%------------------------------------------------------------------------------ % Beginning of journal.tex %------------------------------------------------------------------------------ % % AMS-LaTeX version 2 sample file for journals, based on amsart.cls. % % *** DO NOT USE THIS FILE AS A STARTER. *** % *** USE THE JOURNAL-SPECIFIC *.TEMPLATE FILE. *** % % Replace amsart by the documentclass for the target journal, e.g., tran-l. % \documentclass{amsart} \usepackage{fancyhdr} \usepackage{lastpage} \usepackage{stmaryrd,yhmath} \pagestyle{fancy} \fancyhead{} \fancyfoot{} \lhead{\scshape\nouppercase\leftmark} % \rightmark doesn't work either \rhead{Jacob's ladder \dots} \rfoot{Page \thepage\ of \pageref{LastPage}} %\documentclass{gtart} \newcommand{\bdis}{\begin{displaymath}} \newcommand{\edis}{\end{displaymath}} \newcommand{\be}{\begin{equation}} \newcommand{\ee}{\end{equation}} \newcommand{\mbb}{\mathbb} \newcommand{\mcal}{\mathcal} \newcommand{\pd}{\partial} \newcommand{\vp}{\varphi} \newcommand{\tT}{\tilde{T}} \newcommand{\tU}{\tilde{U}} \newcommand{\btau}{\bar{\tau}} \newcommand{\vth}{\vartheta} \newcommand{\tg}{\tilde{\gamma}} \newcommand{\bg}{\bar{\gamma}} \newcommand{\mG}{\mathring{G}} \newcommand{\mT}{\mathring{T}} \newcommand{\mg}{\mathring{g}} \newcommand{\zf}{\zeta\left(\frac{1}{2}+it\right)} \newcommand{\zfvp}{\zeta\left(\frac{1}{2}+i\vp_1(t)\right)} \newtheorem{theorem}{Theorem}%[section] \newtheorem{lemma}[]{Lemma} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} \newtheorem{rem}[]{Remark} \newtheorem{form}[theorem]{Formula} \newtheorem{cor}[]{Corollary} \theoremstyle{remark} \newtheorem{remark}[]{Remark} \newtheorem*{mydef1}{{\bf Theorem}} \newtheorem*{mydef11}{{\bf Theorem 1}} \newtheorem*{mydef12}{{\bf Theorem 2}} \newtheorem*{mydef2}{{\bf Definition}} \newtheorem*{mydef3}{{\bf Problem}} \newtheorem*{mydef4}{{\bf Corollary}} \newtheorem*{mydef5}{{\bf Lemma}} \numberwithin{equation}{section} % Absolute value notation \newcommand{\abs}[1]{\lvert#1\rvert} % Blank box placeholder for figures (to avoid requiring any % particular graphics capabilities for printing this document). \newcommand{\blankbox}[2]{% \parbox{\columnwidth}{\centering % Set fboxsep to 0 so that the actual size of the box will match the % given measurements more closely. \setlength{\fboxsep}{0pt}% \fbox{\raisebox{0pt}[#2]{\hspace{#1}}}% }% } \begin{document} \title{Jacob's ladders and the asymptotic equivalence of some integrals} \author{Jan Moser} \address{Department of Mathematical Analysis and Numerical Mathematics, Comenius University, Mlynska Dolina M105, 842 48 Bratislava, SLOVAKIA} \email{jan.mozer@fmph.uniba.sk} \keywords{Riemann zeta-function} \begin{abstract} In this paper we continue the study of a new class of integrals containing the products of the factors $|\zeta|^2$. Namely, we obtain asymptotic formulae for some class of integrals containing a product of polynomial in the variable $\vp_1^{n+1}(t)$ and the factors of type $|\zeta|^2$. \end{abstract} \maketitle \section{Introduction} \subsection{} Let us remind that Hardy and Littlewood started to study the following integral in 1926 \bdis \int_1^T\left|\zf\right|^4{\rm d}t=\int_1^T Z^4(t){\rm d}t \edis where \be \label{1.1} Z(t)=e^{i\vth(t)}\zf,\ \vth(t)=-\frac t2\ln\pi+\text{Im}\ln\Gamma\left(\frac 14+i\frac t2\right) \ee is the signal generated by the Riemann zeta-function on the critical line. Hardy and Littlewood derived the following estimate (see \cite{1}, pp. 41 -- 59; \cite{8}, p. 124) \bdis \int_1^T\left|\zf\right|^4{\rm d}t=\mcal{O}(T\ln^4T) . \edis In this direction, Ingham derived in 1926 the following asymptotic formula \be \label{1.2} \int_1^T\left|\zf\right|^4{\rm d}t=\frac{1}{2\pi^2}T\ln^4T+\mcal{O}(T\ln^3T) \ee (see \cite{2}, p. 227; \cite{8}, p. 125). Let us remind, finally, the Ingham - Heath-Brown formula (see \cite{3}, p. 129) \be \label{1.3} \int_0^T Z^4(t){\rm d}t=T\sum_{p=0}^4C_p\ln^{4-p}T+\mcal{O}(T^{\frac 78+\epsilon}),\ C_0=\frac{1}{2\pi^2} \ee that improves the Ingham's formula (\ref{1.2}). \begin{remark} The small improvements of the Heath-Brown exponent $\frac 78$ are irrelevant for our purposes. \end{remark} \subsection{} We have obtained in our paper \cite{6} the following formula \be \label{1.4} \begin{split} & \int_T^{T+U}\left|\zeta\left(\frac 12+i\vp_1^{n+1}(t)\right)\right|^4\prod_{k=0}^n\left|\zeta\left(\frac 12+i\vp_1^k(t)\right)\right|^2{\rm d}t\sim \frac{1}{2\pi^2}U\ln^{n+5}T,\\ & U=T^{\frac 78+2\epsilon},\quad T\to\infty , \end{split} \ee that is valid for every fixed $n\in\mbb{N}$ where (see \cite{7}, (2.1)) \bdis \begin{split} & \vp_1^0(t)=1,\ \vp_1^1(t)=\vp_1(t),\ \vp_1^2(t)=\vp_1(\vp_1(t)),\dots , \\ & \vp_1^{k+1}(t)=\vp_1(\vp_1^k(t)), \dots , \end{split} \edis and $\vp_1^k$ stands for the $k$-th iteration of the Jacob's ladder. \\ Next, by the Riemann-Siegel formula (comp. \cite{8}, p. 79) \be \label{1.5} Z(t)=2\sum_{n\leq x}\frac{1}{\sqrt{n}}\cos\{\vth(t)-t\ln n\}+\mcal{O}(t^{-\frac 14}),\ x=\sqrt{\frac{t}{2\pi}} , \ee we have \be \label{1.6} \begin{split} & \left|\zeta\left(\frac 12+i\vp_1^k(t)\right)\right|= \\ & = \left|2\sum_{n\leq x_k}\frac{1}{\sqrt{n}}\cos\{\vth[\vp_1^k(t)]-\vp_1^k(t)\ln n\}+\mcal{O}\left\{\left|\vp_1^k(t)\right|^{-\frac 14}\right\}\right|, \\ & x_k=\sqrt{\frac{\vp_1^k(t)}{2\pi}},\quad k=0,1,\dots,n+1 . \end{split} \ee \begin{remark} Let us remind the graph of the function $Z(\tau),\ t=2\pi \tau$ constructed by the Riemann-Siegel formula (\ref{1.5}) -- in the neighborhood of the Lehmer-Meller second pair of the zeroes of the function $\zf$, $\tau\approx 2728.52$. This graph, in connection with (\ref{1.1}), (\ref{1.6}), gives some information about how complicated is the structure of the signal \bdis P_n(t)=\left|\zeta\left(\frac 12+i\vp_1^{n+1}(t)\right)\right|^4\prod_{k=0}^n\left|\zeta\left(\frac 12+i\vp_1^k(t)\right)\right|^2,\ t\in [T,T+U] , \edis for example, in the case \bdis n=10^{10^{10^{34}}} \edis (Skeewes constant). \end{remark} \subsection{} In this paper we shall show that the integral \be \label{1.7} \int_T^{T+U}\left\{\sum_{p=0}^4 A_p\left[\ln\vp_1^{n+1}(t)\right]^{4-p}\right\}\prod_{k=0}^n\left|\zeta\left(\frac 12+i\vp_1^k(t)\right)\right|^2{\rm d}t \ee is asymptotically equal (by means of the asymptotic equivalence) to the integral (\ref{1.4}) for some vector \bdis A=(A_0,A_1,A_2,A_3,A_4) , \edis and next, the integrals (\ref{1.4}), (\ref{1.7}) generate some new class of asymptotically equivalent integrals. \section{The result} \subsection{} The following Theorem holds true. \begin{mydef1} If \bdis A=(C_0,C_1+4C_0,C_2+3C_1,C_3+2C_2,C_4+C_3) \edis where \be \label{2.1} C=(C_0,C_1,C_2,C_3,C_4) \ee is the Ingham - Heath-Brown vector (see (\ref{1.3}) then \be \label{2.2} \begin{split} & \int_T^{T+U}\left\{\left|\zeta\left(\frac 12+i\vp_1^{n+1}(t)\right)\right|^4-\sum_{p=0}^4 A_p\left[\ln\vp_1^{n+1}(t)\right]^{4-p}\right\}\times \\ & \times\prod_{k=0}^n\left|\zeta\left(\frac 12+i\vp_1^k(t)\right)\right|^2{\rm d}t=\mcal{O}(U\ln^{n+4}T\ln\ln T),\\ & U=T^{\frac 78+2\epsilon} \end{split} \ee for every fixed $n\in\mbb{N}$, i. e. (see (\ref{1.4})) \be \label{2.3} \begin{split} & \int_T^{T+U}\left|\zeta\left(\frac 12+i\vp_1^{n+1}(t)\right)\right|^4\prod_{k=0}^n\left|\zeta\left(\frac 12+i\vp_1^k(t)\right)\right|^2{\rm d}t\sim \\ & \sim \int_T^{T+U}\left\{\sum_{p=0}^4 A_p\left[\ln\vp_1^{n+1}(t)\right]^{4-p}\right\}\prod_{k=0}^n\left|\zeta\left(\frac 12+i\vp_1^k(t)\right)\right|^2{\rm d}t,\ T\to\infty . \end{split} \ee \end{mydef1} \begin{remark} The asymptotic equivalence of the integrals (\ref{1.4}) and (\ref{1.7}) is expressed by the formula (\ref{2.3}). \end{remark} \subsection{} Let us denote \bdis \begin{split} & \left|\zeta\left(\frac 12+i\vp_1^{n+1}(t)\right)\right|=|\zeta^4|, \\ & \sum_{p=0}^4 A_p\left[\ln\vp_1^{n+1}(t)\right]^{4-p}=P_4 . \end{split} \edis Then, by the simple algebraic manipulation, one obtains \be \label{2.4} P_4=\frac{|\zeta^4|^{2^L}-(P_4)^{2^L}}{\sum_{l=0}^{2^L-1}|\zeta^4|^{2^L-l-1}(P_4)^l} , \ee and, consequently, by (\ref{2.2})-(\ref{2.4}) we obtain ($P_4>0$ as $T\to\infty$) the following \begin{cor} \be \label{2.5} \begin{split} & \int_T^{T+U}|\zeta^4|^{2^L}\frac{\prod_{k=0}^n\left|\zeta\left(\frac 12+i\vp_1^k(t)\right)\right|^2} {\sum_{l=0}^{2^L-1}|\zeta^4|^{2^L-l-1}(P_4)^l}{\rm d}t\sim \\ & \sim\int_T^{T+U} (P_4)^{2^L}\frac{\prod_{k=0}^n\left|\zeta\left(\frac 12+i\vp_1^k(t)\right)\right|^2} {\sum_{l=0}^{2^L-1}(P_4)^{2^L-l-1}(P_4)^l}{\rm d}t,\ T\to\infty \end{split} \ee for every fixed $n,L\in\mbb{N}$ . \end{cor} Next, from the formula \bdis |\zeta^4|-P_4=\left[|\zeta^4|^{\frac{1}{2^L}}-(P_4)^{\frac{1}{2^L}}\right]\sum_{l=0}^{2^L-1} \left(|\zeta|^{\frac{4}{2^L}}\right)^{2^L-l-1}\left[(P_4)^{\frac{1}{2^L}}\right]^l \edis we obtain by a similar way the following \begin{cor} \be \label{2.6} \begin{split} & \int_T^{T+U} |\zeta^4|^{\frac{1}{2^L}}\sum_{l=0}^{2^L-1}\left(|\zeta|^{\frac{4}{2^L}}\right)^{2^L-l-1}\left[(P_4)^{\frac{1}{2^L}}\right]^l\prod_{k=0}^n\left|\zeta\left(\frac 12+i\vp_1^k(t)\right)\right|^2{\rm d}t\sim \\ & \sim\int_T^{T+U}(P_4)^{\frac{1}{2^L}} \sum_{l=0}^{2^L-1}\left(|\zeta|^{\frac{4}{2^L}}\right)^{2^L-l-1}\left[(P_4)^{\frac{4}{2^L}}\right]^l\prod_{k=0}^n\left|\zeta\left(\frac 12+i\vp_1^k(t)\right)\right|^2{\rm d}t,\\ & T\to\infty \end{split} \ee for every fixed $n.L\in\mbb{N}$. \end{cor} \begin{remark} There is the following interpretation of the formula (\ref{2.5}): the mapping \bdis |\zeta^4|\ \leftrightarrow \ P_4 \edis transforms the integral \bdis \int_T^{T+U}|\zeta^4|\frac{\prod_{k=0}^n\left|\zeta\left(\frac 12+i\vp_1^k(t)\right)\right|^2} {\sum_{l=0}^{2^L-1}|\zeta^4|^{2^L-l-1}(P_4)^l}{\rm d}t \edis into the asymptotically equivalent integral \bdis \int_T^{T+U}P_4\frac{\prod_{k=0}^n\left|\zeta\left(\frac 12+i\vp_1^k(t)\right)\right|^2} {\sum_{l=0}^{2^L-1}(P_4)^{2^L-l-1}(P_4)^l}{\rm d}t , \edis and vice-versa. A similar property holds true for the formula (\ref{2.6}). \end{remark} \section{Proof of Theorem} \subsection{} From the formula (see \cite{7}, (3.1)) \bdis \int_T^{T+U} F[\vp_1^{n+1}(t)]\prod_{k=0}^n\tilde{Z}^2[\vp_1^k(t)]{\rm d}t=\int_{\vp_1^{n+1}(T)}^{\vp_1^{n+1}(T+U)}F(t){\rm d}t \edis we obtain in the cases \be \label{3.1} F(t)=\left|\zf\right|^4,\ \sum_{p=0}^4 A_p\ln^{4-p}t,\ U=T^{\frac 78+2\epsilon} \ee the following formulae (see (\ref{1.3})) \be \label{3.2} \begin{split} & J_1=\int_T^{T+U}\left|\zeta\left(\frac 12+i\vp_1^{n+1}(t)\right)\right|^4\prod_{k=0}^4\tilde{Z}^2[\vp_1^k(t)]{\rm d}t= \\ & =\int_{\vp_1^{n+1}(T)}^{\vp_1^{n+1}(T+U)}\left|\zf\right|^4{\rm d}t=\left. t\sum_{p=0}^4 C_p\ln^{4-p}t\right|_{t=\vp_1^{n+1}(T)}^{\vp_1^{n+1}(T+U)}+\mcal{O}(T^{\frac 78+\epsilon}) , \end{split} \ee since (see \cite{6}, (3.5)) \be \label{3.3} t\sim \vp_1^{n+1}(t),\ t\in [T,T+U] , \ee and \be \label{3.4} \begin{split} & J_2=\int_T^{T+U}\left\{\sum_{p=0}^4 A_p[\ln\vp_1^{n+1}(t)]^{4-p}\right\}\prod_{k=0}^4\tilde{Z}^2[\vp_1^k(t)]{\rm d}t=\\ & =\int_{\vp_1^{n+1}(T)}^{\vp_1^{n+1}(T+U)}\left\{\sum_{p=0}^4 A_p\ln^{4-p}t\right\}{\rm d}t . \end{split} \ee \subsection{} By elementary formulae (the constant of integration are omited) \bdis \begin{split} & \int 1{\rm d}t=t, \\ & \int \ln t{\rm d}t=t\ln t-t , \\ & \int \ln^2 t{\rm d}t=t\ln^2t-2t\ln t+2t , \\ & \int \ln^3 t{\rm d}t=t\ln^3t-3t\ln^2t+6t\ln t-6t , \\ & \int \ln^4 t{\rm d}t=t\ln^4t-4t\ln^3t+12t\ln^2t-24t\ln t+24t \end{split} \edis we obtain that (comp. \cite{5}, (3.1)-(3.3)) \bdis \int_{\vp_1^{n+1}(T)}^{\vp_1^{n+1}(T+U)}\left\{\sum_{p=0}^4 A_p\ln^{4-p}t\right\}{\rm d}t=\left. t\sum_{p=0}^4 B_p\ln^{4-p}t\right|_{t=\vp_1^{n+1}(T)}^{t=\vp_1^{n+1}(T)} , \edis where \bdis \begin{split} & B_0=A_0 , \\ & B_1=-4A_0+A_1 , \\ & B_2=12A_0-3A_1+A_2, \\ & B_3=-24A_0+6A_1-2A_2+A_3 , \\ & B_4=24A_0-6A_1+2A_2-A_3+A_4 , \end{split} \edis i. e. \be \label{3.5} A=(B_0,B_1+4B_0,B_2+3B_1,B_3+2B_2,B_4+B_3) . \ee By making use of the equality \bdis B=C \edis (see (\ref{2.1})) we obtain by (\ref{3.4}), (\ref{3.5}) the following formula \be \label{3.6} \begin{split} & J_2=\int_T^{T+U}\left\{\sum_{p=0}^4 A_p[\ln\vp_1^{n+1}(t)]^{4-p}\right\}\prod_{k=0}^4\tilde{Z}^2[\vp_1^k(t)]{\rm d}t= \\ & = \left. t\sum_{p=0}^4 C_p\ln^{4-p}t\right|_{t=\vp_1^{n+1}(T)}^{t=\vp_1^{n+1}(T+U)} . \end{split} \ee \subsection{} If \bdis H(t)=t\sum_{p=0}^4 C_p\ln^{4-p}t,\ t\in [\vp_1^{n+1}(T),\vp_1^{n+1}(T+U)] \edis then \be \label{3.7} \begin{split} & H'(t)\sim C_0\ln^4t , \\ & H[\vp_1^{n+1}(T+U)]-H[\vp_1^{n+1}(T)]=H'(\alpha)[\vp_1^{n+1}(T+U)-\vp_1^{n+1}(T)]\sim \\ & \sim C_0U\ln^4\alpha , \\ & \alpha=\vp_1^{n+1}(T)+\theta[\vp_1^{n+1}(T+U)-\vp_1^{n+1}(T)],\ \theta\in (0,1) \end{split} \ee since in the macroscopic case (comp. (\ref{3.1})) we have (see \cite{7}, (2.9)) \be \label{3.8} \vp_1^{n+1}(T+U)-\vp_1^{n+1}(T)\sim U,\ U=T^{\frac 78+2\epsilon}\in \left[ T^{\frac 13+2\epsilon},\frac{T}{\ln^2T}\right] . \ee Next, by (\ref{3.3}), (\ref{3.8}) \be \label{3.9} \begin{split} & \ln\alpha=\ln\{\vp_1^{n+1}(T)+\theta [\vp_1^{n+1}(T+U)-\vp_1^{n+1}(T)]\}= \\ & = \ln\vp_1^{n+1}(T)+\ln\left\{1+\mcal{O}\left(\frac{U}{\vp_1^{n+1}(T)}\right)\right\}\sim\ln T , \end{split} \ee and, consequently, we obtain from (\ref{3.7}) \be \label{3.10} \left. t\sum_{p=0}^4 C_p\ln^{4-p}t\right|_{t=\vp_1^{n+1}(T)}^{t=\vp_1^{n+1}(T+U)}\sim C_0U\ln^4T . \ee \subsection{} It follows from (\ref{3.3}) and (\ref{3.7}) that \bdis J_1-J_2=\mcal{O}(T^{\frac 78+\epsilon}) , \edis and (see (\ref{3.6}), (\ref{3.10})) \bdis \frac{J_1}{J_2}-1=\mcal{O}\left(\frac{T^{\frac 78+\epsilon}}{J_2}\right)=\mcal{O}\left(\frac{T^{\frac 78+\epsilon}}{U\ln^4T}\right) , \edis i. e. \be \label{3.11} \frac{J_1}{J_2}=1+\mcal{O}\left(\frac{1}{T^{\epsilon}\ln^4T}\right) . \ee Next, in order to transform $J_1$ and $J_2$, we use the formula (see \cite{7}, (3.3)) \bdis \begin{split} & \int_T^{T+U} F[\vp_1^{n+1}(t)]\prod_{k=0}^n\tilde{Z}^2[\vp_1^k(t)]{\rm d}t= \\ & =\left\{1+\mcal{O}\left(\frac{\ln\ln T}{\ln T}\right)\right\}\frac{1}{\ln^{n+1}T}\int_T^{T+U}F[\vp_1^{n+1}(t)]\prod_{k=0}^n \left|\zeta\left(\frac 12+i\vp_1^k(t)\right)\right|^2{\rm d}t \end{split} \edis where (comp. \cite{7}, (2.2)) \bdis \tilde{Z}^2(t)=\frac{\left|\zf\right|^2}{\left\{ 1+\mcal{O}\left(\frac{\ln\ln t}{\ln t}\right)\right\}\ln t} , \edis and we obtain (see (\ref{3.11})) \bdis \frac{\bar{J}_1}{\bar{J}_2}\left\{1+\mcal{O}\left(\frac{\ln\ln T}{\ln T}\right)\right\}=1+\mcal{O}\left(\frac{1}{T^{\epsilon}\ln^4T}\right), \edis i. e. \be \label{3.12} \frac{\bar{J}_1}{\bar{J}_2}=1+\mcal{O}\left(\frac{\ln\ln T}{\ln T}\right) \ee where \be \label{3.13} \begin{split} & \bar{J}_1=\int_T^{T+U}\left|\zeta\left(\frac 12+i\vp_1^{n+1}(t)\right)\right|^4\prod_{k=0}^n\left|\zeta\left(\frac 12+i\vp_1^{k}(t)\right)\right|^2{\rm d}t,\\ & \bar{J}_2=\int_T^{T+U}\left\{\sum_{p=0}^4 A_p[\ln\vp_1^{n+1}(t)]^{4-p}\right\}\prod_{k=0}^n\left|\zeta\left(\frac 12+i\vp_1^{k}(t)\right)\right|^2{\rm d}t . \end{split} \ee Since by (\ref{3.13}) and (\ref{1.4}) \bdis \tilde{J}_2\sim\tilde{J}_1\sim\frac{1}{2\pi^2}U\ln^{n+5}T , \edis then we obtain from (\ref{3.12}) \bdis \bar{J}_1=\bar{J}_2+\mcal{O}\left(\bar{J}_2\frac{\ln\ln T}{\ln T}\right)=\bar{J}_2+\mcal{O}(U\ln^{n+4}T\ln\ln T) \edis that verifies the formula (\ref{2.2}). \thanks{I would like to thank Michal Demetrian for helping me with the electronic version of this work.} \begin{thebibliography}{29}% Replace 9 by 99 if 10 or more references % % Please note the use of "\and" between author names below %\bibitem{fock} %V.A. 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