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Article

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

by
Lei Liu
and
Zhefeng Xu
*
School of Mathematics, Northwest University Xi’an, Xi’an 710069, China
*
Author to whom correspondence should be addressed.
Symmetry 2020, 12(12), 2079; https://doi.org/10.3390/sym12122079
Submission received: 4 November 2020 / Revised: 10 December 2020 / Accepted: 11 December 2020 / Published: 15 December 2020

## 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 $[ 1 , q p )$, 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
$S ( h , k ) = ∑ a = 1 k a k a h k ,$
where
$( ( x ) ) = x − [ x ] − 1 2 , if x not be an integer ; 0 , if x be an integer .$
It plays a significant role in the transformation theory of the Dedekind $η$ 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
$S ( h , k ) + S ( k , h ) = h 2 + k 2 + 1 12 h k − 1 4 .$
Conrey, J.B. et al. [2] studied the $2 m$-th power mean of $S ( h , k )$, and proved the following important asymptotic formula
$∑ ′ h = 1 k S 2 m ( h , k ) = f m ( k ) k 12 2 m + O ( k 9 / 5 + k 2 m − 1 + 1 / ( m + 1 ) ) log 3 k ,$
where $∑ h ′$ denotes summation over all h such that $( h , k ) = 1$ and $f m ( k )$ is defined by the Dirichlet series
$∑ k = 1 ∞ f m ( k ) k s = 2 ζ 2 ( 2 m ) ζ ( 4 m ) · ζ ( s + 4 m − 1 ) ζ 2 ( s + 2 m ) · ζ ( s ) .$
For $m ≥ 2$, Jia, C. [3] reduced the error terms to $O ( k 2 m − 1 )$. While for $m = 1$, Zhang, W. [5] showed
$∑ ′ h = 1 k S 2 ( h , k ) = 5 144 k ϕ ( k ) ∏ p α ∥ k 1 + 1 p + 1 p 2 − 1 1 + 1 p 2 − 1 p 3 α + 1 + O k exp 4 log k log log k .$
Zhang, W. and Yi, Y. [6] studied the first mean value of $S ( h , k )$, and obtained an asymptotic formula
$∑ ′ n ≤ N S ( n , k ) = 1 12 ϕ ( k ) log N + γ + ∑ p | k log p p − 1 + O k 2 ω ( k ) N + N k ϵ$
for positive integer k and $1 < N ≤ 1 2 k$.
Zhang, W. [7] defined the general Dedekind sum $S ( h , n , k )$ as follows:
$S ( h , n , k ) = ∑ a = 1 k B n ¯ a k B n ¯ a h k ,$
where
$B n ¯ ( x ) = B n ( x − [ x ] ) , if x is not an integer ; 0 , if x is an integer .$
called the n-th periodic Bernoulli polynomials defined on $0 < x ≤ 1$, and $B n ( x )$ 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
$S n ( m ) ( h , k ) = ∑ a = 1 k a k B ¯ n ( m ) a h k ,$
where $B ¯ n ( m ) ( x ) = B n ( m ) ( x − [ x ] )$ are the type 2 poly-Bernoulli functions of index m, and obtained some interesting identities. Obviously, $S 1 ( 1 ) = S ( h , k )$ 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
$∑ a ≤ N ∑ b ≤ N a l b k S ( a b ¯ , n , q ) ,$
here n and N be two positive integers with $q ϵ ≤ N ≤ q 1 − ϵ$, $l , k$ be two non-negative integers and $b ¯$ 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 $[ 1 , q p )$, 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,
$∑ a < q p ∑ b < q p S ( a b ¯ , n , q ) = ( n ! ) 2 q 2 2 2 n − 2 π 2 n 1 2 π 2 C p , n − ζ 2 ( n ) p 2 + O ( q 1 + ϵ ) ,$
here $C p , n = ∑ u = 1 ∞ γ p 2 ( u , n ) u 2$, $γ p ( u , n ) = ∑ d 1 d 2 = u sin 2 π d 1 p · d 2 1 − n$.
(ii) when n be an odd number,
$∑ a < q p ∑ b < q p S ( a b ¯ , n , q ) = ( n ! ) 2 q 2 2 2 n − 2 π 2 n + 2 T p , n + O ( q 1 + ϵ ) ,$
here
$T p , n = 1 2 ∑ u = 1 ∞ ν p 2 ( u , n ) u 2 + 1 2 1 + 1 p 2 ∑ u = 1 ∞ ν 2 ( u , n ) u 2 + 1 p 2 ∑ u = 1 ∞ ν ( u , n ) ν p ( p u , n ) u 2 − 1 p 2 ∑ u = 1 ∞ ν ( u , n ) ν ( p u , n ) u 2 − ∑ u = 1 ∞ ν ( u , n ) ν p ( u , n ) u 2 ,$
$ν p ( u , n ) = ∑ d 1 d 2 = u cos 2 π d 1 p · d 2 1 − n$, $ν ( u , n ) = ∑ d | u d 1 − n$.
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
$∑ a < q 2 ∑ b < q 2 S ( a b ¯ , 2 , q ) = − q 2 144 + O ( q 1 + ϵ ) ,$
$∑ a < q 2 ∑ b < q 2 S 2 ( 1 ) ( a b ¯ , q ) = π 144 i q 2 + O ( q 1 + ϵ ) .$
Corollary 2.
Let $q > 2$ be a prime, we have
$∑ a < q 2 ∑ b < q 2 S ( a b ¯ , 4 , q ) = − q 2 3600 + O ( q 1 + ϵ ) ,$
$∑ a < q 2 ∑ b < q 2 S 4 ( 1 ) ( a b ¯ , q ) = π 3 i 10800 q 2 + O ( q 1 + ϵ ) .$
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 $S n ( m ) ( h , k )$.

## 2. Some Lemmas

To prove the theorem, We need the following lemmas.
Lemma 1.
Let k and r be integers with $k ≥ 2$ and $( r , k ) = 1$, χ be a Dirichlet character modulo k. Then we have
$∑ * χ mod k χ ( r ) = ∑ d | ( k , r − 1 ) μ k d ϕ ( d )$
where $∑ * χ mod k$ denotes the summation over all primitive characters modulo k.
Proof.
See Lemma 4 of reference [13]. □
Lemma 2.
Let $k ≥ 3$ and h be two integers with $( h , k ) = 1$, n be positive integer. Then we have
$S ( h , n , k ) = ( n ! ) 2 4 n − 1 k 2 n − 1 π 2 n ∑ d ∣ k d 2 n ϕ ( d ) ∑ χ mod d χ ( − 1 ) = − 1 χ ( h ) | L ( n , χ ) | 2$
for odd n and
$S ( h , n , k ) = ( n ! ) 2 4 n − 1 k 2 n − 1 π 2 n ∑ d ∣ k d 2 n ϕ ( d ) ∑ χ mod d χ ( − 1 ) = 1 χ ( h ) | L ( n , χ ) | 2 − ( n ! ) 2 4 n − 1 π 2 n ζ 2 ( n )$
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 $λ ∈ [ 0 , 1 )$ with $λ ≠ r k$, we have
$∑ a = 1 [ λ k ] χ ( a ) = τ ( χ ) π ∑ n = 1 ∞ χ ¯ ( n ) sin ( 2 π n λ ) n , i f χ ( − 1 ) = 1 ; τ ( χ ) π i ∑ n = 1 ∞ χ ¯ ( n ) ( 1 − cos ( 2 π n λ ) ) n , i f χ ( − 1 ) = − 1 .$
where $[ x ]$ denotes the greatest integer less than or equal to x, $τ ( χ ) = ∑ a = 1 k χ ( a ) e ( a k )$ is the Gauss sum and $e ( y ) = e 2 π i y$.
Proof.
See Section 3.1 of [14]. □
Lemma 4.
Let $k > 3$ be an integer and p be a prime with $p ∤ k$ and $p < k$. Then we have the identity
(i) for even primitive character $χ mod k$,
$∑ a < k p χ ( a ) = τ ( χ ) π i ϕ ( p ) ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) L ( 1 , ξ χ ¯ ) .$
(ii) for odd primitive character $χ mod k$,
$∑ a < k p χ ( a ) = τ ( χ ) π i 1 − χ ¯ ( p ) p L ( 1 , χ ¯ ) + i τ ( χ ) π ϕ ( p ) ∑ ξ mod p ξ ( − 1 ) = 1 τ ( ξ ¯ ) L ( 1 , ξ χ ¯ ) .$
Proof.
From Lemma 3, (i) when $χ ( − 1 ) = 1$, we can write
$∑ a < k p χ ( a ) = τ ( χ ) π ∑ n = 1 ∞ χ ¯ ( n ) sin ( 2 π n / p ) n = τ ( χ ) π ∑ h = 1 p − 1 sin 2 π h p ∑ n = 1 ∞ n ≡ h ( mod p ) χ ¯ ( n ) n .$
Now
$∑ n = 1 ∞ n ≡ h ( mod p ) χ ¯ ( n ) n = 1 ϕ ( p ) ∑ ξ mod p ξ ¯ ( h ) ∑ n = 1 ∞ ξ χ ¯ ( n ) n = 1 ϕ ( p ) ∑ ξ mod p ξ ¯ ( h ) L ( 1 , ξ χ ¯ ) .$
Furthermore,
$∑ h = 1 p − 1 ξ ¯ ( h ) sin 2 π h p = 1 2 i ∑ h = 1 p − 1 ξ ¯ ( h ) e h p − e − h p = 1 2 i ( 1 − ξ ( − 1 ) ) τ ( ξ ¯ ) ,$
where $e ( x ) = e 2 π i x$.
Thus, we obtain the identity for $χ ( − 1 ) = 1$,
$∑ a < k p χ ( a ) = τ ( χ ) π i ϕ ( p ) ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) L ( 1 , ξ χ ¯ ) .$
(ii) when $χ ( − 1 ) = − 1$, we can write
$∑ a < k p χ ( a ) = τ ( χ ) π i ∑ n = 1 ∞ χ ¯ ( n ) 1 − cos ( 2 π n / p ) n = τ ( χ ) π i L ( 1 , χ ¯ ) − ∑ n = 1 ∞ χ ¯ ( n ) cos ( 2 π n / p ) n = τ ( χ ) π i L ( 1 , χ ¯ ) + τ ( χ ) π ∑ n = 1 ∞ χ ¯ ( n ) i cos ( 2 π n / p ) n .$
Now
$∑ n = 1 ∞ χ ¯ ( n ) i cos ( 2 π n / p ) n = i ∑ h = 1 p − 1 cos 2 π h p ∑ n = 1 ∞ n ≡ h ( mod p ) χ ¯ ( n ) n = i ∑ h = 1 p − 1 cos 2 π h p · 1 ϕ ( p ) ∑ ξ mod p ξ ¯ ( h ) L ( 1 , ξ χ ¯ ) + i χ ¯ ( p ) p L ( 1 , χ ¯ ) = i ϕ ( p ) ∑ ξ mod p ∑ h = 1 p − 1 ξ ¯ ( h ) cos 2 π h p L ( 1 , ξ χ ¯ ) + i χ ¯ ( p ) p L ( 1 , χ ¯ ) ,$
noting that
$e h p + e − h p = 2 cos 2 π h p ,$
we have
$∑ h = 1 p − 1 ξ ¯ ( h ) cos 2 π h p = 1 2 ∑ h = 1 p − 1 ξ ¯ ( h ) e h p + e − h p = 1 + ξ ¯ ( − 1 ) 2 τ ( ξ ¯ ) .$
So, we can get
$∑ n = 1 ∞ χ ¯ ( n ) i cos ( 2 π n / p ) n = i ϕ ( p ) ∑ ξ mod p ξ ( − 1 ) = 1 τ ( ξ ¯ ) L ( 1 , ξ χ ¯ ) + i χ ¯ ( p ) p L ( 1 , χ ¯ ) .$
Thus, we obtain the identity for $χ ( − 1 ) = − 1$,
$∑ a < k p χ ( a ) = τ ( χ ) π i 1 − χ ¯ ( p ) p L ( 1 , χ ¯ ) + i τ ( χ ) π ϕ ( p ) ∑ ξ mod p ξ ( − 1 ) = 1 τ ( ξ ¯ ) L ( 1 , ξ χ ¯ ) .$
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
$∑ a < q p ∑ b < q p ∑ χ mod q χ ( − 1 ) = 1 χ ( a b ¯ ) L ( n , χ ) 2 = q 2 2 π 2 C p , n + O ( q 1 + ϵ ) ,$
here $C p , n = ∑ u = 1 ∞ γ p 2 ( u , n ) u 2$, $γ p ( u , n ) = ∑ d 1 d 2 = u sin 2 π d 1 p · d 2 1 − n$. and
$∑ a < q p ∑ b < q p ∑ χ mod q χ ( − 1 ) = − 1 χ ( a b ¯ ) L ( n , χ ) 2 = q 2 π 2 T p , n + O ( q 1 + ϵ ) ,$
here
$T p , n = 1 2 ∑ u = 1 ∞ ν p 2 ( u , n ) u 2 + 1 2 1 + 1 p 2 ∑ u = 1 ∞ ν 2 ( u , n ) u 2 + 1 p 2 ∑ u = 1 ∞ ν ( u , n ) ν p ( p u , n ) u 2 − 1 p 2 ∑ u = 1 ∞ ν ( u , n ) ν ( p u , n ) u 2 − ∑ u = 1 ∞ ν ( u , n ) ν p ( u , n ) u 2 ,$
$ν p ( u , n ) = ∑ d 1 d 2 = u cos 2 π d 1 p · d 2 1 − n$, $ν ( u , n ) = ∑ d | u d 1 − n$.
Proof.
From Lemma 4, for $χ ( − 1 ) = 1$, we have
$∑ a < q p ∑ b < q p ∑ χ mod q χ ( − 1 ) = 1 χ ( a b ¯ ) L ( n , χ ) 2 = ∑ χ mod q χ ( − 1 ) = 1 L ( n , χ ) 2 ∑ a < q p χ ( a ) ∑ b < q p χ ( b ¯ ) = ∑ χ mod q χ ( − 1 ) = 1 L ( n , χ ) 2 τ ( χ ) π i ϕ ( p ) ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) L ( 1 , ξ χ ¯ ) τ ( χ ¯ ) π i ϕ ( p ) ∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) L ( 1 , λ ¯ χ ) = − q π 2 ϕ 2 ( p ) ∑ χ mod q χ ( − 1 ) = 1 L ( n , χ ) 2 ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) L ( 1 , ξ χ ¯ ) ∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) L ( 1 , λ ¯ χ ) = − q π 2 ϕ 2 ( p ) ∑ χ mod q χ ( − 1 ) = 1 ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) ∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) × ∑ u = 1 ∞ χ ¯ ( u ) ∑ d 1 d 2 = u ξ ( d 1 ) d 2 1 − n u ∑ v = 1 ∞ χ ( v ) ∑ d 1 ′ d 2 ′ = v λ ¯ ( d 1 ′ ) d 2 ′ 1 − n v : = − q π 2 ϕ 2 ( p ) M .$
For convenience, we put
$A ( χ ¯ , ξ , y ) = ∑ N < u ≤ y χ ¯ ( u ) ∑ d 1 d 2 = u ξ ( d 1 ) d 2 1 − n ,$
where N is a parameter with $q ≤ N < q 5$. Then from Abel’s identity we have
$M = ∑ χ mod q χ ( − 1 ) = 1 ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) ∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) × ∑ u = 1 ∞ χ ¯ ( u ) ∑ d 1 d 2 = u ξ ( d 1 ) d 2 1 − n u ∑ v = 1 ∞ χ ( v ) ∑ d 1 ′ d 2 ′ = v λ ¯ ( d 1 ′ ) d 2 ′ 1 − n v = ∑ χ mod q χ ( − 1 ) = 1 ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) ∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) × ∑ u ≤ N χ ¯ ( u ) ∑ d 1 d 2 = u ξ ( d 1 ) d 2 1 − n u + ∫ N ∞ A ( χ ¯ , ξ , y ) y 2 d y × ∑ v ≤ N χ ( v ) ∑ d 1 ′ d 2 ′ = v λ ¯ ( d 1 ′ ) d 2 ′ 1 − n v + ∫ N ∞ A ( χ , λ ¯ , y ) y 2 d y = ∑ χ mod q χ ( − 1 ) = 1 ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) ∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) × ∑ u ≤ N χ ¯ ( u ) ∑ d 1 d 2 = u ξ ( d 1 ) d 2 1 − n u ∑ v ≤ N χ ( v ) ∑ d 1 ′ d 2 ′ = v λ ¯ ( d 1 ′ ) d 2 ′ 1 − n v + ∑ χ mod q χ ( − 1 ) = 1 ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) ∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) × ∑ u ≤ N χ ¯ ( u ) ∑ d 1 d 2 = u ξ ( d 1 ) d 2 1 − n u ∫ N ∞ A ( χ , λ ¯ , y ) y 2 d y + ∑ χ mod q χ ( − 1 ) = 1 ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) ∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) × ∑ v ≤ N χ ( v ) ∑ d 1 ′ d 2 ′ = v λ ¯ ( d 1 ′ ) d 2 ′ 1 − n v ∫ N ∞ A ( χ ¯ , ξ , y ) y 2 d y + ∑ χ mod q χ ( − 1 ) = 1 ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) ∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) ∫ N ∞ A ( χ ¯ , ξ , y ) y 2 d y ∫ N ∞ A ( χ , λ ¯ , y ) y 2 d y : = M 1 + M 2 + M 3 + M 4 ,$
we shall calculate each term in the above expression.
(i) From Lemma 1 we have
$∑ χ mod q χ ( − 1 ) = 1 χ ( u ¯ v ) = 1 2 ∑ χ mod q ( 1 + χ ( − 1 ) ) χ ( u ¯ v ) = 1 2 ∑ χ mod q χ ( u ¯ v ) + 1 2 ∑ χ mod q χ ( − u ¯ v ) = 1 2 ∑ d | ( q , u ¯ v − 1 ) μ q d ϕ ( d ) + 1 2 ∑ d | ( q , u ¯ v + 1 ) μ q d ϕ ( d ) .$
In addition, we have
$∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) ξ ( d 1 ) = ∑ ξ mod p ξ ( − 1 ) = − 1 ∑ a = 1 p ξ ¯ ( a ) e 2 π i a p ξ ( d 1 ) = ∑ ξ mod p ξ ( − 1 ) = − 1 ∑ a = 1 p ξ ¯ ( a ) e 2 π i a d 1 p = ϕ ( p ) 2 e 2 π i d 1 p − ϕ ( p ) 2 e 2 π i ( p − 1 ) d 1 p = i ϕ ( p ) sin 2 π d 1 p .$
Similarly, we can also get
$∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) λ ¯ ( d 1 ′ ) = i ϕ ( p ) sin 2 π d 1 ′ p .$
So, we have
$∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) ∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) ∑ d 1 d 2 = u ξ ( d 1 ) d 2 1 − n ∑ d 1 ′ d 2 ′ = v λ ¯ ( d 1 ′ ) d 2 ′ 1 − n = − ϕ 2 ( p ) ∑ d 1 d 2 = u sin 2 π d 1 p · d 2 1 − n ∑ d 1 ′ d 2 ′ = v sin 2 π d 1 ′ p · d 2 ′ 1 − n = − ϕ 2 ( p ) γ p ( u , n ) γ p ( v , n ) .$
Hence, we can write
$M 1 = ∑ χ mod q χ ( − 1 ) = 1 ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) ∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) × ∑ u ≤ N χ ¯ ( u ) ∑ d 1 d 2 = u ξ ( d 1 ) d 2 1 − n u ∑ v ≤ N χ ( v ) ∑ d 1 ′ d 2 ′ = v λ ¯ ( d 1 ′ ) d 2 ′ 1 − n v = ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) ∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) × ∑ ′ 1 ≤ u ≤ N ∑ ′ 1 ≤ v ≤ N ∑ d 1 d 2 = u ξ ( d 1 ) d 2 1 − n ∑ d 1 ′ d 2 ′ = v λ ¯ ( d 1 ′ ) d 2 ′ 1 − n u v ∑ χ mod q χ ( − 1 ) = 1 χ ( u ¯ v ) = − 1 2 ∑ ′ 1 ≤ u ≤ N ∑ ′ 1 ≤ v ≤ N ϕ 2 ( p ) γ p ( u , n ) γ p ( v , n ) u v ∑ d | ( q , u ¯ v − 1 ) μ p d ϕ ( d ) − 1 2 ∑ ′ 1 ≤ u ≤ N ∑ ′ 1 ≤ v ≤ N ϕ 2 ( p ) γ p ( u , n ) γ p ( v , n ) u v ∑ d | ( q , u ¯ v + 1 ) μ q d ϕ ( d ) = − 1 2 ∑ d | q μ q d ϕ ( d ) ∑ ′ 1 ≤ u ≤ N ∑ ′ 1 ≤ v ≤ N u ≡ v ( mod d ) ϕ 2 ( p ) γ p ( u , n ) γ p ( v , n ) u v − 1 2 ∑ d | q μ q d ϕ ( d ) ∑ ′ 1 ≤ u ≤ N ∑ ′ 1 ≤ v ≤ N u ≡ − v ( mod d ) ϕ 2 ( p ) γ p ( u , n ) γ p ( v , n ) u v ,$
where $∑ ′ 1 ≤ n ≤ N$ 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) $d ≤ u , v ≤ N$; (ii) $d ≤ u ≤ N$ and $1 ≤ v ≤ d − 1$; (iii) $1 ≤ u ≤ d − 1$ and $d ≤ v ≤ N$; (iv) $1 ≤ u , v ≤ d − 1$. So we have
$∑ d | q μ q d ϕ ( d ) ∑ ′ d ≤ u ≤ N ∑ ′ d ≤ v ≤ N u ≡ v ( mod d ) ϕ 2 ( p ) γ p ( u , n ) γ p ( v , n ) u v ≪ ∑ d | q ϕ ( d ) ∑ 1 ≤ r 1 ≤ N d ∑ 1 ≤ r 2 ≤ N d ∑ ′ l 1 = 1 d − 1 ∑ ′ l 2 = 1 d − 1 l 1 ≡ l 2 ( mod d ) ϕ 2 ( p ) γ p ( r 1 d + l 1 , n ) γ p ( r 2 d + l 2 , n ) ( r 1 d + l 1 ) ( r 2 d + l 2 ) ≪ ϕ 2 ( p ) ∑ d | q ϕ ( d ) ∑ 1 ≤ r 1 ≤ N d ∑ 1 ≤ r 2 ≤ N d ∑ ′ l 1 = 1 d − 1 ∑ d 2 | ( r 1 d + l 1 ) d 2 1 − n ∑ d 2 ′ | ( r 2 d + l 1 ) d 2 ′ 1 − n ( r 1 d + l 1 ) ( r 2 d + l 1 ) ≪ ϕ 2 ( p ) ∑ d | q ϕ ( d ) ∑ 1 ≤ r 1 ≤ N d ∑ 1 ≤ r 2 ≤ N d ∑ ′ l 1 = 1 d − 1 τ 2 ( r 1 d + l 1 ) τ 2 ( r 2 d + l 1 ) ( r 1 d + l 1 ) ( r 2 d + l 1 ) ≪ ϕ 2 ( p ) ∑ d | q ϕ ( d ) d ∑ 1 ≤ r 1 ≤ N d ∑ 1 ≤ r 2 ≤ N d [ ( r 1 d + 1 ) ( r 2 d + 1 ) ] ϵ r 1 r 2 ≪ ϕ 2 ( p ) q ϵ .$
$∑ d | q μ q d ϕ ( d ) ∑ ′ d ≤ u ≤ N ∑ ′ 1 ≤ v ≤ d − 1 u ≡ v ( mod d ) ϕ 2 ( p ) γ p ( u , n ) γ p ( v , n ) u v ≪ ∑ d | q ϕ ( d ) ∑ 1 ≤ r 1 ≤ N d ∑ ′ 1 ≤ v ≤ d − 1 ∑ ′ l 1 = 1 d − 1 v ≡ l 1 ( mod d ) ϕ 2 ( p ) γ p ( r 1 d + l 1 , n ) γ p ( v , n ) ( r 1 d + l 1 ) v ≪ ϕ 2 ( p ) ∑ d | q ϕ ( d ) ∑ 1 ≤ r 1 ≤ N d ∑ ′ 1 ≤ v ≤ d − 1 τ 2 ( r 1 d + v ) τ 2 ( v ) ( r 1 d + v ) v ≪ ϕ 2 ( p ) ∑ d | q ϕ ( d ) ∑ 1 ≤ r 1 ≤ N d ∑ 1 ≤ v ≤ d − 1 ( r 1 v d ) ϵ − 1 ≪ ϕ 2 ( p ) q ϵ .$
and
$∑ d | q μ q d ϕ ( d ) ∑ ′ 1 ≤ u ≤ d − 1 ∑ ′ d ≤ v ≤ N u ≡ v ( mod d ) ϕ 2 ( p ) γ p ( u , n ) γ p ( v , n ) u v ≪ ϕ 2 ( p ) ∑ d | q ϕ ( d ) ∑ 1 ≤ u ≤ d − 1 ∑ 1 ≤ r 2 ≤ N d ( u r 2 d ) ϵ − 1 ≪ ϕ 2 ( p ) q ϵ ,$
where we have used the estimate $τ 2 ( n ) ≪ n ϵ$.
For the case $1 ≤ u , v ≤ d − 1$, the solution of the congruence $u ≡ v ( mod d )$ is $u = v$. Hence,
$∑ d | q μ q d ϕ ( d ) ∑ ′ 1 ≤ u ≤ d − 1 ∑ ′ 1 ≤ v ≤ d − 1 u ≡ v ( mod d ) ϕ 2 ( p ) γ p ( u , n ) γ p ( v , n ) u v = ϕ 2 ( p ) ∑ d | q μ q d ϕ ( d ) ∑ ′ 1 ≤ u ≤ d − 1 γ p 2 ( u , n ) u 2 = ϕ 2 ( p ) ∑ d | q μ q d ϕ ( d ) ∑ u = 1 ∞ ( u , q ) = 1 γ p 2 ( u , n ) u 2 + O ( ϕ 2 ( p ) q ϵ ) = ( q − 2 ) ϕ 2 ( p ) ∑ u = 1 ∞ ( u , q ) = 1 γ p 2 ( u , n ) u 2 + O ( ϕ 2 ( p ) q ϵ ) = ( q − 2 ) ϕ 2 ( p ) ∑ u = 1 ∞ γ p 2 ( u , n ) u 2 − ∑ u = 1 ∞ q | u γ p 2 ( u , n ) u 2 + O ( ϕ 2 ( p ) q ϵ ) = ( q − 2 ) ϕ 2 ( p ) ∑ u = 1 ∞ γ p 2 ( u , n ) u 2 + O ( ϕ 2 ( p ) q ϵ ) .$
Then from (3)–(6), we have
$− 1 2 ∑ d | q μ q d ϕ ( d ) ∑ ′ 1 ≤ u ≤ N ∑ ′ 1 ≤ v ≤ N u ≡ v ( mod d ) ϕ 2 ( p ) γ p ( u , n ) γ p ( v , n ) u v = − ϕ 2 ( p ) ( q − 2 ) 2 ∑ u = 1 ∞ γ p 2 ( u , n ) u 2 + O ( ϕ 2 ( p ) q ϵ ) = − ϕ 2 ( p ) ( q − 2 ) 2 C p , n + O ( ϕ 2 ( p ) q ϵ ) .$
Similarly, we can also get the estimate
$− 1 2 ∑ d | q μ q d ϕ ( d ) ∑ ′ 1 ≤ u ≤ N ∑ ′ 1 ≤ v ≤ N u ≡ − v ( mod d ) ϕ 2 ( p ) γ p ( u , n ) γ p ( v , n ) u v = − 1 2 ∑ d | q μ q d ϕ ( d ) ∑ ′ 1 ≤ u ≤ N ∑ ′ 1 ≤ v ≤ N u + v = d ϕ 2 ( p ) γ p ( u , n ) γ p ( v , n ) u v − 1 2 ∑ d | q μ q d ϕ ( d ) ∑ ′ 1 ≤ u ≤ N ∑ ′ 1 ≤ v ≤ N u + v = l d , l ≥ 2 ϕ 2 ( p ) γ p ( u , n ) γ p ( v , n ) u v ≪ ϕ 2 ( p ) ∑ d | q ϕ ( d ) ∑ 1 ≤ u ≤ d − 1 γ p ( u , n ) γ p ( d − u , n ) u ( d − u ) + ϕ 2 ( p ) ∑ d | q ϕ ( d ) ∑ ′ 1 ≤ u ≤ N ∑ l = u d + 2 N + u d γ p ( u , n ) γ p ( l d − u , n ) l d u − u 2 ≪ ϕ 2 ( p ) ∑ d | q ϕ ( d ) d ∑ 1 ≤ u ≤ d − 1 u ϵ ( d − u ) ϵ u + ϕ 2 ( p ) ∑ d | q ϕ ( d ) d ∑ ′ 1 ≤ u ≤ N ∑ l = u d + 2 N + u d u ϵ ( l d − u ) ϵ l u − u 2 d ≪ ϕ 2 ( p ) q ϵ + ∑ d | q ϕ ( d ) d ∑ u = 1 N ∑ l = 1 N u ϵ l ϵ u l ≪ ϕ 2 ( p ) q ϵ .$
Then combining (2), (7) and (8), we have
$M 1 = − ϕ 2 ( p ) ( q − 2 ) 2 C p , n + O ( ϕ 2 ( p ) q ϵ ) .$
(ii) From Lemma 4 of [15], we have the estimate
$∑ χ ≠ χ 0 | A ( y , χ ) | 2 ≪ y 1 + ϵ q 2 ,$
where $χ 0$ denotes the principal character modulo q, $A ( y , χ ) = ∑ N < n ≤ y χ ( n ) τ 2 ( n )$. Then from the Cauchy inequality we can easily get
$∑ χ ( − 1 ) = 1 | A ( y , χ ) | ≪ ∑ χ ≠ χ 0 | A ( y , χ ) | ≪ y 1 2 + ϵ q 3 2 .$
Using this estimate we have
$M 2 = ∑ χ mod q χ ( − 1 ) = 1 ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) ∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) × ∑ u ≤ N χ ¯ ( u ) ∑ d 1 d 2 = u ξ ( d 1 ) d 2 1 − n u ∫ N ∞ A ( χ , λ ¯ , y ) y 2 d y = − ϕ 2 ( p ) ∑ 1 ≤ u ≤ N χ ¯ ( u ) γ p ( u , n ) u ∫ N ∞ 1 y 2 ∑ χ mod q χ ( − 1 ) = 1 ∑ N < v ≤ y χ ( v ) γ k ( v , n ) d y ≪ ϕ 2 ( p ) ∑ 1 ≤ u ≤ N χ ¯ ( u ) d ( u ) u ∫ N ∞ 1 y 2 ∑ χ mod q χ ( − 1 ) = 1 A ( y , χ ) d y ≪ ϕ 2 ( p ) ∑ 1 ≤ u ≤ N u ϵ − 1 ∫ N ∞ 1 y 2 ∑ χ mod q χ ( − 1 ) = 1 | A ( y , χ ) | d y ≪ ϕ 2 ( p ) N ϵ ∫ N ∞ q 3 2 y 1 2 + ϵ 1 y 2 d y ≪ ϕ 2 ( p ) q 3 2 N 1 2 − ϵ .$
(iii) Similar to (ii), we can also get
$M 3 ≪ ϕ 2 ( p ) q 3 2 N 1 2 − ϵ .$
(iv) Using the same discussion in (ii), and making use of the absolute convergent properties of the integral, we can calculate
$M 4 = ∑ χ mod q χ ( − 1 ) = 1 ∑ ξ mod p ξ ( − 1 ) = − 1 τ ( ξ ¯ ) ∑ λ mod p λ ( − 1 ) = − 1 τ ( λ ) ∫ N ∞ A ( χ ¯ , ξ , y ) y 2 d y ∫ N ∞ A ( χ , λ ¯ , y ) y 2 d y = ϕ 2 ( p ) ∑ χ mod q χ ( − 1 ) = 1 ∫ N ∞ 1 y 2 ∑ N < u ≤ y χ ¯ ( u ) γ p ( u , n ) d y ∫ N ∞ 1 y 2 ∑ N < v ≤ y χ ( v ) γ p ( v , n ) d y ≪ ϕ 2 ( p ) ∑ χ mod p χ ( − 1 ) = 1 ∫ N ∞ A ( y , χ ¯ ) y 2 d y ∫ N ∞ A ( y , χ ) y 2 d y ≤ ϕ 2 ( p ) ∫ N ∞ ∫ N ∞ 1 y 2 z 2 ∑ χ mod q χ ( − 1 ) = 1 | A ( y , χ ¯ ) | | A ( y , χ ) | d y d z ≪ ϕ 2 ( p ) ∫ N ∞ 1 y 2 ∫ N ∞ 1 z 2 ∑ χ ≠ χ 0 | A ( y , χ ¯ ) | 2 1 2 ∑ χ ≠ χ 0 | A ( y , χ ) | 2 1 2 d y d z ≪ ϕ 2 ( p ) ∫ N ∞ 1 y 2 ∑ χ ≠ χ 0 | A ( y , χ ) | 2 1 2 d y 2 ≪ ϕ 2 ( p ) ∫ N ∞ q y 3 2 − ϵ d y 2 ≪ ϕ 2 ( p ) q 2 N 1 − ϵ .$
Now taking $N = q 4$ and $ϵ < 1 2$, combining (1) and (9)–(12), we have
$M = − ϕ 2 ( p ) ( q − 2 ) 2 C p , n + O ( ϕ 2 ( p ) q ϵ ) .$
Thus we obtain the asymptotic formula for $χ ( − 1 ) = 1$,
$∑ a < q p ∑ b < q p ∑ χ mod q χ ( − 1 ) = 1 χ ( a b ¯ ) L ( n , χ ) 2 = − q π 2 ϕ 2 ( p ) M = q 2 2 π 2 C p , n + O ( q 1 + ϵ ) .$
For $χ ( − 1 ) = − 1$, from Lemma 4, we have
$∑ a < q p ∑ b < q p ∑ χ mod q χ ( − 1 ) = − 1 χ ( a b ¯ ) L ( n , χ ) 2 = ∑ χ mod q χ ( − 1 ) = − 1 L ( n , χ ) 2 τ ( χ ) π i 1 − χ ¯ ( p ) p L ( 1 , χ ¯ ) + i τ ( χ ) π ϕ ( p ) ∑ ξ mod p ξ ( − 1 ) = 1 τ ( ξ ¯ ) L ( 1 , ξ χ ¯ ) × τ ( χ ¯ ) π i 1 − χ ( p ) p L ( 1 , χ ) + i τ ( χ ¯ ) π ϕ ( p ) ∑ λ mod p λ ( − 1 ) = 1 τ ( λ ) L ( 1 , λ ¯ χ ) = q π 2 ∑ χ mod q χ ( − 1 ) = − 1 1 − χ ¯ ( p ) p 1 − χ ( p ) p L ( n , χ ) 2 | L ( 1 , χ ) | 2 + − q π 2 ϕ ( p ) ∑ χ mod q χ ( − 1 ) = − 1 1 − χ ¯ ( p ) p L ( n , χ ) 2 L ( 1 , χ ¯ ) ∑ λ mod p λ ( − 1 ) = 1 τ ( λ ) L ( 1 , λ ¯ χ ) + − q π 2 ϕ ( p ) ∑ χ mod q χ ( − 1 ) = − 1 1 − χ ( p ) p L ( n , χ ) 2 L ( 1 , χ ) ∑ ξ mod p ξ ( − 1 ) = 1 τ ( ξ ¯ ) L ( 1 , ξ χ ¯ ) + q π 2 ϕ 2 ( p ) ∑ χ mod q χ ( − 1 ) = − 1 L ( n , χ ) 2 ∑ ξ mod p ξ ( − 1 ) = 1 τ ( ξ ¯ ) L ( 1 , ξ χ ¯ ) ∑ λ mod p λ ( − 1 ) = 1 τ ( λ ) L ( 1 , λ ¯ χ ) : = A + B + C + D .$
Using the same method as proving $χ ( − 1 ) = 1$, we can easily get
$A = q 2 2 π 2 1 + 1 p 2 ∑ u = 1 ∞ ν 2 ( u , n ) u 2 − q 2 π 2 p 2 ∑ u = 1 ∞ ν ( u , n ) ν ( p u , n ) u 2 + O ( q 1 + ϵ ) ,$
$B = C = q 2 2 π 2 p 2 ∑ u = 1 ∞ ν ( u , n ) ν p ( p u , n ) u 2 − p 2 2 π 2 ∑ u = 1 ∞ ν ( u , n ) ν p ( u , n ) u 2 + O ( q 1 + ϵ ) ,$
$D = q 2 2 π 2 ∑ u = 1 ∞ ν p 2 ( u , n ) u 2 + O ( q 1 + ϵ ) .$
Thus we obtain the asymptotic formula for $χ ( − 1 ) = − 1$,
$∑ a < q p ∑ b < q p ∑ χ mod q χ ( − 1 ) = − 1 χ ( a b ¯ ) L ( n , χ ) 2 = q 2 π 2 T p , n + O ( q 1 + ϵ ) .$
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,
$∑ a < q p ∑ b < q p S ( a b ¯ , n , q ) = ( n ! ) 2 q 4 n − 1 π 2 n ϕ ( q ) ∑ a < q p ∑ b < q p ∑ χ mod q χ ( − 1 ) = 1 χ ( a b ¯ ) L ( n , χ ) 2 + ( n ! ) 2 ζ 2 ( n ) 4 n − 1 π 2 n 1 q 2 n − 1 − 1 ∑ a < q p ∑ b < q p 1 = ( n ! ) 2 q 2 2 2 n − 2 π 2 n 1 2 π 2 C p , n − ζ 2 ( n ) p 2 + O ( q 1 + ϵ ) .$
(ii) when n be an odd number,
$∑ a < q p ∑ b < q p S ( a b ¯ , n , q ) = ( n ! ) 2 q 4 n − 1 π 2 n ϕ ( q ) ∑ a < q p ∑ b < q p ∑ χ mod q χ ( − 1 ) = − 1 χ ( a b ¯ ) L ( n , χ ) 2 = ( n ! ) 2 q 4 n − 1 π 2 n ϕ ( q ) q 2 π 2 T p , n + O ( q 1 + ϵ ) = ( n ! ) 2 q 2 2 2 n − 2 π 2 n + 2 T p , n + O ( q 1 + ϵ ) .$
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
$S n ( 1 ) ( h , k ) = ( 2 π ) n − 1 i n + 1 · n ! S ( h , n , k ) .$
Noting that $ζ ( 2 ) = π 2 / 6$ and $ζ ( 4 ) = π 4 / 90$, we easily get Corollarys 1 and 2.

## Author Contributions

Conceptualization, L.L. and Z.X.; methodology, L.L. and Z.X.; software, L.L. and Z.X.; validation, L.L. and Z.X.; formal analysis, L.L. and Z.X.; investigation, L.L. and Z.X.; resources, L.L. and Z.X.; data curation, L.L. and Z.X.; writing—original draft preparation, L.L. and Z.X.; writing—review and editing, L.L. and Z.X.; visualization, L.L. and Z.X.; supervision, L.L. and Z.X.; project administration, L.L. and Z.X.; funding acquisition, L.L. and Z.X. All authors have read and agreed to the published version of the manuscript.

## Funding

This work is supported by the Basic Research Program for Nature Science of the Shaanxi Province (2014JM1001, 2015KJXX-27) and N.S.F. (11971381, 11471258, 11701447) of China.

## Conflicts of Interest

The authors declare no conflict of interest.

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Liu, L.; Xu, Z. Mean Value of the General Dedekind Sums over Interval $${[1,\frac{q}{p})}$$. Symmetry 2020, 12, 2079. https://doi.org/10.3390/sym12122079

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Liu L, Xu Z. Mean Value of the General Dedekind Sums over Interval $${[1,\frac{q}{p})}$$. Symmetry. 2020; 12(12):2079. https://doi.org/10.3390/sym12122079

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Liu, Lei, and Zhefeng Xu. 2020. "Mean Value of the General Dedekind Sums over Interval $${[1,\frac{q}{p})}$$" Symmetry 12, no. 12: 2079. https://doi.org/10.3390/sym12122079

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