Average Size of Ramanujan Sum Associated with Divisor Function

Abstract

Let m and n be positive integers, and let $c(m,n)$ denote the Ramanujan sum. In this paper, we establish an asymptotic formula for the mean value of $c(m,n)$, which is associated with the divisor function, over both m and n.

1. Introduction

The Ramanujan sum function holds significant importance in number theory due to its numerous properties. For any positive integers m and n, the classical Ramanujan sum $c(m,n)$ is defined as

$$\begin{array}{c}\hfill c(m,n):=\sum _{\begin{array}{c}1\le j\le m\\ gcd(j,m)=1\end{array}}e\left({\displaystyle \frac{jn}{m}}\right)=\sum _{d|gcd(m,n)}d\mu \left({\displaystyle \frac{m}{d}}\right),\end{array}$$

(1)

where $e\left(z\right):=e^{2\pi iz}$, $\mu \left(x\right)$ is the Möbius function, and $gcd(m,n)$ denotes the greatest common factor of m and n.

Many authors have conducted a series of mean value studies on the arithmetic function $c(m,n)$. Suppose that $Y\ge X\ge 3$ are real numbers, $k\ge 1$ is a fixed integer, and $\zeta \left(s\right)$ is the Riemann zeta function. Define

$$\begin{array}{c}\hfill C_k(X,Y):=\sum _{1\le n\le Y}{\left(\sum _{1\le m\le X}c(m,n)\right)}^k.\end{array}$$

(2)

Chan and Kumchev [1] proved that

$$\begin{array}{c}\hfill C_1(X,Y)=Y-{\displaystyle \frac{1}{\zeta \left(2\right)}}{X}^2+O(X{Y}^{1/3}logX)+O\left(X^3{Y}^{-1}\right).\end{array}$$

For $k=2$, if $Y\ge X^2{(logX)}^B$, for some fixed $B>0$, it holds that

$$\begin{array}{c}\hfill C_2(X,Y)={\displaystyle \frac{Y{X}^2}{2\zeta \left(2\right)}}+O(X^4+XYlogX);\end{array}$$

if $X\le Y\le X^2{(logX)}^B$, then

$$\begin{array}{c}\hfill C_2(X,Y)={\displaystyle \frac{X^{2}Y}{2\zeta \left(2\right)}}\left(1+\kappa \left(u\right)\right)+O\left(X^{2}Y{(logX)}^{10}\left(X^{-1/2}\right)+{(Y/X)}^{-1/2}\right),\end{array}$$

where $u=log\left(X^{-2}Y\right)$ and

$$\kappa \left(u\right)={\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }{\displaystyle \frac{e^{-itu}\zeta (1-it)}{{(1+it)}^2(1-it)\zeta (1+it)}}\mathrm{d}t.$$

Hereafter, this question has been studied in number fields $\mathbb{F}$, where $\mathbb{F}$ is a finite-degree field extension of the rational number field $\mathbb{Q}$. Let $Y>X\ge 3$ be sufficiently large real numbers. Suppose that $k\ge 1$ is a fixed integer. Define

$$\begin{array}{c}\hfill C_{\mathbb{F},k}(X,Y):=\sum _{1\le N\left(\mathbf{n}\right)\le Y}{\left(\sum _{1\le N\left(\mathbf{m}\right)\le X}c(\mathbf{m},\mathbf{n})\right)}^k.\end{array}$$

W. G. Nowak [2] established the main term of $C_{\mathbb{F},1}(X,Y)$ for quadratic number fields under the condition $Y>X^{\delta }$, where $\delta >1973/820$. W. Zhai [3] obtained a more precise result by deriving an asymptotic formula for $C_{\mathbb{F},1}(X,Y)$ in the case of quadratic number fields. Furthermore, W. Zhai [4] provided the asymptotic formula of $C_{\mathbb{F},2}(X,Y)$ in quadratic number fields. The most recent development is due to C. Sneha and G. Shivani [5] establishing more precise results for $C_{\mathbb{F},2}(X,Y)$ on the quadratic and cubic fields, as well as results for general number fields. For related works involving Ramanujan sums in certain number fields, one may refer to [6,7].

“Sums of sums” is a highly interesting topic that has attracted the attention of numerous researchers. Various functions have been studied in this context, including generalized Ramanujan sums (see [8,9,10]), sums involving the von Mangoldt function (see [11]), sums related to the Liouville function (see [12]), and sums with weighting functions (see [13]).

As is well known, the divisor function plays a fundamental and important role in number theory. For $k\ge 2$, let $d_k\left(n\right)$ denote the number of ways that n can be expressed as a product of k factors. When $k=2$, $d_2\left(n\right)$ is the familiar Dirichlet divisor function. There have been a series of advancements in the study of $d_2\left(n\right)$. Using the hyperbola method, Dirichlet proved that

$$\begin{array}{c}\hfill \sum _{n\le x}{d}_2\left(n\right)=xlogx+(2\gamma -1)x+O(x^{\frac{1}{2}}),\end{array}$$

for $x\ge 1$, where $\gamma$ is Euler’s constant. We denote

$$\begin{array}{c}\hfill \mathrm{\Delta }\left(x\right)=\sum _{n\le x}{d}_2\left(n\right)-xlogx+(2\gamma -1)x.\end{array}$$

It follows that $\mathrm{\Delta }\left(x\right)=O(x^{\frac{1}{2}})$. The exponent ${\displaystyle \frac{1}{2}}$ has been improved by many authors (for example, [14,15,16,17]). Currently, the best result, due to Huxley [18], is

$$\begin{array}{c}\hfill \mathrm{\Delta }\left(x\right)\ll x^{{\displaystyle \frac{131}{416}}}{(logx)}^{{\displaystyle \frac{26497}{8320}}}.\end{array}$$

Furthermore, many other properties have been extensively studied, including the omega results ([19,20]), sign changes ([21]) and moments related to $\mathrm{\Delta }\left(x\right)$ ([22,23,24]). Results concerning $d_k\left(n\right)$ for $k\ge 3$ are presented in [25]. In this paper, we combine the Ramanujan sum with the divisor function. We rewrite (2) in the form of

$$\begin{array}{c}\hfill C_k(X,Y)=\sum _{1\le n\le Y}\prod _{j=1}^k\sum _{1\le m_j\le X}c(m_j,n).\end{array}$$

We multiply the variables $m_1$, $m_2,\dots ,m_k$ together, then we denote, for $k\ge 2$,

$$\begin{array}{c}\hfill D_k(X,Y):=\sum _{1\le n\le Y}\sum _{1\le m_1{m}_2\cdots m_k\le X}\prod _{j=1}^{k}c(m_j,n).\end{array}$$

For the case $k=2$, we obtain the following theorem.

Theorem 1.

Suppose $Y>X\ge 3$, and let ${\mathcal{L}}_1:=logX$ and ${\mathcal{L}}_2:=logY$. For $X{\mathcal{L}}_1^6{\mathcal{L}}_2^2<Y<X^{\frac{3}{2}}{\mathcal{L}}_1^{-1}$, we have

$$\begin{array}{cc}\hfill D_2(X,Y)=& {\displaystyle \frac{XY}{2\zeta \left(2\right)}}-{\displaystyle \frac{\zeta \left(0\right)X^2{\mathcal{L}}_1}{2\zeta \left(2\right)}}+{\displaystyle \frac{\zeta \left(0\right)X^2{\mathcal{L}}_2}{{\zeta }^2\left(2\right)}}+{\displaystyle \frac{X^2}{{\zeta }^2\left(2\right)}}\left({\zeta }^{\prime }\left(0\right)+\zeta \left(0\right)+{\displaystyle \frac{{\zeta }^{\prime }\left(2\right)\zeta \left(0\right)}{\zeta \left(2\right)}}\right)\hfill \\ \hfill & +O\left(X^{\frac{1}{2}}Y{\mathcal{L}}_1+X^{\frac{5}{2}}{Y}^{-\frac{1}{2}}{\mathcal{L}}_1^3{\mathcal{L}}_2\right).\hfill \end{array}$$

For $Y\ge X^{\frac{3}{2}}{\mathcal{L}}_1^{-1}$, we have

$$\begin{array}{c}\hfill D_2(X,Y)={\displaystyle \frac{XY}{2\zeta \left(2\right)}}+{\displaystyle \frac{\zeta \left(0\right)X^2{\mathcal{L}}_1}{2\zeta \left(2\right)}}-{\displaystyle \frac{\zeta \left(0\right)\left(1+\zeta \left(2\right)\right)X^2{\mathcal{L}}_2}{2{\zeta }^2\left(2\right)}}+O(X^{\frac{1}{2}}Y{\mathcal{L}}_1).\end{array}$$

Notations. We use $\mathbb{N}$, $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ to denote the sets of positive integers, rational numbers, real numbers, and complex numbers, respectively. We use expressions $f=O\left(g\right)$ or $f\ll g$ to mean $\left|f\right|\le Cg$ for some constant $C>0$. We write $f\asymp g$ to indicate that $f\ll g$ and $g\ll f$. The greatest common divisor of m and n is denoted by $(m,n)$ and their least common multiple is denoted by $[m,n]$. Let $s=\sigma +it\in \mathbb{C}$, where $\sigma ,t\in \mathbb{R}$. Throughout this paper, $\epsilon$ denotes a sufficiently small real positive number, not necessarily the same at each occurrence.

2. Analytic Continuation

The following lemmas form the foundation for the proof of Theorem 1.

Lemma 1.

Let $k\ge 2$ be a fixed integer. Suppose that $f(n_1,\dots ,n_k)$ is a multivariable arithmetic function whose Dirichlet series is given by

$$F(s_1,\dots ,s_k)=\sum _{n_1=1}^{\infty }\cdots \sum _{n_k=1}^{\infty }{\displaystyle \frac{f(n_1,\dots ,n_k)}{n_1^{s_1}\cdots n_k^{s_k}}}$$

which is absolutely convergent for ${\sigma }_j>{\mu }_j>0$ $(j=1,\dots ,k)$. Assume that $x_j$, $T_j\ge 5$ $(j=1,\dots ,k)$ are parameters satisfying $x_j\in \mathbb{N}+\frac{1}{2}$ $(j=1,\dots ,k)$ and define

$$b_j={\mu }_j+{\displaystyle \frac{1}{log{x}_j}},(j=1,\dots ,k).$$

Then, we have

$$\begin{array}{cc}\hfill & \sum _{n_1\le x_1}\dots \sum _{n_k\le x_k}f(n_1,\cdots ,n_k)\hfill \\ \hfill & ={\displaystyle \frac{1}{{\left(2\pi i\right)}^k}}{\int }_{b_1-i{T}_1}^{b_1+i{T}_1}\cdots {\int }_{b_k-i{T}_k}^{b_k+i{T}_k}{\displaystyle \frac{F(s_1,\dots ,s_k)x_1^{s_1}\cdots x_k^{s_k}}{s_1\cdots s_k}}\mathrm{d}{s}_k\cdots \mathrm{d}{s}_1+O(x_1^{b_1}\cdots x_k^{b_k}E),\hfill \end{array}$$

where

$$E:=\sum _{j=1}^k\sum _{n_1=1}^{\infty }\cdots \sum _{n_k=1}^{\infty }{\displaystyle \frac{\left|f\right(n_1,\dots ,n_k\left)\right|}{n_1^{b_1}\cdots n_k^{b_k}}}\times {\displaystyle \frac{1}{T_j|log\frac{x_j}{n_j}|+1}}.$$

Proof. 

See Lemma 2.6 in [4]. □

Taking

$$\begin{array}{c}\hfill f(n,m_1,m_2,\dots ,m_k)=\underset{1\le m_1{m}_2\cdots m_k\le X}{\prod _{i=1}^k}c(m_i,n),\end{array}$$

its Dirichlet series is given by

$$\begin{array}{c}\hfill {\mathcal{F}}_k(\omega ,s)=\sum _{n=1}^{\infty }\sum _{m_1=1}^{\infty }\cdots \sum _{m_k=1}^{\infty }{\displaystyle \frac{f(n,m_1,m_2,\dots ,m_k)}{n^{\omega }{(m_1{m}_2\cdots m_k)}^s}},\end{array}$$

where s, $\omega \in \mathbb{C}$.

For any $s\in \mathbb{C}$, we define

$${\sigma }_s\left(n\right)=\sum _{d|n}{d}^s.$$

For $\Re s>1$, we derive

$$\begin{array}{c}\hfill \sum _{m=1}^{\infty }{\displaystyle \frac{c(m,n)}{m^s}}=\sum _{m=1}^{\infty }{\displaystyle \frac{1}{m^s}}\sum _{d|m,d|n}d\mu \left({\displaystyle \frac{m}{d}}\right)=\sum _{d|n}d\sum _{t=1}^{\infty }{\displaystyle \frac{1}{{\left(dt\right)}^s}}\mu \left(t\right)={\displaystyle \frac{{\sigma }_{1-s}\left(n\right)}{\zeta \left(s\right)}},\end{array}$$

(3)

which implies

$$\begin{array}{c}\hfill {\mathcal{F}}_k(\omega ,s)={\displaystyle \frac{1}{{\zeta }^k\left(s\right)}}\sum _{n=1}^{\infty }{\sigma }_{1-s}^k\left(n\right)n^{-\omega }.\end{array}$$

(4)

Proposition 1.

Suppose s, $\omega \in \mathbb{C}$. If $\Re \omega >max\{1,2-2\Re s,3-2\Re s\}$ and $\Re s>1$, then

$$\begin{array}{c}\hfill {\mathcal{F}}_2(\omega ,s)={\displaystyle \frac{\zeta \left(\omega \right){\zeta }^2(\omega +s-1)\zeta (\omega +2s-2)}{{\zeta }^2\left(s\right)\zeta (2\omega +2s-2)}}.\end{array}$$

Proof. 

Let $s_1,s_2\in \mathbb{C}$, such that $\Re s_1>1$, $\Re s_2>1$. Also, assume that $\Re \omega >max\{1,2-s_1,2-s_2,3-s_1-s_2\}$. Then, we define the function ${\mathcal{F}}_2(\omega ,s_1,s_2)$ as follows

$$\begin{array}{c}\hfill {\mathcal{F}}_2(\omega ,s_1,s_2)={\displaystyle \frac{1}{\zeta \left(s_1\right)\zeta \left(s_2\right)}}\sum _{n=1}^{\infty }{\sigma }_{1-s_1}\left(n\right){\sigma }_{1-s_2}\left(n\right)n^{-\omega }.\end{array}$$

By applying Ramanujan’s identity (refer to [26] for details), we can transform the series on the right-hand side. Specifically, we obtain

$$\begin{array}{c}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{{\sigma }_{1-s_1}\left(n\right){\sigma }_{1-s_2}\left(n\right)}{n^{\omega }}}={\displaystyle \frac{\zeta \left(\omega \right)\zeta (\omega +s_1-1)\zeta (\omega +s_2-1)\zeta (\omega +s_1+s_2-2)}{\zeta (2\omega +s_1+s_2-2)}}.\end{array}$$

Finally, when we set $s_1=s_2=s$, we arrive at the result that concludes the proof. □

We also require some upper bound estimations of the Riemann zeta function.

Lemma 2.

For σ, $t\in \mathbb{R}$, the following estimates hold:

If $\sigma \le 0$,

$$\begin{array}{c}\hfill \zeta (\sigma +it)\ll {\left(\right|t|+2)}^{\frac{1}{2}-\sigma }log\left(\right|t|+2).\end{array}$$

(5)

If $0\le \sigma \le \frac{1}{2}$,

$$\begin{array}{c}\hfill \zeta (\sigma +it)\ll {\left(\right|t|+2)}^{\frac{1}{2}-{\displaystyle \frac{2}{3}}\sigma }log\left(\right|t|+2).\end{array}$$

(6)

If $\frac{1}{2}\le \sigma \le 1$,

$$\begin{array}{c}\hfill \zeta (\sigma +it)\ll {\left(\right|t|+2)}^{{\displaystyle \frac{1-\sigma }{3}}}log\left(\right|t|+2).\end{array}$$

(7)

If $\sigma \ge 1$,

$$\begin{array}{c}\hfill \zeta (\sigma +it)\ll min\left\{{\displaystyle \frac{1}{\sigma -1}},log\left(\right|t|+2)\right\},\end{array}$$

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\zeta (\sigma +it)}}\ll min\left\{{\displaystyle \frac{1}{\sigma -1}},log\left(\right|t|+2)\right\},\end{array}$$

(8)

$$\begin{array}{c}\hfill {\zeta }^{\left(k\right)}(\sigma +it)\ll min\left\{{\displaystyle \frac{\sigma }{{(1-\sigma )}^k}},{log}^{k+1}\left(\right|t|+2)\right\}.\end{array}$$

(9)

Proof. 

The detailed proofs can be found in Chapter 3 of the work by E. C. Titchmarsh [26] and Chapter 7 of the work by Pan and Pan [27]. □

Lemma 3.

Suppose $V>10$ is a large parameter and $|t-1/2|\le {log}^{-1}V$. Then,

$$\begin{array}{c}\hfill {\int }_{-V}^V{\left|\zeta (\sigma +it)\right|}^4\mathrm{d}t\ll V{log}^{4}V,\end{array}$$

(10)

$$\begin{array}{c}\hfill {\int }_{-V}^V{\left|\zeta (\sigma +it)\right|}^2\mathrm{d}t\ll VlogV.\end{array}$$

(11)

Proof. 

See Chapter 25 of Pan and Pan [27]. □

Lemma 4.

Suppose that T and U are two real large numbers, where U is a half integer and $g\left(n\right)$ is an arithmetic function satisfying $g\left(n\right)\ll n^{\epsilon }$ for any $\epsilon >0$. Define

$$G\left(s\right):=\sum _{n=1}^{\infty }{\displaystyle \frac{g\left(n\right)}{n^s}},s\in \mathbb{C},$$

which is absolutely convergent for $\Re s>1$. For $1<\sigma <2$, define

$$\mathcal{E}(U,T;\sigma )=\sum _{n=1}^{\infty }{\displaystyle \frac{g\left(n\right)}{n^{\sigma }}}\times {\displaystyle \frac{1}{T|log\frac{U}{n}|+1}}.$$

Then, we have

$$\begin{array}{c}\hfill \mathcal{E}(U,T;\sigma )\ll {\displaystyle \frac{G\left(\sigma \right)U^{\epsilon }}{T}}.\end{array}$$

(12)

Proof. 

We express the function $\mathcal{E}(U,T;\sigma )$ as the sum of three components:

$$\mathcal{E}(U,T;\sigma )={\mathcal{E}}_1(U,T;\sigma )+{\mathcal{E}}_2(U,T;\sigma )+{\mathcal{E}}_3(U,T;\sigma ),$$

where

$$\begin{array}{cc}\hfill & {\mathcal{E}}_1(U,T;\sigma )=\sum _{n\le U/2}{\displaystyle \frac{g\left(n\right)}{n^{\sigma }}}\times {\displaystyle \frac{1}{T|log\frac{U}{n}|+1}},\hfill \\ \hfill & {\mathcal{E}}_2(U,T;\sigma )=\sum _{U/2<n\le 2U}{\displaystyle \frac{g\left(n\right)}{n^{\sigma }}}\times {\displaystyle \frac{1}{T|log\frac{U}{n}|+1}},\hfill \\ \hfill & {\mathcal{E}}_3(U,T;\sigma )=\sum _{n>2U}{\displaystyle \frac{g\left(n\right)}{n^{\sigma }}}\times {\displaystyle \frac{1}{T|log\frac{U}{n}|+1}}.\hfill \end{array}$$

Trivially, we can obtain the following inequality

$$\begin{array}{c}\hfill {\mathcal{E}}_1(U,T;\sigma )+{\mathcal{E}}_3(U,T;\sigma )\ll {\displaystyle \frac{G\left(\sigma \right)}{T}}.\end{array}$$

(13)

For the component ${\mathcal{E}}_2(U,T;\sigma )$, we first note that

$$\begin{array}{c}\hfill {\mathcal{E}}_2(U,T;\sigma )\ll U^{\epsilon -\sigma }\sum _{U/2<n\le 2U}{\displaystyle \frac{1}{T\left|log\frac{U}{n}\right|+1}}\ll {\displaystyle \frac{U^{\epsilon }}{T}},\end{array}$$

(14)

By applying a well-known estimate

$$\begin{array}{c}\hfill \sum _{U/2<n\le 2U}{\displaystyle \frac{1}{T|log\frac{U}{n}|+1}}\ll {\displaystyle \frac{UlogU}{T}},\end{array}$$

we further simplify the inequality for

$$\begin{array}{c}\hfill {\mathcal{E}}_2(U,T;\sigma )\ll {\displaystyle \frac{U^{\epsilon }}{T}}.\end{array}$$

(15)

Combining the results of (13)–(15), we derive the inequality (12). □

3. Proof of Theorem 1

Suppose X and Y are half integers. Recall that

$$\begin{array}{c}\hfill {\mathcal{F}}_2(\omega ,s)={\displaystyle \frac{\zeta \left(\omega \right){\zeta }^2(\omega +s-1)\zeta (\omega +2s-2)}{{\zeta }^2\left(s\right)\zeta (2\omega +2s-2)}}.\end{array}$$

Let $s={\sigma }_0+i{t}_0$, $\omega ={\sigma }_1+i{t}_1$, and $b_0=b_1=b=1+{\displaystyle \frac{1}{logY}}$. For $Y>X\ge 3$, assume $T_0=X^2$, $T_1=4Y^2$. Applying Lemma 1, we obtain

$$D_2(X,Y)=M+O\left(XYE\right),$$

where

$$\begin{array}{c}\hfill E=\sum _{n=1}^{\infty }\sum _{m_1=1}^{\infty }\sum _{m_2=1}^{\infty }{\displaystyle \frac{c(m_1,n)c(m_2,n)}{n^b{\left(m_1{m}_2\right)}^b}}\left({\displaystyle \frac{1}{X^2\left|log\frac{X}{m_1{m}_2}\right|+1}}+{\displaystyle \frac{1}{4Y^2\left|log\frac{Y}{n}\right|+1}}\right)\end{array}$$

and

$$M={\displaystyle \frac{1}{{\left(2\pi i\right)}^2}}{\int }_{b-i{T}_0}^{b+i{T}_0}{\int }_{b-i{T}_1}^{b+i{T}_1}g(\omega ,s)\mathrm{d}\omega \mathrm{d}s$$

with

$$\begin{array}{c}\hfill g(\omega ,s):={\displaystyle \frac{\zeta \left(\omega \right){\zeta }^2(\omega +s-1)\zeta (\omega +2s-2)X^s{Y}^{\omega }}{{\zeta }^2\left(s\right)\zeta (2\omega +2s-2)\omega s}}.\end{array}$$

3.1. Estimation of E

First, we handle E. Let

$$\begin{array}{c}\hfill E=E_1+E_2,\end{array}$$

where

$$\begin{array}{c}\hfill E_1=\sum _{n=1}^{\infty }\sum _{m_1=1}^{\infty }\sum _{m_2=1}^{\infty }{\displaystyle \frac{c(m_1,n)c(m_2,n)}{n^b{\left(m_1{m}_2\right)}^b}}{\displaystyle \frac{1}{X^2\left|log\frac{X}{m_1{m}_2}\right|+1}}\end{array}$$

and

$$\begin{array}{c}\hfill E_2=\sum _{n=1}^{\infty }\sum _{m_1=1}^{\infty }\sum _{m_2=1}^{\infty }{\displaystyle \frac{c(m_1,n)c(m_2,n)}{n^b{\left(m_1{m}_2\right)}^b}}{\displaystyle \frac{1}{4Y^2\left|log\frac{Y}{n}\right|+1}}.\end{array}$$

By writing $m_1=d_1{k}_1$ and $m_2=d_2{k}_2$, we obtain

$$\begin{array}{c}\hfill E_1=\sum _{n=1}^{\infty }\sum _{d_1,d_2,k_1,k_2=1}^{\infty }{\displaystyle \frac{{\sum }_{d_1|n}{d}_1\mu \left(k_1\right){\sum }_{d_2|n}{d}_2\mu \left(k_2\right)}{n^b{\left(d_1{k}_1{d}_2{k}_2\right)}^b}}{\displaystyle \frac{1}{X^2\left|log\frac{X}{d_1{k}_1{d}_2{k}_2}\right|+1}}.\end{array}$$

This can be rewritten as

$$\begin{array}{c}\hfill E_1=\sum _{d_1,d_2,k_1,k_2=1}^{\infty }{\displaystyle \frac{\mu \left(k_1\right)\mu \left(k_2\right)d_1{d}_2}{{\left(d_1{k}_1{d}_2{k}_2\right)}^b}}\left(\sum _{\stackrel{n=1}{d_1|n,d_2|n}}^{\infty }{\displaystyle \frac{1}{n^b}}\right){\displaystyle \frac{1}{X^2\left|log\frac{X}{d_1{k}_1{d}_2{k}_2}\right|+1}}.\end{array}$$

Since

$$\begin{array}{c}\hfill \sum _{\stackrel{n=1}{d_1|n,d_2|n}}^{\infty }{\displaystyle \frac{1}{n^b}}=\sum _{k=1}^{\infty }{\displaystyle \frac{1}{{\left(k[d_1,d_2]\right)}^b}}\end{array}$$

and we know the well-known relation

$$d_1{d}_2=[d_1,d_2](d_1,d_2),$$

we can conclude the following inequality:

$$\begin{array}{c}\hfill E_1\le \zeta \left(b\right)\sum _{d_1,d_2,k_1,k_2=1}^{\infty }{\displaystyle \frac{|\mu \left(k_1\right)\mu \left(k_2\right)|(d_1,d_2)}{{\left(d_1{k}_1{d}_2{k}_2\right)}^b}}{\displaystyle \frac{1}{X^2\left|log\frac{X}{d_1{k}_1{d}_2{k}_2}\right|+1}}.\end{array}$$

Now, we make a substitution. Let $d=(d_1,d_2)$, and write $d_1=l_{1}d$, $d_2=l_{2}d$, where $(l_1,l_2)=1$. Therefore, we have

$$\begin{array}{cc}\hfill E_1& \le \zeta \left(b\right)\sum _{\begin{array}{c}d,l_1,l_2,k_1,k_2=1\\ (l_1,l_2)=1\end{array}}^{\infty }{\displaystyle \frac{1}{{\left(k_1{k}_2{l}_1{l}_2\right)}^b{d}^{2b-1}}}{\displaystyle \frac{1}{X^2\left|log\frac{X}{k_1{k}_2{l}_1{l}_2{d}^2}\right|+1}}\hfill \\ \hfill & \le \zeta \left(b\right)\sum _{d,l_1,l_2,k_1,k_2=1}^{\infty }{\displaystyle \frac{1}{{\left(k_1{k}_2{l}_1{l}_2\right)}^b{d}^b}}{\displaystyle \frac{1}{X^2\left|log\frac{X}{k_1{k}_2{l}_1{l}_2{d}^2}\right|+1}}\hfill \\ \hfill & \le \zeta \left(b\right)\sum _{d,l_1,l_2,k_1,k_2=1}^{\infty }{\displaystyle \frac{1}{{\left(k_1{k}_2{l}_1{l}_2\right)}^b{d}^b}}{\displaystyle \frac{1}{X^2}}+{\mathcal{E}}_1\hfill \\ \hfill & =X^{-2}{\mathcal{L}}_2^6+{\mathcal{E}}_1,\hfill \end{array}$$

(16)

where

$$\begin{array}{c}\hfill {\mathcal{E}}_1=\zeta \left(b\right)\sum _{X/2\le k_1{k}_2{l}_1{l}_2{d}^2\le X}{\displaystyle \frac{1}{{\left(k_1{k}_2{l}_1{l}_2\right)}^b{d}^b}}{\displaystyle \frac{1}{X^2\left|log\frac{X}{k_1{k}_2{l}_1{l}_2{d}^2}\right|+1}}.\end{array}$$

Next, we turn our attention to handling the term ${\mathcal{E}}_1$.

$$\begin{array}{cc}\hfill {\mathcal{E}}_1& =\zeta \left(b\right)\sum _{X/2\le k_1{k}_2{l}_1{l}_2{d}^2\le X}{\displaystyle \frac{d^b}{{\left(k_1{k}_2{l}_1{l}_2\right)}^b{d}^{2b}}}{\displaystyle \frac{1}{X^2\left|log\frac{X}{k_1{k}_2{l}_1{l}_2{d}^2}\right|+1}}\hfill \\ \hfill & \ll X^{\frac{1}{2}}\zeta \left(b\right)\sum _{X/2\le k_1{k}_2{l}_1{l}_2{d}^2\le X}{\displaystyle \frac{1}{{\left(k_1{k}_2{l}_1{l}_2\right)}^b{d}^{2b}}}{\displaystyle \frac{1}{X^2\left|log\frac{X}{k_1{k}_2{l}_1{l}_2{d}^2}\right|+1}}.\hfill \end{array}$$

Applying Lemma 4, we have

$$\begin{array}{c}\hfill {\mathcal{E}}_1\ll X^{-\frac{3}{2}+\epsilon }{\mathcal{L}}_2^6.\end{array}$$

(17)

Further from (16) and (17), we conclude that

$$\begin{array}{c}\hfill E_1\ll X^{-\frac{3}{2}+\epsilon }{\mathcal{L}}_2^6.\end{array}$$

(18)

Now, we deal with $E_2$. Recalling (3), from the estimate (8), we have

$$\begin{array}{c}\hfill \sum _{m_1=1}^{\infty }\sum _{m_2=1}^{\infty }{\displaystyle \frac{c(m_1,n)c(m_2,n)}{{\left(m_1{m}_2\right)}^b}}\ll {\displaystyle \frac{{\sigma }_{1-b}^2\left(n\right)}{{\zeta }^2\left(b\right)}}\ll d_2^2\left(n\right){\mathcal{L}}_2^2.\end{array}$$

Then, from the estimate (9), we obtain

$$\begin{array}{c}\hfill E_2\ll {\mathcal{L}}_2^2\sum _{n=1}^{\infty }{\displaystyle \frac{d_2^2\left(n\right)}{n^b}}{\displaystyle \frac{1}{Y^2\left|log\frac{Y}{n}\right|+1}}.\end{array}$$

Applying Lemma 4, we conclude that

$$\begin{array}{c}\hfill E_2\ll Y^{\epsilon -2}.\end{array}$$

(19)

From (18) and (19), finally, we obtain

$$\begin{array}{c}\hfill O\left(XYE\right)=O(X^{\epsilon -\frac{1}{2}}Y{\mathcal{L}}_2^6).\end{array}$$

(20)

3.2. Treatment for M

In the following, we deal with M. Consider a rectangular domain with vertices at the points $\omega =b\pm i4Y^2$ and $\omega ={\displaystyle \frac{1}{2}}\pm i4Y^2$. Within this domain, the integrand $g(\omega ,s)$ has a second-order pole $\omega =2-s$ and two simple poles $\omega =1$ and $\omega =3-2s$. By applying the residue theorem, we can deduce that

$$\begin{array}{cc}\hfill M=& {\displaystyle \frac{1}{2\pi i}}{\int }_{b-i{X}^2}^{b+i{X}^2}\left(R_1+R_2+R_3\right)\mathrm{d}s\hfill \\ \hfill & +{\displaystyle \frac{1}{{\left(2\pi i\right)}^2}}{\int }_{b-i{X}^2}^{b+i{X}^2}\left({\int }_{\frac{1}{2}-i4Y^2}^{\frac{1}{2}+i4Y^2}+{\int }_{\frac{1}{2}+i4Y^2}^{b+i4Y^2}-{\int }_{\frac{1}{2}-i4Y^2}^{b-i4Y^2}\right)g(\omega ,s)\mathrm{d}\omega \mathrm{d}s\hfill \\ \hfill =& {\displaystyle \frac{1}{2\pi i}}{\int }_{b-i{X}^2}^{b+i{X}^2}\left(R_1+R_2+R_3\right)\mathrm{d}s+E_{01}+E_{02}-E_{03},\hfill \end{array}$$

(21)

where

$$\begin{array}{cc}\hfill & R_1:=\underset{\omega =2-s}{Res}g(\omega ,s)={\displaystyle \frac{\zeta (2-s)X^s{Y}^{2-s}}{s(2-s)\zeta \left(2\right)\zeta \left(s\right)}}\left(\frac{1}{2}+{\mathcal{L}}_2+{\displaystyle \frac{1}{2-s}}\right),\hfill \\ \hfill & R_2:=\underset{\omega =1}{Res}g(\omega ,s)={\displaystyle \frac{\zeta (2s-1)X^{s}Y}{s\zeta \left(2s\right)}},\hfill \\ \hfill & R_3:=\underset{\omega =3-2s}{Res}g(\omega ,s)={\displaystyle \frac{\zeta (3-2s){\zeta }^2(2-s)X^s{Y}^{3-2s}}{s(3-2s){\zeta }^2\left(s\right)\zeta (4-2s)}},\hfill \end{array}$$

and

$$\begin{array}{c}\hfill E_{01}:={\displaystyle \frac{1}{{\left(2\pi i\right)}^2}}{\int }_{b-i{X}^2}^{b+i{X}^2}{\int }_{\frac{1}{2}-i4Y^2}^{\frac{1}{2}+i4Y^2}g(\omega ,s)\mathrm{d}\omega \mathrm{d}s,\end{array}$$

$$\begin{array}{c}\hfill E_{02}:={\displaystyle \frac{1}{{\left(2\pi i\right)}^2}}{\int }_{b-i{X}^2}^{b+i{X}^2}{\int }_{\frac{1}{2}+i4Y^2}^{b+i4Y^2}g(\omega ,s)\mathrm{d}\omega \mathrm{d}s,\end{array}$$

$$\begin{array}{c}\hfill E_{03}:={\displaystyle \frac{1}{{\left(2\pi i\right)}^2}}{\int }_{b-i{X}^2}^{b+i{X}^2}{\int }_{\frac{1}{2}-i4Y^2}^{b-i4Y^2}g(\omega ,s)\mathrm{d}\omega \mathrm{d}s.\end{array}$$

We start by estimating the quantity $E_{01}$ to analyze. For $|t_0|\le X^2$, $|t_1|\le 4Y^2$, according to Lemma 2, we have

$$\begin{array}{cc}\hfill & g\left(\frac{1}{2}+i{t}_1,b+i{t}_0\right)\hfill \\ \hfill & \ll X{Y}^{\frac{1}{2}}{\displaystyle \frac{\left|\zeta \left(\frac{1}{2}+i{t}_1\right)\right|{\left|\zeta \left(\frac{1}{2}+\frac{1}{logY}+i{t}_1+i{t}_0\right)\right|}^2\left|\zeta \left(\frac{1}{2}+\frac{2}{logY}+i{t}_1+i2t_0\right)\right|}{{\left|\zeta \left(1+\frac{1}{logY}+i{t}_0\right)\right|}^2\left|\zeta \left(1+\frac{2}{logY}+i{t}_1+i2t_0\right)\right|\left(\right|t_1|+1)\left(\right|t_0|+1)}}\hfill \\ \hfill & \ll X{Y}^{\frac{1}{2}}{\mathcal{L}}_2^3\times \hfill \\ \hfill & {\displaystyle \frac{\left|\zeta \left(\frac{1}{2}+i{t}_1\right)\right|{\left|\zeta \left(\frac{1}{2}+\frac{1}{logY}+i{t}_1+i{t}_0\right)\right|}^2\left|\zeta \left(\frac{1}{2}+\frac{2}{logY}+i{t}_1+i2t_0\right)\right|}{\left(\right|t_1|+1)\left(\right|t_0|+1)}}.\hfill \end{array}$$

Applying Lemmas 2 and 3, we can deduce

$$\begin{array}{cc}\hfill E_{01}& \ll X{Y}^{\frac{1}{2}}{\mathcal{L}}_2^3\hfill \\ \hfill & {\int }_{-X^2}^{X^2}{\int }_{-4Y^2}^{4Y^2}{\displaystyle \frac{\left|\zeta \left(\frac{1}{2}+i{t}_1\right)\right|{\left|\zeta \left(\frac{1}{2}+\frac{1}{logY}+i{t}_1+i{t}_0\right)\right|}^2\left|\zeta \left(\frac{1}{2}+\frac{2}{logY}+i{t}_1+i2t_0\right)\right|}{\left(\right|t_1|+1)\left(\right|t_0|+1)}}\mathrm{d}{t}_1\mathrm{d}{t}_0\hfill \\ \hfill & \ll X{Y}^{\frac{1}{2}}{\mathcal{L}}_2^3\left({\int }_1+{\int }_2\right),\hfill \end{array}$$

(22)

where

$${\int }_1=\underset{|t_1|\le |t_0|}{{\int }_{-X^2}^{X^2}{\int }_{-4Y^2}^{4Y^2}}{\displaystyle \frac{\left|\zeta \left(\frac{1}{2}+i{t}_1\right)\right|{\left|\zeta \left(\frac{1}{2}+\frac{1}{logY}+i{t}_1+i{t}_0\right)\right|}^2\left|\zeta \left(\frac{1}{2}+\frac{2}{logY}+i{t}_1+i2t_0\right)\right|}{\left(\right|t_1|+1)\left(\right|t_0|+1)}}\mathrm{d}{t}_1\mathrm{d}{t}_0$$

and

$${\int }_2=\underset{|t_0|\le |t_1|}{{\int }_{-X^2}^{X^2}{\int }_{-4Y^2}^{4Y^2}}{\displaystyle \frac{\left|\zeta \left(\frac{1}{2}+i{t}_1\right)\right|{\left|\zeta \left(\frac{1}{2}+\frac{1}{logY}+i{t}_1+i{t}_0\right)\right|}^2\left|\zeta \left(\frac{1}{2}+\frac{2}{logY}+i{t}_1+i2t_0\right)\right|}{\left(\right|t_1|+1)\left(\right|t_0|+1)}}\mathrm{d}{t}_1\mathrm{d}{t}_0.$$

Now, we estimate ${\int }_1$. Let $L_1\left(V\right):={\int }_0^V{\left|\zeta (\sigma -it)\right|}^4\mathrm{d}t$ for $\left|\sigma -\frac{1}{2}\right|\le {\displaystyle \frac{1}{logV}}$. By Lemma 3 and partial summation, we obtain

$$\begin{array}{cc}\hfill {\int }_{-T}^T{\displaystyle \frac{{\left|\zeta (\sigma +it)\right|}^4}{\left|t\right|+1}}\mathrm{d}t& \ll {\int }_{-10}^{10}{\displaystyle \frac{{\left|\zeta (\sigma +it)\right|}^4}{\left|t\right|+1}}\mathrm{d}t+{\int }_{10<\left|t\right|<T}{\displaystyle \frac{{\left|\zeta (\sigma +it)\right|}^4}{t}}\mathrm{d}t\hfill \\ \hfill & \ll 1+{\int }_{10}^T{\displaystyle \frac{{\left|\zeta (\sigma +it)\right|}^4}{t}}\mathrm{d}t\ll 1+{\int }_{10}^T{\displaystyle \frac{L_1\left(t\right)}{t}}\mathrm{d}t\hfill \\ \hfill & \ll 1+{\displaystyle \frac{L_1\left(T\right)}{T}}+{\int }_{10}^T{\displaystyle \frac{L_1\left(t\right)}{t^2}}\mathrm{d}t\ll {log}^{4}T{\int }_2^T{\displaystyle \frac{{log}^{4}t}{t}}\mathrm{d}t\hfill \\ \hfill & \ll {log}^{5}T.\hfill \end{array}$$

(23)

Let $L_2\left(V\right):={\int }_0^V{\left|\zeta (\sigma -it)\right|}^2\mathrm{d}t$ for $\left|\sigma -\frac{1}{2}\right|\le {\displaystyle \frac{1}{logV}}$. Similarly, we have

$$\begin{array}{c}\hfill {\int }_{-V}^V{\displaystyle \frac{{\left|\zeta (\sigma +it)\right|}^2}{\left|t\right|+1}}\mathrm{d}t\ll 1+{\displaystyle \frac{L_2\left(V\right)}{V}}+{\int }_{10}^V{\displaystyle \frac{L_2\left(t\right)}{t^2}}\mathrm{d}t\ll {log}^{2}V.\end{array}$$

(24)

From (11) and Cauchy’s inequality, we obtain

$$\begin{array}{c}\hfill {\int }_0^T\left|\zeta \left(\frac{1}{2}+it\right)\right|\mathrm{d}t\ll T{log}^{\frac{1}{2}}T,\end{array}$$

which, by partial summation, yields

$$\begin{array}{c}\hfill {\int }_{-T}^T{\displaystyle \frac{\left|\zeta \left(\frac{1}{2}+it\right)\right|}{\left|t\right|+1}}\mathrm{d}t\ll {log}^{1.5}T.\end{array}$$

Note that for the integral ${\int }_1$, we have

$$\begin{array}{c}\hfill |t_1+t_0\left|+1\le \right|t_1|+|t_0\left|+1\le 2(\right|t_0|+1)\end{array}$$

and

$$\begin{array}{c}\hfill |t_1+2t_0|+1\le \left|t_1\right|+2\left|{t|}_0\right|+1\le 3(\left|t_0\right|+1),\end{array}$$

so that

$$\begin{array}{c}\hfill {\displaystyle \frac{\left(\right|t_1|+1)\left(\right|t_1+t_0{|+1)}^{\frac{1}{2}}\left(\right|t_1+2t_0{|+1)}^{\frac{1}{2}}}{\left(\right|t_1|+1)\left(\right|t_0|+1)}}\le \sqrt{6}.\end{array}$$

Thus, from (23) and Cauchy’s inequality, we obtain

$$\begin{array}{cc}\hfill {\int }_1& \le \sqrt{6}{\int }_{|t_1|\le |t_0|}{\displaystyle \frac{\left|\zeta \left(\frac{1}{2}+i{t}_1\right)\right|{\left|\zeta \left(\frac{1}{2}+\frac{1}{logY}+i{t}_1+i{t}_0\right)\right|}^2\left|\zeta \left(\frac{1}{2}+\frac{2}{logY}+i{t}_1+i2t_0\right)\right|}{\left(\right|t_1|+1)\left(\right|t_1+t_0{|+1)}^{\frac{1}{2}}\left(\right|t_1+2t_0{|+1)}^{\frac{1}{2}}}}\mathrm{d}{t}_1\mathrm{d}{t}_0\hfill \\ \hfill & \ll {\int }_{-X^2}^{X^2}{\displaystyle \frac{\left|\zeta \left(\frac{1}{2}+i{t}_1\right)\right|}{\left(\right|t_1|+1)}}\mathrm{d}{t}_1{\left({\int }_{|t_1|\le |t_0|}{\displaystyle \frac{{\left|\zeta \left(\frac{1}{2}+\frac{1}{logY}+i{t}_1+i{t}_0\right)\right|}^4}{\left(\right|t_1+t_0|+1)}}\mathrm{d}{t}_0\right)}^{\frac{1}{2}}\times \hfill \\ \hfill & {\left({\int }_{|t_1|\le |t_0|}{\displaystyle \frac{{\left|\zeta \left(\frac{1}{2}+\frac{2}{logY}+i{t}_1+i2t_0\right)\right|}^2}{\left(\right|t_1+2t_0|+1)}}\mathrm{d}{t}_0\right)}^{\frac{1}{2}}\hfill \\ \hfill & \ll {\int }_{-X^2}^{X^2}{\displaystyle \frac{\left|\zeta \left(\frac{1}{2}+i{t}_1\right)\right|}{\left(\right|t_1|+1)}}\mathrm{d}{t}_1{\left({\int }_{-2X^2}^{2X^2}{\displaystyle \frac{{\left|\zeta \left(\frac{1}{2}+\frac{1}{logY}+i{y}_1\right)\right|}^4}{\left(\right|y_1|+1)}}\mathrm{d}{y}_1\right)}^{\frac{1}{2}}\times \hfill \\ \hfill & {\left({\int }_{-3X^2}^{3X^2}{\displaystyle \frac{{\left|\zeta \left(\frac{1}{2}+\frac{2}{logY}+i{y}_2\right)\right|}^2}{\left(\right|y_2|+1)}}\mathrm{d}{y}_2\right)}^{\frac{1}{2}}\hfill \\ \hfill & \ll {\mathcal{L}}_1^{1.5}{\left({\int }_{-2Y^2}^{2Y^2}{\displaystyle \frac{{\left|\zeta \left(\frac{1}{2}+\frac{1}{logY}+i{y}_1\right)\right|}^4}{\left(\right|y_1|+1)}}\mathrm{d}{y}_1\right)}^{\frac{1}{2}}{\left({\int }_{-3Y^2}^{3Y^2}{\displaystyle \frac{{\left|\zeta \left(\frac{1}{2}+\frac{2}{logY}+i{y}_2\right)\right|}^2}{\left(\right|y_2|+1)}}\mathrm{d}{y}_2\right)}^{\frac{1}{2}}\hfill \\ \hfill & \ll {\mathcal{L}}_1^{1.5}{\mathcal{L}}_2^{3.5},\hfill \end{array}$$

(25)

where we substitute the variables $y_1=t_1+t_0$ and $y_2=t_1+2t_0$ in the fourth line.

At present, we estimate ${\int }_2$. Note that for the integral ${\int }_2$, we have

$$\begin{array}{c}\hfill |t_1+t_0|+1\le |t_1|+|t_0|+1\le 2(|t_1|+1),\\ \hfill |t_1+2t_0|+1\le |t_1|+2|t_0|+1\le 3(|t_1|+1),\end{array}$$

which imply

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{|t_1|+1}}\le {\displaystyle \frac{2}{|t_1+t_0|+1}},\\ \hfill {\displaystyle \frac{1}{|t_1|+1}}\le {\displaystyle \frac{3}{|t_1+2t_0|+1}}.\end{array}$$

Therefore, through Hölder’s inequality, we have

$$\begin{array}{cc}\hfill {\int }_2& ={\int }_{|t_0|<|t_1|}{\displaystyle \frac{\left|\zeta \left(\frac{1}{2}+i{t}_1\right)\right|{\left|\zeta \left(\frac{1}{2}+\frac{1}{logY}+i{t}_1+i{t}_0\right)\right|}^2\left|\zeta \left(\frac{1}{2}+\frac{2}{logY}+i{t}_1+i2t_0\right)\right|}{\left(\right|t_1|+1)\left(\right|t_0|+1)}}\mathrm{d}{t}_1\mathrm{d}{t}_0\hfill \\ \hfill & \le \sqrt{2}\sqrt[4]{3}{\int }_{-X^2}^{X^2}{\displaystyle \frac{\mathrm{d}{t}_0}{|t_0|+1}}\times \hfill \\ \hfill & {\int }_{|t_0|<|t_1|\le 4Y^2}{\displaystyle \frac{\left|\zeta \left(\frac{1}{2}+i{t}_1\right)\right|{\left|\zeta \left(\frac{1}{2}+\frac{1}{logY}+i{t}_1+i{t}_0\right)\right|}^2\left|\zeta \left(\frac{1}{2}+\frac{2}{logY}+i{t}_1+i2t_0\right)\right|}{\left(\right|t_1{|+1)}^{\frac{1}{4}}\left(\right|t_1+t_0{|+1)}^{\frac{1}{2}}\left(\right|t_1+2t_0{|+1)}^{\frac{1}{4}}}}\mathrm{d}{t}_1\hfill \\ \hfill & \ll {\int }_{-X^2}^{X^2}{\displaystyle \frac{\mathrm{d}{t}_0}{|t_0|+1}}{\left({\int }_{-4Y^2}^{4Y^2}{\displaystyle \frac{{\left|\zeta \left(\frac{1}{2}+i{t}_1\right)\right|}^4}{\left(\right|t_1|+1)}}\mathrm{d}{t}_1\right)}^{{\displaystyle \frac{1}{4}}}{\left({\int }_{-8Y^2}^{8Y^2}{\displaystyle \frac{{\left|\zeta \left(\frac{1}{2}+\frac{1}{logY}+i{y}_1\right)\right|}^4}{\left(\right|y_1|+1)}}\mathrm{d}{y}_1\right)}^{\frac{1}{2}}\hfill \\ \hfill & \times {\left({\int }_{-12Y^2}^{12Y^2}{\displaystyle \frac{{\left|\zeta \left(\frac{1}{2}+\frac{2}{logY}+i{y}_2\right)\right|}^4}{|y_2|+1}}\mathrm{d}{y}_2\right)}^{{\displaystyle \frac{1}{4}}}\hfill \\ \hfill & \ll {\mathcal{L}}_1{\mathcal{L}}_2^5.\hfill \end{array}$$

(26)

Consequently, from (22), (25), and (26), it can be deduced that

$$\begin{array}{c}\hfill E_{01}\ll X{Y}^{\frac{1}{2}}{\mathcal{L}}_1{\mathcal{L}}_2^8.\end{array}$$

(27)

Next, we estimate $E_{02}$. In this case, by Lemma 2, for $|t_0|\le X^2$ and ${\displaystyle \frac{1}{2}}\le \sigma \le b$, we have

$$\begin{array}{cc}\hfill & g(\sigma +i{Y}^2,b+i{t}_0)\hfill \\ \hfill & \ll X{Y}^{\sigma }\left|{\displaystyle \frac{\zeta \left(\sigma +i{Y}^2\right){\zeta }^2\left(\sigma +\frac{1}{logY}+i{Y}^2+i{t}_0\right)\zeta \left(\sigma +\frac{2}{logY}+i{Y}^2+i2t_0\right)}{{\zeta }^2\left(1+\frac{1}{logY}+i{t}_0\right)\zeta \left(2\sigma +\frac{2}{logY}+i{Y}^2+i2t_0\right)Y^2\left(\right|t_0|+2)}}\right|\hfill \\ \hfill & \ll X{Y}^{\sigma }{\displaystyle \frac{Y^{\frac{2(1-\sigma )}{3}}logY\times \left(\right|Y^2+t_0{|+2)}^{\frac{2(1-\sigma )}{3}}{log}^2\left(\right|Y^2+t_0|+2)}{Y^2\left(\right|t_0|+1)}}\hfill \\ \hfill & \times \left(\right|Y^2+2t_0{|+2)}^{{\displaystyle \frac{1-\sigma }{3}}}log\left(\right|Y^2+2t_0|+2)\hfill \\ \hfill & \ll X{Y}^{{\displaystyle \frac{2}{3}}-{\displaystyle \frac{5}{3}}\sigma }\left(\right|t_0{|+1)}^{-1}{\mathcal{L}}_2^5,\hfill \end{array}$$

so that we obtain

$$\begin{array}{c}\hfill E_{02}\ll X{Y}^{\frac{2}{3}}{\mathcal{L}}_2^5{\int }_{-X^2}^{X^2}{\displaystyle \frac{1}{|t_0|+1}}\mathrm{d}{t}_0{\int }_{\frac{1}{2}}^b{Y}^{-\frac{5}{3}\sigma }\mathrm{d}\sigma \ll X{Y}^{-\frac{1}{6}}{\mathcal{L}}_1{\mathcal{L}}_2^5.\end{array}$$

(28)

Similarly, we have

$$\begin{array}{c}\hfill E_{03}\ll X{Y}^{-\frac{1}{6}}{\mathcal{L}}_1{\mathcal{L}}_2^5.\end{array}$$

(29)

From (27)–(29), we deduce that

$$\begin{array}{c}\hfill M={\displaystyle \frac{1}{2\pi i}}{\int }_{b-i{X}^2}^{b+i{X}^2}(R_1+R_2+R_3)\mathrm{d}s+O(X{Y}^{\frac{1}{2}}{\mathcal{L}}_1{\mathcal{L}}_2^8).\end{array}$$

(30)

Then, we calculate the main terms, respectively. Let

$$\begin{array}{c}\hfill M_j:={\displaystyle \frac{1}{2\pi i}}{\int }_{b-i{X}^2}^{b+i{X}^2}{R}_j\mathrm{d}s(j=1,2,3).\end{array}$$

3.3. Treatment for $M_1$

Firstly, we deal with $M_1$. Recall that

$$\begin{array}{c}\hfill R_1={\displaystyle \frac{\zeta (2-s)X^s{Y}^{2-s}}{s(2-s)\zeta \left(2\right)\zeta \left(s\right)}}\left(\frac{1}{2}+{\mathcal{L}}_2+{\displaystyle \frac{1}{2-s}}\right).\end{array}$$

We consider the rectangle domain formed by the four points $s=b\pm i{X}^2$, $s={\displaystyle \frac{5}{2}}\pm i{X}^2$. In this domain, the integrand $R_1$ has a 2-order pole $s=2$. By the residue theorem, we have

$$\begin{array}{cc}\hfill M_1& =-\underset{s=2}{Res}{R}_1-{\displaystyle \frac{1}{2\pi i}}\left({\int }_{{\displaystyle \frac{5}{2}}-i{X}^2}^{{\displaystyle \frac{5}{2}}+i{X}^2}-{\int }_{b-i{X}^2}^{{\displaystyle \frac{5}{2}}-i{X}^2}+{\int }_{b+i{X}^2}^{{\displaystyle \frac{5}{2}}+i{X}^2}\right)R_1\mathrm{d}s\hfill \\ \hfill & =-\underset{s=2}{Res}{R}_1-E_{11}+E_{12}-E_{13},\hfill \end{array}$$

(31)

where

$$\begin{array}{c}\hfill E_{11}:={\displaystyle \frac{1}{2\pi i}}{\int }_{{\displaystyle \frac{5}{2}}-i{X}^2}^{{\displaystyle \frac{5}{2}}+i{X}^2}{R}_1\mathrm{d}s,E_{12}:={\displaystyle \frac{1}{2\pi i}}{\int }_{b-i{X}^2}^{{\displaystyle \frac{5}{2}}-i{X}^2}{R}_1\mathrm{d}s,E_{13}:={\displaystyle \frac{1}{2\pi i}}{\int }_{b+i{X}^2}^{{\displaystyle \frac{5}{2}}+i{X}^2}{R}_1\mathrm{d}s.\end{array}$$

We calculate that

$$\begin{array}{cc}\hfill \underset{s=2}{Res}{R}_1=& {\displaystyle \frac{\zeta \left(0\right)X^2{\mathcal{L}}_1}{2\zeta \left(2\right)}}-{\displaystyle \frac{\zeta \left(0\right)X^2{\mathcal{L}}_2}{{\zeta }^2\left(2\right)}}-{\displaystyle \frac{X^2}{{\zeta }^2\left(2\right)}}\left({\zeta }^{\prime }\left(0\right)+\zeta \left(0\right)+{\displaystyle \frac{{\zeta }^{\prime }\left(2\right)\zeta \left(0\right)}{\zeta \left(2\right)}}\right).\hfill \end{array}$$

From the estimates (5) and (8), for $\left|t\right|\le X^2$, we have

$$\begin{array}{c}\hfill R_1\ll X^{{\displaystyle \frac{5}{2}}}{Y}^{-\frac{1}{2}}{\mathcal{L}}_2{\displaystyle \frac{\left|\zeta \left(-\frac{1}{2}-it\right)\right|}{{\left(\right|t|+1)}^2}}\ll X^{{\displaystyle \frac{5}{2}}}{Y}^{-\frac{1}{2}}{\mathcal{L}}_2{\displaystyle \frac{{log}^2\left(\right|t|+2)}{\left(\right|t|+1)}}.\end{array}$$

Then we have

$$\begin{array}{c}\hfill E_{11}\ll X^{{\displaystyle \frac{5}{2}}}{Y}^{-\frac{1}{2}}{\mathcal{L}}_2{\int }_{-X^2}^{X^2}{\displaystyle \frac{{log}^2\left(\right|t|+2)}{\left(\right|t|+1)}}\mathrm{d}t\ll X^{{\displaystyle \frac{5}{2}}}{Y}^{-\frac{1}{2}}{\mathcal{L}}_1^3{\mathcal{L}}_2.\end{array}$$

(32)

For $b\le \sigma \le {\displaystyle \frac{5}{2}}$, we have

$$\begin{array}{cc}\hfill R_1& \ll {\displaystyle \frac{\left|\zeta (2-\sigma -i{X}^2)\right|X^{\sigma }{Y}^{2-\sigma }}{X^4\left|\zeta (\sigma +i{X}^2)\right|}}\left(\frac{1}{2}+logY+{\displaystyle \frac{1}{X^2}}\right)\hfill \\ \hfill & \ll X^{-4+\sigma }{Y}^{2-\sigma }\left|\zeta (2-\sigma -i{X}^2)\right|{\mathcal{L}}_1{\mathcal{L}}_2.\hfill \end{array}$$

Through Lemma 2, we have

$$\begin{array}{c}\hfill R_1\ll \left\{\begin{array}{cc}{X}^{-\frac{14}{3}+{\displaystyle \frac{5\sigma }{3}}}{Y}^{2-\sigma }{\mathcal{L}}_1^2{\mathcal{L}}_2,\hfill & b<\sigma \le \frac{3}{2};\hfill \\ X^{-\frac{17}{3}+{\displaystyle \frac{7\sigma }{3}}}{Y}^{2-\sigma }{\mathcal{L}}_1^2{\mathcal{L}}_2,\hfill & \frac{3}{2}<\sigma \le 2;\hfill \\ X^{-7+3\sigma }{Y}^{2-\sigma }{\mathcal{L}}_1^2{\mathcal{L}}_2,\hfill & 2<\sigma \le {\displaystyle \frac{5}{2}}.\hfill \end{array}\right.\end{array}$$

It follows that

$$\begin{array}{cc}\hfill E_{12}& \ll \left({\int }_b^{\frac{3}{2}}+{\int }_{\frac{3}{2}}^2+{\int }_2^{\frac{5}{2}}\right)R_1\mathrm{d}\sigma \hfill \\ \hfill & \ll {\mathcal{L}}_1^2{\mathcal{L}}_2\left(X^{-2}Y+X^{\frac{-13}{6}}{Y}^{\frac{1}{2}}+X^{-1}+X^{\frac{1}{2}}{Y}^{-\frac{1}{2}}\right)\hfill \\ \hfill & \ll X^{-2}Y{\mathcal{L}}_1^2{\mathcal{L}}_2+X^{\frac{1}{2}}{Y}^{-\frac{1}{2}}{\mathcal{L}}_1^2{\mathcal{L}}_2.\hfill \end{array}$$

(33)

Similarly, we obtain

$$\begin{array}{c}\hfill E_{13}\ll X^{-2}Y{\mathcal{L}}_1^2{\mathcal{L}}_2+X^{\frac{1}{2}}{Y}^{-\frac{1}{2}}{\mathcal{L}}_1^2{\mathcal{L}}_2.\end{array}$$

(34)

Then, from (32)–(34), we obtain

$$\begin{array}{cc}\hfill M_1=& -{\displaystyle \frac{\zeta \left(0\right)X^2{\mathcal{L}}_1}{2\zeta \left(2\right)}}+{\displaystyle \frac{\zeta \left(0\right)X^2{\mathcal{L}}_2}{{\zeta }^2\left(2\right)}}+{\displaystyle \frac{X^2}{\zeta \left(2\right)}}\left({\zeta }^{\prime }\left(0\right)+\zeta \left(0\right)+{\displaystyle \frac{{\zeta }^{\prime }\left(2\right)\zeta \left(0\right)}{\zeta \left(2\right)}}\right)\hfill \\ \hfill & +O(X^{-2}Y{\mathcal{L}}_1^2{\mathcal{L}}_2+X^{{\displaystyle \frac{5}{2}}}{Y}^{-\frac{1}{2}}{\mathcal{L}}_1^3{\mathcal{L}}_2).\hfill \end{array}$$

(35)

3.4. Treatment for $M_2$

Let

$$h\left(n\right)=\sum _{n=n_1^2{n}_2^2}{n}_1\mu \left(n_2\right),$$

then we have

$$\begin{array}{cc}\hfill \sum _{n\le X}h\left(n\right)& =\sum _{n_2\le X^{\frac{1}{2}}}\mu \left(n_2\right)\sum _{n_1\le X^{\frac{1}{2}}{n}_2^{-1}}{n}_1\hfill \\ \hfill & =\sum _{n_2\le X^{\frac{1}{2}}}\mu \left(n_2\right)\left({\displaystyle \frac{X}{2n_2^2}}+O\left({\displaystyle \frac{X^{\frac{1}{2}}}{n_2}}\right)\right)\hfill \\ \hfill & ={\displaystyle \frac{X}{2}}\sum _{n_2\le X^{\frac{1}{2}}}{\displaystyle \frac{\mu \left(n_2\right)}{n_2^2}}+O\left(X^{\frac{1}{2}}\sum _{n_2\le X^{{\displaystyle \frac{1}{2}}}}{\displaystyle \frac{1}{n_2}}\right)\hfill \\ \hfill & ={\displaystyle \frac{X}{2\zeta \left(2\right)}}+O(X^{\frac{1}{2}}{\mathcal{L}}_1).\hfill \end{array}$$

(36)

Since

$$\begin{array}{c}\hfill \sum _{n=1}^{\infty }{\displaystyle \frac{h\left(n\right)}{n^s}}=\sum _{n_1=1}^{\infty }\sum _{n_2=1}^{\infty }{\displaystyle \frac{n_1\mu \left(n_2\right)}{{\left(n_1{n}_2\right)}^{2s}}}={\displaystyle \frac{\zeta (2s-1)}{\zeta \left(2s\right)}},\end{array}$$

by Perron’s formula, we obtain

$$\begin{array}{c}\hfill \sum _{n\le X}h\left(n\right)={\displaystyle \frac{1}{2\pi i}}{\int }_{b-i{X}^2}^{b+i{X}^2}{\displaystyle \frac{\zeta (2s-1)X^s}{\zeta \left(2s\right)s}}\mathrm{d}s+O\left(X^{-1+\epsilon }\right).\end{array}$$

(37)

Consequently, recalling that

$$\begin{array}{c}\hfill R_2={\displaystyle \frac{\zeta (2s-1)X^{s}Y}{s\zeta \left(2s\right)}},\end{array}$$

collecting (36) and (37), it can be derived that

$$\begin{array}{c}\hfill M_2={\displaystyle \frac{XY}{2\zeta \left(2\right)}}+O(X^{\frac{1}{2}}Y{\mathcal{L}}_1).\end{array}$$

(38)

3.5. Treatment for $M_3$

We write

$$\begin{array}{cc}\hfill M_3& ={\displaystyle \frac{1}{2\pi i}}\left({\int }_{b-i\infty }^{b+i\infty }-{\int }_{b-i\infty }^{b-i{X}^2}-{\int }_{b+i{X}^2}^{b+i\infty }\right)R_3\mathrm{d}s\hfill \\ \hfill & ={\displaystyle \frac{1}{2\pi i}}{\int }_{b-i\infty }^{b+i\infty }{R}_3\mathrm{d}s+O\left({\int }_{\left|t\right|>X^2}\left|R_3\right|\mathrm{d}t\right).\hfill \end{array}$$

Recalling that

$$\begin{array}{c}\hfill R_3={\displaystyle \frac{\zeta (3-2s){\zeta }^2(2-s)X^s{Y}^{3-2s}}{s(3-2s){\zeta }^2\left(s\right)\zeta (4-2s)}},\end{array}$$

when $s=b+it$ for $\left|t\right|>X^2$, we obtain by Lemma 2

$$\begin{array}{c}\hfill R_3\ll {XY\left|t\right|}^{{\displaystyle \frac{4}{3logY}}-2}{log}^6\left|t\right|.\end{array}$$

Then, we have

$$\begin{array}{c}\hfill O\left({\int }_{\left|t\right|>X^2}\left|R_3\right|\mathrm{d}t\right)=O\left(XY{\int }_{\left|t\right|>X^2}{\left|t\right|}^{{\displaystyle \frac{4}{3logY}}-2}{log}^6\left|t\right|\mathrm{d}t\right)=O(X^{-1}Y{\mathcal{L}}_1^6).\end{array}$$

(39)

Since $R_3$ is holomorphic in $1\le \Re s\le {\displaystyle \frac{3}{2}}$, we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2\pi i}}{\int }_{b-i\infty }^{b+i\infty }{R}_3\mathrm{d}s={\displaystyle \frac{1}{2\pi i}}{\int }_{\frac{3}{2}-i\infty }^{\frac{3}{2}+i\infty }{R}_3\mathrm{d}s.\end{array}$$

When $s={\displaystyle \frac{3}{2}}+it$, by Lemma 2, we have

$$\begin{array}{c}\hfill R_3={\displaystyle \frac{\zeta (-2it){\zeta }^2\left(\frac{1}{2}-it\right)X^{\frac{3}{2}}}{\left(\frac{3}{2}+it\right)(-2it)\zeta \left(\frac{3}{2}+it\right)\zeta (1-2it)}}\ll X^{\frac{3}{2}}{\left(\left|t\right|+2\right)}^{-\frac{7}{6}}{log}^6\left(\right|t|+2).\end{array}$$

It follows that

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2\pi i}}{\int }_{\frac{3}{2}-i\infty }^{\frac{3}{2}+i\infty }{R}_3\mathrm{d}s\ll X^{\frac{3}{2}}.\end{array}$$

(40)

Consequently, from (39) and (40), we obtain

$$\begin{array}{c}\hfill M_3=O(X^{\frac{3}{2}}+X^{-1}Y{\mathcal{L}}_1^6).\end{array}$$

(41)

4. Conclusions

Collecting the above computations for (20), (30), (35), (38), and (41), we have

$$\begin{array}{cc}\hfill D_2(X,Y)=& {\displaystyle \frac{XY}{2\zeta \left(2\right)}}-{\displaystyle \frac{\zeta \left(0\right)X^2{\mathcal{L}}_1}{2\zeta \left(2\right)}}+{\displaystyle \frac{\zeta \left(0\right)X^2{\mathcal{L}}_2}{{\zeta }^2\left(2\right)}}+{\displaystyle \frac{X^2}{{\zeta }^2\left(2\right)}}\left({\zeta }^{\prime }\left(0\right)+\zeta \left(0\right)+{\displaystyle \frac{{\zeta }^{\prime }\left(2\right)\zeta \left(0\right)}{\zeta \left(2\right)}}\right)\hfill \\ \hfill & +O(X^{\epsilon -\frac{1}{2}}Y{\mathcal{L}}_2^6+X{Y}^{\frac{1}{2}}{\mathcal{L}}_1{\mathcal{L}}_2^8+X^{-2}Y{\mathcal{L}}_1^2{\mathcal{L}}_2+X^{\frac{5}{2}}{Y}^{-\frac{1}{2}}{\mathcal{L}}_1^3{\mathcal{L}}_2\hfill \\ \hfill & +X^{\frac{1}{2}}Y{\mathcal{L}}_1+X^{\frac{3}{2}}+X^{-1}Y{\mathcal{L}}_1^6)\hfill \\ \hfill =& {\displaystyle \frac{XY}{2\zeta \left(2\right)}}-{\displaystyle \frac{\zeta \left(0\right)X^2{\mathcal{L}}_1}{2\zeta \left(2\right)}}+{\displaystyle \frac{\zeta \left(0\right)X^2{\mathcal{L}}_2}{{\zeta }^2\left(2\right)}}+{\displaystyle \frac{X^2}{{\zeta }^2\left(2\right)}}\left({\zeta }^{\prime }\left(0\right)+\zeta \left(0\right)+{\displaystyle \frac{{\zeta }^{\prime }\left(2\right)\zeta \left(0\right)}{\zeta \left(2\right)}}\right)\hfill \\ \hfill & +O(X^{\frac{1}{2}}Y{\mathcal{L}}_1+X^{\frac{5}{2}}{Y}^{-\frac{1}{2}}{\mathcal{L}}_1^3{\mathcal{L}}_2).\hfill \end{array}$$

It is evident that the main term must dominate the error term. Thus, the proof of Theorem 1 is complete.