On the average number of cyclic subgroups of the groups ${\mathbb {Z}}_{n_1} \times {\mathbb {Z}}_{n_2}\times {\mathbb {Z}}_{n_3}$ with $n_1,n_2,n_3\le x$

Abstract

Let ${{\mathbb {Z}}}_{n}$ be the additive group of residue classes modulo n. Let $c(n_1,n_2,n_3)$ denote the number of cyclic subgroups of the group ${{\mathbb {Z}}}_{n_1}\times {{\mathbb {Z}}}_{n_2}\times {{\mathbb {Z}}}_{n_3}$, where $n_1, n_2$ and $n_3$ are arbitrary positive integers. In this paper we obtain an asymptotic formula for the sum $\sum _{n_1,n_2,n_3\le _x} c(n_1,n_2,n_3).$

1 Introduction

Let ${\mathbb {Z}}_{n}$ denote the additive group of residue classes modulo n. We recall that a finite abelian group G ($|G|>1$) is of rank r if it is isomorphic to ${\mathbb {Z}}_{n_1} \times \cdots \times {\mathbb {Z}}_{n_r}$, where $n_1,\ldots ,n_r \in {\mathbb {N}}{\setminus } \{1\}$ and $n_j \mid n_{j+1}$ ($1\le j\le r-1$). Let $\text {rank}(G)$ denote the rank of G and s(G) be the number of subgroups of G. The one variable level function $t_r(n)=\sum _{|G|=n, \, \text {rank}(G)\le r} s(G)$ and its summatory function $\sum _{n\le x} t_r(n)$ were investigated by Bhowmik and Wu [1]. There are many other asymptotic results on the number of subgroups of abelian groups. See, e.g., the papers [3, 9] and their references.

We are interested in the following different problem, not studied in the above papers. Let $r\ge 1$ be a fixed integer. For arbitrary positive integers $n_1,\ldots ,n_r,$ consider the group ${\mathbb {Z}}_{n_1}\times \cdots \times {\mathbb {Z}}_{n_r}$. Let $s(n_1,\ldots , n_r)$ and $c(n_1,\ldots , n_r)$ denote the total number of subgroups and the number of cyclic subgroups of the given group ${\mathbb {Z}}_{n_1}\times \cdots \times {\mathbb {Z}}_{n_r}$, respectively. Here $s(n_1,\ldots , n_r)$ and $c(n_1,\ldots , n_r)$ are multiplicative functions of r variables.

Suppose $x>0$ is a real number. For each $r\ge 1$ define the multivariable sums

$$\begin{aligned}&S_r(x):=\sum _{n_1\le x, \ldots , n_r\le x} s(n_1,\ldots , n_r), \\&C_r(x):=\sum _{n_1\le x, \ldots , n_r\le x} c(n_1,\ldots , n_r). \end{aligned}$$

When $r=1$, the group ${\mathbb {Z}}_n$ is cyclic and hence $s(n)=c(n)=\tau (n)$, where $\tau (n)$ denotes the number of divisors of n. So, $S_1(x)=C_1(x)= \sum _{n\le x} \tau (n)$. Therefore, the problem of finding estimates for the sums $S_r(x)$ and $C_r(x)$ can be viewed as a generalization of the well-known Dirichlet divisor problem.

Now consider the case $r\ge 2$. If $(n_i, n_j)=1$ for all $1\le i<j\le r$, then ${\mathbb {Z}}_{n_1}\times \cdots \times {\mathbb {Z}}_{n_r}$ is cyclic and isomorphic to ${\mathbb {Z}}_{n_1\cdots n_r}$, which is a consequence of the Chinese remainder theorem. If otherwise, then ${\mathbb {Z}}_{n_1}\times \cdots \times {\mathbb {Z}}_{n_r}$ has rank at least two.

There are several ways to compute $s(n_1,n_2)$ and $c(n_1,n_2).$ For example, one way is using Goursat’s lemma for groups in [2, 8, 13] and another one is using the concept of the fundamental group lattice in [10, 11]. See also the paper [4]. For any $n_1,n_2\in {{\mathbb {N}}}$ we have

$$\begin{aligned} \begin{aligned} s(n_1,n_2)&= \sum _{d \mid n_1, e\mid n_2} (d,e) \\&= \sum _{d \mid (n_1,n_2)} \varphi (d) \tau (n_1/d)\tau (n_2/d) \end{aligned} \end{aligned}$$

(1.1)

and

$$\begin{aligned} \begin{aligned} c(n_1,n_2)&= \sum _{d \mid n_1, e\mid n_2} \varphi ((d,e)) \\&= \sum _{d \mid (n_1,n_2)} (\mu *\varphi )(d) \tau (n_1/d)\tau (n_2/d). \end{aligned} \end{aligned}$$

(1.2)

When $r\ge 3$, we have by Theorem 1 of [12],

$$\begin{aligned} c(n_1,\ldots ,n_r)=\sum _{d_1\mid n_1,\ldots , d_r\mid n_r} \frac{\varphi (d_1)\cdots \varphi (d_r)}{\varphi ([d_1,\ldots , d_r])}. \end{aligned}$$

(1.3)

However, in the case $r\ge 3$ there is no such a closed formula for $s(n_1, \ldots , n_r)$.

W. G. Nowak and L. Tóth [6] studied the average orders of the functions $s(n_1,n_2)$ and $c(n_1,n_2)$. They proved that the asymptotic formulas

$$\begin{aligned} S_2(x)=x^2\left( \sum _{j=0}^3 A_{j} \log ^j x\right) +O(x^{\frac{1117}{701}+\varepsilon }) \end{aligned}$$

(1.4)

and

$$\begin{aligned} C_2(x)=x^2\left( \sum _{j=0}^3 B_{j} \log ^j x\right) +O(x^{\frac{1117}{701}+\varepsilon }) \end{aligned}$$

(1.5)

hold, where $A_{j}$ and $B_j$ ($0\le j\le 3$) are explicit constants, of which definitions are omitted here. Note that the exponent of the error term is $1117/701=1.593437\cdots$. The authors of the present paper proved in [14] that both error terms in the asymptotic formulas (1.4) and (1.5) can be improved into $O(x^{1.5}(\log x)^{6.5})$.

In this paper we shall prove the following theorem, by using a multidimensional Perron formula and the complex integration method.

Theorem 1.1

The asymptotic formula

$$\begin{aligned} C_3(x) = x^3 \left( \sum _{j=0}^7 c_{j} \log ^j x\right) +O(x^{\frac{8}{3}+\varepsilon }) \end{aligned}$$

(1.6)

holds, where $c_j\, (0\le j\le 7)$ are explicit constants.

Notations

Throughout this paper, ${\mathbb {N}}$ denotes the set of all positive integers, ${\mathbb {N}}_0={\mathbb {N}}\cup \{0\},$$(a_1,\ldots ,a_k)$ and $[a_1,\ldots ,a_k]$ are the gcd and lcm of $a_1,\ldots ,a_k\in {\mathbb {N}}$, $\varphi$ is Euler’s totient function, $\mu$ is the Möbius function, $\zeta$ denotes the Riemann zeta-function, $\tau _k(n)$ denotes the number of ways n can be written as a product of k positive integers ($\tau (n)=\tau _2(n)$).

2 Preliminary lemmas

We shall use the following lemmas.

Lemma 2.1

Suppose $\mathfrak {R}z>1, \mathfrak {R}w>1$. Then

$$\begin{aligned} C(z,w):= \sum _{m\ge 1} \sum _{n\ge 1} \frac{c(m,n)}{m^z n^w}=\zeta ^2(z)\zeta ^2(w)\zeta (z+w-1)\zeta ^{-2}(z+w). \end{aligned}$$

Proof

See the paper [6]. $\square$

Lemma 2.2

Suppose that $r\ge 2$ is a fixed integer and $f(n_1,\ldots ,n_r)$ is an arithmetical function of r variables such that it is symmetric in the variables $n_1,\ldots ,n_r$ and its Dirichlet series

$$\begin{aligned} F(z_1,\ldots ,z_r):=\sum _{n_1\ge 1} \cdots \sum _{n_r\ge 1} \frac{f(n_1,\ldots ,n_r)}{n_1^{z_1}\cdots n_r^{z_r}} \end{aligned}$$

is absolutely convergent for $\mathfrak {R}z_j> \sigma _a$ ($1\le j\le r$) with some $\sigma _a>0$. Suppose that $x\notin {\mathbb {N}}$ and T are two large parameters such that $10\le T\le x/2$ and define

$$\begin{aligned} b_j =\sigma _a+\frac{10^{j-1}}{\log x}, \quad \ T_j=10^{j-1} T \ (1\le j\le r). \end{aligned}$$

Then we have

$$\begin{aligned} \sum _{n_1\le x} \cdots \sum _{n_r\le x} f(n_1,\ldots ,n_r) = I_f(x,T) + O(x^{r\sigma _a}E_f(x,T)), \end{aligned}$$

where

$$\begin{aligned} I_f(x,T):= & {} \frac{1}{(2\pi i)^r} \int _{b_1-iT_1}^{b_1+iT_1}\cdots \int _{b_r-iT_r}^{b_r+iT_r} F(z_1,\ldots ,z_r) x^{z_1+\cdots +z_r}\frac{dz_r\cdots dz_1}{z_r\cdots z_1},\\ E_f(x,T):= & {} \sum _{n_1\ge 1} \cdots \sum _{n_r\ge 1} \frac{|f(n_1,\ldots ,n_r)| (n_1\cdots n_r)^{-b_1}}{\min _{1\le j \le r} T\left| \log \frac{x}{n_j}\right| +1}. \end{aligned}$$

Proof

This is a slight modification of Lemma 2.2 in [14]. $\square$

Lemma 2.3

Suppose $\ell \ge 0$ is a fixed integer. For $\sigma >1$ we have

$$\begin{aligned} \zeta ^{(\ell )}(\sigma +it)&\ll \min \left( \frac{1}{(\sigma -1)^{1+\ell }}, \log ^{1+\ell } (|t|+2)\right) . \end{aligned}$$

Proof

The case $\ell =0$ can be found in Chapter 7 of Pan and Pan [7]. The case $\ell \ge 1$ follows from the result of the case $\ell =0$ and Cauchy’s theorem. $\square$

Lemma 2.4

Suppose $\ell \ge 0$ is a fixed integer. Then we have

$$\begin{aligned} \zeta ^{(\ell )}(\sigma +it)\ll \left\{ \begin{array}{ll} (|t|+2)^{\frac{1-\sigma }{3}}\log ^{1+\ell } (|t|+2) ,&{}\hbox {if } 1/2\le \sigma \le 1 ,\\ (|t|+2)^{\frac{3-4\sigma }{6}}\log ^{1+\ell } (|t|+2),&{}\hbox {if } 0\le \sigma \le 1/2. \end{array}\right. \end{aligned}$$

Proof

The estimate for the case $\ell =0$ follows from the bounds

$$\begin{aligned} \zeta (it)\ll (|t|+2)^{1/2},\ \ \zeta (1/2+it)\ll (|t|+2)^{1/6}, \ \ \zeta (1+it)\ll \log (|t|+2) \end{aligned}$$

and the Phragmen-Lindelöf principle. The estimate for the case $\ell \ge 1$ follows from the result of the case $\ell =0$ and Cauchy’s theorem. $\square$

Lemma 2.5

Suppose $V>10$ is a large parameter and $41/60\le u\le 1$. Then we have

$$\begin{aligned} \int _{-V}^{ V} | \zeta (u+iv)|^{10} dv \ll V^{1+\varepsilon }. \end{aligned}$$

Proof

It is proved in Chapter 8 of Ivić [5] that Lemma 2.5 holds for $u=41/60.$ By Lemma 2.3 we see that Lemma 2.5 trivially holds for $u=1.$ The case $41/60<u<1$ follows from the cases $u=41/60$ and $u=1$. $\square$

Lemma 2.6

We have the asymptotic formula

$$\begin{aligned} \sum _{n\le x}\tau _4(n)=xQ_3(\log x)+O(x^{1/2}\log ^5 x), \end{aligned}$$

where $Q_3(u)$ is a polynomial in u of degree 3.

Proof

Well-known. It follows from the Perron formula and the fourth power moment of $\zeta (s)$ over the critical line. $\square$

Lemma 2.7

Suppose $\ell \ge 0$ is a fixed integer, $11/20-\varepsilon \le u\le 11/20+\varepsilon .$ Then we have

$$\begin{aligned} \int _{-V}^V|\zeta ^{(\ell )}(u+it)|^5dt\ll V^{1+\varepsilon }. \end{aligned}$$

Proof

The proof for the case $\ell =0$ and $u=11/20$ can be found in Chapter 8 of [5]. The same method works for an arbitrary u with $11/20-\varepsilon \le u\le 11/20+\varepsilon .$ The case $\ell \ge 1$ follows from the result of the case $\ell =0$ and Cauchy’s theorem. $\square$

Lemma 2.8

Suppose $\alpha _j>2/3,\ \beta _j>1\ (j=1,2,3)$ are fixed real numbers such that $\alpha _1+\alpha _2+\alpha _3>2$, $f_j(t)\ (t\in {\mathbb {R}}, j=1,2,3)$ and $f(t_1,t_2)\ (t_1,t_2\in {\mathbb {R}})$ are complex-valued functions such that

$$\begin{aligned} f_j(t)\ll \frac{\log ^{\beta _j} (|t|+2)}{(|t|+2)^{\alpha _j}} \ (j=1,2,3), \ \ f(t_1,t_2)\ll 1. \end{aligned}$$

(2.1)

Suppose $U>4$ and $U^{\prime }>4$ are positive parameters such that $U\asymp U^{\prime }$, and define

$$\begin{aligned} W&=\int _{-\infty }^\infty \int _{-\infty }^\infty f_1(t_1)f_2(t_2)f_3(t_1+t_2)f(t_1,t_2)dt_1dt_2,\\ W(U, U^{\prime })&=\int _{-U}^U \int _{-U^{\prime }}^{U^{\prime }} f_1(t_1)f_2(t_2)f_3(t_1+t_2)f(t_1,t_2)dt_1dt_2. \end{aligned}$$

Then

(1) we have the estimate

$$\begin{aligned} W-W(U,U^{\prime })\ll \frac{\log ^{\beta _1+\beta _2+\beta _3+1} U}{U^{\min (\alpha _1,1)+\alpha _2+\alpha _3-2}} +\frac{\log ^{\beta _1+\beta _2+\beta _3+1} U}{U^{\alpha _1+\min (\alpha _2,1)+\alpha _3-2}} +\frac{\log ^{\beta _1+\beta _2+\beta _3+1} U}{U^{\alpha _1+\alpha _2+\min (\alpha _3,1)-2}}; \end{aligned}$$

(2) the integral W is absolutely convergent.

Proof

Without loss of generality, we suppose $U\le U^{\prime }.$ Obviously, from (2.1) we have

$$\begin{aligned} W-W(U,U^{\prime })\ll W_1 +W_2 +W_3 +W_4, \end{aligned}$$

(2.2)

where

$$\begin{aligned}&W_1:=\int _{\begin{array}{c} t_1\ge 0, t_2\ge 0\\ \max (|t_1|,|t_2|)\ge U \end{array}} \frac{\log ^{\beta _1} (|t_1|+2)}{(|t_1|+2)^{\alpha _1}} \frac{\log ^{\beta _2} (|t_2|+2)}{(|t_2|+2)^{\alpha _2}} \frac{\log ^{\beta _3} (|t_1+t_2|+2)}{(|t_1+t_2|+2)^{\alpha _3}} dt_1dt_2,\\&W_2:=\int _{\begin{array}{c} t_1\le 0, t_2\ge 0\\ \max (|t_1|,|t_2|)\ge U \end{array}} \frac{\log ^{\beta _1} (|t_1|+2)}{(|t_1|+2)^{\alpha _1}} \frac{\log ^{\beta _2} (|t_2|+2)}{(|t_2|+2)^{\alpha _2}} \frac{\log ^{\beta _3} (|t_1+t_2|+2)}{(|t_1+t_2|+2)^{\alpha _3}} dt_1dt_2,\\&W_3:=\int _{\begin{array}{c} t_1\le 0, t_2\le 0\\ \max (|t_1|,|t_2|)\ge U \end{array}} \frac{\log ^{\beta _1} (|t_1|+2)}{(|t_1|+2)^{\alpha _1}} \frac{\log ^{\beta _2} (|t_2|+2)}{(|t_2|+2)^{\alpha _2}} \frac{\log ^{\beta _3} (|t_1+t_2|+2)}{(|t_1+t_2|+2)^{\alpha _3}} dt_1dt_2,\\&W_4:=\int _{\begin{array}{c} t_1\ge 0, t_2\le 0\\ \max (|t_1|,|t_2|)\ge U \end{array}} \frac{\log ^{\beta _1} (|t_1|+2)}{(|t_1|+2)^{\alpha _1}} \frac{\log ^{\beta _2} (|t_2|+2)}{(|t_2|+2)^{\alpha _2}} \frac{\log ^{\beta _3} (|t_1+t_2|+2)}{(|t_1+t_2|+2)^{\alpha _3}} dt_1dt_2. \end{aligned}$$

So, it suffices for us to prove that for $j=1,2,3,4$,

$$\begin{aligned} W_j\ll \frac{\log ^{\beta _1+\beta _2+\beta _3+1} U}{U^{\min (\alpha _1,1)+\alpha _2+\alpha _3-2}} +\frac{\log ^{\beta _1+\beta _2+\beta _3+1} U}{U^{\alpha _1+\min (\alpha _2,1)+\alpha _3-2}} +\frac{\log ^{\beta _1+\beta _2+\beta _3+1} U}{U^{\alpha _1+\alpha _2+\min (\alpha _3,1)-2}}. \end{aligned}$$

(2.3)

It is easy to see that $W_1=W_3,W_2=W_4.$

I. Estimate of $W_1$:

We write

$$\begin{aligned}&W_1=W_{11}+W_{12},\nonumber \\&W_{11}:=\int _{\begin{array}{c} U\le t_1<\infty \\ 0\le t_2\le t_1 \end{array}} \frac{\log ^{\beta _1} (t_1+2)}{(t_1+2)^{\alpha _1}} \frac{\log ^{\beta _2} (t_2+2)}{(t_2+2)^{\alpha _2}} \frac{\log ^{\beta _3} (t_1+t_2+2)}{(t_1+t_2+2)^{\alpha _3}} dt_1dt_2,\nonumber \\&W_{12}:=\int _{\begin{array}{c} U\le t_2<\infty \\ 0\le t_1\le t_2 \end{array}} \frac{\log ^{\beta _1} (t_1+2)}{(t_1+2)^{\alpha _1}} \frac{\log ^{\beta _2} (t_2+2)}{(t_2+2)^{\alpha _2}} \frac{\log ^{\beta _3} (t_1+t_2+2)}{(t_1+t_2+2)^{\alpha _3}} dt_1dt_2. \end{aligned}$$

(2.4)

It is easily seen that

$$\begin{aligned} W_{11}&\ll \int _{U}^\infty \frac{\log ^{\beta _1+\beta _2+\beta _3} t_1}{t_1^{\alpha _1+\alpha _3}} dt_1 \int _0^{t_1} \frac{1}{(t_2+2)^{\alpha _2}} dt_2\nonumber \\&\ll \int _{U}^\infty \frac{\log ^{\beta _1+\beta _2+\beta _3} t_1}{t_1^{\alpha _1+\alpha _3}} \times t_1^{\max (1-\alpha _2,0)}\log t_1 dt_1 \nonumber \\&\ll \frac{\log ^{\beta _1+\beta _2+\beta _3+1} U}{U^{\alpha _1+\min (\alpha _2,1)+\alpha _3-2}}. \end{aligned}$$

(2.5)

Similarly we have

$$\begin{aligned} W_{12} \ll \frac{\log ^{\beta _1+\beta _2+\beta _3+1} U}{U^{\min (\alpha _1,1)+\alpha _2+\alpha _3-2}}. \end{aligned}$$

(2.6)

From (2.4) to (2.6) we see that (2.3) holds for $j=1.$

II. Estimate of $W_4$:

We write

$$\begin{aligned}&W_4=W_{41}+W_{42},\nonumber \\&W_{41}:=\int _{\begin{array}{c} U\le t_1<\infty \\ 0\le t_2\le t_1 \end{array}} \frac{\log ^{\beta _1} (t_1+2)}{(t_1+2)^{\alpha _1}} \frac{\log ^{\beta _2} (t_2+2)}{(t_2+2)^{\alpha _2}} \frac{\log ^{\beta _3} (t_1-t_2+2)}{(t_1-t_2+2)^{\alpha _3}} dt_1dt_2,\nonumber \\&W_{42}:=\int _{\begin{array}{c} U\le t_2<\infty \\ 0\le t_1\le t_2 \end{array}} \frac{\log ^{\beta _1} (t_1+2)}{(t_1+2)^{\alpha _1}} \frac{\log ^{\beta _2} (t_2+2)}{(t_2+2)^{\alpha _2}} \frac{\log ^{\beta _3} (t_2-t_1+2)}{(t_2-t_1+2)^{\alpha _3}} dt_1dt_2. \end{aligned}$$

(2.7)

We estimate $W_{41}$ first. We have

$$\begin{aligned} W_{41}&\ll \int _{U}^\infty \frac{\log ^{\beta _1+\beta _2+\beta _3} t_1}{t_1^{\alpha _1}}dt_1 \int _0^{t_1} \frac{1}{(t_2+2)^{\alpha _2}} \frac{1}{(t_1-t_2+2)^{\alpha _3}} dt_2 \nonumber \\&=\int _{U}^\infty \frac{\log ^{\beta _1+\beta _2+\beta _3} t_1}{t_1^{\alpha _1}}dt_1 \int _0^{t_1/2} \frac{1}{(t_2+2)^{\alpha _2}} \frac{1}{(t_1-t_2+2)^{\alpha _3}} dt_2\nonumber \\&\ \ \ +\int _{U}^\infty \frac{\log ^{\beta _1+\beta _2+\beta _3} t_1}{t_1^{\alpha _1}}dt_1 \int _{t_1/2}^{t_1} \frac{1}{(t_2+2)^{\alpha _2}} \frac{1}{(t_1-t_2+2)^{\alpha _3}} dt_2\nonumber \\&\ll \int _{U}^\infty \frac{\log ^{\beta _1+\beta _2+\beta _3} t_1}{t_1^{\alpha _1+\alpha _3}}dt_1 \int _0^{t_1/2} \frac{1}{(t_2+2)^{\alpha _2}} dt_2\nonumber \\&\ \ \ +\int _{U}^\infty \frac{\log ^{\beta _1+\beta _2+\beta _3} t_1}{t_1^{\alpha _1+\alpha _2}}dt_1 \int _{t_1/2}^{t_1} \frac{1}{(t_1-t_2+2)^{\alpha _3}} dt_2\nonumber \\&\ll \int _{U}^\infty \frac{\log ^{\beta _1+\beta _2+\beta _3} t_1}{t_1^{\alpha _1+\alpha _3}}\times t_1^{\max (1-\alpha _2,0)}\log t_1dt_1\nonumber \\&\ \ \ +\int _{U}^\infty \frac{\log ^{\beta _1+\beta _2+\beta _3} t_1}{t_1^{\alpha _1+\alpha _2}}dt_1 \times t_1^{\max (1-\alpha _3,0)}\log t_1 \nonumber \\&\ll \frac{\log ^{\beta _1+\beta _2+\beta _3+1} U}{U^{\alpha _1+\min (\alpha _2,1)+\alpha _3-2}}+ \frac{\log ^{\beta _1+\beta _2+\beta _3+1} U}{U^{\alpha _1+\alpha _2+\min (\alpha _3,1)-2}}. \end{aligned}$$

(2.8)

Similarly we have

$$\begin{aligned}&W_{42} \ll \frac{\log ^{\beta _1+\beta _2+\beta _3+1} U}{U^{\min (\alpha _2,1)+\alpha _2+\alpha _3-2}}+ \frac{\log ^{\beta _1+\beta _2+\beta _3+1} U}{U^{\alpha _1+\alpha _2+\min (\alpha _3,1)-2}}. \end{aligned}$$

(2.9)

From (2.7) to (2.9) we see that (2.3) holds for $j=4.$

Now the first assertion of Lemma 2.8 follows from (2.2) and (2.3). The second assertion of Lemma 2.8 is obviously a consequence of the first assertion. $\square$

Lemma 2.9

Suppose $M(s)=AM_1(s)M_2(s)$ such that $M_1(s)$ is meromorphic in the region $|s-s_0|\le \eta ,$$s=s_0$ is a pole of order $r\ge 2,$ and $M_2(s)$ is analytic in $|s-s_0|\le \eta ,$ where $\eta >0$ is a positive real number, $A\not = 0$ is a constant. Write

$$\begin{aligned} M_1(s)=\sum _{n=-r}^\infty a_n(s-s_0)^n. \end{aligned}$$

Then we have

$$\begin{aligned} \text {Res}_{s=s_0}M(s)=A\sum _{m=0}^{r-1}a_{-m-1}\frac{M^{(m)}_2(s_0)}{m!}. \end{aligned}$$

Proof

It follows from the theory of complex variable functions. $\square$

Lemma 2.10

Let $k, \ell \ge 2$ be two fixed integers and $f_j(s) \ (j=1,\ldots ,k)$ be functions which are at least $\ell$-times differentiable. Then for each $0\le j\le \ell$ we have

$$\begin{aligned} (f_1(s)\cdots f_k(s))^{(j)}=\sum _{\begin{array}{c} n_1+\cdots +n_k=j\\ n_1,\ldots ,n_k\ge 0 \end{array}} \frac{j!}{n_1!\cdots n_k!}f_1^{(n_1)}(s)\cdots f_k^{(n_k)}(s). \end{aligned}$$

Proof

Well-known. $\square$

3 On the Dirichlet series of $c(n_1,n_2,n_3)$

In this section we shall study the Dirichlet series

$$\begin{aligned} C(s_1,s_2,s_3):=\sum _{n_1=1}^\infty \sum _{n_2=1}^\infty \sum _{n_3=1}^\infty \frac{c(n_1, n_2, n_3)}{n_1^{s_1} n_2^{s_2} n_3^{s_3} } \ \ (\mathfrak {R}s_j>1, j=1,2,3). \end{aligned}$$

(3.1)

We have the following proposition.

Proposition 3.1

Suppose that $s_j=\sigma _j+it_j, \sigma _j>1 \ (j=1,2,3)$. Then we have

$$\begin{aligned} C(s_1,s_2,s_3)= & {} \zeta ^2(s_1)\zeta ^2(s_2)\zeta ^2(s_3)\zeta (s_1+s_2-1)\zeta (s_1+s_3-1) \nonumber \\&\times \zeta (s_2+s_3-1)\zeta (s_1+s_2+s_3-2)H(s_1,s_2,s_3), \end{aligned}$$

(3.2)

where $H(s_1,s_2,s_3)$ can be written as a triple Dirichlet series, which is absolutely convergent when $\sigma _1, \sigma _2, \sigma _3$ satisfy the following conditions:

$$\begin{aligned}&\sigma _j>1/3 \ (j=1,2,3),\ \ 2\sigma _j+\sigma _{\ell }>2 \ (1\le j\not = \ell \le 3),\nonumber \\&\sigma _1+\sigma _2+\sigma _3+\sigma _j>3 \ (j=1,2,3), \nonumber \\&2(\sigma _1+\sigma _2+\sigma _3)-\sigma _j>4 \ (j=1,2,3). \end{aligned}$$

(3.3)

Proof

Noting that $c(n_1,n_2,n_3)$ is multiplicative and (1.3) holds, we have that

$$\begin{aligned} C(s_1,s_2,s_3)= \zeta (s_1)\zeta (s_2)\zeta (s_3) \prod _{p} \left( \sum _{(\alpha _1, \alpha _2,\alpha _3)\in {\mathbb {N}}_0^3 } \frac{\varphi (p^{\alpha _1})\varphi (p^{\alpha _2}) \varphi (p^{\alpha _3}) }{\varphi (p^{\max (\alpha _1, \alpha _2, \alpha _3)}) p^{ \alpha _1s_1 + \alpha _2 s_2+\alpha _3s_3 }}\right) . \end{aligned}$$

(3.4)

Write

$$\begin{aligned} {\mathbb {N}}_0^3=D_0\bigcup D_1\bigcup D_2 \bigcup D_3 \bigcup D_4 \bigcup D_5, \end{aligned}$$

(3.5)

where

$$\begin{aligned} D_0:= & {} \{(0,0,0)\},\ \ D_1:=\bigcup _{j=1}^7 D_{1j}, \\ D_{11}:= & {} {\mathbb {N}}\times \{0\} \times \{0\},\ \ D_{12}:=\{0\}\times {\mathbb {N}} \times \{0\}, \\ D_{13}:= & {} \{0\}\times \{0\}\times {\mathbb {N}},\ \ D_{14}:=\{(k, k, 0): k\in {\mathbb {N}}\}, \\ D_{15}:= & {} \{(k, 0, k): k\in {\mathbb {N}}\},\ \ D_{16}:=\{(0, k, k): k\in {\mathbb {N}}\}, \\ D_{17}:= & {} \{(k, k, k): k\in {\mathbb {N}}\}, \ \ D_2:={\mathbb {N}}^3{\setminus } D_{17},\\ D_3:= & {} {\mathbb {N}}\times {\mathbb {N}}\times \{0\}{\setminus } D_{14}=\{(\alpha ,\beta ,0):\alpha ,\beta \in {\mathbb {N}}, \alpha \not = \beta \},\\ D_4:= & {} {\mathbb {N}}\times \{0\} \times {\mathbb {N}}{\setminus } D_{15},\ D_5:= \{0\} \times {\mathbb {N}}\times {\mathbb {N}}{\setminus } D_{16}. \end{aligned}$$

For each $1\le j\le 7,$ we define

$$\begin{aligned} D_{1j}(p;s_1,s_2,s_3):= \sum _{(\alpha _1,\alpha _2,\alpha _3)\in D_{1j}} \frac{\varphi (p^{\alpha _1})\varphi (p^{\alpha _2}) \varphi (p^{\alpha _3}) }{\varphi (p^{\max (\alpha _1, \alpha _2, \alpha _3)})p^{ \alpha _1s_1 + \alpha _2 s_2+\alpha _3s_3 }}. \end{aligned}$$

Also we define for $j=2,3,4,5$,

$$\begin{aligned} D_{j}(p;s_1,s_2,s_3):=\sum _{(\alpha _1,\alpha _2,\alpha _3)\in D_{j}} \frac{\varphi (p^{\alpha _1})\varphi (p^{\alpha _2}) \varphi (p^{\alpha _3}) }{\varphi (p^{\max (\alpha _1, \alpha _2, \alpha _3)})p^{ \alpha _1s_1 + \alpha _2 s_2+\alpha _3s_3 }}. \end{aligned}$$

For $j=1,2,3,$ it is easy to see that

$$\begin{aligned} D_{11}(p;s_1,s_2,s_3)= & {} \sum _{\alpha _1\in {\mathbb {N}}}\frac{1}{p^{\alpha _1 s_1}}=\frac{p^{-s_1}}{1-p^{-s_1}}, \nonumber \\ D_{12}(p;s_1,s_2,s_3)= & {} \sum _{\alpha _2\in {\mathbb {N}}}\frac{1}{p^{\alpha _2 s_2}}=\frac{p^{-s_2}}{1-p^{-s_2}}, \nonumber \\ D_{13}(p;s_1,s_2,s_3)= & {} \sum _{\alpha _3\in {\mathbb {N}}}\frac{1}{p^{\alpha _3 s_3}}=\frac{p^{-s_3}}{1-p^{-s_3}}. \end{aligned}$$

(3.6)

For $j=4$ we have

$$\begin{aligned} D_{14}(p;s_1,s_2,s_3)= & {} \sum _{\alpha _1\in {\mathbb {N}}}\frac{\varphi (p^{\alpha _1})}{p^{\alpha _1 s_1+\alpha _1 s_2}}= \sum _{\alpha _1\in {\mathbb {N}}}\frac{p^{\alpha _1}-p^{\alpha _1-1}}{p^{\alpha _1 s_1+\alpha _1 s_2}}\\= & {} \frac{p^{-(s_1+s_2-1)}}{1-p^{-(s_1+s_2-1)}}-\frac{p^{-(s_1+s_2)}}{1-p^{-(s_1+s_2-1)}}.\nonumber \end{aligned}$$

(3.7)

Similarly, we have

$$\begin{aligned} D_{15}(p;s_1,s_2,s_3) =\frac{p^{-(s_1+s_3-1)}}{1-p^{-(s_1+s_3-1)}}-\frac{p^{-(s_1+s_3)}}{1-p^{-(s_1+s_3-1)}} \end{aligned}$$

(3.8)

and

$$\begin{aligned} D_{16}(p;s_1,s_2,s_3) =\frac{p^{-(s_2+s_3-1)}}{1-p^{-(s_2+s_3-1)}}-\frac{p^{-(s_2+s_3)}}{1-p^{-(s_2+s_3-1)}}. \end{aligned}$$

(3.9)

For $j=7$ we have

$$\begin{aligned}&D_{17}(p;s_1,s_2,s_3)\nonumber \\&\quad =\sum _{\alpha \in {\mathbb {N}}}\frac{\varphi ^2(p^{\alpha })}{p^{\alpha ( s_1+ s_2+s_3)}} =\sum _{\alpha \in {\mathbb {N}}}\frac{ p^{2\alpha }-2p^{2\alpha -1}+p^{2\alpha -2}}{p^{\alpha ( s_1+ s_2+s_3)}}\nonumber \\&\quad =\frac{p^{-(s_1+s_2+s_3-2)}}{1-p^{-(s_1+s_2+s_3-2)}}-2\frac{p^{-(s_1+s_2+s_3-1)}}{1-p^{-(s_1+s_2+s_3-2)}} + \frac{p^{-(s_1+s_2+s_3)}}{1-p^{-(s_1+s_2+s_3-2)}}. \end{aligned}$$

(3.10)

For simplicity, we write

$$\begin{aligned} u_1:= & {} s_1,\ u_2:=s_2,\ u_3:=s_3,\ u_4:=s_1+s_2-1,\\ u_5:= & {} s_1+s_3-1,\ u_6:=s_2+s_3-1,\ u_7:=s_1+s_2+s_3-2. \end{aligned}$$

So, from (3.6) to (3.10), we can write

$$\begin{aligned} 1+\sum _{j=1}^7 D_{1j}(p;s_1,s_2,s_3)=1+\sum _{j=1}^7\frac{p^{-u_j}}{1-p^{-u_j}}+B_1(p;s_1,s_2,s_3), \end{aligned}$$

(3.11)

where

$$\begin{aligned} B_1(p;s_1,s_2,s_3):= & {} -\frac{p^{-(s_1+s_2 )}}{1-p^{-(s_1+s_2 -1)}}-\frac{p^{-(s_1+s_3 )}}{1-p^{-(s_1+s_3 -1)}}- \frac{p^{-(s_2+s_3 )}}{1-p^{-(s_2+s_3 -1)}}\\&\quad -2\frac{p^{-(s_1+s_2+s_3-1)}}{1-p^{-(s_1+s_2+s_3-2)}}+\frac{p^{-(s_1+s_2+s_3)}}{1-p^{-(s_1+s_2+s_3-2)}}. \end{aligned}$$

Obviously, we have

$$\begin{aligned} B_1(p;s_1,s_2,s_3)\ll p^{-\sigma _1-\sigma _2 }+p^{-\sigma _1-\sigma _3}+p^{-\sigma _2-\sigma _3 }+p^{-\sigma _1-\sigma _2-\sigma _3+1}. \end{aligned}$$

(3.12)

It is easy to see that

$$\begin{aligned}&\prod _{\ell =1}^7(1-p^{-u_{\ell }})\nonumber \\&\quad =1-\sum _{\ell =1}^7 p^{-u_{\ell }}+\sum _{m_1=2}^7 (-1)^{m_1}\sum _{1\le \ell _1<\ell _2<\cdots <\ell _{m_1} \le 7} p^{-(u_{\ell _1}+ \cdots + u_{\ell _{m_1}})}. \end{aligned}$$

(3.13)

By the expression $1/(1-\xi )=\sum _{m=0}^\infty \xi ^m\ (|\xi |<1),$ we have

$$\begin{aligned} 1+ \sum _{j=1}^7\frac{p^{-u_j}}{1-p^{-u_j}} = 1+\sum _{j=1}^7 p^{-u_j}+\sum _{j=1}^7\sum _{m=2}^\infty p^{-mu_j}. \end{aligned}$$

(3.14)

So, from (3.13) and (3.14) we get

$$\begin{aligned}&\left( \prod _{\ell =1}^7(1-p^{-u_{\ell }})\right) \left( 1+ \sum _{j=1}^7 \frac{p^{-u_j}}{1-p^{-u_j}}\right) \nonumber \\&\quad =1+ \sum _{m_1=2}^7 (-1)^{m_1}\sum _{1\le \ell _1<\ell _2<\cdots<\ell _{m_1} \le 7}p^{-(u_{\ell _1}+\cdots +u_{\ell _{m_1}})} -\sum _{j=1}^7 \sum _{\ell =1}^7p^{-u_{\ell }-u_j} \nonumber \\&\quad \quad +\sum _{j=1}^7\sum _{m_1=2}^7 (-1)^{m_1}\sum _{1\le \ell _1<\ell _2<\cdots <\ell _{m_1}\le 7} p^{-(u_j+u_{\ell _1}+\cdots +u_{\ell _{m_1}})}\nonumber \\&\quad \quad +\sum _{j=1}^7\sum _{m=2}^\infty p^{-mu_j}\times \prod _{\ell =1}^7(1-p^{-u_{\ell }})\nonumber \\&\quad =1-\sum _{j=1}^7\sum _{\ell =1}^7 p^{-u_{\ell }-u_j}+\sum _{j=1}^7\sum _{m=2}^\infty p^{-mu_j}\times \prod _{\ell =1}^7(1-p^{-u_{\ell }})\nonumber \\&\quad \quad +O\left( \sum _{1\le j\not = \ell \le 7} p^{-\mathfrak {R}(u_{\ell }+u_j)}\right) \nonumber \\&\quad =1+O\left( \sum _{1\le j\not = \ell \le 7} p^{-\mathfrak {R}(u_{\ell }+u_j)}+\sum _{j=1}^7 p^{-3\mathfrak {R}u_j}\right) . \end{aligned}$$

(3.15)

Now we estimate $D_2(p;s_1, s_2, s_3).$ We write

$$\begin{aligned} D_2(p;s_1, s_2, s_3)= & {} \sum _{j=1}^3\sum _{\begin{array}{c} (\alpha _1, \alpha _2, \alpha _3)\in {\mathbb {N}}^3 \\ \alpha _j \text { largest} \end{array}} \frac{\varphi (p^{\alpha _1}) \varphi (p^{\alpha _2})\varphi (p^{\alpha _3})}{\varphi (p^{\alpha _j}) p^{\alpha _1s_1+\alpha _2s_2+\alpha _3s_3}}\nonumber \\= & {} \sum _{j=1}^3D_{2j}(p;s_1, s_2, s_3), \end{aligned}$$

(3.16)

say. We only estimate $D_{23}(p;s_1, s_2, s_3).$ The estimates for $D_{21}(p;s_1, s_2, s_3)$ and $D_{22}(p;s_1, s_2, s_3)$ are similar.

We write

$$\begin{aligned} D_{23}(p;s_1, s_2, s_3)=\sum _{j=1}^5 D_{23j}(p;s_1, s_2, s_3), \end{aligned}$$

(3.17)

where

$$\begin{aligned}&D_{231}(p;s_1, s_2, s_3):=\sum _{1\le \alpha _1<\alpha _2<\alpha _3} \frac{\varphi (p^{\alpha _1}) \varphi (p^{\alpha _2}) }{ p^{\alpha _1s_1+\alpha _2s_2+\alpha _3s_3}},\\&D_{232}(p;s_1, s_2, s_3):=\sum _{1\le \alpha _2<\alpha _1<\alpha _3} \frac{\varphi (p^{\alpha _1}) \varphi (p^{\alpha _2}) }{ p^{\alpha _1s_1+\alpha _2s_2+\alpha _3s_3}},\\&D_{233}(p;s_1, s_2, s_3):=\sum _{1\le \alpha _1<\alpha _2=\alpha _3} \frac{\varphi (p^{\alpha _1}) \varphi (p^{\alpha _2}) }{ p^{\alpha _1s_1+\alpha _2s_2+\alpha _3s_3}},\\&D_{234}(p;s_1, s_2, s_3):=\sum _{1\le \alpha _2<\alpha _1=\alpha _3} \frac{\varphi (p^{\alpha _1}) \varphi (p^{\alpha _2}) }{ p^{\alpha _1s_1+\alpha _2s_2+\alpha _3s_3}},\\&D_{235}(p;s_1, s_2, s_3):=\sum _{1\le \alpha _1=\alpha _2<\alpha _3} \frac{\varphi (p^{\alpha _1}) \varphi (p^{\alpha _2}) }{ p^{\alpha _1s_1+\alpha _2s_2+\alpha _3s_3}}. \end{aligned}$$

We have the estimate

$$\begin{aligned} \sum _{m=M}^\infty \xi ^m\ll |\xi |^M,\ \ (M|\ge 0, \ |\xi |<C<1). \end{aligned}$$

(3.18)

So, from (3.18) we get

$$\begin{aligned} D_{231}(p;s_1, s_2, s_3)&=\sum _{\alpha _1=1}^\infty \sum _{\alpha _2=\alpha _1+1}^\infty \sum _{\alpha _3=\alpha _2+1}^\infty \frac{(p^{\alpha _1}-p^{\alpha _1-1})(p^{\alpha _2}-p^{\alpha _2-1}) }{ p^{\alpha _1s_1+\alpha _2s_2+\alpha _3s_3}}\nonumber \\&\ll \sum _{\alpha _1=1}^\infty \frac{1}{p^{\alpha _1\sigma _1 -\alpha _1}}\sum _{\alpha _2=\alpha _1+1}^\infty \frac{1}{p^{\alpha _2\sigma _2-\alpha _2}} \sum _{\alpha _3=\alpha _2+1}^\infty \frac{1}{p^{\alpha _3\sigma _3}}\nonumber \\&\ll \sum _{\alpha _1=1}^\infty \frac{1}{p^{\alpha _1\sigma _1-\alpha _1}}\sum _{\alpha _2=\alpha _1+1}^\infty \frac{1}{p^{\alpha _2\sigma _2-\alpha _2}} \times \frac{1}{p^{(\alpha _2+1)\sigma _3}}\nonumber \\&\ll \frac{1}{p^{\sigma _3}}\sum _{\alpha _1=1}^\infty \frac{1}{p^{\alpha _1\sigma _1-\alpha _1}} \sum _{\alpha _2=\alpha _1+1}^\infty \frac{1}{p^{\alpha _2(\sigma _2+\sigma _3-1)}}\nonumber \\&\ll \frac{1}{p^{\sigma _3}}\sum _{\alpha _1=1}^\infty \frac{1}{p^{\alpha _1\sigma _1-\alpha _1}} \times \frac{1}{p^{(\alpha _1+1)(\sigma _2+\sigma _3-1)}}\nonumber \\&\ll \frac{1}{p^{\sigma _2+2\sigma _3-1}}\sum _{\alpha _1=1}^\infty \frac{1}{p^{\alpha _1(\sigma _1+\sigma _2+\sigma _3-2)}}\nonumber \\&\ll \frac{1}{p^{\sigma _2+2\sigma _3-1}}\times \frac{1}{p^{ \sigma _1+\sigma _2+\sigma _3-2}}= \frac{1}{p^{\sigma _1+2\sigma _2+3\sigma _3-3}}. \end{aligned}$$

(3.19)

Similarly we have

$$\begin{aligned} D_{232}(p;s_1, s_2, s_3)\ll \frac{1}{p^{2\sigma _1+ \sigma _2+3\sigma _3-3}}. \end{aligned}$$

(3.20)

We have

$$\begin{aligned} D_{233}(p;s_1, s_2, s_3)= & {} \sum _{\alpha _1=1}^\infty \frac{\varphi (p^{\alpha _1})}{p^{\alpha _1s_1}}\sum _{\alpha _2=\alpha _1+1}^\infty \frac{\varphi (p^{\alpha _2})}{p^{\alpha _2(s_2+s_3)}}\nonumber \\\ll & {} \sum _{\alpha _1=1}^\infty \frac{1}{p^{\alpha _1(\sigma _1-1)}}\sum _{\alpha _2=\alpha _1+1}^\infty \frac{1}{p^{\alpha _2(\sigma _2+\sigma _3-1)}}\nonumber \\\ll & {} \sum _{\alpha _1=1}^\infty \frac{1}{p^{\alpha _1(\sigma _1-1)}}\times \frac{1}{p^{(\alpha _1+1)(\sigma _2+\sigma _3-1)}}\nonumber \\\ll & {} \frac{1}{p^{ \sigma _2+\sigma _3-1}}\sum _{\alpha _1=1}^\infty \frac{1}{p^{\alpha _1(\sigma _1+\sigma _2+\sigma _3-2)}}\nonumber \\\ll & {} \frac{1}{p^{\sigma _1+ 2\sigma _2+2\sigma _3-3}}. \end{aligned}$$

(3.21)

Similarly we have

$$\begin{aligned} D_{234}(p;s_1, s_2, s_3)\ll \frac{1}{p^{2\sigma _1+ \sigma _2+2\sigma _3-3}}. \end{aligned}$$

(3.22)

Finally, we estimate $D_{235}(p;s_1,s_2,s_3).$ We have

$$\begin{aligned} D_{235}(p;s_1, s_2, s_3)= & {} \sum _{\alpha _1=1}^\infty \frac{\varphi ^2(p^{\alpha _1})}{p^{\alpha _1(s_1+s_2)}}\sum _{\alpha _3=\alpha _1+1}^\infty \frac{1}{p^{\alpha _3s_3}}\nonumber \\\ll & {} \sum _{\alpha _1=1}^\infty \frac{1}{p^{\alpha _1(\sigma _1+\sigma _2-2)}}\sum _{\alpha _3=\alpha _1+1}^\infty \frac{1}{p^{\alpha _3\sigma _3}}\nonumber \\\ll & {} \sum _{\alpha _1=1}^\infty \frac{1}{p^{\alpha _1(\sigma _1+\sigma _2-2)}}\times \frac{1}{p^{(\alpha _1+1)\sigma _3}}\nonumber \\\ll & {} \frac{1}{p^{ \sigma _3}}\sum _{\alpha _1=1}^\infty \frac{1}{p^{\alpha _1(\sigma _1+\sigma _2+\sigma _3-2)}}\nonumber \\\ll & {} \frac{1}{p^{ \sigma _3}}\times \frac{1}{p^{ \sigma _1+\sigma _2+\sigma _3-2}}=\frac{1}{p^{ \sigma _1+\sigma _2+2\sigma _3-2}}. \end{aligned}$$

(3.23)

From (3.17) to (3.23) we get

$$\begin{aligned} D_{23}(p;s_1,s_2,s_3)\ll \frac{1}{p^{ \sigma _1+\sigma _2+2\sigma _3-2}} +\frac{1}{p^{ \sigma _1+2\sigma _2+2\sigma _3-3}}+\frac{1}{p^{ 2\sigma _1+\sigma _2+2\sigma _3-3}}. \end{aligned}$$

(3.24)

Similarly we have

$$\begin{aligned} D_{22}(p;s_1,s_2,s_3)\ll \frac{1}{p^{ \sigma _1+2\sigma _2+\sigma _3-2}} +\frac{1}{p^{ \sigma _1+2\sigma _2+2\sigma _3-3}}+\frac{1}{p^{ 2\sigma _1+2\sigma _2+ \sigma _3-3}}. \end{aligned}$$

(3.25)

and

$$\begin{aligned} D_{21}(p;s_1,s_2,s_3)\ll \frac{1}{p^{ 2\sigma _1+\sigma _2+\sigma _3-2}} +\frac{1}{p^{ 2\sigma _1+2\sigma _2+\sigma _3-3}}+\frac{1}{p^{ 2\sigma _1+\sigma _2+2\sigma _3-3}}. \end{aligned}$$

(3.26)

From (3.16), (3.24–3.26) we have

$$\begin{aligned} D_2(p;s_1,s_2,s_3)\ll \frac{p^{\sigma _1}+p^{\sigma _2}+p^{\sigma _3}}{p^{ 2\sigma _1+2\sigma _2+2\sigma _3-3}} +\frac{p^{-\sigma _1}+p^{-\sigma _2}+p^{-\sigma _3}}{p^{ \sigma _1+ \sigma _2+ \sigma _3-2}}. \end{aligned}$$

(3.27)

We estimate $D_3(p;s_1,s_2,s_3)$ now. We have

$$\begin{aligned} D_3(p;s_1,s_2,s_3)= & {} \sum _{1\le \alpha<\beta }\frac{\varphi (p^\alpha )}{p^{\alpha s_1+\beta s_2}}+ \sum _{1\le \beta < \alpha } \frac{\varphi (p^\beta )}{p^{\alpha s_1+\beta s_2}}\nonumber \\\ll & {} \sum _{\alpha =1}^\infty \frac{p^{\alpha }}{p^{\alpha s_1 }}\sum _{\beta =\alpha +1}^\infty \frac{1}{p^{\beta s_2}}+ \sum _{ \beta =1}^\infty \frac{p^{ \beta }}{p^{ \beta s_2 }}\sum _{ \alpha = \beta +1}^\infty \frac{1}{p^{ \alpha s_1}}\nonumber \\\ll & {} \sum _{\alpha =1}^\infty \frac{p^{\alpha }}{p^{\alpha \sigma _1 }}\times \frac{1}{p^{(\alpha +1)\sigma _2}} +\sum _{ \beta =1}^\infty \frac{p^{ \beta }}{p^{ \beta \sigma _2 }}\times \frac{1}{p^{( \beta +1)\sigma _1}}\nonumber \\\ll & {} \frac{1}{p^{\sigma _1+2\sigma _2-1}}+\frac{1}{p^{2\sigma _1+ \sigma _2-1}}. \end{aligned}$$

(3.28)

Similarly we have

$$\begin{aligned} D_4(p;s_1,s_2,s_3) \ll \frac{1}{p^{\sigma _1+2\sigma _3-1}}+\frac{1}{p^{2\sigma _1+ \sigma _3-1}} \end{aligned}$$

(3.29)

and

$$\begin{aligned} D_5(p;s_1,s_2,s_3) \ll \frac{1}{p^{\sigma _2+2\sigma _3-1}}+\frac{1}{p^{2\sigma _2+ \sigma _3-1}}. \end{aligned}$$

(3.30)

We write

$$\begin{aligned}&\sum _{(\alpha _1, \alpha _2,\alpha _3)\in {\mathbb {N}}_0^3 } \frac{\varphi (p^{\alpha _1})\varphi (p^{\alpha _2}) \varphi (p^{\alpha _3}) }{\varphi (p^{\max (\alpha _1, \alpha _2, \alpha _3)})p^{ \alpha _1s_1 + \alpha _2 s_2+\alpha _3s_3 }} =\prod _{j=1}^7(1-p^{-u_j})^{-1}\times h(p;s_1,s_2,s_3), \nonumber \\ \end{aligned}$$

(3.31)

where

$$\begin{aligned}&h(p;s_1,s_2,s_3)\nonumber \\&\quad =\prod _{j=1}^7(1-p^{-u_j})\times \sum _{(\alpha _1, \alpha _2,\alpha _3)\in {\mathbb {N}}_0^3 } \frac{\varphi (p^{\alpha _1})\varphi (p^{\alpha _2}) \varphi (p^{\alpha _3}) }{\varphi (p^{\max (\alpha _1, \alpha _2, \alpha _3)})p^{ \alpha _1s_1 + \alpha _2 s_2+\alpha _3s_3 }} \nonumber \\&\quad =\prod _{j=1}^7(1-p^{-u_j})\times (1+\sum _{j=1}^7D_{1j}(p;s_1,s_2,s_3)+\sum _{j=2}^5D_j(p;s_1,s_2,s_3))\nonumber \\&\quad =1+h^{*}(p;s_1,s_2,s_3), \end{aligned}$$

(3.32)

say. From (3.11), (3.12), (3.15), (3.27)–(3.30) we get

$$\begin{aligned} h^{*}(p;s_1,s_2,s_3)\ll&\sum _{1\le j\not = \ell \le 7}\frac{1}{p^{\mathfrak {R}(u_j+u_{\ell })}}+\sum _{j=1}^7 p^{-3\mathfrak {R}u_j}+\frac{1}{p^{\sigma _1+\sigma _2+\sigma _3-1}}\nonumber \\&\ \ +\frac{p^{\sigma _1}+p^{\sigma _2}+p^{\sigma _3}}{p^{ 2\sigma _1+2\sigma _2+2\sigma _3-3}} +\frac{p^{-\sigma _1}+p^{-\sigma _2}+p^{-\sigma _3}}{p^{ \sigma _1+ \sigma _2+ \sigma _3-2}}\nonumber \\&\ \ +\sum _{1\le j\not =\ell \le 3}\frac{1}{p^{2\sigma _j+\sigma _{\ell }-1}}. \end{aligned}$$

(3.33)

Now we see from (3.4), (3.31)–(3.33) that Proposition 3.1 holds and $H(s_1,s_2,s_3)$ can be written as a triple Dirichlet series which is absolutely convergent under the condition (3.3). $\square$

4 On an infinite sum

Suppose $x\notin {\mathbb {N}}$ and T are two large parameters such that $5\le T\le x/2$. For each $1\le j\le 3,$ define the infinite sum

$$\begin{aligned} E_j(x,T):=\sum _{n_1,n_2,n_3\ge 1}\frac{c(n_1, n_2, n_3)(n_1n_2n_3)^{-1-1/\log x}}{T|\log \frac{x}{n_j}|+1}. \end{aligned}$$

Since $c(n_1, n_2, n_3)$ is symmetric for $n_1,n_2, n_3,$ we have

$$\begin{aligned} E_1(x,T)=E_2(x,T)=E_3(x,T)=: E(x, T), \end{aligned}$$

say.

Write

$$\begin{aligned} E(x,T)=\sum _{n_3=1}^\infty \frac{n_3^{-1-1/\log x}}{T|\log \frac{x}{n_3}|+1}\times G(n_3, 1/\log x), \end{aligned}$$

(4.1)

where

$$\begin{aligned} G(n_3, \sigma ):=\sum _{n_1, n_2\ge 1}c(n_1, n_2, n_3)(n_1n_2)^{-1-\sigma },\ \ 0<\sigma <1. \end{aligned}$$

4.1 Upper bound of $G(n_3, \sigma )$

In this subsection we shall give an upper bound of $G(n_3, \sigma ).$

If $n_3=1,$ then $c(n_1, n_2, 1)=c(n_1, n_2),$ so from Lemmas 2.1 and 2.3 we have

$$\begin{aligned} G(1, \sigma )\ll \sigma ^{-5}. \end{aligned}$$

(4.2)

Now suppose $n_3\ge 2.$ We begin with the formula (1.3). By noting that $\varphi (n)\gg n/\log n$ and $[m,n]=mn/(m,n)$ we have

$$\begin{aligned} c(n_1,n_2,n_3)= & {} \sum _{d_j|n_j(1\le j\le 3)}\frac{\varphi (d_1)\varphi (d_2)\varphi (d_3)}{\varphi ([d_1, d_2, d_3])}\nonumber \\\ll & {} \sum _{ d_j\mid n_j(1\le j\le 3)}\frac{\varphi (d_1)\varphi (d_2)\varphi (d_3)}{[d_1, d_2, d_3]}\times \log [d_1, d_2, d_3]\nonumber \\\ll & {} \sum _{ d_j\mid n_j (1\le j\le 3)}\frac{d_1d_2d_3}{[d_1, d_2, d_3]}\times \log (d_1d_2d_3)\nonumber \\= & {} \sum _{ d_j\mid n_j (1\le j\le 3)}\frac{d_1d_2d_3}{[[d_1, d_2], d_3]}\times \log (d_1d_2d_3)\nonumber \\= & {} \sum _{ d_j\mid n_j (1\le j\le 3)}\frac{d_1d_2\ ([d_1,d_2], d_3)}{[d_1, d_2]}\times \log (d_1d_2d_3)\nonumber \\= & {} \sum _{ d_j\mid n_j (1\le j\le 3)}\frac{d_1d_2\ ([d_1,d_2], d_3)}{[d_1, d_2]}\times \log (d_1d_2)\nonumber \\&+\sum _{ d_j\mid n_j(1\le j\le 3)}\frac{d_1d_2\ ([d_1,d_2], d_3)}{[d_1, d_2]} \times \log d_3. \end{aligned}$$

(4.3)

Inserting the estimate (4.3) into the expression of $G(n_3, \sigma ),$ we have

$$\begin{aligned} G(n_3, \sigma )\ll G_1(n_3, \sigma )+G_2(n_3, \sigma ), \end{aligned}$$

(4.4)

where

$$\begin{aligned} G_1(n_3, \sigma ):= & {} \sum _{d_3|n_3}\sum _{n_1,n_2=1}^\infty \sum _{ d_1|n_1,d_2|n_2}\frac{d_1d_2\ ([d_1,d_2], d_3)\log (d_1d_2)}{[d_1, d_2]\ (n_1n_2)^{1+\sigma }},\nonumber \\ G_2(n_3, \sigma ):= & {} \sum _{d_3|n_3}\log d_3 \ \sum _{n_1,n_2=1}^\infty \sum _{ d_1|n_1,d_2|n_2}\frac{d_1d_2\ ([d_1,d_2], d_3)}{[d_1, d_2]\ (n_1n_2)^{1+\sigma }} . \end{aligned}$$

(4.5)

We estimate $G_1(n_3, \sigma )$ first. By writing $n_j=d_j\ell _j$ ($j=1,2$) we have

$$\begin{aligned} G_1(n_3, \sigma )= & {} \sum _{d_3\mid n_3}\sum _{d_1, d_2,\ell _1,\ell _2\ge 1} \frac{d_1d_2\ ([d_1,d_2], d_3) \log (d_1d_2)}{[d_1, d_2]\ \ (d_1\ell _1d_2\ell _2)^{1+\sigma }}\nonumber \\= & {} \sum _{d_3\mid n_3}\sum _{d_1, d_2\ge 1} \frac{d_1d_2\ ([d_1,d_2], d_3) \log (d_1d_2)}{[d_1, d_2]\ \ (d_1d_2)^{1+\sigma }} \sum _{ \ell _1,\ell _2\ge 1} \frac{1}{ (\ell _1\ell _2)^{1+\sigma }}\nonumber \\= & {} \zeta ^2(1+\sigma )\sum _{d_3\mid n_3}\sum _{d_1, d_2\ge 1} \frac{d_1d_2\ ([d_1,d_2], d_3)\log (d_1d_2)}{[d_1, d_2]\ \ (d_1d_2)^{1+\sigma }}\nonumber \\= & {} \zeta ^2(1+\sigma )\sum _{d\mid n_3}d\sum _{\begin{array}{c} d_1, d_2\ge 1\\ d\mid [d_1, d_2] \end{array}} \frac{d_1d_2 \log (d_1d_2)}{[d_1, d_2]\ \ (d_1d_2)^{1+\sigma }}. \end{aligned}$$

(4.6)

Let $a=(d_1,d_2),\ d_1=ab, d_2=ac$. Then $(b,c)=1$ and $[d_1,d_2]=abc.$ So from (4.6) we have

$$\begin{aligned} G_1(n_3, \sigma )= & {} \zeta ^2(1+\sigma )\sum _{d\mid n_3}d\sum _{\begin{array}{c} a,b,c\ge 1, (b,c)=1\\ d\mid abc \end{array}} \frac{a^2bc\ \log a^2bc}{abc(a^2bc)^{1+\sigma }}\nonumber \\\le & {} \zeta ^2(1+\sigma )\sum _{d\mid n_3}d\sum _{\begin{array}{c} a,b,c\ge 1 \\ d\mid abc \end{array}}\frac{a \ \log a^2bc}{ (a^2bc)^{1+\sigma }}\nonumber \\\le & {} 2\zeta ^2(1+\sigma )\sum _{d\mid n_3}d\sum _{\begin{array}{c} a,b,c\ge 1 \\ d\mid abc \end{array}} \frac{ \log abc}{ (abc)^{1+\sigma }}\nonumber \\= & {} 2\zeta ^2(1+\sigma )\sum _{d\mid n_3}d\sum _{\begin{array}{c} n\ge 1 \\ d\mid n \end{array}} \frac{\tau _3(n) \log n}{ n^{1+\sigma }}\nonumber \\= & {} 2\zeta ^2(1+\sigma )\sum _{d\mid n_3}d\sum _{n\ge 1}\frac{\tau _3(nd) \log nd}{ (dn)^{1+\sigma }}\nonumber \\\ll & {} \zeta ^2(1+\sigma )\sum _{d\mid n_3}\tau _3(d)\sum _{n\ge 1}\frac{\tau _3(n) (\log n+\log d)}{ n^{1+\sigma }}\nonumber \\\ll & {} \sigma ^{-2}\ \sum _{d\mid n_3}\tau _3(d)(A(\sigma )+B(\sigma )\log d), \end{aligned}$$

(4.7)

where we used the estimate $\tau _3(nd)\le \tau _3(n)\tau _3(d)$ and Lemma 2.3, and where

$$\begin{aligned} A(\sigma ):=\sum _{n\ge 1}\frac{\tau _3(n)\log n}{ n^{1+\sigma }},\ B(\sigma ):=\sum _{n\ge 1}\frac{\tau _3(n) }{ n^{1+\sigma }}=\zeta ^3(1+\sigma ). \end{aligned}$$

By Lemma 2.3 we get

$$\begin{aligned} B(\sigma )\ll \sigma ^{-3}. \end{aligned}$$

(4.8)

It is easy to see that

$$\begin{aligned} A(\sigma )=-B^{\prime }(\sigma )=-3\zeta ^2(1+\sigma )\zeta ^{\prime }(1+\sigma ). \end{aligned}$$

By Lemma 2.3 again we get

$$\begin{aligned} A(\sigma )\ll \sigma ^{-4}. \end{aligned}$$

(4.9)

From (4.7) to (4.9) we get

$$\begin{aligned} G_1(n_3, \sigma )\ll & {} \sigma ^{-2} \sum _{d\mid n_3}\tau _3(d)(\sigma ^{-4}+\sigma ^{-3}\log d)\nonumber \\\ll & {} \sigma ^{-2}\sum _{d\mid n_3}\tau _3(d)(\sigma ^{-4}+\sigma ^{-3}\log n_3)\nonumber \\\ll & {} \sigma ^{-6}\tau _4(n_3)\log n_3. \end{aligned}$$

(4.10)

$G_2(n_3,\sigma )$ can be bounded in the same way. We have

$$\begin{aligned} G_2(n_3,\sigma )\ll \sigma ^{-5}\tau _4(n_3)\log n_3. \end{aligned}$$

(4.11)

From (4.4), (4.10) and (4.11) we get

$$\begin{aligned} G(n_3,\sigma )\ll \sigma ^{-6}\tau _4(n_3)\log n_3 \ (n_3\ge 2). \end{aligned}$$

(4.12)

4.2 Upper bound of E(x, T)

Inserting (4.2) and (4.12) into (4.1), we get

$$\begin{aligned} E(x,T)\ll \frac{\log ^4 x}{T}+\log ^ 6 x\times \ \sum _{n_3=2}^\infty \frac{n_3^{-1-1/\log x}\tau _4(n_3)\log n_3}{T|\log \frac{x}{n_3}|+1}. \end{aligned}$$

(4.13)

Let

$$\begin{aligned} U_1:= & {} \sum _{n_3\le x/2} \frac{n_3^{-1-1/\log x}\tau _4(n_3)\log n_3}{T|\log \frac{x}{n_3}|+1},\nonumber \\ U_2:= & {} \sum _{x/2<n_3\le 3x/2} \frac{n_3^{-1-1/\log x}\tau _4(n_3)\log n_3}{T|\log \frac{x}{n_3}|+1},\nonumber \\ U_3:= & {} \sum _{n_3>3x/2} \frac{n_3^{-1-1/\log x}\tau _4(n_3)\log n_3}{T|\log \frac{x}{n_3}|+1}. \end{aligned}$$

(4.14)

We have by Lemma 2.3 that

$$\begin{aligned} U_1+U_3\ll & {} \frac{1}{T}\sum _{n=1}^\infty \frac{\tau _4(n)\log n}{ n^{1+1/\log x} } \nonumber \\= & {} \frac{1}{T}\sum _{n=1}^\infty \frac{\tau _4(n)\log n}{ n^{1+\sigma } }|_{\sigma =1/\log x} \nonumber \\= & {} \frac{1}{T}\left( \sum _{n=1}^\infty \frac{-\tau _4(n) }{ n^{1+\sigma } }\right) ^{\prime }|_{\sigma =1/\log x}\nonumber \\= & {} \frac{1}{T}\left( -\zeta ^4(1+\sigma )\right) ^{\prime }|_{\sigma =1/\log x}\nonumber \\= & {} \frac{1}{T}\left( -4\zeta ^3(1+\sigma )\zeta ^{\prime }(1+\sigma )\right) |_{\sigma =1/\log x}\nonumber \\\ll & {} \frac{\log ^5 x}{T}. \end{aligned}$$

(4.15)

Now we bound $U_2.$ We write

$$\begin{aligned} U_2= & {} U_{21}+U_{22}+U_{23},\nonumber \\ U_{21}:= & {} \sum _{x/2<n_3\le x e^{-1/T}} \frac{n_3^{-1-1/\log x}\tau _4(n_3)\log n_3}{T|\log \frac{x}{n_3}|+1},\nonumber \\ U_{22}:= & {} \sum _{x e^{-1/T}x<n_3\le x e^{1/T}} \frac{n_3^{-1-1/\log x}\tau _4(n_3)\log n_3}{T|\log \frac{x}{n_3}|+1},\nonumber \\ U_{23}:= & {} \sum _{xe^{1/T} <n_3\le 3x/2 } \frac{n_3^{-1-1/\log x}\tau _4(n_3)\log n_3}{T|\log \frac{x}{n_3}|+1}. \end{aligned}$$

(4.16)

When $x/2<n_3\le x e^{-1/T},$ we have $T|\log \frac{x}{n_3}|=T\log \frac{x}{n_3}\ge 1.$ Thus

$$\begin{aligned} U_{21}\ll & {} \frac{\log x}{xT}\sum _{x/2<n_3\le xe^{-1/T}} \frac{ \tau _4(n_3) }{ \log \frac{x}{n_3}}\nonumber \\\ll & {} \frac{ \log x}{xT}\sum _{ x/T\ll \ell \le x/2} \frac{ x\tau _4([x]-\ell ) }{ \ell }\ \ (n_3=[x]-\ell )\nonumber \\\ll & {} \frac{ \log ^2 x}{T}\max _{x/T\ll L\le x/4}\frac{1}{L} \sum _{L<\ell \le 2L} \tau _4([x]-\ell ) \nonumber \\= & {} \frac{ \log ^2 x}{T}\max _{x/T\ll L\le x/4}\frac{1}{L} \sum _{[x]-2L\le n< [x]-L} \tau _4(n) \nonumber \\\ll & {} \frac{ \log ^2 x}{T}\max _{x/T\ll L\le x/4}\frac{1}{L} \left( L\log ^3 x+x^{1/2}\log ^5 x\right) \nonumber \\\ll & {} \max _{x/T\ll L\le x/4} \left( \frac{ \log ^5 x}{T}+\frac{ \log ^2 x}{TL}\times x^{1/2}\log ^5 x\right) \nonumber \\\ll & {} \frac{ \log ^5 x}{T}+ x^{-1/2}\log ^7 x, \end{aligned}$$

(4.17)

where in the second line we used the estimate $\log (x/([x]-\ell ))\gg \ell /x$ and in the fifth line we used Lemma 2.6. Similarly we have

$$\begin{aligned} U_{23}\ll \frac{ \log ^5 x}{T}+ x^{-1/2}\log ^7 x. \end{aligned}$$

(4.18)

When $xe^{-1/T}<n_3\le xe^{1/T}x,$ we have $T|\log \frac{x}{n_3}|\le 1.$ Thus by Lemma 2.6 again we get

$$\begin{aligned} U_{22}\ll \frac{\log x}{x}\sum _{x e^{-1/T}x<n_3\le xe^{1/T}x}\tau _4(n_3)\ll \frac{\log ^4 x}{T}+x^{-1/2}\log ^6 x. \end{aligned}$$

(4.19)

Combining (4.13)–(4.19) we get

Proposition 4.1

Suppose $x\notin {\mathbb {N}}$ and T are two large parameters such that $5\le T\le x/2$. Then

$$\begin{aligned} E(x,T)\ll T^{-1}\log ^{11} x+x^{-1/2}\log ^{13} x. \end{aligned}$$

(4.20)

5 Proof of the theorem

In this section, we shall prove Theorem 1.1.

Suppose $x\ge 20, x\notin {\mathbb {N}}$. Let $10\le T\le x/2$ be a parameter to be determined later. Define

$$\begin{aligned} b_j =1+\frac{10^{j-1}}{\log x}, \quad \ T_j=10^{j-1} T \ (1\le j\le 3). \end{aligned}$$

By Lemma 2.2 with $r=3$ and $\sigma _a=1$ we have

$$\begin{aligned} C_3(x) = I(x,T)+O(x^3 E(x,T)), \end{aligned}$$

(5.1)

where

$$\begin{aligned} I(x,T)=\frac{1}{(2\pi i)^3}\int _{b_1-iT_1}^{b_1+iT_1}\int _{b_2-iT_2}^{b_2+iT_2} \int _{b_3-iT_3}^{b_3+iT_3} \frac{C(s_1,s_2,s_3)x^{s_1+s_2+s_3}}{s_1s_2s_3} ds_3 ds_2ds_1, \end{aligned}$$

$C(s_1,s_2,s_3)$ and E(x, T) being defined in the last section. Here E(x, T) can be bounded by Proposition 4.1, so we only need to evaluate I(x, T). We shall evaluate I(x, T) in the order $s_3, s_2,s_1.$

5.1 Evaluation of I(x, t) for $s_3$

In this subsection we shall evaluate the integral I(x, T) for the variable $s_3$. Consider the rectangle domain formed by the four points $s_3=b_3\pm iT_3, s_3=11/20\pm iT_3$. From Proposition 3.1 we see easily that in this domain the integral function

$$\begin{aligned} g(s_1,s_2,s_3):= & {} \frac{C(s_1,s_2,s_3)x^{s_1+s_2+s_3}}{s_1s_2s_3}\\= & {} \zeta ^2(s_1)\zeta ^2(s_2)\zeta ^2(s_3)\zeta (s_1+s_2-1)\zeta (s_1+s_3-1)\nonumber \\&\times \zeta (s_2+s_3-1)\zeta (s_1+s_2+s_3-2)H(s_1,s_2,s_3)\frac{x^{s_1+s_2+s_3}}{s_1s_2s_3}\nonumber \end{aligned}$$

(5.2)

has four poles, which are $s_3^{(0)}=1$, $s_3^{(1)}=2-s_1,$$s_3^{(2)}=2-s_2,$$s_3^{(3)}=3-s_1-s_2,$ respectively. Here $s_3=1$ is a pole of order 2 and the other three poles are simple. By the residue theorem we get

$$\begin{aligned} I(x,T)=\sum _{j=0}^3I_j(x,T)+H_1(x,T)+H_2(x,T)-H_3(x,T), \end{aligned}$$

(5.3)

where

$$\begin{aligned} I_j(x,T)&=\frac{1}{(2\pi i)^2}\int _{b_1-iT_1}^{b_1+iT_1}\int _{b_2-iT_2}^{b_2+iT_2} \text {Res}_{s_3=s_3^{(j)}}g(s_1,s_2,s_3) ds_2ds_1\ (j=0,1,2,3),\\ H_1(x,T)&=\frac{1}{(2\pi i)^3}\int _{b_1-iT_1}^{b_1+iT_1}\int _{b_2-iT_2}^{b_2+iT_2} \int _{11/20+iT_3}^{b_3+iT_3} g(s_1,s_2,s_3) ds_3 ds_2ds_1,\\ H_2(x,T)&= \frac{1}{(2\pi i)^3}\int _{b_1-iT_1}^{b_1+iT_1}\int _{b_2-iT_2}^{b_2+iT_2} \int _{11/20-iT_3}^{11/20+iT_3} g(s_1,s_2,s_3) ds_3 ds_2ds_1,\\ H_3(x,T)&= \frac{1}{(2\pi i)^3}\int _{b_1-iT_1}^{b_1+iT_1}\int _{b_2-iT_2}^{b_2+iT_2} \int _{11/20-iT_3}^{b_3-iT_3} g(s_1,s_2,s_3) ds_3 ds_2ds_1. \end{aligned}$$

5.1.1 Estimates of $H_j(x,T)\ (j=1,2,3)$

We estimate $H_1(x,T)$ first. In this case we have $s_1=b_1+it_1$ with $|t_1|\le T,$$s_2=b_2+it_2$ with $|t_2|\le 10T$ and $s_3=\sigma _3+i100T$ with $11/20\le \sigma _3\le b_3.$ By the condition (3.3) in Proposition 3.1 we see that

$$\begin{aligned} H(s_1,s_2,s_3)\ll 1. \end{aligned}$$

By Lemmas 2.3 and 2.4 we have

$$\begin{aligned}&\zeta ^2(s_1)\zeta ^2(s_2) \zeta (s_1+s_2-1) \ll \log ^5 x,\\&\quad \zeta ^2(s_3) \zeta (s_1+s_3-1) \zeta (s_2+s_3-1)\zeta (s_1+s_2+s_3-2) \ll T^{\frac{5}{3}\max (1-\sigma _3,0)}\log ^5 T. \end{aligned}$$

From the above estimates and (5.2) we have

$$\begin{aligned} g(s_1,s_2,s_3)\ll \frac{x^2\log ^{10} x}{T(|t_1|+1)(|t_2|+1)} \times x^{\sigma _3}T^{\frac{5}{3}\max (1-\sigma _3, 0)}. \end{aligned}$$

(5.4)

So we get

$$\begin{aligned} H_1(x,T)&\ll \frac{x^2\log ^{10} x}{T}\int _{-T}^{T}\frac{dt_1}{|t_1|+1}\int _{-T_2}^{T_2}\frac{dt_2}{|t_2|+1} \left( \int _{11/20}^1x^{\sigma _3}T^{\frac{5}{3}(1-\sigma _3)}d\sigma _3+\int _{1}^{b_3}x^{\sigma _3}d\sigma _3\right) \nonumber \\&\ll \frac{x^3\log ^{12} x}{T}+x^{\frac{51}{20}}T^{-\frac{1}{4}}\log ^{12} x. \end{aligned}$$

(5.5)

Similarly, we have

$$\begin{aligned} H_3(x,T)\ll \frac{x^3\log ^{12} x}{T}+x^{\frac{51}{20}}T^{-\frac{1}{4}}\log ^{12} x. \end{aligned}$$

(5.6)

Now we estimate $H_2(x,T).$ In this case we have $s_1=b_1+it_1$ with $|t_1|\le T,$$s_2=b_2+it_2$ with $|t_2|\le 10T$ and $s_3=11/20+it_3$ with $|t_3|\le T_3=100T.$ Similar to (5.4) we have

$$\begin{aligned} g(s_1,s_2,s_3)\ll & {} x^{\frac{51}{20}}\log ^5 x \nonumber \\&\times \frac{\left| \zeta ^2(s_3) \zeta (s_1+s_3-1) \zeta (s_2+s_3-1)\zeta (s_1+s_2+s_3-2) \right| }{(|t_1|+1)(|t_2|+2)(|t_3|+1)} \end{aligned}$$

(5.7)

So we have

$$\begin{aligned} H_2(x,T) \ll x^{\frac{51}{20}}\log ^5 x \times H_2(T), \end{aligned}$$

(5.8)

where

$$\begin{aligned} H_2(T):= \int _{-T_1}^{T_1}dt_1\int _{-T_2}^{T_2}dt_2\int _{-T_3}^{T_3}\frac{\left| \zeta ^2(s_3) \zeta (s_1+s_3-1) \zeta (s_2+s_3-1)\zeta (s_1+s_2+s_3-2) \right| }{(|t_1|+1)(|t_2|+2)(|t_3|+1)}dt_3. \end{aligned}$$

By Hölder’s inequality we get

$$\begin{aligned} H_2(T)\le (H_{21}(T))^{\frac{2}{5}} (H_{22}(T))^{\frac{1}{5}} (H_{23}(T))^{\frac{1}{5}} (H_{24}(T))^{\frac{1}{5}}, \end{aligned}$$

(5.9)

where

$$\begin{aligned} H_{21}(T):= & {} \int _{-T_1}^{T_1}dt_1\int _{-T_2}^{T_2}dt_2\int _{-T_3}^{T_3}\frac{\left| \zeta (11/20+it_3) \right| ^5}{(|t_1|+1)(|t_2|+2)(|t_3|+1)}dt_3,\\ H_{22}(T):= & {} \int _{-T_1}^{T_1}dt_1\int _{-T_2}^{T_2}dt_2\int _{-T_3}^{T_3}\frac{\left| \zeta (11/20+1/\log x+i(t_1+t_3)) \right| ^5 }{(|t_1|+1)(|t_2|+2)(|t_3|+1)}dt_3,\\ H_{23}(T):= & {} \int _{-T_1}^{T_1}dt_1\int _{-T_2}^{T_2}dt_2\int _{-T_3}^{T_3}\frac{\left| \zeta (11/20+10/\log x+i(t_2+t_3) ) \right| ^5 }{(|t_1|+1)(|t_2|+2)(|t_3|+1)}dt_3,\\ H_{24}(T):= & {} \int _{-T_1}^{T_1}dt_1\int _{-T_2}^{T_2}dt_2\int _{-T_3}^{T_3}\frac{\left| \zeta (11/20+11/\log x+i(t_1+t_2+t_3) ) \right| ^5 }{(|t_1|+1)(|t_2|+2)(|t_3|+1)}dt_3. \end{aligned}$$

By Lemma 2.7 and partial summation we get

$$\begin{aligned} H_{21}(T)=\int _{-T_1}^{T_1}\frac{dt_1}{|t_1|+1}\int _{-T_2}^{T_2}\frac{dt_2}{|t_2|+2} \int _{-T_3}^{T_3}\frac{\left| \zeta (11/20+it_3) \right| ^5}{|t_3|+1}dt_3\ll T^{\varepsilon }. \end{aligned}$$

(5.10)

For $H_{24}(T),$ we have

$$\begin{aligned}&H_{24}(T)\le \int _{-T_3}^{T_3}dt_1\int _{-T_3}^{T_3}dt_2\int _{-T_3}^{T_3}\frac{\left| \zeta (11/20+11/\log x+i(t_1+t_2+t_3) ) \right| ^5 }{(|t_1|+1)(|t_2|+2)(|t_3|+1)}dt_3\nonumber \\&\quad \ll \int _{0\le |t_1|\le |t_2|\le |t_3|\le T_3}\frac{\left| \zeta (11/20+11/\log x+i(t_1+t_2+t_3) ) \right| ^5 }{(|t_1|+1)(|t_2|+2)(|t_3|+1)}dt_3dt_2dt_1. \end{aligned}$$

(5.11)

From the condition $|t_1|\le |t_2|\le |t_3|$ we get

$$\begin{aligned} |t_1+t_2+t_3|+1\le |t_1|+|t_2|+|t_3|+1<3(|t_3|+1). \end{aligned}$$

Thus we get

$$\begin{aligned}&H_{24}(T) \\&\quad = \int _{0\le |t_1|\le |t_2|\le |t_3|\le T_3}\frac{\left| \zeta (11/20+11/\log x+i(t_1+t_2+t_3) ) \right| ^5 }{(|t_1|+1)(|t_2|+2)(|t_1+t_2+t_3|+1)}\\&\qquad \times \frac{|t_1+t_2+t_3|+1}{|t_3|+1}dt_3dt_2dt_1\\&\quad \ll \int _{0\le |t_1|\le |t_2|\le |t_3|\le T_3}\frac{\left| \zeta (11/20+11/\log x+i(t_1+t_2+t_3) ) \right| ^5 }{(|t_1|+1)(|t_2|+2)(|t_1+t_2+t_3|+1)} dt_3dt_2dt_1\\&\quad \le \int _{-T_3}^{T_3} \frac{dt_1}{|t_1|+1} \int _{- T_3}^{T_3} \frac{dt_2}{|t_2|+1} \int _{-T_3}^{T_3} \frac{\left| \zeta (11/20+11/\log x+i(t_1+t_2+t_3) ) \right| ^5 }{ (|t_1+t_2+t_3|+1)} dt_3 \\&\quad \le \int _{-T_3}^{T_3} \frac{dt_1}{|t_1|+1} \int _{- T_3}^{T_3} \frac{dt_2}{|t_2|+1} \int _{-3T_3}^{3T_3} \frac{\left| \zeta (11/20+11/\log x+iv ) \right| ^5 }{ (|v|+1)} dv, \end{aligned}$$

which combining Lemma 2.7 gives

$$\begin{aligned} H_{24}(T) \ll T^{\varepsilon }. \end{aligned}$$

(5.12)

Similarly, we have

$$\begin{aligned} H_{22}(T) \ll T^{\varepsilon },\ \ H_{23}(T) \ll T^{\varepsilon }. \end{aligned}$$

(5.13)

Combining (5.8)–(5.13) we get

$$\begin{aligned} H_2(x,T)\ll x^{\frac{51}{20}+\varepsilon }. \end{aligned}$$

(5.14)

5.1.2 Evaluation of $I_3(x,T)$

Now we evaluate $I_3(x,T).$ Since $s_3^{(3)}=3-s_1-s_2$ is a simple pole of $g(s_1,s_2,s_3),$ we have

$$\begin{aligned} \text {Res}_{s_3=s_3^{(3)}} g(s_1,s_2,s_3)= & {} x^3 \zeta ^2(s_1)\zeta ^2(s_2)\zeta ^2(3-s_1-s_2)\zeta (s_1+s_2-1)\zeta (2-s_2) \\&\times \zeta (2-s_1)H(s_1,s_2,3-s_1-s_2)\frac{1}{s_1s_2(3-s_1-s_2)}. \end{aligned}$$

Define

$$\begin{aligned} f_1(t;\sigma _1)&=\frac{\zeta ^2(\sigma _1+it)\zeta (2-\sigma _1-it)}{\sigma _1+it},\\ f_2(t;\sigma _2)&=\frac{\zeta ^2(\sigma _2+it)\zeta (2-\sigma _2-it)}{\sigma _2+it},\\ f_3(t;\sigma _1,\sigma _2)&=\frac{\zeta ^2(3-\sigma _1-\sigma _2-it)\zeta (\sigma _1+\sigma _2-1+it)}{3-\sigma _1-\sigma _2-it},\\ f(t_1,t_2; \sigma _1, \sigma _2)&=H(\sigma _1+it_1, \sigma _2+it_2, 3-\sigma _1-\sigma _2-i(t_1+t_2)). \end{aligned}$$

From Lemma 2.4 we have

$$\begin{aligned} f_1(t;b_1)\ll \frac{\log ^3 (|t|+2)}{|t|+2},\ f_2(t;b_2)\ll \frac{\log ^3 (|t|+2)}{|t|+2},\ f_3(t;b_1,b_2)\ll \frac{\log ^3 (|t|+2)}{|t|+2}. \end{aligned}$$

From Proposition 3.1 we have

$$\begin{aligned} f(t_1,t_2; b_1, b_2)\ll 1. \end{aligned}$$

So from Lemma 2.8 with $\alpha _j=1, \beta _j=3(j=1,2,3)$ and $U=T,$ we get

$$\begin{aligned} I_3(x,T) =G(b_1, b_2)x^3 +O(x^3T^{-1}\log ^9 x), \end{aligned}$$

(5.15)

where

$$\begin{aligned} G(\sigma _1,\sigma _2)= & {} \frac{1}{(2\pi i)^2}\int _{\sigma _1-\infty }^{\sigma _1+\infty }\int _{\sigma _2-\infty }^{\sigma _2+\infty } \zeta ^2(s_1)\zeta ^2(s_2)\zeta ^2(3-s_1-s_2)\zeta (s_1+s_2-1)\\&\times \zeta (2-s_2) \zeta (2-s_1)H(s_1,s_2,3-s_1-s_2)\frac{1}{s_1s_2(3-s_1-s_2)}ds_2ds_1. \end{aligned}$$

From the condition (3.3) in Proposition 3.1 we see that $H(s_1,s_2, 3-s_1-s_2)$ can be written as a Dirichlet series, which is absolutely convergent when $s_1$ and $s_2$ satisfy the following conditions:

$$\begin{aligned}&1/3<\sigma _1<2,\ 1/3<\sigma _2<2,\ 1<\sigma _1+\sigma _2<8/3, \nonumber \\&\ 2<2\sigma _1+\sigma _2<4,\ 2<2\sigma _2+\sigma _1<4,\ |\sigma _1-\sigma _2|<1. \end{aligned}$$

(5.16)

From Lemma 2.3 and Lemma 2.4 we see that if $1/2\le \sigma _1\le b_1, 1/2\le \sigma _1\le b_2, \sigma _1+\sigma _2\ge 3/2,$ then

$$\begin{aligned} f_1(t;\sigma _1)\ll & {} \frac{(|t|+2)^{\frac{2(1-\sigma _1)}{3}}\log ^3 (|t|+2)}{|t|+2}\ll \frac{\log ^3 (|t|+2)}{(|t|+2)^{\frac{1+2\sigma _1}{3}}},\nonumber \\ f_2(t;\sigma _2)\ll & {} \frac{(|t|+2)^{\frac{2(1-\sigma _2)}{3}}\log ^3 (|t|+2)}{|t|+2}\ll \frac{\log ^3 (|t|+2)}{(|t|+2)^{\frac{1+2\sigma _2}{3}}},\nonumber \\ f_3(t;\sigma _1,\sigma _2)\ll & {} \frac{(|t|+2)^{\frac{1-(\sigma _1+\sigma _2-1)}{3}}\log ^3 (|t|+2)}{|t|+2}\ll \frac{\log ^3 (|t|+2)}{(|t|+2)^{\frac{1+\sigma _1+\sigma _2}{3}}}. \end{aligned}$$

(5.17)

From (5.16), (5.17) and Lemma 2.8 we see that the integral $G(\sigma _1,\sigma _2)$ is absolutely convergent if $2/3\le \sigma _1\le b_1, 2/3\le \sigma _1\le b_2, \sigma _1+\sigma _2\ge 3/2.$ Certainly we can show that the integral $G(\sigma _1,\sigma _2)$ is absolutely convergent in a much wider range. However, the above range is more than enough for our proof.

Now we show that $G(b_1,b_2)$ is a constant. We write

$$\begin{aligned} G(b_1,b_2)= & {} \frac{1}{2\pi i }\int _{b_1-\infty }^{b_1+\infty } \zeta ^2(s_1) \zeta (2-s_1)\frac{ds_1}{s_1} \nonumber \\&\quad \times \frac{1}{2\pi i } \int _{b_2-\infty }^{b_2+\infty } \zeta ^2(s_2)\zeta ^2(3-s_1-s_2)\zeta (s_1+s_2-1)\nonumber \\&\quad \times \zeta (2-s_2) H(s_1,s_2,3-s_1-s_2)\frac{ ds_2}{s_2(3-s_1-s_2)}\nonumber \\&=\frac{1}{2\pi i }\int _{b_1-\infty }^{b_1+\infty } \zeta ^2(s_1) \zeta (2-s_1)\frac{ds_1}{s_1}\frac{1}{2\pi i}\int _{b_2-\infty }^{b_2+\infty } Y(s_1,s_2)ds_2, \end{aligned}$$

(5.18)

say. We move the integral line of the inner integral from $\mathfrak {R}s_2=b_2$ to $\mathfrak {R}s_2=3/4.$ The function $Y(s_1,s_2)$ has two poles in this range, which are $s_2=1$ (degree 3) and $s_2=2-s_1$ (degree 3), respectively. We get

$$\begin{aligned} G(b_1,b_2)= & {} \frac{1}{2\pi i }\int _{b_1-\infty }^{b_1+\infty } \zeta ^2(s_1) \zeta (2-s_1)G_0(s_1)\frac{ds_1}{s_1} \nonumber \\&+\frac{1}{2\pi i }\int _{b_1-\infty }^{b_1+\infty } \zeta ^2(s_1) \zeta (2-s_1)\frac{G_1(s_1)+G_2(s_1)}{s_1}ds_1, \end{aligned}$$

(5.19)

where

$$\begin{aligned} G_0(s_1)= & {} \frac{1}{2\pi i}\int _{3/4-\infty }^{3/4+\infty }Y(s_1,s_2)ds_2,\\ G_1(s_1)= & {} \text {Res}_{s_2=1}Y(s_1,s_2),\ \ G_2(s_1)= \text {Res}_{s_2=2-s_1}Y(s_1,s_2). \end{aligned}$$

Both $G_1(s_1)$ and $G_2(s_1)$ can be computed by Lemmas 2.9 and 2.10, respectively. But the explicit expressions of them are not important here. So we omit the details.

We move both integral lines in (5.19) from $\mathfrak {R}s_1=b_1$ to $\mathfrak {R}s_1=3/4,$ we get

$$\begin{aligned} G(b_1,b_2)= & {} \frac{1}{2\pi i }\int _{3/4-\infty }^{3/4+\infty } \zeta ^2(s_1) \zeta (2-s_1)G_0(s_1)\frac{ds_1}{s_1} \nonumber \\&+\frac{1}{2\pi i }\int _{3/4-\infty }^{3/4+\infty } \zeta ^2(s_1) \zeta (2-s_1)\frac{G_1(s_1)+G_2(s_1)}{s_1}ds_1 +c_1+c_2, \end{aligned}$$

(5.20)

where

$$\begin{aligned} c_1=\text {Res}_{s_1=1} \zeta ^2(s_1) \zeta (2-s_1)\frac{G_0(s_1)}{s_1},\ \ c_2= \text {Res}_{s_1=1} \zeta ^2(s_1) \zeta (2-s_1)\frac{G_1(s_1)+G_2(s_1)}{s_1}. \end{aligned}$$

From (5.20) we see that $G(b_1,b_2)$is an absolute constant. We denote it by C. Hence we have

$$\begin{aligned} I_3(x,T)=Cx^3+O(\frac{x^3\log ^9 x}{T}). \end{aligned}$$

(5.21)

5.2 Evaluation of $I_0(x,T)$

In this subsection, we shall evaluate $I_0(x,T)$. By Lemma 2.9 we get that

$$\begin{aligned} \text {Res}_{s_3=1}g(s_1,s_2,s_3)=\sum _{j=1}^7K_j(s_1,s_2), \end{aligned}$$

(5.22)

where ($\gamma$ is Euler constant)

$$\begin{aligned} K_1(s_1,s_2):= & {} 2\gamma \zeta ^3(s_1)\zeta ^3(s_2)\zeta ^2(s_1+s_2-1)H(s_1,s_2,1)\frac{x^{s_1+s_2+1}}{s_1s_2},\\ K_2(s_1,s_2):= & {} \zeta ^2(s_1)\zeta ^{\prime }(s_1)\zeta ^3(s_2)\zeta ^2(s_1+s_2-1)H(s_1,s_2,1)\frac{x^{s_1+s_2+1}}{s_1s_2},\\ K_3(s_1,s_2):= & {} \zeta ^3(s_1)\zeta ^2(s_2)\zeta ^{\prime }(s_2)\zeta ^2(s_1+s_2-1)H(s_1,s_2,1)\frac{x^{s_1+s_2+1}}{s_1s_2},\\ K_4(s_1,s_2):= & {} \zeta ^3(s_1)\zeta ^3(s_2)\zeta (s_1+s_2-1)\zeta ^{\prime }(s_1+s_2-1)H(s_1,s_2,1)\frac{x^{s_1+s_2+1}}{s_1s_2},\\ K_5(s_1,s_2):= & {} \zeta ^3(s_1)\zeta ^3(s_2)\zeta ^2(s_1+s_2-1)\frac{\partial H(s_1,s_2,s_3)}{\partial s_3}|_{s_3=1}\frac{x^{s_1+s_2+1}}{s_1s_2},\\ K_6(s_1,s_2):= & {} -\zeta ^3(s_1)\zeta ^3(s_2)\zeta ^2(s_1+s_2-1)H(s_1,s_2,1)\frac{x^{s_1+s_2+1}}{s_1s_2},\\ K_7(s_1,s_2):= & {} \zeta ^3(s_1)\zeta ^3(s_2)\zeta ^2(s_1+s_2-1)H(s_1,s_2,1)\frac{x^{s_1+s_2+1}}{s_1s_2}\log x. \end{aligned}$$

So we have

$$\begin{aligned} I_0(x,T)=\sum _{j=1}^7\frac{1}{(2\pi i)^2}\int _{b_1-iT_1}^{b_1+iT_1}\int _{b_2-iT_2}^{b_2+iT_2}K_j(s_1,s_2)ds_2ds_1. \end{aligned}$$

(5.23)

5.2.1 Integration over $s_2$

We consider the domain formed by the four points $s_2=b_2\pm iT_2, s_2=11/20\pm iT_2.$ It is easy to see that for each $1\le j\le 7,$ the function $K_j(s_1,s_2)$ has two poles, which are $s_2^{(1)}=1$ and $s_2^{(2)}=2-s_1,$ respectively. By the residue theorem, we have

$$\begin{aligned} I_0(x,T)=\sum _{j=1}^7 \sum _{\ell =1}^2 I_0^{(j,\ell )}(x,T)+\sum _{j=1}^7(H_j^{(1)}(x,T)+ H_j^{(2)}(x,T)-H_j^{(3)}(x,T)), \end{aligned}$$

(5.24)

where ($1\le j\le 7$)

$$\begin{aligned} I_0^{(j,1)}(x,T):= & {} \frac{1}{2\pi i}\int _{b_1-iT_1}^{b_1+iT_1} \text {Res}_{s_2=1}K_j(s_1,s_2)ds_1, \\ I_0^{(j,2)}(x,T):= & {} \frac{1}{2\pi i}\int _{b_1-iT_1}^{b_1+iT_1} \text {Res}_{s_2=2-s_1}K_j(s_1,s_2)ds_1,\\ H_j^{(1)}(x,T):= & {} \frac{1}{(2\pi i)^2}\int _{b_1-iT_1}^{b_1+iT_1}\int _{11/20+iT_2}^{b_2+iT_2}K_j(s_1,s_2)ds_2ds_1,\\ H_j^{(2)}(x,T):= & {} \frac{1}{(2\pi i)^2}\int _{b_1-iT_1}^{b_1+iT_1}\int _{11/20-iT_2}^{11/20+iT_2}K_j(s_1,s_2)ds_2ds_1,\\ H_j^{(3)}(x,T):= & {} \frac{1}{(2\pi i)^2}\int _{b_1-iT_1}^{b_1+iT_1}\int _{11/20-iT_2}^{b_2-iT_2}K_j(s_1,s_2)ds_2ds_1. \end{aligned}$$

Similar to the estimate of $H_1(x,T)$ we have

$$\begin{aligned} H_j^{(1)}(x,T)\ll \frac{x^3\log ^{10} x}{T},\ \ H_j^{(3)}(x,T)\ll \frac{x^3\log ^{10} x}{T},\ \ (j=1,\ldots ,7). \end{aligned}$$

(5.25)

Similar to the estimate of $H_2(x,T)$ we have

$$\begin{aligned} H_j^{(2)}(x,T)\ll x^{51/20+\varepsilon },\ \ (j=1,\ldots ,7). \end{aligned}$$

(5.26)

5.2.2 Evaluations of $I_0^{(j,1)}(x,T)$

Now we evaluate $I_0^{(j,\ell )}(x,T)\ (1\le j\le 7, \ell =1,2).$ We consider $j=7.$ The proof of other cases are the same.

First consider the pole $s_2=1,$ which is of degree 3 in $K_7(s_1,s_2).$ Define

$$\begin{aligned} A:= & {} \zeta ^3(s_1) \frac{x^{s_1+1}}{s_1}\log x, \ \ M_1(s_2):=\zeta ^3(s_2),\nonumber \\ M_2(s_2):= & {} \zeta ^2(s_1+s_2-1)H^{*}(s_1,s_2) x^{s_2}, \ \ H^{*}(s_1,s_2)=H(s_1,s_2,1)s_2^{-1}. \end{aligned}$$

(5.27)

Write

$$\begin{aligned} M_1(s_2)=\sum _{m=-3}^\infty a_m(s_2-1)^{m}, \ \ a_{-3}=1. \end{aligned}$$

So by Lemmas 2.9 and 2.10 we get

$$\begin{aligned}&\text {Res}_{s_2=1}K_7(s_1,s_2)=A\sum _{m=0}^2\frac{a_{-m-1}}{m!}M_2^{(m)}(1)\nonumber \\&\quad =A\sum _{m=0}^2\frac{a_{-m-1}}{m!}\sum _{\begin{array}{c} n_1+n_2+n_3=m\\ n_1,n_2,n_3\ge 0 \end{array}} (\zeta ^2(s_1+s_2-1))^{(n_1)}\frac{\partial ^{n_2}H^{*}(s_1,s_2) }{\partial s_2^{n_2}}(x^{s_2})^{(n_3)}|_{s_2=1}. \end{aligned}$$

(5.28)

We have

$$\begin{aligned} (\zeta ^2(s_1+s_2-1))^{(0)}|_{s_2=1}= & {} \zeta ^2(s_1),\nonumber \\ (\zeta ^2(s_1+s_2-1))^{(1)}|_{s_2=1}= & {} 2\zeta (s_1)\zeta ^{\prime }(s_1),\nonumber \\ (\zeta ^2(s_1+s_2-1))^{(2)}|_{s_2=1}= & {} 2\zeta (s_1)\zeta ^{\prime \prime }(s_1)+2(\zeta ^{\prime }(s_1))^2. \end{aligned}$$

(5.29)

Suppose $s_1, s_2, s_3$ satisfy the condition (3.3). Then we can write

$$\begin{aligned} H(s_1, s_2, s_3)=\sum _{n_1,n_2,n_3=1}^\infty \frac{h(n_1,n_2,n_3)}{n_1^{s_1}n_2^{s_2}n_3^{s_3} }, \end{aligned}$$

which is absolutely convergent for the variables $s_1, s_2, s_3$. So for any $(\ell _1,\ell _2,\ell _3)\in {\mathbb {N}}_0^3,$ we have

$$\begin{aligned} \frac{\partial ^{\ell _1+\ell _2+\ell _3} H(s_1,s_2,s_3)}{\partial s_1^{\ell _1} \partial s_2^{\ell _2}\partial s_3^{\ell _3} }= (-1)^{\ell _1+\ell _2+\ell _3}\sum _{n_1,n_2,n_3=1}^\infty \frac{h(n_1,n_2,n_3)\prod _{j=1}^3(\log n_j)^{\ell _j} }{n_1^{s_1}n_2^{s_2}n_3^{s_3} }. \end{aligned}$$

(5.30)

Define

$$\begin{aligned} h_0(s_1):= & {} H^{*}(s_1,1)=H(s_1,1,1),\nonumber \\ h_1(s_1):= & {} \frac{\partial H^{*}(s_1,s_2) }{\partial s_2 }|_{s_2=1}, \, h_2(s_1):=\frac{\partial ^2 H^{*}(s_1,s_2) }{\partial s_2^2 }|_{s_2=1}. \end{aligned}$$

(5.31)

From (5.30) it is easy to see that

$$\begin{aligned} h_1(s_1)= & {} -\sum _{n_1,n_2,n_3=1}^\infty \frac{h(n_1,n_2,n_3)(\log n_2-1)}{n_1^{s_1}n_2 n_3 },\\ h_2(s_1)= & {} \sum _{n_1,n_2,n_3=1}^\infty \frac{h(n_1,n_2,n_3)(\log ^2 n_2+2\log n_2+2)}{n_1^{s_1}n_2 n_3 }. \end{aligned}$$

From the condition (3.3) it is easy to see that $h_0(s_1), h_1(s_1)$ and $h_2(s_1)$ are all analytic when $\mathfrak {R}s_1>1/2.$

From (5.27) to (5.31) it is easy to see that

$$\begin{aligned} \text {Res}_{s_2=1}K_7(s_1,s_2)=S_1(s_1)+S_2(s_1)+S_3(s_1)+S_4(s_1), \end{aligned}$$

(5.32)

where

$$\begin{aligned} S_1(s_1):= & {} \zeta ^5(s_1)V_1(s_1,x),\\ S_2(s_1):= & {} \zeta ^4(s_1)\zeta ^{\prime }(s_1)V_2(s_1,x),\nonumber \\ S_3(s_1):= & {} \zeta ^4(s_1)\zeta ^{\prime \prime }(s_1)\frac{x^{2+s_1}\log x}{s_1}h_0(s_1),\nonumber \\ S_4(s_1):= & {} \zeta ^3(s_1)(\zeta ^{\prime }(s_1))^2\frac{x^{2+s_1}\log x}{s_1}h_0(s_1),\nonumber \\ V_1(s_1,x)= & {} \frac{x^{2+s_1}\log x}{s_1} \left( a_{-1}h_0(s_1)+a_{-2}h_1(s_1)+\frac{1}{2} h_2(s_1)\right) \nonumber \\&+ \frac{x^{2+s_1}\log x}{s_1}\left( \frac{1}{2} h_0(s_1)\log ^2 x+\frac{1}{2} h_1(s_1)\log x+a_{-2}h_0(s_1)\log x\right) ,\nonumber \\ V_2(s_1,x)= & {} \frac{x^{2+s_1}\log x}{s_1}\left( 2a_{-2}h_0(s_1)+h_1(s_1)+h_0(s_1)\log x\right) .\nonumber \end{aligned}$$

(5.33)

Hence we have

$$\begin{aligned} I_0^{(7,1)}(x,T)=\sum _{j=1}^4\frac{1}{2\pi i}\int _{b_1-iT}^{b_1+iT}S_j(s_1)ds_1. \end{aligned}$$

(5.34)

We consider the domain formed by the four points $s_1=b_1\pm iT, s_1=11/20\pm iT.$ By the residue theorem, we get

$$\begin{aligned} I_0^{(7,1)}(x,T)=\sum _{j=1}^4 \text {Res}_{s_1=1} S_j(s_1) +\sum _{j=1}^4 (L_{1,j}(x,T)+L_{2,j}(x,T)-L_{3,j}(x,T)), \end{aligned}$$

(5.35)

where

$$\begin{aligned} L_{1,j}(x,T)= & {} \frac{1}{2\pi i}\int _{11/20+iT}^{b_1+iT}S_j(s_1)ds_1,\\ L_{2,j}(x,T)= & {} \frac{1}{2\pi i}\int _{11/20-iT}^{11/20+iT}S_j(s_1)ds_1,\\ L_{3,j}(x,T)= & {} \frac{1}{2\pi i}\int _{11/20-iT}^{b_1-iT}S_j(s_1)ds_1. \end{aligned}$$

Similar to (5.5), by Lemmas 2.3 and 2.4 we have for $1\le j\le 4$ that

$$\begin{aligned} L_{1,j}(x,T)\ll \frac{x^3\log ^{10} x}{T}+x^{\frac{51}{20}}T^{-\frac{1}{4}}\log ^{12} x,\nonumber \\ L_{3,j}(x,T)\ll \frac{x^3\log ^{10} x}{T}+x^{\frac{51}{20}}T^{-\frac{1}{4}}\log ^{12} x. \end{aligned}$$

(5.36)

Similar to the estimate of $H_2(x,T)$, by Lemma 2.7 we have

$$\begin{aligned} L_{2,j}(x,T)\ll x^{51/20+\varepsilon }, \ (j=1,2,3,4). \end{aligned}$$

(5.37)

By (5.33) and Lemma 2.9 we get

$$\begin{aligned} \text {Res}_{s_1=1}S_j(s_1)=x^3Q_j(\log x),(j=1,2,3,4), \end{aligned}$$

(5.38)

where $Q_j(u)(j=1,2,3,4)$ are polynomials in u of degree 7.

From (5.35) to (5.38), we get

$$\begin{aligned} I_0^{(7,1)}(x,T)=x^3Q_{71}(\log x)+O\left( x^{51/20+\varepsilon }+ \frac{x^3\log ^{10} x}{T}\right) , \end{aligned}$$

(5.39)

where $Q_{71}(u)$ is a polynomial in u of degree 7. Similarly for each $1\le j\le 6$, we have

$$\begin{aligned} I_0^{(j,1)}(x,T)=x^3Q_{j1}(\log x)+O\left( x^{51/20+\varepsilon }+ \frac{x^3\log ^{10} x}{T}\right) , \end{aligned}$$

(5.40)

where $Q_{j1}(u)$ is a polynomial in u of degree 7.

Now we consider the pole $s_2=2-s_1,$ which is of degree 2 in $K_7(s_1,s_2).$ Write

$$\begin{aligned} M_1^{*}(s_2)=\zeta ^2(s_1+s_2-1),\ M_2^{*}(s_2)=\zeta ^3(s_2)H(s_1,s_2,1)\frac{x^{s_2}}{s_2}, \end{aligned}$$

(5.41)

where

$$\begin{aligned} M_1^{*}(s_2)=\frac{1}{(s_2-(2-s_1))^2}+ \sum _{n=-1}^\infty b_n (s_2-(2-s_1))^n, \end{aligned}$$

By Lemma 2.9 we have

$$\begin{aligned} \text {Res}_{s_2=2-s_1}K_7(s_1,s_2) = A b_{-1} M_2^{*}(2-s_1)+ A M_2^{*\prime }(2-s_1) = \sum _{j=5}^9S_j(s_1), \end{aligned}$$

(5.42)

where $A=\zeta ^3(s_1)\frac{x^{s_1+1}}{s_1} \log x$ and

$$\begin{aligned} S_5(s_1):= & {} b_{-1} \frac{\zeta ^3(s_1)\zeta ^3(2-s_1)}{s_1(2-s_1)}h_3(s_1)\times x^3\log x,\nonumber \\ S_6(s_1):= & {} 3\frac{\zeta ^3(s_1)\zeta ^2(2-s_1)\zeta ^{\prime }(2-s_1)}{s_1(2-s_1)}h_3(s_1)\times x^3\log x,\nonumber \\ S_7(s_1):= & {} \frac{\zeta ^3(s_1)\zeta ^3(2-s_1)}{s_1(2-s_1)}h_4(s_1)\times x^3\log x,\nonumber \\ S_8(s_1):= & {} -\frac{\zeta ^3(s_1)\zeta ^3(2-s_1)}{s_1(2-s_1)^2}h_3(s_1)\times x^3\log x,\nonumber \\ S_9(s_1):= & {} \frac{\zeta ^3(s_1)\zeta ^3(2-s_1)}{s_1(2-s_1)}h_3(s_1)\times x^3\log ^2 x,\nonumber \\ h_3(s_1):= & {} H(s_1,2-s_1,1)=\sum _{n_1,n_2,n_3=1}^\infty \frac{h(n_1,n_2,n_3)}{n_1^{s_1}n_2^{2-s_1}n_3},\nonumber \\ h_4(s_1):= & {} -\sum _{n_1,n_2,n_3=1}^\infty \frac{h(n_1,n_2,n_3)\log n_2}{n_1^{s_1}n_2^{2-s_1}n_3}. \end{aligned}$$

(5.43)

By the condition (3.3) it is easy to see that both $h_3(s_1)$ and $h_4(s_1)$ are analytic when $1/2<\mathfrak {R}s_1<3/2.$ Similar to (5.21), we have

$$\begin{aligned} \frac{1}{2\pi i}\int _{b_1-iT}^{b_1+iT}S_j(s_1)ds_1= & {} C_j x^3\log x+O\left( \frac{x^3\log ^5 x}{T}\right) ,\ (j=5,6,7,8),\nonumber \\ \frac{1}{2\pi i}\int _{b_1-iT}^{b_1+iT}S_9(s_1)ds_1= & {} C_9 x^3\log ^2 x+O\left( \frac{x^3\log ^6 x}{T}\right) . \end{aligned}$$

(5.44)

where $C_j$$(j= 5, 6, 7, 8, 9)$ are absolute constants.

Thus we get

$$\begin{aligned} I_0^{(7,2)}(x,T)=x^3Q_{72}(\log x)+O\left( \frac{x^3\log ^6 x}{T}\right) , \end{aligned}$$

(5.45)

where $Q_{72}(u)$ is a polynomial in u of degree 2. Similarly for each $1\le j\le 6,$ we have

$$\begin{aligned} I_0^{(j,2)}(x,T)=x^3Q_{j2}(\log x)+O\left( \frac{x^3\log ^6 x}{T}\right) , \end{aligned}$$

(5.46)

where $Q_{j2}(u)$ is a polynomial in u of degree not exceeding 2.

Combining (5.24)–(5.26), (5.39), (5.40), (5.45) and (5.46) we get

$$\begin{aligned} I_0(x,T)=x^3P_0(\log x)+O\left( x^{51/20+\varepsilon }+ \frac{x^3\log ^{10} x}{T}\right) , \end{aligned}$$

(5.47)

where $P_0(u)$ is a polynomial in u of degree 7.

5.3 Evaluation of $I_1(x,T)$

In this subsection we evaluate $I_1(x,T).$ Since $s_3^{(1)}=2-s_1$ is a simple pole of $g(s_1,s_2,s_3),$ we have

$$\begin{aligned} \text {Res}_{s_3=2-s_1}g(s_1,s_2,s_3)= & {} \zeta ^2(s_1)\zeta ^3(s_2)\zeta ^2(2-s_1)\zeta (s_1+s_2-1)\nonumber \\&\times \zeta (s_2-s_1+1)H(s_1,s_2,2-s_1)\frac{x^{s_2+2}}{s_1s_2(2-s_1)}\nonumber \\=: & {} g_2(s_1,s_2), \end{aligned}$$

(5.48)

say. Consider the domain formed by the four points $s_2=b_2\pm iT_2, s_2=11/20\pm iT_2.$ The function $g_2(s_1,s_2)$ has three poles in this domain, which are $s_2^{(1)}=1$ (of degree 3), $s_2^{(2)}=s_1$ (simple) and $s_2^{(3)}=2-s_1$ (simple), respectively. By the residue theorem, we get

$$\begin{aligned} I_1(x,T)=\sum _{j=1}^3I_{1j}(x,T)+O\left( x^{51/20+\varepsilon }+ \frac{x^3\log ^{10} x}{T}\right) , \end{aligned}$$

(5.49)

where

$$\begin{aligned} I_{1j}(x,T)=\frac{1}{2\pi i}\int _{b_1-iT}^{b_1+iT} \text {Res}_{s_2=s_2^{(j)}}g(s_1,s_2)ds_1,\ \ (j=1,2,3). \end{aligned}$$

(5.50)

We omit the details of the proof of (5.49), since it is similar to Sect. 5.1.

5.3.1 Evaluation of $I_{11}(x,T)$

Since $s_2=1$ is a pole (of order 3) of $g_2(s_1,s_2)$, by Lemmas 2.9 and 2.10 we have

$$\begin{aligned} \text {Res}_{s_2=1} g_2(s_1,s_2)= & {} \sum _{m=0}^2\frac{a_{-m-1}}{m!}\sum _{\begin{array}{c} n_1+n_2+n_3+n_4=m\\ n_j\ge 0 \end{array}} x^3\log ^{n_4} x\nonumber \\&\times \frac{\zeta ^2(s_1) \zeta ^2(2-s_1)\zeta ^{(n_1)}(s_1) \zeta ^{(n_2)}(2-s_1)}{s_1(2-s_1)} {\varvec{h}}_{{n_{3}}} (s_1), \end{aligned}$$

(5.51)

where

$$\begin{aligned} {\varvec{h}}_{n_3}(s_1)= \frac{\partial ^{n_3} H^{**} (s_1,s_2)}{\partial s_2^{n_3}}|_{s_2=1},\ \ H^{**}(s_1,s_2):=H(s_1,s_2,2-s_1)s_2^{-1}. \end{aligned}$$

From the condition (3.3) of Proposition 3.1 we see easily that $H(s_1,1,2-s_1)$ is analytic when $1/2<\mathfrak {R}s_1<3/2.$ Hence we have

$$\begin{aligned} {\varvec{h}}_{n_3}(s_1)\ll _\varepsilon 1, \ \ (1/2+\varepsilon<\mathfrak {R}s_1<3/2-\varepsilon ). \end{aligned}$$

Thus similar to (5.21) we can get

$$\begin{aligned} I_{11}(x,T)=x^3P_{11}(\log x)+O(\frac{x^3\log ^{10} x}{T}), \end{aligned}$$

(5.52)

where $P_{11}(u)$ is a polynomial in u of degree 2.

5.3.2 Evaluation of $I_{12}(x,T)$

Since $s_2=s_1$ is a simple pole of $g_2(s_1,s_2)$, we have

$$\begin{aligned} \text {Res}_{s_2=s_1}g_2(s_1,s_2)= & {} \zeta ^5(s_1)\zeta ^2(2-s_1)\zeta (2s_1-1)H(s_1,s_1,2-s_1)\frac{x^{s_1+2}}{s_1^2(2-s_1)}, \end{aligned}$$

(5.53)

which implies that

$$\begin{aligned} I_{12}(x,T)&=\frac{1}{2\pi i}\int _{b_1-i\infty }^{b_1+i\infty }\zeta ^5(s_1)\zeta ^2(2-s_1)\zeta (2s_1-1)H(s_1,s_1,2-s_1)\frac{x^{s_1+2}}{s_1^2(2-s_1)}ds_1\nonumber \\&\quad +\,O(\frac{x^3\log ^8 x}{T^2}). \end{aligned}$$

(5.54)

By the condition (3.3) we see easily that $H(s_1,s_1,2-s_1)$ is analytic if $2/3<\mathfrak {R}s_1<5/3.$ Hence we have $H(s_1,s_1,2-s_1)\ll _\varepsilon 1$ if $2/3+\varepsilon<\mathfrak {R}s_1<5/3-\varepsilon .$ The function $\zeta ^5(s_1)\zeta ^2(2-s_1)\zeta (2s_1-1)H(s_1,s_1,2-s_1)\frac{x^{s_1+2}}{s_1^2(2-s_1)}$ has a pole $s_1=1$ in the region $2/3+\varepsilon <\mathfrak {R}s_1\le b_1$, which is of order 8. So moving the integration line in (5.54) from $\mathfrak {R}s_1=b_1$ to $\mathfrak {R}s_1=2/3+\varepsilon ,$ we get

$$\begin{aligned} I_{12}(x,T)= x^3P_{12}(\log x)+O\left( x^{\frac{8}{3}+\varepsilon }+\frac{x^3\log ^8 x}{T^2}\right) , \end{aligned}$$

(5.55)

where $P_{12}(u)$ is a polynomial in u of degree 7, and where we used the estimate

$$\begin{aligned} \int _{2/3+\varepsilon -i\infty }^{2/3+\varepsilon +i\infty }\left| \zeta ^5 (s_1)\zeta ^2(2-s_1)\zeta (2s_1-1)H(s_1,s_1,2-s_1)\frac{ 1}{s_1^2(2-s_1)}\right| dt_1\ll 1. \end{aligned}$$

The above estimate follows easily from Lemma 2.4.

5.3.3 Evaluation of $I_{13}(x,T)$

Now we consider $I_{13}(x,T).$ Since $s_2=2-s_1$ is a simple pole of $g_2(s_1,s_2),$ we have

$$\begin{aligned} \text {Res}_{s_2=2-s_1}g_2(s_1,s_2)=\zeta ^2(s_1)\zeta ^5 (2-s_1)\zeta (3-2s_1)h_5(s_1)\frac{x^{4-s_1}}{s_1(2-s_1)^2}, \end{aligned}$$

(5.56)

where $h_5(s_1):=H(s_1,2-s_1,2-s_1).$ So we get

$$\begin{aligned} I_{13}(x,T)&=\frac{1}{2\pi i}\int _{b_1-i\infty }^{b_1+i\infty }\zeta ^2(s_1)\zeta ^5(2-s_1)\zeta (3-2s_1)h_5(s_1) \frac{x^{4-s_1}}{s_1(2-s_1)^2}ds_1\nonumber \\&\quad +\,O\left( \frac{x^3\log ^{8} x}{T}\right) . \end{aligned}$$

(5.57)

By the condition (3.3) of Proposition 3.1 it is easy to see that $h_5(s_1)$ is analytic in the region $1/3<\mathfrak {R}s_1<4/3.$ So we have $h_5(s_1)\ll _\varepsilon 1$ when $1\le \mathfrak {R}s_1\le 4/3-\varepsilon .$ Also, the integral function in (5.57) has no poles in the range $1<\mathfrak {R}s_1\le 4/3-\varepsilon .$ Moving the integral line in (5.57) from $\mathfrak {R}s_1=b_1$ to $\mathfrak {R}s_1=4/3-\varepsilon ,$ by the residue theorem and Lemma 2.4 we get

$$\begin{aligned}&I_{13}(x,T)=O\left( \frac{x^3\log ^{6} x}{T}\right) \nonumber \\&\quad + \frac{1}{2\pi i}\int _{4/3-\varepsilon -i\infty }^{4/3-\varepsilon +i\infty }\zeta ^2(s_1)\zeta ^5(2-s_1)\zeta (3-2s_1)h_5(s_1) \frac{x^{4-s_1}}{s_1(2-s_1)^2}ds_1\nonumber \\&\quad \ll \frac{x^3\log ^{8} x}{T}+x^{\frac{8}{3}+\varepsilon } \left| \int _{4/3-\varepsilon -i\infty }^{4/3-\varepsilon +i\infty } \frac{ \zeta ^2(s_1)\zeta ^5(2-s_1)\zeta (3-2s_1) }{s_1(2-s_1)^2}ds_1\right| \nonumber \\&\quad \ll \frac{x^3\log ^{8} x}{T}+x^{\frac{8}{3}+\varepsilon }. \end{aligned}$$

(5.58)

From (5.49), (5.52), (5.55), (5.58) we get

$$\begin{aligned} I_1(x,T)=x^3P_1(\log x)+O\left( \frac{x^3\log ^{8} x}{T}+x^{\frac{8}{3}+\varepsilon }\right) , \end{aligned}$$

(5.59)

where $P_1(u)$ is a polynomial in u of degree 7.

5.4 Evaluation of $I_2(x,T)$

In this subsection we shall evaluate $I_2(x,T)$. Since $s_3^{(2)}=2-s_2$ is a simple pole of $g(s_1,s_2,s_3),$ we have

$$\begin{aligned} \text {Res}_{s_3=2-s_2}g(s_1,s_2,s_3)= & {} \zeta ^3(s_1)\zeta ^2(s_2)\zeta ^2(2-s_2)\zeta (s_1+s_2-1)\nonumber \\&\times \zeta (1+s_1-s_2)H(s_1,s_2,2-s_2)\frac{x^{s_1+2}}{s_1s_2(2-s_2)}\nonumber \\=: & {} \frac{\zeta ^3(s_1)x^{s_1+2}}{s_1}\times g_3(s_1,s_2), \end{aligned}$$

(5.60)

where

$$\begin{aligned} g_3(s_1,s_2):= \zeta ^2(s_2)\zeta ^2(2-s_2)\zeta (s_1+s_2-1) \zeta (1+s_1-s_2)\frac{H(s_1,s_2,2-s_2) }{ s_2(2-s_2)}. \end{aligned}$$

5.4.1 Integration over $s_2$

By the condition (3.3) of Proposition 3.1 it is easy to see that $H(s_1,s_2,2-s_2)$ is analytic when $\sigma _1$ and $\sigma _2$ satisfy

$$\begin{aligned} \sigma _1>1/2,\ \ \max (1/3, 2-2\sigma _1,1-\sigma _1/2)<\sigma _2<\min (5/3,2\sigma _1,1+\sigma _1/2). \end{aligned}$$

(5.61)

We write

$$\begin{aligned} I_2(x,T)=\frac{1}{2\pi i}\int _{b_1-iT}^{b_1+iT}\frac{\zeta ^3(s_1)x^{s_1+2}}{s_1}\left( \frac{1}{2\pi i}\int _{b_2-iT_2}^{b_2+iT_2}g_3(s_1,s_2)ds_2\right) ds_1. \end{aligned}$$

From (5.60) and (5.61) it is easy to check that

$$\begin{aligned} I_2(x,T)=\frac{1}{2\pi i}\int _{b_1-iT}^{b_1+iT}\frac{\zeta ^3(s_1)x^{s_1+2}}{s_1}\left( \frac{1}{2\pi i}\int _{b_2-i\infty }^{b_2+i\infty }g_3(s_1,s_2)ds_2\right) ds_1 +O\left( \frac{x^3\log ^9 x}{T}\right) .\nonumber \\ \end{aligned}$$

(5.62)

We move the integral line of $s_2$ from $\mathfrak {R}s_2=b_2$ to $\mathfrak {R}s_2=1-\varepsilon .$ In the range $1-\varepsilon \le \mathfrak {R}s_2\le b_2$, the function $g_3(s_1,s_2)$ has three poles, which are $s_2=1$ (of order 4), $s_2=s_1$ (simple) and $s_2=2-s_1$ (simple), respectively. By the residue theorem we have

$$\begin{aligned} \frac{1}{2\pi i}\int _{b_2-i\infty }^{b_2+i\infty }g_3(s_1,s_2)ds_2=\sum _{j=3}^6G_j(s_1), \end{aligned}$$

(5.63)

where

$$\begin{aligned}&G_3(s_1)=\text {Res}_{s_2=s_1}g_3(s_1,s_2),\ \ \ G_4(s_1)=\text {Res}_{s_2=2-s_1}g_3(s_1,s_2),\nonumber \\&G_5(s_1)=\text {Res}_{s_2=1}g_3(s_1,s_2),\ \ G_6(s_1)=\frac{1}{2\pi i}\int _{1-\varepsilon -i\infty }^{1-\varepsilon +i\infty }g_3(s_1,s_2)ds_2. \end{aligned}$$

(5.64)

From (5.62)-(5.64) we have

$$\begin{aligned} I_2(x,T)=\sum _{j=3}^6I_{2j}(x,T)+O\left( \frac{x^3\log ^9 x}{T}\right) , \end{aligned}$$

(5.65)

where

$$\begin{aligned} I_{2j}(x,T):=\frac{1}{2\pi i}\int _{b_1-iT}^{b_1+iT}\frac{\zeta ^3(s_1)x^{s_1+2}}{s_1}G_j(s_1)ds_1 \ \ (j=3, 4, 5, 6). \end{aligned}$$

(5.66)

5.4.2 Integration over $s_1$

We first consider $I_{23}(x,T).$ Since $s_2=s_1$ is a simple pole of $g_3(s_1,s_2)$, we have

$$\begin{aligned} G_3(s_1)=-\zeta ^2(s_1)\zeta ^2(2-s_1)\zeta (2s_1-1)\frac{H(s_1,s_1,2-s_1)}{s_1(2-s_1)}. \end{aligned}$$

(5.67)

From (3.3) we see that $H(s_1,s_1,2-s_1)$ can be written as a Dirichlet series, which is absolutely convergent for $2/3<\mathfrak {R}s_1<5/3.$ Hence we have $H(s_1,s_1,2-s_1)\ll _\varepsilon 1$ if $2/3+\varepsilon<\mathfrak {R}s_1<5/3-\varepsilon .$ Note that $s_1=1$ is the pole of $\frac{\zeta ^3(s_1)x^{s_1+2}}{s_1}G_3(s_1)$ of order 8. By the residue theorem, Lemmas 2.3, 2.4 and 2.9 we have

$$\begin{aligned} I_{23}(x,T)= & {} \frac{1}{2\pi i}\int _{b_1-i\infty }^{b_1+i\infty }\frac{\zeta ^3(s_1)x^{s_1+2}}{s_1}G_3(s_1)ds_1+O\left( \frac{x^3\log ^{8} x}{T^2}\right) \nonumber \\= & {} x^3P_{23}(\log x)+\frac{1}{2\pi i}\int _{b_1-i\infty }^{b_1+i\infty }\frac{\zeta ^3(s_1)x^{s_1+2}}{s_1}G_3(s_1)ds_1+ O\left( \frac{x^3\log ^{8} x}{T^2}\right) \nonumber \\= & {} x^3P_{23}(\log x)+O\left( x^{\frac{8}{3}+\varepsilon }+\frac{x^3\log ^{8} x}{T^2}\right) , \end{aligned}$$

(5.68)

where $P_{23}(u)$ is a polynomial in u of degree 7. Similarly we have

$$\begin{aligned} I_{24}(x,T) =x^3P_{24}(\log x)+O\left( x^{\frac{8}{3}+\varepsilon }+\frac{x^3\log ^{8} x}{T^2}\right) \end{aligned}$$

(5.69)

if noting that $G_4(s_1)=\zeta ^2(s_1)\zeta ^2(2-s_1) \zeta (2s_1-1)\frac{H(s_1,2-s_1,s_1)}{s_1(2-s_1)},$ where $P_{24}(u)$ is also a polynomial in u of degree 7.

Now we consider $I_{25}(x,T).$ We compute $G_5(s_1)$ first. Let

$$\begin{aligned}&{\varvec{M}}_1(s_2):=\zeta ^2(s_2)\zeta ^2(2-s_2),\ \ {\varvec{H}}_{s_1}(s_2):=\frac{H(s_1,s_2,2-s_2) }{ s_2(2-s_2)}.\\&{\varvec{M}}_2(s_2):=\zeta (s_1+s_2-1) \zeta (1+s_1-s_2){\varvec{H}}_{s_1}(s_2). \end{aligned}$$

Since $s_2=1$ is the pole of ${\varvec{M}}_1(s_2)$ with order 4,  we can write

$$\begin{aligned} {\varvec{M}}_1(s_2)=\sum _{m=-4}^\infty c_m(s_2-1)^m,\ \ c_{-4}=1. \end{aligned}$$

By Lemmas 2.9 and 2.10 we get

$$\begin{aligned} G_5(s_1)&=\sum _{m=0}^3\frac{c_{-m-1}}{m!}{\varvec{M}}_2^{(m)}(1)\\&=\sum _{m=0}^3\frac{c_{-m-1}}{m!}\sum _{\begin{array}{c} n_1+n_2+n_3=m\\ n_1,n_2,n_3\ge 0 \end{array}}(-1)^{n_2}\zeta ^{(n_1)}(s_1) \zeta ^{(n_2)}(s_1){\varvec{H}}_{s_1}^{(n_3)}(1). \end{aligned}$$

So we have

$$\begin{aligned} I_{25}(x,T)&=\sum _{m=0}^3\frac{c_{-m-1}}{m!}\sum _{\begin{array}{c} n_1+n_2+n_3=m\\ n_1,n_2,n_3\ge 0 \end{array}}(-1)^{n_2}\nonumber \\&\times \frac{1}{2\pi i}\int _{b_1-iT}^{b_1+iT}\frac{\zeta ^3(s_1)x^{s_1+2} \zeta ^{(n_1)}(s_1) \zeta ^{(n_2)}(s_1)}{s_1}{\varvec{H}}_{s_1}^{(n_3)}(1)ds_1. \end{aligned}$$

(5.70)

Suppose $\mathfrak {R}s_1=b_1,|\sigma _2-1|<1/1000.$ Then $s_1,s_2,2-s_2$ satisfy the condition (3.3). Thus

$$\begin{aligned} H(s_1,s_2,2-s_2)=\sum _{n_1,n_2,n_3=1}^\infty \frac{h(n_1,n_2,n_3)}{n_1^{s_1} n_2^{s_2}n_3^{2-s_2}}= \sum _{n_1,n_2,n_3=1}^\infty \frac{h(n_1,n_2,n_3)}{n_1^{s_1} n_3^{2}}(\frac{n_3}{n_2})^{s_2}. \end{aligned}$$

Hence for any $\ell \ge 0$ we have

$$\begin{aligned} \frac{\partial ^{\ell } H(s_1,s_2,2-s_2)}{\partial s_2^{\ell }} =\sum _{n_1,n_2,n_3=1}^\infty \frac{h(n_1,n_2,n_3)}{n_1^{s_1} n_3^{2}}(\frac{n_3}{n_2})^{s_2}\log ^{\ell } \frac{n_3}{n_2}, \end{aligned}$$

which implies that

$$\begin{aligned} \frac{\partial ^{\ell } H(s_1,s_2,2-s_2)}{\partial s_2^{\ell }}|_{s_2=1} =\sum _{n_1,n_2,n_3=1}^\infty \frac{h(n_1,n_2,n_3)(\log n_3-\log n_2)^{\ell }}{n_1^{s_1} n_2 n_3}. \end{aligned}$$

The condition (3.3) implies that the above infinite series is absolutely convergent for $\sigma _1>1/2.$ Thus we get that for any fixed $n_3\ge 0$

$$\begin{aligned} {\varvec{H}}_{s_1}^{(n_3)}(1)\ll _\varepsilon 1,\ \ \sigma _1\ge 1/2+\varepsilon . \end{aligned}$$

(5.71)

The integral function $\frac{\zeta ^3(s_1)x^{s_1+2} \zeta ^{(n_1)}(s_1) \zeta ^{(n_2)}(s_1) }{s_1}{\varvec{H}}_{s_1}^{(n_3)}(1)$ has a pole $s_1=1$ of order $5+n_1+n_2.$ It is easy to see the highest order is 8,  which corresponds to $n_1+n_2=3.$ Consider the domain formed by the four points $s_1=b_1\pm iT, 11/20\pm iT.$ So similar to (5.39), by the residue theorem, Lemmas 2.3, 2.4, 2.7 and 2.9 we get

$$\begin{aligned} I_{25}(x,T)=x^3P_{25}(\log x)+O\left( x^{\frac{51}{20}+\varepsilon }+\frac{x^3\log ^{8} x}{T}\right) , \end{aligned}$$

(5.72)

where $P_{25}(u)$ is a polynomial in u of degree 7.

Finally we consider $I_{26}(x,T).$ From the condition (3.3) we see that if $\mathfrak {R}s_2=1-\varepsilon$, then the function ${\varvec{H}}_{s_1}(s_2)$ is analytic for $\mathfrak {R}s_1>1/2+\varepsilon /2$ and hence is bounded if $\mathfrak {R}s_1>1/2+\varepsilon .$ Later we always suppose $\mathfrak {R}s_1>1/2+\varepsilon .$ We need two estimates of $G_6(s_1).$

Since $\mathfrak {R}s_2=1-\varepsilon ,$ we have $\zeta ^2(2-s_2)\ll 1$ and $H(s_1,s_2,2-s_2)\ll 1.$ By Lemma 2.4 we have

$$\begin{aligned} G_6(s_1)\ll & {} \int _{1-\varepsilon -i\infty }^{1-\varepsilon +i\infty } \left| \frac{ \zeta ^2(s_2) \zeta (s_1+s_2-1) \zeta (1+s_1-s_2) }{ s_2(2-s_2)}\right| dt_2\nonumber \\\ll & {} \int _{-\infty }^{ \infty } \left| \frac{ (|t_2|+1)^\varepsilon (|t_1+t_2|+1)^{\frac{1-\sigma _1}{3}+\varepsilon } (|t_1-t_2|+1)^{\frac{1-\sigma _1}{3}+\varepsilon } }{ (|t_2|+1)^2}\right| dt_2\nonumber \\\ll & {} (|t_1|+1)^{\frac{2(1-\sigma _1)}{3}+\varepsilon }. \end{aligned}$$

(5.73)

The second estimate involving $G_6(s_1)$ is the following:

$$\begin{aligned} \int _1^U |G_6(11/20+it_1) |^{\frac{5}{2}}dt_1\ll U^{1+\varepsilon }. \end{aligned}$$

(5.74)

In order to prove (5.74) suppose $s_1=11/20+it_1$ and $s_2=1-\varepsilon +it_2.$ Then $s_1+s_2-1=11/20-\varepsilon +i(t_1+t_2)$ and $1+s_1-s_2=11/20+\varepsilon +i(t_1-t_2).$ By Hölder’s inequality we get

$$\begin{aligned} G_6(s_1)\ll & {} \int _{-\infty }^{\infty }\frac{|\zeta ^2(s_2)\zeta (s_1+s_2-1)\zeta (1+s_1-s_2)|}{(|t_2|+1)^{\frac{2}{5}+\frac{4}{5}+\frac{4}{5}}}dt_2\nonumber \\\ll & {} \left( \int _{-\infty }^\infty \frac{|\zeta (1-\varepsilon +it_2)|^{10}}{(|t_2|+1)^{2}}dt_2\right) ^{\frac{1}{5}} \times \left( \int _{-\infty }^\infty \frac{|\zeta (11/20-\varepsilon +i(t_1+t_2) )|^{\frac{5}{2}}}{(|t_2|+1)^{2}}dt_2\right) ^{\frac{2}{5}}\nonumber \\&\times \left( \int _{-\infty }^\infty \frac{|\zeta (11/20+\varepsilon +i(t_1-t_2) )|^{\frac{5}{2}}}{(|t_2|+1)^{2}}dt_2\right) ^{\frac{2}{5}}\nonumber \\\ll & {} K_1^{\frac{2}{5}} (t_1)K_2^{\frac{2}{5}}(t_1) \end{aligned}$$

if noting that (by Lemma 2.5)

$$\begin{aligned} \int _{-\infty }^\infty \frac{|\zeta (1-\varepsilon +it_2) |^{10}}{(|t_2|+1)^{2}}dt_2\ll 1, \end{aligned}$$

where

$$\begin{aligned} K_1(t_1):= & {} \int _{-\infty }^\infty \frac{|\zeta (11/20-\varepsilon +i(t_1+t_2) )|^{\frac{5}{2}}}{(|t_2|+1)^{2}}dt_2,\\ K_2(t_1):= & {} \int _{-\infty }^\infty \frac{|\zeta (11/20+\varepsilon +i(t_1-t_2) )|^{\frac{5}{2}}}{(|t_2|+1)^{2}}dt_2. \end{aligned}$$

Hence we have by Cauchy’s inequality that

$$\begin{aligned} \int _1^U|G_6(11/20+it_1)|^{\frac{5}{2}}dt_1\ll & {} \int _1^U K_1(t_1)K_2(t_1) dt_1\nonumber \\\ll & {} \left( \int _1^U K_1^2(t_1) dt_1\right) ^{1/2} \left( \int _1^U K_2^2(t_1) dt_1\right) ^{1/2}. \end{aligned}$$

(5.75)

By Cauchy’s inequality again we have

$$\begin{aligned} K_1^2(t_1)= & {} \left( \int _{-\infty }^\infty \frac{|\zeta (11/20-\varepsilon +i(t_1+t_2) )|^{\frac{5}{2}}}{(|t_2|+1)^{\frac{5}{4}+\frac{3}{4}}}dt_2\right) ^2\nonumber \\\le & {} \int _{-\infty }^\infty \frac{|\zeta (11/20-\varepsilon +i(t_1+t_2) )|^{5}}{(|t_2|+1)^{\frac{5}{2}}}dt_2 \int _{-\infty }^\infty \frac{1}{(|t_2|+1)^{ \frac{3}{2}}}dt_2\nonumber \\\ll & {} \int _{-\infty }^\infty \frac{|\zeta (11/20-\varepsilon +i(t_1+t_2) )|^{5}}{(|t_2|+1)^{\frac{5}{2}}}dt_2. \end{aligned}$$

(5.76)

Hence by Lemma 2.7 we have

$$\begin{aligned} \int _1^U K_1^2(t_1)dt_1\ll & {} \int _{-\infty }^\infty \frac{1}{(|t_2|+1)^{\frac{5}{2}}}dt_2 \int _1^U |\zeta (11/20-\varepsilon +i(t_1+t_2) )|^{5} dt_1\nonumber \\= & {} \int _{-\infty }^\infty \frac{1}{(|t_2|+1)^{\frac{5}{2}}}dt_2 \int _{1+t_2}^{U+t_2} |\zeta (11/20-\varepsilon +iu )|^{5} du\nonumber \\\ll & {} \int _{-\infty }^\infty \frac{|U+t_2|^{1+\varepsilon } +|t_2+1|^{1+\varepsilon }}{(|t_2|+1)^{\frac{5}{2}}}dt_2\nonumber \\\ll & {} U^{1+\varepsilon }. \end{aligned}$$

(5.77)

Similarly we have

$$\begin{aligned} \int _1^U K_2^2(t_1)dt_1 \ll U^{1+\varepsilon }. \end{aligned}$$

(5.78)

The estimate (5.74) follows from (5.75) and (5.78).

Consider the domain formed by the four points $s_1=b_1\pm iT, 11/20\pm iT.$ So by the residue theorem, we get

$$\begin{aligned} I_{26}(x,T)= & {} \text {Res}_{s_1=1}\frac{\zeta ^3(s_1)x^{s_1+2}}{s_1}G_6(s_1)\nonumber \\&+\,H_{261}(x,T)+H_{262}(x,T)-H_{263}(x,T), \end{aligned}$$

(5.79)

where

$$\begin{aligned} H_{261}(x,T):= & {} \frac{1}{2\pi i}\int _{11/20+iT}^{b_1+iT}\frac{\zeta ^3(s_1)x^{s_1+2}}{s_1}G_6(s_1)ds_1,\\ H_{262}(x,T):= & {} \frac{1}{2\pi i}\int _{11/20-iT}^{11/20+iT}\frac{\zeta ^3(s_1)x^{s_1+2}}{s_1}G_6(s_1)ds_1,\\ H_{263}(x,T):= & {} \frac{1}{2\pi i}\int _{11/20-iT}^{b_1-iT}\frac{\zeta ^3(s_1)x^{s_1+2}}{s_1}G_6(s_1)ds_1. \end{aligned}$$

From Lemma 2.4 and (5.73) we get

$$\begin{aligned} H_{261}(x,T)\ll \frac{x^{3+\varepsilon }}{T},\ \ H_{263}(x,T)\ll \frac{x^{3+\varepsilon }}{T}. \end{aligned}$$

(5.80)

By Lemma 2.7 and (5.74) we get via Hölder’s inequality that

$$\begin{aligned} \int _1^U |\zeta ^3(11/20+it_1)G_6(11/20+it_1)|dt_1\ll U^{1+\varepsilon }, \end{aligned}$$

which implies that

$$\begin{aligned} H_{262}(x,T)\ll x^{51/20+\varepsilon }. \end{aligned}$$

(5.81)

From (5.79) to (5.81) and Lemma 2.9 we get that

$$\begin{aligned} I_{26}(x,T)=x^3P_{26}(\log x)+O\left( x^{\frac{51}{20}+\varepsilon }+\frac{x^{3+\varepsilon }}{T}\right) , \end{aligned}$$

(5.82)

where $P_{26}(u)$ is a polynomial in u of degree 2.

From (5.65), (5.68), (5.69), (5.72) and (5.82) we get

$$\begin{aligned} I_{2}(x,T)=x^3P_{2}(\log x)+O\left( x^{\frac{8}{3}+\varepsilon }+\frac{x^{3+\varepsilon }}{T}\right) , \end{aligned}$$

(5.83)

where $P_{2}(u)$ is a polynomial in u of degree 7.

5.5 Completion of the proof

First suppose $10\le x\notin {\mathbb {N}}.$ Suppose T ia a large parameter such that $5\le T\le x/2.$ Combining (5.1), (5.3), (5.5), (5.6), (5.14), (5.21), (5.47), (5.59), (5.83) and Proposition 4.1 we get by taking $T=x^{1/2}$,

$$\begin{aligned} C_3(x)= & {} x^3P(\log x)+O\left( x^{\frac{8}{3}+\varepsilon }+\frac{x^{3+\varepsilon }}{T}\right) \nonumber \\= & {} x^3P(\log x)+O\left( x^{\frac{8}{3}+\varepsilon } \right) . \end{aligned}$$

(5.84)

where P(u) is a polynomial in u of degree 7. Hence our theorem is true when $x\notin {\mathbb {N}}.$

Now suppose $10\le x\in {\mathbb {N}}.$ Then we have by (5.84) that

$$\begin{aligned} C_3(x)= & {} C_3(x+1/2)\nonumber \\= & {} (x+1/2)^3P(\log (x+1/2))+O\left( x^{\frac{8}{3}+\varepsilon } \right) \nonumber \\= & {} x^3P(\log x)+O\left( x^{\frac{8}{3}+\varepsilon } \right) . \end{aligned}$$

(5.85)