Mean value of the Hardy sums over short intervals

Lei Liu; Zhefeng Xu; Lei Liu; Zhefeng Xu

doi:10.3934/math.2020356

AIMS Mathematics

2020, Volume 5, Issue 6: 5551-5563. doi: 10.3934/math.2020356

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Research article

Mean value of the Hardy sums over short intervals

Lei Liu ,
Zhefeng Xu ^,

School of Mathematics, Northwest University, Xi’an 710127, Shaanxi, P. R. China

Received: 06 April 2020 Accepted: 22 June 2020 Published: 29 June 2020
MSC : 11F20, 11L40

The main purpose of this paper is using the mean value of Dirichlet L-functions and character sums to study a kind of mean value of the Hardy sums over short intervals, and give some interesting formulae.

Keywords:

Citation: Lei Liu, Zhefeng Xu. Mean value of the Hardy sums over short intervals[J]. AIMS Mathematics, 2020, 5(6): 5551-5563. doi: 10.3934/math.2020356

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Abstract

1. Introduction

For a positive integer $q$ and an arbitrary integer $h$ , the Dedekind sum $S(h, q)$ is defined by

$S(h, q) = \mathop{\sum\limits_{a = 1}^q}\bigg(\bigg(\frac{a}{q}\bigg)\bigg)\bigg(\bigg(\frac{ah}{q}\bigg)\bigg),$

where

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

It plays an important role in the transformation theory of the Dedekind $\eta$ function. The various properties of $S(h, q)$ were investigated by many authors, related works can be founded in ^[2,3,4] and ^[9,10]. Berndt ^[1] studied the following Hardy sums:

$\begin{align*} &H(h, q) = \mathop{\sum\limits_{j = 1}}^{q-1}(-1)^{j+1+[\frac{hj}{q}]}, \quad\quad\quad\quad\quad s_1(h, q) = \mathop{\sum\limits_{j = 1}}^{q-1}(-1)^{[\frac{hj}{q}]}\bigg(\bigg(\frac{j}{q}\bigg)\bigg), \\ &s_2(h, q) = \mathop{\sum\limits_{j = 1}}^{q-1}(-1)^{j}\bigg(\bigg(\frac{j}{q}\bigg)\bigg)\bigg(\bigg(\frac{hj}{q}\bigg)\bigg), \quad\quad\quad s_3(h, q) = \mathop{\sum\limits_{j = 1}}^{q-1}(-1)^{j}\bigg(\bigg(\frac{hj}{q}\bigg)\bigg), \\ &s_4(h, q) = \mathop{\sum\limits_{j = 1}}^{q-1}(-1)^{[\frac{hj}{q}]}, \quad\quad\quad\quad\quad\quad\quad s_5(h, q) = \mathop{\sum\limits_{j = 1}}^{q-1}(-1)^{j+[\frac{hj}{q}]}\bigg(\bigg(\frac{j}{q}\bigg)\bigg). \end{align*}$

In ^[6], R. Sitaramachandra Rao expressed Hardy sums in terms of Dedekind sum $S(h, q)$ as following:

$\begin{aligned} &H(h, q)=-8 S(h+q, 2 q)+4 S(h, q), \quad \text { if } h+q \text { is odd };\\ &s_{1}(h, q)=2 S(h, q)-4 S(h, 2 q), \quad \text { if } h \text { is even };\\ &s_{2}(h, q)=-S(h, q)+2 S(2 h, q), \quad \text { if } q \text { is even };\\ &\begin{array}{cl} s_{3}(h, q)=2 S(h, q)-4 S(2 h, q), & \text { if } q \text { is odd }; \\ s_{4}(h, q)=-4 S(h, q)+8 S(h, 2 q), & \text { if } h \text { is odd }; \end{array}\\ &s_{5}(h, q)=-10 S(h, q)+4 S(2 h, q)+4 S(h, 2 q), \quad \text { if } h+q \text { is even }. \end{aligned}$

Each one of $H(h, q)$ ( $h+q$ even), $s_1(h, q)$ ( $h$ odd), $s_2(h, q)$ ( $q$ odd), $s_3(h, q)$ ( $q$ even), $s_4(h, q)$ ( $h$ even), $s_5(h, q)$ ( $h+q$ odd) is zero. Z. Xu and W. Zhang ^[8] proved that if $p$ is a prime and $\bar{b}$ denotes the multiplicative inverse of $b$ $\bmod p$ , then

$\sum\limits_{a \lt \frac{p}{4}}\sum\limits_{b \lt \frac{p}{4}}H(2a\bar{b}, p) = \frac{3}{16}p^2+O(p^{1+\epsilon}).$

W. Liu ^[5] proved that if $p\geq5$ is a prime and $\bar{b}$ denotes the multiplicative inverse of $b$ $\bmod p$ , then

$\sum\limits_{a \lt \frac{p}{3}}\sum\limits_{b \lt \frac{p}{3}}H(2a\bar{b}, p) = \frac{1}{5}p^2+O(p^{1+\epsilon}),$

$\sum\limits_{a \lt \frac{p}{3}}\sum\limits_{b \lt \frac{p}{4}}H(2a\bar{b}, p) = \frac{27}{320}p^2+O(p^{1+\epsilon}).$

Let $1 < N < p$ , is there a mean value distribution formula of the Hardy sum in the short interval $[1, N]$ ? In fact, it is very difficult to get a asymptotic formula for the mean value in the case of one variable as

$\sum\limits_{a \lt N}H(a, p).$

In this paper, we use the mean value of $L$ -functions and estimate of character sums to study the mean value

$\mathop{\sum\limits_{a\leq N}\sum\limits_{b\leq N}}a^{l}b^{k}H(2a\overline{b}, p),$

where $l, k$ be two non-negative integers and $\overline{b}$ denote the multiplicative inverse of $b$ modulo $p$ , and obtain the following main conclusion:

Theorem. Let $p > 2$ be a prime and $N$ a positive integer with $1 < N < p$ , $l$ and $k$ be two non-negative integers. Then we have

$\begin{align*} &\mathop{\sum\limits_{a\leq N}\sum\limits_{b\leq N}}a^{l}b^{k}H(2a\overline{b}, p)\\ = &\frac{C(l, k)}{7\cdot2^{l-2}\zeta(3)(l+k+1)}pN^{l+k+1} \\ &+\left\{ \begin{array}{ll} O\left(pN^{l+k}\log^2N+N^{l+k+2}p^{o(1)}\right), & \hbox{if $l = k = 0$; $l = 0, k\geq1$ or $l\geq1, k = 0$;} \\ O\left(pN^{l+k}\log N+N^{l+k+2}p^{o(1)}\right), & \hbox{if $l\geq1, k\geq1$.} \end{array} \right. \end{align*}$

here

$C(l, k) = -\frac{7\zeta(3)}{8}-\mathop{\mathop{\sum\limits_{a = 1}^\infty\sum\limits_{b = a+1}^\infty}}_{(a, 2) = 1, (b, 2) = 1}\frac{a^{l-1}}{b^{l+2}}-\mathop{\mathop{\sum\limits_{b = 1}^\infty\sum\limits_{a = b+1}^{2b}}}_{(a, 2) = 1, (b, 2) = 1}\frac{a^{l-1}}{b^{l+2}} -2^{l+k+1}\mathop{\mathop{\sum\limits_{b = 1}^\infty\sum\limits_{a = 2b+1}^{\infty}}}_{(a, 2) = 1, (b, 2) = 1}\frac{b^{k-1}}{a^{k+2}}$

is a constant depending on $l$ and $k$ .

It is clear that the asymptotic formulas in the Theorem are non-trivial in the range $p^\epsilon < N < p^{1-\epsilon}$ .

From the theorem we may immediately deduce the following corollaries:

Corollary 1. Let $p > 2$ be a prime and $N$ an integer with $1 < N < p$ . Then we have

$\mathop{\sum\limits_{a\leq N}\sum\limits_{b\leq N}}aH(2a\overline{b}, p) = \frac{C(1, 0)}{7\zeta(3)}pN^2+O(pN\log^2N+N^3p^{o(1)}),$

where

$C(1, 0) = \frac{7}{16}\zeta(3)-\frac{\pi^2}{24}-\mathop{\sum\limits_{a = 1}^\infty\sum\limits_{b = 1}^\infty}\frac{1}{(a+2b)^3}-\frac{9}{2}\mathop{\sum\limits_{a = 1}^\infty\sum\limits_{b = 1}^\infty}\frac{1}{(a+2b)^2b}+2\mathop{\sum\limits_{a = 1}^\infty\sum\limits_{b = 1}^\infty}\frac{1}{(a+4b)^2b}$

is a constant.

Corollary 2. Let $p > 2$ be a prime and $N$ an integer with $1 < N < p$ . Then we have

$\mathop{\sum\limits_{a\leq N}\sum\limits_{b\leq N}}abH(2a\overline{b}, p) = \frac{2C(1, 1)}{21\zeta(3)}pN^3+O(pN^2\log N+N^4p^{o(1)}),$

where

$C(1, 1) = \frac{\pi^2}{8}-\frac{25}{16}\zeta(3)-10\mathop{\sum\limits_{a = 1}^\infty\sum\limits_{b = 1}^\infty}\frac{1}{(a+2b)^3}+8\mathop{\sum\limits_{a = 1}^\infty\sum\limits_{b = 1}^\infty}\frac{1}{(a+4b)^3}$

is a constant.

2. Some Lemmas

To complete the proof of the theorem, we need several lemmas.

Lemma 1. Let $q\geq3$ and $h$ be two integers with $(h, q) = 1$ . Then we have

$S(h, q) = \frac{1}{\pi^2q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\mathop {\mathop \sum \limits_{\chi \,\bmod \,d} }\limits_{\chi ( - 1) = - 1}\chi(h)|L(1, \chi)|^2,$

where $\phi(d)$ is the Euler function.

Proof. See Lemma 1 of reference ^[9].

Lemma 2. Let $q > 1$ be an odd integer and $h$ an integer with $(h, q) = 1$ . Then we have

$H(h, q) = \left\{ \begin{array}{ll} \frac{-16}{\pi^2q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\mathop {\mathop \sum \limits_{\chi \,\bmod \,d} }\limits_{\chi ( - 1) = - 1} \chi(h)|L(1, \chi\chi_2^0)|^2, & \hbox{$2\mid h$;} \\ 0, & \hbox{$2\nmid h$.} \end{array} \right.$

where $\chi_2^0$ be a principal Dirichlet character modulo $2$ .

Proof. From ^[1], we have

$\begin{align*} H(h, q) = &-8S(h+q, 2q)+4S(h, q)\\ = &-\frac{4}{\pi^2q}\sum\limits_{d|2q}\frac{d^2}{\phi(d)}\mathop{\mathop{{\sum}}_{\chi\bmod d}}_{\chi(-1) = -1}\chi(h+q)|L(1, \chi)|^2+\frac{4}{\pi^2q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\mathop{\mathop{{\sum}}_{\chi\bmod d}}_{\chi(-1) = -1}\chi(h)|L(1, \chi)|^2. \end{align*}$

Note that $q$ be an odd integer, we can get

$\sum\limits_{d|2q}f(d) = \sum\limits_{d|q}f(d)+\sum\limits_{d|q}f(2d)$

Hence we can write

$\begin{align*} H(h, q) = &-\frac{4}{\pi^2q}\left(\sum\limits_{d|q}\frac{d^2}{\phi(d)}\mathop{\mathop{{\sum}}_{\chi\bmod d}}_{\chi(-1) = -1}\chi(h+q)|L(1, \chi)|^2+\sum\limits_{d|q}\frac{(2d)^2}{\phi(2d)}\mathop{\mathop{{\sum}}_{\chi\bmod 2d}}_{\chi(-1) = -1}\chi(h+q)|L(1, \chi)|^2\right)\\ &+\frac{4}{\pi^2q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\mathop{\mathop{{\sum}}_{\chi\bmod d}}_{\chi(-1) = -1}\chi(h)|L(1, \chi)|^2.\\ = &-\frac{16}{\pi^2q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\mathop{\mathop{{\sum}}_{\chi\bmod d}}_{\chi(-1) = -1}\chi(h+q)\chi_2^0(h+q)|L(1, \chi\chi_2^0)|^2\\ = &\left\{ \begin{array}{ll} \frac{-16}{\pi^2q}\sum\limits_{d|q}\frac{d^2}{\phi(d)}\mathop {\mathop \sum \limits_{\chi \,\bmod \,d} }\limits_{\chi ( - 1) = - 1} \chi(h)|L(1, \chi\chi_2^0)|^2, & \hbox{$2\mid h$;} \\ 0, & \hbox{$2\nmid h$.} \end{array} \right. \end{align*}$

This proves Lemma 2.

Lemma 3. Let $p > 2$ be a prime, $a$ be a positive integer with $(a, p) = 1$ . Then we have the asymptotic formula

$\mathop { \mathop \sum\limits_{\chi \bmod p}}_{\chi(-1) = -1} \chi(2a\bar{b})|L(1, \chi\chi_2^0)|^2 = \mathop {\mathop \sum \limits_{\chi \,\bmod \,p} }\limits_{\chi ( - 1) = - 1} \chi(2a\bar{b})\left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right|^2+O\left(\frac{\log^{2}p}{\sqrt p}\right).$

Proof. For convenience, we let

$A(\chi, y) = \mathop \sum\limits_{p^{2} \lt u\leq y}\chi(u).$

Then applying Abel's identity we have

$L(1, \chi\chi_2^0) = \mathop \sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}+\int_{p^2}^{\infty}\frac{A(\chi\chi_2^0, y)}{y^{2}}dy.$

Hence

$\begin{align*} &\mathop { \mathop \sum\limits_{\chi \bmod p}}_{\chi(-1) = -1} \chi(2a\bar{b})|L(1, \chi\chi_2^0)|^2\\ = &\mathop{\mathop\sum\limits_{\chi \bmod p}}_{\chi(-1) = -1}\chi(2a\bar{b})\left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}+\int_{p^2}^{\infty}\frac{A(\chi\chi_2^0, y)}{y^2}dy\right|^2\\ = &\mathop{\mathop\sum\limits_{\chi \bmod p}}_{\chi(-1) = -1}\chi(2a\bar{b})\left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right|^2+\mathop{\mathop\sum\limits_{\chi \bmod p}}_{\chi(-1) = -1}\chi(2a\bar{b})\left(\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right)\left(\int_{p^2}^{\infty}\frac{A(\overline{\chi\chi_2^0}, y)}{y^2}dy\right)\\ &+\mathop{\mathop\sum\limits_{\chi \bmod p}}_{\chi(-1) = -1}\chi(2a\bar{b})\left(\mathop\sum\limits_{1\leq u\leq p^2}\frac{\overline{\chi\chi_2^0}(u)}{u}\right)\left(\int_{p^2}^{\infty}\frac{A(\chi\chi_2^0, y)}{y^2}dy\right)+\mathop{\mathop\sum\limits_{\chi \bmod p}}_{\chi(-1) = -1}\chi(2a\bar{b})\left|\int_{p^2}^{\infty}\frac{A(\chi\chi_2^0, y)}{y^2}dy\right|^2. \end{align*}$

Using the Cauchy inequality and the estimate $|\sum_{m\leq n\leq M }\chi (n)|\ll\sqrt{p}\log p,$ we have the estimates

$\begin{align*} &\left|\mathop{\mathop\sum\limits_{\chi \bmod p}}_{\chi(-1) = -1}\chi(2a\bar{b})\left(\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right)\left(\int_{p^2}^{\infty}\frac{A(\overline{\chi\chi_2^0}, y)}{y^{2}}dy\right)\right|\\ \ll&\left[\mathop{\mathop\sum\limits_{\chi \bmod p}}_{\chi(-1) = -1}\left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right|^{2}\right]^{\frac{1}{2}}\left[\mathop{\mathop\sum\limits_{\chi \bmod p}}_{\chi(-1) = -1}\left|\int_{p^2}^{\infty}\frac{A(\overline{\chi\chi_2^0}, y)}{y^{2}}dy\right|^{2}\right]^{\frac{1}{2}}\\ \ll&\frac{\log p}{\sqrt p}, \end{align*}$

$\left|\mathop {\mathop \sum \limits_{\chi \,\bmod \,p} }\limits_{\chi ( - 1) = - 1} \chi(2a\bar{b})\left(\mathop\sum\limits_{1\leq u\leq p^2}\frac{\overline{\chi\chi_2^0}(u)}{u}\right)\left(\int_{p^2}^{\infty}\frac{A(\chi\chi_2^0, y)}{y^{2}}dy\right)\right|\ll \frac{\log p}{\sqrt p},$

and

$\left|\mathop {\mathop \sum \limits_{\chi \,\bmod \,p} }\limits_{\chi ( - 1) = - 1} \chi(2a\bar{b})\left|\int_{p^2}^{\infty}\frac{A(\chi\chi_2^0, y)}{y^{2}}dy\right|^2\right|\ll \frac{\log^2 p}{p^2}.$

So we may immediately get the asymptotic formula

$\mathop {\mathop \sum \limits_{\chi \,\bmod \,p} }\limits_{\chi ( - 1) = - 1} \chi(2a\bar{b})|L(1, \chi\chi_2^0)|^2 = \mathop {\mathop \sum \limits_{\chi \,\bmod \,p} }\limits_{\chi ( - 1) = - 1} \chi(2a\bar{b})\left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right|^2+O\left(\frac{\log^{2}p}{\sqrt p}\right).$

This proves Lemma 3.

Lemma 4. Let $p > 2$ be a prime and $N$ an integer with $1 < N < p$ . Then

$\mathop\sum\limits_{a\leq N}\mathop\sum\limits_{b\leq N}\sum\limits_{\chi\bmod p}\chi(-2a\overline{b})\left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right|^2 = O\left(N^2p^{o(1)}\right).$

Proof. From the orthogonality relation for character sums modulo $p$ we have

$\begin{align*} &\quad\mathop\sum\limits_{a\leq N}\mathop\sum\limits_{b\leq N}\sum\limits_{\chi \bmod p}\chi(-2a\overline{b})\left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right|^2\\ & = \mathop\sum\limits_{a\leq N}\mathop\sum\limits_{b\leq N}\sum\limits_{\chi \bmod p}\chi(-2a\overline{b})\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\mathop\sum\limits_{1\leq v\leq p^2}\frac{\overline{\chi\chi_2^0}(v)}{v}\\ & = \phi(p)\mathop{\mathop{\sum}_{a\leq N}\mathop{{\sum}}_{b\leq N}\mathop{{\sum}}_{1\leq u\leq p^2}\mathop{{\sum}}_{1\leq v\leq p^2}}_{-2au\equiv bv(\bmod p), (u, 2p) = 1, (v, 2p) = 1}\frac{1}{uv}, \end{align*}$

Let $J = \lfloor2\log p\rfloor$ . By using the similar method used in ^[7] we can write

$\begin{align*} &\quad\phi(p)\mathop{\mathop{\sum}_{a\leq N}\mathop{{\sum}}_{b\leq N}\mathop{{\sum}}_{1\leq u\leq p^2}\mathop{{\sum}}_{1\leq v\leq p^2}}_{-2au\equiv bv(\bmod p), (u, 2p) = 1, (v, 2p) = 1}\frac{1}{uv}\\ &\leq\phi(p)\mathop\sum\limits_{i, j = 0}^{J}\mathop\sum\limits_{e^i\leq u \lt e^{i+1}}\frac{1}{u}\sum\limits_{e^j\leq v \lt e^{j+1}} \frac{1}{v}\mathop{\mathop\sum\limits_{a, b = 1}^{N}}_{-2au\equiv bv(\bmod p)}1\\ &\leq 2\phi (p)\mathop\sum\limits_{0\leq i\leq j\leq J}\mathop\sum\limits_{e^i\leq u \lt e^{i+1}}\frac{1}{u}\sum\limits_{e^j\leq v \lt e^{j+1}}\frac{1}{v}\mathop{\mathop\sum\limits_{a, b = 1}^{N}}_{-2au\equiv bv(\bmod p)}1\\ &\leq 2\phi (p)\mathop\sum\limits_{0\leq i\leq j\leq J}e^{-(i+j)}\mathop\sum\limits_{e^i\leq u \lt e^{i+1}}\sum\limits_{e^j\leq v \lt e^{j+1}}\mathop{\mathop\sum\limits_{a, b = 1}^{N}}_{-2au\equiv bv(\bmod p)}1 \end{align*}$

Thus,

$\mathop\sum\limits_{a\leq N}\mathop\sum\limits_{b\leq N}\mathop\sum\limits_{\chi \bmod p}\chi(-2a\overline{b})\left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right|^2 \leq 2\phi (p)\mathop\sum\limits_{0\leq i\leq j\leq J}e^{-(i+j)}K_{i, j},$

where $K_{i, j}$ is the number of solutions $(a, b, u, v)$ to the congruence $-2au\equiv bv(\bmod p)$ with $1\leq a, b\leq N, e^{i}\leq u < e^{i+1}$ and $e^{j}\leq v < e^{j+1}$ .

For a solution $(a, b, u, v)$ , write $bv = |k|p-2au$ . We have

$p\leq|k|p\leq \max\{2au+bv\}\leq N(2e^{i+1}+e^{j+1})\leq 3e^{j+1}N.$

So the product $bv$ can take at most

$\frac{3e^{j+1}N}{p}\cdot e^{i+1}\cdot N = \frac{3e^{i+j+2}N^2}{p}$

values. It is clear that if $a, u, k$ are fixed, then $b$ and $v$ can take at most $p^{o(1)}$ possible values. Hence,

$K_{i, j} \leq \frac {3e^{i+j+2}N^2}{p}\cdot p^{o(1)}.$

Noting that

$\sum\limits_{0\leq i\leq j\leq J}e^{-(i+j)} = O(1).$

So we have

This proves Lemma 4.

3. Proof of Theorem

In this section we complete the proof of the Theorem.

From Lemma $2$ and Lemma $3$ we have

$\begin{align*} &\quad\mathop{\sum\limits_{a\leq N}\sum\limits_{b\leq N}}a^{l}b^{k}H(2a\overline{b}, p)\\ & = -\frac{16p}{\pi^{2}\phi(p)}\sum\limits_{a\leq N}\sum\limits_{b\leq N}a^{l}b^{k}\mathop{\mathop \sum\limits_{\chi \bmod p}}_{\chi(-1) = -1}\chi(2a\overline{b}) \big|L(1, \chi\chi_2^0)\big|^2\\ & = -\frac{16p}{\pi^{2}\phi(p)}\sum\limits_{a\leq N}\sum\limits_{b\leq N}a^{l}b^{k}\left(\mathop{\mathop\sum\limits_{\chi \bmod p}}_{\chi(-1) = -1}\chi(2a\bar{b})\left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right|^2+O\left(\frac{\log^{2}p}{\sqrt p}\right)\right)\\ & = -\frac{8p}{\pi^{2}\phi(p)}\sum\limits_{a\leq N}\sum\limits_{b\leq N}a^{l}b^{k}\mathop \sum\limits_{\chi \bmod p}(1-\chi(-1))\chi(2a\overline{b})\left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right|^2\\ &\quad+O\left(\frac{N^{l+k+2}\sqrt p\log^2p}{\phi(p)}\right)\\ & = -\frac{8p}{\pi^{2}\phi(p)}\sum\limits_{a\leq N}\sum\limits_{b\leq N}a^{l}b^{k}\mathop \sum\limits_{\chi \bmod p}\chi(2a\overline{b})\left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right|^2+O\left(\frac{N^{l+k+2}\sqrt p\log^2p}{\phi(p)}\right)\\ &\quad+\frac{8p}{\pi^{2}\phi(p)}\mathop\sum\limits_{a\leq N}\mathop\sum\limits_{b\leq N}a^{l}b^{k}\mathop \sum\limits_{\chi \bmod p}\chi(-2a\overline{b}) \left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right|^2 \\ &: = M_{1}+M_{2}+O\left(\frac{N^{l+k+2}\sqrt p\log^2p}{\phi(p)}\right) \end{align*}$

(1)

From the orthogonality relations for characters modulo $p$ , we have

$\begin{align*} M_{1}& = -\frac{8p}{\pi^{2}\phi(p)}\sum\limits_{a\leq N}\sum\limits_{b\leq N}a^{l}b^{k}\mathop \sum\limits_{\chi \bmod p}\chi(2a\overline{b})\left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right|^2\\ & = -\frac{8p}{\pi^{2}}\mathop{\mathop \sum\limits_{a\leq N}\mathop \sum\limits_{b\leq N}\mathop {{\sum}'}_{1\leq u \leq p^{2}}\mathop {{\sum}'}_{1\leq v \leq p^{2}}}_{2au\equiv bv (\bmod p), (u, 2) = 1, (v, 2) = 1}\frac{a^{l}b^{k}}{uv}\\ & = -\frac{8p}{\pi^{2}}\mathop{\mathop \sum\limits_{a\leq N}\mathop \sum\limits_{b\leq N}\mathop {{\sum}'}_{1\leq u \leq p^{2}}\mathop {{\sum}'}_{1\leq v \leq p^{2}}}_{2au = bv, (u, 2) = 1, (v, 2) = 1}\frac{a^{l}b^{k}}{uv}-\frac{8p}{\pi^{2}}\mathop{\mathop \sum\limits_{a\leq N}\mathop \sum\limits_{b\leq N}\mathop {{\sum}'}_{1\leq u \leq p^{2}}\mathop {{\sum}'}_{1\leq v \leq p^{2}}}_{2au\equiv bv, 2au\neq bv, (u, 2) = 1, (v, 2) = 1 }\frac{a^{l}b^{k}}{uv}\\ &: = M_{11}+M_{12} \end{align*}$

here $\sum\limits_{{{1\leq u\leq p^2}}} {\rm{'}}$ denotes the summation over $u$ from $1$ to $p^2$ with $(u, p) = 1$ .

Now, we calculate $M_{11}$ . First, we write

$\begin{align*} M_{11}& = -\frac{8p}{\pi^{2}}\mathop{\mathop \sum\limits_{a\leq N}\mathop \sum\limits_{b\leq N}\mathop {{\sum}'}_{1\leq u \leq p^{2}}\mathop {{\sum}'}_{1\leq v \leq p^{2}}}_{2au = bv, (u, 2) = 1, (v, 2) = 1}\frac{a^{l}b^{k}}{uv} \\ & = -\frac{2^{k+3}p}{\pi^{2}}\mathop{\mathop{{\sum}}_{d\leq N}\mathop\sum\limits_{a\leq \frac{N}{d}}\mathop\sum\limits_{b\leq\frac{N}{2d}}\mathop{{\sum}'}_{1\leq v\leq\min\left\{\frac{p^2}{a}, \frac{p^2}{b}\right\}}}_{(a, b) = 1, (a, 2) = 1, (v, 2) = 1, (b, 2) = 1}\frac{d^{l+k}a^{l-1}b^{k-1}}{v^{2}}\\ & = -\frac{2^{k+3}p}{\pi^{2}}\mathop{\mathop\sum\limits_{a\leq N}\mathop\sum\limits_{b\leq N}\mathop{{\sum}}_{d\leq\min\left\{\frac{N}{a}, \frac{N}{2b}\right\}}\mathop{{\sum}'}_{1\leq v\leq\min\left\{\frac{p^2}{a}, \frac{p^2}{b}\right\}}}_{(a, b) = 1, (a, 2) = 1, (v, 2) = 1, (b, 2) = 1}\frac{d^{l+k}a^{l-1}b^{k-1}}{v^{2}}\\ & = -\frac{2^{k+3}p}{\pi^{2}}\mathop{\mathop\sum\limits_{1\leq a \lt N}\mathop\sum\limits_{a \lt b\leq N}\mathop{{\sum}}_{d\leq\frac{N}{2b}}\mathop{{\sum}'}_{1\leq v\leq\frac{p^2}{b}}}_{(a, b) = 1, (a, 2) = 1, (v, 2) = 1, (b, 2) = 1}\frac{d^{l+k}a^{l-1}b^{k-1}}{v^{2}}-\frac{2^{k+3}p}{\pi^{2}}\mathop{\mathop\sum\limits_{1\leq b \lt \frac{N}{2}}\mathop\sum\limits_{2b \lt a\leq N}\mathop{{\sum}}_{d\leq\frac{N}{a}}\mathop{{\sum}'}_{1\leq v\leq\frac{p^2}{a}}}_{(a, b) = 1, (a, 2) = 1, (v, 2) = 1, (b, 2) = 1}\frac{d^{l+k}a^{l-1}b^{k-1}}{v^{2}}\\ &\quad-\frac{2^{k+3}p}{\pi^{2}}\mathop{\mathop\sum\limits_{1\leq b\leq\frac{N}{2}}\mathop\sum\limits_{b \lt a \lt 2b}\mathop{{\sum}}_{d\leq\frac{N}{2b}}\mathop{{\sum}'}_{1\leq v\leq\frac{p^2}{a}}}_{(a, b) = 1, (a, 2) = 1, (v, 2) = 1, (b, 2) = 1}\frac{d^{l+k}a^{l-1}b^{k-1}}{v^{2}}-\frac{2^{k+3}p}{\pi^{2}}\mathop{{\sum}}_{d\leq\frac{N}{2}}\mathop{\mathop{{\sum}'}_{1\leq v\leq p^2}}_{(v, 2) = 1}\frac{d^{l+k}}{v^{2}}. \end{align*}$

Note that

$\sum\limits_{1 \leq v \leq p^{2} \atop(v, 2)=1}' v^{-2}=\frac{\pi^{2}}{8}+O\left(p^{-2}\right)$

and $l+k\geq0$ , we can get

$-\frac{2^{k+3}p}{\pi^{2}}\mathop \sum \limits_{d \le \frac{N}{2}} \sum\limits_{1 \leq v \leq p^{2} \atop(v, 2)=1}' \frac{d^{l+k}}{v^{2}} = \frac{-pN^{l+k+1}}{2^{l+1}(l+k+1)}+O(pN^{l+k}).$

So, we can write

$\begin{align*} M_{11}: = A+B+C-\frac{pN^{l+k+1}}{2^{l+1}(l+k+1)}+O(pN^{l+k}). \end{align*}$

(2)

We shall calculate the first three terms in the expression (2). First we calculate $A$ . Write

$\begin{align*} A = &-\frac{2^{k+3}p}{\pi^{2}}\mathop{\mathop\sum\limits_{1\leq a \lt N}\mathop\sum\limits_{a \lt b\leq N}\mathop{{\sum}}_{d\leq\frac{N}{2b}}\mathop{{\sum}'}_{1\leq v\leq\frac{p^2}{b}}}_{(a, b) = 1, (a, 2) = 1, (v, 2) = 1, (b, 2) = 1}\frac{d^{l+k}a^{l-1}b^{k-1}}{v^{2}}\\ = &\frac{-pN^{l+k+1}}{2^{l+1}(l+k+1)}\mathop{\mathop\sum\limits_{1\leq a \lt N}\mathop\sum\limits_{a \lt b\leq N}}_{(a, b) = 1, (a, 2) = 1, (b, 2) = 1}a^{l-1}b^{-l-2}+O\left(pN^{l+k}\mathop{\mathop\sum\limits_{1\leq a \lt N}\mathop\sum\limits_{a \lt b\leq N}}a^{l-1}b^{-l-1}\right), \end{align*}$

note that

$\begin{align*} \mathop{\mathop\sum\limits_{1\leq a \lt N}\mathop\sum\limits_{a \lt b\leq N}}_{(a, b) = 1, (a, 2) = 1, (b, 2) = 1}a^{l-1}b^{-l-2} & = \mathop{\mathop\sum\limits_{1\leq a \lt N}\mathop\sum\limits_{a \lt b\leq N}}_{(a, 2) = 1, (b, 2) = 1}a^{l-1}b^{-l-2}\mathop\sum\limits_{d|(a, b)}\mu(d)\\ & = \mathop{\mathop\sum\limits_{1\leq a \lt N}\mathop\sum\limits_{a \lt b\leq N}}_{(a, 2) = 1, (b, 2) = 1}a^{l-1}b^{-l-2}\mathop{\mathop\sum\limits_{d\leq\frac{N}{b}}}_{(d, 2) = 1}\frac{\mu(d)}{d^{3}}, \end{align*}$

we have

$\begin{align*} A = &\frac{-pN^{l+k+1}}{2^{l+1}(l+k+1)}\mathop{\mathop\sum\limits_{1\leq a \lt N}\mathop\sum\limits_{a \lt b\leq N}}_{(a, 2) = 1, (b, 2) = 1}a^{l-1}b^{-l-2}\left(\mathop{\sum\limits_{d = 1}^\infty}_{(d, 2) = 1}\frac{\mu(d)}{d^{3}}-\mathop{\sum\limits_{d \gt \frac{N}{b}}}_{(d, 2) = 1}\frac{\mu(d)}{d^{3}}\right)\\ &+O\left(pN^{l+k}\mathop\sum\limits_{1\leq a \lt N}\mathop\sum\limits_{a \lt b\leq N}a^{l-1}b^{-l-1}\right). \end{align*}$

By using the identity

$\mathop{\sum\limits_{d = 1}^\infty}_{(d, 2) = 1}\frac{\mu(d)}{d^{s}} = \frac{2^s}{(2^s-1)\zeta(s)},$

we have

(3)

For the case $l = 0$ , we have

$\begin{align*} A = &\frac{-4pN^{k+1}}{7\zeta(3)(k+1)}\mathop{\mathop\sum\limits_{1\leq a \lt N}\mathop\sum\limits_{a \lt b\leq N}}_{(a, 2) = 1, (b, 2) = 1}\frac{1}{ab^2} +O\left(pN^{k}\mathop\sum\limits_{1\leq a \lt N}\mathop\sum\limits_{a \lt b\leq N}\frac{1}{ab}\right)\\ = &\frac{-4pN^{k+1}}{7\zeta(3)(k+1)}\left(\mathop{\mathop\sum\limits_{a = 1}^\infty\mathop\sum\limits_{b = a+1}^\infty}_{(a, 2) = 1, (b, 2) = 1}\frac{1}{ab^2}-\mathop{\mathop\sum\limits_{a \lt N}\mathop\sum\limits_{b \gt N}}_{(a, 2) = 1, (b, 2) = 1}\frac{1}{ab^2} -\mathop{\mathop\sum\limits_{a \gt N}\mathop\sum\limits_{b \gt a}}_{(a, 2) = 1, (b, 2) = 1}\frac{1}{ab^2}\right)+O\left(pN^{k}\log^2N\right)\\ = &\frac{-4pN^{k+1}}{7\zeta(3)(k+1)}\mathop{\mathop\sum\limits_{a = 1}^\infty\mathop\sum\limits_{b = a+1}^\infty}_{(a, 2) = 1, (b, 2) = 1}\frac{1}{ab^2}+O(pN^{k}\log^2N). \end{align*}$

(4)

For the case $l\geq1$ , we have

$\begin{align*} A = &\frac{-pN^{l+k+1}}{7\cdot2^{l-2}\zeta(3)(l+k+1)}\mathop{\mathop\sum\limits_{1\leq a \lt N}\mathop\sum\limits_{a \lt b\leq N}}_{(a, 2) = 1, (b, 2) = 1}a^{l-1}b^{-l-2}+O\left(pN^{l+k}\mathop\sum\limits_{1\leq a \lt N}\mathop\sum\limits_{a \lt b\leq N}a^{l-1}b^{-l-1}\right)\\ = &\frac{-pN^{l+k+1}}{7\cdot2^{l-2}\zeta(3)(l+k+1)}\left(\mathop{\mathop\sum\limits_{a = 1}^\infty\mathop\sum\limits_{b = a+1}^\infty}_{(a, 2) = 1, (b, 2) = 1}\frac{a^{l-1}}{b^{l+2}}-\mathop{\mathop\sum\limits_{a \lt N}\mathop\sum\limits_{b \gt N}}_{(a, 2) = 1, (b, 2) = 1}\frac{a^{l-1}}{b^{l+2}}-\mathop{\mathop\sum\limits_{a \gt N}\mathop\sum\limits_{b \gt a}}_{(a, 2) = 1, (b, 2) = 1}\frac{a^{l-1}}{b^{l+2}}\right)\\ &+O\left(pN^{l+k}\mathop\sum\limits_{1\leq a \lt N}\mathop\sum\limits_{a \lt b\leq N}a^{l-1}b^{-l-1}\right)\\ = &\frac{-pN^{l+k+1}}{7\cdot2^{l-2}\zeta(3)(l+k+1)}\mathop{\mathop\sum\limits_{a = 1}^\infty\mathop\sum\limits_{b = a+1}^\infty}_{(a, 2) = 1, (b, 2) = 1}\frac{a^{l-1}}{b^{l+2}}+O(pN^{l+k}\log N). \end{align*}$

(5)

By using the same method, we can calculate B:

$\begin{align*} B = &-\frac{2^{k+3}p}{\pi^{2}}\mathop{\mathop\sum\limits_{1\leq b \lt \frac{N}{2}}\mathop\sum\limits_{2b \lt a\leq N}\mathop{{\sum}}_{d\leq\frac{N}{a}}\mathop{{\sum}'}_{1\leq v\leq\frac{p^2}{a}}}_{(a, b) = 1, (a, 2) = 1, (v, 2) = 1, (b, 2) = 1}\frac{d^{l+k}a^{l-1}b^{k-1}}{v^{2}}\\ = &\frac{-2^{k+3}pN^{l+k+1}}{7\zeta(3)(l+k+1)}\mathop{\mathop\sum\limits_{1\leq b \lt \frac{N}{2}}\mathop\sum\limits_{2b \lt a\leq N}}_{(a, 2) = 1, (b, 2) = 1}a^{-k-2}b^{k-1} +O\left(pN^{l+k}\mathop\sum\limits_{1\leq b \lt \frac{N}{2}}\mathop\sum\limits_{2b \lt a\leq N}a^{-k-1}b^{k-1}\right). \end{align*}$

For the case $k = 0$ , we can also get

$\begin{align*} B = \frac{-2^{3}pN^{l+1}}{7\zeta(3)(l+1)}\mathop{\mathop\sum\limits_{b = 1}^{\infty}\mathop\sum\limits_{a = 2b+1}^{\infty}}_{(a, 2) = 1, (b, 2) = 1}\frac{1}{a^2b}+O(pN^l\log^2N). \end{align*}$

(6)

For the case $k\geq1$ , we can also get

$\begin{align*} B = \frac{-2^{k+3}pN^{l+k+1}}{7\zeta(3)(l+k+1)}\mathop{\mathop\sum\limits_{b = 1}^{\infty}\mathop\sum\limits_{a = 2b+1}^{\infty}}_{(a, 2) = 1, (b, 2) = 1}\frac{b^{k-1}}{a^{k+2}}+O(pN^{l+k}\log N). \end{align*}$

(7)

By using the same method, we can also calculate C:

$\begin{align*} C = &-\frac{2^{k+3}p}{\pi^{2}}\mathop{\mathop\sum\limits_{1\leq b\leq\frac{N}{2}}\mathop\sum\limits_{b \lt a \lt 2b}\mathop{{\sum}}_{d\leq\frac{N}{2b}}\mathop{{\sum}'}_{1\leq v\leq\frac{p^2}{a}}}_{(a, b) = 1, (a, 2) = 1, (v, 2) = 1, (b, 2) = 1}\frac{d^{l+k}a^{l-1}b^{k-1}}{v^{2}}\\ = &\frac{-pN^{l+k+1}}{7\cdot2^{l-2}\zeta(3)(l+k+1)}\mathop{\mathop\sum\limits_{1\leq b\leq\frac{N}{2}}\mathop\sum\limits_{b \lt a \lt 2b}}_{(a, 2) = 1, (b, 2) = 1}\frac{a^{l-1}}{b^{l+2}} +O\left(pN^{l+k}\mathop\sum\limits_{1\leq b\leq\frac{N}{2}}\mathop\sum\limits_{b \lt a \lt 2b}\frac{a^{l-1}}{b^{l+1}}\right). \end{align*}$

For the case $l = 0$ , we can also get

$\begin{align*} C = \frac{-4pN^{k+1}}{7\zeta(3)(k+1)}\mathop{\mathop\sum\limits_{b = 1}^{\infty}\mathop\sum\limits_{a = b+1}^{2b}}_{(a, 2) = 1, (b, 2) = 1}\frac{1}{ab^2}+O(pN^k\log^2N). \end{align*}$

(8)

For the case $l\geq1$ , we can also get

$\begin{align*} C = \frac{-pN^{l+k+1}}{7\cdot2^{l-2}\zeta(3)(l+k+1)}\mathop{\mathop\sum\limits_{b = 1}^{\infty}\mathop\sum\limits_{a = b+1}^{2b}}_{(a, 2) = 1, (b, 2) = 1}\frac{a^{l-1}}{b^{l+2}}+O(pN^{l+k}\log N). \end{align*}$

(9)

Then from (2)–(9), we can get

$\begin{equation*} M_{11} = \frac{C(l, k)pN^{l+k+1}}{7\cdot2^{l-2}\zeta(3)(l+k+1)}+\left\{ \begin{array}{ll} O\left(pN^{l+k}\log^2N\right), & \hbox{if $l = k = 0$; $l = 0, k\geq1$ or $l\geq1, k = 0$;} \\ O\left(pN^{l+k}\log N\right), & \hbox{if $l\geq1, k\geq1$.} \end{array} \right. \end{equation*}$

(10)

here

$C(l,k) = - \frac{{7\zeta (3)}}{8} - \mathop {\sum\limits_{a = 1}^\infty {\mathop \sum \limits_{b = a + 1}^\infty } }\limits_{(a,2) = 1,(b,2) = 1} \frac{{{a^{l - 1}}}}{{{b^{l + 2}}}} - \mathop {\sum\limits_{b = 1}^\infty {\mathop \sum \limits_{a = b + 1}^{2b} } }\limits_{(a,2) = 1,(b,2) = 1} \frac{{{a^{l - 1}}}}{{{b^{l + 2}}}} - {2^{l + k + 1}}\mathop {\sum\limits_{b = 1}^\infty {\mathop \sum \limits_{a = 2b + 1}^\infty } }\limits_{(a,2) = 1,(b,2) = 1} \frac{{{b^{k - 1}}}}{{{a^{k + 2}}}}$

is a constant related to $l, k$ .

Now, we calculate $M_{12}$ . Similarly to the proof of Lemma $4$ we have

$\begin{align*} M_{12}& = -\frac{8p}{\pi^{2}}\mathop{\mathop \sum\limits_{a\leq N}\mathop \sum\limits_{b\leq N}\mathop {{\sum}'}_{1\leq u \leq p^{2}}\mathop {{\sum}'}_{1\leq v \leq p^{2}}}_{2au\equiv bv, 2au\neq bv, (u, 2) = 1, (v, 2) = 1 }\frac{a^{l}b^{k}}{uv}\\ &\ll p\mathop\sum\limits_{i, j = 0}^{J}\mathop\sum\limits_{e^i\leq u \lt e^{i+1}}\frac{1}{u}\sum\limits_{e^j\leq v \lt e^{j+1}} \frac{1}{v}\mathop{\mathop{\mathop\sum\limits_{a, b = 1}^{N}}_{2au\equiv bv(\bmod p)}}_{2au\neq bv}a^{l}b^{k}\\ &\ll pN^{l+k}\mathop\sum\limits_{0\leq i\leq j\leq J}e^{-i-j}\mathop\sum\limits_{e^i\leq u \lt e^{i+1}}\sum\limits_{e^j\leq v \lt e^{j+1}}\mathop{\mathop{\mathop\sum\limits_{a, b = 1}^{N}}_{2au\equiv bv(\bmod p)}}_{2au\neq bv}1\\ &\ll N^{l+k+2}p^{o(1)}. \end{align*}$

(11)

In addition, from Lemma 4 we have

$\begin{align*} M_{2}& = \frac{8p}{\pi^{2}\phi(p)}\mathop\sum\limits_{a\leq N}\mathop\sum\limits_{b\leq N}a^{l}b^{k}\mathop \sum\limits_{\chi \bmod p}\chi(-2a\overline{b})\left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right|^2\\ &\ll N^{l+k}\mathop\sum\limits_{a\leq N}\mathop\sum\limits_{b\leq N}\mathop \sum\limits_{\chi \bmod p}\chi(-2a\overline{b})\left|\mathop\sum\limits_{1\leq u\leq p^2}\frac{\chi\chi_2^0(u)}{u}\right|^2\\ &\ll N^{l+k+2}p^{o(1)}. \end{align*}$

(12)

Then combining (1) and (10)–(12), we obtain the asymptotic formula

$\begin{align*} &\mathop{\sum\limits_{a\leq N}\sum\limits_{b\leq N}}a^{l}b^{k}H(2a\overline{b}, p)\\ = &\frac{C(l, k)}{7\cdot2^{l-2}\zeta(3)(l+k+1)}pN^{l+k+1}\\ &+\left\{ \begin{array}{ll} O\left(pN^{l+k}\log^2N+N^{l+k+2}p^{o(1)}\right), & \hbox{if $l = k = 0$; $l = 0, k\geq1$ or $l\geq1, k = 0$;} \\ O\left(pN^{l+k}\log N+N^{l+k+2}p^{o(1)}\right), & \hbox{if $l\geq1, k\geq1$.} \end{array} \right. \end{align*}$

This completes the proof of the theorem.

Taking $l = 1, k = 0$ and $l = k = 1$ in the Theorem, we may immediately get the corollaries.

Acknowledgments

The authors would like to thank the anonymous referee for very careful reading of the manuscript and helpful comments. This work is supported by the Basic Research Program for Nature Science of Shaanxi Province (2014JM1001, 2015KJXX-27) and N.S.F.(11971381, 11471258, 11701447) of P. R. China.

Conflict of interest

We declare that we have no conflict of interest.

References

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[2]	L. Carlitz, The reciprocity theorem of Dedekind sums, Pac. J. Math., 3 (1953), 513-522. doi: 10.2140/pjm.1953.3.513
[3]	J. B. Conrey, E. Fransen, R. Klein, et al. Mean values of Dedekind sums, J. Number Theory, 41 (1996), 214-226.
[4]	C. Jia, On the mean values of Dedekind sums, J. Number Theory, 87 (2001), 173-188. doi: 10.1006/jnth.2000.2580
[5]	W. Liu, Mean value of Hardy sums over short intervals, Acta Math. Acad. Paedagogicae Nyiregyhaziensis, 28 (2012), 1-11.
[6]	R. Sitaramachandra Rao, Dedekind and Hardy sums, Acta Arith., 48 (1987), 325-340. doi: 10.4064/aa-48-4-325-340
[7]	I. E. Shparlinski, on some weighted average value of L-functions, Bull. Aust. Math. Soc., 79 (2009), 183-186. doi: 10.1017/S0004972708001020
[8]	Z. Xu, W. Zhang, The mean value of Hardy sums over short intervals, P. Roy. Soc. Edinburgh, 137 (2007), 885-894. doi: 10.1017/S0308210505000648
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AIMS Mathematics

Mean value of the Hardy sums over short intervals

Related Papers:

Abstract

1. Introduction

2. Some Lemmas

3. Proof of Theorem

Acknowledgments

Conflict of interest

References

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AIMS Mathematics

Mean value of the Hardy sums over short intervals

Related Papers:

Abstract

1. Introduction

2. Some Lemmas

3. Proof of Theorem

Acknowledgments

Conflict of interest

References

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