设$k$为正整数, $h$为任意整数.经典Dedekind和的定义如下

$ s(h, k)=\sum\limits_{j=1}^k((\frac{j}{k}))((\frac{hj}{k})), $

此处

$ ((x))=\left\{\begin{array}{ll} x-[x]-\frac{1}{2}, & {\rm{如果}}x {\rm{不为整数}}, \\ 0, & {\rm{如果}}x{\rm{为整数}}, \end{array} \right. $

其中$[y]$表示不超过$y$的最大整数.

关于Dedekind和的最重要的性质, 是互反公式.即对于互素的正整数$h$和$k$, 有

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

(1.1)

Dedekind^[1]基于$\log \eta(\tau)$的变换公式, 给出了上式的第一个证明. Rademacher^[2], Berndt^[3-5]和Dieter^[6]都分别给出了这个互反公式的不同证明.

张文鹏等人研究了Dedekind和的混合均值, 并得到了较强的渐近公式如下.

命题1.1 ^[7] 对于整数$ q \geq 3 $, 有

$ \sum\limits_{a=1}^{q}\vphantom{\sum}^{'} \sum\limits_{b=1}^{q}\vphantom{\sum}^{'} \frac{1}{ab} S(a\overline{b}, q) = \frac{5\pi ^2}{144}q \cdot \prod\limits_{p|q} \frac{(p^2 -1)^2}{p^2(p^2+1)} + O(\exp(\frac{7\ln q}{\ln \ln q})), $

其中$\sum\limits_{a=1}^{q}\vphantom{\sum}^{'}$表示所有与$ p $互素, 且$ 1 \leq a \leq q $的整数$ a $求和, $\prod\limits_{p|q}$表示对$ q $的所有素数$ p $求乘积, $ \exp(y) = {\text{e}}^y $.

命题1.2 ^[8]设整数$ n \geq 1 $, $p$为素数, $k = p^n$.则有

$ \sum\limits_{h=1}^{k}\vphantom{\sum}^{'} \frac{S(h, k)}{h} = \frac{\pi^2}{72} k ( 1-\frac{1}{p^2} ) + O(\sqrt{k}). $

设$ f $为正整数, $ \chi $为模$ f $的Dirichlet特征.定义新型Dedekind和如下

$ s(h, k;\chi)=\sum\limits_{a=0}^{fk-1}\overline{B}_{1, \chi}(\frac{ha}{k}) \overline{B}_{1}(\frac{a}{fk}). $

本文将给出$ s(h, k;\chi) $的混合均值的渐近公式.主要结论如下.

定理1.1 设$ f $为正整数, $ p > 2 $为素数, 满足$ (f, p) = 1 $.设$ \chi $为模$ f $的Dirichlet特征, 满足$ \chi(-1) = 1 $.则有

$ \begin{eqnarray*} \sum\limits_{a = 1}^{fp}\vphantom{\sum}^{'} \sum\limits_{b = 1}^{fp} \vphantom{\sum}^{'} \frac{1}{ab} S(a\overline{b}, p;\chi) &=& \frac{\pi^2}{72} \frac{fp ~\tau(\overline{\chi}) L^2(2, \chi)}{L(4, \chi)} \prod\limits_{q \mid fp} \frac{ ( 1 - \frac{1}{q^2} )^2 (1 - \frac{\chi(q)}{q^2} )^2 }{ 1 - \frac{ \chi(q) }{q^4} } \\ &&+ O( f^{\frac{3}{2}} p^{\frac{1}{2}} \log^2 (fp) ), \end{eqnarray*} $

其中$\sum\limits^{fp}_{a=1}\vphantom{\sum}^{'}$表示所有与$ fp $互素且$ 1 \leq a \leq fp $的整数$ a $求和, $ \overline{b} $表示同余方程$ x \cdot b \equiv 1 (\bmod\ fp) $的解.

定理1.2. 设$ f $为正整数, $ p > 2 $为素数, 满足$ (f, p) = 1 $.设$ \chi $为模$ f $的Dirichlet特征, 满足$ \chi(-1) = 1 $.则有

$ \sum\limits_{h = 1}^{fp}\vphantom{\sum}^{'} \frac{S(h, p;\chi)}{h} = \frac{fp}{2\pi^2} \tau(\overline{\chi}) L^2(2, \chi) \prod\limits_{q | fp} ( 1-\frac{\chi(q)}{q^2} )^2 + O( f^{\frac{3}{2}}p^{\frac{1}{2}}\log^2 (fp) ). $

引理2.1 设$f, k$为正整数, $h$为任意整数, 满足$(fk, h)=1$与$(f, k)=1$.设$\chi$为模$f$的Dirichlet原特征.则有

$ S(h, k; \chi)=\left\{\begin{array}{ll} \displaystyle \frac{f}{\pi^{2}k\phi(f)}\tau(\overline{\chi})\sum\limits_{d\mid k} \frac{d^2}{\phi(d)}\chi(\frac{k}{d})\sum\limits_{\psi \bmod fd \atop \psi(-1)=-1}\psi(h) L(1, \chi\overline{\psi}) L(1, \psi), & {\rm{如果}} \chi(-1)=1, \\ 0, & {\rm{如果}} \chi(-1)=-1, \end{array}\right. $

其中$\tau(\overline{\chi})$为Gauss和, $\displaystyle\sum\limits_{\psi \bmod fd \atop \psi(-1)=-1}$表示对模$fd$的所有奇特征求和.

证 对任意实数$y$, 由文献[9]中的定理3.1可知

$ \overline{B}_{n, \chi}(y)=f^{n-1}\sum\limits_{b=0}^{f-1}\overline{\chi}(b)\overline{B}_{n}(\frac{b+y}{f}). $

此外当$0\leq x< 1$时, 有$ B_{n}(x)=-\frac{n!}{(2\pi i)^{n}}\sum_{r=-\infty \atop r\neq0}^{+\infty}\frac{e(rx)}{r^{n}}, $其中$e(y)=\hbox{e}^{2\pi iy}$.因此

$ \begin{eqnarray*} s(h, k; \chi)&=&\sum\limits_{a=0}^{fk-1}\overline{B}_{1, \chi} (\frac{ha}{k} ) \overline{B}_{1} (\frac{a}{fk} ) =\sum\limits_{a=0}^{fk-1}\sum\limits_{b=0}^{f-1}\overline{\chi}(b)\overline{B}_{1} (\frac{ha+kb}{fk} ) \overline{B}_{1} (\frac{a}{fk} )\\ &=&\mathop{\sum\limits_{a=1}^{fk-1}\sum\limits_{b=0}^{f-1}}_{fk\nmid ha+kb}\overline{\chi}(b)B_{1} (\frac{ha+kb}{fk} )B_{1} (\frac{a}{fk} )\\ &=&-\frac{1}{4\pi^{2}}\mathop{\sum\limits_{a=1}^{fk-1}\sum\limits_{b=0}^{f-1}}_{fk\nmid ha+kb}\overline{\chi}(b) \sum\limits_{r=-\infty \atop r\neq0}^{+\infty}\sum\limits_{s=-\infty \atop s\neq0}^{+\infty}\frac{1}{rs} e (\frac{(rh+s)a+krb}{fk} )\\ &=&-\frac{1}{4\pi^{2}} \sum\limits_{a=0}^{fk-1}\sum\limits_{b=0}^{f-1}\overline{\chi}(b)\sum\limits_{r=-\infty \atop r\neq0}^{+\infty} \sum\limits_{s=-\infty \atop s\neq0}^{+\infty}\frac{1}{rs}~e (\frac{(rh+s)a+krb}{fk} )\\ && +\frac{1}{4\pi^{2}} \sum\limits_{r=-\infty \atop r\neq0}^{+\infty}\sum\limits_{s=-\infty \atop s\neq0}^{+\infty} \frac{1}{rs} \sum\limits_{b=0}^{f-1}\overline{\chi}(b) e (\frac{rb}{f} )\\ &&+\frac{1}{4\pi^{2}}\mathop{\sum\limits_{a=1}^{fk-1}\sum\limits_{b=0}^{f-1}}_{fk\mid ha+kb}\overline{\chi}(b) \sum\limits_{r=-\infty \atop r\neq0}^{+\infty}\sum\limits_{s=-\infty \atop s\neq0}^{+\infty} \frac{1}{rs}e(\frac{(rh+s)a+krb}{fk})\\ &=&-\frac{1}{4\pi^{2}} \sum\limits_{r=-\infty \atop r\neq0}^{+\infty}\sum\limits_{s=-\infty \atop s\neq0}^{+\infty} \frac{1}{rs} \sum\limits_{a=0}^{fk-1} e^{2 \pi i \frac{(rh+s)a}{fk}} \sum\limits_{b=0}^{f-1}\overline{\chi}(b) e^{2 \pi i \frac{rb}{f}}. \end{eqnarray*} $

注意到$\chi$为模$f$的原特征, 以及$(fk, h)=1$, $(f, k)=1$.可得

$ \begin{eqnarray*} S(h, k; \chi)&=&-\frac{fk}{4\pi^{2}}\tau(\overline{\chi})\mathop{\sum\limits_{r=-\infty \atop r\neq0}^{+\infty} \sum\limits_{s=-\infty \atop s\neq0}^{+\infty}}_{rh+s\equiv 0 (\bmod fk)}\frac{\chi(r)}{rs} =-\frac{fk}{4\pi^{2}}\tau(\overline{\chi}) \sum\limits_{d\mid fk}\underset{rh+s\equiv 0\left(\bmod\frac{fk}{d}\right) \atop \left(s, \frac{fk}{d}\right)=1}{\sum^{+\infty}_{r=-\infty \atop r\neq0} \sum^{+\infty}_{s=-\infty \atop s\neq0}}\frac{\chi(rd)}{rdsd} \\ &=&-\frac{fk}{4\pi^{2}}\tau(\overline{\chi})\sum\limits_{d\mid fk}\frac{1}{d^{2}}\sum\limits_{r=-\infty \atop r\neq0}^{+\infty}\sum\limits_{s=-\infty \atop s\neq0}^{+\infty}\frac{\chi(rd)}{rs}\sum\limits_{s\equiv(-rh)(\bmod\frac{fk}{d}) \atop (s, \frac{fk}{d})=1}1. \end{eqnarray*} $

再由特征和的正交性质, 有

$ \begin{eqnarray*} &&S(h, k;\chi)=-\frac{fk}{4\pi^{2}}\tau(\overline{\chi})\sum\limits_{d\mid fk} \frac{d^{-2}}{\phi(\frac{fk}{d})}\sum\limits_{\psi \bmod \frac{fk}{d}}\sum\limits_{r=-\infty \atop r\neq0}^{+\infty}\sum\limits_{s=-\infty \atop s\neq0}^{+\infty} \frac{\chi(rd)}{rs}\psi(s)\overline{\psi}(-rh)\\ &&\;\;\;\;\;\;\;\;=-\frac{fk}{4\pi^{2}}\tau(\overline{\chi})\sum\limits_{d\mid fk} \frac{d^{-2}}{\phi(\frac{fk}{d})}~\chi(d)\sum\limits_{\psi \bmod \frac{fk}{d}}\overline{\psi}(-h)\sum\limits_{r=-\infty \atop r\neq0}^{+\infty}\frac{\chi(r)\overline{\psi}(r)}{r}\sum\limits_{s=-\infty \atop s\neq0}^{+\infty} \frac{\psi(s)}{s}\\ &&=\left\{\begin{array}{ll} \displaystyle\frac{fk}{\pi^{2}}\tau(\overline{\chi})\sum\limits_{d\mid fk} \frac{d^{-2}}{\phi(\frac{fk}{d})}~\chi(d)\sum\limits_{\psi \bmod \frac{fk}{d} \atop \psi(-1)=-1}\overline{\psi}(h) L(1, \chi\overline{\psi}) L(1, \psi), & \;{\rm{当}}\; \chi(-1)=1, \\ 0, & \;{\rm{当}}\; \chi(-1)=-1, \end{array}\right.\\ &&=\left\{\begin{array}{ll} \displaystyle\frac{1}{\pi^{2}fk}\tau(\overline{\chi})\sum\limits_{d\mid fk} \frac{d^2}{\phi(d)}\chi\left(\frac{fk}{d}\right)\sum\limits_{\psi \bmod d \atop \psi(-1)=-1}\psi(h) L(1, \chi\overline{\psi}) L(1, \psi), & \;{\rm{当}}\; \chi(-1)=1, \\ 0, & \;{\rm{当}}\; \chi(-1)=-1, \end{array}\right.\\ &&=\left\{\begin{array}{ll} \displaystyle \frac{f}{\pi^{2}k\phi(f)}\tau(\overline{\chi})\sum\limits_{d\mid k} \frac{d^2}{\phi(d)}\chi\left(\frac{k}{d}\right)\sum\limits_{\psi \bmod fd \atop \psi(-1)=-1}\psi(h) L(1, \chi\overline{\psi}) L(1, \psi), & \;{\rm{如果}}\; \chi(-1)=1, \\ 0, & \;{\rm{如果}}\; \chi(-1)=-1. \end{array}\right. \end{eqnarray*} $

引理2.1证毕.

由引理2.1有

$ S(h, p;\chi) = \frac{fp}{\pi^2 \phi(p) \phi(f)} \tau(\overline{\chi}) \sum\limits_{\psi \bmod fp \atop \psi(-1) = -1} \psi(h) L(1, \chi\overline{\psi}) L(1, \psi)+O\left(\frac{f^{\frac{3}{2}}}{p}\right). $

因此

$ \begin{eqnarray*} &&\sum\limits_{a = 1}^{fp}\vphantom{\sum}^{'} \sum\limits_{b = 1}^{fp}\vphantom{\sum}^{'} \frac{1}{ab} S(a\overline{b}, p;\chi) \\ &=& \frac{fp}{\pi^2 \phi(p) \phi(f)} \tau(\overline{\chi}) \sum\limits_{\psi \bmod fp \atop \psi(-1) = -1} \sum\limits_{a = 1}^{fp}\vphantom{\sum}^{'} \frac{\psi(a)}{a} \sum\limits_{b = 1}^{fp}\vphantom{\sum}^{'} \frac{ \overline{\psi}(b) }{b} L(1, \chi\overline{\psi}) L(1, \psi) + O\left( \frac{f^{\frac{3}{2}} \log^2(fp)}{p} \right). \end{eqnarray*} $

容易证明

$ \begin{eqnarray*} L(1, \psi) = \sum\limits_{n = 1}^{\infty} \frac{\psi(n)}{n} = \sum\limits_{n = 1}^{fp} \frac{\psi(n)}{n} + \int_{fp}^{\infty} \frac{\sum\limits_{fp < n \leq y} \psi(n)}{y^2} dy = \sum\limits_{n = 1}^{fp} \frac{\psi(n)}{n} + O\left( \frac{\log (fp)}{(fp)^{\frac{1}{2}}} \right), \end{eqnarray*} $

从而

$ \begin{eqnarray*} &&\sum\limits_{a = 1}^{fp}\vphantom{\sum}^{'} \sum\limits_{b = 1}^{fp}\vphantom{\sum}^{'} \frac{1}{ab} S(a\overline{b}, p;\chi) \\ &=& \frac{fp}{\pi^2 \phi(p) \phi(f)} \tau(\overline{\chi}) \sum\limits_{\psi \bmod fp \atop \psi(-1) = -1} L(1, \psi)^2 L(1, \chi\overline{\psi}) L(1, \overline{\psi}) + O\left( f^{\frac{3}{2}} p^{\frac{1}{2}} \log^2(fp) \right). \end{eqnarray*} $

取参数$N$满足$fp\leq N \leq (fp)^3$, 有

$ \begin{eqnarray*} && L^2(1, \psi) = \sum\limits_{n=1}^{\infty} \frac{\psi(n) d(n)}{n} = \sum\limits_{1 \leq n \leq N} \frac{\psi(n) d(n)}{n} + \int_{N}^{\infty} \frac{\sum\limits_{N < n \leq y} \psi(n)d(n)}{y^2}dy, \\ && L(1, \chi\overline{\psi}) L(1, \overline{\psi}) = \sum\limits_{n=1}^{\infty} \frac{\overline{\psi}(n) r(n)}{ n } = \sum\limits_{1 \leq n \leq N} \frac{\overline{\psi}(n) r(n)}{ n } + \int_{N}^{\infty} \frac{\sum\limits_{N < n \leq y}\overline{\psi}(n) r(n) }{y^2} dy, \end{eqnarray*} $

其中$ d(n) = \sum\limits_{t | n} 1 $为除数函数, $ r(n) = \sum\limits_{t|n} \chi(t) $.再由Pólya-Vinogradov不等式可得

$ \begin{eqnarray*} \sum\limits_{N < n \leq y} \overline{\psi}(n)r(n) &=& 2\sum\limits_{n \leq \sqrt{y}} \overline{\psi}(n) \sum\limits_{m \leq \frac{y}{n}} \chi(m) \overline{\psi}(m) - ( \sum\limits_{n \leq \sqrt{y}} \overline{\psi}(n) )^2 \\ &&- 2\sum\limits_{n \leq \sqrt{N}} \overline{\psi}(n) \sum\limits_{m \leq \frac{y}{n}} \chi(m) \overline{\psi}(m) + ( \sum\limits_{n \leq \sqrt{N}} \overline{\psi}(n) )^2 \\ &\ll& y^{\frac{1}{2}} (fp)^{\frac{1}{2}} ( \log(fp) )^2, \end{eqnarray*} $

以及

$ \begin{eqnarray*} \sum\limits_{N < n \leq y} \psi(n) d(n) &=& 2\sum\limits_{n \leq \sqrt{y}} \psi(n) \sum\limits_{m \leq \frac{y}{n}} \psi(m) - ( \sum\limits_{n \leq \sqrt{y}} \psi(n) )^2 \\ &&- 2\sum\limits_{n \leq \sqrt{N}} \psi(n) \sum\limits_{m \leq \frac{N}{n}} \psi(m) + ( \sum\limits_{n \leq \sqrt{N}} \psi(n) )^2 \\ &\ll& y^{\frac{1}{2}} (fp)^{\frac{1}{2}} ( \log(fp) )^2. \end{eqnarray*} $

则有

$ \begin{eqnarray} \sum\limits_{a = 1}^{fp}\vphantom{\sum}^{'} \sum\limits_{b = 1}^{fp}\vphantom{\sum}^{'} \frac{1}{ab} S(a\overline{b}, p;\chi) &=& \frac{fp}{\pi^2 \phi(p) \phi(f)} \tau(\overline{\chi}) \sum\limits_{n = 1}^{N} \frac{d(n)}{n} \sum\limits_{m = 1}^{N} \frac{r(m)}{m} \sum\limits_{\psi \bmod fp \atop \psi(-1) = -1} \psi(n) \overline{\psi}(m)\nonumber\\ &&+ O ( \frac{f^{\frac{5}{2}}p^{\frac{3}{2}}\log^2(fp)}{N^{\frac{1}{2}}} ) + O( f^{\frac{3}{2}} p^{\frac{1}{2}} \log^2(fp) ). \end{eqnarray} $

(3.1)

由特征的正交关系可得

$ \begin{eqnarray} && \sum\limits_{n = 1}^{N} \frac{d(n)}{n} \sum\limits_{m = 1}^{N} \frac{r(m)}{m} \sum\limits_{\psi \bmod fp \atop \psi(-1) = -1} \psi(n) \overline{\psi}(m)\nonumber\\ &=& \frac{1}{2} \sum\limits_{n = 1}^{N} \frac{d(n)}{n} \sum\limits_{ m = 1 }^{N} \frac{r(m)}{m} \sum\limits_{\psi \bmod fp} (1 - \psi(-1)) \psi(n) \overline{\psi}(m)\nonumber \\ & =& \frac{ \phi(fp) }{ 2 } \mathop{\sum\limits_{n = 1 \atop (n, fp) = 1}^{N} \sum\limits_{m = 1 \atop (m, fp) = 1}^{N} }_{n \equiv m (\bmod fp)} \frac{d(n)r(m)}{nm} - \frac{ \phi(fp) }{ 2 } \mathop{\sum\limits_{n = 1 \atop (n, fp) = 1}^{N} \sum\limits_{m = 1 \atop (m, fp) = 1}^{N} }_{n \equiv -m (\bmod fp)} \frac{d(n)r(m)}{nm} \nonumber\\ & :=& \frac{ \phi(fp) }{ 2 } M_1 - \frac{ \phi(fp) }{ 2 } M_2. \end{eqnarray} $

(3.2)

首先考虑$ M_1 $, 有

$ \begin{eqnarray} M_1 &=& \mathop{ \sum\limits_{1 \leq n \leq fp -1 \atop (n, fp) = 1} \sum\limits_{1 \leq m \leq fp -1 \atop (m, fp) = 1} }_{n \equiv m (\bmod fp)} \frac{d(n)r(m)}{nm} + \mathop{ \sum\limits_{1 \leq n \leq fp -1 \atop (n, fp) = 1} \sum\limits_{fp \leq m \leq N \atop (m, fp) = 1} }_{n \equiv m (\bmod fp)} \frac{d(n)r(m)}{nm} \nonumber\\ &&+ \mathop{ \sum\limits_{fp \leq n \leq N \atop (n, fp) = 1} \sum\limits_{1 \leq m \leq fp -1 \atop (m, fp) = 1} }_{n \equiv m (\bmod fp)} \frac{d(n)r(m)}{nm} + \mathop{ \sum\limits_{fp \leq n \leq N \atop (n, fp) = 1} \sum\limits_{fp \leq m \leq N \atop (m, fp) = 1} }_{n \equiv m (\bmod fp)} \frac{d(n)r(m)}{nm} \nonumber\\ &:=& \Omega_1 + \Omega_2 + \Omega_3 + \Omega_4. \end{eqnarray} $

(3.3)

易证

$ \begin{eqnarray*} \Omega_1 = \sum\limits_{1 \leq n \leq fp - 1 \atop (n, fp) = 1} \frac{d(n)r(n)}{n^2} = \sum\limits_{n = 1 \atop (n, fp) = 1}^{\infty} \frac{d(n)r(n)}{n^2} + O( (fp)^{\varepsilon - 1} ). \end{eqnarray*} $

由欧拉乘积公式可得

$ \begin{eqnarray*} &&\sum\limits_{n = 1 \atop (n, fp) = 1} ^{\infty} \frac{d(n)r(n)}{n^2}\\ &=&\prod\limits_{q \nmid fp} ( 1 + \frac{d(q)r(q)}{q^2} + \frac{d(q^2)r(q^2)}{q^4} + \cdots +\frac{d(q^k)r(q^k)}{q^{2k}} + \cdots ) \\ &=& \prod\limits_{q \nmid fp} ( \sum\limits_{k = 0}^{\infty} \frac{d(q^k)r(q^k)}{q^{2k}} ) = \prod\limits_{q \nmid fp} ( \sum\limits_{k = 0}^{\infty} \frac{1}{q^{2k}} \frac{(k + 1) ( 1 - \chi(q)^{k + 1} )}{ 1 - \chi(q) } ) \\ &=& \prod\limits_{q \nmid fp} \frac{1}{1 - \chi(q)} \sum\limits_{k = 0}^{\infty} \frac{(k + 1) ( 1 - \chi(q)^{k + 1} )}{ q^{2k} } = \prod\limits_{q \nmid fp} \frac{1}{1 - \chi(q)} [ \sum\limits_{k = 0}^{\infty} (k + 1) ( \frac{1}{q^2} )^k - \sum\limits_{k = 0}^{\infty} \frac{ (k + 1) \chi(q)^{k + 1} }{q^{2k}} ] \\ &=& \prod\limits_{q \nmid fp} \frac{1}{1 - \chi(q)} [ \sum\limits_{k = 0}^{\infty} (k + 1) ( \frac{1}{q^2} )^k - \chi(q) \sum\limits_{k = 0}^{\infty} (k + 1) ( \frac{\chi(q)}{q^2} )^k ] \\ &=& \prod\limits_{q \nmid fp} \frac{1}{1 - \chi(q)} [ \frac{1}{ ( 1 - \frac{1}{q^2} )^2} - \chi(q) \frac{1}{ ( 1 - \frac{\chi(q)}{q^2} )^2} ] \\ &=& \prod\limits_{q \nmid fp} \frac{ q^4 }{ 1 - \chi(q) } [ \frac{ (q^2 - \chi(q))^2 - \chi(q)( q^2 - 1 )^2 }{ ( q^2 - 1 )^2 ( q^2 - \chi(q) )^2 } ] = \prod\limits_{q \nmid fp} \frac{ q^4 }{ 1 - \chi(q) } [ \frac{ q^4 - q^4 \chi(q) - \chi(q) + \chi(q)^2 }{ ( q^2 - 1 )^2 ( q^2 - \chi(q) )^2 } ] \\ &=& \prod\limits_{q \nmid fp} \frac{ q^4(q^4 - \chi(q)) }{ ( q^2 - 1 )^2 ( q^2 - \chi(q) )^2 } = \prod\limits_{q \nmid fp} \frac{ 1 - \frac{ \chi(q) }{q^4} }{ ( 1 - \frac{1}{q^2} )^2 ( 1 - \frac{\chi(q)}{q^2} )^2 } \\ &=& \frac{ \zeta^2(2) L^2(2, \chi) }{ L(4, \chi) } \prod\limits_{q \mid fp} \frac{ ( 1 - \frac{1}{q^2} )^2 (1 - \frac{\chi(q)}{q^2} )^2 }{ 1 - \frac{ \chi(q) }{q^4} } = \frac{\pi^{4}}{36} \frac{ L^2(2, \chi) }{ L(4, \chi) } \prod\limits_{q \mid fp} \frac{ ( 1 - \frac{1}{q^2} )^2 (1 - \frac{\chi(q)}{q^2} )^2 }{ 1 - \frac{ \chi(q) }{q^4} }. \end{eqnarray*} $

因此

$ \begin{eqnarray} \Omega_1=\frac{\pi^{4}}{36} \frac{ L^2(2, \chi) }{ L(4, \chi) } \prod\limits_{q \mid fp} \frac{ \left( 1 - \frac{1}{q^2} \right)^2 \left(1 - \frac{\chi(q)}{q^2} \right)^2 }{ 1 - \frac{ \chi(q) }{q^4} } + O((fp)^{\epsilon-1}). \end{eqnarray} $

(3.4)

类似可得

$ \begin{eqnarray} \Omega_2&\ll& (fp)^{\epsilon}\mathop{\mathop{\sum\limits_{1\leq n\leq fp-1}}_{(n, fp)=1} \mathop{\sum\limits_{fp\leq m\leq N}}_{(m, fp)=1}}_{n \equiv m(\bmod fp)}\frac{1}{nm}=(fp)^{\epsilon}\mathop{\mathop{\sum\limits_{1\leq n\leq fp-1}}_{(n, fp)=1} \sum\limits_{1\leq l\leq\left[\frac{N}{fp}\right]-1}\mathop{\sum\limits_{0\leq r\leq fp-1}}_{(r, fp)=1}}_{n \equiv l fp+r (\bmod fp)}\frac{1}{n(l fp+r)}\nonumber\\ &\ll& (fp)^{\epsilon}\sum\limits_{1\leq l \leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\frac{1}{r(lfp+r)}\ll (fp)^{\epsilon} \sum\limits_{1\leq l\leq \frac{N}{fp}}\frac{1}{lfp}\sum\limits_{1\leq r\leq fp-1}\frac{1}{r} \nonumber\\ &\ll&(fp)^{\epsilon-1}, \end{eqnarray} $

(3.5)

$ \begin{eqnarray} \Omega_3&\ll& (fp)^{\epsilon}\mathop{\mathop{\sum\limits_{fp\leq n\leq N}}_{(n, fp)=1} \mathop{\sum\limits_{1\leq m\leq fp-1}}_{(m, fp)=1}}_{n \equiv m(\bmod fp)}\frac{1}{nm}=(fp)^{\epsilon}\mathop{\sum\limits_{1\leq l\leq\left[\frac{N}{fp}\right]-1}\mathop{\sum\limits_{0\leq r\leq fp-1}}_{(r, fp)=1} \mathop{\sum\limits_{1\leq m\leq fp-1}}_{(m, fp)=1}}_{lfp+r \equiv m (\bmod fp)}\frac{1}{(l fp+r)m}\nonumber\\ &\ll& (fp)^{\epsilon}\sum\limits_{1\leq l \leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\frac{1}{(lfp+r)r}\ll (fp)^{\epsilon} \sum\limits_{1\leq l\leq \frac{N}{fp}}\frac{1}{lfp}\sum\limits_{1\leq r\leq fp-1}\frac{1}{r} \nonumber\\ &\ll&(fp)^{\epsilon-1}, \end{eqnarray} $

(3.6)

以及

$ \begin{eqnarray} \Omega_4 &\ll& (fp)^{\epsilon} \mathop{\mathop{\sum\limits_{fp\leq n\leq N}}_{(n, fp)=1} \mathop{\sum\limits_{fp\leq m\leq N}}_{(m, fp)=1}}_{n \equiv m(\bmod fp)}\frac{1}{nm}\nonumber\\ &=& (fp)^{\epsilon}\mathop{\sum\limits_{1\leq l_1\leq \left[\frac{N}{fp}\right]-1}\sum\limits_{0\leq r_1\leq fp-1 \atop (r_1, fp)=1}\sum\limits_{1\leq l_2\leq \left[\frac{N}{fp}\right]-1}\sum\limits_{0\leq r_2\leq fp-1 \atop (r_2, fp)=1}}_{l_1fp+r_1\equiv l_2fp+r_2 (\bmod fp)}\frac{1}{(l_1fp+r_1)(l_2fp+r_2)}\nonumber\\ &\ll& (fp)^{\epsilon}\sum\limits_{1\leq l_1\leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\sum\limits_{1\leq l_2\leq \frac{N}{fp}}\frac{1}{(l_1fp+r)(l_2fp+r)}\nonumber\\ &\ll& (fp)^{\epsilon}\sum\limits_{1\leq l_1\leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\frac{1}{l_1fp+r}\sum\limits_{1\leq l_2\leq \frac{N}{fp}}\frac{1}{l_2fp}\nonumber\\ &\ll& (fp)^{\epsilon-1}. \end{eqnarray} $

(3.7)

结合(3.3)-(3.7)有

$ \begin{eqnarray} M_1=\frac{\pi^{4}}{36} \frac{ L^2(2, \chi) }{ L(4, \chi) } \prod\limits_{q \mid fp} \frac{ ( 1 - \frac{1}{q^2} )^2 (1 - \frac{\chi(q)}{q^2} )^2 }{ 1 - \frac{ \chi(q) }{q^4} } + O((fp)^{\epsilon-1}). \end{eqnarray} $

(3.8)

接下来考虑$M_2$.可得

$ \begin{eqnarray*} M_{2} &=&\mathop{\mathop{\sum\limits_{n=1}^{N}}_{(n, fp)=1}\mathop{\sum\limits_{m=1}^{N}}_{(m, fp)=1}}_{n \equiv -m(\bmod fp)}\frac{r(n)\overline{r}(m)}{nm}\ll (fp)^{\epsilon}\mathop{\mathop{\sum\limits_{n=1}^{N}}_{(n, fp)=1}\mathop{\sum\limits_{m=1}^{N}}_{(m, fp)=1}}_{n \equiv -m(\bmod fp)}\frac{1}{nm}\nonumber\\ &\ll& (fp)^{\epsilon}\mathop{\sum\limits_{0\leq l_1\leq \frac{N}{fp}}\sum\limits_{1\leq r_1\leq fp-1}\sum\limits_{0\leq l_2\leq \frac{N}{fp}}\sum\limits_{1\leq r_2\leq fp-1}}_{l_1fp+r_1\equiv -(l_2fp+r_2) (\bmod fp)}\frac{1}{(l_1fp+r_1)(l_2fp+r_2)}\nonumber\\ &\ll& (fp)^{\epsilon}\sum\limits_{0\leq l_1\leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\sum\limits_{0\leq l_2\leq \frac{N}{fp}}\frac{1}{(l_1fp+r)(l_2fp+fp-r)}\nonumber\\ &=& (fp)^{\epsilon}\sum\limits_{1\leq r\leq fp-1}\frac{1}{r(fp-r)}+(fp)^{\epsilon}\sum\limits_{1\leq r\leq fp-1}\sum\limits_{1\leq l_2\leq \frac{N}{fp}}\frac{1}{r(l_2fp+fp-r)}\nonumber\\ &&+(fp)^{\epsilon}\sum\limits_{1\leq l_1\leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\frac{1}{(l_1fp+r)(fp-r)}\nonumber\\ &&+(fp)^{\epsilon}\sum\limits_{1\leq l_1\leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\sum\limits_{1\leq l_2\leq \frac{N}{fp}}\frac{1}{(l_1fp+r)(l_2fp+fp-r)}\nonumber\\ &\ll& (fp)^{\epsilon-1}\sum\limits_{1\leq r\leq fp-1}\left(\frac{1}{r}+\frac{1}{fp-r}\right)+(fp)^{\epsilon}\sum\limits_{1\leq r\leq fp-1}\frac{1}{r}\sum\limits_{1\leq l_2\leq \frac{N}{fp}}\frac{1}{l_2fp}\nonumber\\ &&+(fp)^{\epsilon}\sum\limits_{1\leq l_1\leq \frac{N}{fp}}\frac{1}{l_1fp}\sum\limits_{1\leq r\leq fp-1}\frac{1}{fp-r} + (fp)^{\epsilon}\sum\limits_{1\leq l_1\leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\frac{1}{l_1fp+r}\sum\limits_{1\leq l_2\leq \frac{N}{fp}}\frac{1}{l_2fp}\nonumber\\ &\ll& (fp)^{\epsilon-1}. \end{eqnarray*} $

(3.9)

结合(3.1), (3.2), (3.8)和(3.9)式, 并取$N=(fp)^3$, 立即可得

$ \begin{eqnarray*} \sum\limits_{a = 1}^{fp}\vphantom{\sum}^{'} \sum\limits_{b = 1}^{fp} \vphantom{\sum}^{'} \frac{1}{ab} S(a\overline{b}, p;\chi) = \frac{\pi^2}{72} \frac{fp ~\tau(\overline{\chi}) L^2(2, \chi)}{L(4, \chi)} \prod\limits_{q \mid fp} \frac{ \left( 1 - \frac{1}{q^2} \right)^2 \left(1 - \frac{\chi(q)}{q^2} \right)^2 }{ 1 - \frac{ \chi(q) }{q^4} } + O\left( f^{\frac{3}{2}} p^{\frac{1}{2}} \log^2 (fp) \right). \end{eqnarray*} $

定理1.1证毕.

由引理2.1有$ S(h, p;\chi) = \frac{fp}{\pi^2 \phi(p) \phi(f)} \tau(\overline{\chi}) \sum\limits_{\psi \bmod fp \atop \psi(-1) = -1} \psi(h) L(1, \chi\overline{\psi}) L(1, \psi)+O\left(\frac{f^{\frac{3}{2}}}{p}\right). $因此

$ \begin{eqnarray*} \sum\limits_{h = 1}^{fp}\vphantom{\sum}^{'} \frac{S(h, p;\chi)}{h} = \frac{fp}{\pi^2 \phi(p) \phi(f)} \tau(\overline{\chi}) \sum\limits_{\psi \bmod fp \atop \psi(-1) = -1} ( \sum\limits_{h = 1}^{fp}\vphantom{\sum}^{'} \frac{\psi(h)}{h} ) L(1, \chi\overline{\psi}) L(1, \psi) + O( \frac{f^{\frac{3}{2}} \log(fp)}{p} ). \end{eqnarray*} $

容易证明

$ \begin{eqnarray*} L(1, \psi) = \sum\limits_{n = 1}^{\infty} \frac{\psi(n)}{n} = \sum\limits_{n = 1}^{fp} \frac{\psi(n)}{n} + \int_{fp}^{\infty} \frac{\sum\limits_{fp < n \leq y} \psi(n)}{y^2} dy = \sum\limits_{n = 1}^{fp} \frac{\psi(n)}{n} + O\left( \frac{\log (fp)}{(fp)^{\frac{1}{2}}} \right), \end{eqnarray*} $

从而

$ \begin{eqnarray*} \sum\limits_{h = 1}^{fp}\vphantom{\sum}^{'} \frac{S(h, p;\chi)}{h} = \frac{fp}{\pi^2 \phi(p) \phi(f)} \tau(\overline{\chi}) \sum\limits_{\psi \bmod fp \atop \psi(-1) = -1} L(1, \psi)^2 L(1, \chi\overline{\psi}) + O\left( f^{\frac{3}{2}} p^{\frac{1}{2}} \log(fp) \right). \end{eqnarray*} $

取参数$N$满足$fp\leq N \leq (fp)^3$, 有

$ \begin{eqnarray*} && L^2(1, \psi) = \sum\limits_{n=1}^{\infty} \frac{\psi(n) d(n)}{n} = \sum\limits_{1 \leq n \leq N} \frac{\psi(n) d(n)}{n} + \int_{N}^{\infty} \frac{\sum\limits_{N < n \leq y} \psi(n)d(n)}{y^2}dy, \\ && L(1, \chi\overline{\psi}) = \sum\limits_{n=1}^{\infty} \frac{\chi(n)\overline{\psi}(n)}{ n } = \sum\limits_{1 \leq n \leq N} \frac{\chi(n)\overline{\psi}(n)}{ n } + \int_{N}^{\infty} \frac{\sum\limits_{N < n \leq y}\chi(n)\overline{\psi}(n) }{y^2} dy, \end{eqnarray*} $

其中$ d(n) = \sum_{t | n} 1 $为除数函数.再由Pólya-Vinogradov不等式可得

$ \begin{eqnarray*} \sum\limits_{N < n \leq y} \psi(n) d(n)&=& 2\sum\limits_{n \leq \sqrt{y}} \psi(n) \sum\limits_{m \leq \frac{y}{n}} \psi(m) - ( \sum\limits_{n \leq \sqrt{y}} \psi(n) )^2 \\ && - 2\sum\limits_{n \leq \sqrt{N}} \psi(n) \sum\limits_{m \leq \frac{N}{n}} \psi(m) + ( \sum\limits_{n \leq \sqrt{N}} \psi(n) )^2 \\ &\ll& y^{\frac{1}{2}} (fp)^{\frac{1}{2}} ( \log(fp) )^2, \end{eqnarray*} $

以及

$ \begin{equation*} \sum\limits_{N < n \leq y} \chi(n)\overline{\psi}(n) \ll (fp)^{\frac{1}{2}}\log(fp). \end{equation*} $

则有

$ \begin{eqnarray} && \sum\limits_{h = 1}^{fp}\vphantom{\sum}^{'} \frac{S(h, p;\chi)}{h} = \frac{fp}{\pi^2 \phi(p) \phi(f)} \tau(\overline{\chi}) \sum\limits_{n = 1}^{N} \frac{d(n)}{n} \sum\limits_{m = 1}^{N} \frac{\chi(m)}{m} \sum\limits_{\psi \bmod fp \atop \psi(-1) = -1} \psi(n) \overline{\psi}(m)\nonumber\\ &&\qquad + O \left( \frac{f^{\frac{5}{2}}p^{\frac{3}{2}}\log^2(fp)}{N^{\frac{1}{2}}} \right) + O\left( f^{\frac{3}{2}} p^{\frac{1}{2}} \log(fp) \right). \end{eqnarray} $

(4.1)

由特征的正交关系可得

$ \begin{eqnarray} && \sum\limits_{n = 1}^{N} \frac{d(n)}{n} \sum\limits_{m = 1}^{N} \frac{\chi(m)}{m} \sum\limits_{\psi \bmod fp \atop \psi(-1) = -1} \psi(n) \overline{\psi}(m)\nonumber\\ &=& \frac{1}{2} \sum\limits_{n = 1}^{N} \frac{d(n)}{n} \sum\limits_{ m = 1 }^{N} \frac{\chi(m)}{m} \sum\limits_{\psi \bmod fp} (1 - \psi(-1)) \psi(n) \overline{\psi}(m)\nonumber \\ & =& \frac{ \phi(fp) }{ 2 } \mathop{\sum\limits_{n = 1 \atop (n, fp) = 1}^{N} \sum\limits_{m = 1 \atop (m, fp) = 1}^{N} }_{n \equiv m (\bmod fp)} \frac{d(n)\chi(m)}{nm} - \frac{ \phi(fp) }{ 2 } \mathop{\sum\limits_{n = 1 \atop (n, fp) = 1}^{N} \sum\limits_{m = 1 \atop (m, fp) = 1}^{N} }_{n \equiv -m (\bmod fp)} \frac{d(n)\chi(m)}{nm} \nonumber\\ & :=& \frac{ \phi(fp) }{ 2 } M_1 - \frac{ \phi(fp) }{ 2 } M_2. \end{eqnarray} $

(4.2)

首先考虑$ M_1 $, 有

$ \begin{eqnarray} M_1 &=& \mathop{ \sum\limits_{1 \leq n \leq fp -1 \atop (n, fp) = 1} \sum\limits_{1 \leq m \leq fp -1 \atop (m, fp) = 1} }_{n \equiv m (\bmod fp)} \frac{d(n)\chi(m)}{nm} + \mathop{ \sum\limits_{1 \leq n \leq fp -1 \atop (n, fp) = 1} \sum\limits_{fp \leq m \leq N \atop (m, fp) = 1} }_{n \equiv m (\bmod fp)} \frac{d(n)\chi(m)}{nm} \nonumber\\ &&+ \mathop{ \sum\limits_{fp \leq n \leq N \atop (n, fp) = 1} \sum\limits_{1 \leq m \leq fp -1 \atop (m, fp) = 1} }_{n \equiv m (\bmod fp)} \frac{d(n)\chi(m)}{nm} + \mathop{ \sum\limits_{fp \leq n \leq N \atop (n, fp) = 1} \sum\limits_{fp \leq m \leq N \atop (m, fp) = 1} }_{n \equiv m (\bmod fp)} \frac{d(n)\chi(m)}{nm} \nonumber\\ &:=& \Omega_1 + \Omega_2 + \Omega_3 + \Omega_4. \end{eqnarray} $

(4.3)

易证

$ \begin{eqnarray*} \Omega_1 = \sum\limits_{1 \leq n \leq fp - 1 \atop (n, fp) = 1} \frac{d(n)\chi(n)}{n^2} = \sum\limits_{n = 1 \atop (n, fp) = 1}^{\infty} \frac{d(n)\chi(n)}{n^2} + O( (fp)^{\varepsilon - 1} ). \end{eqnarray*} $

由欧拉乘积公式可得

$ \begin{eqnarray*} &&\sum\limits_{n = 1 \atop (n, fp)=1}^{\infty} \frac{d(n)\chi(n)}{n^2} = \prod\limits_{q \nmid fp} ( 1 + \frac{d(q)\chi(q)}{q^2} + \frac{d(q^2)\chi(q^2)}{q^4} + \cdots + \frac{d(q^k)\chi(q^k)}{q^{2k}} + \cdots ) \\ &&= \prod\limits_{q \nmid fp} ( \sum\limits_{k = 0}^{\infty} \frac{d(q^k)\chi(q^k)}{q^{2k}} ) = \prod\limits_{q \nmid fp} ( \sum\limits_{k = 0}^{\infty} \frac{(k+1)\chi(q)^k}{q^{2k}} ) \\ &&= \prod\limits_{q \nmid fp} \frac{1}{ ( 1-\frac{\chi(q)}{q^2} )^2}= L^2(2, \chi) \prod\limits_{q | fp} ( 1-\frac{\chi(q)}{q^2} )^2. \end{eqnarray*} $

因此

$ \begin{eqnarray} \Omega_1=L^2(2, \chi) \prod\limits_{q | fp} ( 1-\frac{\chi(q)}{q^2} )^2 + O((fp)^{\varepsilon-1}). \end{eqnarray} $

(4.4)

类似可得

$ \begin{eqnarray} \Omega_2&\ll& (fp)^{\epsilon}\mathop{\mathop{\sum\limits_{1\leq n\leq fp-1}}_{(n, fp)=1} \mathop{\sum\limits_{fp\leq m\leq N}}_{(m, fp)=1}}_{n \equiv m(\bmod fp)}\frac{1}{nm}=(fp)^{\epsilon}\mathop{\mathop{\sum\limits_{1\leq n\leq fp-1}}_{(n, fp)=1} \sum\limits_{1\leq l\leq\left[\frac{N}{fp}\right]-1}\mathop{\sum\limits_{0\leq r\leq fp-1}}_{(r, fp)=1}}_{n \equiv l fp+r (\bmod fp)}\frac{1}{n(l fp+r)}\nonumber\\ &\ll& (fp)^{\epsilon}\sum\limits_{1\leq l \leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\frac{1}{r(lfp+r)}\ll (fp)^{\epsilon} \sum\limits_{1\leq l\leq \frac{N}{fp}}\frac{1}{lfp}\sum\limits_{1\leq r\leq fp-1}\frac{1}{r} \nonumber\\ &\ll&(fp)^{\epsilon-1}, \end{eqnarray} $

(4.5)

$ \begin{eqnarray} \Omega_3&\ll& (fp)^{\epsilon}\mathop{\mathop{\sum\limits_{fp\leq n\leq N}}_{(n, fp)=1} \mathop{\sum\limits_{1\leq m\leq fp-1}}_{(m, fp)=1}}_{n \equiv m(\bmod fp)}\frac{1}{nm}=(fp)^{\epsilon}\mathop{\sum\limits_{1\leq l\leq\left[\frac{N}{fp}\right]-1}\mathop{\sum\limits_{0\leq r\leq fp-1}}_{(r, fp)=1} \mathop{\sum\limits_{1\leq m\leq fp-1}}_{(m, fp)=1}}_{lfp+r \equiv m (\bmod fp)}\frac{1}{(l fp+r)m}\nonumber\\ &\ll& (fp)^{\epsilon}\sum\limits_{1\leq l \leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\frac{1}{(lfp+r)r}\ll (fp)^{\epsilon} \sum\limits_{1\leq l\leq \frac{N}{fp}}\frac{1}{lfp}\sum\limits_{1\leq r\leq fp-1}\frac{1}{r} \nonumber\\ &\ll&(fp)^{\epsilon-1}, \end{eqnarray} $

(4.6)

以及

$ \begin{eqnarray*} \Omega_4 &\ll& (fp)^{\epsilon} \mathop{\mathop{\sum\limits_{fp\leq n\leq N}}_{(n, fp)=1} \mathop{\sum\limits_{fp\leq m\leq N}}_{(m, fp)=1}}_{n \equiv m(\bmod fp)}\frac{1}{nm}\nonumber\\ &=& (fp)^{\epsilon}\mathop{\sum\limits_{1\leq l_1\leq \left[\frac{N}{fp}\right]-1}\sum\limits_{0\leq r_1\leq fp-1 \atop (r_1, fp)=1}\sum\limits_{1\leq l_2\leq \left[\frac{N}{fp}\right]-1}\sum\limits_{0\leq r_2\leq fp-1 \atop (r_2, fp)=1}}_{l_1fp+r_1\equiv l_2fp+r_2 (\bmod fp)}\frac{1}{(l_1fp+r_1)(l_2fp+r_2)}\nonumber\\ &\ll& (fp)^{\epsilon}\sum\limits_{1\leq l_1\leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\sum\limits_{1\leq l_2\leq \frac{N}{fp}}\frac{1}{(l_1fp+r)(l_2fp+r)}\nonumber\\ &\ll& (fp)^{\epsilon}\sum\limits_{1\leq l_1\leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\frac{1}{l_1fp+r}\sum\limits_{1\leq l_2\leq \frac{N}{fp}}\frac{1}{l_2fp}\nonumber\\ &\ll& (fp)^{\epsilon-1}. \end{eqnarray*} $

(4.7)

结合(4.3)-(4.7)式有

$ \begin{eqnarray} M_1=L^2(2, \chi) \prod\limits_{q | fp} \left( 1-\frac{\chi(q)}{q^2} \right)^2 + O((fp)^{\varepsilon-1}). \end{eqnarray} $

(4.8)

接下来考虑$M_2$.可得

$ \begin{eqnarray} M_{2}&=&\mathop{\mathop{\sum\limits_{n=1}^{N}}_{(n, fp)=1}\mathop{\sum\limits_{m=1}^{N}}_{(m, fp)=1}}_{n \equiv -m(\bmod fp)}\frac{r(n)\chi(m)}{nm}\ll (fp)^{\epsilon}\mathop{\mathop{\sum\limits_{n=1}^{N}}_{(n, fp)=1}\mathop{\sum\limits_{m=1}^{N}}_{(m, fp)=1}}_{n \equiv -m(\bmod fp)}\frac{1}{nm}\nonumber\\ &\ll& (fp)^{\epsilon}\mathop{\sum\limits_{0\leq l_1\leq \frac{N}{fp}}\sum\limits_{1\leq r_1\leq fp-1}\sum\limits_{0\leq l_2\leq \frac{N}{fp}}\sum\limits_{1\leq r_2\leq fp-1}}_{l_1fp+r_1\equiv -(l_2fp+r_2) (\bmod fp)}\frac{1}{(l_1fp+r_1)(l_2fp+r_2)}\nonumber\\ &\ll& (fp)^{\epsilon}\sum\limits_{0\leq l_1\leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\sum\limits_{0\leq l_2\leq \frac{N}{fp}}\frac{1}{(l_1fp+r)(l_2fp+fp-r)}\nonumber\\ &=& (fp)^{\epsilon}\sum\limits_{1\leq r\leq fp-1}\frac{1}{r(fp-r)}+(fp)^{\epsilon}\sum\limits_{1\leq r\leq fp-1}\sum\limits_{1\leq l_2\leq \frac{N}{fp}}\frac{1}{r(l_2fp+fp-r)}\nonumber\\ &&+(fp)^{\epsilon}\sum\limits_{1\leq l_1\leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\frac{1}{(l_1fp+r)(fp-r)}\nonumber\\ &&+(fp)^{\epsilon}\sum\limits_{1\leq l_1\leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\sum\limits_{1\leq l_2\leq \frac{N}{fp}}\frac{1}{(l_1fp+r)(l_2fp+fp-r)}\nonumber\\ &\ll& (fp)^{\epsilon-1}\sum\limits_{1\leq r\leq fp-1}\left(\frac{1}{r}+\frac{1}{fp-r}\right)+(fp)^{\epsilon}\sum\limits_{1\leq r\leq fp-1}\frac{1}{r}\sum\limits_{1\leq l_2\leq \frac{N}{fp}}\frac{1}{l_2fp}\nonumber\\ &&+(fp)^{\epsilon}\sum\limits_{1\leq l_1\leq \frac{N}{fp}}\frac{1}{l_1fp}\sum\limits_{1\leq r\leq fp-1}\frac{1}{fp-r} + (fp)^{\epsilon}\sum\limits_{1\leq l_1\leq \frac{N}{fp}}\sum\limits_{1\leq r\leq fp-1}\frac{1}{l_1fp+r}\sum\limits_{1\leq l_2\leq \frac{N}{fp}}\frac{1}{l_2fp}\nonumber\\ &\ll& (fp)^{\epsilon-1}. \end{eqnarray} $

(4.9)

结合(4.1), (4.2), (4.8)和(4.9)式, 并取$N=(fp)^3$, 立即可得

$ \begin{eqnarray*} \sum\limits_{h = 1}^{fp}\vphantom{\sum}^{'} \frac{S(h, p;\chi)}{h} = \frac{fp}{2\pi^2} \tau(\overline{\chi}) L^2(2, \chi) \prod\limits_{q | fp} \left( 1-\frac{\chi(q)}{q^2} \right)^2 + O\left( f^{\frac{3}{2}}p^{\frac{1}{2}}\log^2 (fp) \right). \end{eqnarray*} $

定理1.2证毕.