On the Symmetric Form of the Three Primes Theorem Weighted by Δ(x)

Abstract

Let x be a large real number, $d\left(n\right)$ be the Dirichlet divisor function and $\Delta \left(x\right)={\sum }_{n⩽x}d\left(n\right)-xlogx-(2\gamma -1)x$. In this paper, we consider a weighted form of the three primes theorem: $S(N;k_1,k_2,k_3):={\sum }_{p_1+p_2+p_3=N}{\Delta }^{k_1}\left(p_1\right){\Delta }^{k_2}\left(p_2\right){\Delta }^{k_3}\left(p_3\right),$ where $p_1,p_2,p_3$ run over primes, $k_1,k_2,k_3\in \{1,2,\cdots ,9\}$, and N is an large odd integer. For the case $k_i\in \{2,3,\cdots ,9\}$ $(i=1,2,3)$, the others two $k_j\in \{2,3,4\}$ with $j\ne i$, an asymptotic formula for $S(N;k_1,k_2,k_3)$ has been derived, along with a non-trivial upper bound estimate for $S(N;k_1,k_2,k_3)$ when $min(k_1,k_2,k_3)=1$.

1. Introduction

The general divisor problem is to study the asymptotic behavior of the well known sum

$$D\left(x\right):=\sum _{n⩽x}d\left(n\right)=xlogx+(2\gamma -1)x+\Delta \left(x\right),$$

(1)

as x tends to infinity, where $\gamma$ is the Euler’s constant and $\Delta \left(x\right)$ is the “error term”. It was first proved by Dirichlet that $\Delta \left(x\right)=O\left(x^{1/2}\right)$. Furthermore, the exponent $1/2$ was improved by many authors [1,2,3,4]. Until now, the best result, that is

$$\Delta \left(x\right)\ll x^{131/416}{(logx)}^{26497/8320},$$

(2)

was given by Huxley [5]. It is conjectured that

$$\Delta \left(x\right)=O\left(x^{1/4+\epsilon }\right)$$

(3)

holds for any $\epsilon >0$, which is supported by the classical mean-square result

$${\int }_1^T{\Delta }^2\left(x\right)dx=C_2{T}^{3/2}+O(T{log}^{3}TloglogT)$$

(4)

proved by Lau and Tsang [6], where $C_2$ is a positive constant which can be written down explicitly by the Riemann zeta-function $\zeta \left(s\right)$, and the upper bound estimate

$${\int }_1^T{|\Delta \left(x\right)|}^{A_0}dx\ll T^{1+{\displaystyle \frac{A_0}{4}}+\epsilon },$$

(5)

where $A_0>2$ is a real number. The estimate of type (5) can be found in Ivić [7] with $A_0=35/4$. In this paper, we use Huxley’s result (2) and Ivić’s [7] method to obtain that we can take $A_0=262/27$ (similar to Lemma 8 in this paper).

For any integer $3⩽k⩽9$, we have derived the asymptotic formula of unified form

$${\int }_1^T{\Delta }^k\left(x\right)dx=C_k{T}^{1+k/4}+O(T^{1+{\displaystyle \frac{k}{4}}-{\delta }_k+\epsilon }),$$

(6)

where $C_k$ and ${\delta }_k>0$ are explicit constants. The exact value of ${\delta }_k$ was studied by many authors [8,9,10,11,12,13,14].

Also, the discrete mean values

$${\mathcal{D}}_k\left(x\right):=\sum _{n⩽x}{\Delta }^k\left(n\right),1⩽k,k\in \mathbb{N}$$

(7)

have been investigated by various authors. Define the continuous mean values

$${\mathcal{C}}_k\left(x\right):={\int }_1^x{\Delta }^k\left(t\right)dt.$$

(8)

Many authors found that the discrete and the continuous mean value formulas are connected deeply with each other and studied the difference between ${\mathcal{D}}_k\left(x\right)$ and ${\mathcal{C}}_k\left(x\right)$ [4,15,16,17,18].

As analogues of the discrete mean values, we can consider the sum of $\Delta (·)$ over a subset of $\mathbb{N}$. For example, the sum over primes, namely

$$\sum _{p⩽x}{\Delta }^k\left(p\right)$$

(9)

for integer $2⩽k⩽9$. The authors [19,20] have proved that, for $2⩽k⩽9$, (9) has the asymptotic formula

$$\sum _{p⩽x}{\Delta }^k\left(p\right)=B_k(\infty )\sum _{p⩽x}{p}^{k/4}+O(x^{1+{\displaystyle \frac{k}{4}}-\theta (k,A_0)+\epsilon }),$$

(10)

where $B_k(\infty )$ and $\theta (k,A_0)>0$ $(2⩽k⩽9)$ are computable constants.

Vinogradov [21] proved that, for every sufficiently large odd integer N, the equation

$$N=p_1+p_2+p_3$$

(11)

is solvable for prime numbers $p_1,p_2,p_3$. More precisely, he proved the asymptotic formula

$$\sum _{p_1+p_2+p_3=N}1={\displaystyle \frac{1}{2}}\mathfrak{S}\left(N\right){\displaystyle \frac{N^2}{{log}^{3}N}}+O\left({\displaystyle \frac{N^2}{{log}^{4}N}}\right),$$

(12)

where $\mathfrak{S}\left(N\right)$ is the singular series

$$\mathfrak{S}\left(N\right)=\prod _{p\mid N}\left(1-{\displaystyle \frac{1}{{(p-1)}^2}}\right)\prod _{p\nmid N}\left(1-{\displaystyle \frac{1}{{(p-1)}^3}}\right).$$

(13)

Formula (12) is called the three primes theorem, or Vinogradov–Goldbach theorem.

2. Main Result

In this paper, we consider the sum of $\Delta (·)$ over primes in arithmetic progression, namely

$$\sum _{\begin{array}{c}p⩽x\\ p\equiv \ell \left(modq\right)\end{array}}{\Delta }^k\left(p\right),$$

(14)

where $q>1$ is an integer, $\ell ⩽q$, $(\ell ,q)=1$. Furthermore, we consider an exponential sum connected with $\Delta (·)$, namely

$$\sum _{p⩽x}\Delta \left(p\right)\mathbf{e}\left(p\alpha \right).$$

(15)

We obtain asymtotic formula for (14) and an upper bound estimate for (15). These results enable us to prove an analogue of Vinogradov’s three primes theorem with coefficients $\Delta (·)$. for $k_1,k_2,k_3\in \{1,2,\cdots ,9\}$, i.e., the asymptotic formula of

$$S(N;k_1,k_2,k_3):=\sum _{p_1+p_2+p_3=N}{\Delta }^{k_1}\left(p_1\right){\Delta }^{k_2}\left(p_2\right){\Delta }^{k_3}\left(p_3\right).$$

(16)

Then, we have the following theorems.

Theorem 1.

Suppose $S(N;k_1,k_2,k_3)$ is defined by (16). When one of $k_i\in \{2,3,\cdots ,9\}$ $(i=1,2,3)$, the other two $k_j\in \{2,3,4\}$ $(j\ne i)$, we have the asymptotic formula

$$\begin{array}{cc}\hfill S(N;k_1,k_2,k_3)=& \left(3+{\displaystyle \frac{k_1+k_2+k_3}{4}}\right){\displaystyle \frac{\Gamma \left(1+\frac{k_1}{4}\right)\Gamma \left(1+\frac{k_2}{4}\right)\Gamma \left(1+\frac{k_2}{4}\right)}{\Gamma \left(3+\frac{k_1+k_2+k_3}{4}\right)}}\hfill \\ \hfill & \times \mathcal{C}(k_1,k_2,k_3)\mathfrak{S}\left(N\right){\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{3}N}}+O\left({\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{4}N}}\right),\hfill \end{array}$$

(17)

where $\mathcal{C}(k_1,k_2,k_3)$ are computable constants, Γ denotes the Gamma function and $\mathfrak{S}\left(N\right)$ is defined in (12).

Theorem 2.

If there exists $k_j=1$ for $j\in \{1,2,3\}$, then

$$S(N;k_1,k_2,k_3)\ll N^{2+{\displaystyle \frac{k_1+k_2+k_3}{4}}-{\displaystyle \frac{1}{16}}+\epsilon }.$$

(18)

By Theorem 1, Equation (12), and the Cauchy–Schwarz’s inequality, we obtain that the trivial upper bound for the case in Theorem 2 is

$$S(N;k_1,k_2,k_3)\ll {\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{3}N}},$$

(19)

and thus, Theorem 2 gives a non-trivial upper bound.

Notations

Throughout this paper, N is a large positive odd number and $\epsilon$ denotes a sufficiently small positive number, not necessarily the same at each occurrence. Let p, with or without subscripts, always denote a prime number. As usual, $d\left(n\right)$, $\Lambda \left(n\right)$, $\mu \left(n\right)$, and $\phi \left(n\right)$ denote the Dirichlet divisor function, the von Mangoldt function, the Möbius function, and the Euler totient function, respectively, $\mathbb{C}$ denotes the set of complex numbers, $\mathbb{R}$ denotes the set of real numbers, and $\mathbb{N}$ denotes the set of natural numbers. We write $\mathbf{e}\left(t\right):=exp\left(2\pi it\right)$. The notation $m\sim M$ means $M<m⩽2M$, $f\left(x\right)\ll g\left(x\right)$ means $f\left(x\right)=O\left(g\right(x\left)\right)$; $f\left(x\right)\asymp g\left(x\right)$ means $f\left(x\right)\ll g\left(x\right)\ll f\left(x\right)$.

3. The Circle Method

For $k_1,k_2,k_3\in \{1,2,\cdots ,9\}$, $\alpha \in \mathbb{R}$ and $N>1$, define

$$F_i\left(\alpha \right):=\sum _{p⩽N}{\Delta }^{k_i}\left(p\right)\mathbf{e}\left(p\alpha \right),i=1,2,3.$$

(20)

By the definition of $S(N;k_1,k_2,k_3)$ and the well-known identity

$${\int }_0^1\mathbf{e}\left(u\alpha \right)d\alpha =\left\{\begin{array}{c}1\mathrm{if}u=0,\hfill \\ 0\mathrm{if}u\in \mathbb{Z},u\ne 0,\hfill \end{array}\right.$$

(21)

we have

$$\begin{array}{cc}\hfill S(N;k_1,k_2,k_3)& ={\int }_0^1{F}_1\left(\alpha \right)F_2\left(\alpha \right)F_3\left(\alpha \right)\mathbf{e}(-N\alpha )d\alpha \hfill \\ \hfill & ={\int }_{{\displaystyle \frac{1}{c}}}^{1+{\displaystyle \frac{1}{c}}}{F}_1\left(\alpha \right)F_2\left(\alpha \right)F_3\left(\alpha \right)\mathbf{e}(-N\alpha )d\alpha \hfill \end{array}$$

(22)

for any $c>0$.

In order to compute the integral, we choose two parameters Q and $\tau$ as follows:

$$Q={log}^{A}N,\tau ={\displaystyle \frac{N}{{log}^{B}N}}$$

(23)

with $A,B$ real, $A>15$ and $B>A+2$.

For any $0⩽a<q⩽Q$ with $(a,q)=1$, define

$$\begin{array}{cc}\hfill I(a,q)& :=\left[{\displaystyle \frac{a}{q}}-{\displaystyle \frac{1}{q\tau }},{\displaystyle \frac{a}{q}}+{\displaystyle \frac{1}{q\tau }}\right],\hfill \\ \hfill E_1& :=\bigcup _{1⩽q⩽Q}\bigcup _{\begin{array}{c}0⩽a<q\\ (a,q)=1\end{array}}I(a,q),E_2=\left[{\displaystyle \frac{1}{\tau }},1+{\displaystyle \frac{1}{\tau }}\right]\setminus E_1.\hfill \end{array}$$

(24)

We call $E_1$ the $major$ $arc$ and $E_2$ the $minor$ $arc$.

We then have

$$S(N;k_1,k_2,k_3)=S_1\left(N\right)+S_2\left(N\right),$$

(25)

where

$$S_1\left(N\right)={\int }_{E_1}{F}_1\left(\alpha \right)F_2\left(\alpha \right)F_3\left(\alpha \right)\mathbf{e}(-N\alpha )d\alpha ,S_2\left(N\right)={\int }_{E_2}{F}_1\left(\alpha \right)F_2\left(\alpha \right)F_3\left(\alpha \right)\mathbf{e}(-N\alpha )d\alpha .$$

(26)

The problem is now reduced to evaluating $S_1\left(N\right)$ and giving an upper bound of $S_2\left(N\right)$.

4. Additional Lemmas

In this section, we present some lemmas.

Lemma 1.

For $x⩾1$, $\alpha \in \mathbb{R}$, we have

$$\left|\sum _{n⩽x}\mathbf{e}\left(n\alpha \right)\right|⩽min\left(x,{\displaystyle \frac{1}{2\parallel \alpha \parallel }}\right),$$

(27)

where $\parallel \alpha \parallel$ denotes the distance from α to the nearest integer.

Proof. 

The proof of this Lemma has been found to be (8.6) in Iwaniec and kowalski [22]. □

Lemma 2.

Let $z⩾1$ be a real number and $k\ge 1$ be an integer. Then, for any $n⩽2z^k$, it holds that

$$\Lambda \left(n\right)=\sum _{j=1}^k{(-1)}^{j-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)\underset{\begin{array}{c}{n}_1{n}_2\cdots n_{2j}=n\\ n_{j+1},\cdots ,n_{2j}⩽z\end{array}}{\sum \cdots \sum }(log{n}_1)\mu \left(n_{j+1}\right)\cdots \mu \left(n_{2j}\right).$$

(28)

Proof. 

See the argument on pp. 1366–1367 of Heath-Brown [23]. □

Lemma 3.

Suppose that

$$L\left(H\right)=\sum _{i=1}^m{A}_i{H}^{a_i}+\sum _{j=1}^n{B}_j{H}^{-b_j},$$

(29)

where $A_i$, $B_j$, $a_i$, and $b_j$ are positive. Assume further that $H_1⩽H_2$. Then there exists some $\mathcal{H}$ with $H_1⩽\mathcal{H}⩽H_2$ and

$$L\left(\mathcal{H}\right)\ll \sum _{i=1}^m{A}_i{H_1}^{a_i}+\sum _{j=1}^n{B}_j{H_2}^{-b_j}+\sum _{i=1}^m\sum _{j=1}^n{\left({A_i}^{b_j}{B_j}^{a_i}\right)}^{1/(a_i+b_j)}.$$

(30)

The implied constant depends only on m and n.

Proof. 

See Lemma 3 of Srinivasan [24]. □

Lemma 4.

For an fixed integer $k\ge 3$, let $m_1,\cdots ,m_k$ are integers, $(i_1,\cdots ,i_{k-1})\in {\{0,1\}}^{k-1}$ such that

$${\alpha }_k=\sqrt{m_1}+{(-1)}^{i_1}\sqrt{m_2}+\cdots +{(-1)}^{i_{k-1}}\sqrt{m_k}\ne 0,$$

(31)

then

$$|{\alpha }_k|\gg max{(m_1,\cdots ,m_k)}^{-(2^{k-2}-{\displaystyle \frac{1}{2}})}.$$

(32)

Proof. 

See Lemma 2.2 of Zhai [12]. □

Lemma 5.

Suppose $k⩾2$ is a fixed integer, $y>1$ is a large parameter. Define

$$\begin{array}{cc}\hfill s_{k,l}(\infty )& :=\sum _{\sqrt{n_1}+\cdots +\sqrt{n_l}=\sqrt{n_{l+1}}+\cdots +\sqrt{n_k}}{\displaystyle \frac{d\left(n_1\right)\cdots d\left(n_k\right)}{{(n_1\cdots n_k)}^{3/4}}}(1⩽l<k),\hfill \\ \hfill s_{k,l}\left(y\right)& :=\sum _{\begin{array}{c}\sqrt{n_1}+\cdots +\sqrt{n_l}=\sqrt{n_{l+1}}+\cdots +\sqrt{n_k}\\ n_1,\cdots ,n_k⩽y\end{array}}{\displaystyle \frac{d\left(n_1\right)\cdots d\left(n_k\right)}{{(n_1\cdots n_k)}^{3/4}}}(1⩽l<k).\hfill \end{array}$$

(33)

Then, we have

(i) $s_{k,l}(\infty )$ is convergent.

(ii)

$$|s_{k,l}(\infty )-s_{k,l}\left(y\right)|\ll y^{-1/2+\epsilon },1⩽l<k.$$

(34)

Proof. 

See Lemma 3.1 of Zhai [12]. □

Lemma 6.

Suppose $N_1,N_2⩾1$ are given real numbers, $\alpha ,\beta$ are real numbers such that $0<\alpha$, $\beta ⩽1$. Define

$$T(N_1,N_2,\alpha ,\beta )=\sum _{\begin{array}{c}{n}_1\sim N_1,n_2\sim N_2\\ n_1\ne n_2\end{array}}{\displaystyle \frac{1}{|{n_1}^{\alpha }-{n_2}^{\alpha }{|}^{\beta }}}.$$

(35)

Then, we have

$$T(N_1,N_2,\alpha ,\beta )\ll {\left(N_1{N}_2\right)}^{1-\alpha \beta /2}log{N}_{1}log{N}_2.$$

(36)

Proof. 

See Lemma 2.11 of [19]. □

Lemma 7.

Suppose that $f\left(x\right):[a,b]\to \mathbb{R}$ has continuous derivatives of arbitrary order on $[a,b]$, where $1⩽a<b⩽2a$. Suppose further that for either $j=1$ or 2 we have

$$|f^{\left(j\right)}\left(x\right)|\asymp {\lambda }_j,x\in [a,b].$$

(37)

Then

$$\sum _{a<n⩽b}\mathbf{e}\left(f\left(n\right)\right)\ll {{\lambda }_1}^{-1}.$$

(38)

$$\sum _{a<n⩽b}\mathbf{e}\left(f\left(n\right)\right)\ll a{{\lambda }_2}^{1/2}+{{\lambda }_2}^{-1/2}.$$

(39)

Proof. 

See Lemma 2.1 of Graham and Kolesnik [25]. □

Lemma 8.

For any fixed real number $0<A⩽262/27=9.70...$ and a large real number x, we have the estimate

$$\sum _{x<p⩽2x}{|\Delta \left(p\right)|}^A\ll x^{1+A/4+\epsilon }.$$

(40)

Proof. 

See Lemma 2.11 of [20]. □

Lemma 9.

For real numbers N and x satisfying $1\ll N\ll x$, we define

$${\delta }_1(x,N)={\displaystyle \frac{x^{1/4}}{\sqrt{2}\pi }}\sum _{n⩽N}{\displaystyle \frac{d\left(n\right)}{n^{3/4}}}cos\left(4\pi \sqrt{nx}-{\displaystyle \frac{\pi }{4}}\right)$$

(41)

Then, uniformly for $N\ll x^{1/4}$, we have

$$\sum _{x<p⩽2x}{\left|{\delta }_1(p,N)\right|}^{10}\ll x^{7/2+\epsilon }.$$

(42)

Proof. 

See Lemma 2.12 of the authors [20]. □

Lemma 10.

For real numbers N and x satisfying $1\ll N\ll x$, we define

$${\delta }_2(x,N):=\Delta \left(x\right)-{\delta }_1(x,N).$$

(43)

We have

$${\delta }_2(x,N)\ll x^{1/2+\epsilon }{N}^{-1/2}.$$

(44)

We also have, for $N\ll x^{1/3}$,

$$\sum _{x<p⩽2x}{\delta }_2{(p,N)}^2\ll x^{3/2+\epsilon }{N}^{-1/2}.$$

(45)

Moreover, if $N\ll x^{1/8}$, then for any fixed real number $2<A⩽A_0=262/27$ we have the estimate

$$\sum _{x<p⩽2x}{\left|{\delta }_2(p,N)\right|}^A\ll \left\{\begin{array}{cc}{x}^{1+{\displaystyle \frac{A}{4}}+\epsilon }{N}^{-{\displaystyle \frac{A}{4}}},\hfill & \hfill 2<A⩽4,\\ x^{1+{\displaystyle \frac{A}{4}}+\epsilon }{N}^{-{\displaystyle \frac{A_0-A}{A_0-4}}},\hfill & \hfill 4<A<A_0.\end{array}\right.$$

(46)

Proof. 

See Lemma 2.13 of the authors [20]. □

For real number $\sigma \in [0,1]$ and fixed integer $1⩽k⩽9$, we state a multiple exponential sums of the following form:

$$\underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}d\left(m_1\right)\cdots d\left(m_k\right)\underset{\begin{array}{c}RS\asymp u\end{array}}{\sum _{r\sim R}\sum _{s\sim S}}{\xi }_r{\eta }_s\mathbf{e}\left(2{\alpha }_k{\left(rs\right)}^{1/2}+\sigma rs\right),$$

(47)

with $q\in \mathbb{Z}$ fixed, $q>0$, ${\xi }_r,{\eta }_s\in \mathbb{C},|{\xi }_r|\ll u^{\epsilon },\left|{\eta }_s\right|\ll u^{\epsilon }$, and $M_1,\cdots ,M_k,R,S,u$ are given real numbers, such that $1\ll M_1,\cdots ,M_k\ll u$, ${\alpha }_1=\sqrt{m_1}$, ${\alpha }_2=\sqrt{m_1}\pm \sqrt{m_2}$ and ${\alpha }_k(3⩽k⩽9)$ are given in Lemma 4. In this sum, without loss of generality we assume $m_1=max(m_1,\cdots ,m_k)$. It is called a “Type I” sum, denoted by $S_I\left(k\right)$, if ${\eta }_s=1$ or ${\eta }_s=logs$; otherwise, it is called a “Type II” sum, denoted by $S_{II}\left(k\right)$.

Lemma 11.

Suppose that ${\xi }_r\ll u^{\epsilon }$, ${\eta }_s=1$ or ${\eta }_s=log\ell$, $RS\asymp u$. Then for $R\ll u^{1/4}$, it holds that

$$\begin{array}{cc}\hfill u^{-\epsilon }{S}_I\left(k\right)& \ll \left\{\begin{array}{cc}{u}^{1/2}{M_1}^{5/4}+u^{3/4}{M_1}^{3/4}\hfill & \hfill k=1,\\ u^{1/2}{M_1}^{5/4}{M}_2+u^{3/4}{\left(M_1{M}_2\right)}^{7/8}\hfill & \hfill k=2,\\ u^{3/4}{M_1}^{(2^{k-3}+3/4)}(M_2\cdots M_k)\hfill & \hfill 3⩽k⩽9,\end{array}\right.\hfill \end{array}$$

(48)

Proof. 

Set $f\left(s\right)=2{\alpha }_k\sqrt{rs}+\sigma rs$. It is easy to see that

$$f^{″}\left(s\right)=-{\displaystyle \frac{1}{2}}{\alpha }_k{r}^{1/2}{s}^{-3/2}\asymp \left|{\alpha }_k\right|R^{1/2}{S}^{-3/2}.$$

(49)

If $R\ll u^{1/4}$, then by Lemma 7, we deduce that

$$\begin{array}{cc}\hfill u^{-\epsilon }{S}_I\left(k\right)& \ll \underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}\sum _{r\sim R}\left|\sum _{s\sim S}\mathbf{e}\left(f\left(s\right)\right)\right|\hfill \\ \hfill & \ll \underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}\sum _{r\sim R}\left(S\left(\right|{\alpha }_k|R^{1/2}{S}^{-3/2}{)}^{1/2}+\left(\right|{\alpha }_k{\left|R^{1/2}{S}^{-3/2}\right)}^{-1/2}\right)\hfill \\ \hfill & \ll u^{1/4}R{M_1}^{5/4}(M_2\cdots M_k)+u^{3/4}\underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1},\cdots ,\sum _{m_k\sim M_k}}{\displaystyle \frac{1}{|{\alpha }_k{|}^{1/2}}}.\hfill \end{array}$$

(50)

For $3⩽k⩽9$, by Lemma 4 we have

$$\begin{array}{cc}\hfill u^{-\epsilon }{S}_I\left(k\right)& \ll u^{1/4}R{M_1}^{5/4}(M_2\cdots M_k)+u^{3/4}{M_1}^{(2^{k-3}+3/4)}(M_2\cdots M_k)\hfill \\ \hfill & \ll u^{1/2}{M_1}^{5/4}(M_2\cdots M_k)+u^{3/4}{M_1}^{(2^{k-3}+3/4)}(M_2\cdots M_k)\hfill \\ \hfill & \ll u^{3/4}{M_1}^{(2^{k-3}+3/4)}(M_2\cdots M_k)\hfill \end{array}$$

(51)

For $k=2$, when ${\alpha }_2=\sqrt{m_1}-\sqrt{m_2}$, by Lemma 6 we have

$$\begin{array}{cc}\hfill u^{-\epsilon }{S}_I\left(2\right)& \ll u^{1/4}R{M_1}^{5/4}{M}_2+u^{3/4}{\left(M_1{M}_2\right)}^{7/8}\hfill \\ \hfill & \ll u^{1/2}{M_1}^{5/4}{M}_2+u^{3/4}{\left(M_1{M}_2\right)}^{7/8},\hfill \end{array}$$

(52)

when ${\alpha }_2=\sqrt{m_1}+\sqrt{m_2}$, it is easier to get the same estimate by using $a+b⩾2\sqrt{ab}$, we omit the details.

When $k=1$, ${\alpha }_1=\sqrt{m_1}$, we can easily obtain

$$\begin{array}{cc}\hfill u^{-\epsilon }{S}_I\left(1\right)& \ll u^{1/4}R{M_1}^{5/4}+u^{3/4}{M_1}^{3/4}\hfill \\ \hfill & \ll u^{1/2}{M_1}^{5/4}+u^{3/4}{M_1}^{3/4}.\hfill \end{array}$$

(53)

Combining (51)–(53) we complete the proof. □

Lemma 12.

Suppose $|{\xi }_r|\ll u^{\epsilon },\left|{\eta }_s\right|\ll u^{\epsilon }$ with $r\sim R$, $s\sim S$, $RS\asymp u$. Then for $u^{1/4}\ll R\ll u^{1/2}$, it holds that

$$\begin{array}{c}\hfill u^{-\epsilon }{S}_{II}\left(k\right)\ll \left\{\begin{array}{cc}{u}^{3/4}{M_1}^{9/8}+u^{7/8}+u^{11/12}{M_1}^{13/12}\hfill & \hfill k=1,\\ u^{3/4}{M_1}^{9/8}{M}_2+u^{7/8}{M}_1{M}_2+u^{5/6}{M_1}^{13/12}{M}_2\hfill & \hfill k=2,\\ u^{3/4}{M_1}^{9/8}{M}_2\cdots M_k+u^{7/8}{M_1}^{2^{k-4}+7/8}{M}_2\cdots M_k\hfill & \hfill 3⩽k⩽9.\end{array}\right.\end{array}$$

(54)

Proof. 

Let Y, which satisfies $1<Y<R$, be a parameter which will be chosen later. By the Weyl–van der Corput inequality (see Lemma 2.5 of Graham and Kolesnik [25]), we have

$$\begin{array}{cc}\hfill & {\left|\underset{\begin{array}{c}RS\asymp u\end{array}}{\sum _{r\sim R}\sum _{s\sim S}}{\xi }_r{\eta }_s\mathbf{e}\left(2{\alpha }_k{\left(rs\right)}^{1/2}+\sigma rs\right)\right|}^2\ll R^2{S}^2{Y}^{-1}\hfill \\ \hfill & RS{Y}^{-1}\sum _{r\sim R}\sum _{0<y⩽Y}\left|\mathfrak{T}(y;r)\right|,\hfill \end{array}$$

(55)

where

$$\mathfrak{T}(y;r)=\sum _{s\sim S}\mathbf{e}\left(g\left(s\right)\right)$$

(56)

with

$$g\left(s\right)=2{\alpha }_k\sqrt{s}\left({(r+y)}^{1/2}-r^{1/2}\right)-\sigma ys.$$

(57)

It is easy to see that

$$g^{″}\left(s\right)=-{\displaystyle \frac{1}{2}}{\alpha }_k{S}^{-3/2}\left({(r+y)}^{1/2}-r^{1/2}\right)\asymp |{\alpha }_k|S^{-3/2}{R}^{-1/2}\left|y\right|$$

(58)

By (39) of Lemma 7, we have

$$\mathfrak{S}(y;r)\ll S{\left(|{\alpha }_k|S^{-3/2}{R}^{-1/2}\left|y\right|\right)}^{1/2}+{\left(|{\alpha }_k|S^{-3/2}{R}^{-1/2}\left|y\right|\right)}^{-1/2}.$$

(59)

Putting (59) into (55), we derive that

$$\begin{array}{cc}\hfill & {\left|\underset{\begin{array}{c}RS\asymp u\end{array}}{\sum _{r\sim R}\sum _{s\sim S}}{\xi }_r{\eta }_s\mathbf{e}\left(2{\alpha }_k{\left(rs\right)}^{1/2}+\sigma rs\right)\right|}^2\ll R^2{S}^2{Y}^{-1}+RS{Y}^{-1}\hfill \\ \hfill & \times \sum _{r\sim R}\sum _{0<y⩽Y}\left(|{\alpha }_k{|}^{1/2}{S}^{1/4}{R}^{-1/4}{\left|y\right|}^{1/2}+|{\alpha }_k{|}^{-1/2}{S}^{3/4}{R}^{1/4}{\left|y\right|}^{-1/2}\right)\hfill \\ \hfill & \ll R^2{S}^2{Y}^{-1}+RS{Y}^{-1}\left(|{\alpha }_k{|}^{1/2}{S}^{1/4}{R}^{3/4}{Y}^{3/2}+{\left|{\alpha }_k\right|}^{-1/2}{S}^{3/4}{R}^{5/4}{Y}^{1/2}\right)\hfill \\ \hfill & \ll R^2{S}^2{Y}^{-1}+|{\alpha }_k{|}^{1/2}{S}^{5/4}{R}^{7/4}{Y}^{1/2}+{\left|{\alpha }_k\right|}^{-1/2}{S}^{7/4}{R}^{9/4}{Y}^{-1/2}.\hfill \end{array}$$

(60)

By noting that $1⩽Y⩽R$, it follows from Lemma 3 that there exists an optimal Y such that

$$\begin{array}{cc}\hfill & {\left|\underset{\begin{array}{c}RS\asymp u\end{array}}{\sum _{r\sim R}\sum _{s\sim S}}{\xi }_r{\eta }_s\mathbf{e}\left(2{\alpha }_k{\left(rs\right)}^{1/2}+\sigma rs\right)\right|}^2\hfill \\ \hfill & \ll u^{5/4}{R}^{1/2}|{\alpha }_k{|}^{1/2}+u^{7/4}|{\alpha }_k{|}^{-1/2}+u^2{R}^{-1}+u^{3/2}{R}^{1/2}+u^{3/2}{R}^{1/3}{\left|{\alpha }_k\right|}^{1/3},\hfill \end{array}$$

(61)

which implies

$$\begin{array}{cc}\hfill & \left|\underset{\begin{array}{c}RS\asymp u\end{array}}{\sum _{r\sim R}\sum _{s\sim S}}{\xi }_r{\eta }_s\mathbf{e}\left(2{\alpha }_k{\left(rs\right)}^{1/2}+\sigma rs\right)\right|\hfill \\ \hfill & \ll u^{5/8}{R}^{1/4}|{\alpha }_k{|}^{1/4}+u^{7/8}|{\alpha }_k{|}^{-1/4}+u{R}^{-1/2}+u^{3/4}{R}^{1/4}+u^{3/4}{R}^{1/6}{\left|{\alpha }_k\right|}^{1/6}.\hfill \end{array}$$

(62)

Therefore, from the above estimate and the condition $u^{1/4}\ll R\ll u^{1/2}$, we obtain

$$\begin{array}{cc}\hfill u^{-\epsilon }{S}_{II}\left(k\right)& \ll \underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}\hfill \\ \hfill & \times \left(u^{5/8}{R}^{1/4}|{\alpha }_k{|}^{1/4}+u^{7/8}|{\alpha }_k{|}^{-1/4}+u{R}^{-1/2}+u^{3/4}{R}^{1/4}+u^{3/4}{R}^{1/6}{\left|{\alpha }_k\right|}^{1/6}\right)\hfill \\ \hfill & \ll \underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}\left(u^{3/4}|{\alpha }_k{|}^{1/4}+u^{7/8}|{\alpha }_k{|}^{-1/4}+u^{7/8}+u^{5/6}{\left|{\alpha }_k\right|}^{1/6}\right)\hfill \end{array}$$

(63)

using Lemma 4, for $3⩽k⩽9$ we have

$$\begin{array}{cc}\hfill u^{-\epsilon }{S}_{II}\left(k\right)& \ll u^{3/4}{M_1}^{9/8}{M}_2\cdots M_k+u^{7/8}{M_1}^{2^{k-4}+7/8}{M}_2\cdots M_k.\hfill \end{array}$$

(64)

For $k=2$, when ${\alpha }_2=\sqrt{m_1}-\sqrt{m_2}$, by Lemma 6 and the condition we derive that

$$\begin{array}{c}\hfill u^{-\epsilon }{S}_{II}\left(2\right)\ll u^{3/4}{M_1}^{9/8}{M}_2+u^{7/8}{M}_1{M}_2+u^{5/6}{M_1}^{13/12}{M}_2.\end{array}$$

(65)

For the case ${\alpha }_2=\sqrt{m_1}+\sqrt{m_2}$, we can use the similar argument in Lemma 11. When $k=1$, ${\alpha }_1=m_1$, we obtain

$$\begin{array}{c}\hfill u^{-\epsilon }{S}_{II}\left(1\right)\ll u^{3/4}{M_1}^{9/8}+u^{7/8}{M}_1+u^{5/6}{M_1}^{13/12}.\end{array}$$

(66)

Thus, combining (64)–(66), we complete the proof. □

5. Asymptotic Formula for ${\mathit{S}}_{\mathbf{1}}\left(\mathit{N}\right)$

Suppose $\alpha =a/q+\lambda \in E_1$ with $0⩽a<q⩽Q$, $(a,q)=1$, $\left|\lambda \right|⩽1/q\tau$. Without loss of generality, we only consider $F_1\left(\alpha \right)$, and for convenience we set $k=k_1$. For $2⩽k⩽9$, we have

$$\begin{array}{cc}\hfill F_1\left(\alpha \right)& =\sum _{p⩽N}{\Delta }^k\left(p\right)\mathbf{e}\left(p\alpha \right)\hfill \\ \hfill & =\sum _{p⩽N}{\Delta }^k\left(p\right)\mathbf{e}\left(\left({\displaystyle \frac{a}{q}}+\lambda \right)p\right)\hfill \\ \hfill & =\sum _{\begin{array}{c}p⩽N\\ (p,q)=1\end{array}}{\Delta }^k\left(p\right)\mathbf{e}\left(\left({\displaystyle \frac{a}{q}}+\lambda \right)p\right)+O\left(\sum _{\begin{array}{c}p⩽N\\ p|q\end{array}}{|\Delta \left(p\right)|}^k\right)\hfill \\ \hfill & =\sum _{\begin{array}{c}\ell =1\\ (\ell ,q)=1\end{array}}^q\mathbf{e}\left({\displaystyle \frac{a\ell }{q}}\right)\sum _{\begin{array}{c}p⩽N\\ p\equiv \ell \left(modq\right)\end{array}}{\Delta }^k\left(p\right)\mathbf{e}\left(p\lambda \right)+O\left(Q^{1+k/2}\right)\hfill \\ \hfill & =\sum _{\begin{array}{c}\ell =1\\ (\ell ,q)=1\end{array}}^q\mathbf{e}\left({\displaystyle \frac{a\ell }{q}}\right)\sum _{\begin{array}{c}p⩽N\\ p\equiv \ell \left(modq\right)\end{array}}{\Delta }^k\left(p\right)\mathbf{e}\left(p\lambda \right)+O\left(N^{\epsilon }\right)\hfill \end{array}$$

(67)

where we use $Q={log}^{A}N\ll N^{\epsilon }$ and the trivial estimate $\Delta \left(x\right)\ll x^{1/2}$.

By partial summation we have

$$F_1\left(\alpha \right)=\sum _{\begin{array}{c}\ell =1\\ (\ell ,q)=1\end{array}}^q\mathbf{e}\left({\displaystyle \frac{a\ell }{q}}\right){\int }_2^N\mathbf{e}\left(u\lambda \right)d\left(\sum _{\begin{array}{c}p⩽u\\ p\equiv \ell \left(modq\right)\end{array}}{\Delta }^k\left(p\right)\right).$$

(68)

We are going to evaluate

$$\pi (u;q,\ell ;k):=\sum _{\begin{array}{c}p⩽u\\ p\equiv \ell \left(modq\right)\end{array}}{\Delta }^k\left(p\right).$$

(69)

5.1. Asymptotic Formula for $\pi (u;q,\mathit{\ell };k)$

Using the notations in Lemmas 9 and 10, let M be a real number such that $1\ll M\ll x$; thus, we have

$$\sum _{\begin{array}{c}p⩽u\\ p\equiv \ell \left(modq\right)\end{array}}{\Delta }^k\left(p\right):={\sum }_{11}+O\left({\sum }_{12}+{\sum }_{22}\right),$$

(70)

where

$$\begin{array}{cc}\hfill {\sum }_{11}& =\sum _{\begin{array}{c}p⩽u\\ p\equiv \ell \left(modq\right)\end{array}}{\delta }_1{(p,M)}^k,\hfill \\ \hfill {\sum }_{12}& =\sum _{p⩽u}|{\delta }_1{(p,M)|}^{k-1}\left|{\delta }_2(p,M)\right|,\hfill \\ \hfill {\sum }_{22}& =\sum _{p⩽u}{\left|{\delta }_2(p,M)\right|}^k.\hfill \end{array}$$

(71)

For ${\sum }_{22}$, using (46) and (45), for $M\ll u^{1/8}$ one has

$${\sum }_{22}\ll \left\{\begin{array}{cc}{u}^{1+{\displaystyle \frac{k}{4}}+\epsilon }{M}^{-{\displaystyle \frac{k}{4}}},\hfill & \hfill 2⩽k⩽4,\\ u^{1+{\displaystyle \frac{k}{4}}+\epsilon }{M}^{-{\displaystyle \frac{A_0-k}{A_0-4}}},\hfill & \hfill 4<k⩽9,\end{array}\right.$$

(72)

where $A_0=262/27$.

For ${\sum }_{12}$, using Lemma 9, Equations (46) and (45), and Hölder’s inequality, we obtain

$${\sum }_{12}\ll \left\{\begin{array}{cc}{u}^{1+{\displaystyle \frac{k}{4}}+\epsilon }{M}^{-{\displaystyle \frac{1}{4}}},\hfill & \hfill 2⩽k⩽8,\\ u^{13/4+\epsilon }{M}^{-{\displaystyle \frac{A_0-9}{A_0-4}}},\hfill & \hfill k=9.\end{array}\right.$$

(73)

Next, we turn to evaluate ${\sum }_{11}$. Using $cos\alpha cos\beta ={\displaystyle \frac{1}{2}}(cos(\alpha -\beta )+cos(\alpha +\beta ))$, we have

$$\begin{array}{cc}\hfill {\sum }_{11}& =\sum _{\begin{array}{c}p⩽u\\ p\equiv \ell \left(modq\right)\end{array}}{\displaystyle \frac{p^{k/4}}{{\left(\sqrt{2}\pi \right)}^k}}\sum _{m_1⩽M}\cdots \sum _{m_k⩽M}{\displaystyle \frac{d\left(m_1\right)\cdots d\left(m_k\right)}{{(m_1\cdots m_k)}^{3/4}}}\prod _{j=1}^{k}cos\left(4\pi \sqrt{m_{j}p}-{\displaystyle \frac{\pi }{4}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{{\left(\sqrt{2}\pi \right)}^k{2}^{k-1}}}\sum _{{\mathbf{i}}_k\in {\{0,1\}}^{k-1}}\sum _{m_1⩽M}\cdots \sum _{m_k⩽M}{\displaystyle \frac{d\left(m_1\right)\cdots d\left(m_k\right)}{{(m_1\cdots m_k)}^{3/4}}}\hfill \\ \hfill & \times \sum _{\begin{array}{c}p⩽u\\ p\equiv \ell \left(modq\right)\end{array}}{p}^{k/4}cos\left(4\pi {\alpha }_k\sqrt{p}-{\displaystyle \frac{\pi }{4}}{\beta }_k\right),\hfill \end{array}$$

(74)

where

$$\begin{array}{cc}\hfill {\alpha }_k& =\sqrt{m_1}+{(-1)}^{i_1}\sqrt{m_2}+\cdots +{(-1)}^{i_{k-1}}\sqrt{m_k},\hfill \\ \hfill {\beta }_k& =1+{(-1)}^{i_1}+\cdots +{(-1)}^{i_{k-1}},\hfill \\ \hfill {\mathbf{i}}_k& =(i_1,\cdots ,i_{k-1}).\hfill \end{array}$$

(75)

We divide the ${\sum }_{11}$ into two parts:

$${\sum }_{11}:={\displaystyle \frac{1}{{\left(\sqrt{2}\pi \right)}^k{2}^{k-1}}}\left({\sum }_{111}+{\sum }_{112}\right),$$

(76)

where

$$\begin{array}{cc}\hfill {\sum }_{111}& =\sum _{{\mathbf{i}}_k\in {\{0,1\}}^{k-1}}cos(-{\displaystyle \frac{\pi }{4}}{\beta }_k)\underset{\begin{array}{c}{\alpha }_k=0\end{array}}{\sum _{m_1⩽M}\cdots \sum _{m_k⩽M}}{\displaystyle \frac{d\left(m_1\right)\cdots d\left(m_k\right)}{{(m_1\cdots m_k)}^{3/4}}}\sum _{\begin{array}{c}p⩽u\\ p\equiv \ell \left(modq\right)\end{array}}{p}^{k/4},\hfill \\ \hfill {\sum }_{112}& =\sum _{{\mathbf{i}}_k\in {\{0,1\}}^{k-1}}\underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1⩽M}\cdots \sum _{m_k⩽M}}{\displaystyle \frac{d\left(m_1\right)\cdots d\left(m_k\right)}{{(m_1\cdots m_k)}^{3/4}}}\sum _{\begin{array}{c}p⩽u\\ p\equiv \ell \left(modq\right)\end{array}}{p}^{k/4}cos\left(4\pi {\alpha }_k\sqrt{p}-{\displaystyle \frac{\pi }{4}}{\beta }_k\right).\hfill \end{array}$$

(77)

First, we consider the contribution of ${\sum }_{111}$. For the inner sum, by partial summation and Siegel–Walfisz theorem, we have

$$\begin{array}{cc}\hfill \sum _{\begin{array}{c}p⩽u\\ p\equiv \ell \left(modq\right)\end{array}}{p}^{k/4}& ={\int }_2^u{t}^{k/4}d\left(\sum _{\begin{array}{c}p⩽t\\ p\equiv \ell \left(modq\right)\end{array}}1\right)\hfill \\ \hfill & ={\int }_2^u{t}^{k/4}d\left({\displaystyle \frac{1}{\phi \left(q\right)}}{\int }_2^t{\displaystyle \frac{1}{logr}}dr+O\left(t{e}^{-c_1\sqrt{logt}}\right)\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{\phi \left(q\right)}}{\int }_2^u{\displaystyle \frac{t^{k/4}}{logt}}dt+O\left((1+|\lambda \left|u\right)u^{1+k/4}{e}^{-c_1\sqrt{logu}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{\phi \left(q\right)}}{\int }_2^u{\displaystyle \frac{t^{k/4}}{logt}}dt+O\left(u^{1+k/4}{e}^{-c_2\sqrt{logu}}\right),\hfill \end{array}$$

(78)

where $c_1,c_2>0$ are constants.

Using a similar argument on (4.1)–(4.4) as made by Zhai [12], we obtain

$${\sum }_{111}={\displaystyle \frac{B_k(\infty )}{\phi \left(q\right)}}{\int }_2^u{\displaystyle \frac{t^{k/4}}{logt}}dt+O\left(u^{1+k/4}{e}^{-c_2\sqrt{logu}}\right)+O\left(u^{1+k/4+\epsilon }{M}^{-1/2}\right),$$

(79)

where

$$B_k(\infty )=\sum _{l=1}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k-1}{l}}\right)s_{k,l}(\infty )cos{\displaystyle \frac{\pi (k-2l)}{4}}.$$

(80)

Next, we consider the contribution of ${\sum }_{112}$. One can see ${\sum }_{112}$ can be written as a linear combination of $O({log}^{k}Mlogu)$ sums of the form

$$\begin{array}{cc}\hfill & \sum _{{\mathbf{i}}_k\in {\{0,1\}}^{k-1}}\underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}{\displaystyle \frac{d\left(m_1\right)\cdots d\left(m_k\right)}{{(m_1\cdots m_k)}^{3/4}}}\sum _{\begin{array}{c}p\sim u\\ p\equiv \ell \left(modq\right)\end{array}}{p}^{k/4}cos\left(4\pi {\alpha }_k\sqrt{p}-{\displaystyle \frac{\pi }{4}}{\beta }_k\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{2}}\sum _{{\mathbf{i}}_k\in {\{0,1\}}^{k-1}}\underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}{\displaystyle \frac{d\left(m_1\right)\cdots d\left(m_k\right)}{{(m_1\cdots m_k)}^{3/4}}}\hfill \\ \hfill \times & \sum _{\begin{array}{c}p\sim u\\ p\equiv \ell \left(modq\right)\end{array}}{p}^{k/4}\left(\mathbf{e}\left(2{\alpha }_k\sqrt{p}-{\displaystyle \frac{{\beta }_k}{8}}\right)-\mathbf{e}\left(-2{\alpha }_k\sqrt{p}+{\displaystyle \frac{{\beta }_k}{8}}\right)\right),\hfill \end{array}$$

(81)

where we use $cos2\pi a=\left(\mathbf{e}\right(a)-\mathbf{e}(-a\left)\right)/2$, and where $1⩽M_1,\cdots ,M_k⩽M$. Without loss of generality, we assume that $max(M_1,\cdots ,M_k)=M_1$. By a splitting argument we only need to estimate

$${\mathcal{S}}_1=\sum _{{\mathbf{i}}_k\in {\{0,1\}}^{k-1}}\underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}{\displaystyle \frac{d\left(m_1\right)\cdots d\left(m_k\right)}{{(m_1\cdots m_k)}^{3/4}}}\sum _{\begin{array}{c}p\sim u\\ p\equiv \ell \left(modq\right)\end{array}}{p}^{k/4}\mathbf{e}\left(2{\alpha }_k\sqrt{p}\right).$$

(82)

Trivially we have

$${\mathcal{S}}_1\ll {\displaystyle \frac{u^{k/4}}{{(M_1\cdots M_k)}^{3/4}}}\left|{{\mathcal{S}}_1}^{*}\right|,$$

(83)

where

$${{\mathcal{S}}_1}^{*}=\sum _{{\mathbf{i}}_k\in {\{0,1\}}^{k-1}}\underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}d\left(m_1\right)\cdots d\left(m_k\right)\sum _{\begin{array}{c}p\sim u\\ p\equiv \ell \left(modq\right)\end{array}}\mathbf{e}\left(2{\alpha }_k\sqrt{p}\right).$$

(84)

It follows from partial summation that

$$\begin{array}{cc}\hfill {{\mathcal{S}}_1}^{*}& =\sum _{{\mathbf{i}}_k\in {\{0,1\}}^{k-1}}\underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}d\left(m_1\right)\cdots d\left(m_k\right){\int }_u^{2u}{\displaystyle \frac{d{\mathcal{A}}_k\left(x\right)}{logx}}\hfill \\ \hfill & =\sum _{{\mathbf{i}}_k\in {\{0,1\}}^{k-1}}\underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}d\left(m_1\right)\cdots d\left(m_k\right)\left({\displaystyle \frac{{\mathcal{A}}_k\left(x\right)}{logx}}{|}_u^{2u}-{\int }_u^{2u}{\displaystyle \frac{{\mathcal{A}}_k\left(x\right)}{x{log}^{2}x}}dx\right),\hfill \end{array}$$

(85)

where

$${\mathcal{A}}_k\left(x\right)=\sum _{\begin{array}{c}p⩽x\\ p\equiv \ell \left(modq\right)\end{array}}(logp)\mathbf{e}\left(2{\alpha }_k\sqrt{p}\right).$$

(86)

Moreover, by the definition of $\Lambda \left(n\right)$, we easily obtain

$$\begin{array}{cc}\hfill {\mathcal{A}}_k\left(x\right)=& \sum _{\begin{array}{c}p⩽x\\ p\equiv \ell \left(modq\right)\end{array}}\Lambda \left(n\right)\mathbf{e}\left(2{\alpha }_k\sqrt{n}\right)-\sum _{\begin{array}{c}{p}^2⩽x\\ p\equiv \ell \left(modq\right)\end{array}}\Lambda \left(n\right)\mathbf{e}\left(2{\alpha }_k\sqrt{n}\right)\hfill \\ \hfill =& \sum _{\begin{array}{c}p⩽x\\ p\equiv \ell \left(modq\right)\end{array}}\Lambda \left(n\right)\mathbf{e}\left(2{\alpha }_k\sqrt{n}\right)+O\left(\sum _{p^2⩽x}1\right)\hfill \\ \hfill =& \sum _{\begin{array}{c}p⩽x\\ p\equiv \ell \left(modq\right)\end{array}}\Lambda \left(n\right)\mathbf{e}\left(2{\alpha }_k\sqrt{n}\right)+O\left(x^{1/2}\right).\hfill \end{array}$$

(87)

Combining (81)–(87), it is sufficient to give the upper bound of the following sum

$$\begin{array}{cc}\hfill {{\mathcal{S}}_1}^{**}& =\sum _{{\mathbf{i}}_k\in {\{0,1\}}^{k-1}}\underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}d\left(m_1\right)\cdots d\left(m_k\right)\sum _{\begin{array}{c}n\sim u\\ n\equiv \ell \left(modq\right)\end{array}}\Lambda \left(n\right)\mathbf{e}\left(2{\alpha }_k\sqrt{n}\right).\hfill \end{array}$$

(88)

It is known that

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{q}}\sum _{m=1}^q\mathbf{e}\left({\displaystyle \frac{(n-\ell )m}{q}}\right)=\left\{\begin{array}{c}1\mathrm{if}q\mid n-\ell ,\hfill \\ 0\mathrm{if}q\nmid n-\ell ,\hfill \end{array}\right.\end{array}$$

(89)

we can represent the innermost sum in (88) as

$$\sum _{\begin{array}{c}n\sim u\\ n\equiv \ell \left(modq\right)\end{array}}\Lambda \left(n\right)\mathbf{e}\left(2{\alpha }_k\sqrt{n}\right)={\displaystyle \frac{1}{q}}\sum _{m=1}^q\sum _{n\sim u}\Lambda \left(n\right)\mathbf{e}\left(2{\alpha }_k\sqrt{n}+{\displaystyle \frac{(n-\ell )m}{q}}\right)$$

(90)

From (88) and (90), we know that it suffices to estimate

$$\sum _{{\mathbf{i}}_k\in {\{0,1\}}^{k-1}}\underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}d\left(m_1\right)\cdots d\left(m_k\right)\left|\sum _{n\sim u}\Lambda \left(n\right)\mathbf{e}\left(2{\alpha }_k\sqrt{n}+nm{q}^{-1}\right)\right|.$$

(91)

After using Heath-Brown’s identity, i.e., Lemma 2 with $k=3$, one can see that the exponential sum

$$\sum _{{\mathbf{i}}_k\in {\{0,1\}}^{k-1}}\underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}d\left(m_1\right)\cdots d\left(m_k\right)\left|\sum _{n\sim u}\Lambda \left(n\right)\mathbf{e}\left(2{\alpha }_k\sqrt{n}+nm{q}^{-1}\right)\right|.$$

(92)

can be written as linear combination of $O\left({log}^{6}x\right)$ sums, each of which is of the form

$$\begin{array}{cc}\hfill {\mathcal{T}}^{*}:=& \sum _{{\mathbf{i}}_k\in {\{0,1\}}^{k-1}}\underset{\begin{array}{c}{\alpha }_k\ne 0\end{array}}{\sum _{m_1\sim M_1}\cdots \sum _{m_k\sim M_k}}d\left(m_1\right)\cdots d\left(m_k\right)\hfill \\ \hfill & \times \left|\sum _{n_1\sim N_1}\cdots \sum _{n_6\sim N_6}(log{n}_1)\mu \left(n_4\right)\mu \left(n_5\right)\mu \left(n_6\right)\mathbf{e}\left(2{\alpha }_k\sqrt{n_1{n}_2\cdots n_6}+(n_1{n}_2\cdots n_6)m{q}^{-1}\right)\right|,\hfill \end{array}$$

(93)

where $N_1\cdots N_6\asymp u$; $2N_i⩽{\left(2u\right)}^{1/3}$, $i=4,5,6$ and some $n_i$ may only take value 1. Therefore, it is sufficient for us to estimate for each ${\mathcal{T}}^{*}$ defined as in (93). Next, we will consider four cases.

5.1.1. Case 1

If there exists an $N_j$ such that $N_j⩾u^{3/4}$, then we must have $j⩽3$ for the fact that $N_j\ll u^{1/3}$ with $j=4,5,6$. Let

$$r=\prod _{\begin{array}{c}1⩽i⩽6\\ i\ne j\end{array}}{n}_i,s=n_j,R=\prod _{\begin{array}{c}1⩽i⩽6\\ i\ne j\end{array}}{N}_i,S=N_j.$$

(94)

In this case, we can see that ${\mathcal{T}}^{*}$ is a sum of “Type I” satisfying $R\ll u^{1/4}$. By Lemma 11, we have

$$\begin{array}{cc}\hfill u^{-\epsilon }{S}_I\left(k\right)& \ll \left\{\begin{array}{cc}{u}^{1/2}{M_1}^{5/4}{M}_2+u^{3/4}{\left(M_1{M}_2\right)}^{7/8},\hfill & \hfill k=2,\\ u^{3/4}{M_1}^{(2^{k-3}+3/4)}(M_2\cdots M_k),\hfill & \hfill 3⩽k⩽9,\end{array}\right.\hfill \end{array}$$

(95)

5.1.2. Case 2

If there exists an $N_j$ such that $u^{1/2}⩽N_j<u^{3/4}$, then we take

$$r=\prod _{\begin{array}{c}1⩽i⩽6\\ i\ne j\end{array}}{n}_i,s=n_j,R=\prod _{\begin{array}{c}1⩽i⩽6\\ i\ne j\end{array}}{N}_i,S=N_j.$$

(96)

Thus, ${\mathcal{T}}^{*}$ is a sum of “Type II” satisfying $u^{1/4}\ll R\ll u^{1/2}$. By Lemma 12, we have

$$\begin{array}{c}\hfill u^{-\epsilon }{S}_{II}\left(k\right)\ll \left\{\begin{array}{cc}{u}^{3/4}{M_1}^{9/8}{M}_2+u^{7/8}{M}_1{M}_2+u^{5/6}{M_1}^{13/12}{M}_2\hfill & \hfill k=2,\\ u^{3/4}{M_1}^{9/8}{M}_2\cdots M_k+u^{7/8}{M_1}^{2^{k-4}+7/8}{M}_2\cdots M_k\hfill & \hfill 3⩽k⩽9.\end{array}\right.\end{array}$$

(97)

5.1.3. Case 3

If there exists an $N_j$ such that $u^{1/4}⩽N_j<u^{1/2}$, then we take

$$r=n_j,s=\prod _{\begin{array}{c}1⩽i⩽6\\ i\ne j\end{array}}{n}_i,R=N_j,S=\prod _{\begin{array}{c}1⩽i⩽6\\ i\ne j\end{array}}{N}_i.$$

(98)

Thus, ${\mathcal{T}}^{*}$ is a sum of "Type II" satisfying $u^{1/4}\ll R\ll u^{1/2}$. By Lemma 12, we have

$$\begin{array}{c}\hfill u^{-\epsilon }{S}_{II}\left(k\right)\ll \left\{\begin{array}{cc}{u}^{3/4}{M_1}^{9/8}{M}_2+u^{7/8}{M}_1{M}_2+u^{5/6}{M_1}^{13/12}{M}_2\hfill & \hfill k=2,\\ u^{3/4}{M_1}^{9/8}{M}_2\cdots M_k+u^{7/8}{M_1}^{2^{k-4}+7/8}{M}_2\cdots M_k\hfill & \hfill 3⩽k⩽9.\end{array}\right.\end{array}$$

(99)

5.1.4. Case 4

If $N_j<u^{1/4}(j=1,2,3,4,5,6)$, without loss of generality, we assume that $N_1⩾N_2⩾\cdots ⩾N_6$. Let ℓ denote the natural number j such that

$$N_1{N}_2\cdots N_{j-1}<u^{1/4},N_1{N}_2\cdots N_j⩾u^{1/4}.$$

(100)

Since $N_1<u^{1/4}$ and $N_6<u^{1/4}$, then $2⩽\ell ⩽5$. Thus, we have

$$u^{1/4}⩽N_1{N}_2\cdots N_{\ell }=(N_1{N}_2\cdots N_{\ell -1})·N_{\ell }<u^{1/4}·u^{1/4}=u^{1/2}.$$

(101)

Let

$$r=\prod _{i=1}^{\ell }{n}_i,s=\prod _{i=\ell +1}^6{n}_i,R=\prod _{i=1}^{\ell }{N}_i,S=\prod _{i=\ell +1}^6{N}_i.$$

(102)

At this time, ${\mathcal{T}}^{*}$ is a sum of “Type II” satisfying $u^{1/4}\ll R\ll u^{1/2}$. By Lemma 12, we have

$$\begin{array}{c}\hfill u^{-\epsilon }{S}_{II}\left(k\right)\ll \left\{\begin{array}{cc}{u}^{3/4}{M_1}^{9/8}{M}_2+u^{7/8}{M}_1{M}_2+u^{5/6}{M_1}^{13/12}{M}_2\hfill & \hfill k=2,\\ u^{3/4}{M_1}^{9/8}{M}_2\cdots M_k+u^{7/8}{M_1}^{2^{k-4}+7/8}{M}_2\cdots M_k\hfill & \hfill 3⩽k⩽9.\end{array}\right.\end{array}$$

(103)

Combining the above four cases, we derive that

$$\begin{array}{cc}\hfill u^{-\epsilon }·{\mathcal{T}}^{*}& \ll \left\{\begin{array}{c}{u}^{1/2}{M_1}^{5/4}{M}_2+u^{3/4}{M_1}^{9/8}{M}_2\hfill \\ +u^{7/8}{M}_1{M}_2+u^{5/6}{M_1}^{13/12}{M}_2\hfill & \hfill k=2,\\ u^{3/4}{M_1}^{(2^{k-3}+3/4)}(M_2\cdots M_k)+u^{7/8}{M_1}^{2^{k-4}+7/8}{M}_2\cdots M_k\hfill & \hfill 3⩽k⩽9.\end{array}\right.\hfill \end{array}$$

(104)

which, combined with (83) and (85), yields

$$\begin{array}{cc}\hfill & u^{-\epsilon }·{\sum }_{112}\hfill \\ \hfill \ll & \left\{\begin{array}{c}{u}^{1/2+k/4}{M_1}^{1/2}{M_2}^{1/4}+u^{3/4+k/4}{M_1}^{3/8}{M_2}^{1/4}\hfill \\ +u^{7/8+k/4}{\left(M_1{M}_2\right)}^{1/4}+u^{5/6+k/4}{M_1}^{1/3}{M_2}^{1/4}\hfill & \hfill k=2,\\ u^{3/4+k/4}{M_1}^{2^{k-3}}{(M_2\cdots M_k)}^{1/4}+u^{7/8+k/4}{M_1}^{2^{k-4}+{\displaystyle \frac{1}{8}}}{(M_2\cdots M_k)}^{1/4}\hfill & \hfill 3⩽k⩽9.\end{array}\right.\hfill \end{array}$$

(105)

Above all, since $M_1,\cdots ,M_k⩽M$, by Equations (76), (79) and (105), we conclude that

$$\begin{array}{cc}\hfill & {\sum }_{11}\hfill \\ \hfill =& {\displaystyle \frac{B_k(\infty )}{{\left(\sqrt{2}\pi \right)}^k{2}^{k-1}}}{\int }_2^u{\displaystyle \frac{t^{k/4}}{logt}}dt+O\left(u^{1+k/4}{e}^{-c_2\sqrt{logu}}\right)\hfill \\ \hfill & +O\left(u^{1+k/4+\epsilon }{M}^{-1/2}+\left\{\begin{array}{c}{u}^{1/2+k/4}{M}^{3/4}+u^{3/4+k/4}{M}^{5/8}\hfill \\ +u^{7/8+k/4}{M}^{1/2}+u^{5/6+k/4}{M}^{7/12},\hfill & \hfill k=2,\\ u^{3/4+k/4}{M}^{2^{k-3}+(k-1)/4}+u^{7/8+k/4}{M}^{2^{k-4}+(2k-1)/8},\hfill & \hfill 3⩽k⩽9.\end{array}\right.\right),\hfill \end{array}$$

(106)

where $B_k(\infty )$ is defined in (79).

Combining (70), (72), (73) and (106), we can find real constants ${\theta }_k>0(2⩽k⩽9)$, such that if we take

$$M=\left\{\begin{array}{cc}{u}^{4{\theta }_k},\hfill & \hfill 2<k⩽8,\\ u^{{\displaystyle \frac{A_0-4}{A_0-9}}{\theta }_k},\hfill & \hfill k=9,\end{array}\right.$$

(107)

where $A_0=262/27$, then

$$\sum _{\begin{array}{c}p⩽u\\ p\equiv \ell \left(modq\right)\end{array}}{\Delta }^k\left(p\right)={\displaystyle \frac{1}{\phi \left(q\right)}}{\displaystyle \frac{B_k(\infty )}{{\left(\sqrt{2}\pi \right)}^k{2}^{k-1}}}{\int }_2^u{\displaystyle \frac{t^{k/4}}{logt}}dt+O\left(u^{1+k/4}{e}^{-c_2\sqrt{logu}}\right)+O\left(u^{1+k/4-{\theta }_k+\epsilon }\right),$$

(108)

and it is easy to see that the latter O-term can be ignored; thus, we have

$$\pi (u;q,\ell ;k)={\displaystyle \frac{1}{\phi \left(q\right)}}{\displaystyle \frac{B_k(\infty )}{{\left(\sqrt{2}\pi \right)}^k{2}^{k-1}}}{\int }_2^u{\displaystyle \frac{t^{k/4}}{logt}}dt+O\left(u^{1+k/4}{e}^{-c_2\sqrt{logu}}\right).$$

(109)

5.2. Asymptotic Formula for $S_1\left(N\right)$

Since $k_1$, $k_2$ and $k_3$ are symmetric in $S_1\left(N\right)$, without loss of generality, we will only evaluate $F_1\left(\alpha \right)$.

Lemma 13.

Using the notations in Section 3. For $\alpha ={\displaystyle \frac{a}{q}}+\lambda$, $\left|\lambda \right|⩽{\displaystyle \frac{1}{q\tau }}$ with $q⩽Q$, we have

$$F_1\left(\alpha \right)={\displaystyle \frac{B_{k_1}(\infty )}{{\left(\sqrt{2}\pi \right)}^{k_1}{2}^{k_1-1}}}{\displaystyle \frac{\mu \left(q\right)}{\phi \left(q\right)}}{\upsilon }_1\left(\lambda \right)+O(N^{1+{\displaystyle \frac{k_1}{4}}}{e}^{-c_3\sqrt{logN}}),$$

(110)

where $c_3>0$ is a constant, $B_{k_1}(\infty )$ is defined in Equation (79), and

$${\upsilon }_1\left(\lambda \right)={\int }_2^N{\displaystyle \frac{u^{\frac{k_1}{4}}\mathbf{e}\left(u\lambda \right)}{logu}}du.$$

(111)

In addition, $F_i\left(\alpha \right)$, $B_{k_i}(\infty )$, and ${\upsilon }_i\left(\lambda \right)$ $(i=2,3)$ are defined similarly and have the similar asymptotic formulas, which are depended only on $k_2$ and $k_3$.

Proof. 

At the begining of Section 5, we obtain

$$\begin{array}{cc}\hfill F_1\left(\alpha \right)& =\sum _{\begin{array}{c}\ell =1\\ (\ell ,q)=1\end{array}}^q\mathbf{e}\left({\displaystyle \frac{a\ell }{q}}\right){\int }_2^N\mathbf{e}\left(u\lambda \right)d\left(\pi (u;q,\ell ;k_1)\right)\hfill \\ \hfill & =\sum _{\begin{array}{c}\ell =1\\ (\ell ,q)=1\end{array}}^q\mathbf{e}\left({\displaystyle \frac{a\ell }{q}}\right){\int }_2^N\mathbf{e}\left(u\lambda \right)d\left({\displaystyle \frac{1}{\phi \left(q\right)}}{\displaystyle \frac{B_{k_1}(\infty )}{{\left(\sqrt{2}\pi \right)}^{k_1}{2}^{k_1-1}}}{\int }_2^u{\displaystyle \frac{t^{\frac{k_1}{4}}}{logt}}dt+O\left(u^{1+{\displaystyle \frac{k_1}{4}}}{e}^{-c_2\sqrt{logu}}\right)\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{\phi \left(q\right)}}{\displaystyle \frac{B_{k_1}(\infty )}{{\left(\sqrt{2}\pi \right)}^{k_1}{2}^{k_1-1}}}\sum _{\begin{array}{c}\ell =1\\ (\ell ,q)=1\end{array}}^q\mathbf{e}\left({\displaystyle \frac{a\ell }{q}}\right){\int }_2^N{\displaystyle \frac{u^{\frac{k_1}{4}}\mathbf{e}\left(u\lambda \right)}{logu}}du+O\left((1+|\lambda \left|N\right)N^{1+{\displaystyle \frac{k_1}{4}}}{e}^{-c_2\sqrt{logN}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{\phi \left(q\right)}}{\displaystyle \frac{B_{k_1}(\infty )}{{\left(\sqrt{2}\pi \right)}^{k_1}{2}^{k_1-1}}}\sum _{\begin{array}{c}\ell =1\\ (\ell ,q)=1\end{array}}^q\mathbf{e}\left({\displaystyle \frac{a\ell }{q}}\right){\int }_2^N{\displaystyle \frac{u^{\frac{k_1}{4}}\mathbf{e}\left(u\lambda \right)}{logu}}du+O\left(N^{1+{\displaystyle \frac{k_1}{4}}}{e}^{-c_3\sqrt{logN}}\right),\hfill \end{array}$$

(112)

where $c_3>0$ is a constant. Noting that

$$\sum _{\begin{array}{c}\ell =1\\ (\ell ,q)=1\end{array}}^q\mathbf{e}\left({\displaystyle \frac{a\ell }{q}}\right)=\mu \left(q\right),$$

(113)

we get

$$F_1\left(\alpha \right)={\displaystyle \frac{B_{k_1}(\infty )}{{\left(\sqrt{2}\pi \right)}^{k_1}{2}^{k_1-1}}}{\displaystyle \frac{\mu \left(q\right)}{\phi \left(q\right)}}{\upsilon }_1\left(\lambda \right)+O\left(N^{1+{\displaystyle \frac{k_1}{4}}}{e}^{-c_3\sqrt{logN}}\right).$$

(114)

□

Then, we turn to evaluate $S_1\left(N\right)$.

Lemma 14.

For $2⩽k_1,k_2,k_3⩽9$, we have

$$\begin{array}{cc}\hfill S_1\left(N\right)=& \left(3+{\displaystyle \frac{k_1+k_2+k_3}{4}}\right){\displaystyle \frac{\Gamma \left(1+\frac{k_1}{4}\right)\Gamma \left(1+\frac{k_2}{4}\right)\Gamma \left(1+\frac{k_2}{4}\right)}{\Gamma \left(3+\frac{k_1+k_2+k_3}{4}\right)}}\mathcal{C}(k_1,k_2,k_3)\mathfrak{S}\left(N\right){\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{3}N}}\hfill \\ \hfill & +O\left({\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{4}N}}\right),\hfill \end{array}$$

(115)

where $\mathfrak{S}\left(N\right)$ is defined in, Γ is the Euler Gamma function, and $\mathcal{C}(k_1,k_2,k_3)$ are constants defined by

$$\mathcal{C}(k_1,k_2,k_3)=\prod _{i=1}^3{\displaystyle \frac{B_{k_i}(\infty )}{{\left(\sqrt{2}\pi \right)}^{k_i}{2}^{k_i-1}}}.$$

(116)

Proof. 

By Equation (109), Lemma 112, and using the trival estimate $|{\upsilon }_i\left(\lambda \right)|\ll N^{1+{\displaystyle \frac{k_i}{4}}}/logN$ $(i=1,2,3)$, we derive that

$$\prod _{i=1}^3{F}_i\left(\alpha \right)=\mathcal{C}(k_1,k_2,k_3){\displaystyle \frac{\mu \left(q\right)}{{\phi }^3\left(q\right)}}\prod _{i=1}^3{\upsilon }_i\left(\lambda \right)+O\left(N^{3+{\displaystyle \frac{k_1+k_2+k_3}{4}}}{e}^{-c_4\sqrt{logN}}\right),$$

(117)

where $c_4>0$ is a constant. Then, we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{\mathcal{C}(k_1,k_2,k_3)}}{\int }_{E_1}\left(\prod _1^3{F}_i\left(\alpha \right)\right)\mathbf{e}(-N\alpha )d\alpha \hfill \\ \hfill =& \sum _{1⩽q⩽Q}\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^q\mathbf{e}\left(-{\displaystyle \frac{aN}{q}}\right){\int }_{-{\displaystyle \frac{1}{q\tau }}}^{{\displaystyle \frac{1}{q\tau }}}\left({\displaystyle \frac{\mu \left(q\right)}{{\phi }^3\left(q\right)}}\left(\prod _{i=1}^3{\upsilon }_i\left(\lambda \right)\right)+O\left(N^{3+{\displaystyle \frac{k_1+k_2+k_3}{4}}}{e}^{-c_4\sqrt{logN}}\right)\right)\mathbf{e}(-N\lambda )d\lambda \hfill \\ \hfill =& \sum _{1⩽q⩽Q}{\displaystyle \frac{\mu \left(q\right)}{{\phi }^3\left(q\right)}}\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^q\mathbf{e}\left(-{\displaystyle \frac{aN}{q}}\right){\int }_{-{\displaystyle \frac{1}{q\tau }}}^{{\displaystyle \frac{1}{q\tau }}}\left(\prod _{i=1}^3{\upsilon }_i\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda +O\left(N^{3+{\displaystyle \frac{k_1+k_2+k_3}{4}}}{e}^{-c_4\sqrt{logN}}Q{\tau }^{-1}\right)\hfill \\ \hfill =& \sum _{1⩽q⩽Q}{\displaystyle \frac{\mu \left(q\right)}{{\phi }^3\left(q\right)}}\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^q\mathbf{e}\left(-{\displaystyle \frac{aN}{q}}\right){\int }_{-{\displaystyle \frac{1}{q\tau }}}^{{\displaystyle \frac{1}{q\tau }}}\left(\prod _{i=1}^3{\upsilon }_i\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda +O\left(N^{2+{\displaystyle \frac{k_1+k_2+k_3}{4}}}{e}^{-c_5\sqrt{logN}}\right),\hfill \end{array}$$

(118)

where $c_5>0$ is a constant.

Define

$${{\upsilon }_i}^{*}\left(\lambda \right):=\sum _{n=3}^N{\displaystyle \frac{n^{\frac{k_i}{4}}\mathbf{e}\left(n\lambda \right)}{logn}},{{\upsilon }_i}^{**}\left(\lambda \right):=\sum _{n=3}^N{\displaystyle \frac{n^{\frac{k_i}{4}}\mathbf{e}\left(n\lambda \right)}{logN}},i=1,2,3.$$

(119)

Next, for $i=1,2,3$, we have

$$\begin{array}{cc}\hfill & {\int }_{n-1}^n{\displaystyle \frac{u^{\frac{k_i}{4}}\mathbf{e}\left(u\lambda \right)}{logu}}du-{\displaystyle \frac{n^{\frac{k_i}{4}}\mathbf{e}\left(n\lambda \right)}{logn}}={\int }_{n-1}^n\left({\displaystyle \frac{u^{\frac{k_i}{4}}\mathbf{e}\left(u\lambda \right)}{logu}}-{\displaystyle \frac{n^{\frac{k_i}{4}}\mathbf{e}\left(n\lambda \right)}{logn}}\right)du\hfill \\ \hfill =& {\int }_{n-1}^n{\int }_n^u{\left({\displaystyle \frac{x^{\frac{k_i}{4}}\mathbf{e}\left(\lambda x\right)}{logx}}\right)}^{\prime }dxdu={\int }_{n-1}^n{\int }_n^u\left({\displaystyle \frac{2\pi i\lambda \mathbf{e}\left(\lambda x\right)x^{\frac{k_i}{4}}}{logx}}+{\displaystyle \frac{x^{\frac{k_i}{4}-1}\mathbf{e}\left(\lambda x\right)}{logx}}-{\displaystyle \frac{x^{\frac{k_i}{4}-1}\mathbf{e}\left(\lambda x\right)}{{log}^{2}x}}\right)dxdu\hfill \\ \hfill \ll & {\displaystyle \frac{\left|\lambda \right|n^{\frac{k_i}{4}}+n^{\frac{k_i}{4}-1}}{logn}},\hfill \end{array}$$

(120)

and thus

$$\begin{array}{cc}\hfill {\upsilon }_i\left(\lambda \right)=& {\int }_2^N{\displaystyle \frac{u^{\frac{k_1}{4}}\mathbf{e}\left(u\lambda \right)}{logu}}du=\sum _{n=3}^N{\int }_{n-1}^n{\displaystyle \frac{u^{\frac{k_1}{4}}\mathbf{e}\left(u\lambda \right)}{logu}}du\hfill \\ \hfill =& \sum _{n=3}^N{\displaystyle \frac{n^{\frac{k_i}{4}}\mathbf{e}\left(n\lambda \right)}{logn}}+O\left(\sum _{n=3}^N\left({\displaystyle \frac{\left|\lambda \right|n^{\frac{k_i}{4}}+n^{\frac{k_i}{4}-1}}{logn}}\right)\right)\hfill \\ \hfill =& \sum _{n=3}^N{\displaystyle \frac{n^{\frac{k_i}{4}}\mathbf{e}\left(n\lambda \right)}{logn}}+O\left(N^{{\displaystyle \frac{k_i}{4}}}+\left|\lambda \right|N^{1+{\displaystyle \frac{k_i}{4}}}\right)\hfill \\ \hfill =& {{\upsilon }_i}^{*}\left(\lambda \right)+O\left(N^{{\displaystyle \frac{k_i}{4}}}{(logN)}^B\right)\hfill \end{array}$$

(121)

Therefore, we get

$$\prod _{i=1}^3{\upsilon }_i\left(\lambda \right)=\prod _{i=1}^3{{\upsilon }_i}^{*}\left(\lambda \right)+O\left(N^{2+{\displaystyle \frac{k_1+k_2+k_3}{4}}}{(logN)}^{B-2}\right),$$

(122)

where we use the trivial estimate $|{\upsilon }_i\left(\lambda \right)|\ll N^{1+{\displaystyle \frac{k_i}{4}}}{(logN)}^{-1}$. Hence, one has

$$\begin{array}{cc}\hfill & {\int }_{-{\displaystyle \frac{1}{q\tau }}}^{{\displaystyle \frac{1}{q\tau }}}\left(\prod _{i=1}^3{\upsilon }_i\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda \hfill \\ \hfill =& {\int }_{-{\displaystyle \frac{1}{q\tau }}}^{{\displaystyle \frac{1}{q\tau }}}\left(\prod _{i=1}^3{{\upsilon }_i}^{*}\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda +O\left(N^{2+{\displaystyle \frac{k_1+k_2+k_3}{4}}}{(logN)}^{B-2}{\left(q\tau \right)}^{-1}\right)\hfill \\ \hfill =& {\int }_{-{\displaystyle \frac{1}{q\tau }}}^{{\displaystyle \frac{1}{q\tau }}}\left(\prod _{i=1}^3{{\upsilon }_i}^{*}\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda +O\left(N^{1+{\displaystyle \frac{k_1+k_2+k_3}{4}}}{(logN)}^{2B-2}\right).\hfill \end{array}$$

(123)

By Lemma 1 and partial summation, we derive that

$$\begin{array}{cc}\hfill & {{\upsilon }_i}^{*}\left(\lambda \right)=\sum _{2<n⩽N}{\displaystyle \frac{n^{\frac{k_i}{4}}\mathbf{e}\left(n\lambda \right)}{logn}}={\int }_2^N{\displaystyle \frac{u^{\frac{k_i}{4}}}{logu}}d\left(\sum _{2<n⩽u}\mathbf{e}\left(\lambda n\right)\right)\hfill \\ \hfill \ll & {\displaystyle \frac{N^{\frac{k_i}{4}}}{logN}}\left|\sum _{2<n⩽N}\mathbf{e}\left(\lambda n\right)\right|+{\int }_2^N\left(\sum _{2<n⩽u}\mathbf{e}\left(\lambda n\right)\right)\left(u^{{\displaystyle \frac{k_i}{4}}-1}{\displaystyle \frac{1}{logu}}-u^{{\displaystyle \frac{k_i}{4}}-1}{\displaystyle \frac{1}{{log}^{2}u}}\right)du\hfill \\ \hfill \ll & N^{{\displaystyle \frac{k_i}{4}}}\underset{2<u⩽N}{max}\left|\sum _{2<n⩽u}\mathbf{e}\left(\lambda n\right)\right|\ll N^{{\displaystyle \frac{k_i}{4}}}\underset{2<u⩽N}{max}min\left(u,{\displaystyle \frac{1}{\left|\lambda \right|}}\right)\hfill \\ \hfill \ll & N^{{\displaystyle \frac{k_i}{4}}}min\left(N,{\displaystyle \frac{1}{\left|\lambda \right|}}\right),\hfill \end{array}$$

(124)

which implies that

$$\begin{array}{cc}\hfill {\int }_{{\displaystyle \frac{1}{q\tau }}}^{{\displaystyle \frac{1}{2}}}\left(\prod _{i=1}^3{{\upsilon }_i}^{*}\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda & \ll N^{{\displaystyle \frac{k_1+k_2+k_3}{4}}}{\int }_{{\displaystyle \frac{1}{q\tau }}}^{{\displaystyle \frac{1}{2}}}{\displaystyle \frac{d\lambda }{{\left|\lambda \right|}^3}}\ll N^{{\displaystyle \frac{k_1+k_2+k_3}{4}}}{\left(q\tau \right)}^2\hfill \\ \hfill & \ll N^{{\displaystyle \frac{k_1+k_2+k_3}{4}}}{\left(Q\tau \right)}^2\ll N^{2+{\displaystyle \frac{k_1+k_2+k_3}{4}}}{(logN)}^{-2(B-A)}.\hfill \end{array}$$

(125)

Therefore, we have

$$\begin{array}{cc}\hfill {\int }_{-{\displaystyle \frac{1}{q\tau }}}^{{\displaystyle \frac{1}{q\tau }}}\left(\prod _{i=1}^3{{\upsilon }_i}^{*}\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda \ll & {\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}\left(\prod _{i=1}^3{{\upsilon }_i}^{*}\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda \hfill \\ \hfill & +O\left(N^{2+{\displaystyle \frac{k_1+k_2+k_3}{4}}}{(logN)}^{-2(B-A)}\right).\hfill \end{array}$$

(126)

On the other hand, we get

$$\begin{array}{cc}\hfill |{{\upsilon }_i}^{*}\left(\lambda \right)-{{\upsilon }_i}^{**}\left(\lambda \right)|& =\left|\sum _{n=3}^N{\displaystyle \frac{n^{\frac{k_i}{4}}\mathbf{e}\left(n\lambda \right)}{logn}}-\sum _{n=3}^N{\displaystyle \frac{n^{\frac{k_i}{4}}\mathbf{e}\left(n\lambda \right)}{logN}}\right|⩽N^{{\displaystyle \frac{k_i}{4}}}\sum _{n=3}^N\left({\displaystyle \frac{1}{logn}}-{\displaystyle \frac{1}{logN}}\right)\hfill \\ \hfill & ⩽N^{{\displaystyle \frac{k_i}{4}}}{\int }_2^N\left({\displaystyle \frac{1}{logu}}-{\displaystyle \frac{1}{logN}}\right)du=N^{{\displaystyle \frac{k_i}{4}}}{\int }_2^N{\int }_u^N{\displaystyle \frac{dt}{t{log}^{2}t}}du\hfill \\ \hfill & =N^{{\displaystyle \frac{k_i}{4}}}{\int }_2^N{\displaystyle \frac{dt}{t{log}^{2}t}}{\int }_2^{t}du\ll N^{{\displaystyle \frac{k_i}{4}}}{\int }_2^N{\displaystyle \frac{dt}{{log}^{2}t}}\ll {\displaystyle \frac{N^{1+\frac{k_i}{4}}}{{log}^{2}N}},\hfill \end{array}$$

(127)

which implies

$${{\upsilon }_i}^{*}\left(\lambda \right)={{\upsilon }_i}^{**}\left(\lambda \right)+O\left({\displaystyle \frac{N^{1+\frac{k_i}{4}}}{{log}^{2}N}}\right).$$

(128)

We have

$$\begin{array}{cc}\hfill a_1{a}_2{a}_3-b_1{b}_2{b}_3& =(a_1-b_1)a_2{a}_3+(a_2-b_2)a_1{b}_3+(a_3-b_3)b_1{b}_2\hfill \\ \hfill & \ll |a_1-b_1\left|\right|a_2{a}_3|+|a_2-b_2\left|\right|a_1{b}_3|+|a_3-b_3\left|\right|b_1{b}_2|,\hfill \end{array}$$

(129)

thus

$$\begin{array}{cc}\hfill \left|\prod _{i=1}^3{{\upsilon }_i}^{*}\left(\lambda \right)-\prod _{i=1}^3{{\upsilon }_i}^{**}\left(\lambda \right)\right|\ll & {\displaystyle \frac{N^{1+\frac{k_1}{4}}}{{log}^{2}N}}\left|{{\upsilon }_2}^{*}\left(\lambda \right){{\upsilon }_3}^{*}\left(\lambda \right)\right|+{\displaystyle \frac{N^{1+\frac{k_2}{4}}}{{log}^{2}N}}\left|{{\upsilon }_3}^{*}\left(\lambda \right){{\upsilon }_3}^{**}\left(\lambda \right)\right|\hfill \\ \hfill & +{\displaystyle \frac{N^{1+\frac{k_3}{4}}}{{log}^{2}N}}\left|{{\upsilon }_1}^{**}\left(\lambda \right){{\upsilon }_2}^{**}\left(\lambda \right)\right|,\hfill \end{array}$$

(130)

and without loss of generality, we derive that

$$\begin{array}{cc}\hfill & {\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}\left(\prod _{i=1}^3{{\upsilon }_i}^{*}\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda \hfill \\ \hfill =& {\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}\left(\prod _{i=1}^3{{\upsilon }_i}^{**}\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda +O\left({\displaystyle \frac{N^{1+\frac{k_1}{4}}}{{log}^{2}N}}{\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}\left|{{\upsilon }_2}^{*}\left(\lambda \right){{\upsilon }_3}^{*}\left(\lambda \right)\right|d\lambda \right)\hfill \\ \hfill & +O\left({\displaystyle \frac{N^{1+\frac{k_2}{4}}}{{log}^{2}N}}{\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}\left|{{\upsilon }_3}^{*}\left(\lambda \right){{\upsilon }_3}^{**}\left(\lambda \right)\right|d\lambda \right)+O\left({\displaystyle \frac{N^{1+\frac{k_3}{4}}}{{log}^{2}N}}{\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}\left|{{\upsilon }_1}^{**}\left(\lambda \right){{\upsilon }_2}^{**}\left(\lambda \right)\right|d\lambda \right).\hfill \end{array}$$

(131)

Noting the fact that

$$\begin{array}{c}\hfill {\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}{\left|{{\upsilon }_i}^{*}\left(\lambda \right)\right|}^{2}d\lambda =\sum _{n_1=3}^N\sum _{n_2=3}^N{\displaystyle \frac{{\left(n_1{n}_2\right)}^{\frac{k_i}{4}}}{log{n}_{1}log{n}_2}}{\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}\mathbf{e}\left(\lambda (n_1-n_2)\right)d\lambda \ll {\displaystyle \frac{N^{1+\frac{k_i}{2}}}{{log}^{2}N}},\end{array}$$

(132)

and

$$\begin{array}{c}\hfill {\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}{\left|{{\upsilon }_i}^{**}\left(\lambda \right)\right|}^{2}d\lambda ={\displaystyle \frac{1}{{log}^{2}N}}\sum _{n_1=3}^N\sum _{n_2=3}^N{\left(n_1{n}_2\right)}^{{\displaystyle \frac{k_i}{4}}}{\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}\mathbf{e}\left(\lambda (n_1-n_2)\right)d\lambda \ll {\displaystyle \frac{N^{1+\frac{k_i}{2}}}{{log}^{2}N}},\end{array}$$

(133)

using the Cauchy–Schwarz’s inequality, we deduce that

$$\begin{array}{cc}\hfill & {\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}\left(\prod _{i=1}^3{{\upsilon }_i}^{*}\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda \hfill \\ \hfill =& {\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}\left(\prod _{i=1}^3{{\upsilon }_i}^{**}\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda +O\left({\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{4}N}}\right).\hfill \end{array}$$

(134)

And then we have

$$\begin{array}{cc}\hfill & {\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}\left(\prod _{i=1}^3{{\upsilon }_i}^{*}\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda \hfill \\ \hfill =& {\displaystyle \frac{1}{{(logN)}^3}}\sum _{n_1=3}^N\sum _{n_2=3}^N\sum _{n_3=3}^N{n_1}^{{\displaystyle \frac{k_1}{4}}}{n_2}^{{\displaystyle \frac{k_2}{4}}}{n_3}^{{\displaystyle \frac{k_3}{4}}}{\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}\mathbf{e}\left((n_1+n_2+n_3-N)\lambda \right)d\lambda +O\left({\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{4}N}}\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{{(logN)}^3}}\sum _{\begin{array}{c}N=n_1+n_2+n_3\\ 3<n_i⩽N\\ i=1,2,3\end{array}}{n_1}^{{\displaystyle \frac{k_1}{4}}}{n_2}^{{\displaystyle \frac{k_2}{4}}}{n_3}^{{\displaystyle \frac{k_3}{4}}}+O\left({\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{4}N}}\right).\hfill \end{array}$$

(135)

Using partial summation and Theorem 3.6 in Karatsuba [26], we derive that

$$\begin{array}{cc}\hfill & {\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}\left(\prod _{i=1}^3{{\upsilon }_i}^{*}\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda \hfill \\ \hfill =& {\displaystyle \frac{1}{N}}\underset{\begin{array}{c}{u}_1+u_2+u_3⩽N\\ 0<u_1,u_2,u_3⩽N\end{array}}{\int \int \int }{u_1}^{{\displaystyle \frac{k_1}{4}}}{u_2}^{{\displaystyle \frac{k_2}{4}}}{u_3}^{{\displaystyle \frac{k_3}{4}}}d{u}_{1}d{u}_{2}d{u}_3+O\left({\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{4}N}}\right)\hfill \\ \hfill =& {\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{3}N}}\underset{\begin{array}{c}{u}_1+u_2+u_3⩽1\\ 0<u_1,u_2,u_3⩽1\end{array}}{\int \int \int }{u_1}^{{\displaystyle \frac{k_1}{4}}}{u_2}^{{\displaystyle \frac{k_2}{4}}}{u_3}^{{\displaystyle \frac{k_3}{4}}}d{u}_{1}d{u}_{2}d{u}_3+O\left({\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{4}N}}\right)\hfill \\ \hfill =& \left(3+{\displaystyle \frac{k_1+k_2+k_3}{4}}\right){\displaystyle \frac{\Gamma \left(1+\frac{k_1}{4}\right)\Gamma \left(1+\frac{k_2}{4}\right)\Gamma \left(1+\frac{k_2}{4}\right)}{\Gamma \left(3+\frac{k_1+k_2+k_3}{4}\right)}}{\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{3}N}}+O\left({\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{4}N}}\right).\hfill \end{array}$$

(136)

By Equations (123), (126) and (136), we can see that the innermost integral in Equation (118) is

$$\begin{array}{cc}\hfill & {\int }_{-{\displaystyle \frac{1}{q\tau }}}^{{\displaystyle \frac{1}{q\tau }}}\left(\prod _{i=1}^3{\upsilon }_i\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda \hfill \\ \hfill =& {\int }_{-{\displaystyle \frac{1}{q\tau }}}^{{\displaystyle \frac{1}{q\tau }}}\left(\prod _{i=1}^3{{\upsilon }_i}^{*}\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda +O\left(N^{1+{\displaystyle \frac{k_1+k_2+k_3}{4}}}{(logN)}^{2B-2}\right)\hfill \\ \hfill =& {\int }_{-{\displaystyle \frac{1}{2}}}^{{\displaystyle \frac{1}{2}}}\left(\prod _{i=1}^3{{\upsilon }_i}^{*}\left(\lambda \right)\right)\mathbf{e}(-N\lambda )d\lambda +O\left(N^{2+{\displaystyle \frac{k_1+k_2+k_3}{4}}}{(logN)}^{-2(B-A)}+N^{1+{\displaystyle \frac{k_1+k_2+k_3}{4}}}{(logN)}^{2B-2}\right)\hfill \\ \hfill =& \left(3+{\displaystyle \frac{k_1+k_2+k_3}{4}}\right){\displaystyle \frac{\Gamma \left(1+\frac{k_1}{4}\right)\Gamma \left(1+\frac{k_2}{4}\right)\Gamma \left(1+\frac{k_2}{4}\right)}{\Gamma \left(3+\frac{k_1+k_2+k_3}{4}\right)}}{\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{3}N}}+O\left({\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{4}N}}\right).\hfill \end{array}$$

(137)

Putting Equation (137) into Equation (118), and using the conclusion of the three primes theorem, we have

$$\begin{array}{cc}\hfill S_1\left(N\right)=& \left(3+{\displaystyle \frac{k_1+k_2+k_3}{4}}\right){\displaystyle \frac{\Gamma \left(1+\frac{k_1}{4}\right)\Gamma \left(1+\frac{k_2}{4}\right)\Gamma \left(1+\frac{k_2}{4}\right)}{\Gamma \left(3+\frac{k_1+k_2+k_3}{4}\right)}}\mathcal{C}(k_1,k_2,k_3)\mathfrak{S}\left(N\right){\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{3}N}}\hfill \\ \hfill & +O\left({\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{4}N}}\right).\hfill \end{array}$$

(138)

Hence, we have completed the proof of Lemma 14. □

6. Upper Bound Estimate for ${\mathit{S}}_{\mathbf{2}}\left(\mathit{N}\right)$

Since $k_1$, $k_2$, $k_3$ are symmetric, without loss of generality, we assume $max(k_1,k_2,k_3)=k_1$, then

$$\begin{array}{cc}\hfill S_2\left(N\right)& ={\int }_{E_2}\left(\prod _{i=1}^3{F}_i\left(\alpha \right)\right)\mathbf{e}(-N\alpha )d\alpha \ll \underset{\alpha \in E_2}{sup}|F_1\left(\alpha \right)|{\int }_0^1\left|F_2\left(\alpha \right)F_3\left(\alpha \right)\right|d\alpha .\hfill \end{array}$$

(139)

Using Cauchy–Schwarz’s inequality, we derive that

$$\begin{array}{cc}\hfill S_2\left(N\right)\ll & \underset{\alpha \in E_2}{sup}\left|F_1\left(\alpha \right)\right|{\left({\int }_0^1{\left|F_2\left(\alpha \right)\right|}^{2}d\alpha \right)}^{1/2}{\left({\int }_0^1{\left|F_3\left(\alpha \right)\right|}^{2}d\alpha \right)}^{1/2}\hfill \\ \hfill \ll & \underset{\alpha \in E_2}{sup}\left|F_1\left(\alpha \right)\right|{\left(\sum _{p_1⩽N}\sum _{p_2⩽N}{\Delta }^{k_2}\left(p_1\right){\Delta }^{k_2}\left(p_2\right){\int }_0^1\mathbf{e}\left((p_1-p_2)\alpha \right)d\alpha \right)}^{1/2}\hfill \\ \hfill & \times {\left(\sum _{p_1⩽N}\sum _{p_2⩽N}{\Delta }^{k_3}\left(p_1\right){\Delta }^{k_3}\left(p_2\right){\int }_0^1\mathbf{e}\left((p_1-p_2)\alpha \right)d\alpha \right)}^{1/2}\hfill \\ \hfill \ll & \underset{\alpha \in E_2}{sup}\left|F_1\left(\alpha \right)\right|{\left(\sum _{p⩽N}{\Delta }^{2k_2}\left(p\right)\right)}^{1/2}{\left(\sum _{p⩽N}{\Delta }^{2k_3}\left(p\right)\right)}^{1/2}\hfill \\ \hfill \ll & \underset{\alpha \in E_2}{sup}\left|F_1\left(\alpha \right)\right|{\displaystyle \frac{N^{1+\frac{k_2+k_3}{4}}}{logN}}\hfill \end{array}$$

(140)

provided that $2⩽k_2⩽4$ and $2⩽k_3⩽4$, where we use Equation (10). Then, in order to prove the theorem, we need to show

$$\underset{\alpha \in E_2}{sup}\left|F_1\left(\alpha \right)\right|\ll {\displaystyle \frac{N^{1+\frac{k_1}{4}}}{{log}^{3}N}}.$$

(141)

For $i=1,2,3$, using partial summation, we obtain

$$\begin{array}{cc}\hfill F_i\left(\alpha \right)& =\sum _{p⩽N}{\Delta }^{k_i}\left(p\right)\mathbf{e}\left(p\alpha \right)={\int }_2^N\mathbf{e}\left(u\alpha \right)d\left(\sum _{p⩽u}{\Delta }^{k_i}\left(p\right)\right)\hfill \\ \hfill & ={\int }_2^N\mathbf{e}\left(u\alpha \right)d\left({\displaystyle \frac{B_{k_i}(\infty )}{{\left(\sqrt{2}\pi \right)}^{k_i}{2}^{k_i-1}}}{\displaystyle \frac{1}{\phi \left(q\right)}}{\int }_2^u{\displaystyle \frac{t^{\frac{k_i}{4}}}{logt}}dt+O\left(u^{1+{\displaystyle \frac{k_i}{4}}}{e}^{-c_6\sqrt{logu}}\right)\right)\hfill \\ \hfill & ={\displaystyle \frac{B_{k_i}(\infty )}{{\left(\sqrt{2}\pi \right)}^{k_i}{2}^{k_i-1}}}{\displaystyle \frac{1}{\phi \left(q\right)}}{\int }_2^N{\displaystyle \frac{u^{\frac{k_i}{4}}\mathbf{e}\left(u\alpha \right)}{logu}}du+O\left(u^{1+{\displaystyle \frac{k_i}{4}}}{e}^{-c_6\sqrt{logu}}\right).\hfill \end{array}$$

(142)

where $c_6>0$ is a constant. For $\alpha =a/q+\lambda$ with $Q<q⩽\tau$, $\left|\lambda \right|⩽{\displaystyle \frac{1}{q\tau }}$ and $(a,q=1)$, $1⩽a⩽q$, and thus, we obtain

$$F_i\left(\alpha \right)\ll {\displaystyle \frac{N^{\frac{k_i}{4}}}{\left|\alpha \right|logN}}\ll {\displaystyle \frac{N^{\frac{k_i}{4}}\tau }{logN}}\ll {\displaystyle \frac{N^{1+\frac{k_i}{4}}}{{log}^{B+1}N}}$$

(143)

by using the first derivative test (see (2.3) of Ivić [7]). If $B⩾2$, Equation (141) holds, and hence

$$S_2\left(N\right)\ll {\displaystyle \frac{N^{2+\frac{k_1+k_2+k_3}{4}}}{{log}^{4}N}}.$$

(144)

Combining Equation (25), Lemma 14, and Equation (144), we complete the proof of Theorem 1.

7. Proof of Theorem 2

We are going to give the upper bound for $F_i\left(\alpha \right)$ $(i=1,2,3)$ when $k_i=1$. We have

$$\begin{array}{c}\hfill \sum _{p⩽N}\Delta \left(p\right)\mathbf{e}\left(p\alpha \right)=\sum _{p⩽N}{\delta }_1(p,M)\mathbf{e}\left(p\alpha \right)+\sum _{p⩽N}{\delta }_2(p,M)\mathbf{e}\left(p\alpha \right):={\Phi }_1+{\Phi }_2.\end{array}$$

(145)

For ${\Phi }_2$, by Equation (45) and Cauchy–Schwarz’s inequality,

$$\sum _{p⩽N}{\delta }_2(p,M)\mathbf{e}\left(p\alpha \right)\ll {\left(\sum _{p⩽N}{{\delta }_2}^2(p,M)\right)}^{1/2}{\left(\sum _{p⩽N}1\right)}^{1/2}\ll N^{5/4+\epsilon }{M}^{-1/4}$$

(146)

holds for $M\ll N^{1/3}$.

In order to estimate ${\Phi }_1$, by Lemma 9 and the similar argument in Section 5.1, it is sufficient to estimate

$${\displaystyle \frac{N^{1/4}}{M^{3/4}}}\sum _{m\sim M}d\left(m\right)\left|\sum _{n_1\sim N_1}\cdots \sum _{n_6\sim N_6}(log{n}_1)\mu \left(n_4\right)\mu \left(n_5\right)\mu \left(n_6\right)\mathbf{e}\left(2m^{1/2}\sqrt{n_1{n}_2\cdots n_6}+(n_1{n}_2\cdots n_6)\alpha \right)\right|,$$

(147)

where $N_1\cdots N_6\asymp u$; $2N_i⩽{\left(2u\right)}^{1/3}$, $i=4,5,6$ and some $n_i$ may only take value 1, $\alpha \in [0,1]$. Furthermore, using Lemmas 11 and 12 of the condition $k=1$, similar to the four cases in Section 4, we have

$$N^{-\epsilon }{\Phi }_1\ll N^{3/4}{M}^{1/2}+N{M}^{3/8}+N^{7/8}{M}^{1/4}+N^{13/12}{M}^{1/3}.$$

(148)

Take $M=N^{1/4}$, combining Equations (145), (146) and (148) we obtain

$$F_i\left(\alpha \right)\ll N^{19/16+\epsilon }.$$

(149)

Thus, if there exists $j\in \{1,2,3\}$ such that $k_j=1$ (without loss of generality we take $j=1$), then

$$\begin{array}{cc}\hfill S(N;k_1,k_2,k_3)& ={\int }_0^1{F}_1\left(\alpha \right)F_2\left(\alpha \right)F_3\left(\alpha \right)\mathbf{e}(-N\alpha )d\alpha \hfill \\ \hfill & \ll \underset{\alpha \in [0,1]}{sup}|F_1\left(\alpha \right)|{\int }_0^1\left|F_2\left(\alpha \right)F_3\left(\alpha \right)\right|d\alpha .\hfill \end{array}$$

(150)

By Equation (140) we obtain

$$\begin{array}{cc}\hfill S(N;k_1,k_2,k_3)& \ll \underset{\alpha \in [0,1]}{sup}\left|F_1\left(\alpha \right)\right|{\displaystyle \frac{N^{1+\frac{k_2+k_3}{4}}}{logN}}\hfill \\ \hfill & \ll N^{2+{\displaystyle \frac{k_1+k_2+k_3}{4}}-{\displaystyle \frac{1}{16}}+\epsilon }.\hfill \end{array}$$

(151)

Thus, we have completed the proof of Theorem 2.

8. Conclusions

This article gives an analogue of the three primes theorem, which is weighted by $\Delta (·)$. The methods in this paper can be used for further research. For example, the three primes theorem weighted by the error term of the sum of other arithmetic functions may only need to give new estimate for the exponential sum generated by new error terms. Moreover, it may be used in other problems which should use the circle method.