短区间上四元混合幂的除数和

On the Sum of Divisors of Mixed Powers of Four Variables in Short Intervals

摘要: 设d(n)为Dirichlet除数函数。本文考虑了短区间上四元混合幂除数函数的均值问题，并得到了一个有效的渐进公式。

Abstract: Let d(n) denote the Dirichlet divisor function. In this paper, we consider the mean value of the mixed power of four variables of the divisor function in short intervals, and obtain an effective asymptotic formula.

1. 引言与主要结果

设d(n)为Dirichlet除数函数，许多学者围绕除数函数的均值问题做了相关的研究工作。Gafurov [1] [2] 首先研究了与二元二次型除数和问题，并得到了以下渐进公式：

${\displaystyle \underset{1⩽m_1,m_2⩽x}{\sum }d}\left(m_1^2+m_2^2\right)=A_1{x}^2\log x+A_2{x}^2+O\left(x^{5/3}{\log }^{9}x\right),$

其中 $A_1$， $A_2$ 都是常数。2000年，Yu [3] 将上式余项改进为 $O\left(x^{3/2+\epsilon }\right)$， $\epsilon$ 为任意小的正实数。

2000年，Calderán和Velasco [4] 研究了三元二次型除数和问题，并得到了以下渐进公式：

$G_2\left(x\right)={\displaystyle \underset{1⩽m_1,m_2,m_3⩽x}{\sum }d}\left(m_1^2+m_2^2+m_3^2\right)=\frac{8\zeta \left(3\right)}{5\zeta \left(5\right)}{x}^3\log x+O\left(x^3\right).$

2012年，Guo和Zhai [5] 利用圆法，以及Heath-Brown [6] 在算术级数中除数和函数的估计，得到了三元二次型除数和问题更加精确的主项和余项：

$G_2\left(x\right)=2C_1{I}_1{x}^3\log x+\left(C_1{I}_2+C_2{I}_1\right)x^3+O\left(x^{8/3+\epsilon }\right),$

其中 $C_1$， $C_2$， $I_1$， $I_2$ 都是常数。2014年，Zhao [7] 将上式余项改进为 $O\left(x^2{\log }^{7}x\right)$。

2015年，Hu和Yao [8] 研究了短区间上的三元二次型除数和问题，并得到了以下渐进公式：

${\displaystyle \underset{\left|m_i-x\right|⩽y}{\sum }d}\left(m_1^2+m_2^2+m_3^2\right)=K_1{L}_1\left(x,y\right)+2\left(\gamma K_1-K_2\right)L_2\left(x,y\right)+O\left(y^{3-\epsilon }\right),$

其中 $y=x^{1/2+2\epsilon }$， $K_1$， $K_2$ 都是常数， $\gamma$ 是Euler常数， $L_1\left(x,y\right)$， $L_2\left(x,y\right)$ 满足 $L_1\left(x,y\right)\asymp y^3\log y$， $L_2\left(x,y\right)\asymp y^3$。

2018年，Zhang和Li [9] 考虑了短区间上混合幂的三元除数和问题，并得到了以下结论：

${\displaystyle \underset{\begin{array}{c}x-y<m_i^2⩽x+y\\ x-y<m_3^k⩽x+y\\ i=1,2\end{array}}{\sum }d}\left(m_1^2+m_2^2+m_3^k\right)=K_3{L}_3\left(x,y\right)+2\left(\gamma K_3-K_4\right)L_4\left(x,y\right)+O\left(y^3{x}^{-2+1/k-\epsilon }\right),$

其中 $3\le k\in \mathbb{Z}$， $y=x^{{\theta }_k}$， ${\theta }_k$ 仅与k有关， $K_3$， $K_4$ 都是常数， $L_3\left(x,y\right)$， $L_4\left(x,y\right)$ 满足 $L_3\left(x,y\right)\asymp y^3{x}^{-2+1/k}\log x$， $L_4\left(x,y\right)\asymp y^3{x}^{-2+1/k}$。

2014年，Hu [10] 研究了四元二次型除数和问题，并得到了以下渐进公式：

${\displaystyle \underset{1⩽m_1,m_2,m_3,m_4⩽x}{\sum }d}\left(m_1^2+m_2^2+m_3^2+m_4^2\right)=2K_5{L}_5{x}^4\log x+\left(K_5{L}_6+K_6{L}_5\right)x^4+O\left(x^{7/2+\epsilon }\right),$

其中 $K_5$， $K_6$， $L_5$， $L_6$ 都是常数。

2016年，Hu [11] 考虑了短区间上的四元二次型除数和问题，并得到了以下渐进公式：

${\displaystyle \underset{\begin{array}{c}\left|m_i-x\right|⩽y\\ 1⩽i⩽4\end{array}}{\sum }d}\left(m_1^2+m_2^2+m_3^2+m_4^2\right)=K_7{L}_7\left(x,y\right)+2\left(\gamma K_7-K_8\right)L_8\left(x,y\right)+O\left(y^{4-\epsilon }\right),$

其中 $y=x^{1/3+\epsilon }$， $K_7$， $K_8$ 都是常数， $L_7\left(x,y\right)$， $L_8\left(x,y\right)$ 满足 $L_7\left(x,y\right)\asymp y^4\log y$， $L_8\left(x,y\right)\asymp y^4$。

在本文中，我们考虑了短区间上四元混合幂的除数和问题，并得到了以下结果：

定理1.1设

$S_k\left(x,y\right):={\displaystyle \underset{\begin{array}{c}x-y<m_i^2⩽x+y\\ x-y<m_4^k⩽x+y\\ i=1,2,3\end{array}}{\sum }d}\left(m_1^2+m_2^2+m_3^2+m_4^k\right),$

那么对于 $3\le k\in \mathbb{Z}$， $y=x^{1-{\delta }_k+4\epsilon }$，

${\delta }_k=\{\begin{array}{ll}\frac{7}{36},\hfill & k=3;\hfill \\ \frac{1}{2^{k-1}+k+1},\hfill & 4⩽k⩽5;\hfill \\ \frac{2}{k\left(k^2-3k+5\right)},\hfill & k⩾6,\hfill \end{array}$

我们有

$S_k\left(x,y\right)={\mathcal{K}}_1{\mathfrak{L}}_1\left(x,y\right)+2\left(\gamma {\mathcal{K}}_1-{\mathcal{K}}_2\right){\mathfrak{L}}_2\left(x,y\right)+O\left(y^4{x}^{-\frac{5}{2}+\frac{1}{k}-\epsilon }\right),$

其中 ${\mathcal{K}}_1$， ${\mathcal{K}}_2$ 是收敛的奇异级数，对于j = 1，2，有

${\mathfrak{L}}_j\left(x,y\right)=\frac{1}{8k}{\displaystyle \underset{4\left(x-y\right)<n⩽4\left(x+y\right)}{\sum }{\left(\log n\right)}^{2-j}}{\displaystyle \underset{\begin{array}{c}x-y<m_i⩽x+y\\ i=1,2,3,4\\ m_1+m_2+m_3+m_4=n\end{array}}{\sum }{\left(m_1{m}_2{m}_3\right)}^{-\frac{1}{2}}}{\left(m_4\right)}^{-1+\frac{1}{k}},$

并且满足 ${\mathfrak{L}}_1\left(x,y\right)\asymp y^4{x}^{-5/2+1/k}\log x$， ${\mathfrak{L}}_2\left(x,y\right)\asymp y^4{x}^{-5/2+1/k}$。

本文主要运用圆法来证明定理1.1。在处理主区间上的积分时，主要运用Zhang和Li [9] 文中的结果，以及相关的指数和估计。在处理余区间上的积分时，主要运用Vaughan [12] 和Wooley [13] 的相关结论。

2. 记号说明

我们约定一些在本文中通用的符号： $\mathbb{N}$ 为全体自然数集； $\mathbb{Z}$ 为全体整数集； $ℝ$ 为全体实数集；e(t)为 ${\text{e}}^{2\pi it}$ 的缩写； $\epsilon$ 为任意小的正实数； $\Vert \alpha \Vert$ 表示 $\alpha$ 到整数的最短距离；d(n)为除数函数； $\phi \left(n\right)$ 为Euler函数； $\gamma$ 为Euler常数；(a，b)为a，b的最大公约数； $f\left(x\right)=O\left(g\left(x\right)\right)$，即存在常数C，使得 $\left|f\left(x\right)\right|⩽Cg\left(x\right)$ ； $f\left(x\right)\ll g\left(x\right)$，即 $f\left(x\right)=O\left(g\left(x\right)\right)$ ； $f\left(x\right)\asymp g\left(x\right)$，即 $g\left(x\right)\ll f\left(x\right)\ll g\left(x\right)$。

在本文中x，y都表示充分大的正整数，且 $y\ll x$ ；记

$N_1=x-y,N_2=x+y,$

对于k ≥ 2， $v>u>0$， $\lambda \in ℝ$，定义

$f_k\left(\alpha \right)={\displaystyle \underset{N_1<m^k⩽N_2}{\sum }e}\left(m^k\alpha \right),f\left(-\alpha \right)={\displaystyle \underset{4N_1<m⩽4N_2}{\sum }d}\left(m\right)e\left(-m\alpha \right),$

以及

$\begin{array}{l}{T}_1\left(u,v\right)={\displaystyle \underset{u<m⩽v}{\sum }e}\left(-m\lambda \right),T_1^{*}\left(u,v\right)={\displaystyle \underset{u<m⩽v}{\sum }\left(\log m\right)e}\left(-m\lambda \right),\\ T_k\left(u,v\right)=\frac{1}{k}{\displaystyle \underset{u^k<m⩽v^k}{\sum }{m}^{\frac{1}{k}-1}}e\left(m\lambda \right).\end{array}$

3. 主要引理

我们将首先证明以下渐进公式，它将在主区间上积分的估计中起到重要作用。

引理3.1设 $\alpha =\frac{a}{q}+\lambda$， $\left|\lambda \right|\le \frac{1}{q\tau }$， $f\left(-\alpha \right)$ 的定义同上，则有

$f\left(-\alpha \right)=\frac{T_1^{*}\left(4N_1,4N_2\right)}{q}+\frac{-2\log q+2\gamma }{q}{T}_1\left(4N_1,4N_2\right)+O\left(x^{\epsilon }\left(q^{\frac{3}{2}}+q^{\frac{2}{3}}{x}^{\frac{1}{3}}+q^{\frac{3}{2}}\left|\lambda \right|x+q^{\frac{3}{2}}{\left|\lambda \right|}^{2}xy\right)\right).$

证明：由 $f\left(-\alpha \right)$ 的定义以及Able分部求和公式，有

$\begin{array}{c}f\left(-\alpha \right)={\displaystyle \underset{4N_1<m⩽4N_2}{\sum }d}\left(m\right)e\left(-m\left(\frac{a}{q}+\lambda \right)\right)={\displaystyle \underset{r=1}{\overset{q}{\sum }}e}\left(-\frac{ar}{q}\right){\displaystyle \underset{\begin{array}{c}4N_1<m⩽4N_2\\ m\equiv r\left(modq\right)\end{array}}{\sum }d}\left(m\right)e\left(-m\lambda \right)\\ ={\displaystyle \underset{r=1}{\overset{q}{\sum }}e}\left(-\frac{ar}{q}\right){\displaystyle {\int }_{4N_1}^{4N_2}e}\left(-u\lambda \right)d\left({\displaystyle \underset{\begin{array}{c}m⩽u\\ m\equiv r\left(modq\right)\end{array}}{\sum }d}\left(m\right)\right).\end{array}$ (1)

对于 $r,q\in \mathbb{Z}$， $1⩽r⩽q$， $u>0$，定义

$D\left(u;q,r\right)={\displaystyle \underset{\begin{array}{c}n⩽u\\ n\equiv r\left(modq\right)\end{array}}{\sum }d}\left(n\right).$ (2)

首先引用Heath-Brown [6] 关于 $D\left(u;q,r\right)$ 的结论，有

$D\left(u;q,r\right)=R\left(u;q,r\right)+\Delta \left(u;q,r\right)$ (3)

$R\left(u;q,r\right)=\frac{u\cdot A\left(q,r\right)}{q^2}\left(\log u-2\log q+2\gamma -1\right)+\frac{2u\cdot B\left(q,r\right)}{q^2},$ (4)

其中

$A\left(q,r\right)={\displaystyle \underset{d|\left(q,r\right)}{\sum }{\displaystyle \underset{l|q/d}{\sum }d}l}\mu \left(\frac{q}{dl}\right),B\left(q,r\right)={\displaystyle \underset{d|\left(q,r\right)}{\sum }{\displaystyle \underset{l|q/d}{\sum }d}l}\mu \left(\frac{q}{dl}\right)\log l.$

由郭汝庭和翟文广 [5] 关于 $R\left(u;q,r\right)$ 和 $\Delta \left(u;q,r\right)$ 的一些结果，令

$c_1\left(q,r\right)=\frac{A\left(q,r\right)}{q^2},c_2\left(q,r\right)=\frac{A\left(q,r\right)\cdot \left(-2\log q+2\gamma -1\right)}{q^2}+\frac{2B\left(q,r\right)}{q^2},$

$B_1\left(q\right)={\displaystyle \underset{r=1}{\overset{q}{\sum }}{c}_1}\left(q,r\right)e\left(-\frac{ar}{q}\right),B_2\left(q\right)={\displaystyle \underset{r=1}{\overset{q}{\sum }}{c}_2}\left(q,r\right)e\left(-\frac{ar}{q}\right).$

我们有

$R\left(u;q,r\right)=c_1\left(q,r\right)u\log u+c_2\left(q,r\right)u,B_1\left(q\right)=\frac{1}{q},B_2\left(q\right)=\frac{-2\log q+2\gamma -1}{q},$ (5)

且 $\Delta \left(u;q,r\right)$ 满足

$F\left(u;q,a\right)={\displaystyle \underset{r=1}{\overset{q}{\sum }}\Delta }\left(u;q,r\right)e\left(-\frac{ar}{q}\right)\ll \left(q^{\frac{3}{2}}+q^{\frac{2}{3}}{u}^{\frac{1}{3}}\right){\left(qu\right)}^{\epsilon },$ (6)

$H\left(T;q,a\right)={\displaystyle {\int }_0^{T}F}\left(u;q,a\right)\text{d}u\ll q^{\frac{3}{2}+\epsilon }T.$ (7)

由公式(1)~(5)，以及Zhang和Li [9] 文中相关的结论，

${\displaystyle {\int }_u^{v}e}\left(-t\lambda \right)\text{d}t=T_1\left(u,v\right)+O\left(1+\left(v-u\right)\left|\lambda \right|\right),$

${\displaystyle {\int }_u^v\left(\log t\right)e\left(-t\lambda \right)\text{d}t}=T_1^{*}\left(u,v\right)+O\left(\left(\log v\right)\left(1+\left(v-u\right)\left|\lambda \right|\right)\right),$

我们有

$f\left(-\alpha \right)={\displaystyle \underset{r=1}{\overset{q}{\sum }}e}\left(-\frac{ar}{q}\right){\displaystyle {\int }_{4N_1}^{4N_2}e}\left(-u\lambda \right)\text{d}\left(D\left(u;q,r\right)\right)=J_1+J_2,$ (8)

其中

$\begin{array}{c}{J}_1={\displaystyle \underset{r=1}{\overset{q}{\sum }}e}\left(-\frac{ar}{q}\right){\displaystyle {\int }_{4N_1}^{4N_2}e}\left(-u\lambda \right)\text{d}\left(R\left(u;q,r\right)\right)\\ =\frac{T_1^{*}\left(4N_1,4N_2\right)}{q}+\frac{-2\log q+2\gamma }{q}{T}_1\left(4N_1,4N_2\right)+O\left(\left(\log x\right)\left(1+\left|\lambda \right|y\right)\right),\end{array}$ (9)

由两次分部积分运算以及公式(6)~(7)，我们有

$\begin{array}{c}{J}_2={\displaystyle \underset{r=1}{\overset{q}{\sum }}e}\left(-\frac{ar}{q}\right){\displaystyle {\int }_{4N_1}^{4N_2}e}\left(-u\lambda \right)\text{d}\left(\Delta \left(u;q,r\right)\right)\\ =F\left(4N_2;q,a\right)e\left(-4N_2\lambda \right)-F\left(4N_1;q,a\right)e\left(-4N_1\lambda \right)+2\pi i\lambda {\displaystyle \underset{r=1}{\overset{q}{\sum }}e}\left(-\frac{ar}{q}\right){\displaystyle {\int }_{4N_1}^{4N_2}e}\left(-u\lambda \right)\Delta \left(u;q,r\right)\text{d}u\\ =F\left(4N_2;q,a\right)e\left(-4N_2\lambda \right)-F\left(4N_1;q,a\right)e\left(-4N_1\lambda \right)+2\pi i\lambda H\left(4N_2;q,a\right)e\left(-4N_2\lambda \right)\\ -2\pi i\lambda H\left(4N_1;q,a\right)e\left(-4N_1\lambda \right)+{\left(2\pi i\lambda \right)}^2{\displaystyle {\int }_{4N_1}^{4N_2}H}\left(u;q,a\right)e\left(-u\lambda \right)\text{d}u\\ \ll x^{\epsilon }\left(q^{\frac{3}{2}}+q^{\frac{2}{3}}{x}^{\frac{1}{3}}+q^{\frac{3}{2}}\left|\lambda \right|x+q^{\frac{3}{2}}{\left|\lambda \right|}^{2}xy\right),\end{array}$ (10)

结合公式(8)~(10)，我们证明了引理3.1。

其次我们将证明以下积分均值，它将在余区间上积分的估计中起到重要作用。

引理3.2设 $2\le k\in \mathbb{Z}$， $f_k\left(\alpha \right)$ 的定义同上，且 $y\gg x^{1-1/k}$，则有

${\displaystyle {\int }_0^1{\left|f_k\left(\alpha \right)\right|}^6\text{d}\alpha }\ll y^4{x}^{-4+\frac{4}{k}+\epsilon }.$

证明：由复指数函数的正交性

${\displaystyle {\int }_0^{1}e}\left(u\alpha \right)\text{d}\alpha =\{\begin{array}{l}1,若u=0,\hfill \\ 0,若u\in \mathbb{Z},u\ne 0,\hfill \end{array}$

以及 $f_k\left(\alpha \right)$ 的定义，我们有

${\displaystyle {\int }_0^1{\left|f_k\left(\alpha \right)\right|}^6\text{d}\alpha }={\displaystyle \underset{\begin{array}{c}{m}_1^k-m_2^k=m_3^k-m_4^k+m_5^k-m_6^k\\ N_1^{1/k}<m_i⩽N_2^{1/k}\\ i=1,\cdots ,6\end{array}}{\sum }1},$ (11)

即上式左端积分等于方程

$m_1^k-m_2^k=m_3^k-m_4^k+m_5^k-m_6^k$ (12)

的解数R，其中 $N_1^{1/k}<m_i\le N_2^{1/k}$， $i=1,\cdots ,6$。

我们将按照以下三种情况分类考察解数R：

情形一： $m_1^k-m_2^k=m_3^k-m_4^k+m_5^k-m_6^k=0$， $m_3=m_4$。

此时有 $m_1=m_2$， $m_3=m_4$， $m_5=m_6$，方程(12)的变量只有三个，则其解数R满足

$R\ll {\left(N_2^{\frac{1}{k}}-N_1^{\frac{1}{k}}\right)}^3\ll y^3{x}^{-3+\frac{3}{k}}.$

情形二： $m_1^k-m_2^k=m_3^k-m_4^k+m_5^k-m_6^k=0$， $m_3\ne m_4$。

此时有 $m_1=m_2$，且 $m_5$， $m_6$ 满足

$m_4^k-m_3^k=m_5^k-m_6^k=\left(m_5-m_6\right)\left(m_5^{k-1}+m_5^{k-2}{m}_6+\cdots +m_6^{k-1}\right),$

若固定 $m_3$， $m_4$ 的取值，则 $m_5$， $m_6$ 的取法也随之确定，且其取法数 $R_{5,6}$ 满足

$R_{5,6}\ll d\left(m_4^k-m_3^k\right)\ll x^{\epsilon },$

因此解数R满足

$R\ll R_{5,6}\cdot {\left(N_2^{\frac{1}{k}}-N_1^{\frac{1}{k}}\right)}^3\ll y^3{x}^{-3+\frac{3}{k}+\epsilon }.$

情形三： $m_1^k-m_2^k=m_3^k-m_4^k+m_5^k-m_6^k\ne 0$。

此时 $m_1$， $m_2$ 满足

$m_3^k-m_4^k+m_5^k-m_6^k=m_1^k-m_2^k=\left(m_1-m_2\right)\left(m_1^{k-1}+m_1^{k-2}{m}_2+\cdots +m_2^{k-1}\right),$

若固定 $m_3$， $m_4$， $m_5$， $m_6$ 的取值，则 $m_1$， $m_2$ 的取法也随之确定，且其取法数 $R_{1,2}$ 满足

$R_{1,2}\ll d\left(m_3^k-m_4^k+m_5^k-m_6^k\right)\ll x^{\epsilon },$

因此解数R满足

$R\ll R_{1,2}\cdot {\left(N_2^{\frac{1}{k}}-N_1^{\frac{1}{k}}\right)}^4\ll y^4{x}^{-4+\frac{4}{k}+\epsilon }.$

结合以上三种情况的讨论，并由条件 $y\gg x^{1-1/k}$ 可得解数R满足：

$R\ll y^3{x}^{-3+\frac{3}{k}}+y^3{x}^{-3+\frac{3}{k}+\epsilon }+y^4{x}^{-4+\frac{4}{k}+\epsilon }\ll y^4{x}^{-4+\frac{4}{k}+\epsilon },$ (13)

结合公式(11)~(13)，我们有

${\displaystyle {\int }_0^1{\left|f_k\left(\alpha \right)\right|}^6\text{d}\alpha }\ll y^4{x}^{-4+\frac{4}{k}+\epsilon },$

由此我们证明了引理3.2。

此外还需要用到以下圆法中的经典结论：

引理3.3设 $a,q\in \mathbb{Z}$， $\alpha \in ℝ$，且 $\left(a,q\right)=1$， $\left|\alpha -\frac{a}{q}\right|⩽\frac{1}{q^2}$， $\varphi \left(x\right)=\alpha x^k+{\alpha }_1{x}^{k-1}+\cdots +{\alpha }_{k-1}x+{\alpha }_k$，则有

${\displaystyle \underset{1⩽x⩽X}{\sum }e}\left(\varphi \left(x\right)\right)\ll X^{1+\epsilon }{\left(q^{-1}+X^{-1}+q{X}^{-k}\right)}^{\frac{1}{2^{k-1}}}.$

证明：参见Vaughan [12]。

引理3.4设 $2\le k\in \mathbb{Z}$， $\left({\alpha }_1,\cdots ,{\alpha }_k\right)\in {ℝ}^k$，若存在整数 $j\left(2⩽j⩽k\right)$，使得对于某些 $a\in \mathbb{Z}$ ， $q\in \mathbb{N}$ ，有 $\left|{\alpha }_j-\frac{a}{q}\right|⩽q^{-2}$， $q⩽Y^j$，那么我们有

${\displaystyle \underset{1⩽m⩽Y}{\sum }e}\left({\alpha }_{1}m+\cdots +{\alpha }_k{m}^k\right)\ll Y^{1+\epsilon }{\left(q^{-1}+Y^{-1}+q{Y}^{-j}\right)}^{\frac{1}{2k\left(k-1\right)}}.$

证明：参见Wooley [13]。

4. 定理1.1的证明

设k ≥ 3， $y=x^{1-{\delta }_k+4\epsilon }$，其中

${\delta }_k=\{\begin{array}{ll}\frac{7}{36},\hfill & k=3;\hfill \\ \frac{1}{2^{k-1}+k+1},\hfill & 4⩽k⩽5;\hfill \\ \frac{2}{k\left(k^2-3k+5\right)},\hfill & k⩾6,\hfill \end{array}$

并令

$x^{\epsilon }\ll 2Q<\tau ,Q\tau \asymp y^{\beta }{x}^{-\beta +1},Q\ll y^{\frac{\beta }{2}}{x}^{\frac{-\beta +1}{2}-\frac{\epsilon }{2}},$ (14)

其中

$\beta =\{\begin{array}{l}2,k=3,\hfill \\ k,k⩾4.\hfill \end{array}$

由复指数函数的正交性，对于满足公式(14)的 $\tau$，我们有

${\displaystyle \underset{\begin{array}{c}x-y<m_i^2⩽x+y\\ x-y<m_4^k⩽x+y\\ i=1,2,3\end{array}}{\sum }d}\left(m_1^2+m_2^2+m_3^2+m_4^k\right)={\displaystyle {\int }_0^1{f}_2^3}\left(\alpha \right)f_k\left(\alpha \right)f\left(-\alpha \right)\text{d}\alpha ={\displaystyle {\int }_{\frac{1}{\tau }}^{1+\frac{1}{\tau }}{f}_2^3}\left(\alpha \right)f_k\left(\alpha \right)f\left(-\alpha \right)\text{d}\alpha ,$

其中 $f_k\left(\alpha \right)$， $f\left(-\alpha \right)$ 的定义同上。由Dirichlet有理逼近定理，每一个 $\alpha \in \left[\frac{1}{\tau },1+\frac{1}{\tau }\right]$ 都可以写成

$\alpha =\frac{a}{q}+\lambda ,\left|\lambda \right|⩽\frac{1}{q\tau },$

其中整数a，q满足 $1⩽a⩽q⩽\tau$， $\left(a,q\right)=1$，定义优弧

$\mathfrak{M}\left(a,q\right)=\left[\frac{a}{q}-\frac{1}{q\tau },\frac{a}{q}+\frac{1}{q\tau }\right],$

并定义主区间和余区间分别为

$\mathfrak{M}={\displaystyle \underset{q⩽Q}{\cup }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\cup }\mathfrak{M}}}\left(a,q\right),\mathfrak{m}=\left[\frac{1}{\tau },1+\frac{1}{\tau }\right]\backslash \mathfrak{M},$ (15)

对于不同的优弧 $\mathfrak{M}\left(a_1,q_1\right)$， $\mathfrak{M}\left(a_2,q_2\right)$， $\frac{a_1}{q_1}\ne \frac{a_2}{q_2}$，由公式(14)，我们有

$\left|\frac{a_1}{q_1}-\frac{a_2}{q_2}\right|=\left|\frac{a_1{q}_2-a_2{q}_1}{q_1{q}_2}\right|⩾\frac{1}{q_1{q}_2}⩾\frac{1}{2}\left(\frac{1}{q_{1}Q}+\frac{1}{q_{2}Q}\right)>\frac{1}{q_1\tau }+\frac{1}{q_2\tau },$

即任意两个不同的优弧是不相交的，由以上定义把积分区间 $\left[\frac{1}{\tau },1+\frac{1}{\tau }\right]$ 划分为主区间 $\mathfrak{M}$ 和余区间 $\mathfrak{m}$，因此我们有

$S_k\left(x,y\right)={\displaystyle \underset{\begin{array}{c}x-y<m_i^2⩽x+y\\ x-y<m_4^k⩽x+y\\ i=1,2,3\end{array}}{\sum }d}\left(m_1^2+m_2^2+m_3^2+m_4^k\right)=I_1\left(x,y\right)+I_2\left(x,y\right)$ (16)

其中

$I_1\left(x,y\right)={\displaystyle {\int }_{\mathfrak{M}}{f}_2^3}\left(\alpha \right)f_k\left(\alpha \right)f\left(-\alpha \right)\text{d}\alpha ,I_2\left(x,y\right)={\displaystyle {\int }_{\mathfrak{m}}{f}_2^3}\left(\alpha \right)f_k\left(\alpha \right)f\left(-\alpha \right)\text{d}\alpha .$

4.1. 主区间上积分的估计

由主区间的定义，我们有

$I_1\left(x,y\right)={\displaystyle \underset{1⩽q⩽Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{\displaystyle {\int }_{\mathfrak{M}\left(a,q\right)}{f}_2{}^3}}}\left(\alpha \right)f_k\left(\alpha \right)f\left(-\alpha \right)\text{d}\alpha ={\displaystyle \underset{1⩽q⩽Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{\displaystyle {\int }_{-\frac{1}{q\tau }}^{\frac{1}{q\tau }}{f}_2^3}}}\left(\frac{a}{q}+\lambda \right)f_k\left(\frac{a}{q}+\lambda \right)f\left(-\frac{a}{q}-\lambda \right)\text{d}\lambda ,$ (17)

考察被积函数 $f_2^3\left(\alpha \right)f_k\left(\alpha \right)f\left(-\alpha \right)$，由引理3.1以及Zhang和Li [9] 文中关于 $f_k\left(\alpha \right)$ 的结论

$f_k\left(\alpha \right)=\frac{S_k\left(q,a\right)}{q}{T}_k\left(N_1^{\frac{1}{k}},N_2^{\frac{1}{k}}\right)+O\left(q^{1-1/k}\left(1+\left|\lambda \right|y\right)\right),$

$S_k\left(q,a\right)={\displaystyle \underset{r=1}{\overset{q}{\sum }}e}\left(\frac{a{r}^k}{q}\right)\ll q^{1-1/k},$ (18)

我们有

$f_2^3\left(\alpha \right)f_k\left(\alpha \right)f\left(-\alpha \right)=m\left(q;x,y\right)+r_1\left(q;x,y\right)+r_2\left(q;x,y\right)+r_3\left(q;x,y\right),$

其中

$m\left(q;x,y\right)=\frac{S_2^3\left(q,a\right)S_k\left(q,a\right)}{q^5}{T}_2^3\left(N_1^{\frac{1}{2}},N_2^{\frac{1}{2}}\right)T_k\left(N_1^{\frac{1}{k}},N_2^{\frac{1}{k}}\right)\left(T_1^{*}\left(4N_1,4N_2\right)+2\left(\gamma -\log q\right)T_1\left(4N_1,4N_2\right)\right),$ (19)

$r_1\left(q;x,y\right)\ll q^{-\frac{3}{2}-\frac{1}{k}}\left(1+\left|\lambda \right|y\right)T_2^2\left(N_1^{\frac{1}{2}},N_2^{\frac{1}{2}}\right)T_k\left(N_1^{\frac{1}{k}},N_2^{\frac{1}{k}}\right)\left(T_1^{*}\left(4N_1,4N_2\right)+2\left(\gamma -\log q\right)T_1\left(4N_1,4N_2\right)\right),$

$r_2\left(q;x,y\right)\ll q^{-\frac{3}{2}-\frac{1}{k}}\left(1+\left|\lambda \right|y\right)T_2^3\left(N_1^{\frac{1}{2}},N_2^{\frac{1}{2}}\right)\left(T_1^{*}\left(4N_1,4N_2\right)+2\left(\gamma -\log q\right)T_1\left(4N_1,4N_2\right)\right),$

$r_3\left(q;x,y\right)\ll q^{-\frac{3}{2}-\frac{1}{k}}\left(q^{\frac{3}{2}}{x}^{\epsilon }+q^{\frac{2}{3}}{x}^{\frac{1}{3}+\epsilon }+q^{\frac{3}{2}}\left|\lambda \right|x^{1+\epsilon }+q^{\frac{3}{2}}{\left|\lambda \right|}^2{x}^{1+\epsilon }y\right)T_2^3\left(N_1^{\frac{1}{2}},N_2^{\frac{1}{2}}\right)T_k\left(N_1^{\frac{1}{k}},N_2^{\frac{1}{k}}\right).$

将上式代入到公式(17)中，我们有

$I_1\left(x,y\right)=M\left(x,y\right)+R_1\left(q;x,y\right)+R_2\left(q;x,y\right)+R_3\left(q;x,y\right),$ (20)

其中

$M\left(x,y\right)={\displaystyle \underset{1⩽q⩽Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{\displaystyle {\int }_{-\frac{1}{q\tau }}^{\frac{1}{q\tau }}m\left(q;x,y\right)\text{d}\lambda }}},R_1\left(q;x,y\right)={\displaystyle \underset{1⩽q⩽Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{\displaystyle {\int }_{-\frac{1}{q\tau }}^{\frac{1}{q\tau }}{r}_1\left(q;x,y\right)\text{d}\lambda }}}$

$R_2\left(q;x,y\right)={\displaystyle \underset{1⩽q⩽Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{\displaystyle {\int }_{-\frac{1}{q\tau }}^{\frac{1}{q\tau }}{r}_2\left(q;x,y\right)\text{d}\lambda }}},R_3\left(q;x,y\right)={\displaystyle \underset{1⩽q⩽Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{\displaystyle {\int }_{-\frac{1}{q\tau }}^{\frac{1}{q\tau }}{r}_3\left(q;x,y\right)\text{d}\lambda }}}$

由Zhang和Li [9] 文中相关的结论

$T_1\left(u,v\right)\ll \min \left(v-u,\frac{1}{\Vert \lambda \Vert }\right),T_1^{*}\left(u,v\right)\ll \left(\log v\right)\min \left(v-u,\frac{1}{\Vert \lambda \Vert }\right),$

$T_k\left(N_1^{1/k},N_2^{1/k}\right)\ll \min \left(y{x}^{-1+1/k},\frac{1}{x^{1-1/k}\Vert \lambda \Vert }\right),$

我们有

$\begin{array}{c}{R}_1\left(q;x,y\right)\ll {\displaystyle \underset{1⩽q⩽Q}{\sum }{q}^{-\frac{1}{2}-\frac{1}{k}}{\displaystyle {\int }_{-\frac{1}{q\tau }}^{\frac{1}{q\tau }}\left(1+\left|\lambda \right|y\right)T_2^2}}\left(N_1^{\frac{1}{2}},N_2^{\frac{1}{2}}\right)T_k\left(N_1^{\frac{1}{k}},N_2^{\frac{1}{k}}\right)\times \left(T_1^{*}\left(4N_1,4N_2\right)+2\left(\gamma -\log q\right)T_1\left(4N_1,4N_2\right)\right)\text{d}\lambda \\ \ll {\displaystyle \underset{1⩽q⩽Q}{\sum }{q}^{-\frac{1}{2}-\frac{1}{k}}{\displaystyle {\int }_{-\frac{1}{q\tau }}^{\frac{1}{q\tau }}\left(1+\left|\lambda \right|y\right)}}\left(\log x\right)\min \left(y,\frac{1}{\Vert \lambda \Vert }\right)\min {\left(y{x}^{-\frac{1}{2}},\frac{1}{x^{\frac{1}{2}}\Vert \lambda \Vert }\right)}^2\min \left(y{x}^{-1+\frac{1}{k}},\frac{1}{x^{1-\frac{1}{k}}\Vert \lambda \Vert }\right)\text{d}\lambda \\ \ll {\displaystyle \underset{1⩽q⩽Q}{\sum }{q}^{-\frac{1}{2}-\frac{1}{k}}}\left(\log x\right)\left\{{\displaystyle {\int }_0^{\frac{1}{y}}{y}^4}{x}^{-2+\frac{1}{k}}\left(1+\lambda y\right)\text{d}\lambda +{\displaystyle {\int }_{\frac{1}{y}}^{\frac{1}{q\tau }}{x}^{-2+\frac{1}{k}}}{\lambda }^{-4}\left(1+\lambda y\right)\text{d}\lambda \right\}\\ \ll {\displaystyle \underset{1⩽q⩽Q}{\sum }{q}^{-\frac{1}{2}-\frac{1}{k}}}\left(\log x\right)y^3{x}^{-2+\frac{1}{k}}\\ \ll y^3{x}^{-2+\frac{1}{k}}{Q}^{\frac{1}{2}-\frac{1}{k}}\log x,\end{array}$ (21)

同理有

$R_2\left(q;x,y\right)={\displaystyle \underset{1⩽q⩽Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{\displaystyle {\int }_{-\frac{1}{q\tau }}^{\frac{1}{q\tau }}{r}_2\left(q;x,y\right)\text{d}\lambda }}}\ll y^3{x}^{-\frac{3}{2}}{Q}^{\frac{1}{2}-\frac{1}{k}}\log x,$ (22)

$R_3\left(q;x,y\right)={\displaystyle \underset{1⩽q⩽Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{\displaystyle {\int }_{-\frac{1}{q\tau }}^{\frac{1}{q\tau }}{r}_3\left(q;x,y\right)\text{d}\lambda }}}\ll y^3{x}^{-\frac{5}{2}+\frac{1}{k}+\epsilon }{Q}^{2-\frac{1}{k}}+y^3{x}^{-\frac{13}{6}+\frac{1}{k}+\epsilon }{Q}^{\frac{7}{6}-\frac{1}{k}}+y^2{x}^{-\frac{3}{2}+\frac{1}{k}+\epsilon }{Q}^{2-\frac{1}{k}}.$ (23)

对于 $M\left(x,y\right)$，我们有

$M\left(x,y\right)={\displaystyle \underset{1⩽q⩽Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{\displaystyle {\int }_{-\frac{1}{q\tau }}^{\frac{1}{q\tau }}m\left(q;x,y\right)\text{d}\lambda }}}=E_1\left(x,y\right)+E_2\left(x,y\right),$ (24)

其中

$E_1\left(x,y\right)={\displaystyle \underset{1⩽q⩽Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{\displaystyle {\int }_{-\frac{1}{2}}^{\frac{1}{2}}m\left(q;x,y\right)\text{d}\lambda }}},$

$E_2\left(x,y\right)={\displaystyle \underset{1⩽q⩽Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{\displaystyle {\int }_{-\frac{1}{2}}^{-\frac{1}{q\tau }}m}}}\left(q;x,y\right)\text{d}\lambda +{\displaystyle \underset{1⩽q⩽Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{\displaystyle {\int }_{\frac{1}{q\tau }}^{\frac{1}{2}}m}}}\left(q;x,y\right)\text{d}\lambda .$

对于 $E_2\left(x,y\right)$，我们有以下估计：

$\begin{array}{c}{E}_2\left(x,y\right)\ll {\displaystyle \underset{1⩽q⩽Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{\displaystyle {\int }_{\frac{1}{q\tau }}^{\frac{1}{2}}{x}^{-\frac{5}{2}+\frac{1}{k}}}}}\left(\log x\right){\lambda }^{-5}\text{d}\lambda \ll {\displaystyle \underset{1⩽q⩽Q}{\sum }{x}^{-\frac{5}{2}+\frac{1}{k}}}\left(\log x\right)q^{\frac{5}{2}-\frac{1}{k}}{\tau }^4\\ \ll x^{-\frac{5}{2}+\frac{1}{k}}\left(\log x\right)Q^{\frac{7}{2}-\frac{1}{k}}{\tau }^4\ll y^4{x}^{-\frac{5}{2}+\frac{1}{k}-\epsilon }.\end{array}$ (25)

对于 $E_1\left(x,y\right)$，由公式(19)，我们有

$\begin{array}{c}{E}_1\left(x,y\right)={\displaystyle \underset{q=1}{\overset{\infty }{\sum }}{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }\frac{S_2^3\left(q,a\right)S_k\left(q,a\right)}{q^5}}}\left({\mathfrak{L}}_1\left(x,y\right)+2\left(\gamma -\log q\right){\mathfrak{L}}_2\left(x,y\right)\right)\\ +{\displaystyle \underset{q>Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }\frac{S_2^3\left(q,a\right)S_k\left(q,a\right)}{q^5}}}\left({\mathfrak{L}}_1\left(x,y\right)+2\left(\gamma -\log q\right){\mathfrak{L}}_2\left(x,y\right)\right),\end{array}$ (26)

其中

$\begin{array}{c}{\mathfrak{L}}_1\left(x,y\right)={\displaystyle {\int }_{-\frac{1}{2}}^{\frac{1}{2}}{T}_2^3}\left(N_1^{\frac{1}{2}},N_2^{\frac{1}{2}}\right)T_k\left(N_1^{\frac{1}{k}},N_2^{\frac{1}{k}}\right)T_1^{*}\left(4N_1,4N_2\right)\text{d}\lambda \\ =\frac{1}{8k}{\displaystyle \underset{\begin{array}{c}{N}_1<m_i⩽N_2\\ i=1,2,3,4\end{array}}{\sum }\frac{1}{m_1^{1/2}{m}_2^{1/2}{m}_3^{1/2}{m}_4^{1-1/k}}}{\displaystyle \underset{\begin{array}{c}4N_1<n⩽4N_2\\ m_1+m_2+m_3+m_4=n\end{array}}{\sum }\log }\left(n\right)\\ \asymp \left(\log x\right){\displaystyle \underset{\begin{array}{c}{N}_1<m_i⩽N_2\\ i=1,2,3,4\end{array}}{\sum }\frac{1}{m_1^{1/2}{m}_2^{1/2}{m}_3^{1/2}{m}_4^{1-1/k}}}\asymp y^4{x}^{-\frac{5}{2}+\frac{1}{k}}\log x,\end{array}$ (27)

$\begin{array}{c}{\mathfrak{L}}_2\left(x,y\right)={\displaystyle {\int }_{-\frac{1}{2}}^{\frac{1}{2}}{T}_2^3}\left(N_1^{\frac{1}{2}},N_2^{\frac{1}{2}}\right)T_k\left(N_1^{\frac{1}{k}},N_2^{\frac{1}{k}}\right)T_1\left(4N_1,4N_2\right)\text{d}\lambda \\ =\frac{1}{8k}{\displaystyle \underset{\begin{array}{c}{N}_1<m_i⩽N_2\\ i=1,2,3,4\end{array}}{\sum }\frac{1}{m_1^{1/2}{m}_2^{1/2}{m}_3^{1/2}{m}_4^{1-1/k}}}{\displaystyle \underset{\begin{array}{c}4N_1<n⩽4N_2\\ m_1+m_2+m_3+m_4=n\end{array}}{\sum }1}\\ \asymp {\displaystyle \underset{\begin{array}{c}{N}_1<m_i⩽N_2\\ i=1,2,3,4\end{array}}{\sum }\frac{1}{m_1^{1/2}{m}_2^{1/2}{m}_3^{1/2}{m}_4^{1-1/k}}}\asymp y^4{x}^{-\frac{5}{2}+\frac{1}{k}}.\end{array}$ (28)

对于公式(26)中的求和项，我们有以下估计

$\begin{array}{l}{\displaystyle \underset{q>Q}{\sum }{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }\frac{S_2^3\left(q,a\right)S_k\left(q,a\right)}{q^5}}}\left({\mathfrak{L}}_1\left(x,y\right)+2\left(\gamma -\log q\right){\mathfrak{L}}_2\left(x,y\right)\right)\\ \ll {\displaystyle \underset{q⩾Q}{\sum }{q}^{-\frac{3}{2}-\frac{1}{k}}}{y}^4{x}^{-\frac{5}{2}+\frac{1}{k}}\log x\ll y^4{x}^{-\frac{5}{2}+\frac{1}{k}}\left(\log x\right)Q^{-\frac{1}{2}-\frac{1}{k}}\ll y^4{x}^{-\frac{5}{2}+\frac{1}{k}-\epsilon }.\end{array}$ (29)

综合公式(20)~(29)，我们有

$\begin{array}{c}{I}_1\left(x,y\right)=M\left(x,y\right)+R_1\left(q;x,y\right)+R_2\left(q;x,y\right)+R_3\left(q;x,y\right)\\ ={\mathcal{K}}_1{\mathfrak{L}}_1\left(x,y\right)+2\left(\gamma {\mathcal{K}}_1-{\mathcal{K}}_2\right){\mathfrak{L}}_2\left(x,y\right)+O\left(y^4{x}^{-\frac{5}{2}+\frac{1}{k}-\epsilon }\right)\\ +O\left(y^3{x}^{-\frac{3}{2}}{Q}^{\frac{1}{2}-\frac{1}{k}}\log x+y^3{x}^{-\frac{13}{6}+\frac{1}{k}+\epsilon }{Q}^{\frac{7}{6}-\frac{1}{k}}+y^2{x}^{-\frac{3}{2}+\frac{1}{k}+\epsilon }{Q}^{2-\frac{1}{k}}\right),\end{array}$ (30)

其中

${\mathcal{K}}_1={\displaystyle \underset{q=1}{\overset{\infty }{\sum }}\frac{1}{q^5}}{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{S}_2^3}\left(q,a\right)S_k\left(q,a\right),{\mathcal{K}}_2={\displaystyle \underset{q=1}{\overset{\infty }{\sum }}\frac{\log q}{q^5}}{\displaystyle \underset{\begin{array}{c}1⩽a⩽q\\ \left(a,q\right)=1\end{array}}{\sum }{S}_2^3}\left(q,a\right)S_k\left(q,a\right).$

接下来我们需要证明级数 ${\mathcal{K}}_1$， ${\mathcal{K}}_2$ 均收敛，定义Dirichlet级数

$L\left(s,\Phi \right)={\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\Phi \left(n\right)}{n^s}},$

其中

$\Phi \left(n\right)={\displaystyle \underset{\begin{array}{c}1⩽a⩽n\\ \left(a,n\right)=1\end{array}}{\sum }{S}_2^3}\left(n,a\right)S_k\left(n,a\right),$

由公式(18)得 $\Phi \left(n\right)\ll n^{\frac{5}{2}-\frac{1}{k}}\phi \left(n\right)\ll n^{\frac{7}{2}-\frac{1}{k}}$，则当 $Res>\frac{9}{2}-\frac{1}{k}$ 时，级数 $L\left(s,\Phi \right)$ 收敛。此时

$-L^{\prime }\left(s,\Phi \right)={\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\Phi \left(n\right)\log n}{n^s}},$

同理得 $\Phi \left(n\right)\log n\ll n^{\frac{5}{2}-\frac{1}{k}}\phi \left(n\right)\log n\ll n^{4-\frac{1}{k}}$，则当 $Res>5-\frac{1}{k}$ 时，级数 $-L^{\prime }\left(s,\Phi \right)$ 收敛。因此由 ${\mathcal{K}}_1=L\left(5,\Phi \right)$， ${\mathcal{K}}_2=-L^{\prime }\left(5,\Phi \right)$ 知 ${\mathcal{K}}_1$， ${\mathcal{K}}_2$ 均为收敛的级数。

4.2. 余区间上积分的估计

由余区间的定义，当 $\alpha \in \mathfrak{m}$ 时， $\alpha =\frac{a}{q}+\lambda$， $\left|\lambda \right|⩽\frac{1}{q\tau }$， $1⩽a⩽q$， $\left(a,q\right)=1$，且q满足

$Q<q⩽\tau .$ (31)

设 $X-Y=N_1^{\frac{1}{k}}$， $X+Y=N_1^{\frac{1}{k}}$， $\mathcal{X}=X-Y$， $m=n+\mathcal{X}$，改写 $f_k\left(\alpha \right)$ 为

$f_k\left(\alpha \right)={\displaystyle \underset{N_1<m^k⩽N_2}{\sum }e}\left(m^k\alpha \right)={\displaystyle \underset{X-Y<m⩽X+Y}{\sum }e}\left(m^k\alpha \right)=e\left(\alpha {\mathcal{X}}^k\right){\displaystyle \underset{1⩽n⩽2Y}{\sum }e}\left({\alpha }_k{n}^k+{\alpha }_{k-1}{n}^{k-1}+\cdots +{\alpha }_{1}n\right),$ (32)

其中

${\alpha }_j=\left(\begin{array}{c}k\\ j\end{array}\right){\mathcal{X}}^{k-j}\alpha ,\left(1⩽j⩽k\right).$

4.2.1. k = 3时，定理1.1的证明

当k = 3时，对 $f_2^2\left(\alpha \right)$ 进行估计，我们有

${\left|f_2\left(\alpha \right)\right|}^2={\displaystyle \underset{N_1^{\frac{1}{2}}<m,n⩽N_2^{\frac{1}{2}}}{\sum }e}\left(\left(m^2-n^2\right)\alpha \right)=\left(N_2^{\frac{1}{2}}-N_1^{\frac{1}{2}}\right)+T\left(x\right)+\overline{T\left(x\right)},$ (33)

其中

$T\left(x\right)={\displaystyle \underset{N_1^{\frac{1}{2}}<n<m⩽N_2^{\frac{1}{2}}}{\sum }e}\left(\left(m^2-n^2\right)\alpha \right).$

对于T(x)，我们有以下估计

$\begin{array}{c}T\left(x\right)={\displaystyle \underset{1⩽v⩽N_2^{\frac{1}{2}}-N_1^{\frac{1}{2}}}{\sum }e}\left(v^2\alpha \right){\displaystyle \underset{N_1^{\frac{1}{2}}<n⩽N_2^{\frac{1}{2}}-v}{\sum }e}\left(2nv\alpha \right)\ll {\displaystyle \underset{1⩽v⩽y{x}^{-\frac{1}{2}}}{\sum }\min }\left(y{x}^{-\frac{1}{2}},\frac{1}{\Vert 2v\alpha \Vert }\right)\ll \left(y{x}^{-\frac{1}{2}}+q\log q\right)\left(\frac{y{x}^{-\frac{1}{2}}}{q}+1\right)\\ \ll y^2{x}^{-1}{q}^{-1}+y{x}^{-\frac{1}{2}}\log q+q\log q,\end{array}$

结合上式以及公式(31)、(33)，我们有

$\underset{\alpha \in \mathfrak{m}}{\sup }\left|f_2\left(\alpha \right)\right|\ll y{x}^{-\frac{1}{2}}{Q}^{-\frac{1}{2}}+y^{\frac{1}{2}}{x}^{-\frac{1}{4}}{\log }^{\frac{1}{2}}x+{\tau }^{\frac{1}{2}}{\log }^{\frac{1}{2}}x.$ (34)

由Hölder不等式，引理3.2，公式(34)，以及Zhang和Li [9] 对于 $f\left(-\alpha \right)$ 积分均值的估计

${\displaystyle {\int }_0^1{\left|f\left(-\alpha \right)\right|}^2\text{d}\alpha }\ll y{\log }^{3}x,$ (35)

我们有

$\begin{array}{c}{I}_2\left(x,y\right)\ll \underset{\alpha \in \mathfrak{m}}{\sup }\left|f_2\left(\alpha \right)\right|{\left({\displaystyle {\int }_0^1{\left|f_2\left(\alpha \right)\right|}^6\text{d}\alpha }\right)}^{\frac{1}{3}}{\left({\displaystyle {\int }_0^1{\left|f_3\left(\alpha \right)\right|}^6\text{d}\alpha }\right)}^{\frac{1}{6}}{\left({\displaystyle {\int }_0^1{\left|f\left(-\alpha \right)\right|}^2\text{d}\alpha }\right)}^{\frac{1}{2}}\\ \ll y^{\frac{5}{2}}{x}^{-\frac{10}{9}+\frac{\epsilon }{2}}{\log }^{\frac{3}{2}}x\left(y{x}^{-\frac{1}{2}}{Q}^{-\frac{1}{2}}+y^{\frac{1}{2}}{x}^{-\frac{1}{4}}{\log }^{\frac{1}{2}}x+{\tau }^{\frac{1}{2}}{\log }^{\frac{1}{2}}x\right)\\ \ll y^3{x}^{-\frac{49}{36}+\epsilon }+y^{\frac{7}{2}}{x}^{-\frac{29}{18}+\epsilon }{Q}^{-\frac{1}{2}}.\end{array}$ (36)

结合公式(16)、(30)、(36)，我们有

$\begin{array}{c}{S}_3\left(x,y\right)={\mathcal{K}}_1{\mathfrak{L}}_1\left(x,y\right)+2\left(\gamma {\mathcal{K}}_1-{\mathcal{K}}_2\right){\mathfrak{L}}_2\left(x,y\right)+O\left(y^4{x}^{-\frac{13}{6}-\epsilon }\right)\\ +O\left(y^3{x}^{-\frac{3}{2}}{Q}^{\frac{1}{6}}\log x+y^3{x}^{-\frac{11}{6}+\epsilon }{Q}^{\frac{5}{6}}+y^2{x}^{-\frac{7}{6}+\epsilon }+y^{\frac{7}{2}}{x}^{-\frac{29}{18}+\epsilon }{Q}^{-\frac{1}{2}}\right),\end{array}$

且当 $x^{\frac{11}{36}}\ll Q\ll x^{\frac{11}{36}+\frac{7}{2}\epsilon }$，我们有

$S_3\left(x,y\right)={\mathcal{K}}_1{\mathfrak{L}}_1\left(x,y\right)+2\left(\gamma {\mathcal{K}}_1-{\mathcal{K}}_2\right){\mathfrak{L}}_2\left(x,y\right)+O\left(y^4{x}^{-\frac{13}{6}-\epsilon }\right).$

4.2.2. 4 ≤ k ≤ 5时，定理1.1的证明

当4 ≤ k ≤ 5时，由引理3.3以及公式(32)，我们有

$\begin{array}{c}{f}_k\left(\alpha \right)=e\left(\alpha {\mathcal{X}}^k\right){\displaystyle \underset{1⩽n⩽2Y}{\sum }e}\left({\alpha }_k{n}^k+{\alpha }_{k-1}{n}^{k-1}+\cdots +{\alpha }_{1}n\right)\\ \ll {\displaystyle \underset{1⩽n⩽2Y}{\sum }e}\left({\alpha }_k{n}^k+{\alpha }_{k-1}{n}^{k-1}+\cdots +{\alpha }_{1}n\right)\\ \ll Y^{1+\epsilon }{\left(q^{-1}+Y^{-1}+q{Y}^{-k}\right)}^{\frac{1}{2^{k-1}}},\end{array}$

由上式以及公式(31)，我们有

$\begin{array}{c}\underset{\alpha \in \mathfrak{m}}{\sup }\left|f_k\left(\alpha \right)\right|\ll Y^{1+\epsilon }{\left(Q^{-1}+Y^{-1}+\tau Y^{-k}\right)}^{\frac{1}{2^{k-1}}}\\ \ll y{x}^{-1+\frac{1}{k}+\epsilon }{\left(Q^{-1}+y^{-1}{x}^{1-\frac{1}{k}}+\tau y^{-k}{x}^{k-1}\right)}^{\frac{1}{2^{k-1}}}\\ \ll y{x}^{-1+\frac{1}{k}+\epsilon }{\left(Q^{-1}+y^{-1}{x}^{1-\frac{1}{k}}\right)}^{\frac{1}{2^{k-1}}}.\end{array}$ (37)

由Hölder不等式，引理3.2以及公式(35)、(37)，我们有

$\begin{array}{c}{I}_2\left(x,y\right)\ll \underset{\alpha \in \mathfrak{m}}{\sup }\left|f_k\left(\alpha \right)\right|{\left({\displaystyle {\int }_0^1{\left|f_2\left(\alpha \right)\right|}^6\text{d}\alpha }\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_0^1{\left|f\left(-\alpha \right)\right|}^2\text{d}\alpha }\right)}^{\frac{1}{2}}\\ \ll y^{\frac{7}{2}}{x}^{-2+\frac{1}{k}+2\epsilon }{\left(Q^{-1}+y^{-1}{x}^{1-\frac{1}{k}}\right)}^{\frac{1}{2^{k-1}}}\\ \ll y^{\frac{7}{2}}{x}^{-2+\frac{1}{k}+2\epsilon }{Q}^{-\frac{1}{2^{k-1}}}+y^{\frac{7}{2}-\frac{1}{2^{k-1}}}{x}^{-2+\frac{1}{k}+\frac{k-1}{k\cdot 2^{k-1}}+2\epsilon }.\end{array}$ (38)

结合公式(16)、(30)、(38)，我们有

$\begin{array}{c}{S}_k\left(x,y\right)={\mathcal{K}}_1{\mathfrak{L}}_1\left(x,y\right)+2\left(\gamma {\mathcal{K}}_1-{\mathcal{K}}_2\right){\mathfrak{L}}_2\left(x,y\right)+O\left(y^4{x}^{-\frac{5}{2}+\frac{1}{k}-\epsilon }\right)\\ +O\left(y^3{x}^{-\frac{3}{2}}{Q}^{\frac{1}{2}-\frac{1}{k}}+y^3{x}^{-\frac{11}{6}+\frac{1}{k}+\epsilon }{Q}^{\frac{7}{6}-\frac{1}{k}}+y^2{x}^{-\frac{3}{2}+\frac{1}{k}+\epsilon }{Q}^{2-\frac{1}{k}}+y^{\frac{7}{2}}{x}^{-2+\frac{1}{k}+2\epsilon }{Q}^{-\frac{1}{2^{k-1}}}\right),\end{array}$

且当 $x^{\frac{2^{k-2}}{2^{k-1}+k+1}+2^{k-1}\epsilon }\ll Q\ll x^{\frac{2^{k-2}+1/2}{2^{k-1}+k+1}+\left(2k-1/2\right)\epsilon }$，我们有

$S_k\left(x,y\right)={\mathcal{K}}_1{\mathfrak{L}}_1\left(x,y\right)+2\left(\gamma {\mathcal{K}}_1-{\mathcal{K}}_2\right){\mathfrak{L}}_2\left(x,y\right)+O\left(y^4{x}^{-\frac{5}{2}+\frac{1}{k}-\epsilon }\right).$

4.2.3. k ≥ 6时，定理1.1的证明

当k ≥ 6时，由引理3.4以及公式(32)，我们有

$\begin{array}{c}{f}_k\left(\alpha \right)=e\left(\alpha {\mathcal{X}}^k\right){\displaystyle \underset{1⩽n⩽2Y}{\sum }e}\left({\alpha }_k{n}^k+{\alpha }_{k-1}{n}^{k-1}+\cdots +{\alpha }_{1}n\right)\\ \ll {\displaystyle \underset{1⩽n⩽2Y}{\sum }e}\left({\alpha }_k{n}^k+{\alpha }_{k-1}{n}^{k-1}+\cdots +{\alpha }_{1}n\right)\\ \ll Y^{1+\epsilon }{\left(q^{-1}+Y^{-1}+q{Y}^{-k}\right)}^{\frac{1}{2k\left(k-1\right)}},\end{array}$

由上式以及公式(31)，我们有

$\begin{array}{c}\underset{\alpha \in \mathfrak{m}}{\sup }\left|f_k\left(\alpha \right)\right|\ll Y^{1+\epsilon }{\left(Q^{-1}+Y^{-1}+\tau Y^{-k}\right)}^{\frac{1}{2k\left(k-1\right)}}\\ \ll y{x}^{-1+\frac{1}{k}+\epsilon }{\left(Q^{-1}+y^{-1}{x}^{1-\frac{1}{k}}+\tau y^{-k}{x}^{k-1}\right)}^{\frac{1}{2k\left(k-1\right)}}\\ \ll y{x}^{-1+\frac{1}{k}+\epsilon }{\left(Q^{-1}+y^{-1}{x}^{1-\frac{1}{k}}\right)}^{\frac{1}{2k\left(k-1\right)}}.\end{array}$ (39)

由Hölder不等式，引理3.1以及公式(35)、(39)，我们有

$\begin{array}{c}{I}_2\left(x,y\right)\ll \underset{\alpha \in \mathfrak{m}}{\sup }\left|f_k\left(\alpha \right)\right|{\left({\displaystyle {\int }_0^1{\left|f_2\left(\alpha \right)\right|}^6\text{d}\alpha }\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_0^1{\left|f\left(-\alpha \right)\right|}^2\text{d}\alpha }\right)}^{\frac{1}{2}}\\ \ll y^{\frac{7}{2}}{x}^{-2+\frac{1}{k}+2\epsilon }{\left(Q^{-1}+y^{-1}{x}^{1-\frac{1}{k}}\right)}^{\frac{1}{2k\left(k-1\right)}}\\ \ll y^{\frac{7}{2}}{x}^{-2+\frac{1}{k}+2\epsilon }{Q}^{-\frac{1}{2k\left(k-1\right)}}+y^{\frac{7}{2}-\frac{1}{2k\left(k-1\right)}}{x}^{-2+\frac{1}{k}+\frac{1}{2k^2}+2\epsilon }.\end{array}$ (40)

结合公式(16)、(30)、(40)，我们有

$\begin{array}{c}{S}_k\left(x,y\right)={\mathcal{K}}_1{\mathfrak{L}}_1\left(x,y\right)+2\left(\gamma {\mathcal{K}}_1-{\mathcal{K}}_2\right){\mathfrak{L}}_2\left(x,y\right)+O\left(y^4{x}^{-\frac{5}{2}+\frac{1}{k}-\epsilon }\right)\\ +O\left(y^3{x}^{-\frac{3}{2}}{Q}^{\frac{1}{2}-\frac{1}{k}}+y^3{x}^{-\frac{11}{6}+\frac{1}{k}+\epsilon }{Q}^{\frac{7}{6}-\frac{1}{k}}+y^2{x}^{-\frac{3}{2}+\frac{1}{k}+\epsilon }{Q}^{2-\frac{1}{k}}+y^{\frac{7}{2}}{x}^{-2+\frac{1}{k}+2\epsilon }{Q}^{-\frac{1}{2k\left(k-1\right)}}\right),\end{array}$

且当 $x^{\frac{2k-2}{k^2-3k+5}+2k\left(k-1\right)\epsilon }\ll Q\ll x^{\frac{2k^2-6k+6}{\left(k-2\right)\left(k^2-3k+5\right)}+\frac{6k}{k-2}\epsilon }$，我们有

$S_k\left(x,y\right)={\mathcal{K}}_1{\mathfrak{L}}_1\left(x,y\right)+2\left(\gamma {\mathcal{K}}_1-{\mathcal{K}}_2\right){\mathfrak{L}}_2\left(x,y\right)+O\left(y^4{x}^{-\frac{5}{2}+\frac{1}{k}-\epsilon }\right).$

结合以上三种情况的讨论，我们最终证明了定理1.1。