Mean Value of the General Dedekind Sums over Interval \({[1,\frac{q}{p})}\)

Abstract

Let $q>2$ be a prime, p be a given prime with $p<q$. The main purpose of this paper is using transforms, the hybrid mean value of Dirichlet L-functions with character sums and the related properties of character sums to study the mean value of the general Dedekind sums over interval $[1,\frac{q}{p})$, and give some interesting asymptotic formulae.

1. Introduction

For a positive integer k and an arbitrary integer h with $(h,k)=1$, the classical Dedekind sum $S(h,k)$ is defined by

$$S(h,k)=\sum _{a=1}^k\left(\left(\frac{a}{k}\right)\right)\left(\left(\frac{ah}{k}\right)\right),$$

where

$$\left(\left(x\right)\right)=\left\{\begin{array}{cc}x-\left[x\right]-\frac{1}{2},\hfill & \mathrm{if}x\mathrm{not}\mathrm{be}\mathrm{an}\mathrm{integer};\hfill \\ 0,\hfill & \mathrm{if}x\mathrm{be}\mathrm{an}\mathrm{integer}.\hfill \end{array}\right.$$

It plays a significant role in the transformation theory of the Dedekind $\eta$ function. In [1,2,3,4,5], many researchers have investigated the various properties of $S(h,k)$. Perhaps the most well-known property of the Dedekind sums is the reciprocity formula

$$S(h,k)+S(k,h)=\frac{h^2+k^2+1}{12hk}-\frac{1}{4}.$$

Conrey, J.B. et al. [2] studied the $2m$-th power mean of $S(h,k)$, and proved the following important asymptotic formula

$$\underset{h=1}{\overset{k}{{\sum }^{\prime }}}{S}^{2m}(h,k)=f_m\left(k\right){\left(\frac{k}{12}\right)}^{2m}+O\left((k^{9/5}+k^{2m-1+1/(m+1)}){log}^{3}k\right),$$

where ${\sum }_h^{\prime }$ denotes summation over all h such that $(h,k)=1$ and $f_m\left(k\right)$ is defined by the Dirichlet series

$$\sum _{k=1}^{\infty }\frac{f_m\left(k\right)}{k^s}=2\frac{{\zeta }^2\left(2m\right)}{\zeta \left(4m\right)}·\frac{\zeta (s+4m-1)}{{\zeta }^2(s+2m)}·\zeta \left(s\right).$$

For $m\ge 2$, Jia, C. [3] reduced the error terms to $O\left(k^{2m-1}\right)$. While for $m=1$, Zhang, W. [5] showed

$$\underset{h=1}{\overset{k}{{\sum }^{\prime }}}{S}^2(h,k)=\frac{5}{144}k\varphi \left(k\right)\prod _{p^{\alpha }\parallel k}{\left(1+\frac{1}{p}+\frac{1}{p^2}\right)}^{-1}\left[{\left(1+\frac{1}{p}\right)}^2-\frac{1}{p^{3\alpha +1}}\right]+O\left(kexp\left(\frac{4logk}{loglogk}\right)\right).$$

Zhang, W. and Yi, Y. [6] studied the first mean value of $S(h,k)$, and obtained an asymptotic formula

$$\underset{n\le N}{{\sum }^{\prime }}S(n,k)=\frac{1}{12}\varphi \left(k\right)\left(logN+\gamma +\sum _{p|k}\frac{logp}{p-1}\right)+O\left(\frac{k{2}^{\omega \left(k\right)}}{N}+N{k}^{ϵ}\right)$$

for positive integer k and $1<N\le \frac{1}{2}k$.

Zhang, W. [7] defined the general Dedekind sum $S(h,n,k)$ as follows:

$$S(h,n,k)=\sum _{a=1}^k\overline{B_n}\left(\frac{a}{k}\right)\overline{B_n}\left(\frac{ah}{k}\right),$$

where

$$\overline{B_n}\left(x\right)=\left\{\begin{array}{cc}{B}_n(x-\left[x\right]),\hfill & \mathrm{if}x\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{integer};\hfill \\ 0,\hfill & \mathrm{if}x\mathrm{is}\mathrm{an}\mathrm{integer}.\hfill \end{array}\right.$$

called the n-th periodic Bernoulli polynomials defined on $0<x\le 1$, and $B_n\left(x\right)$ is the n-th Bernoulli polynomials. Clearly, $S(h,1,k)=S(h,k)$ is the classical Dedekind sum.

Recently, Kim et al. [8,9,10] studied the poly-Dedekind sums given by

$$S_n^{\left(m\right)}(h,k)=\sum _{a=1}^k\left(\frac{a}{k}\right){\overline{B}}_n^{\left(m\right)}\left(\frac{ah}{k}\right),$$

where ${\overline{B}}_n^{\left(m\right)}\left(x\right)=B_n^{\left(m\right)}(x-\left[x\right])$ are the type 2 poly-Bernoulli functions of index m, and obtained some interesting identities. Obviously, $S_1^{\left(1\right)}=S(h,k)$ is the classical Dedekind sum.

Let $q>2$ be a prime, p be a given prime with $p<q$. Using the similar method of Shparlinski, I.E. [11] and combining with the mean value of L-functions and estimate of character sums, the authors and Wang, N. [12] studied the mean value distribution of the general Dedekind sums over short interval, that is

$$\sum _{a\le N}\sum _{b\le N}{a}^l{b}^{k}S(a\overline{b},n,q),$$

here n and N be two positive integers with $q^{ϵ}\le N\le q^{1-ϵ}$, $l,k$ be two non-negative integers and $\overline{b}$ denote the multiplicative inverse of b modulo q. However, in the final remarks, Shparlinski, I.E. [11] pointed out that “the author sees no reason why an appropriate asymptotic formula cannot hold for even larger values of N, up to $q/2$”. In this paper we can take N to $q/p$, then through transform, mean value of Dirichlet L-functions and the properties of character sums to study the mean value of the general Dedekind sums over interval $[1,\frac{q}{p})$, and obtain some sharper asymptotic formulae for it.

Now we give the main conclusion.

Theorem 1.

Let $q>2$ be a prime, p be a given prime with $p<q$, n be a positive integer. Then we have

(i) when n be an even number,

$$\sum _{a<\frac{q}{p}}\sum _{b<\frac{q}{p}}S(a\overline{b},n,q)=\frac{{(n!)}^2{q}^2}{2^{2n-2}{\pi }^{2n}}\left(\frac{1}{2{\pi }^2}{C}_{p,n}-\frac{{\zeta }^2\left(n\right)}{p^2}\right)+O\left(q^{1+ϵ}\right),$$

here ${\displaystyle C_{p,n}=\sum _{u=1}^{\infty }\frac{{\gamma }_p^2(u,n)}{u^2}}$, ${\displaystyle {\gamma }_p(u,n)=\sum _{d_1{d}_2=u}sin\frac{2\pi d_1}{p}·d_2^{1-n}}$.

(ii) when n be an odd number,

$$\sum _{a<\frac{q}{p}}\sum _{b<\frac{q}{p}}S(a\overline{b},n,q)=\frac{{(n!)}^2{q}^2}{2^{2n-2}{\pi }^{2n+2}}{T}_{p,n}+O\left(q^{1+ϵ}\right),$$

here

$$\begin{array}{ccc}\hfill T_{p,n}=& & \frac{1}{2}\sum _{u=1}^{\infty }\frac{{\nu }_p^2(u,n)}{u^2}+\frac{1}{2}\left(1+\frac{1}{p^2}\right)\sum _{u=1}^{\infty }\frac{{\nu }^2(u,n)}{u^2}\hfill \\ & +& \frac{1}{p^2}\sum _{u=1}^{\infty }\frac{\nu (u,n){\nu }_p(pu,n)}{u^2}-\frac{1}{p^2}\sum _{u=1}^{\infty }\frac{\nu (u,n)\nu (pu,n)}{u^2}\hfill \\ & -& \sum _{u=1}^{\infty }\frac{\nu (u,n){\nu }_p(u,n)}{u^2},\hfill \end{array}$$

${\displaystyle {\nu }_p(u,n)=\sum _{d_1{d}_2=u}cos\frac{2\pi d_1}{p}·d_2^{1-n}}$, ${\displaystyle \nu (u,n)=\sum _{d|u}{d}^{1-n}}$.

It is clear that $C_{p,n}$ and $T_{p,n}$ are constants depending on p and n. From our theorem we may immediately deduce the following corollaries:

Corollary 1.

Let $q>2$ be a prime, we have

$$\sum _{a<\frac{q}{2}}\sum _{b<\frac{q}{2}}S(a\overline{b},2,q)=-\frac{q^2}{144}+O\left(q^{1+ϵ}\right),$$

$$\sum _{a<\frac{q}{2}}\sum _{b<\frac{q}{2}}{S}_2^{\left(1\right)}(a\overline{b},q)=\frac{\pi }{144i}{q}^2+O\left(q^{1+ϵ}\right).$$

Corollary 2.

Let $q>2$ be a prime, we have

$$\sum _{a<\frac{q}{2}}\sum _{b<\frac{q}{2}}S(a\overline{b},4,q)=-\frac{q^2}{3600}+O\left(q^{1+ϵ}\right),$$

$$\sum _{a<\frac{q}{2}}\sum _{b<\frac{q}{2}}{S}_4^{\left(1\right)}(a\overline{b},q)=\frac{{\pi }^{3}i}{10800}{q}^2+O\left(q^{1+ϵ}\right).$$

For the general index m, the method of our article does not obtain the expected result. It would be an interesting question to continue to study the mean value of $S_n^{\left(m\right)}(h,k)$.

2. Some Lemmas

To prove the theorem, We need the following lemmas.

Lemma 1.

Let k and r be integers with $k\ge 2$ and $(r,k)=1$, χ be a Dirichlet character modulo k. Then we have

$$\underset{\chi modk}{{\sum }^{*}}\chi \left(r\right)=\sum _{d\left|\right(k,r-1)}\mu \left(\frac{k}{d}\right)\varphi \left(d\right)$$

where ${{\sum }^{*}}_{\chi modk}$ denotes the summation over all primitive characters modulo k.

Proof. 

See Lemma 4 of reference [13]. □

Lemma 2.

Let $k\ge 3$ and h be two integers with $(h,k)=1$, n be positive integer. Then we have

$$S(h,n,k)=\frac{{(n!)}^2}{4^{n-1}{k}^{2n-1}{\pi }^{2n}}\sum _{d\mid k}\frac{d^{2n}}{\varphi \left(d\right)}\underset{\chi (-1)=-1}{\sum _{\chi modd}}\chi \left(h\right)|L(n,\chi ){|}^2$$

for odd n and

$$S(h,n,k)=\frac{{(n!)}^2}{4^{n-1}{k}^{2n-1}{\pi }^{2n}}\sum _{d\mid k}\frac{d^{2n}}{\varphi \left(d\right)}\underset{\chi (-1)=1}{\sum _{\chi modd}}\chi \left(h\right)|L(n,\chi ){|}^2-\frac{{(n!)}^2}{4^{n-1}{\pi }^{2n}}{\zeta }^2\left(n\right)$$

for even n.

Proof. 

See Theorem of reference [7]. □

Lemma 3.

Let χ be a primitive Dirichlet character modulo k. Then for any real number $\lambda \in [0,1)$ with $\lambda \ne \frac{r}{k}$, we have

$$\sum _{a=1}^{\left[\lambda k\right]}\chi \left(a\right)=\left\{\begin{array}{cc}\frac{\tau \left(\chi \right)}{\pi }{\displaystyle \sum _{n=1}^{\infty }}\frac{\overline{\chi }\left(n\right)sin\left(2\pi n\lambda \right)}{n},\hfill & if\chi (-1)=1;\hfill \\ \frac{\tau \left(\chi \right)}{\pi i}{\displaystyle \sum _{n=1}^{\infty }}\frac{\overline{\chi }\left(n\right)(1-cos\left(2\pi n\lambda \right))}{n},\hfill & if\chi (-1)=-1.\hfill \end{array}\right.$$

where $\left[x\right]$ denotes the greatest integer less than or equal to x, $\tau \left(\chi \right)={\sum }_{a=1}^k\chi \left(a\right)e\left(\frac{a}{k}\right)$ is the Gauss sum and $e\left(y\right)=e^{2\pi iy}$.

Proof. 

See Section 3.1 of [14]. □

Lemma 4.

Let $k>3$ be an integer and p be a prime with $p\nmid k$ and $p<k$. Then we have the identity

(i) for even primitive character $\chi modk$,

$$\sum _{a<\frac{k}{p}}\chi \left(a\right)=\frac{\tau \left(\chi \right)}{\pi i\varphi \left(p\right)}\underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)L(1,\xi \overline{\chi }).$$

(ii) for odd primitive character $\chi modk$,

$$\sum _{a<\frac{k}{p}}\chi \left(a\right)=\frac{\tau \left(\chi \right)}{\pi i}\left(1-\frac{\overline{\chi }\left(p\right)}{p}\right)L(1,\overline{\chi })+\frac{i\tau \left(\chi \right)}{\pi \varphi \left(p\right)}\underset{\xi (-1)=1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)L(1,\xi \overline{\chi }).$$

Proof. 

From Lemma 3, (i) when $\chi (-1)=1$, we can write

$$\sum _{a<\frac{k}{p}}\chi \left(a\right)=\frac{\tau \left(\chi \right)}{\pi }\sum _{n=1}^{\infty }\overline{\chi }\left(n\right)\frac{sin(2\pi n/p)}{n}=\frac{\tau \left(\chi \right)}{\pi }\sum _{h=1}^{p-1}sin\frac{2\pi h}{p}\underset{n\equiv h(modp)}{\sum _{n=1}^{\infty }}\frac{\overline{\chi }\left(n\right)}{n}.$$

Now

$$\underset{n\equiv h(modp)}{\sum _{n=1}^{\infty }}\frac{\overline{\chi }\left(n\right)}{n}=\frac{1}{\varphi \left(p\right)}\sum _{\xi modp}\overline{\xi }\left(h\right)\sum _{n=1}^{\infty }\frac{\xi \overline{\chi }\left(n\right)}{n}=\frac{1}{\varphi \left(p\right)}\sum _{\xi modp}\overline{\xi }\left(h\right)L(1,\xi \overline{\chi }).$$

Furthermore,

$$\begin{array}{ccc}\hfill \sum _{h=1}^{p-1}\overline{\xi }\left(h\right)sin\frac{2\pi h}{p}& =& \frac{1}{2i}\sum _{h=1}^{p-1}\overline{\xi }\left(h\right)\left(e\left(\frac{h}{p}\right)-e\left(\frac{-h}{p}\right)\right)\hfill \\ & =& \frac{1}{2i}(1-\xi (-1))\tau \left(\overline{\xi }\right),\hfill \end{array}$$

where $e\left(x\right)=e^{2\pi ix}$.

Thus, we obtain the identity for $\chi (-1)=1$,

$$\sum _{a<\frac{k}{p}}\chi \left(a\right)=\frac{\tau \left(\chi \right)}{\pi i\varphi \left(p\right)}\underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)L(1,\xi \overline{\chi }).$$

(ii) when $\chi (-1)=-1$, we can write

$$\begin{array}{ccc}\hfill \sum _{a<\frac{k}{p}}\chi \left(a\right)& =& \frac{\tau \left(\chi \right)}{\pi i}\sum _{n=1}^{\infty }\overline{\chi }\left(n\right)\frac{1-cos(2\pi n/p)}{n}\hfill \\ & =& \frac{\tau \left(\chi \right)}{\pi i}\left(L(1,\overline{\chi })-\sum _{n=1}^{\infty }\overline{\chi }\left(n\right)\frac{cos(2\pi n/p)}{n}\right)\hfill \\ & =& \frac{\tau \left(\chi \right)}{\pi i}L(1,\overline{\chi })+\frac{\tau \left(\chi \right)}{\pi }\sum _{n=1}^{\infty }\overline{\chi }\left(n\right)\frac{icos(2\pi n/p)}{n}.\hfill \end{array}$$

Now

$$\begin{array}{ccc}& & \sum _{n=1}^{\infty }\overline{\chi }\left(n\right)\frac{icos(2\pi n/p)}{n}\hfill \\ & =& i\sum _{h=1}^{p-1}cos\frac{2\pi h}{p}\underset{n\equiv h(modp)}{\sum _{n=1}^{\infty }}\frac{\overline{\chi }\left(n\right)}{n}\hfill \\ & =& i\sum _{h=1}^{p-1}cos\frac{2\pi h}{p}·\frac{1}{\varphi \left(p\right)}\sum _{\xi modp}\overline{\xi }\left(h\right)L(1,\xi \overline{\chi })+\frac{i\overline{\chi }\left(p\right)}{p}L(1,\overline{\chi })\hfill \\ & =& \frac{i}{\varphi \left(p\right)}\sum _{\xi modp}\left(\sum _{h=1}^{p-1}\overline{\xi }\left(h\right)cos\frac{2\pi h}{p}\right)L(1,\xi \overline{\chi })+\frac{i\overline{\chi }\left(p\right)}{p}L(1,\overline{\chi }),\hfill \end{array}$$

noting that

$$e\left(\frac{h}{p}\right)+e\left(\frac{-h}{p}\right)=2cos\frac{2\pi h}{p},$$

we have

$$\sum _{h=1}^{p-1}\overline{\xi }\left(h\right)cos\frac{2\pi h}{p}=\frac{1}{2}\sum _{h=1}^{p-1}\overline{\xi }\left(h\right)\left[e\left(\frac{h}{p}\right)+e\left(\frac{-h}{p}\right)\right]=\frac{1+\overline{\xi }(-1)}{2}\tau \left(\overline{\xi }\right).$$

So, we can get

$$\sum _{n=1}^{\infty }\overline{\chi }\left(n\right)\frac{icos(2\pi n/p)}{n}=\frac{i}{\varphi \left(p\right)}\underset{\xi (-1)=1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)L(1,\xi \overline{\chi })+\frac{i\overline{\chi }\left(p\right)}{p}L(1,\overline{\chi }).$$

Thus, we obtain the identity for $\chi (-1)=-1$,

$$\sum _{a<\frac{k}{p}}\chi \left(a\right)=\frac{\tau \left(\chi \right)}{\pi i}\left(1-\frac{\overline{\chi }\left(p\right)}{p}\right)L(1,\overline{\chi })+\frac{i\tau \left(\chi \right)}{\pi \varphi \left(p\right)}\underset{\xi (-1)=1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)L(1,\xi \overline{\chi }).$$

This proves Lemma 4. □

Lemma 5.

Let $q>2$ be a prime, p be a given prime with $p<q$, n be a positive integer. Then we have

$$\sum _{a<\frac{q}{p}}\sum _{b<\frac{q}{p}}\underset{\chi (-1)=1}{\sum _{\chi modq}}\chi \left(a\overline{b}\right){\left|L(n,\chi )\right|}^2=\frac{q^2}{2{\pi }^2}{C}_{p,n}+O\left(q^{1+ϵ}\right),$$

here ${\displaystyle C_{p,n}=\sum _{u=1}^{\infty }\frac{{\gamma }_p^2(u,n)}{u^2}}$, ${\displaystyle {\gamma }_p(u,n)=\sum _{d_1{d}_2=u}sin\frac{2\pi d_1}{p}·d_2^{1-n}}$. and

$$\sum _{a<\frac{q}{p}}\sum _{b<\frac{q}{p}}\underset{\chi (-1)=-1}{\sum _{\chi modq}}\chi \left(a\overline{b}\right){\left|L(n,\chi )\right|}^2=\frac{q^2}{{\pi }^2}{T}_{p,n}+O\left(q^{1+ϵ}\right),$$

here

$$\begin{array}{ccc}\hfill T_{p,n}& =& {\displaystyle \frac{1}{2}\sum _{u=1}^{\infty }\frac{{\nu }_p^2(u,n)}{u^2}+\frac{1}{2}\left(1+\frac{1}{p^2}\right)\sum _{u=1}^{\infty }\frac{{\nu }^2(u,n)}{u^2}}\hfill \\ & & +\frac{1}{p^2}\sum _{u=1}^{\infty }\frac{\nu (u,n){\nu }_p(pu,n)}{u^2}-\frac{1}{p^2}\sum _{u=1}^{\infty }\frac{\nu (u,n)\nu (pu,n)}{u^2}\hfill \\ & & -\sum _{u=1}^{\infty }\frac{\nu (u,n){\nu }_p(u,n)}{u^2},\hfill \end{array}$$

${\displaystyle {\nu }_p(u,n)=\sum _{d_1{d}_2=u}cos\frac{2\pi d_1}{p}·d_2^{1-n}}$, ${\displaystyle \nu (u,n)=\sum _{d|u}{d}^{1-n}}$.

Proof. 

From Lemma 4, for $\chi (-1)=1$, we have

$$\begin{array}{ccc}& & \sum _{a<\frac{q}{p}}\sum _{b<\frac{q}{p}}\underset{\chi (-1)=1}{\sum _{\chi modq}}\chi \left(a\overline{b}\right){\left|L(n,\chi )\right|}^2\hfill \\ & =& \underset{\chi (-1)=1}{\sum _{\chi modq}}{\left|L(n,\chi )\right|}^2\sum _{a<\frac{q}{p}}\chi \left(a\right)\sum _{b<\frac{q}{p}}\chi \left(\overline{b}\right)\hfill \\ & =& \underset{\chi (-1)=1}{\sum _{\chi modq}}{\left|L(n,\chi )\right|}^2\frac{\tau \left(\chi \right)}{\pi i\varphi \left(p\right)}\underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)L(1,\xi \overline{\chi })\frac{\tau \left(\overline{\chi }\right)}{\pi i\varphi \left(p\right)}\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)L(1,\overline{\lambda }\chi )\hfill \\ & =& \frac{-q}{{\pi }^2{\varphi }^2\left(p\right)}\underset{\chi (-1)=1}{\sum _{\chi modq}}{\left|L(n,\chi )\right|}^2\underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)L(1,\xi \overline{\chi })\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)L(1,\overline{\lambda }\chi )\hfill \\ & =& \frac{-q}{{\pi }^2{\varphi }^2\left(p\right)}\underset{\chi (-1)=1}{\sum _{\chi modq}}\underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)\hfill \\ & & \times \left(\sum _{u=1}^{\infty }\frac{\overline{\chi }\left(u\right){\displaystyle \sum _{d_1{d}_2=u}}\xi \left(d_1\right)d_2^{1-n}}{u}\right)\left(\sum _{v=1}^{\infty }\frac{\chi \left(v\right){\displaystyle \sum _{d_1^{\prime }{d}_2^{\prime }=v}}\overline{\lambda }\left(d_1^{\prime }\right){d_2^{\prime }}^{1-n}}{v}\right)\hfill \\ & :=& \frac{-q}{{\pi }^2{\varphi }^2\left(p\right)}M.\hfill \end{array}$$

For convenience, we put

$$A(\overline{\chi },\xi ,y)=\sum _{N<u\le y}\overline{\chi }\left(u\right)\sum _{d_1{d}_2=u}\xi \left(d_1\right)d_2^{1-n},$$

where N is a parameter with $q\le N<q^5$. Then from Abel’s identity we have

$$\begin{array}{ccc}\hfill M& =& \underset{\chi (-1)=1}{\sum _{\chi modq}}\underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)\hfill \\ \hfill & & \times \left(\sum _{u=1}^{\infty }\frac{\overline{\chi }\left(u\right){\displaystyle \sum _{d_1{d}_2=u}}\xi \left(d_1\right)d_2^{1-n}}{u}\right)\left(\sum _{v=1}^{\infty }\frac{\chi \left(v\right){\displaystyle \sum _{d_1^{\prime }{d}_2^{\prime }=v}}\overline{\lambda }\left(d_1^{\prime }\right){d_2^{\prime }}^{1-n}}{v}\right)\hfill \\ \hfill & =& \underset{\chi (-1)=1}{\sum _{\chi modq}}\underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)\hfill \\ \hfill & & \times \left(\sum _{u\le N}\frac{\overline{\chi }\left(u\right){\displaystyle \sum _{d_1{d}_2=u}}\xi \left(d_1\right)d_2^{1-n}}{u}+{\int }_N^{\infty }\frac{A(\overline{\chi },\xi ,y)}{y^2}dy\right)\hfill \\ \hfill & & \times \left(\sum _{v\le N}\frac{\chi \left(v\right){\displaystyle \sum _{d_1^{\prime }{d}_2^{\prime }=v}}\overline{\lambda }\left(d_1^{\prime }\right){d_2^{\prime }}^{1-n}}{v}+{\int }_N^{\infty }\frac{A(\chi ,\overline{\lambda },y)}{y^2}dy\right)\hfill \\ \hfill & =& \underset{\chi (-1)=1}{\sum _{\chi modq}}\underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)\hfill \\ \hfill & & \times \left(\sum _{u\le N}\frac{\overline{\chi }\left(u\right){\displaystyle \sum _{d_1{d}_2=u}}\xi \left(d_1\right)d_2^{1-n}}{u}\right)\left(\sum _{v\le N}\frac{\chi \left(v\right){\displaystyle \sum _{d_1^{\prime }{d}_2^{\prime }=v}}\overline{\lambda }\left(d_1^{\prime }\right){d_2^{\prime }}^{1-n}}{v}\right)\hfill \\ \hfill & & +\underset{\chi (-1)=1}{\sum _{\chi modq}}\underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)\hfill \\ \hfill & & \times \left(\sum _{u\le N}\frac{\overline{\chi }\left(u\right){\displaystyle \sum _{d_1{d}_2=u}}\xi \left(d_1\right)d_2^{1-n}}{u}\right)\left({\int }_N^{\infty }\frac{A(\chi ,\overline{\lambda },y)}{y^2}dy\right)\hfill \\ \hfill & & +\underset{\chi (-1)=1}{\sum _{\chi modq}}\underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)\hfill \\ \hfill & & \times \left(\sum _{v\le N}\frac{\chi \left(v\right){\displaystyle \sum _{d_1^{\prime }{d}_2^{\prime }=v}}\overline{\lambda }\left(d_1^{\prime }\right){d_2^{\prime }}^{1-n}}{v}\right)\left({\int }_N^{\infty }\frac{A(\overline{\chi },\xi ,y)}{y^2}dy\right)\hfill \\ \hfill & & +\underset{\chi (-1)=1}{\sum _{\chi modq}}\underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)\left({\int }_N^{\infty }\frac{A(\overline{\chi },\xi ,y)}{y^2}dy\right)\left({\int }_N^{\infty }\frac{A(\chi ,\overline{\lambda },y)}{y^2}dy\right)\hfill \\ & :=& M_1+M_2+M_3+M_4,\hfill \end{array}$$

(1)

we shall calculate each term in the above expression.

(i) From Lemma 1 we have

$$\begin{array}{ccc}\hfill \underset{\chi (-1)=1}{\sum _{\chi modq}}\chi \left(\overline{u}v\right)& =& \frac{1}{2}\sum _{\chi modq}(1+\chi (-1))\chi \left(\overline{u}v\right)\hfill \\ & =& \frac{1}{2}\sum _{\chi modq}\chi \left(\overline{u}v\right)+\frac{1}{2}\sum _{\chi modq}\chi (-\overline{u}v)\hfill \\ & =& \frac{1}{2}\sum _{d\left|\right(q,\overline{u}v-1)}\mu \left(\frac{q}{d}\right)\varphi \left(d\right)+\frac{1}{2}\sum _{d\left|\right(q,\overline{u}v+1)}\mu \left(\frac{q}{d}\right)\varphi \left(d\right).\hfill \end{array}$$

In addition, we have

$$\begin{array}{ccc}\hfill \underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)\xi \left(d_1\right)& =& \underset{\xi (-1)=-1}{\sum _{\xi modp}}\sum _{a=1}^p\overline{\xi }\left(a\right)e^{2\pi i\frac{a}{p}}\xi \left(d_1\right)\hfill \\ & =& \underset{\xi (-1)=-1}{\sum _{\xi modp}}\sum _{a=1}^p\overline{\xi }\left(a\right)e^{2\pi i\frac{a{d}_1}{p}}\hfill \\ & =& \frac{\varphi \left(p\right)}{2}{e}^{2\pi i\frac{d_1}{p}}-\frac{\varphi \left(p\right)}{2}{e}^{2\pi i\frac{(p-1)d_1}{p}}\hfill \\ & =& i\varphi \left(p\right)sin\frac{2\pi d_1}{p}.\hfill \end{array}$$

Similarly, we can also get

$$\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)\overline{\lambda }\left(d_1^{\prime }\right)=i\varphi \left(p\right)sin\frac{2\pi d_1^{{}^{\prime }}}{p}.$$

So, we have

$$\begin{array}{ccc}& & \underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)\sum _{d_1{d}_2=u}\xi \left(d_1\right)d_2^{1-n}\sum _{d_1^{\prime }{d}_2^{\prime }=v}\overline{\lambda }\left(d_1^{\prime }\right){d_2^{\prime }}^{1-n}\hfill \\ & =& -{\varphi }^2\left(p\right)\sum _{d_1{d}_2=u}sin\frac{2\pi d_1}{p}·d_2^{1-n}\sum _{d_1^{\prime }{d}_2^{\prime }=v}sin\frac{2\pi d_1^{\prime }}{p}·{d_2^{\prime }}^{1-n}\hfill \\ & =& -{\varphi }^2\left(p\right){\gamma }_p(u,n){\gamma }_p(v,n).\hfill \end{array}$$

Hence, we can write

$$\begin{array}{ccc}\hfill M_1& =& \underset{\chi (-1)=1}{\sum _{\chi modq}}\underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)\hfill \\ \hfill & & \times \left(\sum _{u\le N}\frac{\overline{\chi }\left(u\right){\displaystyle \sum _{d_1{d}_2=u}}\xi \left(d_1\right)d_2^{1-n}}{u}\right)\left(\sum _{v\le N}\frac{\chi \left(v\right){\displaystyle \sum _{d_1^{\prime }{d}_2^{\prime }=v}}\overline{\lambda }\left(d_1^{\prime }\right){d_2^{\prime }}^{1-n}}{v}\right)\hfill \\ \hfill & =& \underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)\hfill \\ \hfill & & \times \underset{1\le u\le N}{{\sum }^{\prime }}\underset{1\le v\le N}{{\sum }^{\prime }}\frac{{\displaystyle \sum _{d_1{d}_2=u}}\xi \left(d_1\right)d_2^{1-n}{\displaystyle \sum _{d_1^{\prime }{d}_2^{\prime }=v}}\overline{\lambda }\left(d_1^{\prime }\right){d_2^{\prime }}^{1-n}}{uv}\underset{\chi (-1)=1}{\sum _{\chi modq}}\chi \left(\overline{u}v\right)\hfill \\ \hfill & =& -\frac{1}{2}\underset{1\le u\le N}{{\sum }^{\prime }}\underset{1\le v\le N}{{\sum }^{\prime }}\frac{{\varphi }^2\left(p\right){\gamma }_p(u,n){\gamma }_p(v,n)}{uv}\sum _{d\left|\right(q,\overline{u}v-1)}\mu \left(\frac{p}{d}\right)\varphi \left(d\right)\hfill \\ \hfill & & -\frac{1}{2}\underset{1\le u\le N}{{\sum }^{\prime }}\underset{1\le v\le N}{{\sum }^{\prime }}\frac{{\varphi }^2\left(p\right){\gamma }_p(u,n){\gamma }_p(v,n)}{uv}\sum _{d\left|\right(q,\overline{u}v+1)}\mu \left(\frac{q}{d}\right)\varphi \left(d\right)\hfill \\ \hfill & =& -\frac{1}{2}\sum _{d|q}\mu \left(\frac{q}{d}\right)\varphi \left(d\right)\underset{u\equiv v(modd)}{\underset{1\le u\le N}{{\sum }^{\prime }}\underset{1\le v\le N}{{\sum }^{\prime }}}\frac{{\varphi }^2\left(p\right){\gamma }_p(u,n){\gamma }_p(v,n)}{uv}\hfill \\ & & -\frac{1}{2}\sum _{d|q}\mu \left(\frac{q}{d}\right)\varphi \left(d\right)\underset{u\equiv -v(modd)}{\underset{1\le u\le N}{{\sum }^{\prime }}\underset{1\le v\le N}{{\sum }^{\prime }}}\frac{{\varphi }^2\left(p\right){\gamma }_p(u,n){\gamma }_p(v,n)}{uv},\hfill \end{array}$$

(2)

where ${\displaystyle \underset{1\le n\le N}{{\sum }^{\prime }}}$ denotes the summation over n from 1 to N such that $(n,q)=1$.

For calculation convenience, we divide the sum over u or v into four cases: (i) $d\le u,v\le N$; (ii) $d\le u\le N$ and $1\le v\le d-1$; (iii) $1\le u\le d-1$ and $d\le v\le N$; (iv) $1\le u,v\le d-1$. So we have

$$\begin{array}{ccc}\hfill & & \sum _{d|q}\mu \left(\frac{q}{d}\right)\varphi \left(d\right)\underset{u\equiv v(modd)}{\underset{d\le u\le N}{{\sum }^{\prime }}\underset{d\le v\le N}{{\sum }^{\prime }}}\frac{{\varphi }^2\left(p\right){\gamma }_p(u,n){\gamma }_p(v,n)}{uv}\hfill \\ \hfill & \ll & \sum _{d|q}\varphi \left(d\right)\sum _{1\le r_1\le \frac{N}{d}}\sum _{1\le r_2\le \frac{N}{d}}\underset{l_1\equiv l_2(modd)}{\underset{l_1=1}{\overset{d-1}{{\sum }^{\prime }}}\underset{l_2=1}{\overset{d-1}{{\sum }^{\prime }}}}\frac{{\varphi }^2\left(p\right){\gamma }_p(r_{1}d+l_1,n){\gamma }_p(r_{2}d+l_2,n)}{(r_{1}d+l_1)(r_{2}d+l_2)}\hfill \\ \hfill & \ll & {\varphi }^2\left(p\right)\sum _{d|q}\varphi \left(d\right)\sum _{1\le r_1\le \frac{N}{d}}\sum _{1\le r_2\le \frac{N}{d}}\underset{l_1=1}{\overset{d-1}{{\sum }^{\prime }}}\frac{{\displaystyle \sum _{d_2|(r_{1}d+l_1)}}{d}_2^{1-n}{\displaystyle \sum _{d_2^{\prime }|(r_{2}d+l_1)}}{d_2^{\prime }}^{1-n}}{(r_{1}d+l_1)(r_{2}d+l_1)}\hfill \\ \hfill & \ll & {\varphi }^2\left(p\right)\sum _{d|q}\varphi \left(d\right)\sum _{1\le r_1\le \frac{N}{d}}\sum _{1\le r_2\le \frac{N}{d}}\underset{l_1=1}{\overset{d-1}{{\sum }^{\prime }}}\frac{{\tau }_2(r_{1}d+l_1){\tau }_2(r_{2}d+l_1)}{(r_{1}d+l_1)(r_{2}d+l_1)}\hfill \\ \hfill & \ll & {\varphi }^2\left(p\right)\sum _{d|q}\frac{\varphi \left(d\right)}{d}\sum _{1\le r_1\le \frac{N}{d}}\sum _{1\le r_2\le \frac{N}{d}}\frac{{\left[(r_{1}d+1)(r_{2}d+1)\right]}^{ϵ}}{r_1{r}_2}\hfill \\ & \ll & {\varphi }^2\left(p\right)q^{ϵ}.\hfill \end{array}$$

(3)

$$\begin{array}{ccc}& & \sum _{d|q}\mu \left(\frac{q}{d}\right)\varphi \left(d\right)\underset{u\equiv v(modd)}{\underset{d\le u\le N}{{\sum }^{\prime }}\underset{1\le v\le d-1}{{\sum }^{\prime }}}\frac{{\varphi }^2\left(p\right){\gamma }_p(u,n){\gamma }_p(v,n)}{uv}\hfill \\ \hfill & \ll & \sum _{d|q}\varphi \left(d\right)\sum _{1\le r_1\le \frac{N}{d}}\underset{v\equiv l_1(modd)}{\underset{1\le v\le d-1}{{\sum }^{\prime }}\underset{l_1=1}{\overset{d-1}{{\sum }^{\prime }}}}\frac{{\varphi }^2\left(p\right){\gamma }_p(r_{1}d+l_1,n){\gamma }_p(v,n)}{(r_{1}d+l_1)v}\hfill \\ \hfill & \ll & {\varphi }^2\left(p\right)\sum _{d|q}\varphi \left(d\right)\sum _{1\le r_1\le \frac{N}{d}}\underset{1\le v\le d-1}{{\sum }^{\prime }}\frac{{\tau }_2(r_{1}d+v){\tau }_2\left(v\right)}{(r_{1}d+v)v}\hfill \\ \hfill & \ll & {\varphi }^2\left(p\right)\sum _{d|q}\varphi \left(d\right)\sum _{1\le r_1\le \frac{N}{d}}\sum _{1\le v\le d-1}{\left(r_{1}vd\right)}^{ϵ-1}\hfill \\ & \ll & {\varphi }^2\left(p\right)q^{ϵ}.\hfill \end{array}$$

(4)

and

$$\begin{array}{ccc}& & \sum _{d|q}\mu \left(\frac{q}{d}\right)\varphi \left(d\right)\underset{u\equiv v(modd)}{\underset{1\le u\le d-1}{{\sum }^{\prime }}\underset{d\le v\le N}{{\sum }^{\prime }}}\frac{{\varphi }^2\left(p\right){\gamma }_p(u,n){\gamma }_p(v,n)}{uv}\hfill \\ \hfill & \ll & {\varphi }^2\left(p\right)\sum _{d|q}\varphi \left(d\right)\sum _{1\le u\le d-1}\sum _{1\le r_2\le \frac{N}{d}}{\left(u{r}_{2}d\right)}^{ϵ-1}\hfill \\ & \ll & {\varphi }^2\left(p\right)q^{ϵ},\hfill \end{array}$$

(5)

where we have used the estimate ${\tau }_2\left(n\right)\ll n^{ϵ}$.

For the case $1\le u,v\le d-1$, the solution of the congruence $u\equiv v(modd)$ is $u=v$. Hence,

$$\begin{array}{ccc}\hfill & & \sum _{d|q}\mu \left(\frac{q}{d}\right)\varphi \left(d\right)\underset{u\equiv v(modd)}{\underset{1\le u\le d-1}{{\sum }^{\prime }}\underset{1\le v\le d-1}{{\sum }^{\prime }}}\frac{{\varphi }^2\left(p\right){\gamma }_p(u,n){\gamma }_p(v,n)}{uv}\hfill \\ \hfill & =& {\varphi }^2\left(p\right)\sum _{d|q}\mu \left(\frac{q}{d}\right)\varphi \left(d\right)\underset{1\le u\le d-1}{{\sum }^{\prime }}\frac{{\gamma }_p^2(u,n)}{u^2}\hfill \\ \hfill & =& {\varphi }^2\left(p\right)\sum _{d|q}\mu \left(\frac{q}{d}\right)\varphi \left(d\right)\underset{(u,q)=1}{\sum _{u=1}^{\infty }}\frac{{\gamma }_p^2(u,n)}{u^2}+O\left({\varphi }^2\left(p\right)q^{ϵ}\right)\hfill \\ \hfill & =& (q-2){\varphi }^2\left(p\right)\underset{(u,q)=1}{\sum _{u=1}^{\infty }}\frac{{\gamma }_p^2(u,n)}{u^2}+O\left({\varphi }^2\left(p\right)q^{ϵ}\right)\hfill \\ \hfill & =& (q-2){\varphi }^2\left(p\right)\left(\sum _{u=1}^{\infty }\frac{{\gamma }_p^2(u,n)}{u^2}-\underset{q|u}{\sum _{u=1}^{\infty }}\frac{{\gamma }_p^2(u,n)}{u^2}\right)+O\left({\varphi }^2\left(p\right)q^{ϵ}\right)\hfill \\ & =& (q-2){\varphi }^2\left(p\right)\sum _{u=1}^{\infty }\frac{{\gamma }_p^2(u,n)}{u^2}+O\left({\varphi }^2\left(p\right)q^{ϵ}\right).\hfill \end{array}$$

(6)

Then from (3)–(6), we have

$$\begin{array}{ccc}\hfill & & -\frac{1}{2}\sum _{d|q}\mu \left(\frac{q}{d}\right)\varphi \left(d\right)\underset{u\equiv v(modd)}{\underset{1\le u\le N}{{\sum }^{\prime }}\underset{1\le v\le N}{{\sum }^{\prime }}}\frac{{\varphi }^2\left(p\right){\gamma }_p(u,n){\gamma }_p(v,n)}{uv}\hfill \\ \hfill & =& -\frac{{\varphi }^2\left(p\right)(q-2)}{2}\sum _{u=1}^{\infty }\frac{{\gamma }_p^2(u,n)}{u^2}+O\left({\varphi }^2\left(p\right)q^{ϵ}\right)\hfill \\ & =& -\frac{{\varphi }^2\left(p\right)(q-2)}{2}{C}_{p,n}+O\left({\varphi }^2\left(p\right)q^{ϵ}\right).\hfill \end{array}$$

(7)

Similarly, we can also get the estimate

$$\begin{array}{ccc}\hfill & & -\frac{1}{2}\sum _{d|q}\mu \left(\frac{q}{d}\right)\varphi \left(d\right)\underset{u\equiv -v(modd)}{\underset{1\le u\le N}{{\sum }^{\prime }}\underset{1\le v\le N}{{\sum }^{\prime }}}\frac{{\varphi }^2\left(p\right){\gamma }_p(u,n){\gamma }_p(v,n)}{uv}\hfill \\ \hfill & =& -\frac{1}{2}\sum _{d|q}\mu \left(\frac{q}{d}\right)\varphi \left(d\right)\underset{u+v=d}{\underset{1\le u\le N}{{\sum }^{\prime }}\underset{1\le v\le N}{{\sum }^{\prime }}}\frac{{\varphi }^2\left(p\right){\gamma }_p(u,n){\gamma }_p(v,n)}{uv}\hfill \\ \hfill & & -\frac{1}{2}\sum _{d|q}\mu \left(\frac{q}{d}\right)\varphi \left(d\right)\underset{u+v=ld,l\ge 2}{\underset{1\le u\le N}{{\sum }^{\prime }}\underset{1\le v\le N}{{\sum }^{\prime }}}\frac{{\varphi }^2\left(p\right){\gamma }_p(u,n){\gamma }_p(v,n)}{uv}\hfill \\ \hfill & \ll & {\varphi }^2\left(p\right)\sum _{d|q}\varphi \left(d\right)\sum _{1\le u\le d-1}\frac{{\gamma }_p(u,n){\gamma }_p(d-u,n)}{u(d-u)}\hfill \\ \hfill & & +{\varphi }^2\left(p\right)\sum _{d|q}\varphi \left(d\right)\underset{1\le u\le N}{{\sum }^{\prime }}\sum _{l=\left[\frac{u}{d}\right]+2}^{\left[\frac{N+u}{d}\right]}\frac{{\gamma }_p(u,n){\gamma }_p(ld-u,n)}{ldu-u^2}\hfill \\ \hfill & \ll & {\varphi }^2\left(p\right)\sum _{d|q}\frac{\varphi \left(d\right)}{d}\sum _{1\le u\le d-1}\frac{u^{ϵ}{(d-u)}^{ϵ}}{u}\hfill \\ \hfill & & +{\varphi }^2\left(p\right)\sum _{d|q}\frac{\varphi \left(d\right)}{d}\underset{1\le u\le N}{{\sum }^{\prime }}\sum _{l=\left[\frac{u}{d}\right]+2}^{\left[\frac{N+u}{d}\right]}\frac{u^{ϵ}{(ld-u)}^{ϵ}}{lu-\frac{u^2}{d}}\hfill \\ \hfill & \ll & {\varphi }^2\left(p\right)\left(q^{ϵ}+\sum _{d|q}\frac{\varphi \left(d\right)}{d}\sum _{u=1}^N\sum _{l=1}^N\frac{u^{ϵ}{l}^{ϵ}}{ul}\right)\hfill \\ & \ll & {\varphi }^2\left(p\right)q^{ϵ}.\hfill \end{array}$$

(8)

Then combining (2), (7) and (8), we have

$$\begin{array}{cc}\hfill & M_1=-\frac{{\varphi }^2\left(p\right)(q-2)}{2}{C}_{p,n}+O\left({\varphi }^2\left(p\right)q^{ϵ}\right).\end{array}$$

(9)

(ii) From Lemma 4 of [15], we have the estimate

$$\sum _{\chi \ne {\chi }_0}{\left|A(y,\chi )\right|}^2\ll y^{1+ϵ}{q}^2,$$

where ${\chi }_0$ denotes the principal character modulo q, ${\displaystyle A(y,\chi )=\sum _{N<n\le y}\chi \left(n\right){\tau }_2\left(n\right)}$. Then from the Cauchy inequality we can easily get

$$\sum _{\chi (-1)=1}\left|A(y,\chi )\right|\ll \sum _{\chi \ne {\chi }_0}\left|A(y,\chi )\right|\ll y^{\frac{1}{2}+ϵ}{q}^{\frac{3}{2}}.$$

Using this estimate we have

$$\begin{array}{ccc}\hfill M_2& =& \underset{\chi (-1)=1}{\sum _{\chi modq}}\underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)\hfill \\ \hfill & & \times \left(\sum _{u\le N}\frac{\overline{\chi }\left(u\right){\displaystyle \sum _{d_1{d}_2=u}}\xi \left(d_1\right)d_2^{1-n}}{u}\right)\left({\int }_N^{\infty }\frac{A(\chi ,\overline{\lambda },y)}{y^2}dy\right)\hfill \\ \hfill & =& -{\varphi }^2\left(p\right)\sum _{1\le u\le N}\frac{\overline{\chi }\left(u\right){\gamma }_p(u,n)}{u}{\int }_N^{\infty }\frac{1}{y^2}\left(\underset{\chi (-1)=1}{\sum _{\chi modq}}\sum _{N<v\le y}\chi \left(v\right){\gamma }_k(v,n)\right)dy\hfill \\ \hfill & \ll & {\varphi }^2\left(p\right)\sum _{1\le u\le N}\frac{\overline{\chi }\left(u\right)d\left(u\right)}{u}{\int }_N^{\infty }\frac{1}{y^2}\left(\underset{\chi (-1)=1}{\sum _{\chi modq}}A(y,\chi )\right)dy\hfill \\ \hfill & \ll & {\varphi }^2\left(p\right)\sum _{1\le u\le N}{u}^{ϵ-1}{\int }_N^{\infty }\frac{1}{y^2}\left(\underset{\chi (-1)=1}{\sum _{\chi modq}}\left|A(y,\chi )\right|\right)dy\hfill \\ & \ll & {\varphi }^2\left(p\right)N^{ϵ}{\int }_N^{\infty }\frac{q^{\frac{3}{2}}{y}^{\frac{1}{2}+{ϵ}_1}}{y^2}dy\ll {\varphi }^2\left(p\right)\frac{q^{\frac{3}{2}}}{N^{\frac{1}{2}-ϵ}}.\hfill \end{array}$$

(10)

(iii) Similar to (ii), we can also get

$$\begin{array}{c}\hfill M_3\ll {\varphi }^2\left(p\right)\frac{q^{\frac{3}{2}}}{N^{\frac{1}{2}-ϵ}}.\end{array}$$

(11)

(iv) Using the same discussion in (ii), and making use of the absolute convergent properties of the integral, we can calculate

$$\begin{array}{ccc}\hfill M_4& =& \underset{\chi (-1)=1}{\sum _{\chi modq}}\underset{\xi (-1)=-1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)\underset{\lambda (-1)=-1}{\sum _{\lambda modp}}\tau \left(\lambda \right)\left({\int }_N^{\infty }\frac{A(\overline{\chi },\xi ,y)}{y^2}dy\right)\left({\int }_N^{\infty }\frac{A(\chi ,\overline{\lambda },y)}{y^2}dy\right)\hfill \\ \hfill & =& {\varphi }^2\left(p\right)\underset{\chi (-1)=1}{\sum _{\chi modq}}\left({\int }_N^{\infty }\frac{1}{y^2}\sum _{N<u\le y}\overline{\chi }\left(u\right){\gamma }_p(u,n)dy\right)\left({\int }_N^{\infty }\frac{1}{y^2}\sum _{N<v\le y}\chi \left(v\right){\gamma }_p(v,n)dy\right)\hfill \\ \hfill & \ll & {\varphi }^2\left(p\right)\underset{\chi (-1)=1}{\sum _{\chi modp}}\left({\int }_N^{\infty }\frac{A(y,\overline{\chi })}{y^2}dy\right)\left({\int }_N^{\infty }\frac{A(y,\chi )}{y^2}dy\right)\hfill \\ \hfill & \le & {\varphi }^2\left(p\right){\int }_N^{\infty }{\int }_N^{\infty }\frac{1}{y^2{z}^2}\underset{\chi (-1)=1}{\sum _{\chi modq}}|A(y,\overline{\chi })\left|\right|A(y,\chi )|dydz\hfill \\ \hfill & \ll & {\varphi }^2\left(p\right){\int }_N^{\infty }\frac{1}{y^2}{\int }_N^{\infty }\frac{1}{z^2}{\left(\sum _{\chi \ne {\chi }_0}{\left|A(y,\overline{\chi })\right|}^2\right)}^{\frac{1}{2}}{\left(\sum _{\chi \ne {\chi }_0}{\left|A(y,\chi )\right|}^2\right)}^{\frac{1}{2}}dydz\hfill \\ \hfill & \ll & {\varphi }^2\left(p\right){\left({\int }_N^{\infty }\frac{1}{y^2}{\left(\sum _{\chi \ne {\chi }_0}{\left|A(y,\chi )\right|}^2\right)}^{\frac{1}{2}}dy\right)}^2\hfill \\ & \ll & {\varphi }^2\left(p\right){\left({\int }_N^{\infty }\frac{q}{y^{\frac{3}{2}-ϵ}}dy\right)}^2\ll {\varphi }^2\left(p\right)\frac{q^2}{N^{1-ϵ}}.\hfill \end{array}$$

(12)

Now taking $N=q^4$ and $ϵ<\frac{1}{2}$, combining (1) and (9)–(12), we have

$$M=-\frac{{\varphi }^2\left(p\right)(q-2)}{2}{C}_{p,n}+O\left({\varphi }^2\left(p\right)q^{ϵ}\right).$$

Thus we obtain the asymptotic formula for $\chi (-1)=1$,

$$\sum _{a<\frac{q}{p}}\sum _{b<\frac{q}{p}}\underset{\chi (-1)=1}{\sum _{\chi modq}}\chi \left(a\overline{b}\right){\left|L(n,\chi )\right|}^2=\frac{-q}{{\pi }^2{\varphi }^2\left(p\right)}M=\frac{q^2}{2{\pi }^2}{C}_{p,n}+O\left(q^{1+ϵ}\right).$$

For $\chi (-1)=-1$, from Lemma 4, we have

$$\begin{array}{ccc}& & \sum _{a<\frac{q}{p}}\sum _{b<\frac{q}{p}}\underset{\chi (-1)=-1}{\sum _{\chi modq}}\chi \left(a\overline{b}\right){\left|L(n,\chi )\right|}^2\hfill \\ & =& \underset{\chi (-1)=-1}{\sum _{\chi modq}}{\left|L(n,\chi )\right|}^2\left[\frac{\tau \left(\chi \right)}{\pi i}\left(1-\frac{\overline{\chi }\left(p\right)}{p}\right)L(1,\overline{\chi })+\frac{i\tau \left(\chi \right)}{\pi \varphi \left(p\right)}\underset{\xi (-1)=1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)L(1,\xi \overline{\chi })\right]\hfill \\ & & \times \left[\frac{\tau \left(\overline{\chi }\right)}{\pi i}\left(1-\frac{\chi \left(p\right)}{p}\right)L(1,\chi )+\frac{i\tau \left(\overline{\chi }\right)}{\pi \varphi \left(p\right)}\underset{\lambda (-1)=1}{\sum _{\lambda modp}}\tau \left(\lambda \right)L(1,\overline{\lambda }\chi )\right]\hfill \\ & =& \frac{q}{{\pi }^2}\underset{\chi (-1)=-1}{\sum _{\chi modq}}\left(1-\frac{\overline{\chi }\left(p\right)}{p}\right)\left(1-\frac{\chi \left(p\right)}{p}\right){\left|L(n,\chi )\right|}^2{\left|L(1,\chi )\right|}^2\hfill \\ & & +\frac{-q}{{\pi }^2\varphi \left(p\right)}\underset{\chi (-1)=-1}{\sum _{\chi modq}}\left(1-\frac{\overline{\chi }\left(p\right)}{p}\right){\left|L(n,\chi )\right|}^{2}L(1,\overline{\chi })\underset{\lambda (-1)=1}{\sum _{\lambda modp}}\tau \left(\lambda \right)L(1,\overline{\lambda }\chi )\hfill \\ & & +\frac{-q}{{\pi }^2\varphi \left(p\right)}\underset{\chi (-1)=-1}{\sum _{\chi modq}}\left(1-\frac{\chi \left(p\right)}{p}\right){\left|L(n,\chi )\right|}^{2}L(1,\chi )\underset{\xi (-1)=1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)L(1,\xi \overline{\chi })\hfill \\ & & +\frac{q}{{\pi }^2{\varphi }^2\left(p\right)}\underset{\chi (-1)=-1}{\sum _{\chi modq}}{\left|L(n,\chi )\right|}^2\underset{\xi (-1)=1}{\sum _{\xi modp}}\tau \left(\overline{\xi }\right)L(1,\xi \overline{\chi })\underset{\lambda (-1)=1}{\sum _{\lambda modp}}\tau \left(\lambda \right)L(1,\overline{\lambda }\chi )\hfill \\ & :=& A+B+C+D.\hfill \end{array}$$

Using the same method as proving $\chi (-1)=1$, we can easily get

$$A=\frac{q^2}{2{\pi }^2}\left(1+\frac{1}{p^2}\right)\sum _{u=1}^{\infty }\frac{{\nu }^2(u,n)}{u^2}-\frac{q^2}{{\pi }^2{p}^2}\sum _{u=1}^{\infty }\frac{\nu (u,n)\nu (pu,n)}{u^2}+O\left(q^{1+ϵ}\right),$$

$$B=C=\frac{q^2}{2{\pi }^2{p}^2}\sum _{u=1}^{\infty }\frac{\nu (u,n){\nu }_p(pu,n)}{u^2}-\frac{p^2}{2{\pi }^2}\sum _{u=1}^{\infty }\frac{\nu (u,n){\nu }_p(u,n)}{u^2}+O\left(q^{1+ϵ}\right),$$

$$D=\frac{q^2}{2{\pi }^2}\sum _{u=1}^{\infty }\frac{{\nu }_p^2(u,n)}{u^2}+O\left(q^{1+ϵ}\right).$$

Thus we obtain the asymptotic formula for $\chi (-1)=-1$,

$$\sum _{a<\frac{q}{p}}\sum _{b<\frac{q}{p}}\underset{\chi (-1)=-1}{\sum _{\chi modq}}\chi \left(a\overline{b}\right){\left|L(n,\chi )\right|}^2=\frac{q^2}{{\pi }^2}{T}_{p,n}+O\left(q^{1+ϵ}\right).$$

This proves Lemma 5. □

3. Proof of Theorem and Corollaries

In this section we will accomplish the proof of the theorem and corollaries. From Lemmas 2 and 5, we have:

(i) when n be an even number,

$$\begin{array}{ccc}& & \sum _{a<\frac{q}{p}}\sum _{b<\frac{q}{p}}S(a\overline{b},n,q)\hfill \\ & =& \frac{{(n!)}^{2}q}{4^{n-1}{\pi }^{2n}\varphi \left(q\right)}\sum _{a<\frac{q}{p}}\sum _{b<\frac{q}{p}}\underset{\chi (-1)=1}{\sum _{\chi modq}}\chi \left(a\overline{b}\right){\left|L(n,\chi )\right|}^2+\frac{{(n!)}^2{\zeta }^2\left(n\right)}{4^{n-1}{\pi }^{2n}}\left(\frac{1}{q^{2n-1}}-1\right)\sum _{a<\frac{q}{p}}\sum _{b<\frac{q}{p}}1\hfill \\ & =& \frac{{(n!)}^2{q}^2}{2^{2n-2}{\pi }^{2n}}\left(\frac{1}{2{\pi }^2}{C}_{p,n}-\frac{{\zeta }^2\left(n\right)}{p^2}\right)+O\left(q^{1+ϵ}\right).\hfill \end{array}$$

(ii) when n be an odd number,

$$\begin{array}{ccc}\hfill \sum _{a<\frac{q}{p}}\sum _{b<\frac{q}{p}}S(a\overline{b},n,q)& =& \frac{{(n!)}^{2}q}{4^{n-1}{\pi }^{2n}\varphi \left(q\right)}\sum _{a<\frac{q}{p}}\sum _{b<\frac{q}{p}}\underset{\chi (-1)=-1}{\sum _{\chi modq}}\chi \left(a\overline{b}\right){\left|L(n,\chi )\right|}^2\hfill \\ & =& \frac{{(n!)}^{2}q}{4^{n-1}{\pi }^{2n}\varphi \left(q\right)}\left(\frac{q^2}{{\pi }^2}{T}_{p,n}+O\left(q^{1+ϵ}\right)\right)\hfill \\ & =& \frac{{(n!)}^2{q}^2}{2^{2n-2}{\pi }^{2n+2}}{T}_{p,n}+O\left(q^{1+ϵ}\right).\hfill \end{array}$$

This completes the proof of the Theorem.

Taking $p=2$ and $n=2$ or 4 in the Theorem, we get $C_{2,2}=C_{2,4}=0$. When n be an even number, from reference [7], we can easily calculate

$$S_n^{\left(1\right)}(h,k)=\frac{{\left(2\pi \right)}^{n-1}}{i^{n+1}·n!}S(h,n,k).$$

Noting that $\zeta \left(2\right)={\pi }^2/6$ and $\zeta \left(4\right)={\pi }^4/90$, we easily get Corollarys 1 and 2.