt \int_1^\infty dx\, \frac{\phi(2\sigma,x)(\sigma+{i}t)\exp(-(\delta+{i}\ln(x)) t)}{x^{\sigma+1}} \biggr). \end{aligned}
\end{equation}
\tag{3.3}
$$
Changing the order of integration is possible in view of the absolute convergence of the integrals, both with respect to $t$ and with respect to $x$, so
$$
\begin{equation*}
\begin{aligned} \, &\int_0^\infty dt \int_1^\infty dx\, \frac{\phi(2\sigma,x)(\sigma+{i}t)\exp(-(\delta+{i}\ln(x)) t)}{x^{\sigma+1}} \\ &\qquad =\int_1^\infty dx \int_0^\infty dt \, \frac{\phi(2\sigma,x)(\sigma+{i}t)\exp(-(\delta+{i}\ln(x)) t)}{x^{\sigma+1}}. \end{aligned}
\end{equation*}
\notag
$$
Now we perform integration with respect to $t$ and find the real part:
$$
\begin{equation*}
\begin{aligned} \, &\operatorname{Re}\biggl(\int_1^\infty dx \int_0^\infty dt \, \frac{\phi(2\sigma,x)(\sigma+{i}t)\exp(-(\delta+{i}\ln(x)) t)}{x^{\sigma+1}}\biggr) \\ &\qquad =\sigma\delta\int_1^\infty \frac{\phi(2\sigma,x)}{\delta^2+\ln^2(x)}\, \frac{dx}{x^{\sigma+1}} +2\delta\int_1^\infty\frac{\phi(2\sigma,x)\ln(x)}{(\delta^2+\ln^2(x))^2}\, \frac{dx}{x^{\sigma+1}} \\ &\qquad =\sigma\int_0^\infty\frac{\phi(2\sigma,\exp(\delta y))}{1+y^2} \exp(-\sigma\delta y)\,dy +\frac{2}{\delta}\int_0^\infty\frac{\phi(2\sigma,\exp(\delta y)) y}{(1+y^2)^2} \exp(-\sigma\delta y)\,dy \end{aligned}
\end{equation*}
\notag
$$
(the substitution $x=\exp(\delta y)$ was used in the last equality). Then
$$
\begin{equation*}
\begin{aligned} \, \sigma\int_0^\infty\frac{\phi(2\sigma,\exp(\delta y))}{1+y^2}\exp(-\sigma\delta y)\,dy &=\sigma\int_0^{1/\delta}\frac{\phi(2\sigma,\exp(\delta y))}{1+y^2} \exp(-\sigma\delta y)\,dy \\ &\qquad+\sigma\int_{1/\delta}^\infty\frac{\phi(2\sigma,\exp(\delta y))}{1+y^2} \exp(-\sigma\delta y)\,dy. \end{aligned}
\end{equation*}
\notag
$$
From Lemma 1 we obtain
$$
\begin{equation*}
\begin{gathered} \, \begin{aligned} \, \int_0^{1/\delta}\frac{\phi(2\sigma,\exp(\delta y))}{1+y^2} \exp(-\sigma\delta y)\,dy &=O\biggl(\int_0^{1/\delta}\frac{(\exp(\delta y)-1)^{2\sigma-1}} {1+y^2} \exp(-\sigma\delta y)\,dy\biggr) \\ &=O(\delta^{2\sigma-1}), \end{aligned} \\ \sigma\int_{1/\delta}^\infty \frac{\phi(2\sigma,\exp(\delta y))}{1+y^2} \exp(-\sigma\delta y)\,dy <\sigma\int_{1/\delta}^\infty \frac{\zeta(2\sigma)}{y^2}\,dy=O(\delta). \end{gathered}
\end{equation*}
\notag
$$
By analogy,
$$
\begin{equation*}
\begin{aligned} \, \frac{2}{\delta}\int_0^\infty\frac{\phi(2\sigma,\exp(\delta y)) y}{(1+y^2)^2} \exp(-\sigma\delta y)\,dy &=\frac{2}{\delta}\int_0^{1/\delta}\frac{\phi(2\sigma,\exp(\delta y)) y}{(1+y^2)^2} \exp(-\sigma\delta y)\,dy \\ &\qquad+\frac{2}{\delta}\int_{1/\delta}^\infty \frac{\phi(2\sigma,\exp(\delta y)) y}{(1+y^2)^2} \exp(-\sigma\delta y)\,dy, \end{aligned}
\end{equation*}
\notag
$$
and using Lemma 1 again, we obtain
$$
\begin{equation*}
\begin{aligned} \, &\frac{2}{\delta}\int_0^{1/\delta}\frac{\phi(2\sigma,\exp(\delta y))y}{(1+y^2)^2} \exp(-\sigma\delta y)\,dy \\ &\qquad =-\frac{2\zeta(2\sigma-1)}{\delta(2\sigma-1)}\int_0^{1/\delta}\frac{(\exp(\delta y)-1)^{2\sigma-1}y}{(1+y^2)^2}\exp(-\sigma\delta y)\,dy \\ &\qquad\qquad +\frac{2\zeta(2\sigma-1)}{\delta}\int_0^{1/\delta}\frac{(\exp(\delta y)-1) y}{(1+y^2)^2} \exp(-\sigma\delta y)\,dy \\ &\qquad\qquad +O\biggl(\delta^{-1}\int_0^{1/\delta}\frac{(\exp(\delta y)-1)^{2\sigma}y}{(1+y^2)^2} \exp(-\sigma\delta y)\,dy\biggr) \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, &\int_0^{1/\delta}\frac{(\exp(\delta y)-1)^{2\sigma-1}y}{(1+y^2)^2} \exp(-\sigma\delta y)\,dy =\int_0^{1/\delta}\frac{\delta^{2\sigma-1} y^{2\sigma}(1+O(\delta y))}{(1+y^2)^2}\,dy \\ &\qquad =\delta^{2\sigma-1}\int_0^\infty\frac{ y^{2\sigma}}{(1+y^2)^2}\,dy -\delta^{2\sigma-1}\int_{1/\delta}^\infty \frac{ y^{2\sigma}}{(1+y^2)^2}\,dy \\ &\qquad\qquad+O\biggl(\delta^{2\sigma}\int_0^{1/\delta} \frac{y^{2\sigma+1}}{(1+y^2)^2}\,dy \biggr) \\ &\qquad =-\delta^{2\sigma-1}\frac{\pi(2\sigma-1)}{4\cos(\sigma\pi)}+O(\delta^2)+O(\delta^{2\sigma}),\qquad \delta \to 0. \end{aligned}
\end{equation*}
\notag
$$
Similarly,
$$
\begin{equation*}
\begin{aligned} \, &\int_0^{1/\delta}\frac{(\exp(\delta y)-1)y}{(1+y^2)^2} \exp(-\sigma\delta y)\,dy =\int_0^{1/\delta}\frac{\delta y^2(1+O(\delta y))}{(1+y^2)^2}\,dy \\ &\qquad =\delta\int_0^\infty\frac{y^2}{(1+y^2)^2}\,dy -\delta\int_{1/\delta}^\infty \frac{y^2}{(1+y^2)^2}\,dy +O\biggl(\delta^2\int_0^{1/\delta} \frac{y^3}{(1+y^2)^2}\,dy\biggr) \\ &\qquad =\delta\frac{\pi}{4}+O\biggl(\delta^2\ln\biggl(1+\frac{1}{\delta^2}\biggr)\biggr) \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, &\int_0^{1/\delta}\frac{(\exp(\delta y)-1)^{2\sigma}y}{(1+y^2)^2} \exp(-\sigma\delta y)\,dy =O\biggl( \delta^{2\sigma}\int_0^\infty\frac{y^{2\sigma+1}}{(1+y^2)^2}\,dy\biggr) \\ &\qquad=O(\delta^{2\sigma}),\qquad \delta \to 0. \end{aligned}
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
\begin{aligned} \, &\frac{2}{\delta}\int_0^{1/\delta}\frac{\phi(2\sigma,\exp(\delta y))y}{(1+y^2)^2} \exp(-\sigma\delta y)\,dy \\ &\qquad =\frac{\pi\zeta(2\sigma-1)\delta^{2\sigma-2}}{2\cos(\sigma\pi)} +\frac{\pi\zeta(2\sigma-1)}{2}+O(\delta^{2\sigma-1}) \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\frac{1}{\delta}\int_{1/\delta}^\infty \frac{\phi(2\sigma,\exp(\delta y))}{(1+y^2)^2 y} \exp(-\sigma\delta y)\,dy <\frac{1}{\delta}\int_{1/\delta}^\infty \frac{\zeta(2\sigma)}{y^3}\,dy=O(\delta).
\end{equation*}
\notag
$$
Thus, taking into account that $\sigma<1$ and $O(\delta)=O(\delta^{2\sigma-1})$, we obtain
$$
\begin{equation}
\begin{aligned} \, \notag &2\operatorname{Re}\biggl(\int_0^\infty dt \int_1^\infty dx\, \frac{\phi(2\sigma,x)(\sigma+{i}t)\exp(-(\delta+{i}\ln(x)) t)}{x^{\sigma+1}}\biggr) \\ &\qquad =\frac{\pi\zeta(2\sigma-1)\delta^{2\sigma-2} }{\cos(\sigma\pi)} +\pi\zeta(2\sigma-1)+O(\delta^{2\sigma-1}),\qquad \delta \to 0. \end{aligned}
\end{equation}
\tag{3.4}
$$
After the substitution of (3.4) into (3.3) we have
$$
\begin{equation}
\begin{aligned} \, &\frac{1}{\delta}\int_0^\infty|\zeta(\sigma{+}\,it)|^2\exp(-\delta t)\,dt =\frac{\zeta(2\sigma)}{\delta^2}-\frac{\pi\zeta(2\sigma-1)\delta^{2\sigma-3} }{\cos(\sigma\pi)} \nonumber \\ &\qquad-\frac{2\pi\zeta(2\sigma-1)}{\delta}+O(\delta^{2\sigma-2}),\qquad \delta \to 0. \end{aligned}
\end{equation}
\tag{3.5}
$$
Because
$$
\begin{equation*}
\begin{aligned} \, \Gamma(3-2\sigma)\Gamma(2\sigma-1)&=(2-2\sigma)\Gamma(1-2\sigma)\Gamma(2\sigma-1) \\ &=\frac{2\pi(1-\sigma)}{\sin(\pi(2\sigma-1))} =-\frac{\pi(1-\sigma)}{\sin(\pi\sigma)\cos(\pi\sigma)}, \end{aligned}
\end{equation*}
\notag
$$
it follows that
$$
\begin{equation*}
\frac{\zeta(2\sigma-1)\Gamma(2\sigma-1)\Gamma(3-2\sigma)\sin(\pi\sigma)}{1-\sigma} =-\frac{\pi\zeta(2\sigma-1) }{\cos(\sigma\pi)}.
\end{equation*}
\notag
$$
The substitution of (3.5) into (3.2) leads to (3.1), and Theorem 2 is proved. Let us estimate the integral $\displaystyle\int_{-\infty}^{\infty} |\zeta(\sigma+{i}(T+t))|^2\exp(-t^2/\Delta)\,dt$. Theorem 3. For $1/2<\sigma<1$ and sufficiently large $\Delta$ the following estimate holds (nonuniformly in $\sigma$): Proof. The first two terms in (3.6) are obtained directly from the first two terms in representation (2.1), and the third term in (2.1) gives
$$
\begin{equation*}
\begin{aligned} \, I_3 &=-\frac{2}{\sqrt{\pi}\,\Delta}\operatorname{Re} \biggl(\int_{-\infty}^{\infty}dt\int_0^\infty dy\,\phi(2\sigma,\exp(y))(\sigma+iT+it) \\ &\qquad\qquad\qquad\qquad\times \exp\biggl(-\frac{t^2}{\Delta^2}-(\sigma+iT+it)y\biggr) \biggr) \end{aligned}
\end{equation*}
\notag
$$
(the substitution $y=\log(x)$ has been used). As the integrals are absolutely convergent, the order of integration can be changed, so integration with respect to $t$ yields
$$
\begin{equation*}
I_3=2\int_0^\infty\phi(2\sigma,\exp(y)) \biggl(\cos(Ty)\exp\biggl(-\frac{y^2\Delta^2}4-\sigma y\biggr) \biggr)'_y\,dy.
\end{equation*}
\notag
$$
Using Lemma 1 and the expansion $\exp(y)=1+y+O(y^2)$ as $y\to0$ for $y<1$, we represent $\phi(2\sigma,\exp(y))$ in the form
$$
\begin{equation*}
\phi(2\sigma,\exp(y))=-\frac{\zeta(2\sigma-1)}{2\sigma-1}y^{2\sigma-1}+\zeta(2\sigma-1)y+r(y),
\end{equation*}
\notag
$$
where $r(y)=O(y^{2\sigma})$ as $y\to0$. This is also trivially true for $y \geqslant 1$ since $\phi(2\sigma,\exp(y))$ is bounded for each $y$ by definition. Thus,
$$
\begin{equation*}
\begin{aligned} \, I_3 &=-2\zeta(2\sigma-1)\int_0^\infty\biggl(\frac{y^{2\sigma-1}}{2\sigma-1}-y\biggr) \biggl(\cos(Ty)\exp\biggl(-\frac{y^2\Delta^2}4-\sigma y\biggr) \biggr)'_y\, dy \\ &\qquad -2\int_0^\infty r(y)\biggl(T\sin(Ty)+\biggl(\frac{y\Delta^2}2+\sigma\biggr)\cos(Ty)\biggr) \exp\biggl(-\frac{y^2\Delta^2}4-\sigma y\biggr) \, dy. \end{aligned}
\end{equation*}
\notag
$$
We transform the first integral by integration by parts:
$$
\begin{equation*}
\begin{aligned} \, &2\zeta(2\sigma-1) \int_0^\infty(y^{2\sigma-2}-1)\cos(yT) \exp\biggl(-\frac{y^2\Delta^2}4-\sigma y\biggr) \, dy \\ &\ =2\zeta(2\sigma-1) \int_0^\infty(y^{2\sigma-2}-1)\cos(yT)\exp\biggl(-\frac{y^2\Delta^2}4\biggr) \, dy +O(\Delta^{-2\sigma}),\qquad \Delta \to \infty \end{aligned}
\end{equation*}
\notag
$$
(the estimate is not uniform in $\sigma$), and the second can be estimated as
$$
\begin{equation*}
\begin{aligned} \, &\biggl|2\int_0^\infty r(y)\biggl(T\sin(Ty)+\biggl(\frac{y\Delta^2}2+\sigma\biggr)\cos(Ty)\biggr) \exp\biggl(-\frac{y^2\Delta^2}4-\sigma y\biggr)\, dy\biggr| \\ &\qquad < 2\int_0^\infty|r(y)|\biggl(T+\frac{y\Delta^2}2+\sigma\biggr) \exp\biggl(-\frac{y^2\Delta^2}4\biggr)\, dy \\ &\qquad=O(T\Delta^{-2\sigma-1})+O(\Delta^{-2\sigma}),\qquad \Delta \to \infty. \end{aligned}
\end{equation*}
\notag
$$
Now the theorem is proved. One consequence of this theorem is the following statement: if $\Delta(T)\!>\!T^{1\mkern-1mu/\mkern-1mu(\mkern-1.5mu2\sigma+1\mkern-2mu)+\varepsilon}$ for $1/2<\sigma<1$, then
$$
\begin{equation*}
\lim_{T\to \infty}\frac{1}{\sqrt{\pi}\Delta(T)}\int_{-\infty}^{\infty} |\zeta(\sigma+i(T+t))|^2 \exp\biggl(-\frac{t^2}{\Delta(T)^2}\biggr)\, dt=\zeta(2\sigma).
\end{equation*}
\notag
$$
The results presented in this paper can be useful in refining $O$- and $\Omega$-estimates for the Riemann zeta function in the critical strip. 
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