On moments of the error term of the multivariable k-th divisor functions

1. Introduction and Results

Let $k\geqslant 2$ be an integer, $\tau_{k}(n)$ denote the number of ways $n$ can be written as a product of $k$ fixed factors. When $k=2$ , $\tau_{2}(n)=\tau(n)$ is the Dirichlet divisor function. The problems about it are important in analytic number theory and hence have a long history [6, 8, 12].

For $k\geqslant 3$ , suppose $x$ is a large real positive number, we have

\sum_{n\leqslant x}\tau_{k}(n)=xP_{k-1}(\log x)+\Delta_{k}(x):=M_{k}(x)+\Delta_{k}(x),

(1.1)

where $P_{k-1}(t)$ is a given polynomial in $t$ of degree $k-1$ and $\Delta_{k}(x)$ is the error term. We denote

\alpha_{k}=\inf\left\{a:\Delta_{k}(x)=O(x^{a})\right\},\qquad\beta_{k}=\inf\left\{b:\int_{1}^{x}{\Delta_{k}}^{2}(t)dt=O(x^{1+2b})\right\}.

(1.2)

There are many results about the upper bounds and lower bounds for $\alpha_{k}$ and $\beta_{k}$ . For unified conclusions on $k$ , Voronoi [13] proved that $\alpha_{k}\leqslant(k-1)/(k+1)$ for $k\geqslant 3$ in 1903. In 1916 Hardy [3] showed that $\alpha_{k}\geqslant\beta_{k}\geqslant(k-1)/2k$ holds for $k\geqslant 3$ . Hardy and Littlewood [4] proved that $\alpha_{k}\leqslant(k-1)/(k+2)$ for $k\geqslant 4$ in 1923. Ivić [6] gave a summary, namely

\displaystyle\alpha_{3}\leqslant 43/96,\qquad\alpha_{k}\leqslant(3k-4)/4k\qquad(4\leqslant k\leqslant 8),\cdots

(1.3)

and

\displaystyle\beta_{k}=(k-1)/2k,\qquad(k=2,3,4),\qquad\beta_{5}\leqslant 119/260,\qquad\beta_{6}\leqslant 1/2,\qquad\beta_{7}\leqslant 39/70,\cdots

(1.4)

For the results on $k=3$ , in 1956, Tong [10] developed a new method of deriving an asymptotic formula for the mean square for $\Delta_{3}(x)$ , which can be stated as follows:

Suppose $T$ is a large real number, $\varepsilon$ is a sufficiently small real positive number and $\Delta_{3}(x)$ is defined in (1.1), then

\int_{1}^{T}{\Delta_{3}}^{2}(x)dx=\frac{1}{10\pi^{2}}\sum_{n=1}^{\infty}\frac{{\tau_{3}}^{2}(n)}{n^{4/3}}T^{5/3}+O(T^{5/3-1/14+\varepsilon}).

(1.5)

Let $r\geqslant 2$ be a fixed integer. In this paper we consider the sum

\sum_{n_{1},\cdots,n_{r}\leqslant x}\tau_{k}(n_{1}\cdots n_{r})=M_{r,k}(x)+\Delta_{r,k}(x),

(1.6)

where $M_{r,k}(x)$ is the main term and $\Delta_{r,k}(x)$ is the error term. Tóth and Zhai [11] studied the condition $k=2$ . In this paper we first show that

Theorem 1.1.

Let $r\geqslant 2$ , $k\geqslant 3$ be fixed integers. Suppose $\alpha_{k}$ are expressed in (1.3), then for any real number $x\geqslant 2$ , the asymptotic formula

\sum_{n_{1},\cdots,n_{r}\leqslant x}\tau_{k}(n_{1}\cdots n_{r})=M_{r,k}(x)+O(x^{r-1+\alpha_{k}+\varepsilon})

holds for any $\varepsilon>0$ , and $M_{r,k}(x)$ is expressed by

M_{r,k}(x)=x^{r}\sum_{\ell=0}^{r(k-1)}d_{r,k,\ell}(\log x)^{\ell},

where $d_{r,k,\ell}$ $(0\leqslant\ell\leqslant r(k-1))$ are computable constants.

The first author [2] have studied the mean square for $\Delta_{r,2}(x)$ and have got an asymptotic formula. In this paper we concentrate on the integral

\int_{1}^{T}{\Delta_{r,k}}^{2}(x)dx

for $k\geqslant 3$ and large real $T$ . We give an asymptotic formula for $k=3$ and the upper bounds for $k\geqslant 4$ . The results are stated as follows.

Theorem 1.2.

Let $T\geqslant 2$ be a large real number and $r\geqslant 2$ be a fixed integer. Then the asymptotic formula

\int_{1}^{T}{\Delta_{r,3}}^{2}(x)dx=\frac{r^{2}}{6\pi^{2}}T^{2r-1/3}L_{4r-4}(\log T)+O(T^{2r-7/18+\varepsilon})

holds for any $\varepsilon>0$ , where $L_{4r-4}(u)$ is a polynomial in $u$ of degree $(4r-4)$ denoted by

\displaystyle L_{4r-4}(u)=\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}D_{r,3,\ell_{1},\ell_{2}}\sum_{t=0}^{\ell_{1}+\ell_{2}}\frac{(-1)^{t}(\ell_{1}+\ell_{2})!}{(2r-\frac{1}{3})^{t+1}(\ell_{1}+\ell_{2}-t)!}u^{\ell_{1}+\ell_{2}-t},

and $D_{r,3,\ell_{1},\ell_{2}}$ $(0\leqslant\ell_{1},\ell_{2}\leqslant 2(r-1))$ are computable constants.

Theorem 1.2 can be viewed as an analogue of Tong’s result.

For higher power moments of $\Delta_{3}(x)$ , in 1992 Heath-Brown [5] proved that for a large real positive $T$ , the upper bound estimate

\int_{1}^{T}|{\Delta_{3}}(x)|^{3}dx\ll T^{2+\varepsilon}

(1.7)

holds for any $\varepsilon>0$ . This is the best upper bound since the average order of $\Delta_{3}(x)$ is $1/3$ , which can be obtained by (1.5). In this paper we give a similar result for $\Delta_{r,3}(x)$ .

Theorem 1.3.

Let $T\geqslant 2$ be a large real number, we have

\int_{1}^{T}|\Delta_{r,3}(x)|^{3}dx\ll T^{3r-1+\varepsilon}.

Using Theorem 1.2 we have the average order of $\Delta_{r,3}(x)$ is $(r-2/3)$ . Thus this is also the best upper bound.

Moreover, we obtain the following result.

Theorem 1.4.

Let $T\geqslant 2$ be a large real number, we have

\int_{1}^{T}\Delta_{r,3}(x)dx\ll T^{r+1/6+\varepsilon}.

Corollary 1.5.

For a large real number $T$ , $\Delta_{r,3}(x)$ has at least $T^{5/96-\varepsilon}$ sign changes in $[T,2T]$ .

Theorem 1.6.

Let $T\geqslant 2$ be a large real number, for fixed integer $k\geqslant 4$ ,

\int_{1}^{T}{\Delta_{r,k}}^{2}(x)dx\ll T^{2r-1+2\beta_{k}+\varepsilon}

holds for any $\varepsilon>0$ , where $\beta_{k}$ are expressed by (1.4).

Notation.

Throughout this paper, $\varepsilon$ denotes a sufficiently small real positive number, not necessarily the same at each occurrence. As usual, $\mathbb{R}$ denotes the set of real numbers, $\mathbb{N}$ denotes the set of natural number, $\zeta$ denotes the Riemann-zeta function. For a positive integer $k$ , $\tau_{k}$ denotes the $k$ th Dirichlet divisor function. For a complex number $z$ , $\Re z$ denotes the real part of $z$ . For integers $m$ and $n$ , $\binom{m}{n}$ denotes a binomial coefficient ( $m>n$ ). For a complex number $\theta$ , the function $e(\theta)$ denotes $e^{2\pi i\theta}$ .

2. Some lemmas

We present some lemmas.

Lemma 2.1.

Let $T\geqslant 10$ be a given large real number and $N$ be a real number such that $T^{\varepsilon}\ll N\ll T^{2/3}$ , and $\Delta_{3}(\cdot)$ be defined in (1.1). For any $T\leqslant x\leqslant 2T$ , define

	$\displaystyle\delta_{31}(x,N)$	$\displaystyle=\frac{x^{1/3}}{\sqrt{3}\pi}\sum_{n\leqslant N}\frac{\tau_{3}(n)}{n^{2/3}}\cos\left(6\pi(nx)^{1/3}\right),$
	$\displaystyle\delta_{32}(x,N)$	$\displaystyle=\Delta_{3}(x)-\delta_{31}(x,N).$

Then we have

\int_{T}^{2T}{\delta_{32}}^{2}(x,N)dx\ll T^{5/3+\varepsilon}N^{-1/3}+T^{14/9+\varepsilon}.

Proof.

See Lemma 2.12 in Cao, Tanigawa and Zhai [1]. ∎

Lemma 2.2.

Let $k\geqslant 3$ , $r\geqslant 2$ be integers, and $t_{1},\cdots,t_{r}$ be complex numbers such that $0<|t_{1}|,\cdots,|t_{r}|<1$ . Then

		$\displaystyle-(r-1)(1-t_{1})^{k}\cdots(1-t_{r})^{k}+\sum_{j=1}^{r}\prod_{\begin{subarray}{c}i=1\\ i\neq j\end{subarray}}^{r}(1-t_{i})^{k}$
	$\displaystyle=$	$\displaystyle 1-\sum_{j=2}^{r}(j-1)\sum_{1\leqslant a_{1}<a_{2}<\cdots<a_{j}\leqslant r}\sum_{1\leqslant b_{1}<b_{2}<\cdots<b_{j}\leqslant k}(-1)^{b_{1}+\cdots+b_{j}}\binom{k}{b_{1}}\cdots\binom{k}{b_{j}}{t_{a_{1}}}^{b_{1}}\cdots{t_{a_{j}}}^{b_{j}},$

which means this polynomial does not contain terms of the form $c{t_{j}}^{l}$ $(l>0$ and $c\neq 0)$ for every $1\leqslant j\leqslant r$ .

Proof.

For any given $1\leqslant j\leqslant r$ , there are $(r-j)$ terms in the latter series which contains $(1-t_{a_{1}})^{k}\cdots(1-t_{a_{j}})^{k}$ . By comparing the coefficients of $(1-t_{a_{1}})^{k}\cdots(1-t_{a_{j}})^{k}$ in the first term and in the latter series, and using the binomial theorem we can finish the proof. ∎

Lemma 2.3.

Let $k\geqslant 2$ be a fixed integer and let $s_{1},\cdots,s_{k}$ be complex numbers. Then for $\Re{s_{j}}>1(j=1,2,\cdots,k)$ , we have

\sum_{n_{1},\cdots,n_{r}=1}^{\infty}\frac{\tau_{k}(n_{1}\cdots n_{r})}{n_{1}^{s_{1}}\cdots n_{r}^{s_{r}}}=\zeta^{k}(s_{1})\cdots\zeta^{k}(s_{r})F_{r,k}(s_{1},\cdots,s_{r}),

where

F_{r,k}(s_{1},\cdots,s_{r})=\sum_{n_{1},\cdots,n_{r}=1}^{\infty}\frac{f_{r,k}(n_{1},\cdots,n_{r})}{n_{1}^{s_{1}}\cdots n_{r}^{s_{r}}}.

This series is absolutely convergent provided that $\Re s_{j}>0$ and $\Re(s_{j}+s_{l})>1(1\leqslant j,l\leqslant r)$ , and $f_{r,k}(n_{1},\cdots,n_{r})$ is multiplicative and symmetric in all variables.
Moreover, for any function $g(n_{1},\cdots,n_{r})$ satisfying $g(n_{1},\cdots,n_{r})\ll\left(\prod_{j=1}^{r}n_{j}\right)^{\varepsilon}$ , the series

\sum_{n_{1},\cdots,n_{r}=1}^{\infty}\frac{f_{r,k}(n_{1},\cdots,n_{r})g(n_{1},\cdots,n_{r})}{n_{1}^{s_{1}}\cdots n_{r}^{s_{r}}}

(2.1)

is absolutely convergent provided that $\Re s_{j}>0$ and $\Re(s_{j}+s_{l})>1(1\leqslant j,l\leqslant r)$ .

Proof.

The function $n\mapsto\tau_{k}(n)$ is multiplicative and $\tau_{k}(p^{\nu})=\binom{\nu+k-1}{k-1}$ for every prime power $p^{\nu}(\nu\geqslant 0)$ . The function $(n_{1},\cdots,n_{r})\mapsto\tau_{k}(n_{1}\cdots n_{r})$ is multiplicative, viewed as a function of $r$ variables. Therefore its multiple Dirichlet series can be expanded into an Euler product. We obtain

	$\displaystyle\sum_{n_{1},\cdots,n_{r}=1}^{\infty}\frac{\tau_{k}(n_{1}\cdots n_{r})}{{n_{1}}^{s_{1}}\cdots{n_{r}}^{s_{r}}}$	$\displaystyle=\prod_{p}\sum_{\nu_{1}\cdots\nu_{r}=0}^{\infty}\frac{\tau_{k}(p^{\nu_{1}+\cdots+\nu_{r}})}{p^{\nu_{1}s_{1}+\cdots+\nu_{r}s_{r}}}$
		$\displaystyle=\zeta^{k}(s_{1})\cdots\zeta^{k}(s_{r})F_{r,k}(s_{1},\cdots,s_{r}),$

where

	$\displaystyle F_{r,k}(s_{1},\cdots,s_{r})$	$\displaystyle=\prod_{p}\left(1-\frac{1}{p^{s_{1}}}\right)^{k}\cdots\left(1-\frac{1}{p^{s_{r}}}\right)^{k}\sum_{\nu_{1}\cdots\nu_{r}=0}^{\infty}\frac{\tau_{k}(p^{\nu_{1}+\cdots+\nu_{r}})}{p^{\nu_{1}s_{1}+\cdots+\nu_{r}s_{r}}}$
		$\displaystyle=\prod_{p}\left(1-\frac{1}{p^{s_{1}}}\right)^{k}\cdots\left(1-\frac{1}{p^{s_{r}}}\right)^{k}$
		$\displaystyle\times\left(1-r+\sum_{\nu_{1}=0}^{\infty}\frac{\tau_{k}(p^{\nu_{1}})}{p^{\nu_{1}s_{1}}}+\cdots+\sum_{\nu_{1}=0}^{\infty}\frac{\tau_{k}(p^{\nu_{r}})}{p^{\nu_{r}s_{r}}}+\sum_{\begin{subarray}{c}\nu_{1}\cdots\nu_{r}=0\\ \#A(\nu_{1},\cdots,\nu_{r})\geqslant 2\end{subarray}}^{\infty}\frac{\tau_{k}(p^{\nu_{1}+\cdots+\nu_{r}})}{p^{v_{1}s_{1}+\cdots+v_{r}s_{r}}}\right)$
		$\displaystyle=\prod_{p}\left(1-\frac{1}{p^{s_{1}}}\right)^{k}\cdots\left(1-\frac{1}{p^{s_{r}}}\right)^{k}$
		$\displaystyle\times\left(1-r+\left(1-\frac{1}{p^{s_{1}}}\right)^{-k}+\cdots+\left(1-\frac{1}{p^{s_{r}}}\right)^{-k}+\sum_{\begin{subarray}{c}\nu_{1}\cdots\nu_{r}=0\\ \#A(\nu_{1},\cdots,\nu_{r})\geqslant 2\end{subarray}}^{\infty}\frac{\tau_{k}(p^{\nu_{1}+\cdots+\nu_{r}})}{p^{v_{1}s_{1}+\cdots+v_{r}s_{r}}}\right),$

where $\#A(\nu_{1},\cdots,\nu_{r}):=\{j:1\leqslant j\leqslant r,\nu_{j}\neq 0\}$ . And the terms $1/p^{\ell s_{j}}$ (the case $\nu_{t}=0$ for all $t\neq j$ and $\nu_{j}=\ell$ ) only appear in

\left(1-\frac{1}{p^{s_{1}}}\right)^{k}\cdots\left(1-\frac{1}{p^{s_{r}}}\right)^{k}\left(1-r+\left(1-\frac{1}{p^{s_{1}}}\right)^{-k}+\cdots+\left(1-\frac{1}{p^{s_{r}}}\right)^{-k}\right).

Using Lemma 2.2 in the case that $t_{j}=1/p^{s_{j}}(1\leqslant j\leqslant r)$ we obtain that the coefficient of $1/p^{\ell s_{j}}$ is zero for every $1\leqslant j\leqslant r$ and $\ell>0$ .

Hence if $\Re s_{j}>0$ and $\Re(s_{j}+s_{l})>1(1\leqslant j,l\leqslant r)$ , then $F_{r,k}(s_{1},\cdots,s_{r})$ is absolutely convergent. And the convergence of (2.1) is a direct corollary. ∎

Lemma 2.4.

Suppose $G_{0},m_{0}$ are fixed real positive numbers, let G(x) be a monotonic function defined on $[a,b]$ such that $|G(x)|\leqslant G_{0}$ and m(x) be a differentiable real function such that $|m^{{}^{\prime}}(x)|\geqslant m_{0}$ on $[a,b]$ , $F(\cdot)=\cos(\cdot)$ or $\sin(\cdot)$ or $e(\cdot)$ , then

\int_{a}^{b}G(x)F\left(m(x)\right)dx\ll G_{0}{m_{0}}^{-1}.

Proof.

See Lemma 2.1 in Ivić [6]. ∎

Lemma 2.5.

Suppose $N$ is a given large real number, $1\leqslant N_{1},N_{2}\leqslant N$ are given real numbers, $a,b$ are integers such that $a,b\ll N$ . Define

T(a,b)=\sum_{\begin{subarray}{c}n_{1}\sim N_{1},n_{2}\sim N_{2}\\ an_{1}\neq bn_{2}\end{subarray}}\frac{1}{|an_{1}-bn_{2}|}.

Then we have

T(a,b)\ll(N_{1}N_{2})^{1/2}\log N.

Proof.

Let $an_{1}-bn_{2}=\eta$ , we have $\eta\equiv an_{1}(mod\hskip 2.84544ptb)$ , so we can find a constant $c_{0}$ such that $1\leqslant c_{0}<b$ , $r=bt+c_{0}$ , where $t$ is an integer such that $0<t<2N^{2}$ , thus

	$\displaystyle T(a,b)$	$\displaystyle=\sum_{n_{1}\sim N_{1}}\sum_{\begin{subarray}{c}n_{2}\sim N_{2}\\ an_{1}\neq bn_{2}\end{subarray}}\frac{1}{\|an_{1}-bn_{2}\|}$
		$\displaystyle\leqslant\sum_{n_{1}\sim N_{1}}\left(1+\sum_{1\leqslant t\leqslant 2N^{2}}\frac{1}{bt+c_{0}}\right)\ll N_{1}\log N,$

similarly we have $T(a,b)\ll N_{2}\log N$ , thus $T(a,b)\ll(N_{1}N_{2})^{1/2}\log N$ . ∎

Lemma 2.6.

Suppose $x,y$ are large real numbers, $r\geqslant 2$ is a fixed integer, $s,w$ are given real numbers such that $0<s<1/2<w<1$ , $f_{r,3}$ is defined in Lemma 2.3 in the case k=3. Let $\mathbf{M}_{1}$ and $\mathbf{M}_{2}$ denote the vectors $(m_{1},\cdots,m_{r})$ and $(m_{k+1},\cdots,m_{2r})$ , $D_{1}$ and $D_{2}$ denote $\left(\prod_{j=1}^{r-1}m_{j}\right)$ and $\left(\prod_{j=r+1}^{2r-1}m_{j}\right)$ , respectively. Let

		$\displaystyle T_{g,r,3}(x,y;s,w)=\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant x\\ n_{1},n_{2}\leqslant y\\ \frac{m_{r}}{m_{2r}}=\frac{n_{1}}{n_{2}}\end{subarray}}\frac{f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})g(\mathbf{M}_{1},\mathbf{M}_{2})}{D_{1}D_{2}(m_{r}m_{2r})^{s}}\cdot\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{w}},$
		$\displaystyle T_{g,r,3}(s,w)=\sum_{\begin{subarray}{c}\mathbf{M}_{1},\mathbf{M}_{2}\in\mathbb{N}^{r}\\ n_{1},n_{2}\in\mathbb{N}\\ \frac{m_{r}}{m_{2r}}=\frac{n_{1}}{n_{2}}\end{subarray}}\frac{f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})g(\mathbf{M}_{1},\mathbf{M}_{2})}{D_{1}D_{2}(m_{r}m_{2r})^{s}}\cdot\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{w}},$

where $g(\mathbf{M}_{1},\mathbf{M}_{2})$ is any function which satisfies $g(\mathbf{M}_{1},\mathbf{M}_{2})\ll\left(\prod_{j=1}^{2k}m_{j}\right)^{\varepsilon}$ , then we have

(i) $T_{g,r,3}(s,w)$ is absolutely convergent.

(ii) We have

T_{g,r,3}(s,w)-T_{g,r,3}(x,y;s,w)\ll x^{-2s+\varepsilon}+y^{1-2w+\varepsilon}.

Proof.

Use the same argument of Lemma 2.5 in the first author [2]. ∎

Lemma 2.7.

Suppose $s=\sigma+it$ is a complex number, $f(n)$ is an arithmetic function with Dirichlet series

F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^{s}},

which is convergent for $\sigma>1$ . Then for any real number $x\geqslant 2$ and fixed real number $c>1$ , we have

\sum_{n\leqslant x}f(n)\left(1-\frac{n}{x}\right)=\frac{1}{2\pi i}\int_{(c)}F(s)\frac{x^{s}}{s(s+1)}ds.

Proof.

Using Theorem 5.1(the Perron’s formula) in Karatsuba [7] we can finish the proof. ∎

Lemma 2.8.

Let $T\geqslant 2$ be a real number and $\Delta_{3}(\cdot)$ be defined in (1.1), then we have

\int_{1}^{T}\Delta_{3}(x)dx\ll T^{7/6+\varepsilon}.

Proof.

Taking $f(n)=\tau_{3}(n)$ and $x=T$ in Lemma we have

\sum_{n\leqslant T}\tau_{3}(n)\left(T-n\right)=\frac{1}{2\pi i}\int_{(c)}\zeta^{3}(s)\frac{T^{s+1}}{s(s+1)}ds.

Then using (1.1) the left side becomes

TM_{3}(T)+T\Delta_{3}(T)-\sum_{n\leqslant T}n\tau_{3}(n)=TM_{3}(T)-\int_{1}^{T}u{M_{3}}^{\prime}(u)du+\int_{1}^{T}\Delta_{3}(u)du,

where we use partial summation to get

\sum_{n\leqslant T}n\tau_{3}(n)=\int_{1}^{T}u{M_{3}}^{\prime}(u)du+T\Delta_{3}(T)-\int_{1}^{T}\Delta_{3}(u)du.

Thus

	$\displaystyle\int_{1}^{T}\Delta_{3}(u)du=$	$\displaystyle\frac{1}{2\pi i}\int_{(c)}\zeta^{3}(s)\frac{T^{s+1}}{s(s+1)}ds-TM_{3}(T)+\int_{1}^{T}u{M_{3}}^{\prime}(u)du$
		$\displaystyle=\frac{1}{2\pi i}\int_{(c)}\zeta^{3}(s)\frac{T^{s+1}}{s(s+1)}ds-\int_{1}^{T}{M_{3}}(u)du$
		$\displaystyle=\frac{1}{2\pi i}\int_{(\sigma)}\zeta^{3}(s)\frac{T^{s+1}}{s(s+1)}ds,$

where $\sigma$ is a real number such that $0<\sigma<1$ .

It is well known that $\zeta(s)$ has the functional equation

\zeta(s)=\chi(s)\zeta(1-s),

where $\chi(s)$ satisfies $\chi(s)\ll t^{1/2-\sigma}$ for $s=\sigma+it$ and $0<\sigma<1/2$ . Thus for $0<\sigma<1/2$ ,

	$\displaystyle\int_{(\sigma)}\zeta^{3}(s)\frac{T^{s+1}}{s(s+1)}ds$	$\displaystyle=\int_{(\sigma)}\chi^{3}(s)\zeta^{3}(1-s)\frac{T^{s+1}}{s(s+1)}ds$
		$\displaystyle\ll T^{\sigma+1}\int_{-i\infty}^{i\infty}t^{-1/2-3\sigma}\zeta^{3}(1-\sigma+it)dt$
		$\displaystyle\ll T^{7/6+\varepsilon}\int_{-i\infty}^{i\infty}t^{-1-\varepsilon}\zeta^{3}(5/6-\varepsilon+it)dt$
		$\displaystyle\ll T^{7/6+\varepsilon}$

holds by taking $\sigma=1/6+\varepsilon$ , where the convergence of the latter integral can be obtained by integration by parts and the fourth power moment result of $\zeta(s)$ :

\int_{-i\infty}^{i\infty}\zeta^{4}(5/6-\varepsilon+it)dt\ll 1,

which can be found in Titchmarsh [9]. Hence we complete the proof. ∎

3. Proof of Theorem 1.1 and Expression of ${\Delta_{r,k}}(x)$

According to Lemma 2.3,

\tau_{k}(n_{1}\cdots n_{r})=\sum_{n_{1}=m_{1}d_{1},\cdots,n_{r}=m_{r}d_{r}}f_{r,k}(m_{1},\cdots,m_{r})\tau_{k}(d_{1})\cdots\tau_{k}(d_{r})

holds for $n_{1},\cdots,n_{r}\in\mathbb{N}$ , where $f_{r,k}$ is defined in Lemma 2.3.

Then we deduce by (1.1) that

$\displaystyle\sum_{n_{1},\cdots,n_{r}\leqslant x}\tau_{k}(n_{1}\cdots n_{r})$	$\displaystyle=\sum_{m_{1},\cdots,m_{r}\leqslant x}f_{r,k}(m_{1},\cdots,m_{r})\prod_{j=1}^{r}\left(\sum_{d_{j}\leqslant x/m_{j}}\tau_{k}(d_{j})\right)$	(3.1)
	$\displaystyle=\sum_{m_{1},\cdots,m_{r}\leqslant x}f_{r,k}(m_{1},\cdots,m_{r})\prod_{j=1}^{r}\left(M_{k}\left(\frac{x}{m_{j}}\right)+\Delta_{k}\left(\frac{x}{m_{j}}\right)\right)$
	$\displaystyle=\sum_{m_{1},\cdots,m_{r}\leqslant x}f_{r,k}(m_{1},\cdots,m_{r})\sum_{i=0}^{r}\binom{r}{i}\prod_{j=1}^{r-i}{M_{k}}\left(\frac{x}{m_{j}}\right)\prod_{j=1}^{i}{\Delta_{k}}\left(\frac{x}{m_{j}}\right).$

We evaluate the main term

\displaystyle M_{r,k}(x):=\sum_{m_{1},\cdots,m_{r}\leqslant x}f_{r,k}(m_{1},\cdots,m_{r})\prod_{j=1}^{r}{M_{k}}\left(\frac{x}{m_{j}}\right).

Since $M_{k}(u)=uP_{k-1}(\log u)$ with $P_{t}(u)$ a polynomial in u of degree $t$ , we have

\displaystyle\prod_{j=1}^{r}{M_{k}}\left(\frac{x}{m_{j}}\right)=\frac{x^{r}}{m_{1}\cdots m_{r}}\sum_{\ell=0}^{r(k-1)}C_{\ell}(\log m_{1},\cdots,\log m_{r})(\log x)^{\ell},

(3.2)

where

\displaystyle C_{\ell}(\log m_{1},\cdots,\log m_{r})=\sum_{j_{1},\cdots,j_{r}}c(j_{1},\cdots,j_{r})(\log m_{1})^{j_{1}}\cdots(\log m_{r})^{j_{r}},

the sum being over $0<j_{t}\leqslant k-1(1\leqslant j\leqslant r)$ and $c(j_{1},\cdots,j_{r})$ are computable constants. Thus we have

$\displaystyle M_{r,k}(x)$	$\displaystyle=x^{r}\sum_{\ell=0}^{r(k-1)}(\log x)^{\ell}\sum_{m_{1},\cdots,m_{r}\leqslant x}\frac{f_{r,k}(m_{1},\cdots,m_{r})C_{\ell}(\log m_{1},\cdots,\log m_{r})}{m_{1}\cdots m_{r}}$	(3.3)
	$\displaystyle=x^{r}\sum_{\ell=0}^{r(k-1)}d_{r,k,\ell}(\log x)^{\ell}$
	$\displaystyle\qquad-x^{r}\sum_{\ell=0}^{r(k-1)}(\log x)^{\ell}{\sum_{m_{1},\cdots,m_{r}}}^{\prime}\frac{f_{r,k}(m_{1},\cdots,m_{r})C_{\ell}(\log m_{1},\cdots,\log m_{r})}{m_{1}\cdots m_{r}},$

where

\displaystyle d_{r,k,\ell}:=\sum_{m_{1},\cdots,m_{r}=1}^{\infty}\frac{f_{r,k}(m_{1},\cdots,m_{r})C_{\ell}(\log m_{1},\cdots,\log m_{r})}{m_{1}\cdots m_{r}}

is convergent by choosing $g(m_{1},\cdots,m_{r})=C_{\ell}(\log m_{1},\cdots,\log m_{r})$ in Lemma 2.3, and where $\sum^{\prime}$ means there is at least one $j(1\leqslant j\leqslant r)$ such that $m_{j}>x$ . Without loss of generality, we suppose $m_{r}>x$ .

Since $\log n\ll n^{\varepsilon}$ , we obtain

		$\displaystyle x^{r}\sum_{\ell=0}^{r(k-1)}(\log x)^{\ell}{\sum_{m_{1},\cdots,m_{r}}}^{\prime}\frac{f_{r,k}(m_{1},\cdots,m_{r})C_{\ell}(\log m_{1},\cdots,\log m_{r})}{m_{1}\cdots m_{r}}$		(3.4)
		$\displaystyle\ll x^{r}\sum_{\ell=0}^{r(k-1)}(\log x)^{\ell}\sum_{m_{1},\cdots,m_{r-1}=1}^{\infty}\sum_{m_{r}>x}\frac{\|f_{r,k}(m_{1},\cdots,m_{r})\|(m_{1}\cdots m_{r})^{\varepsilon}}{m_{1}\cdots m_{r}}$
		$\displaystyle\ll x^{r}\sum_{\ell=0}^{r(k-1)}(\log x)^{\ell}\sum_{m_{1},\cdots,m_{r-1}=1}^{\infty}\sum_{m_{r}>x}\frac{\|f_{r,k}(m_{1},\cdots,m_{r})\|(m_{1}\cdots m_{r})^{\varepsilon}}{m_{1}\cdots m_{r-1}{m_{r}}^{3\varepsilon}}\times\frac{1}{{m_{r}}^{1-3\varepsilon}}$
		$\displaystyle\ll x^{r+\varepsilon-(1-3\varepsilon)}\sum_{m_{1},\cdots,m_{r}=1}^{\infty}\frac{\|f_{r,k}(m_{1},\cdots,m_{r})\|}{(m_{1}\cdots m_{r-1})^{1-\varepsilon}{m_{r}}^{2\varepsilon}}$
		$\displaystyle\ll x^{r-1+\varepsilon},$

the convergence of the latter series is obtained by Lemma 2.3.

From (3.3) and (3.4) we get

M_{r,k}(x)=x^{r}\sum_{\ell=0}^{r(k-1)}d_{r,k,\ell}(\log x)^{\ell}+O(x^{r-1+\varepsilon}).

(3.5)

By (1.1), (1.2) we obtain $M_{k}(x/m_{j})\ll(x/m_{j})^{1+\varepsilon}$ and $\Delta_{k}(x/m_{j})\ll(x/m_{j})^{\alpha_{k}+\varepsilon}$ for every $1\leqslant j\leqslant r$ , thus the terms in (3.1) which contains two or more $\Delta_{k}(x/m_{j})$ are

		$\displaystyle\ll\sum_{m_{1},\cdots,m_{r}\leqslant x}\|f_{r,k}(m_{1},\cdots,m_{r})\|\left(\prod_{j=1}^{r-2}M_{k}\left(\frac{x}{m_{j}}\right)\right)\left\|\Delta_{k}\left(\frac{x}{m_{r-1}}\right)\right\|\left\|\Delta_{k}\left(\frac{x}{m_{r}}\right)\right\|$		(3.6)
		$\displaystyle\ll x^{r-2+2\alpha_{k}+\varepsilon}\sum_{m_{1},\cdots,m_{r}\leqslant x}\frac{\|f_{r,k}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-2}(m_{r-1}m_{r})^{\alpha_{k}}}$
		$\displaystyle=x^{r-2+2\alpha_{k}+\varepsilon}\sum_{m_{1},\cdots,m_{r}\leqslant x}\frac{\|f_{r,k}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-2}(m_{r-1}m_{r})^{\alpha_{k}}}\times\frac{(m_{r-1}m_{r})^{\frac{1}{2}-\alpha_{k}+\varepsilon}}{(m_{r-1}m_{r})^{\frac{1}{2}-\alpha_{k}+\varepsilon}}$
		$\displaystyle\ll x^{r-2+2\alpha_{k}+2(\frac{1}{2}-\alpha_{k})+\varepsilon}\sum_{m_{1},\cdots,m_{r}=1}^{\infty}\frac{\|f_{r,k}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-2}(m_{r-1}m_{r})^{\frac{1}{2}+\varepsilon}}$
		$\displaystyle\ll x^{r-1+\varepsilon}$

by using Lemma 2.3.

Above all,

\Delta_{r,k}(x)={\Delta_{r,k}}^{*}(x)+O(x^{r-1+\varepsilon})

(3.7)

holds from (3.1), (3.3), (3.5) and (3.6), where

{\Delta_{r,k}}^{*}(x)=r\sum_{m_{1},\cdots,m_{r}\leqslant x}f_{r,k}(m_{1},\cdots,m_{r})\prod_{j=1}^{r-1}M_{k}\left(\frac{x}{m_{j}}\right)\Delta_{k}\left(\frac{x}{m_{r}}\right).

Similar to (3.6) we obtain

	$\displaystyle{\Delta_{r,k}}^{*}(x)$	$\displaystyle\ll\sum_{m_{1},\cdots,m_{r}\leqslant x}\|f_{r,k}(m_{1},\cdots,m_{r})\|\left(\prod_{j=1}^{r-1}M_{k}\left(\frac{x}{m_{j}}\right)\right)\left\|\Delta_{k}\left(\frac{x}{m_{r}}\right)\right\|$
		$\displaystyle\ll x^{r-1+\alpha_{k}+\varepsilon}\sum_{m_{1},\cdots,m_{r}\leqslant x}\frac{\|f_{r,k}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}{m_{r}}^{\alpha_{k}}}$
		$\displaystyle\ll x^{r-1+\alpha_{k}+\varepsilon}\sum_{m_{1},\cdots,m_{r}=1}^{\infty}\frac{\|f_{r,k}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}{m_{r}}^{\alpha_{k}}}$
		$\displaystyle\ll x^{r-1+\alpha_{k}+\varepsilon},$

by using Lemma 2.3. Hence we proof the Theorem 1.1.

4. Proof of Theorem 1.6

4.1. Mean Square of ${\Delta_{r,k}}^{*}(x)$

Let $T\geqslant 2$ be a large real number, and $\mathbf{M}_{1}$ , $\mathbf{M}_{2}$ , $D_{1}$ and $D_{2}$ be defined in Lemma 2.6.

For $k\geqslant 4$ ,

	$\displaystyle\int_{T}^{2T}({\Delta_{r,k}}^{*}(x))^{2}dx$	$\displaystyle\ll\int_{T}^{2T}\sum_{m_{1},\cdots,m_{2r}\leqslant 2T}\|f_{r,k}(\mathbf{M}_{1})f_{r,k}(\mathbf{M}_{2})\|\prod_{j=1}^{r-1}M_{k}\left(\frac{x}{m_{j}}\right)\prod_{j=r+1}^{2r-1}M_{k}\left(\frac{x}{m_{j}}\right)$
		$\displaystyle\qquad\times\left\|\Delta_{k}\left(\frac{x}{m_{r}}\right)\Delta_{k}\left(\frac{x}{m_{2r}}\right)\right\|dx.$

Change the order of integration and summation, and use (1.1) we obtain

	$\displaystyle\int_{T}^{2T}({\Delta_{r,k}}^{*}(x))^{2}dx$	$\displaystyle\ll\sum_{m_{1},\cdots,m_{2r}\leqslant 2T}\|f_{r,k}(\mathbf{M}_{1})f_{r,k}(\mathbf{M}_{2})\|\int_{T}^{2T}\prod_{j=1}^{r-1}M_{k}\left(\frac{x}{m_{j}}\right)\prod_{j=r+1}^{2r-1}M_{k}\left(\frac{x}{m_{j}}\right)$
		$\displaystyle\qquad\times\left\|\Delta_{k}\left(\frac{x}{m_{r}}\right)\Delta_{k}\left(\frac{x}{m_{2r}}\right)\right\|dx$
		$\displaystyle\ll\sum_{m_{1},\cdots,m_{2r}\leqslant 2T}\frac{\|f_{r,k}(\mathbf{M}_{1})f_{r,k}(\mathbf{M}_{2})\|}{D_{1}D_{2}}\int_{T}^{2T}x^{2r-2+\varepsilon}\left\|\Delta_{k}\left(\frac{x}{m_{r}}\right)\Delta_{k}\left(\frac{x}{m_{2r}}\right)\right\|dx$
		$\displaystyle\ll T^{2r-2+\varepsilon}\sum_{m_{1},\cdots,m_{2r}\leqslant 2T}\frac{\|f_{r,k}(\mathbf{M}_{1})f_{r,k}(\mathbf{M}_{2})\|}{D_{1}D_{2}}\int_{T}^{2T}\left\|\Delta_{k}\left(\frac{x}{m_{r}}\right)\Delta_{k}\left(\frac{x}{m_{2r}}\right)\right\|dx.$

By the Cauchy-Schwarz’s inequality and (1.2) we deduce that

		$\displaystyle\int_{T}^{2T}\left\|\Delta_{k}\left(\frac{x}{m_{r}}\right)\Delta_{k}\left(\frac{x}{m_{2r}}\right)\right\|dx$
	$\displaystyle\ll$	$\displaystyle\left(\int_{T}^{2T}{\Delta_{k}}^{2}\left(\frac{x}{m_{r}}\right)dx\right)^{1/2}\left(\int_{T}^{2T}{\Delta_{k}}^{2}\left(\frac{x}{m_{2r}}\right)dx\right)^{1/2}$
	$\displaystyle\ll$	$\displaystyle\left(m_{r}\int_{\frac{T}{m_{r}}}^{\frac{2T}{m_{r}}}{\Delta_{k}}^{2}\left(u\right)du\right)^{1/2}\left(m_{2r}\int_{\frac{T}{m_{2r}}}^{\frac{2T}{m_{2r}}}{\Delta_{k}}^{2}\left(u\right)du\right)^{1/2}$
	$\displaystyle\ll$	$\displaystyle\frac{T^{1+2\beta_{k}+\varepsilon}}{{m_{r}}^{\beta_{k}}{m_{2r}}^{\beta_{k}}}.$

Thus

	$\displaystyle\int_{T}^{2T}({\Delta_{r,k}}^{*}(x))^{2}dx$	$\displaystyle\ll T^{2r-1+2\beta_{k}+\varepsilon}\sum_{m_{1},\cdots,m_{2r}\leqslant 2T}\frac{\|f_{r,k}(\mathbf{M}_{1})f_{r,k}(\mathbf{M}_{2})\|}{D_{1}{m_{r}}^{\beta_{k}}D_{2}{m_{2r}}^{\beta_{k}}}$
		$\displaystyle\ll T^{2r-1+2\beta_{k}+\varepsilon}\left(\sum_{m_{1},\cdots,m_{r}=1}^{\infty}\frac{\|f_{r,k}(\mathbf{M}_{1})\|}{D_{1}{m_{r}}^{\beta_{k}}}\right)^{2}$
		$\displaystyle\ll T^{2r-1+2\beta_{k}+\varepsilon},$

which means we complete the proof of Theorem in the case of $k\geqslant 4$ by replacing $T$ by $T/2$ , $T/2^{2}$ , and so on, and adding up all the results.

5. Proof of Theorem 1.2

5.1. Mean Square of ${\Delta_{r,3}}^{*}(x)$

In order to evaluate

\int_{T}^{2T}({\Delta_{r,3}}^{*}(x))^{2}dx,

we divide the ${\Delta_{r,3}}^{*}(x)$ into three parts, namely

{\Delta_{r,3}}^{*}(x)=M_{1}+O(M_{2}+M_{3}),

where

		$\displaystyle M_{1}=M_{1}(x,y):=r\sum_{m_{1},\cdots,m_{r}\leqslant y}f_{r,3}(m_{1},\cdots,m_{r})\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\Delta_{3}\left(\frac{x}{m_{r}}\right),$
		$\displaystyle M_{2}=M_{2}(x,y):=\sum_{\begin{subarray}{c}m_{1},\cdots,m_{r}\leqslant x\\ m_{1}>y\end{subarray}}\|f_{r,3}(m_{1},\cdots,m_{r})\|\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\right\|,$
		$\displaystyle M_{3}=M_{3}(x,y):=\sum_{\begin{subarray}{c}m_{1},\cdots,m_{r}\leqslant x\\ m_{r}>y\end{subarray}}\|f_{r,3}(m_{1},\cdots,m_{r})\|\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\right\|,$

and $y$ is a parameter such that $T^{\varepsilon}\ll y\ll T$ . Thus

\int_{T}^{2T}({\Delta_{r,3}}^{*}(x))^{2}dx=\int_{T}^{2T}{M_{1}}^{2}dx+O\left(\int_{T}^{2T}(M_{1}M_{2}+M_{1}M_{3})dx\right)+O\left(\int_{T}^{2T}({M_{2}}^{2}+{M_{3}}^{2})dx\right).

(5.1)

Let $\mathbf{M}_{1},\mathbf{M}_{2},D_{1},D_{2}$ be defined in Lemma 2.6, then change the order of summation and integration, by (1.1) we obtain

	$\displaystyle\int_{T}^{2T}{M_{3}}^{2}dx=$	$\displaystyle\int_{T}^{2T}\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant 2T\\ m_{r},m_{2r}>y\end{subarray}}\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\prod_{j=r+1}^{2r-1}M_{3}\left(\frac{x}{m_{j}}\right)$
		$\displaystyle\times\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\Delta_{3}\left(\frac{x}{m_{2r}}\right)\right\|dx$
	$\displaystyle=$	$\displaystyle\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant 2T\\ m_{r},m_{2r}>y\end{subarray}}\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|\int_{T}^{2T}\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\prod_{j=r+1}^{2r-1}M_{3}\left(\frac{x}{m_{j}}\right)$
		$\displaystyle\times\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\Delta_{3}\left(\frac{x}{m_{2r}}\right)\right\|dx$
	$\displaystyle\ll$	$\displaystyle\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant 2T\\ m_{r},m_{2r}>y\end{subarray}}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}}\int_{T}^{2T}x^{2r-2+\varepsilon}\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\Delta_{3}\left(\frac{x}{m_{2r}}\right)\right\|dx$
	$\displaystyle\ll$	$\displaystyle T^{2r-2+\varepsilon}\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant 2T\\ m_{r},m_{2r}>y\end{subarray}}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}}\int_{T}^{2T}\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\Delta_{3}\left(\frac{x}{m_{2r}}\right)\right\|dx,$

using Cauchy-Schwarz’s inequality and (1.2) we deduce that

	$\displaystyle\int_{T}^{2T}\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\Delta_{3}\left(\frac{x}{m_{2r}}\right)\right\|dx$	$\displaystyle\ll\left(\int_{T}^{2T}{\Delta_{3}}^{2}\left(\frac{x}{m_{r}}\right)dx\right)^{1/2}\left(\int_{T}^{2T}{\Delta_{3}}^{2}\left(\frac{x}{m_{2r}}\right)dx\right)^{1/2}$
		$\displaystyle\ll\left(m_{r}\int_{\frac{T}{m_{r}}}^{\frac{2T}{m_{r}}}{\Delta_{3}}^{2}\left(u\right)du\right)^{1/2}\left(m_{2r}\int_{\frac{T}{m_{2r}}}^{\frac{2T}{m_{2r}}}{\Delta_{3}}^{2}\left(u\right)du\right)^{1/2}$
		$\displaystyle\ll\frac{T^{5/3+\varepsilon}}{{m_{r}}^{1/3}{m_{2r}}^{1/3}},$

then by Lemma 2.3 we have

$\displaystyle\int_{T}^{2T}{M_{3}}^{2}dx$	$\displaystyle\ll T^{2r-1/3+\varepsilon}\left(\sum_{\begin{subarray}{c}m_{1},\cdots,m_{r}\leqslant 2T\\ m_{r}>y\end{subarray}}\frac{\|f_{r,3}(\mathbf{M}_{1})\|}{D_{1}{m_{r}}^{1/3}}\right)^{2}$	(5.2)
	$\displaystyle\ll T^{2r-1/3+\varepsilon}\left(\sum_{\begin{subarray}{c}m_{1},\cdots,m_{r}\leqslant 2T\\ m_{r}>y\end{subarray}}\frac{\|f_{r,3}(\mathbf{M}_{1})\|}{D_{1}{m_{r}}^{\varepsilon}}\times\frac{1}{{m_{r}}^{1/3-\varepsilon}}\right)^{2}$
	$\displaystyle\ll T^{2r-1/3+\varepsilon}y^{-2/3+\varepsilon}\left(\sum_{\begin{subarray}{c}m_{1},\cdots,m_{r}\leqslant 2T\\ m_{r}>y\end{subarray}}\frac{\|f_{r,3}(\mathbf{M}_{1})\|}{D_{1}{m_{r}}^{\varepsilon}}\right)^{2}$
	$\displaystyle\ll T^{2r-1/3+\varepsilon}y^{-2/3}.$

Let ${D_{1}}^{\prime}=D_{1}/m_{1}$ , similarly we get

$\displaystyle\int_{T}^{2T}{M_{2}}^{2}dx$	$\displaystyle\ll T^{2r-1/3+\varepsilon}\left(\sum_{\begin{subarray}{c}m_{1},\cdots,m_{r}\leqslant 2T\\ m_{1}>y\end{subarray}}\frac{\|f_{r,3}(\mathbf{M}_{1})\|}{D_{1}{m_{r}}^{1/3}}\right)^{2}$	(5.3)
	$\displaystyle\ll T^{2r-1/3+\varepsilon}\left(\sum_{\begin{subarray}{c}m_{1},\cdots,m_{r}\leqslant 2T\\ m_{1}>y\end{subarray}}\frac{\|f_{r,3}(\mathbf{M}_{1})\|}{{m_{1}}^{2/3+\varepsilon}{D_{1}}^{\prime}{m_{r}}^{1/3}}\times\frac{1}{{m_{1}}^{1/3-\varepsilon}}\right)^{2}$
	$\displaystyle\ll T^{2r-1/3+\varepsilon}y^{-2/3+\varepsilon}\left(\sum_{\begin{subarray}{c}m_{1},\cdots,m_{r}\leqslant 2T\\ m_{1}>y\end{subarray}}\frac{\|f_{r,3}(\mathbf{M}_{1})\|}{{m_{1}}^{2/3+\varepsilon}{D_{1}}^{\prime}{m_{r}}^{1/3}}\right)^{2}$
	$\displaystyle\ll T^{2r-1/3+\varepsilon}y^{-2/3}.$

By (5.1), (5.2), (5.3) we conclude that

\int_{T}^{2T}({\Delta_{r,k}}^{*}(x))^{2}dx=\int_{T}^{2T}{M_{1}}^{2}dx+O\left(\int_{T}^{2T}(M_{1}M_{2}+M_{1}M_{3})dx\right)+O(T^{2r-1/3+\varepsilon}y^{-2/3}).

It remains to evaluate $\int_{T}^{2T}{M_{1}}^{2}dx$ , and then it will follows that $\int_{T}^{2T}M_{1}M_{2}dx$ and $\int_{T}^{2T}M_{1}M_{3}dx$ can be estimated by the Cauchy-Schwarz’s inequality.

5.2. Mean Square of $M_{1}(x,y)$

Let $N$ be a parameter such that $1\ll N\ll x/m_{r}\ll x/y$ , then

	$\displaystyle M_{1}(x,y)$	$\displaystyle=r\sum_{m_{1},\cdots,m_{r}\leqslant y}f_{r,3}(\mathbf{M_{1}})\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\Delta_{3}\left(\frac{x}{m_{r}}\right)$		(5.4)
		$\displaystyle:=M_{11}(x,y,N)+O(M_{12}(x,y,N)),$		(5.4)

where

		$\displaystyle\qquad M_{11}(x,y,N)$
		$\displaystyle=r\sum_{m_{1},\cdots,m_{r}\leqslant y}f_{r,3}(\mathbf{M_{1}})\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\frac{x^{1/3}}{\sqrt{3}\pi m_{r}^{1/3}}\sum_{n\leqslant N}\frac{\tau_{3}(n)}{n^{2/3}}\cos\left(6\pi\left(\frac{nx}{m_{r}}\right)^{\frac{1}{3}}\right),$

and

\displaystyle M_{12}(x,y,N)=\sum_{m_{1},\cdots,m_{r}\leqslant y}|f_{r,3}(\mathbf{M}_{1})|\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\left|\delta_{32}\left(\frac{x}{m_{r}},N\right)\right|,

where $\delta_{32}(\cdot,\cdot)$ is defined in Lemma 2.1. Thus

		$\displaystyle\qquad\int_{T}^{2T}{M_{1}}^{2}(x,y)dx$		(5.5)
		$\displaystyle=\int_{T}^{2T}{M_{11}}^{2}(x,y,N)dx+O\left(\int_{T}^{2T}(M_{11}(x,y,N)M_{12}(x,y,N)+{M_{12}}^{2}(x,y,N))dx\right).$		(5.5)

We are going to evaluate the mean square of $M_{11}$ . Using (3.2) we obtain

		$\displaystyle\qquad M_{11}^{2}(x,y,N)$
		$\displaystyle=\frac{r^{2}x^{2/3}}{3\pi^{2}}\sum_{m_{1},\cdots,m_{2r}\leqslant y}\frac{f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})}{(m_{r}m_{2r})^{1/3}}\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\prod_{j=r+1}^{2r-1}M_{3}\left(\frac{x}{m_{j}}\right)\sum_{n_{1},n_{2}\leqslant N}\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}}$
		$\displaystyle\qquad\times\cos\left(6\pi\left(\frac{n_{1}x}{m_{r}}\right)^{1/3}\right)\cos\left(6\pi\left(\frac{n_{2}x}{m_{2r}}\right)^{1/3}\right)$
		$\displaystyle=\frac{r^{2}x^{2r-4/3}}{3\pi^{2}}\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}(\log x)^{\ell_{1}+\ell_{2}}$
		$\displaystyle\qquad\times\sum_{m_{1},\cdots,m_{2r}\leqslant y}\frac{f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})C_{\ell_{1}}(\log m_{1},\cdots,\log m_{r-1})C_{\ell_{2}}(\log m_{m_{r+1}},\cdots,\log m_{2r-1})}{D_{1}D_{2}(m_{r}m_{2r})^{1/3}}$
		$\displaystyle\qquad\times\sum_{n_{1},n_{2}\leqslant N}\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}}\cos\left(6\pi\left(\frac{n_{1}x}{m_{r}}\right)^{1/3}\right)\cos\left(6\pi\left(\frac{n_{2}x}{m_{2r}}\right)^{1/3}\right).$

We denote $C_{\ell_{1}}(\log m_{1},\cdots,\log m_{r-1})C_{\ell_{2}}(\log m_{m_{r+1}},\cdots,\log m_{2r-1})$ by $C_{\ell_{1},\ell_{2}}(\mathbf{M}_{1},\mathbf{M}_{2})$ , then using $\cos\alpha\cos\beta=\frac{1}{2}(\cos(\alpha-\beta)+\cos(\alpha+\beta))$ we get

M_{11}^{2}(x,y,N)=S_{0}(x,y,N)+S_{1}(x,y,N)+S_{2}(x,y,N),

(5.6)

where

	$\displaystyle S_{0}(x,y,N)$	$\displaystyle=\frac{r^{2}x^{2r-4/3}}{6\pi^{2}}\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}(\log x)^{\ell_{1}+\ell_{2}}$
		$\displaystyle\qquad\times\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant y\\ n_{1},n_{2}\leqslant N\\ \frac{n_{1}}{m_{r}}=\frac{n_{2}}{m_{2r}}\end{subarray}}\frac{f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})C_{\ell_{1},\ell_{2}}(\mathbf{M}_{1},\mathbf{M}_{2})}{D_{1}D_{2}(m_{r}m_{2r})^{1/3}}\cdot\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}},$
	$\displaystyle S_{1}(x,y,N)$	$\displaystyle=\frac{r^{2}x^{2r-4/3}}{6\pi^{2}}\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}(\log x)^{\ell_{1}+\ell_{2}}$
		$\displaystyle\qquad\times\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant y\\ n_{1},n_{2}\leqslant N\\ n_{1}m_{2r}\neq n_{2}m_{r}\end{subarray}}\frac{f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})C_{\ell_{1},\ell_{2}}(\mathbf{M}_{1},\mathbf{M}_{2})}{D_{1}D_{2}(m_{r}m_{2r})^{1/3}}\cdot\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}}$
		$\displaystyle\qquad\times\cos\left(6\pi\left(\left(\frac{n_{1}x}{m_{r}}\right)^{1/3}-\left(\frac{n_{2}x}{m_{2r}}\right)^{1/3}\right)\right),$

and

	$\displaystyle S_{2}(x,y,N)$	$\displaystyle=\frac{r^{2}x^{2r-4/3}}{6\pi^{2}}\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}(\log x)^{\ell_{1}+\ell_{2}}$
		$\displaystyle\qquad\times\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant y\\ n_{1},n_{2}\leqslant N\end{subarray}}\frac{f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})C_{\ell_{1},\ell_{2}}(\mathbf{M}_{1},\mathbf{M}_{2})}{D_{1}D_{2}(m_{r}m_{2r})^{1/3}}\cdot\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}}$
		$\displaystyle\qquad\times\cos\left(6\pi\left(\left(\frac{n_{1}x}{m_{r}}\right)^{1/3}+\left(\frac{n_{2}x}{m_{2r}}\right)^{1/3}\right)\right).$

Then it turns to deal with

\displaystyle\int_{0}=\int_{T}^{2T}S_{0}(x,y,N)dx,\qquad\int_{1}=\int_{T}^{2T}S_{1}(x,y,N)dx,\qquad\int_{2}=\int_{T}^{2T}S_{2}(x,y,N)dx.

5.2.1. Evaluation of $\int_{0}$

Since $C_{\ell_{1},\ell_{2}}(\mathbf{M}_{1},\mathbf{M}_{2})$ satisfies

C_{\ell_{1},\ell_{2}}(\mathbf{M}_{1},\mathbf{M}_{2})\ll\left(\prod_{j=1}^{2r}m_{j}\right)^{\varepsilon},

we choose $g(\mathbf{M}_{1},\mathbf{M}_{2}))=g_{\ell_{1},\ell_{2}}(\mathbf{M}_{1},\mathbf{M}_{2})=C_{\ell_{1},\ell_{2}}(\mathbf{M}_{1},\mathbf{M}_{2})$ , $s=1/3$ , $w=2/3$ in Lemma 2.6, then

	$\displaystyle S_{0}(x,y,N)$	$\displaystyle=\frac{r^{2}x^{2r-4/3}}{6\pi^{2}}\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}(\log x)^{\ell_{1}+\ell_{2}}T_{g,r,3}\left(y,N;\frac{1}{3},\frac{2}{3}\right)$
		$\displaystyle=\frac{r^{2}x^{2r-4/3}}{6\pi^{2}}\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}(\log x)^{\ell_{1}+\ell_{2}}\left(T_{g,r,3}\left(\frac{1}{3},\frac{2}{3}\right)+O(y^{-2/3+\varepsilon}+N^{-1/3+\varepsilon})\right).$

Since $g(\mathbf{M}_{1},\mathbf{M}_{2})$ is related to $\ell_{1},\ell_{2}$ , we denote $T_{g,r,3}\left(\frac{1}{3},\frac{2}{3}\right)$ by $D_{r,3,\ell_{1},\ell_{2}}$ , therefore

S_{0}(x,y,N)=\frac{r^{2}x^{2r-4/3}}{6\pi^{2}}Q_{4r-4}(\log x)+O(x^{2r-4/3+\varepsilon}(y^{-2/3}+N^{-1/3})),

where

Q_{4r-4}(t)=\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}D_{r,3,\ell_{1},\ell_{2}}t^{\ell_{1}+\ell_{2}}

is a polynomial of degree $4r-4$ . Then it follows that

\int_{0}=\frac{r^{2}}{6\pi^{2}}\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}D_{r,3,\ell_{1},\ell_{2}}\int_{T}^{2T}x^{2r-4/3}(\log x)^{\ell_{1}+\ell_{2}}dx+O(T^{2r-1/3+\varepsilon}(y^{-2/3}+N^{-1/3})).

(5.7)

5.2.2. Estimates of $\int_{1}$ and $\int_{2}$

For $\int_{2}$ , change the order of integration and summation we have

	$\displaystyle\int_{2}$	$\displaystyle=\frac{r^{2}}{6\pi^{2}}\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant y\\ n_{1},n_{2}\leqslant N\end{subarray}}\frac{f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})C_{\ell_{1},\ell_{2}}(\mathbf{M}_{1},\mathbf{M}_{2})}{D_{1}D_{2}(m_{r}m_{2r})^{1/3}}\cdot\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}}$
		$\displaystyle\qquad\times\int_{T}^{2T}x^{2r-4/3}(\log x)^{\ell_{1}+\ell_{2}}\cos\left(6\pi\left(\left(\frac{n_{1}x}{m_{r}}\right)^{1/3}+\left(\frac{n_{2}x}{m_{2r}}\right)^{1/3}\right)\right)dx.$

Choosing $F(\cdot)=\cos(\cdot)$ , $G(x)=x^{2r-4/3}(\log x)^{\ell_{1}+\ell_{2}}$ and $m(x)=(n_{1}x/m_{r})^{1/3}+(n_{2}x/m_{2r})^{1/3}$ in Lemma 2.4, then

	$\displaystyle\int_{2}$	$\displaystyle\ll\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant y\\ n_{1},n_{2}\leqslant N\end{subarray}}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})C_{\ell_{1},\ell_{2}}(\mathbf{M}_{1},\mathbf{M}_{2})\|}{D_{1}D_{2}(m_{r}m_{2r})^{1/3}}\cdot\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}}$
		$\displaystyle\qquad\times T^{2r-4/3+\varepsilon}\cdot\frac{T^{2/3}}{(n_{1}/m_{r})^{1/3}+(n_{2}/m_{2r})^{1/3}}.$

We use $C_{\ell_{1},\ell_{2}}(\mathbf{M}_{1},\mathbf{M}_{2})\ll\left(\prod_{j=1}^{2r}m_{j}\right)^{\varepsilon}\ll y^{\varepsilon}\ll T^{\varepsilon}$ and $a^{2}+b^{2}\geqslant 2ab$ to get

$\displaystyle\int_{2}$	$\displaystyle\ll T^{2r-2/3+\varepsilon}\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant y\\ n_{1},n_{2}\leqslant N\end{subarray}}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}(m_{r}m_{2r})^{1/3}}\cdot\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}}\left(\frac{m_{r}m_{2r}}{n_{1}n_{2}}\right)^{1/6}$	(5.8)
	$\displaystyle\ll T^{2r-2/3+\varepsilon}\sum_{n_{1},n_{2}\leqslant N}\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{5/6}}\sum_{m_{1},\cdots,m_{2r}\leqslant y}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}(m_{r}m_{2r})^{1/6}}$
	$\displaystyle\ll T^{2r-2/3+\varepsilon}N^{1/3}\left(\sum_{m_{1},\cdots,m_{r}=1}^{\infty}\frac{\|f_{r,3}(\mathbf{M}_{1})\|}{D_{1}{m_{r}}^{1/6}}\right)^{2}$
	$\displaystyle\ll T^{2r-2/3+\varepsilon}N^{1/3},$

where we use partial summation on $n_{1},n_{2}$ and the convergence of the latter series is obtained by Lemma 2.3.

Then we turn to estimate $\int_{1}$ , similar to $\int_{2}$ , we have

$\displaystyle\int_{1}$	$\displaystyle\ll\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant y\\ n_{1},n_{2}\leqslant N\\ n_{1}m_{2r}\neq n_{2}m_{r}\end{subarray}}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})C_{\ell_{1},\ell_{2}}(\mathbf{M}_{1},\mathbf{M}_{2})\|}{D_{1}D_{2}(m_{r}m_{2r})^{1/3}}\cdot\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}}$	(5.9)
	$\displaystyle\qquad\times T^{2r-4/3+\varepsilon}\cdot\frac{T^{2/3}}{\left\|(n_{1}/m_{r})^{1/3}-(n_{2}/m_{2r})^{1/3}\right\|}$
	$\displaystyle\ll T^{2r-2/3+\varepsilon}\sum_{m_{1},\cdots,m_{2r}\leqslant y}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}(m_{r}m_{2r})^{1/3}}\sum_{\begin{subarray}{c}n_{1},n_{2}\leqslant N\\ n_{1}m_{2r}\neq n_{2}m_{r}\end{subarray}}\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}\left\|(n_{1}/m_{r})^{1/3}-(n_{2}/m_{2r})^{1/3}\right\|}$
	$\displaystyle:=T^{2r-2/3+\varepsilon}\sum_{m_{1},\cdots,m_{2r}\leqslant y}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}(m_{r}m_{2r})^{1/3}}(R_{1}+R_{2}),$

where

	$\displaystyle R_{1}$	$\displaystyle=\sum_{\begin{subarray}{c}n_{1},n_{2}\leqslant N\\ n_{1}m_{2r}\neq n_{2}m_{r}\\ \left\|\left(\frac{n_{1}}{m_{r}}\right)^{1/3}-\left(\frac{n_{2}}{m_{2r}}\right)^{1/3}\right\|<\frac{1}{10}\left(\frac{n_{1}n_{2}}{m_{r}m_{2r}}\right)^{1/6}\end{subarray}}\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}\left\|(n_{1}/m_{r})^{1/3}-(n_{2}/m_{2r})^{1/3}\right\|}$
	$\displaystyle R_{2}$	$\displaystyle=\sum_{\begin{subarray}{c}n_{1},n_{2}\leqslant N\\ n_{1}m_{2r}\neq n_{2}m_{r}\\ \left\|\left(\frac{n_{1}}{m_{r}}\right)^{1/3}-\left(\frac{n_{2}}{m_{2r}}\right)^{1/3}\right\|\geqslant\frac{1}{10}\left(\frac{n_{1}n_{2}}{m_{r}m_{2r}}\right)^{1/6}\end{subarray}}\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}\left\|(n_{1}/m_{r})^{1/3}-(n_{2}/m_{2r})^{1/3}\right\|}.$

Then we have

R_{2}\ll\sum_{n_{1},n_{2}\leqslant N}\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}}\cdot\left(\frac{m_{r}m_{2r}}{n_{1}n_{2}}\right)^{1/6}\ll(m_{r}m_{2r})^{1/6}N^{1/3+\varepsilon}

(5.10)

by using partial summation.

For $R_{1}$ , using the Lagrange’s Mean Value Theorem, for $r_{1},r_{2}\in\mathbb{R}$ ,

|{r_{1}}^{1/3}-{r_{2}}^{1/3}|\asymp(r_{1}r_{2})^{-1/3}|r_{1}-r_{2}|

holds when $r_{1}\asymp r_{2}$ . Thus choosing $r_{1}=n_{1}/m_{r}$ , $r_{2}=n_{2}/m_{2r}$ we get

	$\displaystyle R_{1}$	$\displaystyle\ll\sum_{\begin{subarray}{c}n_{1},n_{2}\leqslant N\\ n_{1}m_{2r}\neq n_{2}m_{r}\end{subarray}}\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}\left\|n_{1}/m_{r}-n_{2}/m_{2r}\right\|}\cdot\left(\frac{m_{r}m_{2r}}{n_{1}n_{2}}\right)^{-1/3}$
		$\displaystyle=(m_{r}m_{2r})^{2/3}\sum_{\begin{subarray}{c}n_{1},n_{2}\leqslant N\\ n_{1}m_{2r}\neq n_{2}m_{r}\end{subarray}}\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{1/3}\left\|n_{1}m_{2r}-n_{2}m_{r}\right\|}.$

Let $N_{1},N_{2}$ satisfy $1\leqslant N_{1},N_{2}\leqslant N$ . Then using $\tau_{3}(n)\ll n^{\varepsilon}$ we deduce that

$\displaystyle R_{1}$	$\displaystyle\ll(m_{r}m_{2r})^{2/3}\log^{2}N\sum_{\begin{subarray}{c}n_{1}\sim N_{1},n_{2}\sim N_{2}\\ n_{1}m_{2r}\neq n_{2}m_{r}\end{subarray}}\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{1/3}\left\|n_{1}m_{2r}-n_{2}m_{r}\right\|}$	(5.11)
	$\displaystyle\ll\frac{(m_{r}m_{2r})^{2/3}}{(N_{1}N_{2})^{1/3}}N^{\varepsilon}\sum_{\begin{subarray}{c}n_{1}\sim N_{1},n_{2}\sim N_{2}\\ n_{1}m_{2r}\neq n_{2}m_{r}\end{subarray}}\frac{1}{\left\|n_{1}m_{2r}-n_{2}m_{r}\right\|}$
	$\displaystyle\ll\frac{(m_{r}m_{2r})^{2/3}}{(N_{1}N_{2})^{1/3}}N^{\varepsilon}\cdot(N_{1}N_{2})^{1/2}\log N$
	$\displaystyle\ll(m_{r}m_{2r})^{2/3}N^{1/3+\varepsilon},$

where we use Lemma 2.5 in the case that $a=m_{2r},b=m_{r}$ .

Therefore we conclude that

R_{1}+R_{2}\ll(m_{r}m_{2r})^{2/3}N^{1/3+\varepsilon}

by (5.10) and (5.11). Thus we obtain by (5.9) and Lemma 2.3 that

$\displaystyle\int_{1}$	$\displaystyle\ll T^{2r-2/3+\varepsilon}N^{1/3}\sum_{m_{1},\cdots,m_{2r}\leqslant y}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}(m_{r}m_{2r})^{1/3}}\cdot(m_{r}m_{2r})^{2/3}$	(5.12)
	$\displaystyle\ll T^{2r-2/3+\varepsilon}N^{1/3}\sum_{m_{1},\cdots,m_{2r}\leqslant y}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|(m_{r}m_{2r})^{1/3+\varepsilon}}{D_{1}D_{2}(m_{r}m_{2r})^{\varepsilon}}$
	$\displaystyle\ll T^{2r-2/3+\varepsilon}N^{1/3}y^{2/3}\sum_{m_{1},\cdots,m_{2r}\leqslant y}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}(m_{r}m_{2r})^{\varepsilon}}$
	$\displaystyle\ll T^{2r-2/3+\varepsilon}N^{1/3}y^{2/3}\left(\sum_{m_{1},\cdots,m_{r}=1}^{\infty}\frac{\|f_{r,3}(\mathbf{M}_{1})\|}{D_{1}{m_{r}}^{\varepsilon}}\right)^{2}$
	$\displaystyle\ll T^{2r-2/3+\varepsilon}N^{1/3}y^{2/3}.$

Above all, combining (5.6), (5.7), (5.8), (5.12) and taking $y=N^{1/2}$ we obtain

	$\displaystyle\int_{T}^{2T}{M_{11}}^{2}(x,y,N)dx=$	$\displaystyle\frac{r^{2}}{6\pi^{2}}\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}D_{r,3,\ell_{1},\ell_{2}}\int_{T}^{2T}x^{2r-4/3}(\log x)^{\ell_{1}+\ell_{2}}dx$		(5.13)
		$\displaystyle\qquad+O(T^{2r-1/3+\varepsilon}N^{-1/3})+O(T^{2r-2/3+\varepsilon}N^{2/3}).$		(5.13)

5.2.3. Other terms in (5.5)

From (1.1) and (5.4) we get

		$\displaystyle\qquad\int_{T}^{2T}{M_{12}}^{2}(x,y,N)dx$
		$\displaystyle\ll\sum_{m_{1},\cdots,m_{2r}\leqslant y}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}}\int_{T}^{2T}x^{2r-2+\varepsilon}\left\|\delta_{32}\left(\frac{x}{m_{r}},N\right)\delta_{32}\left(\frac{x}{m_{2r}},N\right)\right\|dx$
		$\displaystyle\ll T^{2r-2+\varepsilon}\sum_{m_{1},\cdots,m_{2r}\leqslant y}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}}\int_{T}^{2T}\left\|\delta_{32}\left(\frac{x}{m_{r}},N\right)\delta_{32}\left(\frac{x}{m_{2r}},N\right)\right\|dx.$

Suppose that $N\ll(T/y)^{2/3}$ , using the Cauchy-Schwarz’s inequality and Lemma 2.1 we have

		$\displaystyle\qquad\int_{T}^{2T}\left\|\delta_{32}\left(\frac{x}{m_{r}},N\right)\delta_{32}\left(\frac{x}{m_{2r}},N\right)\right\|dx$
		$\displaystyle\ll\left(\int_{T}^{2T}{\delta_{32}}^{2}\left(\frac{x}{m_{r}},N\right)dx\right)^{1/2}\left(\int_{T}^{2T}{\delta_{32}}^{2}\left(\frac{x}{m_{2r}},N\right)dx\right)^{1/2}$
		$\displaystyle\ll\left(m_{r}\int_{\frac{T}{m_{r}}}^{\frac{2T}{m_{r}}}{\delta_{32}}^{2}\left(u,N\right)du\right)^{1/2}\left(m_{2r}\int_{\frac{T}{m_{2r}}}^{\frac{2T}{m_{2r}}}{\delta_{32}}^{2}\left(u,N\right)du\right)^{1/2}$
		$\displaystyle\ll\left(\frac{T^{5/3+\varepsilon}N^{-1/3}}{{m_{r}}^{2/3}}+\frac{T^{14/9+\varepsilon}}{m_{r}^{5/9}}\right)^{1/2}\left(\frac{T^{5/3+\varepsilon}N^{-1/3}}{{m_{2r}}^{2/3}}+\frac{T^{14/9+\varepsilon}}{m_{2r}^{5/9}}\right)^{1/2}$
		$\displaystyle\ll\frac{T^{5/3+\varepsilon}N^{-1/3}}{(m_{r}m_{2r})^{1/3}}+\frac{T^{29/18+\varepsilon}N^{-1/6}}{{m_{r}}^{5/18}{m_{2r}}^{1/3}}+\frac{T^{29/18+\varepsilon}N^{-1/6}}{{m_{r}}^{1/3}{m_{2r}}^{5/18}}+\frac{T^{14/9+\varepsilon}}{(m_{r}m_{2r})^{5/18}}.$

Thus

		$\displaystyle\qquad\int_{T}^{2T}{M_{12}}^{2}(x,y,N)dx$		(5.14)
		$\displaystyle\ll T^{2r-2+\varepsilon}\sum_{m_{1},\cdots,m_{2r}\leqslant y}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}}$
		$\displaystyle\qquad\times\left(\frac{T^{5/3+\varepsilon}N^{-1/3}}{(m_{r}m_{2r})^{1/3}}+\frac{T^{29/18+\varepsilon}N^{-1/6}}{{m_{r}}^{5/18}{m_{2r}}^{1/3}}+\frac{T^{29/18+\varepsilon}N^{-1/6}}{{m_{r}}^{1/3}{m_{2r}}^{5/18}}+\frac{T^{14/9+\varepsilon}}{(m_{r}m_{2r})^{5/18}}\right)$
		$\displaystyle\ll T^{2r-1/3+\varepsilon}N^{-1/3}+T^{2r-7/18+\varepsilon}N^{-1/6}+T^{2r-4/9+\varepsilon},$

where the convergence of all the series can be obtained by Lemma 2.3.

Since $T^{2r-7/18+\varepsilon}N^{-1/6}=(T^{2r-1/3+\varepsilon}N^{-1/3})^{1/2}(T^{2r-4/9+\varepsilon})^{1/2}$ , the term $T^{2r-7/18+\varepsilon}N^{-1/6}$ is superfluous.

It remains to estimate

\int_{T}^{2T}M_{11}(x,y,N)M_{12}(x,y,N)dx.

Using (5.13), (5.14) and the Cauchy-Schwarz’s inequality, we deduce that

		$\displaystyle\qquad\int_{T}^{2T}M_{11}(x,y,N)M_{12}(x,y,N)dx$		(5.15)
		$\displaystyle\ll\left(\int_{T}^{2T}{M_{11}}^{2}(x,y,N)dx\right)^{1/2}\left(\int_{T}^{2T}{M_{12}}^{2}(x,y,N)dx\right)^{1/2}$
		$\displaystyle\ll T^{r-1/6+\varepsilon}(T^{r-1/6+\varepsilon}N^{-1/6}+T^{r-2/9+\varepsilon})$
		$\displaystyle=T^{2r-1/3+\varepsilon}N^{-1/6}+T^{2r-7/18+\varepsilon}.$

Combining (5.5), (5.13), (5.14) and (5.15) we conclude that

	$\displaystyle\int_{T}^{2T}{M_{1}}^{2}(x,y)dx=$	$\displaystyle\frac{r^{2}}{6\pi^{2}}\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}D_{r,3,\ell_{1},\ell_{2}}\int_{T}^{2T}x^{2r-4/3}(\log x)^{\ell_{1}+\ell_{2}}dx$		(5.16)
		$\displaystyle\qquad+O(T^{2r-1/3+\varepsilon}N^{-1/6}+T^{2r-2/3+\varepsilon}N^{2/3}+T^{2r-7/18+\varepsilon}).$		(5.16)

5.3. Conclusion

Taking $y=N^{1/2}$ , using (5.3), (5.16) and the Cauchy-Schwarz’s inequality we easily have

$\displaystyle\int_{T}^{2T}M_{1}(x,y)M_{2}(x,y)dx$	$\displaystyle\ll\left(\int_{T}^{2T}{M_{1}}^{2}(x,y)dx\right)^{1/2}\left(\int_{T}^{2T}{M_{2}}^{2}(x,y)dx\right)^{1/2}$	(5.17)
	$\displaystyle\ll T^{r-1/6+\varepsilon}\times T^{r-1/6+\varepsilon}N^{-1/6}$
	$\displaystyle=T^{2r-1/3+\varepsilon}N^{-1/6}.$

And similarly we obtain

\displaystyle\int_{T}^{2T}M_{1}(x,y)M_{3}(x,y)dx

\displaystyle\ll T^{2r-1/3+\varepsilon}N^{-1/6}.

(5.18)

Above all, choosing $N=T^{1/3}$ which indeed satisfies $N\ll(T/y)^{2/3}=T^{1/2}$ ,

\int_{T}^{2T}{\Delta_{r,3}}^{2}(x)dx=\frac{r^{2}}{6\pi^{2}}\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}D_{r,3,\ell_{1},\ell_{2}}\int_{T}^{2T}x^{2r-4/3}(\log x)^{\ell_{1}+\ell_{2}}dx+O(T^{2r-7/18+\varepsilon})

holds from (3.7), (5.1), (5.2), (5.3), (5.16), (5.17) and (5.18). Then replacing $T$ by $T/2$ , $T/2^{2}$ and so on, and adding up all the results, we obtain

	$\displaystyle\int_{1}^{T}{\Delta_{r,3}}^{2}(x)dx$	$\displaystyle=\frac{r^{2}}{6\pi^{2}}\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}D_{r,3,\ell_{1},\ell_{2}}\int_{1}^{T}x^{2r-4/3}(\log x)^{\ell_{1}+\ell_{2}}dx+O(T^{2r-7/18+\varepsilon})$
		$\displaystyle=\frac{r^{2}}{6\pi^{2}}T^{2r-1/3}L_{4r-4}(\log T)+O(T^{2r-7/18+\varepsilon}),$

where we use integration by part several times to get $L_{4r-4}(u)$ is a polynomial in $u$ of degree $4r-4$ denoted by

\displaystyle L_{4r-4}(u)=\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}D_{r,3,\ell_{1},\ell_{2}}\sum_{t=0}^{\ell_{1}+\ell_{2}}\frac{(-1)^{t}(\ell_{1}+\ell_{2})!}{(2r-\frac{1}{3})^{t+1}(\ell_{1}+\ell_{2}-t)!}u^{\ell_{1}+\ell_{2}-t}.

Hence we have completed the proof of the Theorem.

6. Proof of Theorem 1.3

By (3.7) we obtain

	$\displaystyle\int_{T}^{2T}\|\Delta_{r,3}(x)\|^{3}dx$	$\displaystyle\ll\int_{T}^{2T}\|{\Delta_{r,3}}^{*}(x)\|^{3}dx$
		$\displaystyle\ll\int_{T}^{2T}\left(\sum_{m_{1},\cdots,m_{r}\leqslant 2T}\|f_{r,3}(m_{1},\cdots,m_{r})\|\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\right\|\right)^{3}dx$
		$\displaystyle\ll T^{3r-3+\varepsilon}\int_{T}^{2T}\left(\sum_{m_{1},\cdots,m_{r}\leqslant 2T}\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}}\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\right\|\right)^{3}dx.$

Using the Hölder’s inequality we have

		$\displaystyle\left(\sum_{m_{1},\cdots,m_{r}\leqslant 2T}\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}}\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\right\|\right)^{3}$
	$\displaystyle\ll$	$\displaystyle\left(\sum_{m_{1},\cdots,m_{r}\leqslant 2T}\left(\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}}\right)^{3/2}\right)^{2}\left(\sum_{m_{1},\cdots,m_{r}\leqslant 2T}\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\right\|^{3}\right),$

so we obtain

		$\displaystyle\int_{T}^{2T}\|\Delta_{r,3}(x)\|^{3}dx$
	$\displaystyle\ll$	$\displaystyle T^{3r-3+\varepsilon}\int_{T}^{2T}\left(\sum_{m_{1},\cdots,m_{r}\leqslant 2T}\left(\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}}\right)^{3/2}\right)^{2}\left(\sum_{m_{1},\cdots,m_{r}\leqslant 2T}\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\right\|^{3}\right)dx.$

Changing the order of integration and summation we have

	$\displaystyle\int_{T}^{2T}\|\Delta_{r,3}(x)\|^{3}dx$	$\displaystyle\ll T^{3r-3+\varepsilon}\left(\sum_{m_{1},\cdots,m_{r}\leqslant 2T}\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}}\right)^{3}\int_{T}^{2T}\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\right\|^{3}dx$
		$\displaystyle\ll T^{3r-3+\varepsilon}\left(\sum_{m_{1},\cdots,m_{r}\leqslant 2T}\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}}\right)^{3}\left(m_{r}\int_{\frac{T}{m_{r}}}^{\frac{2T}{m_{r}}}\left\|\Delta_{3}\left(u\right)\right\|^{3}du\right)$
		$\displaystyle\ll T^{3r-1+\varepsilon}\left(\sum_{m_{1},\cdots,m_{r}\leqslant 2T}\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}{m_{r}}^{1/3}}\right)^{3}$
		$\displaystyle\ll T^{3r-1+\varepsilon},$

where we use (1.7) and Lemma 2.3. And Theorem 1.3 holds by replacing $T$ by $T/2$ , $T/2^{2}$ and so on, and adding up all the results.

7. Proof of Theorem 1.4

By (3.7) we have

	$\displaystyle\int_{1}^{T}\Delta_{r,3}(x)dx$	$\displaystyle=\int_{1}^{T}{\Delta_{r,3}}^{*}(x)dx+O(T^{r+\varepsilon})$
		$\displaystyle=r\int_{1}^{T}\sum_{m_{1},\cdots,m_{r}\leqslant x}f_{r,3}(m_{1},\cdots,m_{r})\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\Delta_{3}\left(\frac{x}{m_{r}}\right)dx+O(T^{r+\varepsilon})$
		$\displaystyle:=I_{1}+I_{2}+O(T^{r+\varepsilon}),$

where

	$\displaystyle I_{1}$	$\displaystyle=r\int_{1}^{T}\sum_{m_{1},\cdots,m_{r}\leqslant 2T}f_{r,3}(m_{1},\cdots,m_{r})\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\Delta_{3}\left(\frac{x}{m_{r}}\right)dx,$
	$\displaystyle I_{2}$	$\displaystyle=r\int_{1}^{T}\left(\sum_{m_{1},\cdots,m_{r}\leqslant 2T}-\sum_{m_{1},\cdots,m_{r}\leqslant x}\right)f_{r,3}(m_{1},\cdots,m_{r})\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\Delta_{3}\left(\frac{x}{m_{r}}\right)dx.$

We are going to estimate $I_{2}$ . Since $m_{1},\cdots,m_{r-1}$ are symmetric in $I_{2}$ , we have

I_{2}\ll I_{21}+I_{22},

where

	$\displaystyle I_{21}$	$\displaystyle=\int_{1}^{T}\sum_{\begin{subarray}{c}m_{1},\cdots,m_{r}\leqslant 2T\\ m_{1}>x\end{subarray}}\|f_{r,3}(m_{1},\cdots,m_{r})\|\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\right\|dx,$
	$\displaystyle I_{22}$	$\displaystyle=\int_{1}^{T}\sum_{\begin{subarray}{c}m_{1},\cdots,m_{r}\leqslant 2T\\ m_{r}>x\end{subarray}}\|f_{r,3}(m_{1},\cdots,m_{r})\|\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\right\|dx.$

Changing the order of integration and summation, using (3.2) and $\Delta_{3}(x/m_{r})\ll(x/m_{r})^{\alpha_{3}+\varepsilon}$ we have

	$\displaystyle I_{22}$	$\displaystyle\ll\sum_{\begin{subarray}{c}m_{1},\cdots,m_{r}\leqslant 2T\\ m_{r}>x\end{subarray}}\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}{m_{r}}^{\alpha_{3}}}\int_{1}^{T}x^{r-1+\alpha_{3}+\varepsilon}dx$
		$\displaystyle\ll T^{r+\alpha_{3}+\varepsilon}\sum_{\begin{subarray}{c}m_{1},\cdots,m_{r}\leqslant 2T\\ m_{r}>x\end{subarray}}\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}{m_{r}}^{\varepsilon}}\cdot\frac{1}{{m_{r}}^{\alpha_{3}-\varepsilon}}$
		$\displaystyle\ll T^{r+\varepsilon}\sum_{m_{1},\cdots,m_{r}=1}^{\infty}\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}{m_{r}}^{\varepsilon}}$
		$\displaystyle\ll T^{r+\varepsilon},$

and the convergence of the latter series is obtained by Lemma 2.3.

For $I_{21}$ , similar to $I_{22}$ and using Lemma 2.3 we have

	$\displaystyle I_{21}$	$\displaystyle\ll\sum_{\begin{subarray}{c}m_{1},\cdots,m_{r}\leqslant 2T\\ m_{1}>x\end{subarray}}\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}{m_{r}}^{\alpha_{3}}}\int_{1}^{T}x^{r-1+\alpha_{3}+\varepsilon}dx$
		$\displaystyle\ll T^{r+\alpha_{3}+\varepsilon}\sum_{\begin{subarray}{c}m_{1},\cdots,m_{r}\leqslant 2T\\ m_{1}>x\end{subarray}}\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{{m_{1}}^{1-\alpha_{3}}m_{2}\cdots m_{r-1}{m_{r}}^{\alpha_{3}}}\cdot\frac{1}{{m_{1}}^{\alpha_{3}-\varepsilon}}$
		$\displaystyle\ll T^{r+\varepsilon}\sum_{m_{1},\cdots,m_{r}=1}^{\infty}\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{{m_{1}}^{1-\alpha_{3}}m_{2}\cdots m_{r-1}{m_{r}}^{\alpha_{3}}}$
		$\displaystyle\ll T^{r+\varepsilon}.$

Then we turns to estimate $I_{1}$ . By (3.7) we have

\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\ll\frac{x^{r-1+\varepsilon}}{m_{1}\cdots m_{r-1}},

and its derivative satisfies

\left(\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\right)^{\prime}\ll\frac{x^{r-2+\varepsilon}}{m_{1}\cdots m_{r-1}}.

For a real number $u\geqslant 2$ , define

S(u)=\int_{1}^{u}\Delta_{3}(x)dx,

and we denote $\prod_{j=1}^{r-1}M_{3}\left(x/m_{j}\right)$ by $\mathcal{M}(x;m_{1},\cdots,m_{r-1})$ , then changing the order of integration and summation we have

	$\displaystyle I_{1}=$	$\displaystyle r\sum_{m_{1},\cdots,m_{r}\leqslant 2T}f_{r,3}(m_{1},\cdots,m_{r})\int_{1}^{T}\mathcal{M}(x;m_{1},\cdots,m_{r-1})\Delta_{3}\left(\frac{x}{m_{r}}\right)dx$
	$\displaystyle=$	$\displaystyle r\sum_{m_{1},\cdots,m_{r}\leqslant 2T}f_{r,3}(m_{1},\cdots,m_{r})$
		$\displaystyle\times\left(S(T)\mathcal{M}(T;m_{1},\cdots,m_{r-1})-\int_{1}^{T}S(u)\mathcal{M}^{\prime}(u;m_{1},\cdots,m_{r-1})du\right)$
	$\displaystyle\ll$	$\displaystyle\sum_{m_{1},\cdots,m_{r}\leqslant 2T}\|f_{r,3}(m_{1},\cdots,m_{r})\|\left(\frac{T^{r+1/6+\varepsilon}}{m_{1}\cdots m_{r-1}}+\int_{1}^{T}\|S(u)\|\|\mathcal{M}^{\prime}(u;m_{1},\cdots,m_{r-1})\|du\right)$
	$\displaystyle\ll$	$\displaystyle\sum_{m_{1},\cdots,m_{r}\leqslant 2T}\|f_{r,3}(m_{1},\cdots,m_{r})\|\left(\frac{T^{r+1/6+\varepsilon}}{m_{1}\cdots m_{r-1}}+\frac{T^{7/6+\varepsilon}}{m_{1}\cdots m_{r-1}}\int_{1}^{T}u^{r-2+\varepsilon}du\right)$
	$\displaystyle\ll$	$\displaystyle T^{r+1/6+\varepsilon}\sum_{m_{1},\cdots,m_{r}\leqslant 2T}\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}}\cdot\frac{{m_{r}}^{\varepsilon}}{{m_{r}}^{\varepsilon}}$
	$\displaystyle\ll$	$\displaystyle T^{r+1/6+\varepsilon}\sum_{m_{1},\cdots,m_{r}=1}^{\infty}\frac{\|f_{r,3}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}{m_{r}}^{\varepsilon}}$
	$\displaystyle\ll$	$\displaystyle T^{r+1/6+\varepsilon},$

where we use integration by parts, Lemma 2.3 and Lemma 2.8. Hence we complete the proof.

7.1. Proof of the Corollary

Suppose $T^{\varepsilon}\ll H\ll T$ is a parameter, by Theorem 1.2 we get

	$\displaystyle\int_{T}^{T+H}{\Delta_{r,3}}^{2}(x)dx$	(7.1)
$\displaystyle=$	$\displaystyle\frac{r^{2}}{6\pi^{2}}\left((T+H)^{2r-1/3}L_{4r-4}(\log(T+H))-T^{2r-1/3}L_{4r-4}(\log T)\right)+O(T^{2r-7/18+\varepsilon})$
$\displaystyle\asymp$	$\displaystyle HT^{2r-4/3}$

for $H\gg T^{17/18+\varepsilon}$ . Trivally we have

\int_{T}^{T+H}{\Delta_{r,3}}^{2}(x)dx\ll|\Delta_{r,3}(T)|\int_{T}^{T+H}|\Delta_{r,3}(x)|dx.

By (1.3) and theorem 1.1 we have $\Delta_{r,3}(T)\ll T^{r-1+43/96+\varepsilon}$ . Thus for $H\gg T^{17/18+\varepsilon}$ ,

\int_{T}^{T+H}|\Delta_{r,3}(x)|dx\gg HT^{r-25/32}.

(7.2)

Taking $H\gg T^{91/96+\varepsilon}$ , combining Theorem 1.4 we obtain

\int_{T}^{T+H}|\Delta_{r,3}(x)|dx\gg\int_{T}^{T+H}\Delta_{r,3}(x)dx.

Hence $\Delta_{r,3}(x)$ has at least one sign change in the interval $[T,T+H]$ for $T^{91/96+\varepsilon}\ll H\ll T$ , and has at least $T^{5/96-\varepsilon}$ sign changes in $[T,2T]$ .

Acknowledgement

The authors would like to appreciate the referee for his/her patience in refereeing this paper. This work is supported by Beijing Natural Science Foundation (Grant No.1242003), and the National Natural Science Foundation of China (Grant No.11971476).

		$\displaystyle\ll\sum_{m_{1},\cdots,m_{r}\leqslant x}\|f_{r,k}(m_{1},\cdots,m_{r})\|\left(\prod_{j=1}^{r-2}M_{k}\left(\frac{x}{m_{j}}\right)\right)\left\|\Delta_{k}\left(\frac{x}{m_{r-1}}\right)\right\|\left\|\Delta_{k}\left(\frac{x}{m_{r}}\right)\right\|$		(3.6)
		$\displaystyle\ll x^{r-2+2\alpha_{k}+\varepsilon}\sum_{m_{1},\cdots,m_{r}\leqslant x}\frac{\|f_{r,k}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-2}(m_{r-1}m_{r})^{\alpha_{k}}}$
		$\displaystyle=x^{r-2+2\alpha_{k}+\varepsilon}\sum_{m_{1},\cdots,m_{r}\leqslant x}\frac{\|f_{r,k}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-2}(m_{r-1}m_{r})^{\alpha_{k}}}\times\frac{(m_{r-1}m_{r})^{\frac{1}{2}-\alpha_{k}+\varepsilon}}{(m_{r-1}m_{r})^{\frac{1}{2}-\alpha_{k}+\varepsilon}}$
		$\displaystyle\ll x^{r-2+2\alpha_{k}+2(\frac{1}{2}-\alpha_{k})+\varepsilon}\sum_{m_{1},\cdots,m_{r}=1}^{\infty}\frac{\|f_{r,k}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-2}(m_{r-1}m_{r})^{\frac{1}{2}+\varepsilon}}$
		$\displaystyle\ll x^{r-1+\varepsilon}$

	$\displaystyle{\Delta_{r,k}}^{*}(x)$	$\displaystyle\ll\sum_{m_{1},\cdots,m_{r}\leqslant x}\|f_{r,k}(m_{1},\cdots,m_{r})\|\left(\prod_{j=1}^{r-1}M_{k}\left(\frac{x}{m_{j}}\right)\right)\left\|\Delta_{k}\left(\frac{x}{m_{r}}\right)\right\|$
		$\displaystyle\ll x^{r-1+\alpha_{k}+\varepsilon}\sum_{m_{1},\cdots,m_{r}\leqslant x}\frac{\|f_{r,k}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}{m_{r}}^{\alpha_{k}}}$
		$\displaystyle\ll x^{r-1+\alpha_{k}+\varepsilon}\sum_{m_{1},\cdots,m_{r}=1}^{\infty}\frac{\|f_{r,k}(m_{1},\cdots,m_{r})\|}{m_{1}\cdots m_{r-1}{m_{r}}^{\alpha_{k}}}$
		$\displaystyle\ll x^{r-1+\alpha_{k}+\varepsilon},$

	$\displaystyle\int_{T}^{2T}({\Delta_{r,k}}^{*}(x))^{2}dx$	$\displaystyle\ll\sum_{m_{1},\cdots,m_{2r}\leqslant 2T}\|f_{r,k}(\mathbf{M}_{1})f_{r,k}(\mathbf{M}_{2})\|\int_{T}^{2T}\prod_{j=1}^{r-1}M_{k}\left(\frac{x}{m_{j}}\right)\prod_{j=r+1}^{2r-1}M_{k}\left(\frac{x}{m_{j}}\right)$
		$\displaystyle\qquad\times\left\|\Delta_{k}\left(\frac{x}{m_{r}}\right)\Delta_{k}\left(\frac{x}{m_{2r}}\right)\right\|dx$
		$\displaystyle\ll\sum_{m_{1},\cdots,m_{2r}\leqslant 2T}\frac{\|f_{r,k}(\mathbf{M}_{1})f_{r,k}(\mathbf{M}_{2})\|}{D_{1}D_{2}}\int_{T}^{2T}x^{2r-2+\varepsilon}\left\|\Delta_{k}\left(\frac{x}{m_{r}}\right)\Delta_{k}\left(\frac{x}{m_{2r}}\right)\right\|dx$
		$\displaystyle\ll T^{2r-2+\varepsilon}\sum_{m_{1},\cdots,m_{2r}\leqslant 2T}\frac{\|f_{r,k}(\mathbf{M}_{1})f_{r,k}(\mathbf{M}_{2})\|}{D_{1}D_{2}}\int_{T}^{2T}\left\|\Delta_{k}\left(\frac{x}{m_{r}}\right)\Delta_{k}\left(\frac{x}{m_{2r}}\right)\right\|dx.$

	$\displaystyle\int_{T}^{2T}{M_{3}}^{2}dx=$	$\displaystyle\int_{T}^{2T}\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant 2T\\ m_{r},m_{2r}>y\end{subarray}}\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\prod_{j=r+1}^{2r-1}M_{3}\left(\frac{x}{m_{j}}\right)$
		$\displaystyle\times\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\Delta_{3}\left(\frac{x}{m_{2r}}\right)\right\|dx$
	$\displaystyle=$	$\displaystyle\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant 2T\\ m_{r},m_{2r}>y\end{subarray}}\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|\int_{T}^{2T}\prod_{j=1}^{r-1}M_{3}\left(\frac{x}{m_{j}}\right)\prod_{j=r+1}^{2r-1}M_{3}\left(\frac{x}{m_{j}}\right)$
		$\displaystyle\times\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\Delta_{3}\left(\frac{x}{m_{2r}}\right)\right\|dx$
	$\displaystyle\ll$	$\displaystyle\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant 2T\\ m_{r},m_{2r}>y\end{subarray}}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}}\int_{T}^{2T}x^{2r-2+\varepsilon}\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\Delta_{3}\left(\frac{x}{m_{2r}}\right)\right\|dx$
	$\displaystyle\ll$	$\displaystyle T^{2r-2+\varepsilon}\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant 2T\\ m_{r},m_{2r}>y\end{subarray}}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}}\int_{T}^{2T}\left\|\Delta_{3}\left(\frac{x}{m_{r}}\right)\Delta_{3}\left(\frac{x}{m_{2r}}\right)\right\|dx,$

$\displaystyle\int_{1}$	$\displaystyle\ll\sum_{\ell_{1},\ell_{2}=0}^{2(r-1)}\sum_{\begin{subarray}{c}m_{1},\cdots,m_{2r}\leqslant y\\ n_{1},n_{2}\leqslant N\\ n_{1}m_{2r}\neq n_{2}m_{r}\end{subarray}}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})C_{\ell_{1},\ell_{2}}(\mathbf{M}_{1},\mathbf{M}_{2})\|}{D_{1}D_{2}(m_{r}m_{2r})^{1/3}}\cdot\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}}$	(5.9)
	$\displaystyle\qquad\times T^{2r-4/3+\varepsilon}\cdot\frac{T^{2/3}}{\left\|(n_{1}/m_{r})^{1/3}-(n_{2}/m_{2r})^{1/3}\right\|}$
	$\displaystyle\ll T^{2r-2/3+\varepsilon}\sum_{m_{1},\cdots,m_{2r}\leqslant y}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}(m_{r}m_{2r})^{1/3}}\sum_{\begin{subarray}{c}n_{1},n_{2}\leqslant N\\ n_{1}m_{2r}\neq n_{2}m_{r}\end{subarray}}\frac{\tau_{3}(n_{1})\tau_{3}(n_{2})}{(n_{1}n_{2})^{2/3}\left\|(n_{1}/m_{r})^{1/3}-(n_{2}/m_{2r})^{1/3}\right\|}$
	$\displaystyle:=T^{2r-2/3+\varepsilon}\sum_{m_{1},\cdots,m_{2r}\leqslant y}\frac{\|f_{r,3}(\mathbf{M}_{1})f_{r,3}(\mathbf{M}_{2})\|}{D_{1}D_{2}(m_{r}m_{2r})^{1/3}}(R_{1}+R_{2}),$

On moments of the error term of the multivariable k-th divisor functions

Abstract.

1. Introduction and Results

Theorem 1.1.

Theorem 1.2.

Theorem 1.3.

Theorem 1.4.

Corollary 1.5.

Theorem 1.6.

Notation.

2. Some lemmas

Lemma 2.1.

Proof.

Lemma 2.2.

Proof.

Lemma 2.3.

Proof.

Lemma 2.4.

Proof.

Lemma 2.5.

Proof.

Lemma 2.6.

Proof.

Lemma 2.7.

Proof.

Lemma 2.8.

Proof.

3. Proof of Theorem 1.1 and Expression of Δr,k​(x){\Delta_{r,k}}(x)

4. Proof of Theorem 1.6

4.1. Mean Square of Δr,k∗​(x){\Delta_{r,k}}^{*}(x)

5. Proof of Theorem 1.2

5.1. Mean Square of Δr,3∗​(x){\Delta_{r,3}}^{*}(x)

5.2. Mean Square of M1​(x,y)M_{1}(x,y)

5.2.1. Evaluation of ∫0\int_{0}

5.2.2. Estimates of ∫1\int_{1} and ∫2\int_{2}

5.2.3. Other terms in (5.5)

5.3. Conclusion

6. Proof of Theorem 1.3

7. Proof of Theorem 1.4

7.1. Proof of the Corollary

Acknowledgement

References

3. Proof of Theorem 1.1 and Expression of ${\Delta_{r,k}}(x)$

4.1. Mean Square of ${\Delta_{r,k}}^{*}(x)$

5.1. Mean Square of ${\Delta_{r,3}}^{*}(x)$

5.2. Mean Square of $M_{1}(x,y)$

5.2.1. Evaluation of $\int_{0}$

5.2.2. Estimates of $\int_{1}$ and $\int_{2}$