On Some Mean Value Results for the Zeta-Function and a Rankin–Selberg Problem

Abstract

Denote by ${\Delta }_1(x;\phi )$ the error term in the classical Rankin–Selberg problem. Denote by $\zeta \left(s\right)$ the Riemann zeta-function. We establish an upper bound for this integral ${\int }_0^T{\Delta }_1(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt.$ In addition, when $2\le k\le 4$ is a fixed integer, we will derive an asymptotic formula for the integral ${\int }_1^T{\Delta }_1^k(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt.$ The results rely on the power moments of ${\Delta }_1(t;\phi )$ and $E\left(t\right)$, the classical error term in the asymptotic formula for the mean square of $\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|$.

1. Introduction

Denote

$$\begin{array}{c}\hfill E\left(T\right):={\int }_0^T{\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt-T\left(log\left({\displaystyle \frac{T}{2\pi }}\right)+2\gamma -1\right)(T\ge 2)\end{array}$$

(1)

the error term in the mean square formula for $\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|$. Here, $\zeta \left(s\right)$ denotes the Riemann zeta-function, and $\gamma =-{\mathrm{\Gamma }}^{\prime }\left(1\right)=0.577215\cdots$ represents Euler’s constant. In [1], Ivić, established several theorems concerning the mean values of $\Delta \left(x\right)$, $E\left(t\right)$ and

$$\begin{array}{cc}\hfill \Delta *\left(x\right)& :=-\Delta \left(x\right)+2\Delta \left(2x\right)-{\displaystyle \frac{1}{2}}\Delta \left(4x\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\sum _{n\le 4x}{(-1)}^{n}d\left(n\right)-x(logx+2\gamma -1).\hfill \end{array}$$

The function $\Delta *\left(x\right)$ is the “modified” divisor function, introduced and considered by M. Jutila in [2,3]. Much earlier, F. V. Atkinson [4] derived a classical explicit formula for $E\left(T\right)$, revealing a deep analogy between $\Delta \left(x\right)$ and $E\left(T\right)$. It turns out, however, that $\Delta *\left(x\right)$ serves as a better analogue to $E\left(T\right)$ than $\Delta \left(x\right)$ itself. Specifically, M. Jutila (op.cit.) examined both the local and global behavior of the difference

$$\begin{array}{c}\hfill E*\left(t\right):=E\left(t\right)-2\pi \Delta *\left({\displaystyle \frac{1}{2\pi }}\right),\end{array}$$

and in particular, in [3], he obtained that

$$\begin{array}{c}\hfill {\int }_T^{T+H}{\left(E*\left(t\right)\right)}^{2}dt{\ll }_{\epsilon }H{T}^{1/3}{log}^{3}T+T^{1+\epsilon }(1\le H\le T).\end{array}$$

(2)

Throughout this paper, the notation $\epsilon$ denotes an arbitrarily small positive constant, not necessarily the same at each occurrence. The Vinogradov notation $a\left(x\right){\ll }_{\epsilon }b\left(x\right)$ (equivalent to $a\left(x\right)=O_{\epsilon }\left(b\left(x\right)\right)$) means there exists a constant $C\left(\epsilon \right)>0$, such that $\left|a\right(x\left)\right|\le C\left(\epsilon \right)b\left(x\right)$ holds for all $x\ge x_0\left(\epsilon \right)$.

The importance of (2) becomes apparent when we consider (see, for instance, [5] (Chapter 15))

$$\begin{array}{c}\hfill {\int }_0^T{\left(\Delta *\left(t\right)\right)}^{2}dt\sim A{T}^{3/2},{\int }_0^T{E}^2\left(t\right)dt\sim B{T}^{3/2}(A,B>0,T\to \infty ),\end{array}$$

it transpires that $E*\left(t\right)$ is, in the mean square sense, of a lower order of magnitude than both $\Delta *\left(t\right)$ and $E\left(t\right)$. Similar mean square estimates hold for $\Delta \left(t\right)$ as well, with even sharper bounds known in all three cases—for these results, we refer to the work of Lau and Tsang [6]. For a comprehensive survey of this subject, we direct the reader to K.-M. Tsang’s review article [7].

Motivated by these connections, it becomes sensible to investigate mean value results for $\Delta *\left(t\right)$ (and $\Delta \left(t\right)$) in relation to $\zeta \left(s\right)$, particularly how these quantities interact. The study of mean values (or moments) for $\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|$ constitutes one of the central themes in zeta function theory, having been extensively researched. This importance is reflected in the existence of two specialized monographs: Ivić’s comprehensive treatment [8] and K. Ramachandra’s work [9]. Building on this foundation, Ivić established in [1] the following key estimate:

$$\begin{array}{c}\hfill {\int }_T^{T+H}\Delta *\left({\displaystyle \frac{t}{2\pi }}\right){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt\ll H{T}^{1/6}{log}^{7/2}T\end{array}$$

valid for the range $T^{2/3+\epsilon }\le H=H\left(T\right)\le T$. For C, a suitable positive constant, he obtained that

$$\begin{array}{c}\hfill {\int }_0^T{\left(\Delta *\left({\displaystyle \frac{t}{2\pi }}\right)\right)}^2\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|dt={\displaystyle \frac{C}{4{\pi }^2}}{T}^{3/2}\left(log{\displaystyle \frac{T}{2\pi }}+2\gamma -{\displaystyle \frac{2}{3}}\right)+O_{\epsilon }\left(T^{17/12+\epsilon }\right),\end{array}$$

(3)

and for D, another suitable positive constant, he obtained that

$$\begin{array}{c}\hfill {\int }_0^T{\left(\Delta *\left({\displaystyle \frac{t}{2\pi }}\right)\right)}^3\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|dt=D{T}^{7/4}\left(log{\displaystyle \frac{T}{2\pi }}+2\gamma -{\displaystyle \frac{4}{7}}\right)+O_{\epsilon }\left(T^{27/16+\epsilon }\right).\end{array}$$

(4)

The proofs of (3) and (4) in [1] crucially rely on the special structure of $\Delta *\left({\displaystyle \frac{t}{2\pi }}\right)$ and do not extend easily to handle $\Delta *\left(\alpha t\right)$ or $\Delta \left(\alpha t\right)$ for arbitrary $\alpha >0$.

Subsequently, Ivić and Zhai [10] investigated a related but more complex problem: the asymptotic evaluation of integrals involving ${\Delta }^k\left(t\right){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^2$ for fixed $k\in \mathbb{N}$.

The Dirichlet divisor problem remains a central topic in analytic number theory. In recent years, building upon this classical problem, growing numbers of scholars have turned to investigating the Rankin–Selberg problem—a natural generalization of the divisor problem that significantly expands the scope of research in analytic number theory.

Let $\phi \left(z\right)$ be a holomorphic cusp form of weight $\kappa$ for the full modular group $SL(2,\mathbb{Z})$, with normalized Fourier coefficients $a\left(n\right)$ satisfying $a\left(1\right)=1$ and $T\left(n\right)\phi =a\left(n\right)\phi$ for all $n\in \mathbb{N}$, where $T\left(n\right)$ denotes the n-th Hecke operator. Rankin [11] and Selberg [12] independently introduced the function

$$Z\left(s\right)=\zeta \left(2s\right)\sum _{n=1}^{\infty }{\left|a\left(n\right)\right|}^2{n}^{1-\kappa -s},$$

where $\zeta \left(s\right)$ is the Riemann zeta-function. For $\sigma =\Re \left(s\right)>1$, $Z\left(s\right)$ admits the absolutely convergent Dirichlet series expansion

$$Z\left(s\right)=\sum _{n=1}^{\infty }{c}_n{n}^{-s},$$

with coefficients given by

$$c_n=n^{1-\kappa }\sum _{m^2|n}{m}^{2(\kappa -1)}{\left|a\left({\displaystyle \frac{n}{m^2}}\right)\right|}^2.$$

Deligne’s bound [13] on the Fourier coefficients, $\left|a\left(n\right)\right|\le n^{{\displaystyle \frac{\kappa -1}{2}}}d\left(n\right)$ (where $d\left(n\right)$ is the divisor function), implies that $c_n\ll n^{\epsilon }$ for any $\epsilon >0$.

In 1999, Ivić, Matsumoto, and Tanigawa [14] investigated Riesz means associated with the Rankin–Selberg coefficients $c_n$, defined for $\rho \ge 0$ by

$$D_{\rho }(x;\phi ):={\displaystyle \frac{1}{\mathrm{\Gamma }(\rho +1)}}\sum _{n\le x}{(x-n)}^{\rho }{c}_n.$$

The error term ${\Delta }_{\rho }(x;\phi )$ is introduced through the decomposition

$$D_{\rho }(x;\phi )={\displaystyle \frac{{\pi }^2\kappa R_0}{6\mathrm{\Gamma }(\rho +2)}}{x}^{\rho +1}+{\displaystyle \frac{Z\left(0\right)}{\mathrm{\Gamma }(\rho +1)}}{x}^{\rho }+{\Delta }_{\rho }(x;\phi ).$$

For the specific case $\rho =1$, they established the pointwise bound

$${\Delta }_1(x;\phi )=O(x^{{\displaystyle \frac{6}{5}}})$$

and the mean square estimate

$${\int }_1^T{\Delta }_1^2(x;\phi )dx={\displaystyle \frac{2}{13}}{\left(2\pi \right)}^{-4}\left(\sum _{n=1}^{\infty }{c}_n^2{n}^{-{\displaystyle \frac{7}{4}}}\right)T^{{\displaystyle \frac{13}{4}}}+O\left(T^{3+\epsilon }\right).$$

(5)

Matsumoto [15] refined the error term in the mean square formula, replacing $O\left(T^{3+\epsilon }\right)$ with the sharper bound $O\left(T^3{(logT)}^{3+\epsilon }\right)$. Subsequently, Ivić [5] established an upper bound for the fourth moment of ${\Delta }_1(x;\phi )$. In a comprehensive study, Tanigawa, Zhai, and Zhang [16] developed a unified approach to derive asymptotic formulas for the third, fourth, and fifth power moments of ${\Delta }_1(x;\phi )$ as follows:

$$\begin{array}{cc}\hfill {\int }_T^{2T}{\Delta }_1^k(x;\phi )dx& ={\displaystyle \frac{B_k\left(c\right)}{(8+9k)2^{3k-4}{\pi }^{2k}}}{T}^{1+{\displaystyle \frac{9k}{8}}}+O\left(T^{1+{\displaystyle \frac{9k}{8}}-{\delta }_k+\epsilon }\right),\hfill \end{array}$$

(6)

where

$$\begin{array}{cc}\hfill & B_k\left(f\right):=\sum _{l=1}^{k-1}\left(\begin{array}{c}k-1\\ l\end{array}\right)s_{k;l}\left(f\right)cos{\displaystyle \frac{\pi (k-2l)}{4}},\hfill \\ & s_{k;l}:=\sum _{\sqrt[4]{n_1}+\cdots +\sqrt[4]{n_l}=\sqrt[4]{n_{l+1}}+\cdots +\sqrt[4]{n_k}}{\displaystyle \frac{f\left(n_1\right)\cdots f\left(n_k\right)}{{(n_1\cdots n_k)}^{7/8}}},1\le l\le k,\hfill \\ \hfill & {\delta }_3={\displaystyle \frac{1}{36}},{\delta }_4={\displaystyle \frac{1}{221}},{\delta }_5={\displaystyle \frac{1}{1731}}.\hfill \end{array}$$

The study of power moments for error terms crucially relies on the truncated Voronoi summation formula. In [17], the authors applied splitting arguments to deal with the integral mean value for the truncated summation and improved the above results to ${\delta }_3={\displaystyle \frac{3}{62}},{\delta }_4={\displaystyle \frac{3}{256}},{\delta }_5={\displaystyle \frac{1}{680}}$, respectively. Later, in [18], the authors improved ${\delta }_4$ and ${\delta }_5$ to ${\displaystyle \frac{1}{84}}$ and ${\displaystyle \frac{11}{4080}}$, respectively.

In this paper, to investigate the properties of the Riemann zeta-function and the error terms in the Rankin–Selberg problems more deeply, we consider the asymptotic evaluation of the integrals of ${\Delta }_1^k(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^2$ when $2\le k\le 4$ is fixed. We succeeded in applying the existing results on the moments of ${\Delta }_1(t;\phi )$ and $E\left(t\right)$ to the evaluation of the integrals of ${\Delta }_1^k(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^2$. Throughout our proof, we primarily employ the method of large value estimation. The main technical challenge arises when applying the dyadic decomposition approach to handle the summation of $J_1$ terms, which requires careful analysis and delicate treatment. Specifically, we need to overcome the difficulties in controlling the error terms and maintaining the necessary precision during the decomposition process. Our results are as follows:

Theorem 1. 

Suppose $T\ge 2$ is a large parameter. We have

$${\int }_1^T{\Delta }_1(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt\ll T^{2+\epsilon }$$

(7)

for any $\epsilon >0$.

Theorem 2. 

If k is a fixed integer for which $2\le k\le 4$, then we have

$${\int }_1^T{\Delta }_1^k(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt=c_1\left(k\right)T^{1+{\displaystyle \frac{9k}{8}}}logT+c_2\left(k\right)T^{1+{\displaystyle \frac{9k}{8}}}+O\left(T^{1+{\displaystyle \frac{9k}{8}}-{\eta }_k+\epsilon }\right),$$

(8)

where $c_1\left(k\right)$ and $c_2\left(k\right)$ are explicit constants, and where ${\eta }_k={\displaystyle \frac{1}{4}}(2\le k\le 4)$.

Note that ${\eta }_2$, ${\eta }_3$ and ${\eta }_4$ in Theorem 2 are identical, owing to the broad reasoning applied in the proof in Section 4. However, we observe that the values of ${\eta }_2$ and ${\eta }_3$ could potentially be improved based on the results of Theorems 1 and 2. Due to space limitations, we will not explore this further, but interested readers may consider it.

2. Some Preliminary Lemmas

In this paper, the constants in O-term and ≪-symbol depend most on $\epsilon$. To prove the main theorem, we need the following lemmas:

Lemma 1. 

There exists a constant θ, such that ${\displaystyle \frac{1}{4}}\le \theta <{\displaystyle \frac{1}{3}}$ and

$$E\left(t\right){\ll }_{\epsilon }{t}^{\theta +\epsilon }.$$

(9)

Particularly, we can take $\theta ={\displaystyle \frac{131}{416}}=0.3149\cdots$.

The proof of the bound in (9) is due to N. Watt [19]. It is generally assumed that $\theta ={\displaystyle \frac{1}{4}}$ is permissible, but this is out of reach at present. We all know that $\theta <{\displaystyle \frac{1}{4}}$ cannot hold (see [20], Chapter 13 and Chapter 15).

Lemma 2. 

Let θ be the constant in Lemma 1. Then, we have

$${\int }_1^T{\left|{\Delta }_1(x;\phi )\right|}^{A}dt{\ll }_{\epsilon }{T}^{1+{\displaystyle \frac{9A}{8}}+\epsilon }$$

(10)

and

$${\int }_1^T{\left|E\left(t\right)\right|}^{B}dt{\ll }_{\epsilon }{T}^{1+M\left(B\right)+\epsilon }$$

(11)

for any A and B satisfying $0\le A\le {\displaystyle \frac{16}{3}}$ and $0\le B\le 11$, respectively, where

$$\begin{array}{c}\hfill M\left(B\right):=max\left({\displaystyle \frac{A}{4}},\theta (A-2)\right).\end{array}$$

(12)

When $0\le k\le 9$ is real, the limits

$$\begin{array}{c}\hfill E_k:=\underset{T\to \infty }{lim}{T}^{-1-{\displaystyle \frac{k}{4}}}{\int }_0^T{\left|E\left(t\right)\right|}^{k}dt\end{array}$$

(13)

exist. The first result is Theorem 1 in [16]. In [21], Heath-Brown obtained that the limits of moments without absolute values also exist when $k=1,3,5,7,9$. The merit of (13) that it gets rid of “$\epsilon$” and establishes the existence of the limit (but without an error term).

Lemma 3. 

We have

$${\int }_0^T{\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{4}dt=T{Q}_4(logT)+O\left(T^{{\displaystyle \frac{2}{3}}}{log}^{8}T\right),$$

(14)

where $Q_4\left(x\right)$ is an explicit polynomial of degree four in x with leading coefficient ${\displaystyle \frac{1}{2{\pi }^2}}$.

Ivić and Motohashi [22] first proved this result with error term $O\left(T^{{\displaystyle \frac{2}{3}}}{log}^{c}T\right)$. In [23], Motohashi obtained the value $C=8$.

Lemma 4. 

We have

$$\begin{array}{c}\hfill {\Delta }_1(x;\phi )={\displaystyle \frac{x^{9/8}}{{\left(2\pi \right)}^2}}\sum _{n\le N}{\displaystyle \frac{c_n}{n^{7/8}}}cos\left(8\pi \sqrt[4]{nx}-{\displaystyle \frac{\pi }{4}}\right)+O_{\epsilon }\left(x^{1+\epsilon }+x^{3/2+\epsilon }{N}^{-1/2}\right)\end{array}$$

(15)

for $1\le N\ll x^2$.

This is Lemma 2.1 in [16].

Lemma 5. 

We have

$$\begin{array}{c}\hfill {\Delta }_1(x;\phi )={\displaystyle \frac{x^{9/8}}{{\left(2\pi \right)}^2}}\sum _{n\le Q}{\displaystyle \frac{c_n}{n^{7/8}}}cos\left(8\pi \sqrt[4]{nx}-{\displaystyle \frac{\pi }{4}}\right)+F\left(x\right)\end{array}$$

(16)

for $Q\gg x\gg 1$, where $F\left(x\right)\ll x^{{\displaystyle \frac{7}{8}}}$ if $\parallel x\parallel \gg x^{{\displaystyle \frac{5}{8}}+\epsilon }{Q}^{-{\displaystyle \frac{1}{2}}}$, and we always have $F\left(x\right){\ll }_{\epsilon }x{Q}^{\epsilon }$.

This result is due to A. Ivić, K. Matsumoto, and Y. Tanigawa [14]; here, $\parallel x\parallel$ denotes as usual the distance of x to the nearest integer.

Lemma 6. 

Let $c>0$ be a non-integer real number. Suppose $N\ge 2$, $\Delta >0$ is any real number. Let $\mathcal{A}(N,\Delta ;c)$ denote the number of solutions of inequality

$$\left|n_1^c+n_2^c-n_3^c-n_4^c\right|<\delta N^c,n_j\sim N_j.$$

Then

$$\begin{array}{c}\hfill \mathcal{A}(N,\Delta ;c)\ll N^{\epsilon }\left(N^4\delta +N^2\right).\end{array}$$

(17)

This is Theorem 2 in [24].

Lemma 7. 

For the convolution function $c_n$, we have

$$\sum _{n\le x}{c}_n^2\ll x^{1+\epsilon }$$

for any $\epsilon >0$.

Due to $c_n\ll n^{\epsilon }$, it is easy to obtain the bound.

Lemma 8. 

For $1\le r\ll x$, we have

$$\sum _{n\le x}{c}_n{c}_{n+r}\ll x^{1+\epsilon }.$$

It is easy to obtain the bound by $c_n\ll n^{\epsilon }$.

Lemma 9. 

Suppose that $0<A<A^{\prime }$ are any two fixed constants such that $AT<N<A^{\prime }T$, and let $N^{\prime }=N^{\prime }\left(T\right)={\displaystyle \frac{T}{2\pi }}+{\displaystyle \frac{N}{2}}-{\left({\displaystyle \frac{N^2}{4}}+{\displaystyle \frac{NT}{2\pi }}\right)}^{{\displaystyle \frac{1}{2}}}$. Then, we have

$$E\left(T\right)={\sum }_1\left(T\right)+{\sum }_2\left(T\right)+O\left({log}^{2}T\right),$$

where

$$\begin{array}{cc}\hfill {\sum }_1\left(T\right)=& \sqrt{2}{(T/\left(2\pi \right))}^{1/4}\sum _{n\le N}{(-1)}^{n}d\left(n\right)n^{-3/4}e(T,n)cos\left(f(T,n)\right),\hfill \\ \hfill {\sum }_2\left(T\right)=& -2\sum _{n\le N^{\prime }}d\left(n\right)n^{-1/2}{\left(log{\displaystyle \frac{T}{2\pi n}}\right)}^{-1}cos\left(Tlog\left({\displaystyle \frac{T}{2\pi n}}\right)-T+{\displaystyle \frac{\pi }{4}}\right)\hfill \end{array}$$

with

$$\begin{array}{cc}\hfill f(T,n)=& 2T\mathrm{ar}sinh\left(\sqrt{{\displaystyle \frac{\pi n}{2T}}}\right)+\sqrt{2\pi nT+{\pi }^2{n}^2}-{\displaystyle \frac{\pi }{4}}\hfill \\ \hfill =& -{\displaystyle \frac{\pi }{4}}+2\sqrt{2\pi nT}+{\displaystyle \frac{1}{6}}\sqrt{2{\pi }^3}{n}^{{\displaystyle \frac{3}{2}}}{T}^{-{\displaystyle \frac{1}{2}}}+a_5{n}^{{\displaystyle \frac{5}{2}}}{T}^{-{\displaystyle \frac{3}{2}}}+a_7{n}^{{\displaystyle \frac{7}{2}}}{T}^{-{\displaystyle \frac{5}{2}}}+\cdots ,\hfill \\ \hfill e(T,n)=& {\left(1+{\displaystyle \frac{\pi n}{2T}}\right)}^{-{\displaystyle \frac{1}{4}}}{\left({\left({\displaystyle \frac{2T}{\pi n}}\right)}^{{\displaystyle \frac{1}{2}}}\mathrm{ar}sinh\left(\sqrt{{\displaystyle \frac{\pi n}{2T}}}\right)\right)}^{-1}\hfill \\ \hfill =& 1+O\left({\displaystyle \frac{n}{T}}\right)(1\le n<T)\hfill \end{array}$$

and $\mathrm{ar}sinhx=log\left(x+\sqrt{1+x^2}\right)$.

This is the famous formula of F. V. Atkinson [4]; proofs can also be found in [8,20].

3. Proof of Theorem 1

It suffices to study in (3) the integral over $[T,2T]$, to replace then T by $T{2}^{-j}(j=1,2,\cdots )$ and sum the resulting estimates. Let $T\le t\le 2T$. We take $Q=T^7$ in Lemma 5 and write

$$\begin{array}{c}\hfill {\Delta }_1(t;\phi )={\Delta }_{11}(t;\phi )+{\Delta }_{12}(t;\phi )+F\left(t\right),\end{array}$$

(18)

where $F\left(t\right)$ is as in Lemma 5, and

$$\begin{array}{cc}\hfill & {\Delta }_{11}(t;\phi ):={\displaystyle \frac{t^{9/8}}{{\left(2\pi \right)}^2}}\sum _{n\le T}{c}_n{n}^{-7/8}cos\left(8\pi \sqrt[4]{t}-{\displaystyle \frac{\pi }{4}}\right),\hfill \\ \hfill & {\Delta }_{12}(t;\phi ):={\displaystyle \frac{t^{9/8}}{{\left(2\pi \right)}^2}}\sum _{T<n\le Q}{c}_n{n}^{-7/8}cos\left(8\pi \sqrt[4]{t}-{\displaystyle \frac{\pi }{4}}\right).\hfill \end{array}$$

(19)

Thus, we have

$${\int }_T^{2T}{\Delta }_1(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt={\int }_T^{2T}({\Delta }_{11}(t;\phi )+{\Delta }_{12}(t;\phi )+F\left(t\right)){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt.$$

From the Cauchy–Schwarz inequality, we can find that the form $F\left(t\right)$ in (18) makes a contribution of $O(T^{15/8}logT)$. From the first derivative test (see, e.g., Lemma 2.1 of [5]), Lemmas 3 and 7, we can obtain that the contribution of ${\Delta }_{12}(t;\phi )$ is

$$\begin{array}{cc}\hfill & \ll T^{9/8}{\left\{{\int }_T^{2T}{\left|\sum _{T<n\le Q}{c}_n{n}^{-7/8}cos\left(8\pi \sqrt[4]{nt}-{\displaystyle \frac{\pi }{4}}\right)\right|}^{2}dt{\int }_T^{2T}{\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{4}dt\right\}}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \ll T^{{\displaystyle \frac{13}{8}}}{log}^{2}T{\left\{T^{{\displaystyle \frac{3}{4}}+\epsilon }+T^{{\displaystyle \frac{3}{4}}}\sum _{T<m\ne n\le Q}{\displaystyle \frac{c_m{c}_n}{{\left(mn\right)}^{7/8}\left|\sqrt[4]{m}-\sqrt[4]{n}\right|}}\right\}}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

(20)

In the double sum in (20), when $m\ge 16n$ or $n\ge 16m$, the contribution is $\ll T^{\epsilon }$. We suppose that $m>n$, take $m=n+r$ and use Lemma 8, then we have the contribution of the remaining terms, which is

$$\ll \sum _{r\ll Q}{\displaystyle \frac{1}{r}}\sum _{n\le Q}{\displaystyle \frac{c_n{c}_{n+r}}{n}}\ll T^{\epsilon }.$$

Thus, we have the contribution of ${\Delta }_{12}(t;\phi )$, which is

$$T^{13/8}{log}^{2}T{\left(T^{{\displaystyle \frac{3}{4}}+\epsilon }\right)}^{{\displaystyle \frac{1}{2}}}\ll T^{2+\epsilon }.$$

In addition, by (2.9) in [10], we have

$$\begin{array}{cc}\hfill {\int }_T^{2T}& {\Delta }_{11}(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt\hfill \\ \hfill & ={\int }_T^{2T}{\displaystyle \frac{t^{9/8}}{{\left(2\pi \right)}^2}}\sum _{n\le T}{c}_n{n}^{-7/8}cos\left(8\pi \sqrt[4]{nt}-{\displaystyle \frac{\pi }{4}}\right)\left(log{\displaystyle \frac{t}{2\pi }}+2\gamma +E^{\prime }\left(t\right)\right)dt\hfill \\ \hfill & =I_1\left(T\right)+I_2\left(T\right).\hfill \end{array}$$

(21)

For $I_1\left(T\right)$, by the derivative test, we have

$$\begin{array}{cc}\hfill I_1\left(T\right)& :={\int }_T^{2T}{\displaystyle \frac{t^{9/8}}{{\left(2\pi \right)}^2}}\sum _{n\le T}{c}_n{n}^{-7/8}cos\left(8\pi \sqrt[4]{nt}-{\displaystyle \frac{\pi }{4}}\right)\left(log{\displaystyle \frac{t}{2\pi }}+2\gamma \right)dt\hfill \\ \hfill & \ll T^{9/8}logT\sum _{n\le T}{c}_n{n}^{-7/8}{T}^{3/4}{n}^{-1/4}\hfill \\ \hfill & \ll T^{15/8}logT.\hfill \end{array}$$

The integral $I_2\left(T\right)$, namely

$$I_2\left(T\right):={\int }_T^{2T}{E}^{\prime }\left(t\right){\displaystyle \frac{t^{9/8}}{{\left(2\pi \right)}^2}}\sum _{n\le T}{c}_n{n}^{-7/8}cos\left(8\pi \sqrt[4]{nt}-{\displaystyle \frac{\pi }{4}}\right)dt,$$

is integrated by parts. Since $E\left(t\right)\ll t^{1/3}$ (see, e.g., Chapter 15 of [5], also follows trivially from Lemma 1), the integrated terms are trivially

$$\ll T^{{\displaystyle \frac{1}{3}}+{\displaystyle \frac{9}{8}}+{\displaystyle \frac{1}{8}}+\epsilon }\ll T^{{\displaystyle \frac{19}{12}}+\epsilon }.$$

There remains a multiple of

$$\begin{array}{cc}\hfill & -{\displaystyle \frac{9}{8}}{\int }_T^{2T}{t}^{1/8}E\left(t\right)\sum _{n\le T}{c}_n{n}^{-7/8}cos\left(8\pi \sqrt[4]{nt}-{\displaystyle \frac{\pi }{4}}\right)dt\hfill \\ \hfill & +2\pi {\int }_T^{2T}{t}^{3/8}E\left(t\right)\sum _{n\le T}{c}_n{n}^{-5/8}sin\left(8\pi \sqrt[4]{nt}-{\displaystyle \frac{\pi }{4}}\right)dt.\hfill \end{array}$$

(22)

The estimates of both integrals in (22) are similar, and it is easy to see that the latter is larger. We replace $E\left(t\right)$ with the expression given by Lemma 9. Therefore, by taking $N=T$ in Lemma 9, we have

$${\int }_T^{2T}{t}^{3/8}E\left(t\right)\sum _{n\le T}{c}_n{n}^{-5/8}sin\left(8\pi \sqrt[4]{nt}-{\displaystyle \frac{\pi }{4}}\right)dt=J_1\left(T\right)+J_2\left(T\right)+J_3\left(T\right),$$

where

$$\begin{array}{cc}\hfill J_1\left(T\right):=& {\int }_T^{2T}{t}^{3/8}{\sum }_1\left(t\right)\sum _{n\le T}{c}_n{n}^{-5/8}sin\left(8\pi \sqrt[4]{nt}-{\displaystyle \frac{\pi }{4}}\right)dt,\hfill \\ \hfill J_2\left(T\right):=& {\int }_T^{2T}{t}^{3/8}{\sum }_2\left(t\right)\sum _{n\le T}{c}_n{n}^{-5/8}sin\left(8\pi \sqrt[4]{nt}-{\displaystyle \frac{\pi }{4}}\right)dt,\hfill \\ \hfill J_3\left(T\right):=& {\int }_T^{2T}{t}^{3/8}O\left({log}^{2}T\right)\sum _{n\le T}{c}_n{n}^{-5/8}sin\left(8\pi \sqrt[4]{nt}-{\displaystyle \frac{\pi }{4}}\right)dt.\hfill \end{array}$$

Through the Cauchy–Schwarz inequality, we have

$$\begin{array}{c}\hfill J_2\left(T\right)+J_3\left(T\right)\ll T^{3/8}{\left\{{\int }_T^{2T}\left({\sum }_2^2\left(t\right)+{log}^{4}T\right)dt{\int }_T^{2T}{\left|\sum _{n\le T}{c}_n{n}^{-5/8}{e}^{8\pi i\sqrt[4]{nt}}\right|}^{2}dt\right\}}^{{\displaystyle \frac{1}{2}}}.\end{array}$$

(23)

However, from Chapter 15 of [20], we have

$${\int }_T^{2T}{\sum }_2^2\left(t\right)dt\ll T{log}^{4}T,$$

since ${\sum }_2\left(t\right)$ is essentially a Dirichlet polynomial of length $\asymp T$. The contribution of the other integral in (23) is

$$\begin{array}{cc}\hfill & \ll T\sum _{n\le T}{c}_n^2{n}^{-5/4}+\sum _{m\ne n\le T}{\displaystyle \frac{c_m{c}_n}{{\left(mn\right)}^{5/8}}}{\int }_T^{2T}{e}^{8\pi i(\sqrt[4]{m}-\sqrt[4]{n})\sqrt[4]{t}}dt\hfill \\ \hfill & \ll T{log}^{3}T+T^{3/4}\sum _{m\ne n\le T}{\displaystyle \frac{c_m{c}_n}{{\left(mn\right)}^{5/8}|\sqrt[4]{m}-\sqrt[4]{n}|}},\hfill \end{array}$$

where we have used the first derivative test. Note that if $m\le {\displaystyle \frac{n}{2}}$, then we have $|\sqrt[4]{m}-\sqrt[4]{n}{|}^{-1}\ll n^{-1/4}$, while if $m>2n$, then we have $|\sqrt[4]{m}-\sqrt[4]{n}{|}^{-1}\ll m^{-1/4}$. When $m\asymp n$, the contribution is estimated, as in (20), by Lemma 8. In this way, we can easy to see that

$$\begin{array}{c}\hfill {\int }_T^{2T}{\left|\sum _{n\le T}{c}_n{n}^{-5/8}{e}^{8\pi i\sqrt[4]{nt}}\right|}^{2}dt\ll T^{5/4+\epsilon },\end{array}$$

(24)

and we obtain

$$J_2+J_3\ll T^{3/8}{T}^{1/2}{log}^{2}T·T^{5/8+\epsilon }\ll T^{3/2+\epsilon }.$$

Next, we consider $J_1\left(T\right)$.

$$J_1\left(T\right)=c{\int }_T^{2T}\sum _{m\le T}{T}^{5/8}{\displaystyle \frac{{(-1)}^{m}d\left(m\right)}{m^{3/4}}}e(t,m)cosf(t,m)\sum _{n\le T}{\displaystyle \frac{c_n}{n^{5/8}}}cos\left(8\pi \sqrt[4]{nt}-{\displaystyle \frac{\pi }{4}}\right)dt,$$

where c is a constant. We split the sums over m, n into $O\left({log}^{2}T\right)$ subsums with the ranges of summation $M<m\le M^{\prime }\le 2M$, $N<n\le N^{\prime }\le 2N$, respectively. We write the cosines as exponentials and then obtain $\ll {log}^{2}T$ sums of the form

$$\begin{array}{cc}\hfill T^{5/8}\sum _{M<m\le M^{\prime }}& {\displaystyle \frac{{(-1)}^{m}d\left(m\right)}{m^{3/4}}}\sum _{N<n\le N^{\prime }}{\displaystyle \frac{c_n}{n^{5/8}}}\hfill \\ \hfill & \times {\int }_T^{2T}e(t,m)exp\left(8\pi i\sqrt[4]{nt}-i\sqrt{8\pi mt}-i{a}_3{m}^{3/2}{t}^{-1/2}-\cdots \right)dt.\hfill \end{array}$$

(25)

There is also the expression with + in place of − in the exponential, and their conjugates, but it is (25) that is the relevant sum. The smooth function $e(t,m)$ ($=1+O(m/T)$) may be removed on applying integration by parts. Furthermore, when $N\ge 100M$, the contribution of (25) is

$$\begin{array}{c}\hfill \ll T^{5/8}·T^{3/4}\sum _{M<m\le 2M}d\left(m\right)m^{-3/4}\sum _{N<n\le 2N}{c}_n{n}^{-7/8}\ll T^{7/4+\epsilon },\end{array}$$

(26)

where we have used the first derivative test. Furthermore, the same bound as in (26) holds when $M\ge 100N$. These sums in total make a contribution which is $\ll T^{7/4+\epsilon }$.

Now we consider the case $N/100<M<100N$. The contribution is

$$\begin{array}{c}\hfill \ll T^{{\displaystyle \frac{5}{8}}}{\left\{{\int }_T^{2T}{\left|\sum _{N<n\le N^{\prime }}{\displaystyle \frac{c_n}{n^{5/8}}}{e}^{8\pi i\sqrt[4]{nt}}\right|}^{2}dt\times \sum _{M<m\le M^{\prime }}{\left|{\displaystyle \frac{{(-1)}^{m}d\left(m\right)}{m^{3/4}}}e(t,m)e^{if(t,m)}\right|}^{2}dt\right\}}^{{\displaystyle \frac{1}{2}}}\end{array}$$

(27)

by using the Cauchy–Schwarz inequality. Here, the first integral is estimated as in (24), more precisely by

$$O\left(T^{1+\epsilon }{N}^{-{\displaystyle \frac{1}{4}}}+T^{{\displaystyle \frac{3}{4}}+\epsilon }{N}^{{\displaystyle \frac{1}{2}}}\right).$$

From the first derivative test and Lemma 7, we can obtain that the second integral is

$$\ll T\sum _{m\ge M}{\displaystyle \frac{d^2\left(m\right)}{m^{3/2}}}+\sum _{M<k\ne n\le M^{\prime }}{\displaystyle \frac{d\left(k\right)d\left(m\right)e(t,k)e(t,m)}{{\left(km\right)}^{3/4}}}\underset{t\in [T,2T]}{max}{\displaystyle \frac{1}{|f^{\prime }(t,m)-f^{\prime }(t,k)|}}.$$

We have

$$f^{\prime }(t,\mathsf{\ell })={\displaystyle \frac{\partial f(t,\mathsf{\ell })}{\partial t}}=2\mathrm{ar}sinh\sqrt{{\displaystyle \frac{ß\mathsf{\ell }}{2\mathrm{t}}}},$$

therefore by the mean value theorem we can obtain

$$|f^{\prime }(t,m)-f^{\prime }(t,k)|\asymp {\displaystyle \frac{\sqrt{k}-\sqrt{m}}{\sqrt{T}}}(k\ne m,T\le t\le 2T).$$

Thus, the last expression above is

$$\ll T{M}^{-1/2}{log}^{3}T+T^{1/2}{log}^{4}T.$$

It is easily to see that the expression in (27) is

$$\begin{array}{cc}\hfill & \ll T^{5/8}{\left(\left(T^{1+\epsilon }{M}^{-1/4}+T^{3/4+\epsilon }{M}^{1/2}\right)\left(T{M}^{-1/2}{log}^{3}T+T^{1/2}{log}^{4}T\right)\right)}^{1/2}\hfill \\ \hfill & \ll T^{5/8}{\left(T^{2+\epsilon }{M}^{-3/4}+T^{7/4+\epsilon }+T^{5/4+\epsilon }{M}^{1/2}\right)}^{1/2},\hfill \end{array}$$

since $M\asymp N$.

Taking $M=T{2}^{-j}$ and summing over j we obtain that the contribution of $J_1\left(T\right)$ is $O\left(T^{7/4+\epsilon }\right)$, since $M\asymp N$ in the relevant cases. This gives

$${\int }_T^{2T}{\Delta }_1(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt\ll T^{2+\epsilon },$$

and thus we complete the proof of Theorem 1.

4. Proof of Theorem 2

Similarly to the proof of Theorem 1, it suffices to prove the result for the integral over $[T,2T]$, where $T\ge 10$ is large. Let

$$\begin{array}{c}\hfill T\le t\le 2T,T^{3/4}\ll y=y\left(T\right)\ll T,\end{array}$$

(28)

where y will be determined later. Let

$$\begin{array}{c}\hfill {\Delta }_1(t;\phi )={\Delta }_{11}(t,y;\phi )+{\Delta }_{12}(t,y;\phi ),\end{array}$$

(29)

where

$$\begin{array}{c}\hfill {\Delta }_{11}(t,y;\phi ):={\displaystyle \frac{t^{9/8}}{{\left(2\pi \right)}^2}}\sum _{n\le y}{\displaystyle \frac{c_n}{n^{7/8}}}cos\left(8\pi \sqrt[4]{nt}-{\displaystyle \frac{\pi }{4}}\right),\end{array}$$

(30)

and from Lemma 4 with $N=y$, we have

$$\begin{array}{c}\hfill {\Delta }_{12}(t,y;\phi ){\ll }_{\epsilon }{T}^{3/2+\epsilon }{y}^{-1/2}\left({\ll }_{\epsilon }{T}^{9/8+\epsilon }\right).\end{array}$$

(31)

Then

$$\begin{array}{cc}\hfill {\int }_T^{2T}{\Delta }_1^k(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt& ={\int }_T^{2T}{\left({\Delta }_{11}(t,y;\phi )+{\Delta }_{12}(t,y;\phi )\right)}^k{\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt\hfill \\ \hfill & ={\int }_1+O\left({\int }_2+{\int }_3\right),\hfill \end{array}$$

(32)

where

$$\begin{array}{cc}\hfill & {\int }_1:={\int }_T^{2T}{\Delta }_{11}^k(t,y;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt,\hfill \\ \hfill & {\int }_2:={\int }_T^{2T}\left|{\Delta }_{11}^{k-1}(t,y;\phi ){\Delta }_{12}(t,y;\phi )\right|{\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt,\hfill \\ \hfill & {\int }_3:={\int }_T^{2T}{\Delta }_{12}^k(t,y;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt.\hfill \end{array}$$

In order to estimate ${\int }_3$ we need (31) and

$$\begin{array}{c}\hfill {\int }_T^{2T}{\left|{\Delta }_{12}(t,y;\phi )\right|}^{2}dt{\ll }_{\epsilon }{T}^{13/4+\epsilon }{y}^{-3/4},\end{array}$$

(33)

which is (4.8) in [16]. By (31) and (33), the fourth power moment of $\zeta \left({\displaystyle \frac{1}{2}}+it\right)$ and the Cauchy–Schwarz inequality, we can obtain

$$\begin{array}{cc}\hfill {\int }_3& {\ll }_{\epsilon }{\left({\displaystyle \frac{T^{3/2+\epsilon }}{y^{1/2}}}\right)}^{k-1}{\int }_T^{2T}\left|{\Delta }_{12}(t,y;\phi )\right|{\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt\hfill \\ \hfill & {\ll }_{\epsilon }{\left({\displaystyle \frac{T^{3/2+\epsilon }}{y^{1/2}}}\right)}^{k-1}{\left({\int }_T^{2T}{\left|{\Delta }_{12}(t,y;\phi )\right|}^{2}dt\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_T^{2T}{\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{4}dt\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & {\ll }_{\epsilon }{T}^{3k/2+5/8+\epsilon }{y}^{1/8-k/2}.\hfill \end{array}$$

(34)

Now we consider ${\int }_1$. We write (2.9) in [10] as

$${\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^2=logt+C+E^{\prime }\left(t\right),$$

where, henceforth, we set $C=2\gamma -log2\pi$ for brevity. Thus, we have

$$\begin{array}{cc}\hfill {\int }_1=& {\int }_T^{2T}{\Delta }_{11}^k(t,y;\phi )(logt+C)dt+{\int }_T^{2T}{\Delta }_{11}^k(t,y;\phi )E^{\prime }\left(t\right)dt\hfill \\ \hfill =& {\int }_{11}+{\int }_{12}.\hfill \end{array}$$

(35)

We first consider ${\int }_{12}$. By using integration by parts and Lemma 1 we can obtain

$$\begin{array}{cc}\hfill {\int }_{12}& ={\Delta }_{11}^k(t,y;\phi )E\left(t\right){|}_T^{2T}-k{\int }_T^{2T}{\Delta }_{11}^{k-1}{\Delta }_{11}^{\prime }(t,y;\phi )(t,y;\phi )E\left(t\right)dt\hfill \\ \hfill & \ll T^{{\displaystyle \frac{6k}{5}}+{\displaystyle \frac{131}{416}}+\epsilon }+\left|{\int }_{12}^{*}\right|,\hfill \end{array}$$

(36)

where

$$\begin{array}{c}\hfill {\int }_{12}^{*}:={\int }_T^{2T}{\Delta }_{11}^{k-1}{\Delta }_{11}^{\prime }(t,y;\phi )(t,y;\phi )E\left(t\right)dt.\end{array}$$

In order to estimate ${\int }_{12}^{*}$, we need estimation of the second and the fourth moment of ${\Delta }_{11}^{\prime }(t,y;\phi )$. We can easily see that

$$\begin{array}{cc}\hfill {\Delta }_{11}^{\prime }(t,y;\phi )=& {\displaystyle \frac{9t^{1/8}}{2^5{\pi }^2}}\sum _{n\le y}{\displaystyle \frac{c_n}{n^{7/8}}}cos\left(8\pi \sqrt[4]{nt}-{\displaystyle \frac{\pi }{4}}\right)-{\displaystyle \frac{t^{3/8}}{2\pi }}sin\left(8\pi \sqrt[4]{nt}-{\displaystyle \frac{\pi }{4}}\right)\hfill \\ \hfill \ll & t^{-1}\left|{\Delta }_{11}(t,y;\phi )\right|+t^{3/8}\left|\sum _{n\le y}{\displaystyle \frac{c_n}{n^{7/8}}}e\left(4\sqrt[4]{nt}\right)\right|.\hfill \end{array}$$

(37)

From ${\Delta }_{12}(t,y;\phi )\ll T^{9/8+\epsilon }$ and (31), we have

$${\Delta }_{11}(t,y;\phi ){\ll }_{\epsilon }\left|{\Delta }_1(t;\phi )\right|+T^{9/8+\epsilon }.$$

Thus, by Lemma 2 we have, for any $0\le A\le {\displaystyle \frac{16}{3}}$,

$$\begin{array}{cc}\hfill {\int }_T^{2T}{\left|{\Delta }_{11}(t,y;\phi )\right|}^{A}dt& \ll {\int }_T^{2T}{\left|{\Delta }_1(t;\phi )\right|}^{A}dt+T^{1+9A/8+\epsilon }\hfill \\ \hfill & \ll T^{1+9A/8+\epsilon }.\hfill \end{array}$$

(38)

For the mean square of ${\Delta }_{11}^{\prime }(t,y;\phi )$, by (37) and (38) with $A=2$ and the first derivative test, we have

$$\begin{array}{cc}\hfill {\int }_T^{2T}{\left|{\Delta }_{11}^{\prime }(t,y;\phi )\right|}^{2}dt\ll & {\int }_T^{2T}{t}^{-2}{\left|{\Delta }_{11}(t,y;\phi )\right|}^{2}dt+{\int }_T^{2T}{t}^{3/4}{\left|\sum _{n\le y}{\displaystyle \frac{c_n}{n^{5/8}}}e\left(4{\left(nt\right)}^{{\displaystyle \frac{1}{4}}}\right)\right|}^{2}dt\hfill \\ \hfill \ll & T^{5/4+\epsilon }+T^{3/4}\sum _{m,n\le y}{\displaystyle \frac{c_m{c}_n}{{\left(mn\right)}^{5/8}}}\left|{\int }_T^{2T}e\left(4\left(\sqrt[4]{m}-\sqrt[4]{n}\right)\sqrt[4]{t}\right)dt\right|\hfill \\ \hfill \ll & T^{7/4}\sum _{n\le y}{\displaystyle \frac{c_n^2}{n^{5/4}}}+T^{3/4}\sum _{m\ne n\le y}{\displaystyle \frac{c_m{c}_n}{{\left(mn\right)}^{5/8}}}\left|{\int }_T^{2T}e\left(4\left(\sqrt[4]{m}-\sqrt[4]{n}\right)\sqrt[4]{t}\right)dt\right|\hfill \\ \hfill \ll & T^{7/4}\sum _{n\le y}{\displaystyle \frac{c_n^2}{n^{5/4}}}+T^{3/2}\sum _{m\ne n\le y}{\displaystyle \frac{c_m{c}_n}{{\left(mn\right)}^{5/8}\left|\sqrt[4]{m}-\sqrt[4]{n}\right|}}\hfill \\ \hfill \ll & T^{7/4+\epsilon }{y}^{-1/4}+T^{3/2+\epsilon }{y}^{1/2}.\hfill \end{array}$$

(39)

For the fourth moment of ${\Delta }_{11}^{\prime }(t,y;\phi )$, by (37), we have

$$\begin{array}{cc}\hfill {\int }_T^{2T}{\left|{\Delta }_{11}^{\prime }(t,y;\phi )\right|}^{4}dt\ll & {\int }_T^{2T}{t}^{-4}{\left|{\Delta }_{11}(t,y;\phi )\right|}^{4}dt+{\int }_T^{2T}{t}^{3/2}{\left|\sum _{n\le y}{\displaystyle \frac{c_n}{n^{5/8}}}e\left(4{\left(nt\right)}^{{\displaystyle \frac{1}{4}}}\right)\right|}^{4}dt\hfill \\ \hfill \ll & T^{3/2+\epsilon }+T^{3/2+\epsilon }{\int }_T^{2T}{\left|\sum _{n\sim N}{\displaystyle \frac{c_n}{n^{5/8}}}e\left(4{\left(nt\right)}^{{\displaystyle \frac{1}{4}}}\right)\right|}^{4}dt\hfill \end{array}$$

for some $1\ll N\ll y$. Thus, we have

$$\begin{array}{cc}\hfill {\int }_T^{2T}{\left|{\Delta }_{11}^{\prime }(t,y;\phi )\right|}^{4}dt\ll & T^{3/2+\epsilon }+T^{3/2+\epsilon }{\int }_T^{2T}{\left|\sum _{n\sim N}{\displaystyle \frac{c_n}{n^{5/8}}}e\left(4{\left(nt\right)}^{{\displaystyle \frac{1}{4}}}\right)\right|}^{4}dt\hfill \\ \hfill \ll & T^{3/2+\epsilon }+T^{3/2+\epsilon }\sum _{n_1,n_2,n_3,n_4\sim N}{\displaystyle \frac{c_{n_1}{c}_{n_2}{c}_{n_3}{c}_{n_4}}{{\left(n_1{n}_2{n}_3{n}_4\right)}^{5/8}}}\hfill \\ \hfill & \times \left|{\int }_T^{2T}e\left(2\left(\sqrt[4]{n_1}+\sqrt[4]{n_2}-\sqrt[4]{n_3}-\sqrt[4]{n_4}\right)\sqrt[4]{t}\right)dt\right|\hfill \\ \hfill \ll & T^{3/2+\epsilon }+{\displaystyle \frac{T^{3/2+\epsilon }}{N^{5/2}}}\sum _{n_1,n_2,n_3,n_4\sim N}min\left(T,{\displaystyle \frac{T^{3/4}}{\left|\mathrm{\Omega }\right|}}\right),\hfill \end{array}$$

(40)

where we used trivial estimation and the first derivative test, and we set

$$\mathrm{\Omega }:=\sqrt[4]{n_1}+\sqrt[4]{n_2}-\sqrt[4]{n_3}-\sqrt[4]{n_4}.$$

Note that $min\left(T,{\displaystyle \frac{T^{3/4}}{\left|\mathrm{\Omega }\right|}}\right)=T$ if $\left|\mathrm{\Omega }\right|\le T^{-1/4}$. By (17) in Lemma 6, we have the contribution to the last sum in (40), which is

$$\begin{array}{cc}\hfill \ll & {\displaystyle \frac{T^{3/2+\epsilon }}{N^{5/2}}}T\left(T^{-1/4}{N}^{15/4}+N^2\right)\hfill \\ \hfill \ll & \left(T^{9/4}{N}^{5/4}+T^{5/2}{N}^{-1/2}\right)T^{\epsilon }\hfill \\ \hfill \ll & T^{9/4+\epsilon }{y}^{5/4}\hfill \end{array}$$

on noting that $y\gg T^{3/4}$. If $\left|\mathrm{\Omega }\right|>T^{-1/4}$, then $min\left(T,{\displaystyle \frac{T^{3/4}}{\left|\mathrm{\Omega }\right|}}\right)={\displaystyle \frac{T^{3/4}}{\left|\mathrm{\Omega }\right|}}$. In this case the contribution to the last sum in (40) is, by Lemma 6 again,

$$\begin{array}{cc}\hfill \ll & \underset{T^{-1/4}<\eta \ll N^{1/4}}{max}{\displaystyle \frac{T^{9/4+\epsilon }}{N^{5/2}\eta }}\sum _{\left|\mathrm{\Omega }\right|\sim \eta }1\hfill \\ \hfill \ll & \underset{T^{-1/4}<\eta \ll N^{1/4}}{max}{\displaystyle \frac{T^{9/4+\epsilon }}{N^{5/2}\eta }}\left(\eta N^{15/4}+N^2\right)\hfill \\ \hfill \ll & \left(T^{9/4}{y}^{5/4}+T^{5/2}{y}^{-1/2}\right)T^{\epsilon }\hfill \\ \hfill \ll & T^{9/4+\epsilon }{y}^{5/4}.\hfill \end{array}$$

Inserting the above two estimates into (40), we obtain

$$\begin{array}{c}\hfill {\int }_T^{2T}{\left|{\Delta }_{11}^{\prime }(t,y;\phi )\right|}^{4}dt\ll T^{9/4+\epsilon }{y}^{5/4}.\end{array}$$

(41)

Now, we estimate ${\int }_{12}^{*}$. When $k=2,3,4$, by (38) and (39), Lemma 2 and Hölder’s inequality, we have

$$\begin{array}{cc}\hfill {\int }_{12}^{*}=& {\int }_T^{2T}{\Delta }_1^{k-1}(t,y;\phi ){\Delta }_1^{\prime }(t,y)E\left(t\right)dt\hfill \\ \hfill \ll & {\left({\int }_T^{2T}{\left|{\Delta }_{11}^{\prime }(t,y)\right|}^{2}dt\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_T^{2T}{\left|{\Delta }_{11}(t,y;\phi )\right|}^{2k}dt\right)}^{{\displaystyle \frac{k-1}{2k}}}{\left({\int }_T^{2T}{\left|E\left(t\right)\right|}^{2k}dt\right)}^{{\displaystyle \frac{1}{2k}}}\hfill \\ \hfill \ll & {\left(T^{3/2+\epsilon }{y}^{1/2}\right)}^{1/2}{\left(T^{(1+9k/8+\epsilon )·2k}\right)}^{(k-1)/2k}{\left(T^{1+k/2+\epsilon }\right)}^{1/2}\hfill \\ \hfill \ll & T^{3/8+9k/8+\epsilon }{y}^{1/4}.\hfill \end{array}$$

(42)

Inserting (42) into (36), then by Lemma 2 we have, for $k=2,3,4$,

$$\begin{array}{c}\hfill {\int }_{12}\ll T^{6k/5+13/416+\epsilon }+T^{9k/8+\epsilon }{y}^{1/4}.\end{array}$$

(43)

Now we consider ${\int }_{11}$. By using the formula ${\Delta }_{11}(t,y;\phi )={\Delta }_1(t;\phi )-{\Delta }_{12}(t,y;\phi )$, we can obtain

$$\begin{array}{cc}\hfill {\int }_{11}=& {\int }_T^{2T}{\Delta }_{11}^k(t,y;\phi )(logt+C)dt\hfill \\ \hfill =& {\int }_T^{2T}{\Delta }_1^k(t;\phi )(logt+C)dt\hfill \\ \hfill & +O\left({\int }_T^{2T}\left|{\Delta }_1^{k-1}\left(t\right){\Delta }_{12}(t,y;\phi )\right|(logt+C)dt+{\int }_T^{2T}\left|{\Delta }_{12}^k(t,y;\phi )\right|(logt+C)dt\right)\hfill \\ \hfill =& {\int }_4+O\left({\int }_5+{\int }_6\right).\hfill \end{array}$$

(44)

By (31) and (33), we have

$$\begin{array}{cc}\hfill {\int }_6\ll & {\left({\displaystyle \frac{T^{3/2+\epsilon }}{y^{1/2}}}\right)}^{k-2}{\int }_T^{2T}{\left|{\Delta }_{12}(t,y;\phi )\right|}^{2}dt\hfill \\ \hfill \ll & {\left({\displaystyle \frac{T^{3/2+\epsilon }}{y^{1/2}}}\right)}^{k-2}{T}^{13/4+\epsilon }{y}^{-1/4}\hfill \\ \hfill \ll & T^{3k/2+1/4+\epsilon }{y}^{-k/2+1/4}.\hfill \end{array}$$

When $k=2,3,4$, we have

$$\begin{array}{cc}\hfill {\int }_5\ll & logT{\left({\int }_T^{2T}{\left|{\Delta }_{12}(t,y;\phi )\right|}^{2}dt\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_T^{2T}{\left|{\Delta }_1\left(t\right)\right|}^{2k-2}dt\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill \ll & T^{1+9k/8+\epsilon }{y}^{-3/8},\hfill \end{array}$$

(45)

where Cauchy’s inequality, (33) and Lemma 2 are used. Combining the above estimates, we can obtain

$$\begin{array}{c}\hfill {\int }_5+{\int }_6\ll T^{1+9k/8+\epsilon }{y}^{-3/8}\end{array}$$

(46)

for $k=2,3,4$.

By (28), (43) and (46) we have

$$\begin{array}{c}\hfill {\int }_1={\int }_T^{2T}{\Delta }_1^k(t;\phi )(logt+C)dt+O\left(G_{k1}(T,y;\phi )T^{\epsilon }+G_{k2}(T,y;\phi )T^{\epsilon }\right),\end{array}$$

(47)

where

$$\begin{array}{c}\hfill G_{k1}(T,y;\phi ):=T^{1+9k/8+\epsilon }{y}^{-3/8}\mathrm{for}k=2,3,4\end{array}$$

(48)

and

$$\begin{array}{c}\hfill G_{k2}(T,y;\phi ):=T^{6k/5+131/416+\epsilon }+T^{3/8+9k/8+\epsilon }{y}^{1/4}\mathrm{for}k=2,3,4.\end{array}$$

(49)

Now we consider ${\int }_2$. Taking $k=2,4$ in the estimate (47)–(49), we have

$$\begin{array}{c}\hfill {\int }_T^{2T}{\left|{\Delta }_{11}(t,y;\phi )\right|}^k{\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt\ll T^{1+9k/8+\epsilon }(k=2,4);\end{array}$$

(50)

when $k=2,3,4$, from (34) with $k=2$, through (50) and the Cauchy’s inequality, we have

$$\begin{array}{cc}\hfill {\int }_2\ll & {\left({\int }_T^{2T}{\left|{\Delta }_{12}(t,y;\phi )\right|}^2{\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_T^{2T}{\left|{\Delta }_{11}(t,y;\phi )\right|}^{2k-2}{\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill \ll & {\left(T^{29/8+\epsilon }{y}^{-7/8}\right)}^{{\displaystyle \frac{1}{2}}}{\left(T^{1+9(2k-2)/8}\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill \ll & T^{19/16+9k/8+\epsilon }{y}^{-7/16}.\hfill \end{array}$$

Therefore, for $k=2,3,4$, we have

$$\begin{array}{c}\hfill {\int }_2\ll G_{k3}(T,y):=T^{25/16+9k/8+\epsilon }{y}^{-9/16}.\end{array}$$

(51)

By (32), (34), (47), and (51) we obtain

$$\begin{array}{cc}\hfill {\int }_T^{2T}{\Delta }_1^k(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt=& {\int }_T^{2T}{\Delta }_1^k(t;\phi )(logt+C)dt\hfill \\ \hfill & +O\left(\sum _{j=1}^3{G}_{kj}(T,y)T^{\epsilon }+T^{18k-1/8+\epsilon }{y}^{3-6k/8}\right),\hfill \end{array}$$

(52)

where $G_{kj}(T,y)(j=1,2,3)$ was defined in (48), (49) and (51), respectively. It is easy to see that

$$\begin{array}{cc}\hfill \sum _{j=1}^3{G}_{kj}(T,y)\ll & T^{1+9k/8+\epsilon }{y}^{-3/8}+T^{6k/5+131/416+\epsilon }\hfill \\ \hfill & +T^{3/8+9k/8+\epsilon }{y}^{1/4}+T^{19/16+9k/8+\epsilon }{y}^{-7/16}\hfill \end{array}$$

(53)

for $k=2,3,4$.

Now, taking $y=T$, we have

$$\begin{array}{c}\hfill \sum _{j=1}^3{G}_{kj}(T,y)+T^{3k/2+5/8}{y}^{1/8-k/2}\ll T^{1+9k/8-{\eta }_k^{*}},\end{array}$$

(54)

where ${\eta }_k^{*}={\displaystyle \frac{1}{4}}$. By (52)–(54) we have

$$\begin{array}{c}\hfill {\int }_T^{2T}{\Delta }_1^k(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt={\int }_T^{2T}{\Delta }_1^k(t;\phi )(logt+C)dt+O\left(T^{1+9k/8-{\eta }_k^{*}+\epsilon }\right),\end{array}$$

which implies that

$$\begin{array}{c}\hfill {\int }_1^T{\Delta }_1^k(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt={\int }_1^T{\Delta }_1^k(t;\phi )(logt+C)dt+O\left(T^{1+9k/8-{\eta }_k^{*}+\epsilon }\right).\end{array}$$

Thus, we have

$$\begin{array}{c}\hfill {\int }_1^T{\Delta }_1^k(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^{2}dt=c_1\left(k\right)T^{1+9k/8}logT+c_2\left(k\right)T^{1+9k/8}+O\left(T^{1+9k/8-{\eta }_k+\epsilon }\right),\end{array}$$

where

$$c_1\left(k\right)=C_k,c_2\left(k\right)=C_k\left(C-{\displaystyle \frac{8}{8+9k}}\right),{\eta }_k={\displaystyle \frac{1}{4}}(2\le k\le 4).$$

Therefore, we complete the proof of Theorem 2.

5. Conclusions

In this paper, we studied the asymptotic evaluation of the integrals of ${\Delta }_1^k(t;\phi ){\left|\zeta \left({\displaystyle \frac{1}{2}}+it\right)\right|}^2$ when $2\le k\le 4$ is fixed. The main method is large-value estimation. With the results of this paper, we can further understand the properties of the zeta-function and the error term in the classical Rankin–Selberg problem.