Sums of logarithmic weights involving r-full numbers

Abstract

Let (n, q) denote the greatest common divisor of positive integers n and q, and let $f_{r}$ denote the characteristic function of r-full numbers. We consider several asymptotic formulas for sums of the modified square-full ($r=2$) and cube-full numbers ($r=3$), which is $\sum _{n\le y}\sum _{q\le x}\sum _{d|(n,q)}df_{r}\left( \frac{q}{d}\right) \log \frac{x}{q}$ with any positive real numbers x and y. Moreover, we derive the asymptotic formula of the above with $r=2$ under the Riemann Hypothesis.

1 Introduction

Let $s=\sigma +it$ be the complex variable, and let $\zeta (s)$ be the Riemann zeta-function. Let $r (\ge 2)$ be an integer, we call n an r-full or r-free integer if $p|n \Rightarrow p^{r}|n$ or $p|n \Rightarrow p^{r}\not \mid n$, respectively. In the special case when $r=2$ or 3 integer. We call n a square-full or cube-full numbers, respectively. Let G(r) denote the set of r-full numbers, and let (n, q) denote the greatest common divisor of positive integers n and q. Define

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

and

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

(1.1)

It is worth mentioning that the above sum is an analogue of the Ramanujan sum $c_{q}(n)=\sum _{d|(n,q)}d\mu \left( {q}/{d}\right) ,$ with $\mu$ being the Möbius function. For the case $r=2$ and $r=3$, the Dirichlet series of the function $s_{q}^{(r)}(n)$ is given by

$$\begin{aligned} \sum _{q=1}^{\infty }\frac{s_{q}^{(2)}(n)}{q^s} = \sigma _{1-s}(n)\frac{\zeta (2s)\zeta (3s)}{\zeta (6s)} \end{aligned}$$

(1.2)

for $\textrm{Re}~s >\frac{1}{2}$, and

$$\begin{aligned} \sum _{q=1}^{\infty }\frac{s_{q}^{(3)}(n)}{q^s} = \sigma _{1-s}(n)\frac{\zeta (3s)\zeta (4s)\zeta (5s)\kappa _{9}(s)}{\zeta (8s)} \end{aligned}$$

(1.3)

for $\textrm{Re}~s >\frac{1}{3}$. Here $\sigma _{1-s}(n)=\sum _{d|n}d^{1-s},$ and the function $\kappa _{9}(s)$ is absolutely convergent for $\textrm{Re}~s >\frac{1}{9}$, that is

$$\begin{aligned} \kappa _{9}(s)=\frac{\zeta (13s)\zeta (14s)\zeta (21s)\zeta ^{2}(22s)\zeta ^{2}(23s)\zeta (24s)\cdots }{\zeta (9s)\zeta (10s)\zeta (17s)\zeta ^{}(18s)\zeta ^{}(19s)\zeta (25s)\zeta ^{3}(26s)\cdots } \end{aligned}$$

(see (1.96), (1.97) and (1.98) in [5]). For any large positive real numbers x and y, and any non-negative integer k, we are interested by studying the double sums

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

(1.4)

In this paper, we shall consider the asymptotic formulas for $S_{k}^{(r)}(x,y)$ when $r=2,3$. In the case $k=0$ and $r=2$, the author [6] used the method of Chan and Kumchev [1] (see also [9, 11])¹ and the theory of exponent pairs (see [3, 5])) to deduce the asymptotic formula to $S_{0}^{(2)}(x,y)$. It is shown that

$$\begin{aligned} S_{0}^{(2)}(x,y)&=\frac{\zeta (2)\zeta (3)}{\zeta (6)} xy - \frac{\zeta (4)\zeta (6)}{4\zeta (12)} x^{2} + O\left( x^{\frac{1}{2}} y + x^{}y^{\frac{1}{3}} + \frac{x^3}{y} \right) \end{aligned}$$

(1.5)

holds, where x and y are large real numbers such that $x \ll y \ll x^{\frac{3}{2}}$.

Recently, the author [7] gave a more precise asymptotic for $S_{0}^{(2)}(x,y)$ by using Lemma 2.2 below and some properties of the Riemann zeta-function. He proved that

$$\begin{aligned} S_{0}^{(2)}(x,y)&=\frac{\zeta (2)\zeta (3)}{\zeta (6)} xy + \frac{\zeta (\frac{1}{2})\zeta (\frac{3}{2})}{\zeta (3)} x^{\frac{1}{2}}y - \frac{\zeta (4)\zeta (6)}{4\zeta (12)} x^{2} \nonumber \\&\qquad + O\left( x^{\frac{4}{9}} y \log ^{4}x + x^{}y^{\frac{1}{3}}\log ^{2}y + x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\log ^{\frac{3}{2}}x \right) \end{aligned}$$

(1.6)

holds, where x and y are large real numbers such that $x^{\frac{4}{3}}\log x \ll y \ll \frac{x^{\frac{14}{9}}}{\log ^{4}x}$. Moreover, for $k=0$ and $r=3$, the author [8] showed that

$$\begin{aligned} S_{0}^{(3)}(x,y)&=\frac{\zeta (3)\zeta (4)\zeta (5)\kappa _{9}(1)}{\zeta (8)} xy - \frac{\zeta (6)\zeta (8)\zeta (10)\kappa _{9}(2)}{4\zeta (16)} x^{2} \nonumber \\&\qquad + O\left( x^{\frac{1}{3}} y + x^{}y^{\frac{1}{3}} + \frac{x^3}{y} \right) \end{aligned}$$

(1.7)

holds, where x and y denote large real numbers such that $x \ll y \ll x^{\frac{5}{3}}$. From the above, we notice that it is difficult to improve the error because the term $O\left( x^{\frac{1}{3}}y\right)$ is absorbed into all error terms. For this reason, in this paper, we consider asymptotic formulas for $S_{1}^{(r)}(x,y)$ and give the interesting relation between $S_{0}^{(r)}(x,y)$ and $S_{1}^{(r)}(x,y)$ for $r=2,3$. It is the most interesting problem for us to derive asymptotic formulas of (1.4) when $k=0,1$, and by a similar argument, we may prove that any cases $k(\ge 2)$. Before going into the statements of our theorems, we denote the Fourier integrals $\nu (u)$ and $\xi (u)$ defined by

$$\begin{aligned} \nu (u):=\frac{1}{2\pi }\int _{-\infty }^{\infty } \frac{\zeta (-\frac{1}{2}-it)\zeta (5+2it)\zeta (\frac{15}{2}+3it)}{\zeta (15+6it)}\frac{\textrm{e}^{itu}}{(\frac{5}{2}+it)^{2}(\frac{1}{2}+it)}dt, \end{aligned}$$

(1.8)

and

$$\begin{aligned}&\xi (u):=\nonumber \\&\frac{1}{2\pi }\int _{-\infty }^{\infty } \frac{\zeta (-\frac{1}{2}-it)\zeta (\frac{15}{2}+3it)\zeta (10+4it)\zeta (\frac{25}{2}+5it)\kappa _{9}(\frac{5}{2}+it)}{\zeta (20+8it)}\frac{\textrm{e}^{itu}}{(\frac{5}{2}+it)^{2}(\frac{1}{2}+it)}dt \end{aligned}$$

(1.9)

with $u=\log \frac{x}{y}$, respectively. It follows from (3.6) and (5.5) below that

$$\begin{aligned} |\nu (u)|&\le \frac{4}{(2\pi )^{2}}\frac{\zeta (\frac{3}{2})\zeta (5)\zeta (\frac{15}{2})\zeta (15)}{\zeta (30)}\left( \frac{\pi }{4}+1\right) \end{aligned}$$

and

$$\begin{aligned} |\xi (u)|&\le \frac{4}{(2\pi )^{2}}\frac{\zeta (\frac{3}{2})\zeta (\frac{15}{2})\zeta (10)\zeta (\frac{25}{2})\zeta (20)\kappa _{9}(\frac{5}{2})}{\zeta (40)} \left( \frac{\pi }{4}+1\right) \end{aligned}$$

hold. Here the integrals are computable constants, and, strictly speaking, that is enough for the purpose of this paper. Then we have the following results:

Theorem 1

Let the notation be as above. Let x and y be large real numbers such that $x\log ^{3}x \ll y \ll x^{\frac{14}{9}}$. Then we have

$$\begin{aligned} S_{1}^{(2)}(x,y)&=\frac{\zeta (2)\zeta (3)}{\zeta (6)} xy +2\frac{\zeta (\frac{1}{2})\zeta (\frac{3}{2})}{\zeta (3)} x^{\frac{1}{2}}y +3\frac{\zeta (\frac{1}{3})\zeta (\frac{2}{3})}{\zeta (2)} x^{\frac{1}{3}}y \nonumber \\&\quad - \frac{1}{8} \frac{\zeta (4)\zeta (6)}{\zeta (12)} x^{2} - x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\nu (u) + E_{1}^{(2)}(x,y), \end{aligned}$$

(1.10)

where the function $\nu (u)$ is given by (1.8) and the error term $E_{1}^{(2)}(x,y)$ is estimated by

$$\begin{aligned} E_{1}^{(2)}(x,y)&= O\left( yx^{\frac{1}{6}}~\textrm{exp}\left( -C\frac{(\log x)^{{\frac{1}{3}} }}{(\log \log x)^{\frac{1}{3}} }\right) + x^{}y^{\frac{1}{3}}\log ^{2}y \right) \end{aligned}$$

(1.11)

with C being a positive constant.

Remark 1.1

Using (1.5) and (1.10) we deduce the relation

$$\begin{aligned}&\frac{1}{xy}\left( S_{1}^{(2)}(x,y)-S_{0}^{(2)}(x,y)\right) = \frac{1}{8}\frac{\zeta (4)\zeta (6)}{\zeta (12)}\frac{x}{y} +O\left( x^{-\frac{1}{2}} + \frac{x^2}{y^{2}} \right) \end{aligned}$$

for $x\log ^{3}x \ll y \ll x^{\frac{3}{2}}$. It follows from (1.10) and (1.11) that

$$\begin{aligned} \frac{1}{xy}\sum _{n\le y}&\sum _{q\le x} {s}_{q}^{(2)}(n)\log \frac{x}{q} =\frac{\zeta (2)\zeta (3)}{\zeta (6)} +2\frac{\zeta (\frac{1}{2})\zeta (\frac{3}{2})}{\zeta (3)} x^{-\frac{1}{2}} +3\frac{\zeta (\frac{1}{3})\zeta (\frac{2}{3})}{\zeta (2)} x^{-\frac{2}{3}} \\&\quad - \frac{1}{8}\frac{\zeta (4)\zeta (6)}{\zeta (12)} \frac{x}{y} + O\left( x^{-\frac{5}{6}}\log ^{5} x + y^{-\frac{2}{3}}\log ^{2}y \right) \end{aligned}$$

holds. This means that the logarithmic average order of $s_{q}^{(2)}(n)$ is $\frac{\zeta (2)\zeta (3)}{\zeta (6)}$ where q and n satisfying the condition $q\log ^{3}q \ll n \ll q^{\frac{14}{9}}.$

In fact, it is suspected that there is a deep relationship between a zero-free region of the Riemann zeta-function and the order of magnitude of the error term (1.11). Then we immediately obtain

Conjecture 1

We may conjecture that

$$\begin{aligned} E_{1}^{(2)}(x,y)&= O\left( yx^{\frac{1}{6}}~\textrm{exp}\left( -C\frac{(\log x)^{{\frac{3}{5}} }}{(\log \log x)^{\frac{1}{5}}}\right) + x^{}y^{\frac{1}{3}}\log ^{2}y \right) \end{aligned}$$

holds with an absolute constant $C>0$.

Theorem 2

Let the notation be as above. Let x and y be large real numbers such that $x^{\frac{6}{5}}\log ^{3}x \ll y \ll x^{\frac{19}{12}}$. Then we have

$$\begin{aligned} S_{1}^{(3)}(x,y)&=\frac{\zeta (3)\zeta (4)\zeta (5)\kappa _{9}(1)}{\zeta (8)} xy + 3\frac{\zeta (\frac{1}{3})\zeta (\frac{4}{3})\zeta (\frac{5}{3})\kappa _{9}\left( \frac{1}{3}\right) }{\zeta (\frac{8}{3})}x^{\frac{1}{3}}y + 4\frac{\zeta (\frac{1}{4})\zeta (\frac{3}{4})\zeta (\frac{5}{4})\kappa _{9}\left( \frac{1}{4}\right) }{\zeta (\frac{8}{4})}x^{\frac{1}{4}}y \nonumber \\&\ \ \ + 5\frac{\zeta (\frac{1}{5})\zeta (\frac{3}{5})\zeta (\frac{4}{5})\kappa _{9}\left( \frac{1}{5}\right) }{\zeta (\frac{8}{5})}x^{\frac{1}{5}}y - \frac{1}{8} \frac{\zeta (6)\zeta (8)\zeta (10)\kappa _{9}(2)}{\zeta (16)} x^{2} - x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\xi (u) \nonumber \\&\qquad + O\left( yx^{\frac{1}{8}}~\textrm{exp}\left( -C\frac{(\log x)^{{\frac{1}{3}} }}{(\log \log x)^{\frac{1}{3}} }\right) + xy^{\frac{1}{3}}\log ^{2}y \right) , \end{aligned}$$

(1.12)

where the function $\xi (u)$ is given by (1.9) and C is a positive constant.

Remark 1.2

Similarly as in Remark 1.1, we have

$$\begin{aligned} S_{1}^{(3)}(x,y)-S_{0}^{(3)}(x,y)&= \frac{\zeta (6)\zeta (8)\zeta (10)\kappa _{9}(2)}{8\zeta (16)} x^{2} + O\left( yx^{\frac{1}{3}} + xy^{\frac{1}{3}}\log ^{2}y + \frac{x^3}{y}\right) \end{aligned}$$

for $x^{\frac{6}{5}}\log ^{3}x \ll y \ll x^{\frac{5}{3}}$, and the logarithmic average order of $s_{q}^{(3)}(n)$ is derived by $\frac{\zeta (3)\zeta (4)\zeta (5)\kappa _{9}(1)}{\zeta (8)}$ under q and n satisfying the condition $q^{\frac{6}{5}}\log ^{3}q \ll n \ll q^{\frac{19}{12}}.$

Next, we assume the truth of the unproved Riemann Hypothesis, that all the complex zeros of the Riemann zeta-function $\zeta (s)$ lie on the line $\sigma =\frac{1}{2}$. We consider the precise asymptotic formula concerning ${S}_{1}^{(2)}(x,y)$. Then we derive the following

Theorem 3

Assume that the Riemann Hypothesis is true. Let x and y be large real numbers such that $x\log ^{3}x \ll y \ll x^{\frac{29}{18}}\textrm{exp}\left( -A\frac{\log x}{\log \log x}\right)$. Then the error term $E_{1}^{(2)}(x,y)$ of (1.10) is estimated by

$$\begin{aligned} E_{1}^{(2)}(x,y)&= O\left( yx^{\frac{1}{12}}\textrm{exp}\left( A\frac{\log x}{\log \log x}\right) + x^{}y^{\frac{1}{3}}\log ^{2}y \right) \end{aligned}$$

(1.13)

with A being a positive constant.

In addition, we assume that all the zeros $\rho$ of the Riemann zeta-function $\zeta (s)$ on the critical line are simple, where $\rho =\frac{1}{2}+i\gamma$ denotes a nontrivial zero of the Riemann zeta-function, and $\gamma$ denotes the imaginary part of zero on the critical line. Then we may derive a sum involving the zeros $\rho$ of $\zeta (s)$ concerning $E_{1}^{(2)}(x,y)$. To improve the order of magnitude of its sum, we make use of the Gonek-Hejhal Hypothesis (Gonek [2], and Hejhal [4] independently conjectured), namely

$$\begin{aligned} J_{-\lambda }(T):= \sum _{0 < \gamma \le T}\frac{1}{|\zeta '(\rho )|^{2\lambda }} \asymp T(\log {T})^{(\lambda -1)^2} \end{aligned}$$

for real number $\lambda < \frac{3}{2}$, where $\zeta '(s)$ is the first derivative of $\zeta (s)$, then we may deduce a new estimate of $E_{1}^{(2)}(x,y)$, which will be done elsewhere.

Notations. Throughout this paper, we use the following notations: The Riemann zeta-function $\zeta (s)$, defined by $\sum _{n=1}^{\infty }\frac{1}{n^s}$ for $\sigma >1$, admits of analytic continuation over the whole complex plane having as its only singularity a simple pole with residue 1 at $s=1$. In what follows, C donotes any arbitrarily positive number, not necessarily the same ones at each occurrence.

2 Some Lemmas

Lemma 2.1

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

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

Proof

This is Riesz typical means of Perron’s formula. For more details, see (5.20)–(5.22) in [10]. $\square$

Lemma 2.2

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

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

• (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 can be found in [Theorem 1.4, [11]]. $\square$

Lemma 2.3

There is an absolute constant $C>0$ such that $\zeta (s)\ne 0$ for

$$\begin{aligned} \sigma \ge 1- C(\log t)^{-\frac{2}{3}}(\log \log t)^{-\frac{1}{3}} \quad (t\ge t_0). \end{aligned}$$

Proof

This lemma is given by Theorem 6.1 in [5]. $\square$

Lemma 2.4

For $|t|\ge 2$ and $\sigma \ge 1-C(\log t)^{-\frac{2}{3}}(\log \log t)^{-\frac{1}{3}}$ we have

$$\begin{aligned} \zeta (\sigma +it) \ll (\log t)^{\frac{2}{3}}(\log \log t)^{\frac{1}{3}} \qquad \textrm{and} \qquad \frac{1}{\zeta (\sigma +it)} \ll (\log t)^{\frac{2}{3}}(\log \log t)^{\frac{1}{3}}. \end{aligned}$$

Proof

The first term of this lemma is a well-known result. The second term of this lemma is given by Lemma 12.3 in [5]. $\square$

Lemma 2.5

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

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

Proof

The proof of this lemma follows from Theorem II.3.8 in [12] (see also [5, 13]). $\square$

Lemma 2.6

For any positive number $T>1$ we have

$$\begin{aligned} \int _{1}^{T}|\zeta (\sigma +it)|^{4}dt&\ll \left\{ \begin{array}{cl} T^{3-4\sigma } &{} \ \ \left( 0< \sigma < \frac{1}{2}\right) , \ \\ T^{}\log ^{4}T &{} \ \ \left( \sigma = \frac{1}{2}\right) . \ \\ T &{} \ \ \left( \sigma > \frac{1}{2}\right) , \end{array} \right. \end{aligned}$$

(2.1)

Proof

The second and third terms of (2.1) are due to Theorem 5.1 and Theorem 8.5 in [5]. We use (2.2) below and the formula $\int _{1}^{T}|\zeta (\sigma +it)|^{4}dt=O(T)$ for $\frac{1}{2}<\sigma \le 1$ to deduce (2.1). $\square$

Lemma 2.7

Assume that the Riemann hypothesis is true. Then we have

$$\begin{aligned} \zeta (\sigma +it) \ll t^{\varepsilon } \qquad \textrm{and} \qquad \frac{1}{\zeta (\sigma +it)} \ll t^{\varepsilon } \end{aligned}$$

for every $\sigma$$(\frac{1}{2}+\delta \le \sigma \le 2)$ and $t\ge t_{0}$ being a sufficiently large real number.

Proof

The first and second terms of this lemma are given by (14.2.5), (14.2.6), (14.14.1) and (14.16.2) in [13], respectively. $\square$

The next lemma is a well-known result (see [5, 13]), that is

Lemma 2.8

The functional equation of the Riemann zeta-function is given by

$$\begin{aligned} \zeta (s)=\chi (s)\zeta (1-s), \end{aligned}$$

(2.2)

where $\chi (s)=2^{s}\pi ^{s-1}\sin \left( \frac{\pi s}{2}\right) \Gamma (1-s).$ Thus in any bounded vertical strip, we have

$$\begin{aligned} |\chi (s)| \asymp \left( \frac{t}{2\pi }\right) ^{\frac{1}{2}-\sigma }\left( 1+O\left( \frac{1}{t}\right) \right) . \end{aligned}$$

3 Proof of Theorem 1

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 2.1 with (1.2) to get

$$\begin{aligned} \sum _{q\le x}&{s}_{q}^{(2)}(n)\log \frac{x}{q}= \frac{1}{2\pi i} \int _{\alpha -iT}^{\alpha +iT} \sigma _{1-s}(n) \frac{\zeta (2s)\zeta (3s)}{\zeta (6s)}\frac{x^s}{s^{2}}ds + O\left( \sigma _{0}(n)\frac{x}{T^{}}\right) , \end{aligned}$$

where $\alpha = 1 +\frac{1}{\log x}$. Let T be a real parameter at our disposal. We have

$$\begin{aligned} \sum _{n\le y}\sum _{q\le x} {s}_{q}^{(2)}(n) \log \frac{x}{q}&= \frac{1}{2\pi i} \int _{\alpha _{}-iT}^{\alpha _{}+iT} \sum _{n\le y}\sigma _{1-s_{}}(n) \frac{\zeta (2s)\zeta (3s)}{\zeta (6s)} \frac{x^{s}}{s^{2}}ds_{} + O\left( \frac{x}{T}\sum _{n\le y}\sigma _{0}(n)\right) . \end{aligned}$$

Taking $b=1$ and $z=1-s$ into Lemma 2.2 and using the estimate $\sum _{n\le y}\sigma _{0}(n)\ll y\log y$ we have

$$\begin{aligned} S_{1}^{(2)}(x,y)&= K_{1} + K_{2} + O\left( xy^{\frac{1}{3}}\log ^{2}y\right) + O\left( \frac{xy}{T}\log ^{}y\right) , \end{aligned}$$

(3.1)

where

$$\begin{aligned} K_{1}:=\frac{y}{2\pi i} \int _{\alpha _{}-iT}^{\alpha _{}+iT} \frac{\zeta (s)\zeta (2s)\zeta (3s)}{\zeta (6s_{})}\frac{x^{s}}{s^{2}}ds_{}, \end{aligned}$$

and

$$\begin{aligned} \qquad \quad \ \ K_{2}:=\frac{y^2}{2\pi i} \int _{\alpha _{}-iT}^{\alpha _{}+iT} \frac{\zeta (2-s)\zeta (2s)\zeta (3s)}{\zeta (6s_{})}\frac{(x/y)^{s}}{s^{2}(2-s)}ds_{}. \end{aligned}$$

Define

$$\begin{aligned} \varepsilon (T):=\frac{C}{100}(\log T)^{-\frac{2}{3}}(\log \log T)^{-\frac{1}{3}} \end{aligned}$$

(3.2)

with C being the same as that in Lemma 2.3 and $T=x^{2}.$ 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 $K_{1}$, we move the integration, with respect to s, to $\Gamma (\alpha ,\beta ,T)$ with $\beta =\frac{1}{6}-\varepsilon (T)$. We denote the integrals over the horizontal line segments by $K_{1,1}$ and $K_{1,3}$, and the integral over the vertical line segment by $K_{1,2}$, respectively. We use Lemmas 2.3–2.5 to deduce

$$\begin{aligned}&K_{1,1}, K_{1,3} \nonumber \\&\ll \frac{y}{T^{2}\varepsilon (T)} \left( \int _{\beta }^{\frac{1}{3}} + \int _{\frac{1}{3}}^{\frac{1}{2}} + \int _{\frac{1}{2}}^{\alpha }\right) |\zeta (\sigma +iT)||\zeta (2(\sigma +iT))||\zeta (3(\sigma +iT))| x^{\sigma }d\sigma \nonumber \\&\ll \frac{y}{T^2}\frac{\log ^{3}T}{\varepsilon (T)} \left( T^{\frac{3}{2}}\int _{\beta }^{\frac{1}{3}}\left( \frac{x}{T^{4}}\right) ^{\sigma } d\sigma + T^{\frac{5}{6}}\int _{\frac{1}{3}}^{\frac{1}{2}}\left( \frac{x}{T^{}}\right) ^{\sigma } d\sigma + T^{\frac{1}{3}} \int _{\frac{1}{2}}^{\alpha } \left( \frac{x}{T^{\frac{1}{3}}}\right) ^{\sigma } d\sigma \right) \nonumber \\&\ll \frac{y}{T}\frac{\log ^{3}T}{\varepsilon (T)} \left( \frac{x^{\frac{1}{6}}}{T^{\frac{1}{6}}} + \frac{x^{\frac{1}{3}}}{T^\frac{1}{2}} + \frac{x^{\frac{1}{2}}}{T^{\frac{2}{3}}} + \frac{x}{T} \right) \end{aligned}$$

and

$$\begin{aligned}&K_{1,2} \ll yx^{\frac{1}{6}-\varepsilon (T)} \left( \int _{|t|\le T_{0}} + \int _{T_{0}<|t|\le T} \right) \times \\&\qquad \times \frac{|\zeta (\frac{1}{6}-\varepsilon (T)+it)||\zeta (\frac{1}{3}-2\varepsilon (T)+2it)||\zeta (\frac{1}{2}-3\varepsilon (T)+3it)|}{|\zeta (1-6\varepsilon (T)+ 6it)|(1+|t|)^2} dt \nonumber \\&\ll \frac{yx^{\frac{1}{6}-\varepsilon (T)}}{\varepsilon (T)} + yx^{\frac{1}{6}-\varepsilon (T)} \int _{T_{0}<|t|\le T} \frac{t^{\frac{5}{6}+4\varepsilon (T)}}{|\zeta (1-6\varepsilon (T)+ 6it)| t^2}dt \nonumber \\&\ll \frac{yx^{\frac{1}{6}-\varepsilon (T)}}{\varepsilon (T)}. \end{aligned}$$

It remains to evaluate the residues of the poles of the integrand, and there exist three simple poles at $s=1$, $\frac{1}{2}$ and $\frac{1}{3}$ with residues $\frac{\zeta (2)\zeta (3)}{\zeta (6)}x,$ $\frac{2\zeta (\frac{1}{2})\zeta (\frac{3}{2})}{\zeta (3)}x^{\frac{1}{2}},$ and $\frac{3\zeta (\frac{1}{3})\zeta (\frac{2}{3})}{\zeta (2)}x^{\frac{1}{3}},$ respectively. Therefore, we have

$$\begin{aligned} K_{1}&= \frac{\zeta (2)\zeta (3)}{\zeta (6)}xy +\frac{2\zeta (\frac{1}{2})\zeta (\frac{3}{2})}{\zeta (3)}x^{\frac{1}{2}}y +\frac{3\zeta (\frac{1}{3})\zeta (\frac{2}{3})}{\zeta (2)}x^{\frac{1}{3}}y \nonumber \\&\qquad + O\left( yx^{\frac{1}{6}}\textrm{exp}\left( -C\frac{(\log x)^{\frac{1}{3}}}{(\log \log x)^{\frac{1}{3}}}\right) \right) \end{aligned}$$

(3.3)

by setting $T=x^2$, where C is a positive constant.

In $K_{2}$, we move the integration, with respect to s, to $\Gamma (\alpha ,\frac{5}{2},T)$. 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. Using Lemmas 2.5 and 2.8 we have

$$\begin{aligned}&K_{2,1}, K_{2,3} \ll \frac{y^2}{T^3}\int _{\alpha }^{\frac{5}{2}}{|\zeta (2-\sigma -iT)| \left( \frac{x}{y}\right) ^{\sigma }}d\sigma \nonumber \\&\ll \frac{y^2}{T^3} \left( \int _{\alpha }^{2}|\zeta (2-\sigma -iT)|\left( \frac{x}{y}\right) ^{\sigma } d\sigma + \int _{2}^{\frac{5}{2}}|\zeta (\sigma -1+iT)\chi (2-\sigma -iT)|\left( \frac{x}{y}\right) ^{\sigma } d\sigma \right) \nonumber \\&\ll \frac{y^2}{T^3}\log ^{}T \left( T^{-\frac{1}{2}}\int _{\alpha }^{2}\left( \frac{T^{\frac{1}{2}}x}{y}\right) ^{\sigma } d\sigma + T^{-\frac{3}{2}}\int _{2}^{\frac{5}{2}}\left( \frac{T^{}x}{y}\right) ^{\sigma } d\sigma \right) \nonumber \\&\ll \frac{y^2\log ^{}T}{T^{3}} \left( \frac{x}{y} + T^{\frac{1}{2}}\left( \frac{x}{y}\right) ^{2} +T^{}\left( \frac{x}{y}\right) ^{\frac{5}{2}} \right) , \end{aligned}$$

(3.4)

and

$$\begin{aligned} K_{2,2}&=\frac{y^2}{2\pi i}\int _{\frac{5}{2}-i\infty }^{\frac{5}{2}+i\infty } \frac{\zeta (2-s)\zeta (2s)\zeta (3s)}{\zeta (6s)}\frac{\left( \frac{x}{y}\right) ^{s}}{s^{2}(2-s)}ds \nonumber \\&+ O\left( x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\int _{T}^{\infty } \frac{|\zeta \left( \frac{3}{2}+it\right) ||\zeta (5+2it)||\zeta \left( \frac{15}{2}+3it\right) |}{|\zeta \left( 15+6it\right) |} \frac{|\chi \left( -\frac{1}{2}-it\right) |}{(1+t)^{3}}dt \right) \nonumber \\&=- x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\nu (u) + O\left( x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\frac{1}{T}\right) , \end{aligned}$$

(3.5)

where $\nu (u)$ is given by

$$\begin{aligned} \nu (u):=\frac{1}{2\pi }\int _{-\infty }^{\infty } \frac{\zeta (-\frac{1}{2}-it)\zeta (5+2it)\zeta (\frac{15}{2}+3it)}{\zeta (15+6it)}\frac{\textrm{e}^{itu}}{(\frac{5}{2}+it)^{2}(\frac{1}{2}+it)}dt, \end{aligned}$$

with $u=\log \frac{x}{y}$. We use Lemma 2.5 and the inequality $|\frac{1}{\zeta (s)}|\le \frac{\zeta (\sigma )}{\zeta (2\sigma )}$ for $\sigma >1$ to obtain the absolute value of $\nu (u)$, that is

$$\begin{aligned} |\nu (u)|&\le \frac{1}{2\pi }\int _{-\infty }^{\infty } \left| \frac{\zeta (\frac{3}{2}+it)\zeta (5+2it)\zeta (\frac{15}{2}+3it)}{\zeta (15+6it)}\right| \left| \frac{\chi (-\frac{1}{2}-it)}{(\frac{5}{2}+it)^2(\frac{1}{2}+it)}\right| dt \nonumber \\&\le \frac{2}{(2\pi )^{2}}\frac{\zeta (\frac{3}{2})\zeta (5)\zeta (\frac{15}{2})\zeta (15)}{\zeta (30)} \int _{0}^{\infty }\frac{t^{}}{(t^{2}+\left( \frac{5}{2}\right) ^2)\sqrt{t^{2}+\left( \frac{1}{2}\right) ^2}}dt \nonumber \\&\le \frac{2}{(2\pi )^{2}}\frac{\zeta (\frac{3}{2})\zeta (5)\zeta (\frac{15}{2})\zeta (15)}{\zeta (30)} \int _{0}^{\infty }\frac{t^{\frac{1}{2}}}{t^{2}+\left( \frac{1}{2}\right) ^2}dt \nonumber \\&\le \frac{4}{(2\pi )^{2}}\frac{\zeta (\frac{3}{2})\zeta (5)\zeta (\frac{15}{2})\zeta (15)}{\zeta (30)}\left( \frac{\pi }{4}+1\right) , \end{aligned}$$

(3.6)

hence $|\nu (u)|$ is an upper bound. Next, there exists a simple pole at $s=2$ of the integral $K_2$ with residue $\frac{\zeta (4)\zeta (6)}{8\zeta (12)}\left( \frac{x}{y}\right) ^{2}$ by using the value $\zeta (0)=-\frac{1}{2}.$ Hence, we have

$$\begin{aligned} K_{2}= - \frac{\zeta (4)\zeta (6)}{8\zeta (12)}x^2 - x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\nu (u) + O\left( \left( \frac{x}{y}\right) ^{\frac{1}{2}}\right) \end{aligned}$$

(3.7)

with $T=x^2$.

Combining (3.3) and (3.7) with (3.1) and taking $T=x^2$, we obtain the formula (1.10).

4 Proof of Theorem 3

In this section we assume that the Riemann Hypothesis is true, and $1 \le y \le x^M$ for some constant M. Without loss of generality we can assume that $x,\ y\in \mathbb {Z}+\frac{1}{2}$. The proof of this theorem follows by the same method as in Theorem 1, in addition to the Riemann Hypothesis. We start with (3.1), and set $\beta =\frac{1}{12}+\delta$ with $\delta =\frac{10}{\log \log T}$, and $\varepsilon =\frac{1}{\log \log T}$ in Lemma 2.7 and $T=x^2$.

In $K_{1}$, we move the integration, with respect to s, to $\Gamma (\alpha ,\beta ,T)$ with $\beta =\frac{1}{12}+\delta$. We denote the integrals over the horizontal line segments by $K_{1,1}$ and $K_{1,3}$, and the integral over the vertical line segment by $K_{1,2}$, respectively. We use Lemmas 2.5, 2.7, and 2.8 to deduce

$$\begin{aligned}&K_{1,2} \ll yx^{\frac{1}{12}+\delta } \left( \int _{|t|\le T_{0}} + \int _{T_{0}<|t|\le T} \right) \\&\quad \times \frac{|\chi (\frac{1}{12}+\delta +it)||\chi (\frac{1}{6}+2\delta +2it)||\chi (\frac{1}{4}+3\delta +3it)|}{|\zeta (\frac{1}{2}+6\delta +6it)|} \times \nonumber \\&\qquad \times \frac{|\zeta (\frac{11}{12}-\delta -it)||\zeta (\frac{5}{6}-2\delta -2it)||\zeta (\frac{3}{4}-3\delta -3it)|}{(1+|t|)^2} dt \nonumber \\&\ll yx^{\frac{1}{12}+\delta } + yx^{\frac{1}{12}+\delta } \int _{T_{0}<|t|\le T} t^{-1-6\delta +4\varepsilon }dt \nonumber \\&\ll yx^{\frac{1}{12}}\textrm{exp}\left( A\frac{\log x}{\log \log x}\right) \end{aligned}$$

with A being a positive constant, and $T=x^2$. We use Lemmas 2.5, 2.7, 2.8, and the estimates of $K_{1,1}$ and $K_{1,3}$ in the proof of Theorem 1 to deduce

$$\begin{aligned}&K_{1,1}, K_{1,3} \nonumber \\&\ll \frac{y}{T^{2}} \left( \int _{\frac{1}{12}+\delta }^{\frac{1}{6}} + \int _{\frac{1}{6}}^{\alpha }\right) \frac{|\zeta (\sigma +iT)||\zeta (2(\sigma +iT))||\zeta (3(\sigma +iT))|}{|\zeta (6(\sigma +iT))|} x^{\sigma }d\sigma \nonumber \\&\ll \frac{y}{T^2} \left( T^{\frac{3}{2}+4\varepsilon }\int _{\frac{1}{12}+\delta }^{\frac{1}{6}}\left( \frac{x}{T^{6}}\right) ^{\sigma } d\sigma + \frac{x^{\frac{1}{6}}}{T^{\frac{1}{6}}}\log ^{4}T \right) \nonumber \\&\ll \frac{y}{T} \left( \frac{x^{\frac{1}{6}}}{T^{\frac{1}{6}}}\log ^{4}T + \frac{x^{\frac{1}{12}+\delta }}{T^{6\delta -4\varepsilon }} \right) . \end{aligned}$$

Therefore, by using Cauchy’s residue theorem we have

$$\begin{aligned} K_{1}= \frac{\zeta (2)\zeta (3)}{\zeta (6)}xy +\frac{2\zeta (\frac{1}{2})\zeta (\frac{3}{2})}{\zeta (3)}x^{\frac{1}{2}}y +\frac{3\zeta (\frac{1}{3})\zeta (\frac{2}{3})}{\zeta (2)}x^{\frac{1}{3}}y + O\left( yx^{\frac{1}{12}}\textrm{exp}\left( A\frac{\log x}{\log \log x}\right) \right) \end{aligned}$$

(4.1)

by setting $T=x^2$, where A is a positive constant.

As for $K_{2}$, we make use of the same result in the proof of Theorem 1, that is

$$\begin{aligned} K_{2}= - \frac{\zeta (4)\zeta (6)}{8\zeta (12)}x^2 - x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\nu (u) + O\left( \left( \frac{x}{y}\right) ^{\frac{1}{2}}\right) \end{aligned}$$

(4.2)

with $T=x^2$.

Combining (4.1) and (4.2) with (3.1) and taking $T=x^2$, we obtain the formula (1.13).

5 Proof of Theorem 2

Assume that $1 \le y \le x^M$ for some constant M. Without loss of generality we can assume that $x,\ y\in \mathbb {Z}+\frac{1}{2}$. The proof of this theorem follows by the same method as in Theorem 1. We apply Lemma 2.1 with (1.3) and substitute $b=1$ and $z=1-s$ into Lemma 2.2 to deduce

$$\begin{aligned} S_{1}^{(3)}(x,y)&= L_{1} + L_{2} + O\left( xy^{\frac{1}{3}}\log ^{2}y\right) + O\left( \frac{xy}{T}\log ^{}y\right) , \end{aligned}$$

(5.1)

where

$$\begin{aligned} L_{1}:=\frac{y}{2\pi i} \int _{\alpha _{}-iT}^{\alpha _{}+iT} \frac{\zeta (s)\zeta (3s)\zeta (4s)\zeta (5s)\kappa _{9}(s)}{\zeta (8s_{})}~ \frac{x^{s}}{s^{2}}ds_{}, \end{aligned}$$

and

$$\begin{aligned} \qquad \qquad L_{2}:=-\frac{y^2}{2\pi i} \int _{\alpha _{}-iT}^{\alpha _{}+iT} \frac{\zeta (2-s)\zeta (3s)\zeta (4s)\zeta (5s)\kappa _{9}(s)}{\zeta (8s_{})}~ \frac{(x/y)^{s}}{s^{2}(s-2)}ds_{}. \end{aligned}$$

Here $\alpha = 1 +\frac{1}{\log x}$ and T is a real parameter at our disposal.

We move the integration, with respect to s, to $\Gamma (\alpha ,\beta ,T)$ with $\beta =\frac{1}{8}-\varepsilon (T)$, where $\varepsilon (T)$ is given by (3.2). We denote the integrals over the horizontal line segments by $L_{1,1}$ and $L_{1,3}$, and the integral over the vertical line segment by $L_{1,2}$, respectively. Since the estimate $\kappa _{9}(\sigma +iT)\ll 1$ for $\sigma \ge \beta$, it follows from Lemmas 2.3–2.5 that

$$\begin{aligned}&L_{1,1}, L_{1,3} \\&\ll \frac{y}{T^{2}} \left( \int _{\beta }^{\frac{1}{5}} + \int _{\frac{1}{5}}^{\frac{1}{4}} + \int _{\frac{1}{4}}^{\frac{1}{3}}+ \int _{\frac{1}{3}}^{\alpha }\right) \\&\qquad \frac{|\zeta (\sigma +iT)||\zeta (3(\sigma +iT))||\zeta (4(\sigma +iT))||\zeta (5(\sigma +iT))|}{|\zeta (8(\sigma +iT))|} x^{\sigma }d\sigma \\&\quad \ll \frac{y\log ^{5}T}{T^{2}} \left( T^{\frac{5}{3}}\int _{\beta }^{\frac{1}{5}}\left( \frac{x}{T^{\frac{17}{3}}}\right) ^{\sigma } d\sigma \right. \\&\qquad \left. + T^{\frac{7}{6}} \int _{\frac{1}{5}}^{\frac{1}{4}}\left( \frac{x}{T^{3}}\right) ^{\sigma } d\sigma + T^{\frac{5}{6}} \int _{\frac{1}{4}}^{\frac{1}{3}}\left( \frac{x}{T^{\frac{5}{3}}}\right) ^{\sigma } d\sigma + T^{\frac{1}{2}} \int _{\frac{1}{3}}^{\alpha } \left( \frac{x}{T^{\frac{2}{3}}}\right) ^{\sigma } d\sigma \right) \\&\quad \ll \frac{y\log ^{5}T}{T} \left( \frac{x^{\frac{1}{8}}}{T^{\frac{1}{24}}} +\frac{x^{\frac{1}{5}}}{T^{\frac{13}{30}}} +\frac{x^{\frac{1}{4}}}{T^{\frac{7}{12}}} +\frac{x^{\frac{1}{3}}}{T^{\frac{13}{18}}} + \frac{x^{}}{T^{}} \right) . \end{aligned}$$

For $L_{1,2}$, we use integration by parts, Cauchy–Schwarz’s inequality twice, Lemmas 2.3–2.6, and the estimate $\kappa _{9}(\frac{1}{8}+it)\ll 1$ to deduce

$$\begin{aligned} L_{1,2}&\ll \frac{yx^{\frac{1}{8}-\varepsilon (T)}}{\varepsilon (T)} + yx^{\frac{1}{8}-\varepsilon (T)} \int _{T_{0}<|t|\le T} \frac{|\chi (\beta +it)||\chi (3\beta +3it)||\chi (4\beta +4it)|}{|\zeta (8\beta +8it)|} \times \nonumber \\&\ \ \ \ \ \times \frac{|\zeta (1-\beta -it)||\zeta (1-3\beta -3it)||\zeta (1-4\beta -4it)||\zeta (5\beta +5it)|}{(1+|t|)^2} dt \nonumber \\&\ll \frac{yx^{\frac{1}{8}-\varepsilon (T)}}{\varepsilon (T)} + \frac{yx^{\frac{1}{8}-\varepsilon (T)}}{\varepsilon (T)} \int _{T_{0}<|t|\le T} \nonumber \\&\quad \times \frac{|\zeta (\frac{7}{8}+\varepsilon (T)-it)||\zeta (\frac{5}{8}+3\varepsilon (T)-3it)||\zeta (\frac{1}{2}+4\varepsilon (T)-4it)||\zeta (\frac{5}{8}-5\varepsilon (T)+5it)|}{(1+|t|)^{\frac{3}{2}+8\varepsilon (T)}} dt \nonumber \\&\ll \frac{yx^{\frac{1}{8}-\varepsilon (T)}}{\varepsilon (T)} + \frac{yx^{\frac{1}{8}-\varepsilon (T)}}{\varepsilon (T)} \left( \int _{1}^{T}\frac{\left| \zeta \left( \frac{7}{8}+iu\right) \right| ^4}{(1+|u|)^{\frac{3}{2}+8\varepsilon (T)}} du\right) ^{\frac{1}{4}} \left( \int _{1}^{3T}\frac{\left| \zeta \left( \frac{5}{8}+iu\right) \right| ^4}{(1+|u|)^{\frac{3}{2}+8\varepsilon (T)}} du\right) ^{\frac{1}{4}}\\&\qquad \times \left( \int _{1}^{4T} \frac{\left| \zeta \left( \frac{1}{2}+iu\right) \right| ^4}{(1+|u|)^{\frac{3}{2}+8\varepsilon (T)}}du\right) ^{\frac{1}{4}} \left( \int _{1}^{5T}\frac{\left| \zeta \left( \frac{9}{16}+iu\right) \right| ^4}{(1+|u|)^{\frac{3}{2}+8\varepsilon (T)}}du\right) ^{\frac{1}{4}} \\&\ll \frac{yx^{\frac{1}{8}-\varepsilon (T)}}{\varepsilon (T)}. \end{aligned}$$

It remains to evaluate the residues of the poles of the integrand, and there exist four simple poles at $s=1$, $\frac{1}{3}$, $\frac{1}{4}$ and $\frac{1}{5}$ with residues $\frac{\zeta (3)\zeta (4)\zeta (5)\kappa _{9}(1)}{\zeta (8)}x,$ $\frac{3\zeta (\frac{1}{3})\zeta (\frac{4}{3})\zeta (\frac{5}{3})\kappa _{9}\left( \frac{1}{3}\right) }{\zeta (\frac{8}{3})}x^{\frac{1}{3}},$ $\frac{4\zeta (\frac{1}{4})\zeta (\frac{3}{4})\zeta (\frac{5}{4})\kappa _{9}\left( \frac{1}{4}\right) }{\zeta (2)}x^{\frac{1}{4}},$ and $\frac{5\zeta (\frac{1}{5})\zeta (\frac{3}{5})\zeta (\frac{4}{5})\kappa _{9}\left( \frac{1}{5}\right) }{\zeta (\frac{8}{5})}x^{\frac{1}{5}},$ respectively. Therefore, we have

$$\begin{aligned}&L_{1}= \frac{\zeta (3)\zeta (4)\zeta (5)\kappa _{9}(1)}{\zeta (8)}xy +\frac{3\zeta (\frac{1}{3})\zeta (\frac{4}{3})\zeta (\frac{5}{3})\kappa _{9}\left( \frac{1}{3}\right) }{\zeta (\frac{8}{3})}x^{\frac{1}{3}}y + \frac{4\zeta (\frac{1}{4})\zeta (\frac{3}{4})\zeta (\frac{5}{4})\kappa _{9}\left( \frac{1}{4}\right) }{\zeta (2)}x^{\frac{1}{4}}y \nonumber \\&+ \frac{5\zeta (\frac{1}{5})\zeta (\frac{3}{5})\zeta (\frac{4}{5})\kappa _{9}\left( \frac{1}{5}\right) }{\zeta (\frac{8}{5})}x^{\frac{1}{5}}y + O\left( yx^{\frac{1}{8}}\textrm{exp}\left( -C\frac{(\log x)^{\frac{1}{3}}}{(\log \log x)^{\frac{1}{3}}}\right) \right) \end{aligned}$$

(5.2)

by setting $T=x^2$, with C being a positive constant.

For $L_{2}$, we move the integration, with respect to s, to $\Gamma (\alpha ,\frac{5}{2},T)$. We denote the integrals over the horizontal line segments by $L_{2,1}$ and $L_{2,3}$, and the integral over the vertical line segment by $L_{2,2}$, respectively. Following the same method as in (3.4) and (3.5) we have

$$\begin{aligned} L_{2,1}, L_{2,3}&\ll \frac{y^2\log ^{}T}{T^{3}} \left( \frac{x}{y} + T^{\frac{1}{2}}\left( \frac{x}{y}\right) ^{2} +T^{}\left( \frac{x}{y}\right) ^{\frac{5}{2}} \right) , \end{aligned}$$

and

$$\begin{aligned} L_{2,2}&= - \frac{y^2}{2\pi i}\int _{\frac{5}{2}-i\infty }^{\frac{5}{2}+i\infty } \frac{\zeta (2-s)\zeta (3s)\zeta (4s)\zeta (5s)\kappa _{9}(s)}{\zeta (8s_{})}~ \frac{(\frac{x}{y})^{s}}{s^{2}(s-2)}ds_{} \nonumber \\&+ O\left( x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\int _{T}^{\infty } \frac{|\zeta \left( -\frac{1}{2}-it\right) ||\zeta (\frac{15}{2}+3it)||\zeta \left( 10+4it\right) ||\zeta \left( \frac{25}{2}+5it\right) ||\kappa _{9}(\frac{5}{2}+it)|}{|\zeta (20+5it)|(1+t)^{3}}dt \right) \nonumber \\&=- x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\xi (u) + O\left( x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\frac{1}{T}\right) , \end{aligned}$$

(5.3)

where $\xi (u)$ is given by

$$\begin{aligned} \xi (u)&:=\frac{1}{2\pi }\int _{-\infty }^{\infty }\nonumber \\&\quad \frac{\zeta (-\frac{1}{2}-it)\zeta (\frac{15}{2}+3it)\zeta (10+4it)\zeta (\frac{25}{2}+5it)\kappa _{9}(\frac{5}{2}+it)}{\zeta (20+8it)}\frac{\textrm{e}^{itu}}{(\frac{5}{2}+it)^{2}(\frac{1}{2}+it)}dt \end{aligned}$$

(5.4)

with $u=\log \frac{x}{y}$. Similarly as in (3.6), we have

$$\begin{aligned} |\xi (u)|&\le \frac{4}{(2\pi )^{2}}\frac{\zeta (\frac{3}{2})\zeta (\frac{15}{2})\zeta (10)\zeta (\frac{25}{2})\zeta (20)\kappa _{9}(\frac{5}{2})}{\zeta (40)} \left( \frac{\pi }{4}+1\right) . \end{aligned}$$

(5.5)

There exists a simple pole at $s=2$ of the integral $L_2$ with residue $\frac{\zeta (6)\zeta (8)\zeta (10)\kappa _{9}(2)}{8\zeta (16)}\left( \frac{x}{y}\right) ^2.$ Hence, we have

$$\begin{aligned} L_{2}= - \frac{\zeta (6)\zeta (8)\zeta (10)\kappa _{9}(2)}{8\zeta (16)}x^2 - x^{2}\left( \frac{x}{y}\right) ^{\frac{1}{2}}\xi (u) + O\left( \left( \frac{x}{y}\right) ^{\frac{1}{2}}\right) \end{aligned}$$

(5.6)

by using $T=x^2$.

Combining (5.2) and (5.6) with (5.1) and taking $T=x^2$, the formula (1.12) is proved.