On Sums of Sums Involving the Von Mangoldt Function

Abstract

Let $\Lambda$ denote the von Mangoldt function, and (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 involving the von Mangoldt function; $S_{k}(x,y):=\sum _{n\le y}\left( \sum _{q\le x}\right. \left. \sum _{d|(n,q)}d\Lambda \left( \frac{q}{d}\right) \right) ^{k}$ for $k=1,2$.

1 Introduction

For any integers $n\ge 1$, we define

$$\begin{aligned} \Lambda (n)&= \left\{ \begin{array}{cl} \displaystyle \log p & \ \ {\textrm{if}}\ n=p^{m} \ {\textrm{some prime}}\ p \ {\textrm{and some}} \ m\ge 1, \ \\ \displaystyle 0 & \ \ {\textrm{otherwise}}, \ \end{array} \right. \end{aligned}$$

which is the von Mangoldt function. Let $s=\sigma +it$ be the complex variable, where $\sigma$ and t are real, and let $\zeta (s)$ denote the Riemann zeta-function defined by

$$\begin{aligned}\zeta (s)=\sum _{n=1}^\infty \frac{1}{n^s},\end{aligned}$$

and $\zeta '(s)$ its first derivative. The Riemann zeta function can be analytically continued to the whole plane. We define the following sum over the von Mangoldt function

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

(1.1)

where (n, q) denotes the greatest common divisor of integers n and q. This sum is a special type of Anderson–Apostol sum defined by $\sum _{d|(n,q)} f(d) g\left( {q}/{d}\right)$ with any arithmetical functions f and g (see [1, 2]). We use the Dirichlet series

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{\Lambda (n)}{n^s} = - \frac{\zeta '(s)}{\zeta (s)} \ \ \ ({{\textrm{Re}}}~s > 1) \end{aligned}$$

(1.2)

to deduce the Dirichlet series with the coefficients $s_{q}(n)$, namely

$$\begin{aligned} \sum _{q=1}^{\infty }\frac{s_{q}(n)}{q^s} = -\sigma _{1-s}(n)\frac{\zeta '(s)}{\zeta (s)} \end{aligned}$$

(1.3)

for ${{{\textrm{Re}}}~{s} > 1}$ with the divisor function $\sigma _{1-s}(n)=\sum _{d|n}d^{1-s}.$ For any large positive real numbers x and y, we let the double sums

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

(1.4)

The double sum of the type (1.4) was first considered by Chan and Kumchev [3], who proved several interesting asymptotic formulas concerning the Ramanujan sum $c_{q}(n)$, defined by $c_{q}(n)=\sum _{d|(n,q)}d\mu \left( {q}/{d}\right)$ with $\mu$ being the Möbius function, instead of $s_{q}(n)$. In 2015, Minamide, Tanigawa, and the first author [7] were inspired by their work, and considered square-free numbers instead of the Möbius function in the Ramanujan sum, and derived the precise asymptotic formulas. Robles and Roy [16] studied an analogue of the type (1.4) concerning the generalized Ramanujan sums, known as the Cohen–Ramanujan sums. Moreover, the first author considered some sums of the type (1.4) concerning square-full numbers [8], cube-full numbers [10], the Liouville function [11] and others (see [9, 12, 13, 15]). This study aims to derive several asymptotic formulas for (1.4) with $k=1$ and 2.

1.1 Evaluation of $S_{1}(x,y)$

Following the same procedure as in [3] (see also [7, 8, 10, 13, 15, 16]), we obtain some interesting theorems for the double sum $S_k(x,y)$. First, the case $k=1$ implies the following theorem, namely

Theorem 1.5

Let the notation be as above. Let x and y be large real numbers such that $x \log x \ll y \ll \frac{x^{2}}{\log x}$. Then, we have

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

(1.6)

Remark 1.7

Substituting $y=x^{\frac{3}{2}}$ and $y=x\log x$ into (1.6), we obtain

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

and

$$\begin{aligned} S_{1}(x,x\log x )=x^{2}\log x (\log x -1) + \frac{\zeta '(2)}{4\zeta (2)}x^{2} + O\left( \frac{x^{2}}{\log x}\right) , \end{aligned}$$

(1.8)

respectively.

If we could use an alternative method to investigate an asymptotic behavior for $S_{1}(x,y)$ under the condition $y \ll x \log x$, then we may use some analytic method to study the asymptotic formulas for (1.4) for $k=1$. We use analytic properties between the Riemann zeta-function and the von Mangoldt function to investigate the asymptotic behavior of sharp approximate formulas for (1.4), and whose form yields an interesting formula. Before elucidating the statement, let $\kappa (u)$ denote the Fourier integral given by

$$\begin{aligned} \kappa (u):= \frac{1}{2\pi }\int _{-\infty }^{\infty } \frac{\zeta (-\frac{1}{4}-it)\zeta '(\frac{9}{4}+it)}{\zeta (\frac{9}{4}+it)} \frac{\textrm{e}^{itu}}{(\frac{1}{4}+it)(\frac{9}{4}+it)}dt \end{aligned}$$

(1.9)

with $u:=\log \frac{x}{y}$. It follows from (3.15) below that $|\kappa (u)|$ is given by the inequality

$$\begin{aligned} |\kappa (u)|&\le \frac{8}{(2\pi )^{\frac{7}{4}}}\frac{\zeta (\frac{5}{4})\zeta (\frac{9}{4})\zeta '(\frac{9}{4})}{\zeta (\frac{9}{2})}\left( \frac{\pi }{9}+1\right) . \end{aligned}$$

(1.10)

Here, the integral is a computable constant. We use a contour integral of the generating Dirichlet series (the method in [9]) and some properties of the Riemann zeta-function to obtain

Theorem 1.11

Let the notation be as above. Let x and y be large real numbers such that $1\ll y \ll \frac{x^{7/5}}{\log ^{2}x}$. Then, we have

$$\begin{aligned} S_{1}(x,y)&= yx(\log x -1) + \frac{\zeta '(2)}{4\zeta (2)} x^{2} + x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{4}}\kappa (u) \nonumber \\&\qquad + O\left( xy^{\frac{1}{3}}\log ^{4} x + yx^{\frac{1}{2}}\log ^{\frac{5}{2}}x \right) , \end{aligned}$$

(1.12)

where $\kappa (u)$ denotes the Fourier integral given by (1.9).

For $y=x$, we have an interesting formula, namely

Remark 1.13

We substitute $y=x$ into (1.12), then we obtain

$$\begin{aligned} S_{1}(x,x)=x^{2}\log x + \left( \frac{\zeta '(2)}{4\zeta (2)} + \kappa (0) -1 \right) x^{2} + O\left( x^{\frac{3}{2}}\log ^{\frac{5}{2}}x \right) , \end{aligned}$$

where $\kappa (0)$ is a computable constant.

Remark 1.14

Furthermore, we substitute $y=x\log x$ into (1.12) to deduce

$$\begin{aligned} \begin{aligned} S_{1}(x,x\log x )=&x^{2}\log x (\log x -1) + \frac{\zeta '(2)}{4\zeta (2)} x^{2} + \frac{x^{2}}{\log ^{\frac{1}{4}}x} \kappa \left( \log \frac{1}{\log x}\right) \\ &\qquad + O\left( x^{\frac{3}{2}}\log ^{\frac{7}{2}}x \right) . \end{aligned} \end{aligned}$$

It follows from (1.8) and the above that

$$\begin{aligned} \kappa \left( \log \frac{1}{\log x} \right) = O\left( (\log x)^{-\frac{3}{4}} \right) . \end{aligned}$$

1.2 Evaluation of $S_{2}(x,y)$

For the case $k=2$, two different methods to handle function $S_{2}(x,y)$ exist. We use an elementary lattice point counting argument to obtain the formula (1.16) below and use the generating Dirichlet series and the properties of the Riemann zeta-function to prove (1.18) below, which we state as

Theorem 1.15

Let x and y denote large real numbers such that $y \gg \frac{x^{2}}{\log ^{3}x}$. We have

$$\begin{aligned} S_2(x,y)&= \frac{1}{3\zeta (2)}yx^{2}\log ^{3}x + O\left( yx^{2}\log ^{2}x + x^{4} \right) . \end{aligned}$$

(1.16)

To establish the precise asymptotic formula of $S_{2}(x,y)$, we let $\gamma$ denote the Euler–Mascheroni constant, and $\gamma _1$, $\gamma _{2}$ denote the Stieltjes constants defined by (5.5) below. Let $c_{1},\ \ldots ,\ c_{6}$ denote the constants given by

$$\begin{aligned} c_{1}= & \frac{1}{\zeta (2)}\left( \gamma -1-\frac{\zeta '(2)}{\zeta (2)}\right) , \\ c_{2}= & \frac{1}{\zeta (2)}\left( 1-2(\gamma +\gamma _{1}) + 2(1-\gamma )\frac{\zeta '(2)}{\zeta (2)}+2\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^{2} -\frac{\zeta ''(2)}{\zeta (2)} \right) ,\\ c_{3}= & \frac{1}{\zeta (2)}\left( c_{0} -\gamma + 2\gamma _{1}+2\gamma _{2} +\left( 2\gamma +2\gamma _{1}-1\right) \frac{\zeta '(2)}{\zeta (2)} + 2(\gamma -1)\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^2 \right) \\ & +\frac{1}{\zeta (2)} \left( (1-\gamma )\frac{\zeta ''(2)}{\zeta (2)} - \frac{\zeta '''(2)}{3\zeta (2)} + 2\frac{\zeta '(2)}{\zeta (2)} \frac{\zeta ''(2)}{\zeta (2)} - 2\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^3\right) ,\\ c_{4}= & \frac{1}{\zeta (2)}\left( \frac{1}{2} +\gamma \right) , \ \ c_{5}= \frac{2}{\zeta (2)} \left( \gamma ^2+\gamma -\gamma _{1}+1\right) , \end{aligned}$$

and

$$\begin{aligned} c_{6}= \frac{2}{\zeta (2)} \left( \gamma ^3 +\gamma ^2 +2\gamma -\gamma _{1}-3\gamma \gamma _{1}+3\gamma _{2}+1\right) , \end{aligned}$$

where $c_{0}$ is given by

$$\begin{aligned} c_{0}:= \frac{1}{2\pi i}\int _{\frac{5}{4}-i\infty }^{\frac{5}{4}+i\infty } \frac{\zeta '(s)\zeta '(2-s)}{s(2-s)} ds \end{aligned}$$

which is a computable constant. In the last section, the value of $c_{0}$ is evaluated by

$$\begin{aligned} c_{0}&\le 0.425\cdots ~\left| \zeta '\left( \frac{5}{4}\right) \right| \int _{0}^{\infty } \frac{\left| \zeta '\left( \frac{3}{4}(1+iy)\right) \right| }{1+y^2}dy. \end{aligned}$$

The integral on the right-hand side of the above is a computable constant. We obtain

Theorem 1.17

Let the notation be as above. Let x and y be large real numbers such that $x \log ^{16}x \ll y \ll \frac{x^2}{\log ^{16}x}$. Then, we have

$$\begin{aligned} S_2(x,y)&= \frac{1}{3\zeta (2)}yx^{2}\log ^{3}x + c_{1} yx^{2}\log ^{2}x + c_{2} yx^{2}\log x + (c_{3}-c_{6}) yx^{2} \nonumber \\&+\frac{1}{6\zeta (2)} y x^2 \log ^{3}\frac{x^2}{y} - c_{4} y x^2 \log ^{2}\frac{x^2}{y} + c_{5} y x^2 \log \frac{x^2}{y} + E(x,y),\nonumber \\ \end{aligned}$$

(1.18)

where the error term E(x, y) is estimated by

$$\begin{aligned} E(x,y)=O\left( x^{5/3}yL^{8} + x^2yL^{10} \left( \left( \frac{x}{y} \right) ^{1/2} + \left( \frac{y}{x^2}\right) ^{1/2}\right) \right) \end{aligned}$$

(1.19)

with $L=\log (xy)$.

1.3 Open Problems

Here we list two open problems concerning some functions discussed above.

• 1. Investigate asymptotic formulas of the type (1.4) for a fixed integers $k\ge 3$.

• 2. Investigate asymptotic formulas of

  $$\begin{aligned}\sum _{n\le y}\left( \sum _{q\le x}\sum _{d|(n,q)}d\cdot f\left( \frac{q}{d}\right) \right) ^k\end{aligned}$$

  for any arithmetic functions f with a fixed integers $k\ge 1$. For example, we may consider the divisor function $\tau (:=\textbf{1}*\textbf{1})$, the sum-of-sum divisors function $\sigma (:=\textbf{1}*\textrm{id})$, and the Euler totient function $\phi (:=\textrm{id}*\mu )$ in place of f, respectively.

2 Proof of Theorem 1.5

2.1 Exponent Pair

To prove Theorem 1.5, we need the following Lemma. Let $\psi (x)=x-[x]-\frac{1}{2}$ denote the first periodic Bernoulli function. Then, we have

Lemma 2.1

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

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

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

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

(2.2)

Proof

This lemma is given by Lemma 2.1 in [3] (see also [4]). $\square$

2.2 Proof of (1.6).

Using (1.1) and (1.4) with $k=1$, we have

$$\begin{aligned} S_{1}(x,y)&=\sum _{n\le y}\sum _{q\le x} {s}_{q}(n) =\sum _{n\le y}\sum _{\begin{array}{c} dk\le x\\ d|n \end{array}}d \Lambda (k). \end{aligned}$$

Changing the order of summation, we find that

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

(2.3)

Consider $S_{1,1}(x,y)$. We use the identity $\sum _{d|n}\Lambda (d)=\log n$ and the summation formula $\sum _{n\le x}\log n =x\log x -x +O\left( \log x\right)$ to obtain

$$\begin{aligned} S_{1,1}(x,y)&= y\sum _{n\le x}\sum _{k|n}\Lambda (k) = y\sum _{n\le x}\log n \nonumber \\&= yx\log x - yx + O\left( y\log x \right) . \end{aligned}$$

(2.4)

We use (1.2) and the summation formula $\sum _{n\le x}\frac{\Lambda (n)}{n}=\log x + O(1)$ to deduce

$$\begin{aligned} S_{1,2}(x,y)&=\frac{1}{2} \sum _{k\le x} \Lambda (k) \sum _{d\le \frac{x}{k}}d \nonumber \\&=\frac{x^2}{4}\sum _{k\le x}\frac{\Lambda (k)}{k^2} + O\left( x\sum _{k\le x}\frac{\Lambda (k)}{k} \right) \nonumber \\&=-\frac{\zeta '(2)}{4\zeta (2)} x^2 + O\left( x\log x\right) . \end{aligned}$$

(2.5)

Let $N_{j}=N_{j,k}=\left( \frac{x}{k}\right) 2^{-j}$. We use the theory of exponent pairs to obtain

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

where $\sup$ is over all subintervals I in $(N_{j},2N_{j}]$. From (2.2) in Lemma 2.1, we have

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

(2.6)

Substituting (2.4), (2.5), and (2.6) into (2.3), we obtain the assertion of Theorem 1.5. $\square$

3 Proof of Theorem 1.11

3.1 Lemmas

To prove theorem 1.11, we utilize the following Lemmas.

Lemma 3.1

Suppose that the Dirichlet series $\alpha (s):=\sum _{n=1}^{\infty }\frac{a_n}{n^s}$ absolutely converges for $\textrm{Re}~s >\sigma _{a}$. 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 the famous Perron’s formula (see Theorem 5.2 and Corollary 5.3 in [14]). $\square$

Lemma 3.2

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

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

(3.3)

and

$$\begin{aligned} \zeta '(\sigma +it)&\ll \left\{ \begin{array}{cl} \displaystyle t^{\frac{1}{6} (3-4\sigma )}\log ^{2} t & \ \ \left( 0\le \sigma \le \frac{1}{2}\right) , \ \\ \displaystyle t^{\frac{1}{3} (1-\sigma )}\log ^{2} t & \ \ \left( \frac{1}{2} \le \sigma \le 1\right) , \ \\ \displaystyle \log ^{2} t & \ \ \left( 1\le \sigma < 2\right) , \ \\ \displaystyle 1 & \ \ \left( \sigma \ge 2\right) . \ \end{array} \right. \nonumber \\ \end{aligned}$$

(3.4)

Furthermore, we have

$$\begin{aligned} \frac{1}{\zeta (\sigma +it)}&\ll \left\{ \begin{array}{cl} \displaystyle \log t & \ \ \left( 1\le \sigma < 2\right) , \ \\ \displaystyle 1 & \ \ \left( \sigma \ge 2\right) . \ \end{array} \right. \nonumber \\ \end{aligned}$$

(3.5)

Proof

The formula (3.3) follows from Theorem II.3.8 in Tenenbaum [17], and Ivić [6]. The formula (3.4) follows from Lemmas 2.3 and 2.4 in Tóth and Zhai [19]. The estimate (3.5) follows from Titchmarsh [18]. $\square$

Lemma 3.6

Let $\textrm{Re}~z \le 0$, and let $\sigma _{z,b}(n)$ denote the generalization of the divisor function defined by $\sigma _{z,b}(n) =\sum _{d^{b}|n}d^{bz}.$ Then, we have

$$\begin{aligned} \sum _{n\le x} ^{'}\sigma _{z,b}(n)&= D_{z,b}(x) + \Delta _{z,b}(x), \end{aligned}$$

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

$$\begin{aligned} \Delta _{z,b}(x) = O\left( x^{\frac{1}{3}}\log ^{2}x \right) \end{aligned}$$

uniformly for $b\ge 1$ and $D_{z,b}(x)$ is given by the following:

• (i) If $b=1,2$ and $-\frac{2}{3b^{2}}<\textrm{Re}~z \le 0$, then

  $$\begin{aligned} D_{z,b}(x) = \zeta (b(1-z))x +\frac{1}{1+bz}\zeta \left( z+\frac{1}{b}\right) x^{z+\frac{1}{b}}. \end{aligned}$$

  (3.7)

• (ii) If $b\ge 3$ and $-1<\textrm{Re}~z \le 0$, then

  $$\begin{aligned} D_{z,b}(x) = \zeta (b(1-z))x. \end{aligned}$$

Proof

The proof of this result is found in Theorem 1.4 in [16]. $\square$

3.2 Proof of (1.12).

We assume that $1 \le y \le x^M$ for some constant M. Without loss of generality, we can assume that $x,\ y\in \mathbb {Z}+\frac{1}{2}$. Suppose that $\alpha \ge 1 +\frac{1}{\log x}$ and T is a real parameter at our disposal. We apply Lemma 3.1 with (1.3) to deduce

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

(3.8)

where $E_{1}(x;n)$ is the error term given by

$$\begin{aligned} E_{1}(x;n)&\ll \sigma _{0}(n) \sum _{\begin{array}{c} \frac{x}{2}<q<2x \\ q\ne x \end{array}} \Lambda (q)\min \left( 1,\frac{x}{T|x-q|}\right) + \sigma _{0}(n)\frac{(4x)^{\alpha }}{T}\sum _{q=1}^{\infty }\frac{\Lambda (q)}{q^{\alpha }} \\&\ll \sigma _{0}(n) \log x \left( 1+\frac{x }{T} \sum _{1\le k \le x}\frac{1}{k}\right) \ll \sigma _{0}(n)\left( 1+ \frac{x}{T}\log x \right) \log x. \end{aligned}$$

We substitute $b=1$ and $z=1-s$ into Lemma 3.6 and use the well-known estimate $\sum _{n\le y}\sigma _{0}(n)\ll y\log y$ to deduce

$$\begin{aligned} S_{1}(x,y)&= - \frac{1}{2\pi i} \int _{\alpha -iT}^{\alpha +iT} \sum _{n\le y}\sigma _{1-s}(n) \frac{\zeta '(s)}{\zeta (s)}\ \frac{x^s}{s}ds \nonumber \\&\quad + O\left( \left( \frac{x\log ^{2} x}{T}+\log x\right) \sum _{n\le y}\sigma _{0}(n)\right) \nonumber \\&=K_{1}+K_{2} + O\left( xy^{\frac{1}{3}}\log ^{2}y\log ^{2}T\right) +O\left( \frac{xy\log ^{3} x}{T}\right) +O\left( y\log ^{2}x\right) , \end{aligned}$$

(3.9)

where

$$\begin{aligned} K_{1}:= - \frac{y}{2\pi i} \int _{\alpha -iT}^{\alpha +iT} \zeta '(s) \frac{x^s}{s}ds, \end{aligned}$$

and

$$\begin{aligned} K_{2}:= \frac{y^2}{2\pi i} \int _{\alpha -iT}^{\alpha +iT} \frac{\zeta (2-s)\zeta '(s)}{\zeta (s)}\ \frac{\left( \frac{x}{y}\right) ^s}{(s-2)s}ds. \end{aligned}$$

3.3 Calculation of $K_1$

Moving the line of integration to $\textrm{Re}~s=c~(:=\frac{1}{2})$, we consider the following rectangular contour formed by the line segments joining the points $\alpha -iT$, $\alpha +iT$, $c+iT$, $c-iT$, and $\alpha -iT$ in the counter-clockwise sense.

We observe that $s=1$ is a double pole of the integrand. Note that the Laurent expansion of the first derivative of the Riemann zeta-function at its pole s = 1 is given by

$$\begin{aligned} \zeta '(s) = - \frac{1}{(s-1)^2} + O\left( 1\right) . \end{aligned}$$

Thus, we obtain the main term from the sum of the residue coming from the pole $s=1$. Hence, using the Cauchy residue theorem, we have

$$\begin{aligned} K_{1}&=-\frac{y}{2\pi i}\left\{ \int _{c+iT}^{\alpha +iT} + \int _{c-iT}^{c+iT} + \int _{\alpha -iT}^{c-iT}\right\} \zeta '(s) \frac{x^{s}}{s}ds + xy(\log x -1). \end{aligned}$$

(3.10)

The second term (the left vertical line segment) on the right-hand side of (3.10) contributes the quantity

$$\begin{aligned}&\frac{y}{2\pi }\int _{-T}^{T}\zeta '\left( \frac{1}{2}+it\right) \frac{x^{\frac{1}{2}+it}}{\frac{1}{2}+it}dt \nonumber \\&\quad \ll yx^{\frac{1}{2}} + yx^{\frac{1}{2}}\left( \int _{2\pi }^{T}\frac{\left| \zeta '\left( \frac{1}{2}+it\right) \right| ^{2}}{1+|t|}dt \right) ^{\frac{1}{2}}\log ^{\frac{1}{2}}T \ll yx^{\frac{1}{2}} \log ^{\frac{5}{2}} T \end{aligned}$$

(3.11)

using $\int _{2\pi }^{T}\left| \zeta '\left( \frac{1}{2}+iv\right) \right| ^2\frac{dv}{v} \ll \log ^{4}T$ (see (172) in Hall [5]) and Cauchy–Schwarz’s inequality. We can estimate the contributions coming from the upper horizontal line (the lower horizontal line is similar), noting that $T = x^{12}$. We define function F(t) as

$$\begin{aligned} F(t)&:= \frac{1}{2\pi }\int _{\frac{1}{2}}^{1+\frac{1}{\log x}} \zeta '(\sigma +it) \frac{x^{\sigma +it}}{\sigma + it}d\sigma . \end{aligned}$$

Then, we set

$$\begin{aligned} Q &:= \int _{\frac{T}{2}}^{T}\left| \int _{\frac{1}{2}}^{1+\frac{1}{\log x}} \zeta '(\sigma +it) \frac{x^{\sigma +it}}{\sigma + it}d\sigma \right| dt. \end{aligned}$$

Using Lemma 3.2, we obtain

$$\begin{aligned} Q &\ll \int _{\frac{1}{2}}^{1+\frac{1}{\log x}}\int _{\frac{T}{2}}^{T} |\zeta '(\sigma +it)|\frac{x^{\sigma }}{t}dt d\sigma \\&\ll T^{\frac{1}{3}} \log ^{2}T \int _{\frac{1}{2}}^{1+\frac{1}{\log x}} \left( \frac{x}{T^{\frac{1}{3}}}\right) ^{\sigma }d\sigma \ll \left( x^{\frac{1}{2}}T^{\frac{1}{6}} + x \right) \log ^{2}T. \end{aligned}$$

Then, $T^{*}\in [\frac{T}{2},T]$ exists such that $|F(T^{*})|$ is minimum and

$$\begin{aligned} |F(T^{*})| \ll \frac{1}{T}\cdot \left( x^{\frac{1}{2}}T^{\frac{1}{6}} + x \right) \log ^{2}T \ll x^{-8} \end{aligned}$$

by setting $T=x^{12}$. Hence, using horizontal lines of height $\pm T^{*}$ to move the line of integration in (3.10), we find that the total contribution of the horizontal lines, in absolute value, is $\ll yx^{-8}.$ Collecting the error estimates (3.11) and the above, we obtain the total contribution of all error terms, that is,

$$\begin{aligned} yx^{\frac{1}{2}} \log ^{\frac{5}{2}} x + yx^{-8} \ll yx^{\frac{1}{2}} \log ^{\frac{5}{2}}x. \end{aligned}$$

Hence, we have

$$\begin{aligned} K_{1}&= xy(\log x -1) + O\left( yx^{\frac{1}{2}}\log ^{\frac{5}{2}}x \right) . \end{aligned}$$

(3.12)

3.4 Calculation of $K_{2}$

We consider the rectangular contour formed by the line segments joining the points $\alpha -iT$, $\alpha +iT$, $\frac{9}{4}+iT$, $\frac{9}{4}-iT$, and $\alpha -iT$ in the clockwise sense, and we observe that $s=2$ is a simple pole of the integrand. We denote the integrals over the horizontal line segments by $K_{2,1}$ and $K_{2,3}$, and the integral over the vertical line segment by $K_{2,2}$, respectively. A simple pole exists at $s=2$ of the integral $K_2$ with the residue $- \frac{\zeta '(2)}{4\zeta (2)}\left( \frac{x}{y}\right) ^{2}$ using $\zeta (0)=-\frac{1}{2}$. For $K_{2.2}$, we use the functional equation of the Riemann zeta-function

$$\begin{aligned} \zeta (s)=\chi (s)\zeta (1-s) \qquad \textrm{with} \qquad \chi (s) \asymp \left( \frac{|t|}{2\pi }\right) ^{\frac{1}{2}-\sigma } \ \ \ \textrm{for} \ \ \ |t|\ge T_{0}\nonumber \\ \end{aligned}$$

(3.13)

and Lemma 3.2 to deduce

$$\begin{aligned} K_{2,2}&= \frac{y^2}{2\pi i} \int _{\frac{9}{4}-i\infty }^{\frac{9}{4}+i\infty } \frac{\zeta (2-s)\zeta '(s)}{\zeta (s)}\frac{\left( \frac{x}{y}\right) ^{s}}{(s-2)s}ds \nonumber \\ &\quad + O\left( x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{4}} \int _{T}^{\infty } \frac{|\zeta \left( -\frac{1}{4}-it\right) |}{(1+|t|)^2} dt \right) \nonumber \\&= x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{4}}\kappa (u) + O\left( x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{4}}\frac{\log T}{T^{\frac{1}{4}}} \right) , \end{aligned}$$

(3.14)

where the Fourier integral $\kappa (u)$ is given by

$$\begin{aligned} \kappa (u):= \frac{1}{2\pi }\int _{-\infty }^{\infty } \frac{\zeta (-\frac{1}{4}-it)\zeta '(\frac{9}{4}+it)}{\zeta (\frac{9}{4}+it)} \frac{\textrm{e}^{itu}}{(\frac{1}{4}+it)(\frac{9}{4}+it)}dt \end{aligned}$$

with $u:=\log \frac{x}{y}$. Because the absolute value of $\kappa (u)$ is

$$\begin{aligned} |\kappa (u)|&\le \frac{1}{2\pi }\int _{-\infty }^{\infty } \left| \frac{\zeta (\frac{5}{4}+it)\zeta '(\frac{9}{4}+it)}{\zeta (\frac{9}{4}+it)}\right| \left| \frac{\chi (-\frac{1}{4}-it)}{(\frac{1}{4}+it)(\frac{9}{4}+it)}\right| dt \nonumber \\&\le \frac{2}{(2\pi )^{\frac{7}{4}}}\frac{\zeta (\frac{5}{4})\zeta (\frac{9}{4})\zeta '(\frac{9}{4})}{\zeta (\frac{9}{2})} \int _{0}^{\infty }\frac{t^{\frac{3}{4}}}{t^{2}+\left( \frac{3}{4}\right) ^2}dt \nonumber \\&\le \frac{2}{(2\pi )^{\frac{7}{4}}}\frac{\zeta (\frac{5}{4})\zeta (\frac{9}{4})\zeta '(\frac{9}{4})}{\zeta (\frac{9}{2})} \left( \int _{0}^{1}\frac{1}{t^{2}+\left( \frac{3}{4}\right) ^2}dt+\int _{1}^{\infty }t^{-\frac{5}{4}}dt \right) \nonumber \\&\le \frac{8}{(2\pi )^{\frac{7}{4}}}\frac{\zeta (\frac{5}{4})\zeta (\frac{9}{4})\zeta '(\frac{9}{4})}{\zeta (\frac{9}{2})}\left( \frac{\pi }{9}+1\right) \end{aligned}$$

(3.15)

using (3.13), the inequalities $\sqrt{\left( t^{2}+\left( \frac{1}{4}\right) ^{2}\right) }\sqrt{\left( t^{2}+\left( \frac{9}{4}\right) ^{2}\right) } \ge t^{2}+\left( \frac{3}{4}\right) ^{2}$ for $t\ge 0$, $|\frac{1}{\zeta (s)}|\le \frac{\zeta (\sigma )}{\zeta (2\sigma )}$ for $\sigma >1,$ and $|\chi (-\frac{1}{4}-it)| \asymp \left( \frac{|t|}{2\pi }\right) ^{\frac{3}{4}}.$ Hence $|\kappa (u)|$ is a bound. We define the function G(t) as

$$\begin{aligned} G(t)&:= \frac{1}{2\pi }\int _{\alpha }^{\frac{9}{4}} \frac{\zeta (2-\sigma -it)\zeta '(\sigma +it)}{\zeta (\sigma +it)}\ \frac{\left( \frac{x}{y}\right) ^{\sigma +it}}{(\sigma -2+it)(\sigma +it)}d\sigma . \end{aligned}$$

Then, we set

$$\begin{aligned} R &:= \int _{\frac{T}{2}}^{T}\left| \int _{\alpha }^{\frac{9}{4}} \frac{\zeta (2-\sigma -it)\zeta '(\sigma +it)}{\zeta (\sigma +it)}\ \frac{\left( \frac{x}{y}\right) ^{\sigma +it}}{(\sigma -2+it)(\sigma +it)}d\sigma \right| dt. \end{aligned}$$

We use Lemma 3.2 and (3.13) to obtain

$$\begin{aligned} R &\ll \log ^{3}T \int _{\alpha }^{\frac{3}{2}} \left( \frac{x}{y}\right) ^{\sigma } \int _{\frac{T}{2}}^{T}\frac{|\zeta (2-\sigma -it)|}{t^2}dt d\sigma \\&\qquad + \log ^{3}T \int _{\frac{3}{2}}^{\frac{9}{4}}\left( \frac{x}{y}\right) ^{\sigma } \int _{\frac{T}{2}}^{T}\frac{|\chi (2-\sigma -it)\zeta (\sigma -1+it)|}{t^2}dt d\sigma \\&\ll \frac{\log ^{3}T}{T^{1+\frac{1}{3}}}\int _{\alpha }^{\frac{3}{2}}\left( \frac{xT^{\frac{1}{3}}}{y}\right) ^{\sigma } d\sigma + \frac{\log ^{3}T}{T^{2-\frac{1}{6}}}\int _{\frac{3}{2}}^{\frac{9}{4}}\left( \frac{xT^{\frac{2}{3}}}{y}\right) ^{\sigma } d\sigma \\&\ll \frac{\log ^{3}T}{T }\cdot \frac{x}{y} \left( 1+T^{\frac{1}{6}}\left( \frac{x}{y}\right) ^{\frac{1}{2}}+{T^{\frac{2}{3}}}\left( \frac{x}{y}\right) ^{\frac{5}{4}}\right) . \end{aligned}$$

Hence, $T^{*}\in [\frac{T}{2},T]$ exists such that $|G(T^{*})|$ is minimum and

$$\begin{aligned} |G(T^{*})|&\ll \frac{1}{T}\cdot \frac{\log ^{3}T}{T }\cdot \frac{x}{y} \left( 1+T^{\frac{1}{6}}\left( \frac{x}{y}\right) ^{\frac{1}{2}}+{T^{\frac{2}{3}}}\left( \frac{x}{y}\right) ^{\frac{5}{4}}\right) \\&\ll \frac{1}{y^2}\left( \frac{x}{y}\right) ^{\frac{1}{4}}{\frac{\log ^{3}x}{x^{15}}} \ll x^{-10} \end{aligned}$$

by setting $T=x^{12}$. For a similar manner as in $K_{1}$, we have the weak estimates, that is, $K_{2,1},\ K_{2,3} \ll x^{-10}.$ Collecting the error estimates (3.14) and the above, we obtain the total contribution of all error terms, that is, $\ll x^{-\frac{3}{4}}.$ Therefore, we obtain

$$\begin{aligned} K_{2}&= \frac{\zeta '(2)}{4\zeta (2)}x^{2} + x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{4}}\kappa (u) + O\left( x^{-\frac{3}{4}} \right) \end{aligned}$$

(3.16)

with $T=x^{12}$.

3.5 Conclusion

Inserting (3.12) and (3.16) into (3.9), we obtain the formula (1.12), which proves Theorem 1.11.

4 Proof of Theorem 1.15

From (1.1) and the identity $(m,n)[m,n]=mn$ for any integers m and n, we have

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

where

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

We use $\sum _{d|n}\phi (d) = n,$ $\sum _{d|n}\Lambda (d) = \log n,$ and $\sum _{d\le x}\log d = (\log x - 1) x + O(\log x)$ to obtain

$$\begin{aligned}&\sum _{d_{1}k_{1}\le x} \sum _{d_{2}k_{2}\le x} (d_{1},d_{2}) \Lambda (k_{1}) \Lambda (k_{2}) =\sum _{d\le x}\phi (d) \sum _{l_{1}k_{1}\le x/d}\sum _{l_{2}k_{2}\le x/d }\Lambda (k_{1})\Lambda (k_{2}) \\&\quad =\sum _{d\le x}\phi (d) \left( \sum _{mk\le x/d} \Lambda (k)\right) ^{2} = \sum _{d\le x}\phi (d) \left( \sum _{n\le x/d}\sum _{k|n} \Lambda (k)\right) ^{2} \\&\quad =x^{2}(\log x -1)^{2}\sum _{d\le x}\frac{\phi (d)}{d^2} -2x^{2}(\log x -1)\sum _{d\le x}\frac{\phi (d)}{d^2}\log d \\&\qquad +x^2 \sum _{d\le x}\frac{\phi (d)}{d^2}\log ^{2}d + O\left( x\log ^{2}x \sum _{d\le x}\frac{\phi (d)}{d} +\log ^{2}x \sum _{d\le x}\phi (d)\right) . \end{aligned}$$

Using well-known formulas $\sum _{n\le x}\frac{\phi (n)}{n^2} = \frac{1}{\zeta (2)}\log x + O(1),$ $\sum _{n\le x}\frac{\phi (n)}{n^2}\log n = \frac{1}{2\zeta (2)}\log ^{2}x +O(1),$ and $\sum _{n\le x}\frac{\phi (n)}{n^2}\log ^{2}n =\frac{1}{3\zeta (2)}\log ^{3}x + O(1)$ we have

$$\begin{aligned}&\sum _{d_{1}k_{1}\le x} \sum _{d_{2}k_{2}\le x} (d_{1},d_{2}) \Lambda (k_{1}) \Lambda (k_{2}) =\frac{1}{3\zeta (2)}x^{2}\log ^{3}x + O\left( x^{2}\log ^{2}x\right) . \end{aligned}$$

Hence, we have

$$\begin{aligned} S_2(x,y)= \frac{1}{3\zeta (2)}yx^{2}\log ^{3}x + O\left( x^{4} + yx^{2}\log ^{2}x \right) , \end{aligned}$$

which completes the proof of Theorem 1.15. $\square$

5 Proof of Theorem 1.17

5.1 Lemmas

We need the following Lemmas to prove Theorem 1.17, namely

Lemma 5.1

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

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

(5.2)

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

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

(5.3)

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

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

Proof

The proof of this lemma follows from (4.12) in [3]. $\square$

To calculate $S_{2,1}(x,y)$ (See (5.15) with $j=1$ below), we use the Laurent expansions of the Riemann zeta-function at $s=1$, namely

$$\begin{aligned} \zeta (s)&= \frac{1}{s-1} + \gamma + \gamma _{1}(s-1) +\gamma _{2}(s-1)^2 + \gamma _{3}(s-1)^3 + \cdots , \end{aligned}$$

(5.4)

where $\gamma$ is the Euler–Mascheroni constant, and

$$\begin{aligned} \gamma _{k}:=\frac{(-1)^k}{k!}\lim _{N\rightarrow \infty }\left( \sum _{m=1}^{N}\frac{\log ^{k}m}{m} - \frac{\log ^{k+1}N}{k+1}\right) \qquad (k=1,2,\ldots ) \end{aligned}$$

(5.5)

are known as Stieltjes constants. Then, we have

$$\begin{aligned} \zeta '(s)&=- \frac{1}{(s-1)^2} +\gamma _{1} + 2\gamma _{2}(s-1) + 3\gamma _{3}(s-1)^2 + \cdots , \end{aligned}$$

(5.6)

$$\begin{aligned} \frac{\zeta '(s)}{\zeta (s)}&=- \frac{1}{s-1} + \gamma + (2\gamma _{1}-\gamma ^2)(s-1) +(\gamma ^3 -3\gamma \gamma _{1}+3\gamma _{2})(s-1)^2 + \cdots \end{aligned}$$

(5.7)

as $s\rightarrow 1$. We need the following residues, namely

Lemma 5.8

Let the notation be as above. We have

$$\begin{aligned}&\underset{s=1}{{\textrm{Res}}}~\frac{\zeta (s)\zeta '(s)}{\zeta (s +1)}\left( \log \frac{x}{e} - \frac{\zeta '(s +1)}{\zeta (s +1)}\right) \frac{x^s}{s} \nonumber \\&\quad = - \frac{1}{2\zeta (2)} x \log ^{3} x - \frac{1}{\zeta (2)}\left( \gamma -\frac{3}{2} -\frac{3}{2} \frac{\zeta '(2)}{\zeta (2)} \right) x \log ^{2} x \nonumber \\&\qquad - \frac{1}{\zeta (2)}\left( 2(1-\gamma ) + (3-2\gamma ) \frac{\zeta '(2)}{\zeta (2)} +3\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^2 - \frac{3}{2} \frac{\zeta ''(2)}{\zeta (2)} \right) x \log x \nonumber \\&\qquad - \frac{1}{\zeta (2)} \left( \gamma -1+2(\gamma -1) \frac{\zeta '(2)}{\zeta (2)} +(2\gamma -3)\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^2 -3\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^3 \right) x \nonumber \\&\qquad - \frac{1}{\zeta (2)}\left( \left( \frac{3}{2}-\gamma \right) \frac{\zeta ''(2)}{\zeta (2)}-\frac{1}{2} \frac{\zeta '''(2)}{\zeta (2)} +3\frac{\zeta '(2)}{\zeta (2)}\frac{\zeta ''(2)}{\zeta (2)}\right) x, \end{aligned}$$

(5.9)

$$\begin{aligned}&\underset{s=1}{\textrm{Res}}~\frac{(\zeta '(s))^2}{\zeta (s+1)}\frac{x^{s}}{s} = \frac{1}{6\zeta (2)} x \log ^{3} x - \frac{1}{2\zeta (2)}\left( 1+\frac{\zeta '(2)}{\zeta (2)}\right) x \log ^{2} x \nonumber \\&\quad + \frac{1}{\zeta (2)}\left( \left( \frac{\zeta '(2)}{\zeta (2)}\right) ^{2}+\frac{\zeta '(2)}{\zeta (2)} - \frac{\zeta ''(2)}{2\zeta (2)} +1-2\gamma _{1}\right) x\log x \nonumber \\&\quad - \frac{1}{\zeta (2)}\left( \left( \frac{\zeta '(2)}{\zeta (2)}\right) ^{3}+\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^{2} +\left( 1-2\gamma _{1}\right) \frac{\zeta '(2)}{\zeta (2)} - \frac{\zeta '(2)}{\zeta (2)}\frac{\zeta ''(2)}{\zeta (2)}\right) x \nonumber \\&\quad + \frac{1}{\zeta (2)}\left( \frac{\zeta ''(2)}{2\zeta (2)} - \frac{\zeta '''(2)}{6\zeta (2)}+2(\gamma _{1}-2\gamma _{2})-1\right) x, \end{aligned}$$

(5.10)

and

$$\begin{aligned}&\underset{s=1}{\textrm{Res}}~\zeta (2-s)\zeta '(s)\frac{\zeta '(s)}{\zeta (s)}\frac{u^s}{(2-s)s^2} \nonumber \\&\quad = -\frac{1}{6} u \log ^{3} u + \left( \frac{1}{2} + \gamma \right) u \log ^{2} u - 2\left( \gamma ^2+\gamma -\gamma _{1}+1\right) u \log u \nonumber \\&\qquad + 2\left( \gamma ^3 + \gamma ^2 +2\gamma -\gamma _{1}-3\gamma \gamma _{1} +3\gamma _{2}+1\right) u \end{aligned}$$

(5.11)

with $u = {x^2}/{y}$.

Proof

Suppose that g(s) is regular in the neighborhood at $s=1$, and f(s) has only a triple pole at $s=1$, then the Laurent expansion of f(s) implies

$$\begin{aligned} f(s):= \frac{a}{(s-1)^3} + \frac{b}{(s-1)^2} + \frac{c}{s-1} + h(s), \end{aligned}$$

where h(s) is regular in the neighborhood of its pole, and a, b, c are computable constants. We use the residue calculation to deduce

$$\begin{aligned}&\underset{s=1}{{\textrm{Res}}} f(s)g(s) =\frac{a}{2} g''(1) + b g'(1) + c g(1). \end{aligned}$$

To prove (5.9), we use (5.4) and (5.6) to obtain

$$\begin{aligned} \zeta (s)\zeta '(s)&= \frac{-1}{(s-1)^3} +\frac{-\gamma }{(s-1)^2} +\gamma _{2}+\gamma \gamma _{1} + O\left( |s-1|\right) \end{aligned}$$

as $s\rightarrow 1$. We set $g(s):=\frac{1}{\zeta (s +1)}\left( \log \frac{x}{e} - \frac{\zeta '(s +1)}{\zeta (s +1)}\right) \frac{x^s}{s},$ then

$$\begin{aligned} g'(1)=&\frac{1}{\zeta (2)}x\log ^{2}x -\frac{2}{\zeta (2)}\left( 1+\frac{\zeta '(2)}{\zeta (2)}\right) x\log x \\&+\frac{1}{\zeta (2)}\left( 1+2\frac{\zeta '(2)}{\zeta (2)}+2\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^2 - \frac{\zeta ''(2)}{\zeta (2)}\right) x, \end{aligned}$$

and

$$\begin{aligned} g''(1)=&\frac{1}{\zeta (2)}x \log ^{3} x - \frac{3}{\zeta (2)}\left( 1+\frac{\zeta '(2)}{\zeta (2)}\right) x\log x \\&+\frac{2}{\zeta (2)} \left( 2+\frac{\zeta '(2)}{\zeta (2)}+3\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^2 - \frac{3}{2} \frac{\zeta ''(2)}{\zeta (2)} \right) x\log x \\&- \frac{1}{\zeta (2)} \left( 2+4\frac{\zeta '(2)}{\zeta (2)} +6\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^2 \right. \\&\left. + 6\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^3 -6\frac{\zeta '(2)}{\zeta (2)} \frac{\zeta ''(2)}{\zeta (2)} -3 \frac{\zeta ''(2)}{\zeta (2)} + \frac{\zeta '''(2)}{\zeta (2)} \right) x. \end{aligned}$$

Hence, we have

$$\begin{aligned}&\underset{s=1}{{\textrm{Res}}}~\frac{\zeta (s)\zeta '(s)}{\zeta (s +1)}\left( \log \frac{x}{e} - \frac{\zeta '(s +1)}{\zeta (s +1)}\right) \frac{x^s}{s} =-\frac{1}{2}g''(1) -\gamma g'(1). \end{aligned}$$

We use the same method as above to prove (5.10) and (5.11). $\square$

5.2 Expressions of $S_{2,j}(x,y)$ for $j=1,2,3,4$

We assume that $1 \le y \le x^M$ for some constant M. Without loss of generality, we can assume that $x,\ y\in \mathbb {Z}+\frac{1}{2}$. Suppose that T is a real parameter at our disposal. Let $\alpha _{1}=1 + \frac{2}{\log x}$ and $\alpha _{2}=1 + \frac{3}{\log x}$. Applying (3.8) with $\alpha =\alpha _j\ (j=1,2)$  we have

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

(5.12)

where

$$\begin{aligned} F(s_{1},s_{2};n) :=\sigma _{1-s_{1}}(n)\sigma _{1-s_{2}}(n) \frac{\zeta '(s_{1})\zeta '(s_{2})}{\zeta (s_{1})\zeta (s_{2})} \frac{x^{s_{1}+s_{2}}}{s_{1}s_{2}} \end{aligned}$$

and

$$\begin{aligned} E_{2}(x;n)&:= E_{1}(x;n)\left( \frac{1}{2\pi i} \int _{\alpha _{1}-iT}^{\alpha _{1}+iT}\sigma _{1-s_{1}}(n)\frac{\zeta '(s_{1})}{\zeta (s_{1})} \frac{x^{s_{1}}}{s_{1}}ds_{1} \right. \\&\quad + \left. \frac{1}{2\pi i} \int _{\alpha _{2}-iT}^{\alpha _{2}+iT}\sigma _{1-s_{2}}(n)\frac{\zeta '(s_{2})}{\zeta (s_{2})} \frac{x^{s_{2}}}{s_{2}}ds_{2} + E_{1}(x;n) \right) \\&\ll \frac{x^2}{T}\sigma _{0}(n)^{2}\log ^{4} T. \end{aligned}$$

Summing (5.12) over n and using the inequality $\sum _{n\le y}\sigma _{0}(n)^{2} \ll y \log ^{3}y,$ we obtain

$$\begin{aligned} S_{2}(x,y)&= \frac{1}{(2\pi i)^2} \int _{\alpha _{1}-iT}^{\alpha _{1}+iT}\int _{\alpha _{2}-iT}^{\alpha _{2}+iT} G(s_{1},s_{2};y)\frac{\zeta '(s_{1})\zeta '(s_{2})}{\zeta (s_{1})\zeta (s_{2})} \frac{x^{s_{1}+s_{2}}}{s_{1}s_{2}}ds_{2}ds_{1} \nonumber \\&\quad + O\left( \frac{x^{2}y L^{7}}{T}\right) , \end{aligned}$$

(5.13)

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 integral of (5.13). Substituting (5.3) into (5.13), we obtain

$$\begin{aligned} S_{2}(x,y)&=\sum _{j=1}^{4}S_{2,j}(x,y) + O\left( x^{2} y L^{10} \left( \frac{1}{T} + y^{-1/2}\right) \right) , \end{aligned}$$

(5.14)

where

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

(5.15)

Note that we substitute $T=x$ with a small positive constant c into the error term on the right-hand side of (5.14) to obtain

$$\begin{aligned}&\ll x y L^{10} \left( 1 + \frac{x}{y^{1/2}}\right) . \end{aligned}$$

(5.16)

5.3 Evaluation of $S_{2,1}(x,y)$.

Let $\alpha _{1}= 1 +\frac{2}{\log x}$ and $\alpha _{2}= 1 + \frac{3}{\log x}$. From the definition of $R_1(s_1,s_2,y)$, we obtain

$$\begin{aligned} S_{2,1}(x,y)&= \frac{y}{(2\pi i)^2}\int _{\alpha _{1}-iT}^{\alpha _{1}+iT}\int _{\alpha _{2}-iT}^{\alpha _{2}+iT} \frac{\zeta '(s_1)\zeta '(s_2)\zeta (s_1 +s_2 -1)}{\zeta (s_1 +s_2)}\frac{x^{s_{1}+s_{2}}}{s_{1}s_{2}}ds_{2}ds_{1}. \end{aligned}$$

(5.17)

Let $\Gamma (\alpha , \beta ,T)$ denote the following contour comprising the line segments $[\alpha -iT, \beta -iT]$, $[\beta -iT, \beta +iT]$, and $[\beta +iT, \alpha +iT]$ (Fig. 1).

Fig. 1

$\Gamma (\alpha ,\beta ,T)$

In (5.17), 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 weak estimate $\int _1^T |\zeta '(\alpha _1+it)|dt \ll T\log T$ and Lemma 3.2, we have

$$\begin{aligned} \begin{aligned} J_{1,1}, J_{1,3}&\ll \frac{xyL}{T}\int _{-T}^{T}\frac{|\zeta '(\alpha _1+it_1)| }{1+|t_1|} \int _{\frac{1}{2}+\frac{1}{\log x}}^{\alpha _2}{|\zeta '(\sigma _2+iT) \zeta (\alpha _1+\sigma _2-1}\\ &{\qquad +i(t_1 + T))| x^{\sigma _2}}d\sigma _2dt_1 \\ &\ll \frac{xyL^4}{T} \int _{-T}^{T}\frac{|\zeta '(\alpha _1+it_1)| }{1+|t_1|} \int _{\frac{1}{2}+\frac{1}{\log x}}^{\alpha _{2}}T^{\frac{2}{3} (1-\sigma _{2})} x^{\sigma _2} d\sigma _2 dt_1 \\ &\ll \frac{x^{2}yL^{6}}{T^{2/3}}\left( x^{-1/2}+T^{-1/3}\right) . \end{aligned} \end{aligned}$$

For the integral along the vertical line, we have

$$\begin{aligned}&J_{1,2} \ll yx^{3/2} L \\&\quad \times \int _{-T}^{T}\int _{-T}^{T} \frac{|\zeta '(\alpha _{1}+it_1) \zeta '\left( \frac{1}{2}+\frac{1}{\log x}+it_2\right) \zeta (\alpha _1+\frac{1}{\log x} - \frac{1}{2} +i(t_1+t_2))|}{(1+|t_1|)(1+|t_2|)} dt_1 dt_2 \\&\quad \ll yx^{3/2}L^{3} \int _{-2T}^{2T} \left| \zeta \left( \frac{1}{2}+\frac{1}{\log x} +iu\right) \right| \int _{-T}^{T} \frac{|\zeta '(\frac{1}{2}+\frac{1}{\log x}+it)|}{(1+|t|)(1+|t-u|)}dt du. \end{aligned}$$

Now, we use Lemma 3.2 to obtain the estimate

$$\begin{aligned} \begin{aligned} \int _{-T}^{T} \frac{|\zeta '(\frac{1}{2}+\frac{1}{\log x}++it)|}{(1+|t|)(1+|t-u|)}dt&\ll T^{1/6}L^2 \left( \int _{|t-u|>\frac{1}{2} |u|} + \int _{|t-u|\le \frac{1}{2} |u|} \right) \frac{1}{(1+|t|)(1+|t-u|)}dt \\ &\ll \frac{T^{1/6}L^3}{1+|u|}, \end{aligned} \end{aligned}$$

and use the Cauchy–Schwarz inequality and the above to deduce

$$\begin{aligned} J_{1,2}&\ll yx^{3/2} T^{1/6} L^{6} \int _{-2T}^{2T} \frac{\left| \zeta \left( \frac{1}{2}+\frac{1}{\log x} +iu\right) \right| }{1+|u|} du \nonumber \\&\ll yx^{3/2}T^{1/6} L^{8}. \end{aligned}$$

(5.18)

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

$$\begin{aligned} \frac{\zeta '(s_1)\zeta '(2-s_1)}{\zeta (2)s_1(2-s_1)}~x^2 =:H_{1}(s_{1})x^{2}, \end{aligned}$$

and a double pole at $s_2=1$ with residue

$$\begin{aligned}&- \frac{(\zeta '(s_1))^2}{\zeta (s_{1}+1)s_{1}}x^{s_{1}+1} - \frac{\zeta (s_1)\zeta '(s_1)}{\zeta (s_{1}+1)s_1}\left( \log \frac{x}{e} - \frac{\zeta '(s_{1}+1)}{\zeta (s_{1}+1)}\right) ~{x^{s_1+1}} \\&\qquad =:H_{2}(s_1) x^{s_{1}+1} + H_{3}(s_1) x^{s_{1}+1}. \end{aligned}$$

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

$$\begin{aligned} \begin{aligned}&\frac{x^2y}{2\pi i}\int _{\alpha _1-iT}^{\alpha _1+iT} H_1(s_1)ds_1 {+} \frac{x y}{2\pi i}\int _{\alpha _1-iT}^{\alpha _1{+}iT}H_2(s_1)x^{s_1}ds_1 {+} \frac{x y}{2\pi i}\int _{\alpha _1-iT}^{\alpha _1+iT}H_3(s_1)x^{s_1}ds_1 \\ &\quad =: I_1 + I_2 + I_{3}, \ {\text {say}}. \end{aligned} \end{aligned}$$

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

$$\begin{aligned} I_1&=\frac{x^2y}{2\pi i}\int _{5/4-i\infty }^{5/4+i\infty } H_1(s_1)ds_1 +O\left( x^2y\int _{T}^{\infty }\left| H_1\left( \frac{5}{4}\pm it_1\right) \right| dt_1\right) \nonumber \\&=\frac{c_{0}}{\zeta (2)}x^2y+O\left( \frac{x^2y L^4}{T^{11/12}}\right) , \end{aligned}$$

where the computable constant $c_0$ is given by

$$\begin{aligned} c_{0}:= \frac{1}{2\pi i}\int _{\frac{5}{4} -i\infty }^{\frac{5}{4} +i\infty } \frac{\zeta '(s_1)\zeta '(2-s_1)}{s_1(2-s_1)}ds_1. \end{aligned}$$

(5.19)

For $I_2$, we move the line of integration to $\Gamma (\alpha _1,\frac{1}{2}+\frac{1}{\log x},T)$. Using Lemma 3.2, the integrals over the horizontal lines are

$$\begin{aligned} \ll \frac{xy L^{5}}{T} \int _{\frac{1}{2}+\frac{1}{\log x}}^{\alpha _1}T^{\frac{2}{3}(1-\sigma _1)}x^{\sigma _1} d\sigma _1 \ll \frac{x^{3/2}y L^{5}}{T} \left( x^{1/2} + T^{1/3}\right) \end{aligned}$$

and that over the vertical line is

$$\begin{aligned}&\ll xy L \int _{-T}^{T}\frac{|\zeta '(\frac{1}{2}+it_1)|^2}{1+|t_1|} x^{1/2}dt_1 \ll x^{3/2}yL^{5} \end{aligned}$$

using $\int _{2\pi }^{T}\left| \zeta '\left( \frac{1}{2}+iv\right) \right| ^2\frac{dv}{v} \ll \log ^{4}T$ (see (172) in Hall [5]). Moving the path of integration, a pole of order 4 exists at $s_1=1$. Hence, we use Cauchy’s theorem and (5.10) to obtain

$$\begin{aligned} I_2&= -\frac{1}{6\zeta (2)} x^2 y \log ^{3} x + \frac{1}{2\zeta (2)}\left( 1+\frac{\zeta '(2)}{\zeta (2)}\right) x^2 y \log ^{2} x \\&\quad - \frac{1}{\zeta (2)}\left( \left( \frac{\zeta '(2)}{\zeta (2)}\right) ^{2}+\frac{\zeta '(2)}{\zeta (2)} - \frac{\zeta ''(2)}{2\zeta (2)} + 2\gamma _{1} +1\right) x^2 y\log x \\&\quad + \frac{1}{\zeta (2)}\left( \left( \frac{\zeta '(2)}{\zeta (2)}\right) ^{3}+\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^{2} +\left( 2\gamma _{1}+1\right) \frac{\zeta '(2)}{\zeta (2)} - \frac{\zeta '(2)}{\zeta (2)}\frac{\zeta ''(2)}{\zeta (2)}\right) x^2 y \\&\quad - \frac{1}{\zeta (2)}\left( \frac{\zeta ''(2)}{2\zeta (2)} - \frac{\zeta '''(2)}{6\zeta (2)}-2(\gamma _{1}+\gamma _{2})-1\right) x^2 y \\ &\quad + O\left( \frac{x^{3/2}y L^5}{T}\left( x^{1/2} + T^{1/3}\right) \right) \\&\quad + O(x^{3/2}yL^5), \end{aligned}$$

where $\gamma _{1}$ and $\gamma _{2}$ are the Stieltjes constants.

Similarly to $I_2$, we move the line of integration to $\Gamma (\alpha _1,\frac{1}{2}+\frac{1}{\log x},T)$ to calculate $I_3$. The integrals over the horizontal lines are

$$\begin{aligned} \ll \frac{xy L^{6}}{T} \int _{\frac{1}{2}+\frac{1}{\log x}}^{\alpha _1}T^{\frac{2}{3}(1-\sigma _1)}x^{\sigma _1} d\sigma _1 \ll \frac{x^{3/2}y L^{6}}{T} \left( x^{1/2} + T^{1/3}\right) \end{aligned}$$

and the integral over the vertical line is

$$\begin{aligned}&\ll xy L^{4} \int _{-T}^{T}\frac{|\zeta (\frac{1}{2}+it_1)\zeta '(\frac{1}{2}+it_1)|}{1+|t_1|} x^{1/2}dt_1 \ll x^{3/2}yL^{7} \end{aligned}$$

using $\int _{2\pi }^{T}\left| \zeta \left( \frac{1}{2}+iv\right) \zeta '\left( \frac{1}{2}+iv\right) \right| \frac{dv}{v} \ll \log ^{3}T$ (see (173) in Hall [5]). Furthermore, when moving the path of integration, a triple pole exists at $s_1=1$. Hence, using Cauchy’s theorem and (5.9) we have

$$\begin{aligned} \begin{aligned} I_3&= \frac{1}{2\zeta (2)} x^2 y \log ^{3} x + \frac{1}{\zeta (2)}\left( \gamma -\frac{3}{2} -\frac{3}{2} \frac{\zeta '(2)}{\zeta (2)} \right) x^2 y \log ^{2} x \\ &\quad + \frac{1}{\zeta (2)}\left( 2(1-\gamma ) + (3-2\gamma ) \frac{\zeta '(2)}{\zeta (2)} +3\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^2 - \frac{3}{2} \frac{\zeta ''(2)}{\zeta (2)} \right) x^2 y \log x \\ &\quad + \frac{1}{\zeta (2)} \left( \gamma -1+2(\gamma -1) \frac{\zeta '(2)}{\zeta (2)} +(2\gamma -3)\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^2 -3\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^3 \right) x^2 y \\ &\quad + \frac{1}{\zeta (2)}\left( \left( \frac{3}{2}-\gamma \right) \frac{\zeta ''(2)}{\zeta (2)}-\frac{1}{2} \frac{\zeta '''(2)}{\zeta (2)} +3\frac{\zeta '(2)}{\zeta (2)}\frac{\zeta ''(2)}{\zeta (2)}\right) x^2 y \\ &\quad + O\left( \frac{x^{3/2}y L^6}{T}\left( x^{1/2} + T^{1/3}\right) \right) + O(x^{3/2}yL^7),\end{aligned} \end{aligned}$$

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

$$\begin{aligned} S_{2,1}(x,y)&= \frac{1}{3\zeta (2)}yx^{2}\log ^{3}x + \frac{1}{\zeta (2)}\left( \gamma -1-\frac{\zeta '(2)}{\zeta (2)}\right) yx^{2}\log ^{2}x \nonumber \\&\quad + \frac{1}{\zeta (2)}\left( 1-2(\gamma +\gamma _{1}) +2(1-\gamma )\frac{\zeta '(2)}{\zeta (2)} +2\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^2 \right. \nonumber \\&\left. \quad -\frac{\zeta ''(2)}{\zeta (2)} \right) yx^{2}\log x + \frac{1}{\zeta (2)}\Bigg (c_{0} +\gamma + 2(\gamma _{1}+\gamma _{2}) +\left( 2\gamma +2\gamma _{1}-1\right) \nonumber \\ &\qquad \frac{\zeta '(2)}{\zeta (2)} + 2(\gamma -1)\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^2 \Bigg ) yx^{2} \nonumber \\&\quad + \frac{1}{\zeta (2)} \left( (1-\gamma )\frac{\zeta ''(2)}{\zeta (2)} - \frac{\zeta '''(2)}{3\zeta (2)} + 2\frac{\zeta '(2)}{\zeta (2)} \frac{\zeta ''(2)}{\zeta (2)} - 2\left( \frac{\zeta '(2)}{\zeta (2)}\right) ^3\right) yx^{2} \nonumber \\&\quad + O(x^{5/3}yL^{8}). \end{aligned}$$

(5.20)

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

5.4 Estimation of $S_{2,4}(x,y)$.

This is determined explicitly by

$$\begin{aligned} S_{2,4}(x,y)&= \frac{y^3}{(2\pi i)^2}\int _{\alpha _1-iT}^{\alpha _1+iT}\int _{\alpha _2-iT}^{\alpha _2+iT} \\&\quad \frac{\zeta (3-s_1-s_2)\zeta (2-s_1)\zeta (2-s_2)}{\zeta (4-s_1-s_2)(3-s_1-s_2){s_1s_2}} \frac{\zeta '(s_1)\zeta '(s_2)}{\zeta (s_1)\zeta (s_2)} \left( \frac{x}{y}\right) ^{s_1+s_2} ds_2ds_1. \end{aligned}$$

For this purpose, we move the line of integral with respect to $s_{2}$ to contour $\Gamma (\beta ,\alpha _{2} , T)$, where $\beta =\frac{5}{2} -\alpha _{1}=\frac{3}{2} - \frac{2}{\log x}$. No poles are present when we deform the path of the integral over $s_2$. The contribution from the horizontal lines is

$$\begin{aligned}&J_{4,1}, J_{4,3} \ll xy^{2}\left( \frac{x}{y}\right) ^{\frac{1}{\log x}} \int _{-T}^{T}\frac{\left| \zeta \left( 1-\frac{2}{\log x} -it_{1}\right) \zeta '\left( 1+\frac{2}{\log x}+it_{1}\right) \right| }{\left| \zeta \left( 1+\frac{2}{\log x}+it_{1}\right) \right| (1+|t_{1}|)} dt_1 \\&\ \ \times \int _{\alpha _{2}}^{\beta } \frac{\left| \zeta \left( 2-\frac{2}{\log x} -\sigma _2 -i(t_{1}+T)\right) \zeta \left( 2-\sigma _{2}-iT\right) \zeta '\left( \sigma _{2}+iT\right) \right| }{\left| \zeta \left( 3-\frac{2}{\log x}-\sigma _{2}-i(t_{1}+T)\right) \zeta \left( \sigma _{2}+iT\right) \right| (1+|t_{1}+T|)T} \left( \frac{x}{y}\right) ^{\sigma _{2}}d\sigma _{2}. \end{aligned}$$

The inner integral is estimated as

$$\begin{aligned}&\ll \frac{L^5}{T(1+|t_1+T|)}\left( \frac{x}{y}\right) \left( 1+T^{1/6}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\right) , \end{aligned}$$

where we have used Lemma 3.2 and assumption $y \ll x^M$. Hence, we have

$$\begin{aligned} J_{4,1},J_{4,3}&\ll \frac{ x^{2}y L^{8}}{T}\left( 1 +T^{\frac{1}{6}}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\right) \int _{-T}^{T}\frac{\left| \zeta \left( 1-\frac{2}{\log x}-it_{1}\right) \right| }{(1+|t_{1}|)(1+|t_{1}|+T)}dt_{1} \\&\ll \frac{x^{2}y L^{10}}{T^{2}}\left( 1+T^{\frac{1}{6}}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\right) . \end{aligned}$$

For the integral on the vertical line, we find that

$$\begin{aligned} J_{4,2}&\ll y^{3}\int _{-T}^{T}\int _{-T}^{T} \frac{\left| \zeta (\frac{1}{2}-i(t_{1}+t_{2}))\zeta (1-\frac{2}{\log x}-it_{1})\zeta (\frac{1}{2}+\frac{2}{\log x}-it_{2})\right| }{(1+|t_1+t_{2}|)(1+|t_{1}|)(1+|t_{2}|)} \\&\qquad \times \frac{\left| \zeta '\left( 1+\frac{2}{\log x}+it_{1}\right) \zeta '\left( \frac{3}{2} -\frac{2}{\log x}-it_{2}\right) \right| }{\left| \zeta \left( 1+\frac{2}{\log x}+it_{1}\right) \zeta \left( \frac{3}{2} -\frac{2}{\log x}-it_{2}\right) \right| } \left( \frac{x}{y}\right) ^{{5}/{2}}dt_{1}dt_{2}\\&\quad \ll y^{3}\left( \frac{x}{y}\right) ^{{5}/{2}} L^{6} \int _{-2T}^{2T} \frac{\left| \zeta \left( \frac{1}{2}-iu\right) \right| }{1+|u|} \int _{-T}^{T}\frac{\left| \zeta \left( \frac{1}{2}+\frac{2}{\log x}-it_{2}\right) \right| }{(1+|t_{2}|)(1+|u-t_{2}|)}dt_{2}du \\&\quad \ll x^{2}y L^{10} \left( \frac{x}{y}\right) ^{{1}/{2}} \end{aligned}$$

using a well-known estimate $\int _{1}^{T}|\zeta (\frac{1}{2}+it)|^{2}\frac{dt}{t} \ll \log ^{2}T.$ Hence, we take $T=x$ to obtain

$$\begin{aligned} S_{2,4}(x,y) \ll x^{2}y L^{10}\left( \frac{x}{y}\right) ^{{1}/{2}}. \end{aligned}$$

(5.21)

5.5 Estimation of $S_{2,3}(x,y)$.

This is determined explicitly by

$$\begin{aligned} S_{2,3}(x,y)&=\frac{y^2}{(2\pi i)^2} \int _{\alpha _1-iT}^{\alpha _1+iT}\int _{\alpha _2-iT}^{\alpha _2+iT}\\ &\quad \frac{\zeta (2-s_2)\zeta (1+s_1-s_2)\zeta '(s_1)\zeta '(s_2)}{\zeta (2+s_1-s_2)\zeta (s_2)(2-s_2)}\frac{x^{s_1+s_2}y^{-s_2}}{s_1s_2}ds_2ds_1. \end{aligned}$$

We move the path of integration with respect to $s_2$ to $\Gamma (\frac{3}{2}, \alpha _2, T)$. No poles with this deformation exist. The contribution from the horizontal lines is

$$\begin{aligned} J_{3,1}, J_{3,3}&\ll \frac{y^2 x L}{T^2} \int _{-T}^{T}\frac{|\zeta '(\alpha _1+it_1)|}{1+|t_1|} \\&\qquad \int _{\alpha _2}^{3/2} \frac{|\zeta (2-\sigma _2-iT)\zeta (1+\alpha _1-\sigma _2+i(t_1-T))\zeta '(\sigma _{2}+iT)|}{|\zeta (\sigma _{2}+iT)|} \\&\qquad \times \left( \frac{x}{y}\right) ^{\sigma _2} d\sigma _2 dt_1 \\&\quad \ll \frac{y^2 x L^{6}}{T^2} \int _{-T}^{T}\frac{|\zeta '(\alpha _1+it_1)|}{1+|t_1|}\\&\qquad \int _{\alpha _2}^{3/2} T^{\frac{1}{3}(-1+\sigma _2)} (1+|t_1-T|)^{\frac{1}{3}(-1+\sigma _2)}\left( \frac{x}{y}\right) ^{\sigma _2}d\sigma _2 dt_1 \\&\quad \ll y x^2 L^{8} \left( T^{-2}+T^{-5/3}\left( \frac{x}{y}\right) ^{1/2}\right) \end{aligned}$$

using Lemma 3.2. In contrast, the contribution from the vertical lines is

$$\begin{aligned} J_{3,2}&\ll y^2 x \int _{-T}^{T}\frac{|\zeta '(\alpha _1+it_1)|}{1+|t_1|} \\&\qquad \int _{-T}^T \frac{|\zeta (\frac{1}{2}-it_2)\zeta (\frac{1}{2}+{\frac{2}{\log x}}+i(t_1-t_2))\zeta '(\frac{3}{2} +it_2)|}{|\zeta (\frac{3}{2}+{\frac{2}{\log x}}+i(t_1-t_2))|(1+|t_2|)^2|\zeta (\frac{3}{2} +it_2)|}\left( \frac{x}{y}\right) ^{3/2}dt_2dt_1 \\&\ll y^{2}{x} \left( \frac{x}{y}\right) ^{3/2}L^{6} \int _{-T}^{T} \int _{-2T}^{2T}\frac{|\zeta (\frac{1}{2}-it_2)|}{(1+|t_2|)^2} \frac{|\zeta (\frac{1}{2}+\frac{1}{\log x}+iu)|}{1+|u+t_{2}|} du dt_{2} \\&\ll y^2 x\left( \frac{x}{y} \right) ^{3/2}L^{8}. \end{aligned}$$

Hence, we substitute $T=x$ into the above to obtain

$$\begin{aligned} S_{2,3}(x,y) \ll x^{2}yL^{8}\left( \frac{x}{y}\right) ^{1/2}. \end{aligned}$$

(5.22)

5.6 Evaluation of $S_{2,2}(x,y)$.

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

$$\begin{aligned} S_{2,2}(x,y)&=\frac{y^2}{(2\pi i)^2}\int _{\alpha _1-iT}^{\alpha _1+iT}\int _{\alpha _2-iT}^{\alpha _2+iT} \nonumber \\&\quad \frac{\zeta (2-s_1)\zeta (1-s_1+s_2)\zeta '(s_1)\zeta '(s_2)}{\zeta (2-s_1+s_2)\zeta (s_1)(2-s_1)} \frac{x^{s_1+s_2}y^{-s_1}}{s_1s_2}ds_2ds_1. \end{aligned}$$

(5.23)

First, we move the integral line from $s_1$ to $\Gamma (\frac{3}{2},\alpha _{1},T)$. The estimates over the horizontal and vertical lines are the same as that of $S_{2,3}(x,y)$, but a simple pole exists at $s_1=s_2$ inside this contour. The residue of the integrand of (5.23) at this pole is

$$\begin{aligned} -\frac{\zeta (2-s_2)(\zeta '(s_2))^{2}}{\zeta (2)\zeta (s_2)(2-s_2)s_2^{2}} x^{2s_2}y^{-s_2}. \end{aligned}$$

The contribution from the horizontal lines is

$$\begin{aligned} J_{2,1}, J_{2,3}&\ll \frac{y^2 x L}{T^2} \int _{-T}^{T}\frac{|\zeta '(\alpha _2+it_2)|}{1+|t_2|} \\&\qquad \int _{\alpha _1}^{3/2} \frac{|\zeta (2-\sigma _1-iT)\zeta (1+\alpha _2-\sigma _1+i(t_2-T))\zeta '(\sigma _{1}+iT)|}{|\zeta (\sigma _{1}+iT)|}\times \\&\quad \times \left( \frac{x}{y}\right) ^{\sigma _1}d\sigma _1 dt_2 \\&\ll \frac{y^2 x L^6}{T^2} \int _{-T}^{T}\frac{|\zeta '(\alpha _2+it_2)|}{1+|t_2|}\\ &\quad \int _{\alpha _1}^{3/2} T^{\frac{1}{3}(-1+\sigma _1)} (1+|t_2-T|)^{\frac{1}{3}(-1+\sigma _1)}\left( \frac{x}{y}\right) ^{\sigma _1}d\sigma _2 dt_1 \\&\ll y x^2 L^8 \left( T^{-2}+T^{-5/3}\left( \frac{x}{y}\right) ^{1/2}\right) \end{aligned}$$

using Lemma 3.2. In contrast, the contribution from the vertical lines is

$$\begin{aligned} J_{2,2}&\ll y^2 x \int _{-T}^{T}\frac{|\zeta '(\alpha _2+it_2)|}{1+|t_2|} \\&\quad \int _{-T}^T \frac{|\zeta (\frac{1}{2}-it_1)\zeta (\frac{1}{2}+\frac{3}{\log x}+i(t_2-t_1))\zeta '(\frac{3}{2} +it_1)|}{|\zeta (\frac{3}{2}+\frac{3}{\log x}+i(t_2-t_1))|(1+|t_1|)^2|\zeta (\frac{3}{2} +it_1)|}\left( \frac{x}{y}\right) ^{3/2}dt_1dt_2 \\&\ll y^{2}{x} \left( \frac{x}{y}\right) ^{3/2}L^{6} \int _{-T}^{T} \int _{-2T}^{2T}\frac{|\zeta (\frac{1}{2}-it_1)|}{(1+|t_1|)^2} \frac{|\zeta (\frac{1}{2}+\frac{3}{\log x}+iu)|}{1+|u+t_{1}|} du dt_{1} \\&\ll y^2 x\left( \frac{x}{y} \right) ^{3/2}L^{8}. \end{aligned}$$

Hence, we substitute $T=x$ into the above to obtain

$$\begin{aligned} J_{2,1}, J_{2,2}, J_{2,3} \ll x^{2}yL^{8}\left( \frac{x}{y}\right) ^{1/2}. \end{aligned}$$

Hence, we have

$$\begin{aligned} S_{2,2}(x,y)&= \frac{y^2}{\zeta (2)}Q(x,y) + O\left( x^{2}yL^8\left( \frac{x}{y}\right) ^{1/2}\right) , \end{aligned}$$

where

$$\begin{aligned} Q(x,y):= -\frac{1}{2\pi i}\int _{\alpha _2-iT}^{\alpha _2+iT} \zeta (2-s_2) \zeta '(s_2) \frac{\zeta '(s_2)}{\zeta (s_2)}\cdot \frac{1}{(2-s_2)s_2^2} \left( \frac{x^2}{y}\right) ^{s_2}ds_2. \end{aligned}$$

It remains to evaluate the integral Q(x, y). We move the integration with respect to $s_2$ to $\Gamma (\alpha _2,\alpha _{0},T)$ with $\alpha _{0}=1-\frac{c}{\log T}$, where c is a small positive constant, and denote the integrals over the horizontal line segments by $Q_{1}(x,y)$ and $Q_{3}(x,y)$, and the integral over the vertical line segment by $Q_{2}(x,y)$, respectively. Using Lemma 3.2 and the estimate $\left| - \frac{\zeta ' (\sigma +iT)}{\zeta (\sigma +iT)}\right| \ll \log T$ for $\sigma \ge \alpha _{0}$, we have

$$\begin{aligned} \begin{aligned} Q_{1}(x,y)&\ll \int _{\alpha _{0}}^{\alpha _{2}}\left| \zeta (2-\sigma -iT) \zeta '(\sigma +iT)\right| \left| - \frac{\zeta '(\sigma +iT)}{\zeta (\sigma +iT)}\right| \cdot \frac{1}{T^3} \left( \frac{x^2}{y}\right) ^{\sigma }d\sigma \\ &\ll \left( \frac{x^2}{y}\right) ^{\alpha _{2}} L^{4}(\alpha _{2}-\alpha _{0})T^{-2} \ll \frac{x^2}{y} \frac{L^{3}}{T^{2}}, \end{aligned} \end{aligned}$$

and similarly, $Q_{3}(x,y) \ll \frac{x^2}{y} \frac{L^{3}}{T^{2}},$ and

$$\begin{aligned} \begin{aligned} Q_{2}(x,y)&\ll \int _{-T}^{T}\left| \zeta \left( 1+\frac{c}{\log T}-it_2\right) \zeta '\left( 1-\frac{c}{\log T}+it_2\right) \right| \\ &\qquad \quad ~\left| - \frac{\zeta '\left( 1-\frac{c}{\log T}+it_2\right) }{\zeta \left( 1-\frac{c}{\log T}+it_2\right) }\right| \frac{1}{1+|t_2|^2} \left( \frac{x^2}{y}\right) ^{\alpha _{0}}dt_2 \ll \left( \frac{x^2}{y}\right) ^{\alpha _{0}}L^4. \end{aligned} \end{aligned}$$

Therefore, using Cauchy’s theorem, (5.11) with $u=x^2/y$ in Lemma 5.8 and taking $T={x}$ in the above we have

$$\begin{aligned} S_{2,2}(x,y)&=\frac{1}{6\zeta (2)} y x^2 \log ^{3}\frac{x^2}{y} - \frac{2\gamma +1}{2\zeta (2)} y x^2 \log ^{2}\frac{x^2}{y} \nonumber \\&\quad + \frac{2\left( \gamma ^2+\gamma -\gamma _{1}+1\right) }{\zeta (2)} y x^2 \log \frac{x^2}{y}\nonumber \\&\quad -\frac{ 2\left( \gamma ^3 +\gamma ^2 +2\gamma -\gamma _{1}-3\gamma \gamma _{1}+3\gamma _{2}+1\right) }{\zeta (2)} y x^2 \nonumber \\&\quad + O\left( x^{2}yL^{8}\left( \frac{x}{y} \right) ^{1/2}\right) . \end{aligned}$$

(5.24)

5.7 Asymptotic Formula of (1.18).

Now, we substitute (5.16), (5.20), (5.21), (5.22), and (5.24) into (5.14) to obtain the assertion of theorem 1.15. $\square$

6 Evaluation of $c_0$

We use (5.19) and Lemma 3.2 to obtain

$$\begin{aligned} c_{0}&= \frac{1}{2\pi i} \int _{{5}/{4}-iT}^{{5}/{4}+iT} \frac{\zeta '(s)\zeta '(2-s)}{s(2-s)}ds + O\left( T^{-1/2} \right) , \end{aligned}$$

then

$$\begin{aligned} |c_{0}|&\le \frac{\left| \zeta '(\frac{5}{4})\right| }{2\pi } \int _{-T}^{T} \frac{\left| \zeta '\left( \frac{3}{4}-it\right) \right| }{\left| \frac{3}{4}-it\right| \left| \frac{5}{4}+it\right| }dt + O\left( T^{-1/2} \right) . \end{aligned}$$

As $T\rightarrow \infty$, then we have

$$\begin{aligned} |c_{0}|&\le \frac{\left| \zeta '(\frac{5}{4})\right| }{\pi } \int _{0}^{\infty } \frac{\left| \zeta '\left( \frac{3}{4}+it\right) \right| }{\sqrt{((\frac{3}{4})^2 +t^2)((\frac{5}{4})^2 +t^2)}}dt \\&\le \frac{4\left| \zeta '(\frac{5}{4})\right| }{3\pi } \int _{0}^{\infty } \frac{\left| \zeta '\left( \frac{3}{4}(1+iy)\right) \right| }{1+y^2}dy \\&\le 0.425\cdots ~\left| \zeta '\left( \frac{5}{4}\right) \right| \int _{0}^{\infty } \frac{\left| \zeta '\left( \frac{3}{4}(1+iy)\right) \right| }{1+y^2}dy. \end{aligned}$$

Here, the integral on the right-hand side of the above is absolutely convergent, and it is a computable constant.