On sums of sums involving cube-full numbers

Abstract

Let $f_{3}$ denote the characteristic function of cube-full numbers, and let (n, q) be the greatest common divisor of positive integers n and q. For any positive real numbers x and y, we shall consider several asymptotic formulas for sums of sums of modified cube-full numbers, which is $\sum _{n\le y}\left( \sum _{q\le x}\sum _{d|(n,q)}df_{3}\left( {q}/{d}\right) \right) ^{k}$ with $k=1, 2$.

1 Introduction

For any integer $r (\ge 2)$, we call n an r-full integer if $p|n \Rightarrow p^{r}|n$ and call n an r-free integer if $p|n \Rightarrow p^{r}\not \mid n$, where the letter p denotes a prime number. If $r=2$ or $r=3$, we use the terms square-full or cube-full. Let G(r) denote the set of r-full numbers, then we set

$$\begin{aligned} f_{r}(n)&:=\left\{ \begin{array}{cl} 1 &{} \ \ \mathrm{{if}}\ \ n\in G(r), \ \\ 0 &{} \ \ \mathrm{{if}}\ \ n\notin G(r). \ \end{array} \right. \end{aligned}$$

Let $s=\sigma +it$ be the complex variable, and let $\zeta (s)$ be the Riemann zeta-function. Denote the Dirichlet series $F_{r}(s)$ defined by $F_{r}(s) := \sum _{n=1}^{\infty }\frac{f_{r}(n)}{n^s}.$ Following (7.3) in Krätzel [7], the representation of $F_{r}(s)$ is more complicated for $k\ge 3$, and it is known that

$$\begin{aligned} F_{r}(s) = \prod _{\ell =r}^{2r-1}\frac{\zeta (\ell s)}{\zeta ((2r+2)s)}~\kappa _{2r+3}(s) \end{aligned}$$

(1.1)

holds. Here, $c_{2r+3}(n)$ denotes a certain arithmetical function whose associated Dirichlet series $\kappa _{2r+3}(s) := \sum _{n=1}^{\infty }\frac{c_{2r+3}(n)}{n^s},$ which is absolutely convergent for $\mathrm{Re}~s >\frac{1}{2r+3}$. We define a sum over the r-full numbers by

$$\begin{aligned} s_{q}^{(r)}(n)&:= \sum _{d|(n,q)}df_{r}\left( \frac{q}{d}\right) , \end{aligned}$$

(1.2)

where (n, q) denotes the greatest common divisor of integers n and q. For any large positive real numbers x and y, we set the double sums

$$\begin{aligned} S_{k}^{(r)}(x,y)&:= \sum _{n\le y}\left( \sum _{q\le x} {s}_{q}^{(r)}(n)\right) ^{k} \qquad (k=1, 2). \end{aligned}$$

(1.3)

For $r=2$, Kiuchi [6] considered the asymptotic formula for the double sum (1.3) concerning square-full numbers, and used the theory of exponent pairs to derive the precise asymptotic formula

$$\begin{aligned} S_{1}^{(2)}(x,y)&=\frac{\zeta (2)\zeta (3)}{\zeta (6)} xy - \frac{\zeta (4)\zeta (6)}{4\zeta (12)} x^{2} + O\left( x^{\frac{1}{2}} y + x^{}y^{\frac{1}{3}} + \frac{x^3}{y} \right) , \end{aligned}$$

(1.4)

where x and y are large real numbers such that $x \ll y \ll x^{\frac{3}{2}}$. When $k=2$, Kiuchi [6] also showed that

$$\begin{aligned} S_{2}^{(2)}(x,y)=\frac{\zeta (2)^{}\zeta ^{2}(3)}{\zeta ^{2}(6)} x^{2} y \log x + O\left( x^{2} y + x^4 \right) \end{aligned}$$

(1.5)

holds, where x and y are large real numbers such that $y \gg \frac{x^2}{\log x}$. Moreover, he used analytic properties of the Riemann zeta-function to obtain the asymptotic formula

$$\begin{aligned} S_{2}^{(2)}(x,y)&= \frac{\zeta (2)^{}\zeta ^{2}(3)}{\zeta ^{2}(6)} x^{2} y \log \frac{x^3}{y} + c_{0} x^{2} y \nonumber \\&\quad + \frac{\zeta (2)\zeta ^{2}(3)}{\zeta ^{2}(6)} \left( 2\gamma -2 + 5 \frac{\zeta ^{'}(2)}{\zeta (2)} + 9 \frac{\zeta ^{'}(3)}{\zeta (3)} - 18 \frac{\zeta ^{'}(6)}{\zeta (6)} \right) x^{2} y \nonumber \\&\quad +O\left( x^2y\left( L^{5}x^{-\frac{1}{14}} + L^{6}y^{-\frac{1}{2}} + L^{2}\left( \frac{x}{y} \right) ^{\frac{1}{2}} + L^2\left( \frac{y}{x^2}\right) ^{\frac{1}{2}}\right) \right) , \end{aligned}$$

(1.6)

where $c_0$ is a computable constant, and x and y are large real numbers such that $x \log ^{4}x \ll y \ll \frac{x^2}{\log ^{6}x}$. To prove the precise asymptotic formulas (1.4), (1.5) and (1.6), we used the method of proofs of Chan and Kumchev [2] (see also Kiuchi, Minamide and Tanigawa [5], Kühn and Robles [8], Robles [10], Robles and Roy [11]).

For $r=3$, it is derived from (1.1) that the Dirichlet series for the generating function $f_{3}(n)$ is

$$\begin{aligned} F(s) := \sum _{n=1}^{\infty }\frac{f_{3}(n)}{n^s} = \frac{\zeta (3s)\zeta (4s)\zeta (5s)\kappa (s)}{\zeta (8s)} \end{aligned}$$

(1.7)

for $\mathrm{{Re}}~s > \frac{1}{3}$, where $\kappa (s)$ is the Dirichlet series generated by a certain arithmetical function $c_{9}(n)$ (see (7.3) in Krätzel [7]), that is $\kappa (s):=\sum _{n=1}^{\infty }\frac{c_{9}(n)}{n^s}$ which is absolutely convergent for $\mathrm{Re}~s > \frac{1}{9}$. Moreover, the asymptotic formula for the sum of $f_{3}(n)$ is also known, and one can see that

$$\begin{aligned} \sum _{n\le x}f_{3}(n)&= \frac{\zeta \left( \frac{4}{3}\right) \zeta \left( \frac{5}{3}\right) \kappa \left( \frac{1}{3}\right) }{\zeta \left( \frac{8}{3}\right) }x^{\frac{1}{3}} + \frac{\zeta \left( \frac{3}{4}\right) \zeta \left( \frac{5}{4}\right) \kappa \left( \frac{1}{4}\right) }{\zeta \left( 2\right) }x^{\frac{1}{4}} \nonumber \\&\quad + \frac{\zeta \left( \frac{3}{5}\right) \zeta \left( \frac{4}{5}\right) \kappa \left( \frac{1}{5}\right) }{\zeta \left( \frac{8}{5}\right) }x^{\frac{1}{5}} + \Delta (x) \end{aligned}$$

(1.8)

holds with the error term $\Delta (x) = O\left( x^{\frac{1}{8}}\log ^{4}x\right)$ for any large positive real number x (see section 7.1.3 in Krätzel [7]). In 1988, Balasubramanian et al. [1] showed that $\Delta (x)=\Omega \left( x^{\frac{1}{12}}\sqrt{\log x}\right)$ holds, and the improvement on the estimate of $\Delta (x)$ has been studied by many authors. Under the Riemann hypothesis, Wu [14] obtained that $\Delta (x)=O\left( x^{\frac{97}{804}+\varepsilon }\right)$ holds for any $\varepsilon >0$. Using (1.7), the Dirichlet series generated by the coefficients $s_{q}(n)$ is expressed by

$$\begin{aligned} \sum _{q=1}^{\infty }\frac{s_{q}^{(3)}(n)}{q^s} = \sigma _{1-s}(n)\frac{\zeta (3s)\zeta (4s)\zeta (5s)\kappa (s) }{\zeta (8s)} \end{aligned}$$

(1.9)

for ${\mathrm{{Re}}~{s} > \frac{1}{3}}$, where $\sigma _{1-s}(n)=\sum _{d|n}d^{1-s}$ is the generalized divisor function.

Now, we shall consider several asymptotic formulas of (1.3) concerning cube-full numbers. Our theorems are proved by the same way as in [6], and we shall deduce several interesting formulas for the double sum $S_{k}^{(3)}(x,y)$. We use the theory of exponent pairs and elementary methods to deal with $S_{1}^{(3)}(x,y)$. Then the case $k=1$ implies the following theorem, namely

Theorem 1

Let x and y be large real numbers such that $x \ll y \ll x^{\frac{5}{3}}$. Then we have

$$\begin{aligned} S_{1}^{(3)}(x,y)&=\frac{\zeta (3)\zeta (4)\zeta (5)\kappa (1) }{\zeta (8)} xy - \frac{\zeta (6)\zeta (8)\zeta (10)\kappa (2)}{4\zeta (16)} x^{2} \nonumber \\&\qquad + O\left( x^{\frac{1}{3}} y + x^{}y^{\frac{1}{3}} + \frac{x^3}{y} \right) . \end{aligned}$$

(1.10)

It follows from (1.10) that

$$\begin{aligned} \frac{1}{xy}\sum _{n\le y}\sum _{q\le x}s_{q}^{(3)}(n)&= \frac{\zeta (3)\zeta (4)\zeta (5)\kappa (1)}{\zeta (8)} - \frac{\zeta (6)\zeta (8)\zeta (10) \kappa (2)}{4\zeta (16)} \frac{x}{y} \\&\quad + O\left( x^{-\frac{2}{3}} + y^{-\frac{2}{3}} + \frac{x^2}{y^2} \right) \end{aligned}$$

holds. This is described by saying that the average order of $s_{q}^{(3)}(n)$ is $\frac{\zeta (3)\zeta (4)\zeta (5)\kappa (1)}{\zeta (8)}$ under q and n satisfying the condition $q^{} \ll n \ll q^{\frac{5}{3}}.$

Remark 1.1

It would be an interesting problem to investigate the asymptotic behaviour of $S_{1}^{(3)}(x,y)$ under the condition $y \ll x^{}$. However, this would require a different method.

For $k=2$, there are two quite different methods to deal with this function $S_{2}^{(3)}(x,y)$. We utilize an elementary lattice point counting argument to obtain the formula (1.11) below, and use the generating Dirichlet series and the properties of the Riemann zeta-function to prove (1.12) below, which we state as

Theorem 2

Let x and y be large real numbers such that $y \gg \frac{x^2}{\log x}$. Then we have

$$\begin{aligned} S_{2}^{(3)}(x,y)= \frac{\zeta ^{2}(3)\zeta ^{2}(4)\zeta ^{2}(5)\kappa ^{2}(1)}{\zeta (2)\zeta ^{2}(8)}x^{2}y \log x + O\left( x^{2}y +x^{4}\right) . \end{aligned}$$

(1.11)

Similarly, as in Theorem 1, we use (1.11) to get

$$\begin{aligned} \frac{1}{y}\sum _{n\le y}\left( \sum _{q\le x}s_{q}^{(3)}(n)\right) ^2 = \frac{\zeta ^{2}(3)\zeta ^{2}(4)\zeta ^{2}(5)\kappa ^{2}(1)}{\zeta (2)\zeta ^{2}(8)} x^2 \log x + O\left( x^2 + \frac{x^{4}}{y}\right) . \end{aligned}$$

This is described by saying that the average order of $s_{q}^{(3)}(n)$ is

$$\begin{aligned} \frac{\zeta ^{}(3)\zeta ^{}(4)\zeta ^{}(5)\kappa ^{}(1)}{\sqrt{\zeta (2)}\zeta ^{}(8)} \sqrt{\log q} \end{aligned}$$

under q and n satisfying the condition $n \gg \frac{q^2}{\log q}.$ We utilize the generating Dirichlet series and the properties of the Riemann zeta-function to prove (1.12) below, which we state as

Theorem 3

Let x and y be large real numbers such that $x \log ^{6}x \ll y \ll \frac{x^2}{\log ^{4}x}$. Then we have

$$\begin{aligned} S_{2}^{(3)}(x,y)&=\frac{\zeta ^{2}(3)\zeta ^{2}(4)\zeta ^{2}(5)\kappa ^{2}(1) }{\zeta (2)\zeta ^{2}(8)} \left( \log \frac{x^3}{y} + c_{1}\right) x^{2}y + \eta x^{2} y \nonumber \\&\quad +O\left( x^2yL^{2}\left( L^{3}x^{-\frac{1}{3}} + L^{4}y^{-\frac{1}{2}} + L\left( \frac{x}{y} \right) ^{\frac{1}{2}} + \left( \frac{y}{x^2}\right) ^{\frac{1}{2}}\right) \right) , \end{aligned}$$

(1.12)

where $\eta$ is a computable constant, which is defined by (5.9) below, and the constant $c_{1}$ is given by

$$\begin{aligned} c_{1}= 2\gamma -2 + 9\frac{\zeta ^{'}(3)}{\zeta (3)} + 12\frac{\zeta ^{'}(4)}{\zeta (4)} + 15\frac{\zeta ^{'}(5)}{\zeta (5)} - 24\frac{\zeta ^{'}(8)}{\zeta (8)} - \frac{\zeta ^{'}(2)}{\zeta (2)} + 3\frac{\kappa '(1)}{\kappa (1)}. \end{aligned}$$

Remark 1.2

It would be an interesting problem to investigate the asymptotic behaviour of $S_{2}^{(3)}(x,y)$ under the condition $y \ll x \log ^{6}x$. However, this would require a different method.

2 Some lemmas

To prove our theorems, we first prepare several lemmas. Let $\psi (x)=x-[x]-\frac{1}{2}$ denote the first periodic Bernoulli function. In the proof of Theorem 1, we need an upper bound of the sum

$$\begin{aligned} \sum _{n\in I}\psi \left( \frac{y}{n}\right) . \end{aligned}$$

An efficient way to estimate these $\psi$-sums is to apply the theory of exponent pairs: An exponent pair $(\kappa , \lambda )$ is a pair of numbers $0 \le \kappa \le \frac{1}{2} \le \lambda \le 1$ such that

$$\begin{aligned} \sum _{n \in I} e^{2\pi if(n)} \ll A^{\kappa }N^{\lambda } \end{aligned}$$

holds, where $I \subset (N,2N]$ and $A \ll |f'(u)| \ll A$ for $u \in I$. For the precise definition and its properties, the reader should consult Graham and Kolesnik [3] and Ivić [4]. Now applying Lemma 4.3 in [3] with $f(n)=\frac{y}{n}$, we have

Lemma 2.1

Let $(\kappa ,\lambda )$ be an exponent pair. If I is a subinterval in (N, 2N], we have

$$\begin{aligned} \sum _{n\in I}\psi \left( \frac{y}{n}\right) \ll y^{\frac{\kappa }{\kappa +1}}N^{\frac{\lambda -\kappa }{\kappa +1}} + \frac{N^{2}}{y}. \end{aligned}$$

In particular, if we take the exponent pair $(\kappa , \lambda )=\left( \frac{1}{2}, \frac{1}{2}\right)$, we get

$$\begin{aligned} \sum _{n\in I}\psi \left( \frac{y}{n}\right) \ll y^{\frac{1}{3}}+\frac{N^{2}}{y}. \end{aligned}$$

(2.1)

The proofs of Theorem 3 need the following lemmas, namely

Lemma 2.2

Suppose that the Dirichlet series $\alpha (s):=\sum _{n=1}^{\infty }\frac{a_n}{n^s}$ absolutely converges for $\mathrm{Re}~s >\sigma _{a}$. If $\sigma _{0}>\max (0,\sigma _{a})$ and $x>0,\ T>0$, then

$$\begin{aligned} \sum _{n\le x}{}^{'}a_{n}&= \frac{1}{2\pi i} \int _{\sigma _{0}-iT}^{\sigma _{0}+iT}\alpha (s)\frac{x^s}{s}ds + R, \end{aligned}$$

where

$$\begin{aligned} R&\ll \sum _{\begin{array}{c} {\frac{x}{2}<n<2x} \\ {n\ne x} \end{array}} |a_n|\min \left( 1,\frac{x}{T|x-n|}\right) + \frac{(4x)^{\sigma _{0}}}{T}\sum _{n=1}^{\infty }\frac{|a_{n}|}{n^{\sigma _{0}}}, \end{aligned}$$

and $\sum {}^{'}$ indicates that the last term is to be halved if x is an integer.

Proof

This is Perron’s famous formula (see Theorem 5.2 and Corollary 5.3 in Montgomery and Vaughan [9]). $\square$

Lemma 2.3

Let $G(s_1,s_2;y)$ be a sum function defined by

$$\begin{aligned} G(s_{1},s_{2};y)&=\sum _{n\le y}{}^{'}\sigma _{1-s_1}(n)\sigma _{1-s_2}(n) \end{aligned}$$

(2.2)

and $L=\log y$. Then we have

$$\begin{aligned} G(s_{1},s_{2};y)=\sum _{j=1}^{4}R_{j}(s_{1},s_{2};y) + O\left( y L^{6}\left( y^{-\frac{1}{2}}+\frac{1}{T}\right) \right) \end{aligned}$$

(2.3)

for $\mathrm{{Re}}~s_{j} \ge 1/2$ and $|\mathrm{{Im}}~s_{j}| \le T \ (j=1,2)$, where

$$\begin{aligned} R_{1}(s_{1},s_{2};y)&=y \frac{\zeta (s_1)\zeta (s_2)\zeta (s_1 +s_2 -1)}{\zeta (s_1 +s_2)}, \\ R_{2}(s_{1},s_{2};y)&=y^{2-s_1} \frac{\zeta (2-s_1)\zeta (1-s_1 +s_2)\zeta (s_2)}{(2-s_1)\zeta (2-s_1 +s_2)}, \\ R_{3}(s_{1},s_{2};y)&=y^{2-s_2} \frac{\zeta (2-s_2)\zeta (1+s_1 -s_2)\zeta (s_1)}{(2-s_2)\zeta (2+s_1 -s_2)}, \\ R_{4}(s_{1},s_{2};y)&=y^{3-s_1 -s_2} \frac{\zeta (3-s_1 -s_2)\zeta (2-s_2)\zeta (2-s_1)}{(3-s_1 -s_2)\zeta (4-s_1 -s_2)}, \end{aligned}$$

where $\sum {}^{'}$ indicates that the last term is to be halved if y is an integer.

Proof

The proof of this lemma follows from (4.12) in Chan and Kumchev [2]. $\square$

Lemma 2.4

For $t\ge t_{0} >0$ uniformly in $\sigma$, we have

$$\begin{aligned} \zeta (\sigma +it)&=\left\{ \begin{array}{ll} \displaystyle t^{\frac{1}{6} (3-4\sigma )}\log t &{} \ \ \left( 0\le \sigma \le \frac{1}{2}\right) , \ \\ \displaystyle t^{\frac{1}{3} (1-\sigma )}\log t &{} \ \ \left( \frac{1}{2} \le \sigma \le 1\right) , \ \\ \displaystyle \log t &{} \ \ \left( 1\le \sigma < 2\right) , \ \\ \displaystyle 1 &{} \ \ \left( \sigma \ge 2\right) . \\ \end{array} \right. \end{aligned}$$

Proof

The proof of this lemma follows from Theorem II.3.8 in Tenenbaum [12], and Ivić [4]. Also see Titchmarsh [13]. $\square$

3 Proof of Theorem 1

We use (1.2) and (1.3) and change the order of summation to obtain

$$\begin{aligned} S_{1}^{(3)}(x,y)&=\sum _{n\le y}\sum _{q\le x} {s}_{q}^{(3)}(n) \nonumber \\&=y\sum _{dk\le x} f_{3}(k) - \frac{1}{2} \sum _{dk\le x}d f_{3}(k) - \sum _{dk\le x}d f_{3}(k) \psi \left( \frac{y}{d}\right) \nonumber \\&=:S_{1,1}^{(3)}(x,y)- S_{1,2}^{(3)}(x,y) - S_{1,3}^{(3)}(x,y). \end{aligned}$$

(3.1)

We consider the first term on the right of (3.1). We use (1.7) to get

$$\begin{aligned} \sum _{k\le x}\frac{f_{3}(k)}{k} = \frac{\zeta (3)\zeta (4)\zeta (5)\kappa (1)}{\zeta (8)}+ O\left( x^{-\frac{2}{3}}\right) . \end{aligned}$$

(3.2)

We obtain from (1.7) and the above

$$\begin{aligned} S_{1,1}^{(3)}(x,y)&= yx\sum _{k\le x}\frac{f_{3}(k)}{k} + O\left( y\sum _{k\le x}f_{3}(k)\right) \nonumber \\&= \frac{\zeta (3)\zeta (4)\zeta (5)\kappa (1)}{\zeta (8)} xy + O\left( x^{\frac{1}{3}}y\right) . \end{aligned}$$

(3.3)

Similarly, we have

$$\begin{aligned} S_{1,2}^{(3)}(x,y)&=\frac{1}{2} \sum _{k\le x} f_{3}(k) \left( \frac{x^2}{2k^2}+O\left( \frac{x}{k}\right) \right) \nonumber \\&=\frac{\zeta (6)\zeta (8)\zeta (10)\kappa (2)}{4\zeta (16)}x^2 + O\left( x\right) . \end{aligned}$$

(3.4)

To estimate $S_{1,3}(x,y)$, we use the theory of exponent pairs. Let $\displaystyle N_{j}=N_{j,k}=\left( \frac{x}{k}\right) 2^{-j}$. Then we have

$$\begin{aligned} S_{1,3}^{(3)}(x,y)&=\sum _{k\le x} f_{3}(k) \sum _{d\le \frac{x}{k}}d \>\! \psi \left( \frac{y}{d}\right) \\&\ll \sum _{k\le x} f_{3}(k)\sum _{j=0}^{\infty }N_{j}\sup _{I}\left| \sum _{d\in {I}}\psi \left( \frac{y}{d}\right) \right| , \end{aligned}$$

where the $\sup$ is over all subintervals I in $(N_{j},2N_{j}]$. From (2.1) of Lemma 2.1 and (3.2), we have

$$\begin{aligned} S_{1,3}^{(3)}(x,y)&\ll \sum _{k\le x} f_{3}(k)\sum _{j=0}^{\infty }\left\{ N_{j}y^{\frac{1}{3}} + \frac{N_{j}^{3}}{y} \right\} \nonumber \\&\ll \sum _{k\le x}\frac{f_{3}(k)}{k} \cdot xy^{1/3} + \sum _{k\le x} \frac{f_{3}(k)}{k^3} \cdot \frac{ x^{3}}{y} \nonumber \\&\ll xy^{{1}/{3}} + \frac{x^{3}}{y}. \end{aligned}$$

(3.5)

Substituting (3.3), (3.4) and (3.5) into (3.1), we get the assertion of Theorem 1.

$\square$

4 Proof of Theorem 2

From (1.2) and (1.3), we have

$$\begin{aligned} S_{2}^{(3)}(x,y)&=\sum _{n\le y}\left( \sum _{\begin{array}{c} {dk\le x} \\ {d|n} \end{array}}d f_{3}(k)\right) ^{2} \nonumber \\&=\sum _{d_{1}k_{1}\le x}d_{1} f_{3}(k_{1}) \sum _{d_{2}k_{2}\le x}d_{2} f_{3}(k_{2})\sum _{\begin{array}{c} {n\le y} \\ {{d_{1}|n, d_{2}|n}} \end{array}}1 \nonumber \\&=\sum _{d_{1}k_{1}\le x}\sum _{d_{2}k_{2}\le x} d_{1}d_{2} f_{3}(k_{1})\ f_{3}(k_{2}) \left[ \frac{y}{[d_{1},d_{2}]}\right] \nonumber \\&=y \sum _{d_{1}k_{1}\le x}\sum _{d_{2}k_{2}\le x} (d_{1},d_{2}) f_{3}(k_{1})\ f_{3}(k_{2}) + O\left( E\right) , \end{aligned}$$

(4.1)

where $[d_1,d_2]$ denotes the least common multiple of $d_1$ and $d_2$. We use (1.7) to get

$$\begin{aligned} E&:= \sum _{d_{1}k_{1}\le x} \sum _{d_{2}k_{2}\le x} d_{1}d_{2} f_{3}(k_{1})f_{3}(k_{2}) \\&\ll x^{2}~ \sum _{k_{1} \le x} \frac{f_{3}(k_1)}{k_{1}^{2}} \cdot x^{2}~\sum _{k_{2} \le x} \frac{f_{3}(k_{2})}{k_{2}^{2}} \ll x^{4}. \end{aligned}$$

To evaluate the main term of (4.1), we use

$$\begin{aligned} \sum _{mk\le x}f_{3}(k)&= \frac{\zeta (3)\zeta (4)\zeta (5)\kappa (1)}{\zeta (8)} {x} + O\left( {x}^{{1}/{3}}\right) , \end{aligned}$$

(4.2)

which follows from (1.7) and (3.2). Using the Gauss identity $\sum _{d|n}\phi (d) = n,$ (4.2) and $\sum _{d\le x}\frac{\phi (d)}{d^2}=\frac{1}{\zeta (2)}\log x +O(1)$, we have

$$\begin{aligned}&\sum _{d_{1}k_{1}\le x} \sum _{d_{2}k_{2}\le x} (d_{1},d_{2}) f_{3}(k_{1})f_{3}(k_{2}) =\sum _{d\le x}\phi (d) \sum _{dl_{1}k_{1}\le x}\sum _{dl_{2}k_{2}\le x }f_{3}(k_{1})f_{3}(k_{2}) \\&\quad =\sum _{d\le x}\phi (d) \left( \sum _{mk\le x/d} f_{3}(k)\right) ^{2} \\&\quad =\frac{\zeta ^{2}(3)\zeta ^{2}(4)\zeta ^{2}(5)\kappa ^{2}(1)}{\zeta ^{2}(8)} x^{2}~\sum _{d\le x}\frac{\phi (d)}{d^2} + O\left( x^{4/3}\sum _{d \le x}\frac{\phi (d)}{d^{4/3}} \right) \\&\quad =\frac{\zeta ^{2}(3)\zeta ^{2}(4)\zeta ^{2}(5)\kappa ^{2}(1)}{\zeta (2)\zeta ^{2}(8)} x^{2}\log x + O\left( x^{2}\right) . \end{aligned}$$

Hence, we have

$$\begin{aligned} S_{2}^{(3)}(x,y)= \frac{\zeta ^{2}(3)\zeta ^{2}(4)\zeta ^{2}(5)\kappa ^{2}(1)}{\zeta (2)\zeta ^{2}(8)} x^{2}y \log x + O\left( x^{2}y +x^{4}\right) . \end{aligned}$$

This completes the proof of Theorem 2. $\square$

5 Proof of Theorem 3

In this section, we assume that $1 \le y \le x^M$ for some constant M. Without loss of generality we can assume that $x,\ y\in {\mathbb {Z}}+\frac{1}{2}$. We apply Lemma 2.2 with (1.9), then

$$\begin{aligned} \sum _{q\le x} {s}_{q}^{(3)}(n)&= \frac{1}{2\pi i} \int _{\alpha -iT}^{\alpha +iT} \sigma _{1-s}(n) \frac{\zeta (3s)\zeta (4s)\zeta (5s)\kappa (s)}{\zeta (8s)} \frac{x^s}{s}ds + E_{1}(x,n) \end{aligned}$$

(5.1)

with $\alpha = 1 +\frac{1}{\log x}$ and T being a real parameter at our disposal, where $E_{1}(x,n)$ is the error term given by

$$\begin{aligned} E_{1}(x,n)&\ll \frac{x}{T} \sum _{q=1}^{\infty }\frac{s_{q}^{(3)}(n)}{q} \ll \frac{x^{}}{T} \sigma _{0}(n) \end{aligned}$$

by using (1.9). Let $\alpha _{1}=1 + \frac{1}{\log x}$ and $\alpha _{2}=1 + \frac{2}{\log x}$. Applying (5.1) with $\alpha =\alpha _j\ (j=1,2)$  we have

$$\begin{aligned} \left( \sum _{q\le x} {s}_{q}^{(3)}(n)\right) ^{2}&= \frac{1}{(2\pi i)^2} \int _{\alpha _{1}-iT}^{\alpha _{1}+iT}\int _{\alpha _{2}-iT}^{\alpha _{2}+iT} F(s_{1},s_{2},n) ds_{2}ds_{1} + E_{2}(x,n), \end{aligned}$$

(5.2)

where

$$\begin{aligned} F(s_{1},s_{2},n)&=\sigma _{1-s_{1}}(n)\sigma _{1-s_{2}}(n)\\ {}&\quad \times \frac{\zeta (3s_1)\zeta (4s_1)\zeta (5s_1)\zeta (3s_2)\zeta (4s_2)\zeta (5s_2)\kappa (s_1)\kappa (s_2)}{\zeta (8s_{1})\zeta (8s_{2})}~ \frac{x^{s_{1}+s_{2}}}{s_{1}s_{2}} \end{aligned}$$

and

$$\begin{aligned}&E_{2}(x,n) \\&\quad = E_{1}(x,n)\left( \frac{1}{2\pi i} \int _{\alpha _{1}-iT}^{\alpha _{1}+iT}\sigma _{1-s_{1}}(n) \frac{\zeta (3s_1)\zeta (4s_{1})\zeta (5s_{1}) \kappa (s_1)}{\zeta (8s_{1})}~ \frac{x^{s_{1}}}{s_{1}}ds_{1} \right. \\&\quad \quad + \left. \frac{1}{2\pi i} \int _{\alpha _{2}-iT}^{\alpha _{2}+iT}\sigma _{1-s_{2}}(n) \frac{\zeta (3s_2)\zeta (4s_{2})\zeta (5s_{2})\kappa (s_2)}{\zeta (8s_{2})}~ \frac{x^{s_{2}}}{s_{2}}ds_{2} + E_{1}(x,n) \right) . \end{aligned}$$

It follows that

$$\begin{aligned} \displaystyle E_2(x, n) \ll \frac{x^2}{T}\sigma _{0}(n)^{2}\log T. \end{aligned}$$

Summing (5.2) over n and using the estimate $\sum _{n\le y}\sigma _{0}(n)^{2} \ll y^{}\log ^{3}y,$ we get

$$\begin{aligned} S_{2}^{(3)}(x,y)&= \frac{1}{(2\pi i)^2} \int _{\alpha _{1}-iT}^{\alpha _{1}+iT}\int _{\alpha _{2}-iT}^{\alpha _{2}+iT} G(s_{1},s_{2};y)\nonumber \\&\quad \times \frac{\zeta (3s_1)\zeta (4s_{1})\zeta (5s_{1})\kappa (s_1)\zeta (3s_2)\zeta (4s_{2})\zeta (5s_{2})\kappa (s_2) }{\zeta (8s_{1})\zeta (8s_{2})} \nonumber \\&\quad \times \frac{x^{s_{1}+s_{2}}}{s_{1}s_{2}}ds_{2}\,ds_{1} + O\left( \frac{x^{2}y^{}L^{4}}{T}\right) , \end{aligned}$$

(5.3)

where $G(s_1,s_2;y) := \sum _{n\le y}\sigma _{1-s_{1}}(n)\sigma _{1-s_{2}}(n)$ and $L=\log (Txy)$.

Now we shall evaluate the integrals in appearing in (5.3). Substituting (2.3) into (5.3), we have

$$\begin{aligned} S_{2}^{(3)}(x,y)&=\sum _{j=1}^{4}S_{2,j}^{(3)}(x,y) + O\left( x^{2} y^{}L^{8} \left( \frac{1}{T} + y^{-1/2}\right) \right) , \end{aligned}$$

(5.4)

where

$$\begin{aligned} S_{2,j}^{(3)}(x,y)&=\frac{1}{(2\pi i)^2}\int _{\alpha _{1}-iT}^{\alpha _{1}+iT}\int _{\alpha _{2}-iT}^{\alpha _{2}+iT} R_{j}(s_{1},s_{2};y) \\&\quad \times \frac{\zeta (3s_1)\zeta (4s_{1})\zeta (5s_{1})\kappa (s_1)\zeta (3s_2)\zeta (4s_{2})\zeta (5s_{2})\kappa (s_2)}{\zeta (8s_1)\zeta (8s_2)} \\&\quad \times \frac{x^{s_{1}+s_{2}}}{s_{1}s_{2}}ds_{2}\,ds_{1}. \end{aligned}$$

Note that we substitute $T=x^{}$ into the error term on the right-hand side of (5.4) to get

$$\begin{aligned}&\ll x^{2} y^{}L^{8} \left( x^{-1} + y^{-1/2}\right) . \end{aligned}$$

(5.5)

5.1 Evaluation of $S_{2,1}^{(3)}(x,y)$

Let $\alpha _{1}= 1 +\frac{1}{\log x}$ and $\alpha _{2}= 1 + \frac{2}{\log x}$. From the definition of $R_1(s_1,s_2,y)$, we get

$$\begin{aligned} S_{2,1}^{(3)}(x,y)&= \frac{y}{(2\pi i)^2}\int _{\alpha _{1}-iT}^{\alpha _{1}+iT}\int _{\alpha _{2}-iT}^{\alpha _{2}+iT} \frac{\zeta (s_1)\zeta (s_2)\zeta (s_1 +s_2 -1)}{\zeta (s_1 +s_2)} \nonumber \\&\quad \times \frac{\zeta (3s_1)\zeta (4s_1)\zeta (5s_1)\kappa (s_1)\zeta (3s_2)\zeta (4s_2)\zeta (5s_2)\kappa (s_2)}{\zeta (8s_1)\zeta (8s_2)}\frac{x^{s_{1}+s_{2}}}{s_{1}s_{2}}ds_{2}ds_{1}. \end{aligned}$$

(5.6)

Let $\Gamma (\alpha , \beta ,T)$ denote the contour consisting of the line segments $[\alpha -iT, \beta -iT]$, $[\beta -iT, \beta +iT]$ and $[\beta +iT, \alpha +iT]$. In (5.6), we move the integration with respect to $s_2$ to $\Gamma (\alpha _2, \frac{1}{2}+\frac{1}{\log x},T)$. We denote the integrals over the horizontal line segments by $J_{1,1}$ and $J_{1,3}$, and the integral over the vertical line segment by $J_{1,2}$, respectively. Then using the estimate $\int _1^T |\zeta (\alpha _1+it)|dt \ll T$ and Lemma 2.4, we have

$$\begin{aligned}&J_{1,1}, J_{1,3} \\&\quad \ll \frac{xyL}{T}\int _{-T}^{T}\frac{|\zeta (\alpha _1+it_1)| }{1+|t_1|}dt_1 \\&\qquad \quad \times \int _{\frac{1}{2}+\frac{1}{\log x}}^{\alpha _2}{|\zeta (\sigma _2+iT) \zeta (\alpha _1+\sigma _2-1+i(t_1 + T))| x^{\sigma _2}}\, d\sigma _2 \\&\quad \ll \frac{xyL^3}{T} \int _{-T}^{T}\frac{|\zeta (\alpha _1+it_1)| }{1+|t_1|}dt_1 \int _{\frac{1}{2}+\frac{1}{\log x}}^{\alpha _{2}}T^{\frac{2}{3} (1-\sigma _{2})} x^{\sigma _2} \, d\sigma _2 \\&\quad \ll \frac{x^{2}yL^{4}}{T^{2/3}}\left( x^{-1/2}+T^{-1/3}\right) . \end{aligned}$$

For the integral along the vertical line we have

$$\begin{aligned} J_{1,2}&\ll yx^{\frac{3}{2}} L \\&\quad \times \int _{-T}^{T}\int _{-T}^{T} \frac{|\zeta (\alpha _{1}+it_1) \zeta \left( \frac{1}{2}+\frac{1}{\log x}+it_2\right) \zeta (\alpha _1+\frac{1}{\log x} - \frac{1}{2} +i(t_1+t_2))|}{(1+|t_1|)(1+|t_2|)} dt_1\, dt_2 \\&\ll yx^{\frac{3}{2}}L^{2} \int _{-2T}^{2T} \left| \zeta \left( \frac{1}{2}+\frac{1}{\log x} +iu\right) \right| \int _{-T}^{T} \frac{|\zeta (\frac{1}{2}+\frac{2}{\log x}+it)|}{(1+|t|)(1+|t-u|)}dt\, du. \end{aligned}$$

Hence we use the estimate

$$\begin{aligned} \int _{-T}^{T} \frac{|\zeta (\frac{1}{2}+it)|^2}{(1+|t|)(1+|t-u|)}dt \ll \frac{|u|^{\frac{1}{3}}}{1+|u|} \end{aligned}$$

(see p.161 in [5]) and the Cauchy–Schwarz inequality to get

$$\begin{aligned} J_{1,2}&\ll yx^{\frac{3}{2}}L^{3} \int _{-2T}^{2T} \left| \zeta \left( \frac{1}{2}+\frac{1}{\log x} +iu\right) \right| \frac{|u|^{\frac{1}{3}}}{1+|u|} du \nonumber \\&\ll yx^{\frac{3}{2}}T^{\frac{1}{3}} L^{5}. \end{aligned}$$

(5.7)

It remains to evaluate the residues of the poles of the integrand when we move the line of integration to $\Gamma (\alpha _2,\frac{1}{2}+\frac{1}{\log x},T)$. There exists a simple pole at $s_2=2-s_1$ with residue

$$\begin{aligned}&\frac{\zeta (s_1)\zeta (2-s_1)\zeta (3s_1)\zeta (4s_1)\zeta (5s_1)\zeta (6-3s_1)\zeta (8-4s_1)\zeta (10-5s_1)\kappa (s_1)\kappa (2-s_1)}{\zeta (2)\zeta (8s_1)\zeta (16-8s_1)s_1(2-s_1)} ~x^2 \\&\quad =:H_{1}(s_{1})x^{2}, \end{aligned}$$

and also a simple pole at $s_2=1$ with residue

$$\begin{aligned} \frac{\zeta (3)\zeta (4)\zeta (5)\kappa (1)}{\zeta (8)} \cdot ~\frac{\zeta ^{2}(s_1)\zeta (3s_1)\zeta (4s_1)\zeta (5s_1)\kappa (s_1)}{\zeta (8s_1)\zeta (s_{1}+1)s_1}~{x^{s_1+1}}&=:H_{2}(s_1) x^{s_{1}+1}. \end{aligned}$$

The contributions to $S_{2,1}^{(3)}(x,y)$ from these residues are

$$\begin{aligned}&\ \frac{x^2y}{2\pi i}\int _{\alpha _1-iT}^{\alpha _1+iT} H_1(s_1)ds_1 + \frac{x^{}y}{2\pi i}\int _{\alpha _1-iT}^{\alpha _1+iT}H_2(s_1)x^{s_1}\, ds_1 \\&\quad := I_1 + I_2, \ \mathrm{say}. \end{aligned}$$

For $I_1$, moving the line of integration to $\Gamma (\alpha _1, \frac{5}{4},T)$, we have

$$\begin{aligned} I_1&=\frac{x^2y}{2\pi i}\int _{\frac{5}{4}-i\infty }^{\frac{5}{4}+i\infty } H_1(s_1)ds_1+O\left( x^2y\int _{T}^{\infty }\left| H_1\left( \frac{5}{4}+it_1\right) \right| dt_1\right) +O\left( \frac{x^{2}yL^{2}}{T^{\frac{23}{12}}}\right) \\&=\eta x^{2}y + O\left( \frac{x^2y L^2}{T^{\frac{11}{12}}}\right) , \end{aligned}$$

where the constant $\eta$ is given by

$$\begin{aligned} \eta := \frac{1}{2\pi i}\int _{\frac{5}{4}-i\infty }^{\frac{5}{4}+i\infty } H_1(s_1)ds_1, \end{aligned}$$

(5.8)

which is an absolutely convergent integral given by

$$\begin{aligned} \eta&:= \frac{1}{2\pi i}\nonumber \\&\int _{\frac{5}{4}-i\infty }^{\frac{5}{4}+i\infty } \frac{\zeta (s)\zeta (2-s)\zeta (3s)\zeta (4s)\zeta (5s)\zeta (6-3s)\zeta (8-4s)\zeta (10-5s)\kappa (s)\kappa (2-s)}{\zeta (2)\zeta (8s)\zeta (16-8s)s(2-s)} ds. \end{aligned}$$

(5.9)

Now, we use the inequalities $|\zeta (s)| \le \zeta (\sigma )$ and $\left| \frac{1}{\zeta (s)}\right| \le \frac{\zeta (\sigma )}{\zeta (2\sigma )}$ for $\sigma >1$ (see (8.4.1), (8.7.1) in [13]) to obtain

$$\begin{aligned} |\eta |&\le \frac{\zeta (3)\zeta (5)\zeta (6)\zeta (10)\zeta (\frac{5}{4}) \zeta (\frac{9}{4}) \zeta ^{2}(\frac{15}{4})\zeta (\frac{25}{4})\kappa \left( \frac{3}{4}\right) \kappa \left( \frac{5}{4}\right) }{\pi \zeta (2) \zeta (12)\zeta (20)}\\&\quad \times \int _{0}^{\infty } \frac{\left| \zeta \left( \frac{3}{4}+it\right) \right| }{\sqrt{(\frac{9}{16} +t^2)(\frac{25}{16} +t^2)}}dt. \end{aligned}$$

Here, the integral on the right-hand side of the above is a computable constant, and that is, strictly speaking, enough for the purpose of this paper.

For $I_2$, we move the line of integration to $\Gamma (\alpha _1,\frac{1}{2}+\frac{1}{\log x},T)$. The integrals over the horizontal lines are

$$\begin{aligned} \ll \frac{xy L^{3}}{T} \int _{\frac{1}{2}+\frac{1}{\log x}}^{\alpha _1}T^{\frac{2}{3}(1-\sigma _1)}x^{\sigma _1} d\sigma _1 \ll \frac{x^{\frac{3}{2}}y L^{3}}{T} \left( x^{\frac{1}{2}} + T^{\frac{1}{3}}\right) \end{aligned}$$

and the integral over the vertical line is

$$\begin{aligned}&\ll xy L^{3} \int _{-T}^{T}\frac{|\zeta (\frac{1}{2}+it_1)|^2}{1+|t_1|} x^{\frac{1}{2}}dt_1 \ll x^{\frac{3}{2}}yL^{5} \end{aligned}$$

by using the estimate $\int _{1}^{T}|\zeta (\frac{1}{2}+it)|^{2}dt \ll T\log T$ and integration by parts. Furthermore, when moving the path of integration there is a double pole at $s_1=1$. Hence, using Cauchy’s theorem, we have

$$\begin{aligned} I_2&=\frac{\zeta ^{2}(3)\zeta ^{2}(4)\zeta ^{2}(5)\kappa ^{2}(1)}{\zeta (2)\zeta ^{2}(8)} x^{2}y \log x + \frac{\zeta ^{2}(3)\zeta ^{2}(4)\zeta ^{2}(5)\kappa ^{2}(1) }{\zeta (2)\zeta ^{2}(8)} \\&\qquad \times \left( 2\gamma -1 + \frac{\kappa '(1)}{\kappa (1)} + 3\frac{\zeta ^{'}(3)}{\zeta (3)} + 4\frac{\zeta ^{'}(4)}{\zeta (4)} + 5\frac{\zeta ^{'}(5)}{\zeta (5)} - 8\frac{\zeta ^{'}(8)}{\zeta (8)} - \frac{\zeta ^{'}(2)}{\zeta (2)} \right) x^{2}y \\&\qquad + O\left( \frac{x^{\frac{3}{2}}y L^3}{T}\left( x^{\frac{1}{2}} + T^{\frac{1}{3}}\right) \right) + O(x^{\frac{3}{2}}yL^5), \end{aligned}$$

where $\gamma$ is the Euler constant. Combining these results we have

$$\begin{aligned} S_{2,1}^{(3)}(x,y)&=\frac{\zeta ^{2}(3)\zeta ^{2}(4)\zeta ^{2}(5)\kappa ^{2}(1)}{\zeta (2)\zeta ^{2}(8)} x^{2}y \log x + \eta x^{2}y \nonumber \\&\quad + \frac{\zeta ^{2}(3)\zeta ^{2}(4)\zeta ^{2}(5)\kappa ^{2}(1) }{\zeta (2)\zeta ^{2}(8)} \left( 2\gamma -1 +\frac{\kappa '(1)}{\kappa (1)} + 3\frac{\zeta ^{'}(3)}{\zeta (3)} + 4\frac{\zeta ^{'}(4)}{\zeta (4)} \right. \nonumber \\&\quad \,\left. +5\frac{\zeta ^{'}(5)}{\zeta (5)} - 8\frac{\zeta ^{'}(8)}{\zeta (8)} - \frac{\zeta ^{'}(2)}{\zeta (2)} \right) x^{2}y + O\left( x^{2}y L^{5}\cdot x^{-\frac{1}{3}}\right) . \end{aligned}$$

(5.10)

Here, we substituted $T=x^{}$ into the error term of $S_{2,1}(x,y)$.

5.2 Estimation of $S_{2,4}^{(3)}(x,y)$

Explicitly we have

$$\begin{aligned} S_{2,4}^{(3)}(x,y)&= \frac{y^3}{(2\pi i)^2}\int _{\alpha _1-iT}^{\alpha _1+iT}\int _{\alpha _2-iT}^{\alpha _2+iT} \frac{\zeta (3-s_1-s_2)\zeta (2-s_1)\zeta (2-s_2)}{\zeta (4-s_1-s_2)(3-s_1-s_2){s_1s_2}} \\&\quad \times \frac{\zeta (3s_1)\zeta (4s_1)\zeta (5s_1)\kappa (s_1)\zeta (3s_2)\zeta (4s_2)\zeta (5s_2)\kappa (s_2)}{\zeta (8s_1)\zeta (8s_2)}\left( \frac{x}{y}\right) ^{s_1+s_2} ds_2\, ds_1. \end{aligned}$$

For this purpose, we move the line of integral with respect to $s_2$ to contour $\Gamma (\alpha _{2}, \beta , T)$, where $\beta =\frac{5}{2} -\alpha _{1}=\frac{3}{2} - \frac{1}{\log x}$. We denote the integrals over the horizontal line segments by $J_{4,1}$ and $J_{4,3}$, and the integral over the vertical line segment by $J_{4,2}$, respectively. There are no poles when we deform the path of integral over $s_2$. The contribution from the horizontal lines are

$$\begin{aligned} J_{4,1}, J_{4,3}&\ll xy^{2}\left( \frac{x}{y}\right) ^{\frac{1}{\log x}} \int _{-T}^{T}\frac{\left| \zeta \left( 1-\frac{1}{\log x} -it_{1}\right) \right| }{1+|t_{1}|} dt_1 \\&\quad \times \int _{\alpha _{2}}^{\beta } \frac{\left| \zeta \left( 2-\frac{1}{\log x} -\sigma _2 -i(t_{1}+T)\right) \zeta \left( 2-\sigma _{2}-iT\right) \right| }{(1+|t_{1}+T|)T}\left( \frac{x}{y}\right) ^{\sigma _{2}}d\sigma _{2}. \end{aligned}$$

The inner integral is estimated as

$$\begin{aligned}&\ll \frac{L^3}{T(1+|t_1+T|)}\left( \frac{x}{y}\right) \left( 1+T^{\frac{1}{3}}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\right) , \end{aligned}$$

where we have used the assumption $y \ll x^M$. Hence, we have

$$\begin{aligned} J_{4,1},J_{4,3}&\ll \frac{ x^{2}y L^{3}}{T}\left( 1 +T^{\frac{1}{3}}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\right) \int _{-T}^{T}\frac{\left| \zeta \left( 1-\frac{1}{\log x}-it_{1}\right) \right| }{(1+|t_{1}|)(1+|t_{1}|+T)}dt_{1} \\&\ll \frac{x^{2}y L^{4}}{T^{2}}\left( 1+T^{\frac{1}{3}}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\right) . \end{aligned}$$

For the integral on the vertical line we find that

$$\begin{aligned} J_{4,2}&\ll y^{3}\int _{-T}^{T}\int _{-T}^{T} \frac{\left| \zeta (\frac{1}{2}-i(t_{1}+t_{2}))\zeta (1-\frac{1}{\log x}-it_{1})\zeta (\frac{1}{2}+\frac{1}{\log x}-it_{2})\right| }{(1+|t_1+t_{2}|)(1+|t_{1}|)(1+|t_{2}|)} \left( \frac{x}{y}\right) ^{\frac{5}{2}}dt_{1}\, dt_{2}\\&\ll y^{3}\left( \frac{x}{y}\right) ^{\frac{5}{2}}\int _{-2T}^{2T} \frac{\left| \zeta \left( \frac{1}{2}-iu\right) \right| }{1+|u|} \int _{-T}^{T}\frac{\left| \zeta \left( \frac{1}{2}+\frac{1}{\log x}-it_{2}\right) \right| }{(1+|t_{2}|)(1+|u-t_{2}|)}dt_{2}\, du \\&\ll x^{2}y \left( \frac{x}{y}\right) ^{\frac{1}{2}}L^{3}. \end{aligned}$$

Hence, we take $T=x^{}$ to get

$$\begin{aligned} S_{2,4}^{(3)}(x,y) \ll x^{2}y \left( \frac{x}{y}\right) ^{\frac{1}{2}}L^{3}. \end{aligned}$$

(5.11)

5.3 Estimation of $S_{2,3}^{(3)}(x,y)$

It is given explicitly by

$$\begin{aligned} S_{2,3}^{(3)}(x,y)&=\frac{y^2}{(2\pi i)^2} \int _{\alpha _1-iT}^{\alpha _1+iT}\int _{\alpha _2-iT}^{\alpha _2+iT} \frac{\zeta (2-s_2)\zeta (1+s_1-s_2)\zeta (s_1)}{\zeta (2+s_1-s_2)(2-s_2)} \\&\quad \times \frac{\zeta (3s_1)\zeta (4s_1)\zeta (5s_1)\kappa (s_1)\zeta (3s_2)\zeta (4s_2)\zeta (5s_2)\kappa (s_2)}{\zeta (8s_1)\zeta (8s_2)} \frac{x^{s_1+s_2}y^{-s_2}}{s_1s_2}ds_2\, ds_1. \end{aligned}$$

We move the path of integration with respect to $s_2$ to $\Gamma (\alpha _2, \frac{3}{2}, T)$. We denote the integrals over the horizontal line segments by $J_{3,1}$ and $J_{3,3}$, and the integral over the vertical line segment by $J_{3,2}$, respectively. Note that there exist no poles with this deformation. The contribution from the horizontal lines are

$$\begin{aligned} J_{3,1}, J_{3,3}&\ll \frac{y^2 x L}{T^2} \int _{-T}^{T}\frac{|\zeta (\alpha _1+it_1)|}{1+|t_1|} \\&\quad \times \int _{\alpha _2}^{\frac{3}{2}} |\zeta (2-\sigma _2-iT)\zeta (1+\alpha _1-\sigma _2+i(t_1-T))| \left( \frac{x}{y}\right) ^{\sigma _2} d\sigma _2\, dt_1 \\&\ll \frac{y^2 x L^2}{T^2} \int _{-T}^{T}\frac{|\zeta (\alpha _1+it_1)|}{1+|t_1|}\\&\quad \times \int _{\alpha _2}^{\frac{3}{2}} T^{\frac{1}{3}(-1+\sigma _2)} (1+|t_1-T|)^{\frac{1}{3}(-1+\sigma _2)}\left( \frac{x}{y}\right) ^{\sigma _2}d\sigma _2\, dt_1 \\&\ll y x^2 L^3 \left( T^{-2}+T^{-\frac{5}{3}}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\right) . \end{aligned}$$

On the other hand, the contribution from the vertical lines is

$$\begin{aligned} J_{3,2}&\ll y^2 x \int _{-T}^{T}\frac{|\zeta (\alpha _1+it_1)|}{1+|t_1|}\\&\quad \times \int _{-T}^T \frac{|\zeta (\frac{1}{2}-it_2)\zeta (\frac{1}{2} +\frac{1}{\log x}+i(t_1-t_2))|}{(1+|t_2|)^2}\left( \frac{x}{y}\right) ^{\frac{3}{2}}dt_2\, dt_1 \\&\ll y^2 x\left( \frac{x}{y} \right) ^{\frac{3}{2}}L. \end{aligned}$$

Hence, we take $T=x^{}$ into the above to obtain

$$\begin{aligned} S_{2,3}^{(3)}(x,y) \ll x^{2} y L \left( \frac{x}{y}\right) ^{\frac{1}{2}}. \end{aligned}$$

(5.12)

5.4 Evaluation of $S_{2,2}^{(3)}(x,y)$

The explicit form of $S_{2,2}^{(3)}(x,y)$ is given by

$$\begin{aligned} S_{2,2}^{(3)}(x,y)&=\frac{y^2}{(2\pi i)^2}\int _{\alpha _1-iT}^{\alpha _1+iT}\int _{\alpha _2-iT}^{\alpha _2+iT} \frac{\zeta (2-s_1)\zeta (1-s_1+s_2)\zeta (s_2)}{\zeta (2-s_1+s_2)(2-s_1)} \nonumber \\&\quad \times \frac{\zeta (3s_1)\zeta (4s_1)\zeta (5s_1)\kappa (s_1)\zeta (3s_2)\zeta (4s_2)\zeta (5s_2)\kappa (s_2)}{\zeta (8s_1)\zeta (8s_2)} \frac{x^{s_1+s_2}y^{-s_1}}{s_1s_2}ds_2\, ds_1. \end{aligned}$$

(5.13)

This time we firstly move the line of the integration over $s_1$ to $\Gamma (\alpha _{1},\frac{3}{2},T)$. The estimates over the horizontal lines and the vertical line are the same as that of $S_{2,3}^{(3)}(x,y)$, but there is a simple pole at $s_1=s_2$ inside this contour. The residue of the integrand of (5.13) at this pole is

$$\begin{aligned} -\frac{\zeta (2-s_2)\zeta (s_2)\zeta ^{2}(3s_2)\zeta ^{2}(4s_2)\zeta ^{2}(5s_2)\kappa ^{2}(s_2)}{\zeta (2)\zeta ^{2}(8s_2)(2-s_{2})s_{2}^{2}}x^{2s_2}y^{-s_2}. \end{aligned}$$

Hence, we have

$$\begin{aligned} S_{2,2}^{(3)}(x,y)&= \frac{x^2y}{2\pi i}\int _{\alpha _2-iT}^{\alpha _2+iT} \frac{\zeta (2-s_2)\zeta (s_2)\zeta ^{2}(3s_2)\zeta ^{2}(4s_2)\zeta ^{2}(5s_2)\kappa ^{2}(s_2)}{\zeta (2)\zeta ^{2}(8s_2)(2-s_{2})s_{2}^{2}} \left( \frac{y}{x^2}\right) ^{1-s_2}ds_2 \\&\quad + O\left( x^{2}yL\left\{ x^{-1} + \left( \frac{x}{y} \right) ^{\frac{1}{2}}\right\} \right) \end{aligned}$$

by taking $T=x^{}$. We move the line of integration to $\Gamma (\alpha _2,\frac{1}{2}+\frac{2}{\log x},T)$. By the same method as before, the integrals over the horizontal lines are estimated as

$$\begin{aligned} \ll \frac{x^2y}{T^3}\left( L^4\left( \frac{y}{x^2}\right) ^{-\frac{2}{\log x}}+L^2T^{\frac{1}{2}}\left( \frac{y}{x^2}\right) ^{\frac{1}{2}}\right) \ll \frac{x^2yL^4}{T^3}\left( 1+T^{\frac{1}{2}}\left( \frac{y}{x^2}\right) ^{\frac{1}{2}}\right) \end{aligned}$$

and the vertical lines are estimated as

$$\begin{aligned} \ll x^2y \left( \frac{y}{x^2}\right) ^{\frac{1}{2}}L^2. \end{aligned}$$

Furthermore, there is a contribution from the pole $s_2=1$ of order 2, hence $S_{2,2}^{(3)}(x,y)$ has the form

$$\begin{aligned} S_{2,2}^{(3)}(x,y)&=\frac{\zeta ^{2}(3)\zeta ^{2}(4)\zeta ^{2}(5)\kappa ^{2}(1)}{\zeta (2)\zeta ^{2}(8)} x^2y \log \frac{x^2}{y} + \frac{\zeta ^{2}(3)\zeta ^{2}(4)\zeta ^{2}(5)\kappa ^{2}(1)}{\zeta (2)\zeta ^{2}(8)}\nonumber \\&\quad \times \left( 6\frac{\zeta '(3)}{\zeta (3)}+8\frac{\zeta '(4)}{\zeta (4)}+10\frac{\zeta '(5)}{\zeta (5)}-16\frac{\zeta '(8)}{\zeta (8)}-1 +2\frac{\kappa '(1)}{\kappa (1)}\right) \nonumber \\&\quad +O\left( x^2y\left( L^{5}x^{-\frac{1}{3}} + L^{6}y^{-\frac{1}{2}} + L^{3}\left( \frac{x}{y} \right) ^{\frac{1}{2}} + L^2\left( \frac{y}{x^2}\right) ^{\frac{1}{2}}\right) \right) \end{aligned}$$

(5.14)

by taking $T=x^{}$.

5.5 Asymptotic formula of (1.12)

Now, we substitute (5.5), (5.10), (5.11), (5.12) and (5.14) into (5.4) to obtain the assertion of Theorem 3. $\square$