1 Introduction
In the early 1970s, Montgomery [Reference Montgomery11] published his famous paper titled “The pair correlation of zeros of the zeta function”. In this paper, assuming the Riemann hypothesis (RH), he investigated the function
$$\begin{eqnarray}F({\it\alpha},T)=\left(\frac{T}{2{\it\pi}}\log T\right)^{-1}\mathop{\sum }_{\substack{ 0<{\it\gamma}\leqslant T \\ 0<{\it\gamma}^{\prime }\leqslant T}}T^{i{\it\alpha}({\it\gamma}-{\it\gamma}^{\prime })}w({\it\gamma}-{\it\gamma}^{\prime }),\end{eqnarray}$$
where
$w(u)=4/(4+u^{2})$
and
${\it\gamma}$
,
${\it\gamma}^{\prime }$
run over the set of the imaginary parts of the nontrivial zeros of the Riemann zeta-function
${\it\zeta}(s)$
in
$0<\text{Im}(s)\leqslant T$
. He obtained an asymptotic formula for
$F({\it\alpha},T)$
$(0<{\it\alpha}\leqslant 1-{\it\epsilon})$
, and using this formula, he obtained several results on the distances of the nontrivial zeros. For example, under the assumption of the RH, he proved that at least
$2/3-{\it\epsilon}$
of the nontrivial zeros are simple, and that
$$\begin{eqnarray}\liminf _{n\rightarrow \infty }\frac{({\it\gamma}_{n+1}-{\it\gamma}_{n})\log {\it\gamma}_{n}}{2{\it\pi}}\leqslant {\it\lambda}<1\end{eqnarray}$$
holds for specific
${\it\lambda}$
, where
${\it\gamma}_{n}$
denotes the imaginary part of the
$n$
th nontrivial zero of
${\it\zeta}(s)$
in the upper half-plane.
Later, Montgomery’s idea, combined with new conceptions or improvements, was extended to many types of
$L$
-functions or other situations. For example, Özlük [Reference Özlük12] investigated the nontrivial zeros of the Dirichlet
$L$
-functions near the real axis. Assuming the generalized Riemann hypothesis (GRH), he proved that at least
$11/12$
of such zeros are simple in some sense. One of the other interesting generalizations is the work of Hejhal [Reference Hejhal7]. From the explicit formula of the Riemann zeta-function introduced in another of his papers [Reference Hejhal6], he constructed a certain asymptotic formula for the function involving the pairs of three distinct zeros of
${\it\zeta}(s)$
. Further, the result of Hejhal was generalized by Rudnick and Sarnak [Reference Rudnick and Sarnak16], and the
$n$
-level correlation of the zeros of principal
$L$
-functions was obtained. In particular, their results agree with the prediction for the Gaussian unitary ensemble of random matrix theory.
Our aim in this paper is to investigate the pair correlation of the zeros of the quadratic
$L$
-functions near the real axis. As prior research, Özlük and Snyder [Reference Özlük and Snyder13] investigated such zeros. Under the assumption of the GRH, they studied the asymptotic behavior of the function
$$\begin{eqnarray}G_{K}({\it\alpha},D)=\left(\frac{1}{2}K\left(\frac{1}{2}\right)D\right)^{-1}\mathop{\sum }_{d\neq 0}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{{\it\rho}\in Z_{d}}K({\it\rho})D^{i{\it\alpha}{\it\gamma}}\end{eqnarray}$$
as
$D\rightarrow \infty$
for
$|{\it\alpha}|<2$
, where
${\it\rho}=1/2+i{\it\gamma}$
runs over the set of all nontrivial zeros of
$L(s,{\it\chi}_{d})$
, the quadratic
$L$
-function associated to the Kronecker symbol
${\it\chi}_{d}=(d/\cdot )$
, and
$K(s)$
is some weight function. From their asymptotic formula, they proved that assuming the GRH, not more than
$6.25\%$
of all integers
$d$
have the property that
$s=1/2$
is a zero of
$L(s,{\it\chi}_{d})$
. Slightly later, by a completely different method, Soundararajan [Reference Soundararajan18] unconditionally proved that
$L(1/2,{\it\chi}_{d})\neq 0$
for at least
$87.5\%$
of all fundamental discriminants
$d$
. On the other hand, there are several researches on the ‘
$n$
-level density’ of the low-lying zeros of quadratic
$L$
-functions. For an odd, square-free integer
$d>0$
,
${\it\chi}_{8d}=\left(\frac{8d}{\cdot }\right)$
becomes a primitive character. Assuming the GRH, we denote the nontrivial zeros of
$L(s,{\it\chi}_{8d})$
by
$$\begin{eqnarray}{\textstyle \frac{1}{2}}+i{\it\gamma}_{8d,j}\quad (j=\pm 1,\pm 2,\ldots ),\end{eqnarray}$$
where
$0\leqslant {\it\gamma}_{8d,1}\leqslant {\it\gamma}_{8d,2}\leqslant \cdots \,$
and
${\it\gamma}_{8d,-j}=-{\it\gamma}_{8d,j}$
. For
$X>0$
, we put
$$\begin{eqnarray}{\mathcal{D}}(X):=\{X\leqslant d\leqslant 2X\mid d:\text{odd, square free}\},\end{eqnarray}$$
and for a Schwartz function
$f\in {\mathcal{S}}(\mathbf{R}^{n})$
, we put
$$\begin{eqnarray}W_{f}^{(n)}(d):=\mathop{\sum }_{\substack{ j_{1},\ldots ,j_{n}=\pm 1,\pm 2,\ldots \\ |j_{k}|:\text{distinct}}}f\left(\frac{{\it\gamma}_{8d,j_{1}}\log X}{2{\it\pi}},\ldots ,\frac{{\it\gamma}_{8d,j_{n}}\log X}{2{\it\pi}}\right).\end{eqnarray}$$
Then, the Katz–Sarnak density conjecture [Reference Katz and Sarnak9] asserts that
$$\begin{eqnarray}\displaystyle \langle W_{f}^{(n)}\rangle _{{\mathcal{D}}(X)}:=\frac{1}{\sharp {\mathcal{D}}(X)}\mathop{\sum }_{d\in {\mathcal{D}}(X)}W_{f}^{(n)}(d){\rm\Phi}\left(\frac{d}{X}\right)\rightarrow \int _{\mathbf{R}^{n}}f(x)W_{\text{USp}}^{(n)}(x)\,dx & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
as
$X\rightarrow \infty$
, where
${\rm\Phi}$
is a nonnegative smooth function supported in
$(1,2)$
satisfying
$\int {\rm\Phi}(x)\,dx=1$
, and
$$\begin{eqnarray}W_{\text{USp}}^{(n)}(x)=\text{det}(K(x_{i},x_{j}))_{i,j=1,\ldots ,n},\end{eqnarray}$$
with
$$\begin{eqnarray}K(x,y)=\frac{\sin {\it\pi}(x-y)}{{\it\pi}(x-y)}-\frac{\sin {\it\pi}(x+y)}{{\it\pi}(x+y)}.\end{eqnarray}$$
Katz and Sarnak [Reference Katz and Sarnak9], assuming the GRH, proved that (1.1) holds if
$n=1$
and
$\hat{f}(u)=\int _{\mathbf{R}}f(x)e^{-2{\it\pi}iux}\,dx$
has a support in
$|u|<2$
. Rubinstein [Reference Rubinstein14], [Reference Rubinstein15], assuming the GRH, established (1.1) under the condition that
$\hat{f}(u)=\int _{\mathbf{R}^{n}}f(u)e^{-2{\it\pi}iu\cdot x}\,dx$
has a support in
$\sum _{j=1}^{n}|u_{j}|<1$
, and later Gao [Reference Gao3], [Reference Gao4], under the GRH, proved that if
$f$
is of the form
$f(x_{1},\ldots ,x_{n})=\prod _{j=1}^{n}f_{j}(x_{j})$
and each
$\hat{f}_{j}$
is supported in
$|u_{j}|<s_{j}$
with
$\sum s_{j}<2$
, then
$$\begin{eqnarray}\lim _{X\rightarrow \infty }\langle W_{f}^{(n)}\rangle _{{\mathcal{D}}(X)}=A(f),\end{eqnarray}$$
where
$A(f)$
is some complicated expression involving
$f_{1},\ldots ,f_{n}$
. Moreover, he confirmed that
$A(f)$
is equal to the right-hand side of (1.1) if
$n=2,3$
. More recently, Levinson and Miller [Reference Levinson and Miller10] proved that this fact is also valid for
$4\leqslant n\leqslant 7$
. Finally, Entin, Roditty-Gershon and Rudnick [Reference Entin, Roditty-Gershon and Rudnick2] proved that assuming the GRH, (1.1) is valid for all
$n$
if
$\hat{f}$
is supported in
$\sum _{j=1}^{n}|u_{j}|<2$
.
If anything, our approach to investigate the pair correlation of low-lying zeros is similar to that of Özlük and Snyder [Reference Özlük and Snyder13], in which they investigated the 1-level density of these zeros. In this paper, assuming the GRH (including the RH), we investigate the function
$F_{K}({\it\alpha},D)$
, defined as follows. Let
$K(s)$
be analytic in
$-1<\text{Re}(s)<2$
and satisfy
$K(1/2-it)=K(1/2+it)$
for any
$t\in \mathbf{R}$
. Moreover, we assume that its Mellin inverse transform
$$\begin{eqnarray}a(x):=\frac{1}{2{\it\pi}i}\int _{c-i\infty }^{c+i\infty }K(s)x^{-s}\,ds\end{eqnarray}$$
converges absolutely for any
$-1<c<2$
,
$x>0$
, and that
$a(x)$
is real, nonnegative, belongs to the
$C^{1}$
-class, and has a support in
$[A,B]$
for some
$0<A<B<\infty$
. Then,
$K(s)$
is given by
$$\begin{eqnarray}K(s)=\int _{0}^{\infty }a(t)t^{s}\,\frac{dt}{t}.\end{eqnarray}$$
For
$d\in \mathbf{Z}$
, let
${\it\chi}_{d}=(d/\cdot )$
be the Kronecker symbol, and let
$L(s,{\it\chi}_{d})$
be the
$L$
-function associated to
${\it\chi}_{d}$
. We denote the set of nontrivial zeros of
$L(s,{\it\chi}_{d})$
by
$Z_{d}$
. For
$x>0$
,
$D>0$
, we put
$$\begin{eqnarray}f_{K}(x,D)=\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}}K({\it\rho}_{1})\overline{K({\it\rho}_{2})}x^{{\it\rho}_{1}+\overline{{\it\rho}_{2}}},\end{eqnarray}$$
and for
${\it\alpha}\in \mathbf{R}$
, we put
$$\begin{eqnarray}\displaystyle F_{K}({\it\alpha},D) & = & \displaystyle \left[\frac{1}{xD\log D}f_{K}(x,D)\right]_{x=D^{{\it\alpha}}}\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}}K({\it\rho}_{1})\overline{K({\it\rho}_{2})}D^{i{\it\alpha}({\it\gamma}_{1}-{\it\gamma}_{2})},\end{eqnarray}$$
where
${\it\rho}_{j}=1/2+i{\it\gamma}_{j}$
for
$j=1,2$
. Then, the main theorem is stated as follows.
Theorem 1.1. Assuming the GRH, for any small
${\it\delta}>0$
, we have
$$\begin{eqnarray}\displaystyle F_{K}({\it\alpha},D) & = & \displaystyle L(1){\it\alpha}+a(D^{-{\it\alpha}})^{2}D^{-{\it\alpha}}\log D+a(D^{-{\it\alpha}})\cdot O({\it\alpha}D^{-{\it\alpha}/2}\log D)\nonumber\\ \displaystyle & & \displaystyle +\,a(D^{-{\it\alpha}})^{2}\cdot O(D^{-{\it\alpha}})+O(\text{min}\{1,{\it\alpha}D^{-{\it\alpha}}\log ^{2}D\})\nonumber\\ \displaystyle & & \displaystyle +\,O(\min \{{\it\alpha}\log D,{\it\alpha}D^{-1/2}\log D+{\it\alpha}(1+{\it\alpha}^{2})D^{-{\it\alpha}/2}\log ^{3}D\})\nonumber\\ \displaystyle & & \displaystyle +\,O(\text{min}\{D^{{\it\alpha}}(\log D)^{-1},{\it\alpha}^{2}D^{-{\it\alpha}}\log ^{3}D\})+o(1)\end{eqnarray}$$
uniformly for
$0<{\it\alpha}<1-{\it\delta}$
as
$D\rightarrow \infty$
, where
$$\begin{eqnarray}L(1)=\int _{0}^{\infty }a(x)^{2}\,dx.\end{eqnarray}$$
The implied constants depend only on
$K(s)$
and
${\it\delta}>0$
.
It should be noticed that the result on the
$1$
-level density by Katz and Sarnak [Reference Katz and Sarnak9] is stronger than that of Özlük and Snyder [Reference Özlük and Snyder13] in some sense. However, the author believes that the asymptotic formula of Theorem 1.1 is more useful than the limit formula (1.1) (
$n=2$
) if we restrict the target only to the study of the pair correlation of low-lying zeros of
$L(s,{\it\chi}_{d})$
, since the factor
${\it\gamma}_{1}-{\it\gamma}_{2}$
is contained in the definition of
$F_{K}({\it\alpha},D)$
, and both the left-hand side and the main terms of the right-hand side of (1.5) are very simple; hence, we can easily compute the integrals involving these terms. In fact, several concrete results on the average gaps of the nontrivial zeros are obtained. Section 4 of this paper is devoted to their study. In Corollary 4.2, we give a certain upper bound for the number of pairs of ‘close zeros’ near the real axis. In Corollary 4.3, we give a lower bound for the weighted sum involving simple zeros of
$L(s,{\it\chi}_{d})$
. Finally, in Corollary 4.4, we prove that there are quite a few pairs of zeros
$(1/2+i{\it\gamma}_{1},1/2+i{\it\gamma}_{2})$
of
$L(s,{\it\chi}_{d})$
$(d\in \mathbf{Z}\setminus \{0\})$
satisfying
$0<|{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant (2{\it\pi}{\it\lambda})/\log D$
, if
${\it\lambda}$
is large to a certain extent.
2 Preliminaries
To prove the main theorem, we prepare several lemmas. The following nine lemmas (Lemmas 2.1–2.9) are all found in [Reference Özlük and Snyder13].
Lemma 2.1. We have
$$\begin{eqnarray}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/y^{2}}=y+o(1)\quad (y\rightarrow \infty ).\end{eqnarray}$$
Here,
$\sum _{d}$
denotes the sum over all nonzero integers
$d$
.
Lemma 2.2.
$$\begin{eqnarray}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/y^{2}}\log |d|=y\log y+O(y)\quad (y\rightarrow \infty ).\end{eqnarray}$$
Lemma 2.3. We have
$$\begin{eqnarray}\mathop{\sum }_{d=\Box }e^{-{\it\pi}d^{2}/y^{2}}=Iy^{1/2}-\frac{1}{2}+O(y^{-1/2})\end{eqnarray}$$
as
$y\rightarrow \infty$
. Here,
$I=(1/4){\it\pi}^{-1/4}{\rm\Gamma}(1/4)$
, and
$\sum _{d=\Box }$
denotes the sum over all positive square integers.
Instead of (2.3), sometimes we use
$$\begin{eqnarray}\mathop{\sum }_{d=\Box }e^{-{\it\pi}d^{2}/y^{2}}=Iy^{1/2}-\frac{1}{2}+O(\text{min}\{1,y^{-1/2}\})\end{eqnarray}$$
or
$$\begin{eqnarray}\mathop{\sum }_{d=\Box }e^{-{\it\pi}d^{2}/y^{2}}=Iy^{1/2}+O(1).\end{eqnarray}$$
Lemma 2.4. Assuming the RH, we have
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)\log p=\frac{1}{2}K\left(\frac{1}{2}\right)x^{1/2}+O(x^{1/4}\log ^{2}x)\quad (x\rightarrow \infty ).\end{eqnarray}$$
Here,
$\sum _{p}$
denotes the sum over all primes
$p$
.
It should be noticed that the implied constant in (2.6) is dependent on
$a(x)$
(and hence on
$K(s)$
). Hereafter, this fact will be valid for almost all asymptotic formulas, although we will not comment again.
Lemma 2.5. Assuming the RH, we have
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\log p=K\left(1\right)x+O(x^{1/2}\log ^{2}x)\quad (x\rightarrow \infty ).\end{eqnarray}$$
Lemma 2.6. Assuming the RH, we have
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\frac{\log p}{\sqrt{p}}=K\left(\frac{1}{2}\right)x^{1/2}+O(\log ^{2}x)\quad (x\rightarrow \infty ).\end{eqnarray}$$
Lemma 2.7. Assuming the RH, we have
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\frac{\log p}{p}=K\left(0\right)+O(x^{-1/2}\log ^{2}x)\quad (x\rightarrow \infty ).\end{eqnarray}$$
Lemma 2.8. Assuming the RH, we have
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)\frac{\log p}{p}=\frac{1}{2}K\left(0\right)+O(x^{-1/4}\log ^{2}x)\quad (x\rightarrow \infty ).\end{eqnarray}$$
Lemma 2.9. We have
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)p^{1/2}\log ^{2}p\ll x^{3/2}\log x\quad (x\rightarrow \infty ).\end{eqnarray}$$
In addition to these nine lemmas quoted from [Reference Özlük and Snyder13], we need several asymptotic formulas.
Lemma 2.10. We have
$$\begin{eqnarray}\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{2}/y^{2}}\log ^{2}d=\frac{1}{2}y\log ^{2}y+O(y\log y)\quad (y\rightarrow \infty ).\end{eqnarray}$$
Proof. The left-hand side is
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{2}/y^{2}}\log ^{2}d & = & \displaystyle \mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{2}/y^{2}}\log ^{2}\left(\frac{d}{y}\right)+2(\log y)\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{2}/y^{2}}\log d\nonumber\\ \displaystyle & & \displaystyle -\,(\log y)^{2}\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{2}/y^{2}}.\end{eqnarray}$$
The first term on the right-hand side of (2.13) is
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{2}/y^{2}}\log ^{2}\left(\frac{d}{y}\right) & = & \displaystyle \int _{1^{-}}^{\infty }e^{-{\it\pi}u^{2}/y^{2}}\log ^{2}\left(\frac{u}{y}\right)d[u]\nonumber\\ \displaystyle & = & \displaystyle \int _{1}^{\infty }e^{-{\it\pi}u^{2}/y^{2}}\log ^{2}\left(\frac{u}{y}\right)du\nonumber\\ \displaystyle & & \displaystyle -\,\int _{1^{-}}^{\infty }e^{-{\it\pi}u^{2}/y^{2}}\log ^{2}\left(\frac{u}{y}\right)d\{u\},\end{eqnarray}$$
where
$[u]$
denotes the integer part of
$u$
, and
$\{u\}:=u-[u]$
. By the change of parameters
$u/y=v$
, we can easily see that
$$\begin{eqnarray}\int _{1}^{\infty }e^{-{\it\pi}u^{2}/y^{2}}\log ^{2}\left(\frac{u}{y}\right)du\ll y,\end{eqnarray}$$
and by partial integration, we have
$$\begin{eqnarray}\int _{1^{-}}^{\infty }e^{-{\it\pi}u^{2}/y^{2}}\log ^{2}\left(\frac{u}{y}\right)d\{u\}\ll \log ^{2}y.\end{eqnarray}$$
Hence, the first term on the right-hand side of (2.13) is
$O(y)$
. Moreover, by Lemma 2.1, we have
$$\begin{eqnarray}\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{2}/y^{2}}=\frac{1}{2}y+o(1),\end{eqnarray}$$
and by Lemma 2.2, we have
$$\begin{eqnarray}\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{2}/y^{2}}\log d=\frac{1}{2}y\log y+O(y).\end{eqnarray}$$
By inserting these into (2.13), we obtain the result.◻
Lemma 2.11. We have
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)p\log p\ll x^{2}\quad (x\rightarrow \infty ).\end{eqnarray}$$
Proof. It should be recalled that the function
$a(u)$
is bounded and has a support in
$[A,B]$
. By the prime number theorem (PNT), the number of primes
$p$
satisfying the condition
$p/x\leqslant B$
is
$O(x/\log x)$
. Therefore,
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)p\log p\ll x\log x\cdot \frac{x}{\log x}=x^{2}.\end{eqnarray}$$
By a similar argument, we obtain the following.
Lemma 2.12. We have
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)p\log p\ll x\quad (x\rightarrow \infty ).\end{eqnarray}$$
Hereafter, we put
$$\begin{eqnarray}L(s):=\int _{0}^{\infty }a(t)^{2}t^{s}\,\frac{dt}{t}.\end{eqnarray}$$
Lemma 2.13. Assuming the RH, we have
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)^{2}(\log p)^{2}=L(1)x\log x+L^{\prime }(1)x+O(\sqrt{x}\log ^{3}x)\quad (x\rightarrow \infty ).\end{eqnarray}$$
Proof. We write
$$\begin{eqnarray}{\it\theta}(u):=\mathop{\sum }_{p\leqslant u}\log p=u+E(u).\end{eqnarray}$$
Then, it is well known that
$E(u)$
is evaluated by
$O(u^{1/2}\log ^{2}u)$
under the assumption of the RH. Now, we have
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)^{2}(\log p)^{2} & = & \displaystyle \int _{0}^{\infty }a\left(\frac{u}{x}\right)^{2}\log u\,d{\it\theta}(u)\nonumber\\ \displaystyle & = & \displaystyle \int _{0}^{\infty }a\left(\frac{u}{x}\right)^{2}\log u\,du+\int _{0}^{\infty }a\left(\frac{u}{x}\right)^{2}\log u\,dE(u).\end{eqnarray}$$
By the change of parameters
$u/x=v$
, we have
$$\begin{eqnarray}\displaystyle \int _{0}^{\infty }a\left(\frac{u}{x}\right)^{2}\log u\,du & = & \displaystyle x\log x\int _{0}^{\infty }a(v)^{2}\,dv+x\int _{0}^{\infty }a(v)^{2}\log v\,dv\nonumber\\ \displaystyle & = & \displaystyle L(1)x\log x+L^{\prime }(1)x.\end{eqnarray}$$
On the other hand, by partial integration, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{\infty }a\left(\frac{u}{x}\right)^{2}\log u\,dE(u)\nonumber\\ \displaystyle & & \displaystyle \quad =-\int _{0}^{\infty }E(u)\left\{\frac{2a\left(\frac{u}{x}\right)a^{\prime }\left(\frac{u}{x}\right)\log u}{x}+\frac{a\left(\frac{u}{x}\right)^{2}}{u}\right\}\,du,\nonumber\end{eqnarray}$$
and by the change of parameters
$u/x=v$
, combining with the estimate
$E(u)\ll \sqrt{u}\log ^{2}u$
, we easily find that
$$\begin{eqnarray}\displaystyle & \displaystyle \int _{0}^{\infty }E(u)\frac{a\left(\frac{u}{x}\right)a^{\prime }\left(\frac{u}{x}\right)\log u}{x}\,du\ll \sqrt{x}\log ^{3}x, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \int _{0}^{\infty }E(u)\frac{a\left(\frac{u}{x}\right)^{2}}{u}\,du\ll \sqrt{x}\log ^{2}x. & \displaystyle \nonumber\end{eqnarray}$$
Hence, we get
$$\begin{eqnarray}\int _{0}^{\infty }a\left(\frac{u}{x}\right)^{2}\log u\,dE(u)\ll \sqrt{x}\log ^{3}x.\end{eqnarray}$$
Lemma 2.14. Assuming the RH, we have
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)^{2}\frac{(\log p)^{2}}{p}=L(0)\log x+L^{\prime }(0)+O(x^{-1/2}\log ^{3}x)\quad (x\rightarrow \infty ). & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
The proof of Lemma 2.14 is almost the same as that of Lemma 2.13; hence, we omit it.
Lemma 2.15. We assume the RH. Then, when
$x^{\frac{1}{l}}\gg 1$
, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p}\mathop{\sum }_{q\neq p}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\nonumber\\ \displaystyle & & \displaystyle \quad =l^{-1}K(1)K\left(\frac{1}{l}\right)x^{1+\frac{1}{l}}+O(l^{-1}x^{1+\frac{1}{2l}}\log ^{2}x).\end{eqnarray}$$
The implied constant is independent of
$l$
.
Proof. The left-hand side is
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p}\mathop{\sum }_{q\neq p}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\nonumber\\ \displaystyle & & \displaystyle \quad =-\mathop{\sum }_{p}a\left(\frac{p}{x}\right)a\left(\frac{p^{l}}{x}\right)\log ^{2}p+\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\log p\mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\log q.\end{eqnarray}$$
By the PNT, the number of primes
$p$
satisfying
$p^{l}/x\leqslant B$
is
$O(lx^{1/l}/\log x)$
. Therefore,
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)a\left(\frac{p^{l}}{x}\right)\log ^{2}p\ll l^{-1}x^{\frac{1}{l}}\log x.\end{eqnarray}$$
By Lemma 2.5,
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\log p=K(1)x+O(x^{1/2}\log ^{2}x).\end{eqnarray}$$
Further,
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\log q & = & \displaystyle \int _{0}^{\infty }a\left(\frac{u^{l}}{x}\right)d{\it\theta}(u)\nonumber\\ \displaystyle & = & \displaystyle \int _{0}^{\infty }a\left(\frac{u^{l}}{x}\right)du+\int _{0}^{\infty }a\left(\frac{u^{l}}{x}\right)dE(u).\nonumber\end{eqnarray}$$
By the change of parameters
$u^{l}/x=v$
, the first term on the right-hand side becomes
$$\begin{eqnarray}\int _{0}^{\infty }a\left(\frac{u^{l}}{x}\right)du=l^{-1}x^{\frac{1}{l}}K\left(\frac{1}{l}\right).\end{eqnarray}$$
Since
$E(u)\ll u^{1/2}\log ^{2}u$
, and since
$a^{\prime }(v)$
is bounded and has a support in
$[A,B]$
, the second integral is
$$\begin{eqnarray}\displaystyle \int _{0}^{\infty }a\left(\frac{u^{l}}{x}\right)dE(u) & = & \displaystyle -\int _{0}^{\infty }E(u)\frac{lu^{l-1}}{x}a^{\prime }\left(\frac{u^{l}}{x}\right)du\nonumber\\ \displaystyle & \ll & \displaystyle lx^{-1}\int _{0}^{\infty }u^{l-\frac{1}{2}}\left|a^{\prime }\left(\frac{u^{l}}{x}\right)\right|\log ^{2}u\,du\nonumber\\ \displaystyle & \ll & \displaystyle lx^{-1}(x^{\frac{1}{l}})^{l-\frac{1}{2}}\log ^{2}(x^{\frac{1}{l}})\cdot x^{\frac{1}{l}}\nonumber\\ \displaystyle & \ll & \displaystyle l^{-1}x^{\frac{1}{2l}}\log ^{2}x.\nonumber\end{eqnarray}$$
Combining these, we have
$$\begin{eqnarray}\mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\log q=l^{-1}x^{\frac{1}{l}}K\left(\frac{1}{l}\right)+O(l^{-1}x^{\frac{1}{2l}}\log ^{2}x).\end{eqnarray}$$
By inserting (2.25)–(2.27) into (2.24), we obtain the result.◻
Lemma 2.16. We assume the RH. Then, when
$x^{\frac{1}{l}}\gg 1$
, we have
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{p}\mathop{\sum }_{q\neq p}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)\frac{(\log p)(\log q)}{\sqrt{pq}} & = & \displaystyle l^{-1}K\left(\frac{1}{2}\right)K\left(\frac{1}{2l}\right)x^{1/2+\frac{1}{2l}}\nonumber\\ \displaystyle & & \displaystyle +\,O(l^{-1}x^{1/2}\log ^{3}x).\end{eqnarray}$$
The implied constant is independent of
$l$
.
The proof of Lemma 2.16 is almost the same as that of Lemma 2.15; hence, we omit it. It should be noticed that the asymptotic formulas of Lemmas 2.15, 2.16 are still valid if we replace the sum with
$p,q\geqslant 3$
,
$p\neq q$
.
To obtain the asymptotic formula of the main theorem, we need the following translation formula for the theta function.
Lemma 2.17. Let
$p,q\geqslant 3$
be distinct primes, and let
$D>0$
. Then,
$$\begin{eqnarray}\mathop{\sum }_{d}\left(\frac{d}{pq}\right)e^{-{\it\pi}d^{2}/D^{2}}=\frac{D}{\sqrt{pq}}\mathop{\sum }_{m}\left(\frac{m}{pq}\right)e^{-{\it\pi}m^{2}D^{2}/p^{2}q^{2}}.\end{eqnarray}$$
Here,
$d$
and
$m$
above run over the set of all nonzero integers.
Proof. If
$(-1/pq)=-1$
, the identity (2.29) clearly holds, since both sides become 0. We assume
$(-1/pq)=1$
, and put
$$\begin{eqnarray}{\it\chi}_{pq}(n)=\left(\frac{n}{pq}\right).\end{eqnarray}$$
Then,
${\it\chi}_{pq}$
becomes a primitive character of modulo
$pq$
. We define the theta function
${\it\psi}(x,{\it\chi}_{pq})$
by
$$\begin{eqnarray}{\it\psi}(x,{\it\chi}_{pq})=\mathop{\sum }_{n=-\infty }^{\infty }{\it\chi}_{pq}(n)e^{-n^{2}{\it\pi}x/pq}\end{eqnarray}$$
for
$x>0$
. Then,
${\it\psi}(x,{\it\chi}_{pq})$
satisfies
$$\begin{eqnarray}{\it\tau}({\it\chi}_{pq}){\it\psi}(x,{\it\chi}_{pq})=\left(\frac{pq}{x}\right)^{1/2}{\it\psi}(x^{-1},{\it\chi}_{pq}),\end{eqnarray}$$
where
${\it\tau}({\it\chi}_{pq})$
is the Gaussian sum associated to
${\it\chi}_{pq}$
(for example, see [Reference Davenport1, p. 67]). Moreover, we have
${\it\tau}({\it\chi}_{pq})=\sqrt{pq}$
. Hence, we get
$$\begin{eqnarray}\mathop{\sum }_{n=-\infty }^{\infty }\left(\frac{n}{pq}\right)e^{-n^{2}{\it\pi}x/pq}=x^{-1/2}\mathop{\sum }_{m=-\infty }^{\infty }\left(\frac{m}{pq}\right)e^{-{\it\pi}m^{2}/pqx}.\end{eqnarray}$$
By putting
$x=pq/D^{2}$
, we obtain (2.29).◻
3 The proof of Theorem 1.1
We start from the explicit formula
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{{\it\rho}\in Z_{d}}K({\it\rho})x^{{\it\rho}} & = & \displaystyle K(1)E({\it\chi}_{d})x-\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right)\nonumber\\ \displaystyle & & \displaystyle +\,a\left(\frac{1}{x}\right)\log \left(\frac{|d|}{{\it\pi}}\right)+O(\text{min}\{x,\log |d|\log x\})\quad (x\geqslant 1),\nonumber\end{eqnarray}$$
introduced in [Reference Özlük and Snyder13]. Here,
$E({\it\chi})=1$
if
${\it\chi}$
is a principal character, and otherwise
$E({\it\chi})=0$
. The error term is interpreted as
$O(1)$
if
$x=1$
. Since the main terms on the right-hand side are real, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}}K({\it\rho}_{1})\overline{K({\it\rho}_{2})}x^{{\it\rho}_{1}+\overline{{\it\rho}_{2}}}\nonumber\\ \displaystyle & & \displaystyle \quad =\left\{K(1)E({\it\chi}_{d})x-\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right)\right.\nonumber\\ \displaystyle & & \displaystyle \left.\quad \quad \quad \quad \quad \quad \quad \quad +\,a\left(\frac{1}{x}\right)\log \left(\frac{|d|}{{\it\pi}}\right)+O(\text{min}\{x,\log |d|\log x\})\right\}^{2}\nonumber\\ \displaystyle & & \displaystyle \quad =K(1)^{2}E({\it\chi}_{d})^{2}x^{2}+\mathop{\sum }_{n,m=1}^{\infty }a\left(\frac{n}{x}\right)a\left(\frac{m}{x}\right){\rm\Lambda}(n){\rm\Lambda}(m)\left(\frac{d}{n}\right)\left(\frac{d}{m}\right)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,a\left(\frac{1}{x}\right)^{2}\log ^{2}\left(\frac{|d|}{{\it\pi}}\right)-2K(1)E({\it\chi}_{d})x\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,2K(1)E({\it\chi}_{d})xa\left(\frac{1}{x}\right)\log \left(\frac{|d|}{{\it\pi}}\right)\nonumber\\ \displaystyle & & \displaystyle \qquad -\,2a\left(\frac{1}{x}\right)\log \left(\frac{|d|}{{\it\pi}}\right)\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O\left(\max \left\{K(1)E({\it\chi}_{d})x,\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right),a\left(\frac{1}{x}\right)\log \left(\frac{|d|}{{\it\pi}}\right)\right\}\right)\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,O\left(\text{min}\{x,\log |d|\log x\}\right)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O\left(\text{min}\{x^{2},\log ^{2}|d|\log ^{2}x\}\right).\nonumber\end{eqnarray}$$
By multiplying both sides by
$e^{-{\it\pi}d^{2}/D^{2}}$
and taking the sum over
$d$
, we have
$$\begin{eqnarray}f_{K}(x,D)=M_{1}+M_{2}+M_{3}+M_{4}+M_{5}+M_{6}+O_{1}+O_{2}+O_{3}+O_{4},\end{eqnarray}$$
where
$$\begin{eqnarray}\displaystyle & \displaystyle M_{1}=K(1)^{2}x^{2}\mathop{\sum }_{d}E({\it\chi}_{d})^{2}e^{-{\it\pi}d^{2}/D^{2}}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle M_{2}=\mathop{\sum }_{n,m=1}^{\infty }a\left(\frac{n}{x}\right)a\left(\frac{m}{x}\right){\rm\Lambda}(n){\rm\Lambda}(m)\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\left(\frac{d}{n}\right)\left(\frac{d}{m}\right), & \displaystyle \nonumber\\ \displaystyle & \displaystyle M_{3}=a\left(\frac{1}{x}\right)^{2}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log ^{2}\left(\frac{|d|}{{\it\pi}}\right), & \displaystyle \nonumber\\ \displaystyle & \displaystyle M_{4}=-2K(1)x\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}E({\it\chi}_{d})\left(\frac{d}{n}\right), & \displaystyle \nonumber\\ \displaystyle & \displaystyle M_{5}=2K(1)xa\left(\frac{1}{x}\right)\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}E({\it\chi}_{d})\log \left(\frac{|d|}{{\it\pi}}\right), & \displaystyle \nonumber\\ \displaystyle & \displaystyle M_{6}=-2a\left(\frac{1}{x}\right)\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\left(\frac{d}{n}\right)\log \left(\frac{|d|}{{\it\pi}}\right), & \displaystyle \nonumber\\ \displaystyle & \displaystyle O_{1}=O\left(\text{min}\left\{O_{11},O_{12}\right\}\right),\qquad O_{2}=O\left(\text{min}\left\{O_{21},O_{22}\right\}\right), & \displaystyle \nonumber\\ \displaystyle & \displaystyle O_{3}=O\left(\text{min}\left\{O_{31},O_{32}\right\}\right),\qquad O_{4}=\left(\text{min}\left\{O_{41},O_{42}\right\}\right), & \displaystyle \nonumber\end{eqnarray}$$
with
$$\begin{eqnarray}\displaystyle & \displaystyle O_{11}=K(1)x^{2}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}E({\it\chi}_{d}), & \displaystyle \nonumber\\ \displaystyle & \displaystyle O_{12}=K(1)x\log x\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}E({\it\chi}_{d})\log |d|, & \displaystyle \nonumber\\ \displaystyle & \displaystyle O_{21}=x\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\left|\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right)\right|, & \displaystyle \nonumber\\ \displaystyle & \displaystyle O_{22}=\log x\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log |d|\left|\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right)\right|, & \displaystyle \nonumber\\ \displaystyle & \displaystyle O_{31}=xa\left(\frac{1}{x}\right)\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\left|\log \left(\frac{|d|}{{\it\pi}}\right)\right|, & \displaystyle \nonumber\\ \displaystyle & \displaystyle O_{32}=a\left(\frac{1}{x}\right)\log x\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log |d|\left|\log \left(\frac{|d|}{{\it\pi}}\right)\right|, & \displaystyle \nonumber\\ \displaystyle & \displaystyle O_{41}=x^{2}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}},\quad O_{42}=\log ^{2}x\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log ^{2}|d|. & \displaystyle \nonumber\end{eqnarray}$$
3.1 Evaluations of
$O_{1}$
,
$O_{2}$
,
$O_{3}$
,
$O_{4}$
First, we evaluate the error terms
$O_{i}$
$(i=1,2,3,4)$
. By Lemma 2.3,
$$\begin{eqnarray}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}E({\it\chi}_{d})=\mathop{\sum }_{d=\Box }e^{-{\it\pi}d^{2}/D^{2}}\ll D^{1/2}.\end{eqnarray}$$
Combining this with abelian summation, we find that
$$\begin{eqnarray}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}E({\it\chi}_{d})\log |d|\ll D^{1/2}\log D.\end{eqnarray}$$
Therefore, we have
$$\begin{eqnarray}O_{1}\ll \text{min}\{x^{2}D^{1/2},xD^{1/2}\log x\log D\}.\end{eqnarray}$$
By Lemmas 2.2 and 2.10, we have
$$\begin{eqnarray}\displaystyle & \displaystyle \mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\left|\log \left(\frac{|d|}{{\it\pi}}\right)\right|\ll D\log D, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log |d|\left|\log \left(\frac{|d|}{{\it\pi}}\right)\right|\ll D\log ^{2}D. & \displaystyle \nonumber\end{eqnarray}$$
Hence, we obtain
$$\begin{eqnarray}O_{3}\ll \text{min}\{xD\log D,D\log x\log ^{2}D\}.\end{eqnarray}$$
Next, we evaluate
$O_{2}$
. In particular, we evaluate
$O_{22}$
in two ways. First, since
$$\begin{eqnarray}\left|\mathop{\sum }_{n}a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right)\right|\leqslant \mathop{\sum }_{n}a\left(\frac{n}{x}\right){\rm\Lambda}(n)\ll x,\end{eqnarray}$$
by Lemma 2.2, we have
$$\begin{eqnarray}O_{22}\ll x\log x\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log |d|\ll xD\log x\log D.\end{eqnarray}$$
On the other hand, we decompose
$$\begin{eqnarray}\begin{array}{@{}c@{}}\displaystyle O_{22}=\log x\left\{\vphantom{+\,\mathop{\sum }_{d\neq \Box }e^{-{\it\pi}d^{2}/D^{2}}\log |d|\left|\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right)\right|}\mathop{\sum }_{d=\Box }e^{-{\it\pi}d^{2}/D^{2}}\log |d|\left|\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right)\right|\right.\\ \displaystyle \qquad \qquad \qquad \quad \qquad \left.+\,\mathop{\sum }_{d\neq \Box }e^{-{\it\pi}d^{2}/D^{2}}\log |d|\left|\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right)\right|\right\}.\end{array}\end{eqnarray}$$
Then, the first term is evaluated by
$$\begin{eqnarray}\displaystyle & & \displaystyle \displaystyle \log x\mathop{\sum }_{d=\Box }e^{-{\it\pi}d^{2}/D^{2}}\log |d|\left|\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right)\right|\nonumber\\ \displaystyle & & \displaystyle \displaystyle \quad \ll x\log x\mathop{\sum }_{d=\Box }e^{-{\it\pi}d^{2}/D^{2}}\log |d|\nonumber\\ \displaystyle & & \displaystyle \displaystyle \quad \ll xD^{1/2}\log x\log D.\end{eqnarray}$$
We evaluate the second term on the right-hand side of (3.5). It is known that, assuming the GRH, for
$d\neq \Box$
, the estimate
$$\begin{eqnarray}\mathop{\sum }_{p\leqslant x}\left(\frac{d}{p}\right)\log p\ll x^{1/2}\log ^{2}|d|x\end{eqnarray}$$
holds (see [Reference Özlük and Snyder13, p. 221]). We decompose
$$\begin{eqnarray}\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right)=\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\left(\frac{d}{p}\right)\log p+\mathop{\sum }_{l\geqslant 2}\mathop{\sum }_{p}a\left(\frac{p^{l}}{x}\right)\left(\frac{d}{p^{l}}\right)\log p.\end{eqnarray}$$
Using the estimate above, the first term is evaluated by
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)\left(\frac{d}{p}\right)\log p & = & \displaystyle \int _{0}^{\infty }a\left(\frac{u}{x}\right)d\left(\mathop{\sum }_{p\leqslant u}\left(\frac{d}{p}\right)\log p\right)\nonumber\\ \displaystyle & = & \displaystyle -x^{-1}\int _{0}^{\infty }\left(\mathop{\sum }_{p\leqslant u}\left(\frac{d}{p}\right)\log p\right)a^{\prime }\left(\frac{u}{x}\right)du\nonumber\\ \displaystyle & \ll & \displaystyle x^{1/2}\log ^{2}|d|x.\end{eqnarray}$$
Next, we evaluate the second term on the right-hand side of (3.7). Since
$a(x)$
has a support in
$[A,B]$
for some
$0<A<B<\infty$
, we may assume that the range of
$l$
is restricted to
$2\leqslant l\ll \log x$
. Therefore,
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{l\geqslant 2}\mathop{\sum }_{p}a\left(\frac{p^{l}}{x}\right)\left(\frac{d}{p^{l}}\right)\log p & \ll & \displaystyle \mathop{\sum }_{l\geqslant 2}\mathop{\sum }_{p}a\left(\frac{p^{l}}{x}\right)\log p\nonumber\\ \displaystyle & \ll & \displaystyle \mathop{\sum }_{2\leqslant l\ll \log x}x^{1/2}\log x\ll x^{1/2}\log ^{2}x.\end{eqnarray}$$
Combining (3.7)–(3.9), we obtain
$$\begin{eqnarray}\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right)\ll x^{1/2}\log ^{2}|d|x\end{eqnarray}$$
for
$d\neq \Box$
, assuming the GRH. Therefore, the second term on the right-hand side of (3.5) is
$$\begin{eqnarray}\displaystyle & & \displaystyle \log x\mathop{\sum }_{d\neq \Box }e^{-{\it\pi}d^{2}/D^{2}}\log |d|\left|\mathop{\sum }_{n=1}^{\infty }a\left(\frac{n}{x}\right){\rm\Lambda}(n)\left(\frac{d}{n}\right)\right|\nonumber\\ \displaystyle & & \displaystyle \quad \ll x^{1/2}\log x\mathop{\sum }_{d\neq 0}e^{-{\it\pi}d^{2}/D^{2}}\log |d|\log ^{2}|d|x\nonumber\\ \displaystyle & & \displaystyle \quad \ll x^{1/2}D\log x\log D(\log ^{2}D+\log ^{2}x).\end{eqnarray}$$
Combining (3.6) and (3.11), we obtain
$$\begin{eqnarray}O_{22}\ll xD^{1/2}\log x\log D+x^{1/2}D\log x\log D(\log ^{2}D+\log ^{2}x).\end{eqnarray}$$
By combining (3.4), (3.12), and the evaluation
$O_{2}\ll O_{22}$
, we obtain
$$\begin{eqnarray}\displaystyle O_{2} & \ll & \displaystyle \min \left\{xD\log x\log D,xD^{1/2}\log x\log D\right.\nonumber\\ \displaystyle & & \displaystyle \qquad \quad \left.+\,x^{1/2}D\log x\log D(\log ^{2}D+\log ^{2}x)\right\}.\end{eqnarray}$$
Finally, since
$$\begin{eqnarray}\displaystyle & \displaystyle x^{2}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\ll x^{2}D, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \log ^{2}x\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log ^{2}|d|\ll D\log ^{2}x\log ^{2}D, & \displaystyle \nonumber\end{eqnarray}$$
we have
$$\begin{eqnarray}O_{4}\ll \text{min}\{x^{2}D,D\log ^{2}x\log ^{2}D\}.\end{eqnarray}$$
3.2 The computations of
$M_{1}$
,
$M_{3}$
and the evaluation of
$M_{5}$
Next, we compute
$M_{1}$
,
$M_{3}$
,
$M_{5}$
. These terms do not involve Kronecker symbols. First, by Lemma 2.3,
$$\begin{eqnarray}\displaystyle M_{1} & = & \displaystyle K(1)^{2}x^{2}\mathop{\sum }_{d=\Box }e^{-{\it\pi}d^{2}/D^{2}}\nonumber\\ \displaystyle & = & \displaystyle IK(1)^{2}x^{2}D^{1/2}-\frac{1}{2}K(1)^{2}x^{2}+O(x^{2}D^{-1/2}).\end{eqnarray}$$
Next, we compute
$M_{3}$
. The sum with respect to
$d$
is
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log ^{2}\left(\frac{|d|}{{\it\pi}}\right) & = & \displaystyle 2\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{2}/D^{2}}\log ^{2}d-4\log {\it\pi}\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{2}/D^{2}}\log d\nonumber\\ \displaystyle & & \displaystyle +\,2(\log {\it\pi})^{2}\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{2}/D^{2}}.\end{eqnarray}$$
By Lemmas 2.1, 2.2, 2.10, we have
$$\begin{eqnarray}\displaystyle & \displaystyle \mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{2}/D^{2}}\log ^{2}d=\frac{1}{2}D\log ^{2}D+O(D\log D), & \displaystyle \nonumber\\ \displaystyle & \displaystyle \mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{2}/D^{2}}\log d=\frac{1}{2}D\log D+O(D),\qquad \mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{2}/D^{2}}=\frac{1}{2}D+O(1). & \displaystyle \nonumber\end{eqnarray}$$
By inserting these into (3.16), we obtain
$$\begin{eqnarray}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log ^{2}\left(\frac{|d|}{{\it\pi}}\right)=D\log ^{2}D+O(D\log D).\end{eqnarray}$$
Hence,
$$\begin{eqnarray}M_{3}=a\left(\frac{1}{x}\right)^{2}\{D\log ^{2}D+O(D\log D)\}.\end{eqnarray}$$
Finally, we evaluate
$M_{5}$
. Since
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}E({\it\chi}_{d})\log \left(\frac{|d|}{{\it\pi}}\right) & \ll & \displaystyle \mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{4}/D^{2}}\log d\nonumber\\ \displaystyle & \ll & \displaystyle D^{1/2}\log D,\nonumber\end{eqnarray}$$
we obtain
$$\begin{eqnarray}M_{5}\ll xa\left(\frac{1}{x}\right)D^{1/2}\log D.\end{eqnarray}$$
3.3 The computation of
$M_{4}$
and the evaluation of
$M_{6}$
Next, we compute
$M_{4}$
. We decompose this by
$$\begin{eqnarray}\displaystyle M_{4} & = & \displaystyle -2K(1)x\left(\mathop{\sum }_{n=p}+\mathop{\sum }_{n=p^{2}}+\mathop{\sum }_{n=p^{k},k\geqslant 3}\right)a\left(\frac{n}{x}\right){\rm\Lambda}(n)\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}E({\it\chi}_{d})\left(\frac{d}{n}\right)\nonumber\\ \displaystyle & =: & \displaystyle -2K(1)x\left(M_{4}^{(1)}+M_{4}^{(2)}+M_{4}^{(3)}\right),\end{eqnarray}$$
say. First,
$$\begin{eqnarray}\displaystyle M_{4}^{(1)} & = & \displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)\log p\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{4}/D^{2}}\left(\frac{d^{2}}{p}\right)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)\log p\left\{\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{4}/D^{2}}-\mathop{\sum }_{\substack{ d=1 \\ p|d}}^{\infty }e^{-{\it\pi}d^{4}/D^{2}}\right\}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)\log p\left\{\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{4}/D^{2}}-\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}p^{4}d^{4}/D^{2}}\right\}.\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)\log p\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{4}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad =\left(K(1)x+O(x^{1/2}\log ^{2}x)\right)\left(ID^{1/2}-\frac{1}{2}+O(D^{-1/2})\right)\nonumber\\ \displaystyle & & \displaystyle \quad =IK(1)xD^{1/2}-\frac{1}{2}K(1)x+O((Dx)^{1/2}\log ^{2}x)+O(xD^{-1/2}).\end{eqnarray}$$
By Lemma 2.3,
$$\begin{eqnarray}\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}p^{4}d^{4}/D^{2}}=\frac{ID^{1/2}}{p}-\frac{1}{2}+O(\text{min}\{1,pD^{-1/2}\}).\end{eqnarray}$$
Therefore, by Lemmas 2.5 and 2.7, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)\log p\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}p^{4}d^{4}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\log p\left\{I\cdot \frac{D^{1/2}}{p}-\frac{1}{2}+O(\text{min}\{1,pD^{-1/2}\})\right\}\nonumber\\ \displaystyle & & \displaystyle \quad =IK(0)D^{1/2}+O(D^{1/2}x^{-1/2}\log ^{2}x)-\frac{1}{2}K(1)x+O(x^{1/2}\log ^{2}x)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O\left(\mathop{\sum }_{p}a\left(\frac{p}{x}\right)(\log p)\;\text{min}\{1,pD^{-1/2}\}\right).\end{eqnarray}$$
By Lemmas 2.5 and 2.11, we have
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\log p\ll x,\qquad D^{-1/2}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)p\log p\ll x^{2}D^{-1/2}.\end{eqnarray}$$
Therefore, the last line of (3.22) is evaluated by
$O(\text{min}\{x,x^{2}D^{-1/2}\})$
. Hence, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)\log p\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}p^{4}d^{4}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad =IK(0)D^{1/2}-\frac{1}{2}K(1)x\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O(D^{1/2}x^{-1/2}\log ^{2}x+x^{1/2}\log ^{2}x+\text{min}\{x,x^{2}D^{-1/2}\}).\end{eqnarray}$$
By inserting (3.21), (3.23) into (3.20), we obtain
$$\begin{eqnarray}M_{4}^{(1)}=IK(1)D^{1/2}x+O((Dx)^{1/2}\log ^{2}x+\text{min}\{x,x^{2}D^{-1/2}\}).\end{eqnarray}$$
Next,
$$\begin{eqnarray}\displaystyle M_{4}^{(2)} & = & \displaystyle \mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)\log p\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{4}/D^{2}}\left(\frac{d^{2}}{p^{2}}\right)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)\log p\left\{\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{4}/D^{2}}-\mathop{\sum }_{\substack{ d=1 \\ p|d}}^{\infty }e^{-{\it\pi}d^{4}/D^{2}}\right\}.\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)\log p\mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{4}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad =\left\{\frac{1}{2}K\left(\frac{1}{2}\right)x^{1/2}+O(x^{1/4}\log ^{2}x)\right\}\left\{ID^{1/2}-\frac{1}{2}+O(D^{-1/2})\right\}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{2}IK\left(\frac{1}{2}\right)D^{1/2}x^{1/2}-\frac{1}{4}K\left(\frac{1}{2}\right)x^{1/2}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O(D^{1/2}x^{1/4}\log ^{2}x+x^{1/2}D^{-1/2}).\end{eqnarray}$$
On the other hand, since
$$\begin{eqnarray}\mathop{\sum }_{\substack{ d=1 \\ p|d}}^{\infty }e^{-{\it\pi}d^{4}/D^{2}}=I\cdot \frac{D^{1/2}}{p}-\frac{1}{2}+O(\text{min}\{1,pD^{-1/2}\}),\end{eqnarray}$$
by Lemmas 2.4 and 2.8, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \displaystyle \mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)\log p\mathop{\sum }_{\substack{ d=1 \\ p|d}}^{\infty }e^{-{\it\pi}d^{4}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle \displaystyle \quad =\mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)\log p\left\{I\cdot \frac{D^{1/2}}{p}-\frac{1}{2}+O(\text{min}\{1,pD^{-1/2}\})\right\}\nonumber\\ \displaystyle & & \displaystyle \displaystyle \quad =ID^{1/2}\left\{\frac{1}{2}K(0)+O(x^{-1/4}\log ^{2}x)\right\}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{1}{2}\left\{\frac{1}{2}K\left(\frac{1}{2}\right)x^{1/2}+O(x^{1/4}\log ^{2}x)\right\}\nonumber\\ \displaystyle & & \displaystyle \displaystyle \qquad +\,O\left(\text{min}\left\{\mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)\log p,D^{-1/2}\mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)p\log p\right\}\right)\nonumber\\ \displaystyle & & \displaystyle \displaystyle \quad =\frac{1}{2}IK(0)D^{1/2}-\frac{1}{4}K\left(\frac{1}{2}\right)x^{1/2}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O(x^{-1/4}D^{1/2}\log ^{2}x)+O(x^{1/4}\log ^{2}x)\nonumber\\ \displaystyle & & \displaystyle \displaystyle \qquad +\,O\left(\text{min}\left\{\mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)\log p,D^{-1/2}\mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)p\log p\right\}\right).\end{eqnarray}$$
By Lemmas 2.4 and 2.12, we have
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)\log p\ll x^{1/2},\qquad \mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)p\log p\ll x.\end{eqnarray}$$
Hence, the last line of (3.27) becomes
$O(\text{min}\{x^{1/2},xD^{-1/2}\})$
. Therefore,
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p}a\left(\frac{p^{2}}{x}\right)\log p\mathop{\sum }_{\substack{ d=1 \\ p|d}}^{\infty }e^{-{\it\pi}d^{4}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{2}IK(0)D^{1/2}-\frac{1}{4}K\left(\frac{1}{2}\right)x^{1/2}+O(x^{-1/4}D^{1/2}\log ^{2}x)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O(x^{1/4}\log ^{2}x)+O(\text{min}\{x^{1/2},xD^{-1/2}\}).\end{eqnarray}$$
By inserting (3.26), (3.28) into (3.25), we obtain
$$\begin{eqnarray}\displaystyle M_{4}^{(2)}=\frac{1}{2}IK\left(\frac{1}{2}\right)D^{1/2}x^{1/2}+O(D^{1/2}x^{1/4}\log ^{2}x)+O(\text{min}\{x^{1/2},xD^{-1/2}\}). & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
Next, we evaluate
$$\begin{eqnarray}M_{4}^{(3)}=\mathop{\sum }_{k\geqslant 3}\mathop{\sum }_{p}a\left(\frac{p^{k}}{x}\right)\log p\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}E({\it\chi}_{d})\left(\frac{d}{p^{k}}\right).\end{eqnarray}$$
Since
$a(u)=0$
for
$u>B$
, we only have to compute the part
$p^{k}/x\leqslant B$
. The number of such
$k$
is
$O(\log x)$
. Since
$p$
must satisfy
$p\leqslant (Bx)^{1/k}\leqslant (Bx)^{1/3}$
, and
$$\begin{eqnarray}\left|\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}E({\it\chi}_{d})\left(\frac{d}{p^{k}}\right)\right|\leqslant \mathop{\sum }_{d=1}^{\infty }e^{-{\it\pi}d^{4}/D^{2}}\ll D^{1/2},\end{eqnarray}$$
using the PNT, we find that
$$\begin{eqnarray}\displaystyle M_{4}^{(3)} & \ll & \displaystyle (\log x)\cdot \left(\mathop{\sum }_{p\leqslant (Bx)^{1/3}}\log p\right)\cdot D^{1/2}\nonumber\\ \displaystyle & \ll & \displaystyle x^{1/3}D^{1/2}\log x.\end{eqnarray}$$
By inserting (3.24), (3.29), (3.30) into (3.19), we obtain
$$\begin{eqnarray}\displaystyle M_{4}=-2IK(1)^{2}D^{1/2}x^{2}+O(D^{1/2}x^{3/2}\log ^{2}x)+O(\text{min}\{x^{2},x^{3}D^{-1/2}\}). & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
Next, we evaluate
$M_{6}$
. We decompose this by
$$\begin{eqnarray}\displaystyle M_{6} & = & \displaystyle \!-2a\left(\frac{1}{x}\right)\left\{\mathop{\sum }_{n=p}+\!\mathop{\sum }_{n=p^{k},k\geqslant 2}\right\}a\left(\frac{n}{x}\right){\rm\Lambda}(n)\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\left(\frac{d}{n}\right)\text{log}\left(\frac{|d|}{{\it\pi}}\right)\nonumber\\ \displaystyle & =: & \displaystyle M_{6}^{(1)}+M_{6}^{(2)},\nonumber\end{eqnarray}$$
say. By the Pólya–Vinogradov inequality, we find that
$$\begin{eqnarray}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\left(\frac{d}{p}\right)\log \left(\frac{|d|}{{\it\pi}}\right)\ll \sqrt{p}\log p\log D.\end{eqnarray}$$
Hence, by Lemma 2.9,
$$\begin{eqnarray}\displaystyle M_{6}^{(1)} & \ll & \displaystyle a\left(\frac{1}{x}\right)\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\log p\cdot \sqrt{p}\log p\log D\nonumber\\ \displaystyle & \ll & \displaystyle a\left(\frac{1}{x}\right)\cdot x^{3/2}\log x\log D.\end{eqnarray}$$
Next, we evaluate
$M_{6}^{(2)}$
. By Lemmas 2.1, 2.2 and the conditions
$k\ll \log x$
,
$p\leqslant \sqrt{Bx}$
, we have
$$\begin{eqnarray}\displaystyle M_{6}^{(2)} & \ll & \displaystyle a\left(\frac{1}{x}\right)\log x\mathop{\sum }_{p\leqslant \sqrt{Bx}}\log p\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log \left(\frac{|d|}{{\it\pi}}\right)\nonumber\\ \displaystyle & \ll & \displaystyle a\left(\frac{1}{x}\right)Dx^{1/2}\log x\log D.\end{eqnarray}$$
Combining (3.32), (3.33), we obtain
$$\begin{eqnarray}M_{6}\ll a\left(\frac{1}{x}\right)\{x^{3/2}\log x\log D+Dx^{1/2}\log x\log D\}.\end{eqnarray}$$
3.4 The computation of
$M_{2}$
Finally, we compute
$$\begin{eqnarray}M_{2}=\mathop{\sum }_{k,l=1}^{\infty }\mathop{\sum }_{p,q\in P}a\left(\frac{p^{k}}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\left(\frac{d}{p^{k}}\right)\left(\frac{d}{q^{l}}\right),\end{eqnarray}$$
where
$P$
denotes the set of all prime numbers. It will be convenient to keep in mind that only
$k,l$
satisfying
$k,l\ll \log x$
contribute to the sum above. First, we evaluate the contribution of the part
$p=2$
to
$M_{2}$
. The contribution of the part
$p=q=2$
is
$$\begin{eqnarray}\displaystyle \ll \mathop{\sum }_{k,l=1}^{\infty }a\left(\frac{2^{k}}{x}\right)a\left(\frac{2^{l}}{x}\right)\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\ll D\log ^{2}x. & & \displaystyle\end{eqnarray}$$
The contribution of the part
$p=2$
,
$q\geqslant 3$
,
$l\geqslant 2$
is
$$\begin{eqnarray}\displaystyle & \ll & \displaystyle \mathop{\sum }_{k}\mathop{\sum }_{l\geqslant 2}\mathop{\sum }_{q\in P}a\left(\frac{2^{k}}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log q)\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\nonumber\\ \displaystyle & \ll & \displaystyle Dx^{1/2}\log ^{2}x.\end{eqnarray}$$
Since
$(\cdot /2^{k}q)$
is a nonprincipal character whose conductor is at most
$2q$
, by the Pólya–Vinogradov inequality, we have
$$\begin{eqnarray}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\left(\frac{d}{2^{k}}\right)\left(\frac{d}{q}\right)\ll q^{1/2}\log q\end{eqnarray}$$
for primes
$q\geqslant 3$
. Therefore, the contribution of the part
$p=2$
,
$q\geqslant 3$
,
$l=1$
is
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{k}\mathop{\sum }_{q\in P_{{\geqslant}3}}a\left(\frac{2^{k}}{x}\right)a\left(\frac{q}{x}\right)(\log 2)(\log q)\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\left(\frac{d}{2^{k}}\right)\left(\frac{d}{q}\right)\nonumber\\ \displaystyle & & \displaystyle \quad \ll (\log x)\mathop{\sum }_{q\in P}a\left(\frac{q}{x}\right)(\log q)\cdot q^{1/2}\log q\nonumber\\ \displaystyle & & \displaystyle \quad \ll x^{3/2}\log ^{2}x,\end{eqnarray}$$
where
$P_{{\geqslant}3}$
denotes the set of all prime numbers greater than 2. By (3.35), (3.36) and (3.37), the contribution of the part
$p=2$
to
$M_{2}$
is at most
$O(Dx^{1/2}\log ^{2}x+x^{3/2}\log ^{2}x)$
. The contribution of the part
$q=2$
is the same. Therefore, we conclude that
$$\begin{eqnarray}\displaystyle M_{2} & = & \displaystyle \mathop{\sum }_{k,l=1}^{\infty }\mathop{\sum }_{p,q\in P_{{\geqslant}3}}a\left(\frac{p^{k}}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\left(\frac{d}{p^{k}}\right)\left(\frac{d}{q^{l}}\right)\nonumber\\ \displaystyle & & \displaystyle +\,O(Dx^{1/2}\log ^{2}x+x^{3/2}\log ^{2}x)\nonumber\\ \displaystyle & =: & \displaystyle \mathop{\sum }_{k,l=1}^{\infty }M_{2}^{(k,l)}+O(Dx^{1/2}\log ^{2}x+x^{3/2}\log ^{2}x),\end{eqnarray}$$
say. Moreover, by the PNT and Lemma 2.1, we have
$$\begin{eqnarray}\displaystyle M_{2}^{(k,l)} & \ll & \displaystyle \mathop{\sum }_{p,q}a\left(\frac{p^{k}}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\nonumber\\ \displaystyle & \ll & \displaystyle klDx^{1/k+\frac{1}{l}}\nonumber\end{eqnarray}$$
for each
$k,l$
. Hence, the total contribution of the part
$k\geqslant 3,l\geqslant 2$
or
$k\geqslant 2,l\geqslant 3$
is at most
$O(Dx^{5/6}\log ^{4}x)$
. Therefore,
$$\begin{eqnarray}\displaystyle M_{2} & = & \displaystyle M_{2}^{(1,1)}+M_{2}^{(2,2)}+2\mathop{\sum }_{l\geqslant 2}M_{2}^{(1,l)}\nonumber\\ \displaystyle & & \displaystyle +\,O(Dx^{5/6}\log ^{4}x+x^{3/2}\log ^{2}x).\end{eqnarray}$$
By the computation above,
$M_{2}^{(2,2)}$
is evaluated by
$$\begin{eqnarray}M_{2}^{(2,2)}\ll Dx.\end{eqnarray}$$
Next, we compute
$M_{2}^{(1,l)}$
for
$l\geqslant 1$
,
$1\ll x\ll D^{1-{\it\delta}}$
.
(A) First, we consider the case that
$l$
is odd. In this case, we decompose
$$\begin{eqnarray}\displaystyle M_{2}^{(1,l)} & = & \displaystyle \left(\mathop{\sum }_{\substack{ p,q\in P_{{\geqslant}3} \\ p=q}}+\mathop{\sum }_{\substack{ p,q\in P_{{\geqslant}3} \\ p\neq q}}\right)a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\left(\frac{d}{p}\right)\left(\frac{d}{q}\right)\nonumber\\ \displaystyle & =: & \displaystyle M_{2,1}^{(1,l)}+M_{2,2}^{(1,l)},\end{eqnarray}$$
say. First, we compute
$M_{2,1}^{(1,1)}$
. This term is given by
$$\begin{eqnarray}\displaystyle M_{2,1}^{(1,1)} & = & \displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)^{2}(\log p)^{2}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle -\,\mathop{\sum }_{p}a\left(\frac{p}{x}\right)^{2}(\log p)^{2}\mathop{\sum }_{\substack{ d \\ p|d}}e^{-{\it\pi}d^{2}/D^{2}}.\end{eqnarray}$$
By Lemmas 2.1, 2.13, the first term on the right-hand side of (3.42) is
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)^{2}(\log p)^{2}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad =\{D+O(1)\}\{L(1)x\log x+L^{\prime }(1)x+O(x^{1/2}\log ^{3}x)\}\nonumber\\ \displaystyle & & \displaystyle \quad =L(1)Dx\log x+O(Dx+x\log x).\end{eqnarray}$$
By Lemmas 2.1, 2.13, 2.14, the second term on the right-hand side of (3.42) is
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)^{2}(\log p)^{2}\mathop{\sum }_{\substack{ d \\ p|d}}e^{-{\it\pi}d^{2}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{p}a\left(\frac{p}{x}\right)^{2}(\log p)^{2}\left\{\frac{D}{p}+O(1)\right\}\nonumber\\ \displaystyle & & \displaystyle \quad \ll D\mathop{\sum }_{p}a\left(\frac{p}{x}\right)^{2}\frac{(\log p)^{2}}{p}+\mathop{\sum }_{p}a\left(\frac{p}{x}\right)^{2}(\log p)^{2}\nonumber\\ \displaystyle & & \displaystyle \quad \ll D\log x+x\log x.\end{eqnarray}$$
By inserting (3.43), (3.44) into (3.42), we obtain
$$\begin{eqnarray}M_{2,1}^{(1,1)}=L(1)Dx\log x+O(Dx+x\log x).\end{eqnarray}$$
If
$l\geqslant 2$
(including the case that
$l$
is even), by the PNT and Lemma 2.1,
$$\begin{eqnarray}\displaystyle M_{2,1}^{(1,l)} & \ll & \displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)a\left(\frac{p^{l}}{x}\right)\log ^{2}p\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\nonumber\\ \displaystyle & \ll & \displaystyle D\cdot \frac{lx^{\frac{1}{l}}}{\log x}\cdot \log ^{2}x\nonumber\\ \displaystyle & \ll & \displaystyle lDx^{\frac{1}{l}}\log x.\end{eqnarray}$$
Next, we compute
$M_{2,2}^{(1,l)}$
.
(a) Let
$x=o(D^{1/2})$
. By the Pólya–Vinogradov inequality, we obtain
$$\begin{eqnarray}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\left(\frac{d}{p}\right)\left(\frac{d}{q}\right)\ll (pq)^{1/2}\log (pq).\end{eqnarray}$$
Hence, by the PNT, we have
$$\begin{eqnarray}\displaystyle M_{2,2}^{(1,l)} & \ll & \displaystyle \mathop{\sum }_{p,q}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)(pq)^{1/2}\log (pq)\nonumber\\ \displaystyle & \ll & \displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)p^{1/2}\log ^{2}p\mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)q^{1/2}\log ^{2}q\nonumber\\ \displaystyle & \ll & \displaystyle \frac{x}{\log x}x^{1/2}\log ^{2}x\cdot \frac{lx^{\frac{1}{l}}}{\log x}x^{\frac{1}{2l}}\log ^{2}x\nonumber\\ \displaystyle & \ll & \displaystyle lx^{3/2(1+\frac{1}{l})}\log ^{2}x.\end{eqnarray}$$
(b) If
$D^{1/2-{\it\delta}}\ll x\ll D^{1-{\it\delta}}$
, by the translation formula of the theta function (Lemma 2.17), we have
$$\begin{eqnarray}M_{2,2}^{(1,l)}=D\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{\substack{ q\geqslant 3 \\ q\neq p}}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\frac{1}{\sqrt{pq}}\mathop{\sum }_{m}\left(\frac{m}{pq}\right)e^{-{\it\pi}m^{2}D^{2}/p^{2}q^{2}}.\end{eqnarray}$$
We decompose this by
$$\begin{eqnarray}M_{2,2}^{(1,l)}=M_{s}^{(1,l)}-M_{p}^{(1,l)}-M_{q}^{(1,l)}+M_{pq}^{(1,l)}+E,\end{eqnarray}$$
where
$$\begin{eqnarray}\displaystyle M_{s}^{(1,l)} & = & \displaystyle D\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{\substack{ q\geqslant 3 \\ q\neq p}}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\frac{1}{\sqrt{pq}}\mathop{\sum }_{m=\Box }e^{-{\it\pi}m^{2}D^{2}/p^{2}q^{2}},\nonumber\\ \displaystyle M_{p}^{(1,l)} & = & \displaystyle D\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{\substack{ q\geqslant 3 \\ q\neq p}}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\frac{1}{\sqrt{pq}}\mathop{\sum }_{\substack{ m=\Box \\ p|m}}e^{-{\it\pi}m^{2}D^{2}/p^{2}q^{2}},\nonumber\\ \displaystyle M_{q}^{(1,l)} & = & \displaystyle D\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{\substack{ q\geqslant 3 \\ q\neq p}}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\frac{1}{\sqrt{pq}}\mathop{\sum }_{\substack{ m=\Box \\ q|m}}e^{-{\it\pi}m^{2}D^{2}/p^{2}q^{2}},\nonumber\\ \displaystyle M_{pq}^{(1,l)} & = & \displaystyle D\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{\substack{ q\neq p \\ q\geqslant 3}}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\frac{1}{\sqrt{pq}}\mathop{\sum }_{\substack{ m=\Box \\ pq|m}}e^{-{\it\pi}m^{2}D^{2}/p^{2}q^{2}},\nonumber\end{eqnarray}$$
and
$$\begin{eqnarray}E=D\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{\substack{ q\geqslant 3 \\ q\neq p}}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\frac{1}{\sqrt{pq}}\mathop{\sum }_{m\neq \Box }e^{-{\it\pi}m^{2}D^{2}/p^{2}q^{2}}.\end{eqnarray}$$
First, by Lemma 2.3,
$$\begin{eqnarray}\displaystyle M_{s}^{(1,l)} & = & \displaystyle D\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{\substack{ q\geqslant 3 \\ q\neq p}}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\frac{1}{\sqrt{pq}}\left(I\sqrt{\frac{pq}{D}}+O(1)\right)\nonumber\\ \displaystyle & = & \displaystyle ID^{1/2}\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{\substack{ q\geqslant 3 \\ q\neq p}}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\nonumber\\ \displaystyle & & \displaystyle +\,O\left(D\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{\substack{ q\geqslant 3 \\ q\neq p}}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)\frac{(\log p)(\log q)}{\sqrt{pq}}\right).\end{eqnarray}$$
By applying Lemmas 2.15, 2.16 to (3.49), we obtain
$$\begin{eqnarray}M_{s}^{(1,l)}\!=\mathit{ID}^{1/2}l^{-1}K(1)K\!\left(\frac{1}{l}\right)\!x^{1+\frac{1}{l}}\!+O(l^{-1}D^{1/2}x^{1+\frac{1}{2l}}\log ^{2}\!x)+O(Dx^{1/2+\frac{1}{2l}}).\end{eqnarray}$$
In the computation above, we used
$\left|K\left(1/2l\right)\right|\ll \int _{A}^{B}t^{1/2l}\,dt/t\ll l$
. Next, by Lemma 2.3,
$$\begin{eqnarray}\displaystyle M_{p}^{(1,l)} & = & \displaystyle D\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{\substack{ q\geqslant 3 \\ q\neq p}}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\frac{1}{\sqrt{pq}}\mathop{\sum }_{m=1}^{\infty }e^{-{\it\pi}p^{2}D^{2}m^{4}/q^{2}}\nonumber\\ \displaystyle & \ll & \displaystyle D\mathop{\sum }_{p}\mathop{\sum }_{q}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)\frac{(\log p)(\log q)}{\sqrt{pq}}\left(\sqrt{\frac{q}{pD}}+O(1)\right)\nonumber\\ \displaystyle & \ll & \displaystyle D^{1/2}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\frac{\log p}{p}\mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)(\log q)\nonumber\\ \displaystyle & & \displaystyle +\,D\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\frac{\log p}{\sqrt{p}}\mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\frac{\log q}{\sqrt{q}}.\nonumber\end{eqnarray}$$
By Lemma 2.7,
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\frac{\log p}{p}\ll 1.\end{eqnarray}$$
By the PNT,
$$\begin{eqnarray}\mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\log q\ll \frac{lx^{\frac{1}{l}}}{\log x}\cdot \log x\ll lx^{\frac{1}{l}}.\end{eqnarray}$$
By Lemma 2.6,
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\frac{\log p}{\sqrt{p}}\ll x^{1/2}.\end{eqnarray}$$
By the PNT and the Stieltjes integral,
$$\begin{eqnarray}\mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\frac{\log q}{\sqrt{q}}\ll \int _{1}^{\infty }a\left(\frac{u^{l}}{x}\right)\,\frac{du}{\sqrt{u}}\ll \int _{1}^{(Bx)^{\frac{1}{l}}}\,\frac{du}{\sqrt{u}}\ll x^{\frac{1}{2l}}.\end{eqnarray}$$
Combining these, we obtain
$$\begin{eqnarray}M_{p}^{(1,l)}\ll lD^{1/2}x^{\frac{1}{l}}+Dx^{1/2+\frac{1}{2l}}.\end{eqnarray}$$
Next, by Lemma 2.3,
$$\begin{eqnarray}\displaystyle M_{q}^{(1,l)} & = & \displaystyle D\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{\substack{ q\geqslant 3 \\ q\neq p}}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\frac{1}{\sqrt{pq}}\mathop{\sum }_{m=1}^{\infty }e^{-{\it\pi}q^{2}D^{2}m^{4}/p^{2}}\nonumber\\ \displaystyle & \ll & \displaystyle D\mathop{\sum }_{p}\mathop{\sum }_{q}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)\frac{(\log p)(\log q)}{\sqrt{pq}}\left(\sqrt{\frac{p}{qD}}+O(1)\right)\nonumber\\ \displaystyle & \ll & \displaystyle D^{1/2}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\log p\mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\frac{\log q}{q}\nonumber\\ \displaystyle & & \displaystyle +\,D\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\frac{\log p}{\sqrt{p}}\mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\frac{\log q}{\sqrt{q}}.\end{eqnarray}$$
By the PNT and the Stieltjes integral,
$$\begin{eqnarray}\displaystyle & \displaystyle \mathop{\sum }_{p}a\left(\frac{p}{x}\right)\log p\ll x, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\frac{\log q}{q}\ll \int _{1}^{\infty }a\left(\frac{u^{l}}{x}\right)\,\frac{du}{u}\ll \int _{1}^{(Bx)^{\frac{1}{l}}}\,\frac{du}{u}\ll l^{-1}\log x. & \displaystyle \nonumber\end{eqnarray}$$
The last line of (3.52) is
$O(Dx^{1/2+1/2l})$
, as we computed in the evaluation of
$M_{p}^{(1,l)}$
. Combining these, we obtain
$$\begin{eqnarray}M_{q}^{(1,l)}\ll l^{-1}D^{1/2}x\log x+Dx^{1/2+\frac{1}{2l}}.\end{eqnarray}$$
The term
$M_{pq}^{(1,l)}$
is evaluated by
$$\begin{eqnarray}\displaystyle M_{pq}^{(1,l)} & = & \displaystyle D\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{\substack{ q\geqslant 3 \\ q\neq p}}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\frac{1}{\sqrt{pq}}\mathop{\sum }_{m=1}^{\infty }e^{-{\it\pi}p^{2}q^{2}D^{2}m^{4}}\nonumber\\ \displaystyle & \ll & \displaystyle D\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\frac{\log p}{\sqrt{p}}\mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\frac{\log q}{\sqrt{q}}\nonumber\\ \displaystyle & \ll & \displaystyle Dx^{1/2+\frac{1}{2l}}.\end{eqnarray}$$
Finally, we evaluate
$E$
. First,
$$\begin{eqnarray}\displaystyle E & = & \displaystyle D\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{\substack{ q\geqslant 3 \\ q\neq p}}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\frac{1}{\sqrt{pq}}\mathop{\sum }_{m\neq \Box }\left(\frac{m}{pq}\right)e^{-{\it\pi}D^{2}m^{2}/p^{2}q^{2}}\nonumber\\ \displaystyle & \ll & \displaystyle D\left|\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{q\geqslant 3}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\frac{1}{\sqrt{pq}}\mathop{\sum }_{m\neq \Box }\left(\frac{m}{p}\right)\left(\frac{m}{q}\right)e^{-{\it\pi}D^{2}m^{2}/p^{2}q^{2}}\right|\nonumber\\ \displaystyle & & \displaystyle \qquad +\,D\mathop{\sum }_{p\geqslant 3}a\left(\frac{p}{x}\right)a\left(\frac{p^{l}}{x}\right)\log ^{2}p\cdot \frac{1}{p}\mathop{\sum }_{m=1}^{\infty }e^{-{\it\pi}m^{2}D^{2}/p^{4}}.\end{eqnarray}$$
By the PNT and Lemma 2.1, the second term on the right-hand side of (3.55) is
$$\begin{eqnarray}\displaystyle & & \displaystyle D\mathop{\sum }_{p\geqslant 3}a\left(\frac{p}{x}\right)a\left(\frac{p^{l}}{x}\right)\log ^{2}p\cdot \frac{1}{p}\mathop{\sum }_{m=1}^{\infty }e^{-{\it\pi}m^{2}D^{2}/p^{4}}\nonumber\\ \displaystyle & & \displaystyle \quad \ll D\mathop{\sum }_{p}a\left(\frac{p}{x}\right)a\left(\frac{p^{l}}{x}\right)\frac{\log ^{2}p}{p}\left(\frac{p^{2}}{D}+O(1)\right)\nonumber\\ \displaystyle & & \displaystyle \quad \ll l^{-1}x^{\frac{2}{l}}\log x+Dx^{\frac{1}{l}}.\end{eqnarray}$$
Next, since
$$\begin{eqnarray}e^{-{\it\pi}m^{2}D^{2}/p^{2}q^{2}}\leqslant \exp (-{\it\pi}m^{2}D^{2}(Bx)^{-2-\frac{2}{l}})\end{eqnarray}$$
holds for
$p\leqslant Bx$
,
$q\leqslant (Bx)^{1/l}$
, the first term on the right-hand side of (3.55) is evaluated by
$$\begin{eqnarray}\displaystyle & & \displaystyle D\left|\mathop{\sum }_{p\geqslant 3}\mathop{\sum }_{q\geqslant 3}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\frac{1}{\sqrt{pq}}\mathop{\sum }_{m\neq \Box }\left(\frac{m}{p}\right)\left(\frac{m}{q}\right)e^{-{\it\pi}D^{2}m^{4}/p^{2}q^{2}}\right|\nonumber\\ \displaystyle & & \displaystyle \qquad \quad \ll D\mathop{\sum }_{m\neq \Box }e^{-{\it\pi}m^{2}D^{2}/(Bx)^{2+\frac{2}{l}}}\left|\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\frac{\log p}{\sqrt{p}}\left(\frac{m}{p}\right)\right|\nonumber\\ \displaystyle & & \displaystyle \qquad \qquad \times \left|\mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\frac{\log q}{\sqrt{q}}\left(\frac{m}{q}\right)\right|.\end{eqnarray}$$
According to [Reference Özlük and Snyder13, p. 221], under the assumption of the GRH,
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\frac{\log p}{\sqrt{p}}\left(\frac{m}{p}\right)\ll \log ^{2}mx\end{eqnarray}$$
holds uniformly for
$m\ll x~m\neq \Box$
. Moreover, since the GRH implies
$$\begin{eqnarray}\mathop{\sum }_{q\leqslant x^{\frac{1}{l}}}\left(\frac{m}{q}\right)\log q\ll x^{\frac{1}{2l}}\log ^{2}(mx^{\frac{1}{l}})\end{eqnarray}$$
uniformly for
$m\ll x^{1/l}$
$m\neq \Box$
, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\frac{\log q}{\sqrt{q}}\left(\frac{m}{q}\right)\nonumber\\ \displaystyle & & \displaystyle \quad =\int _{0}^{\infty }a\left(\frac{u^{l}}{x}\right)\frac{1}{\sqrt{u}}\,d\left(\mathop{\sum }_{q\leqslant u}\left(\frac{m}{q}\right)\log q\right)\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{2}\int _{0}^{\infty }\left\{\mathop{\sum }_{q\leqslant u}\left(\frac{m}{q}\right)\log q\right\}a\left(\frac{u^{l}}{x}\right)u^{-3/2}\,du\nonumber\\ \displaystyle & & \displaystyle \quad \quad -\,\frac{l}{x}\int _{0}^{\infty }\left\{\mathop{\sum }_{q\leqslant u}\left(\frac{m}{q}\right)\log q\right\}a^{\prime }\left(\frac{u^{l}}{x}\right)u^{l-\frac{3}{2}}\,du\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{2}\int _{0}^{\infty }\left\{\mathop{\sum }_{q\leqslant (xv)^{\frac{1}{l}}}\left(\frac{m}{q}\right)\log q\right\}a(v)(xv)^{-\frac{3}{2l}}\cdot x^{\frac{1}{l}}l^{-1}v^{\frac{1}{l}-1}\,dv\nonumber\\ \displaystyle & & \displaystyle \quad \quad -\,\frac{l}{x}\int _{0}^{\infty }\left\{\mathop{\sum }_{q\leqslant (xv)^{\frac{1}{l}}}\left(\frac{m}{q}\right)\log q\right\}a^{\prime }(v)(xv)^{\frac{1}{l}(l-\frac{3}{2})}\cdot x^{\frac{1}{l}}l^{-1}v^{\frac{1}{l}-1}\,dv\nonumber\\ \displaystyle & & \displaystyle \quad \ll \frac{1}{2}\int _{A}^{B}(xv)^{\frac{1}{2l}}\log ^{2}(m(xv)^{\frac{1}{l}})a(v)(xv)^{-\frac{3}{2l}}x^{\frac{1}{l}}l^{-1}v^{\frac{1}{l}-1}\,dv\nonumber\\ \displaystyle & & \displaystyle \quad \quad +\,lx^{-1}\int _{A}^{B}(xv)^{\frac{1}{2l}}\log ^{2}(m(xv)^{\frac{1}{l}})a^{\prime }(v)(xv)^{\frac{1}{l}(l-\frac{3}{2})}x^{\frac{1}{l}}l^{-1}v^{\frac{1}{l}-1}\,dv\nonumber\\ \displaystyle & & \displaystyle \quad \ll \log ^{2}mx\nonumber\end{eqnarray}$$
uniformly for
$m\ll x^{1/l}$
. On the other hand, by forgetting the Kronecker symbols, we have
$$\begin{eqnarray}\mathop{\sum }_{p}a\left(\frac{p}{x}\right)\frac{\log p}{\sqrt{p}}\left(\frac{m}{p}\right)\ll x^{1/2},\qquad \mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\frac{\log q}{\sqrt{q}}\left(\frac{m}{q}\right)\ll x^{\frac{1}{2l}}.\end{eqnarray}$$
Hence, the first term on the right-hand side of (3.55) is
$$\begin{eqnarray}\displaystyle & \ll & \displaystyle D\mathop{\sum }_{m\leqslant x^{\frac{1}{l}}}e^{-{\it\pi}m^{2}D^{2}/(Bx)^{2+\frac{2}{l}}}\log ^{4}mx\nonumber\\ \displaystyle & & \displaystyle +\,D\mathop{\sum }_{m\geqslant x^{\frac{1}{l}}}e^{-{\it\pi}m^{2}D^{2}/(Bx)^{2+\frac{2}{l}}}x^{1/2+\frac{1}{2l}}\nonumber\\ \displaystyle & \ll & \displaystyle x^{1+\frac{1}{l}}\log ^{4}x+D\log ^{4}x\end{eqnarray}$$
for
$D^{1/2-{\it\delta}}\ll x\ll D^{1-{\it\delta}}$
. By inserting (3.56), (3.58) into (3.55), we obtain
$$\begin{eqnarray}E\ll Dx^{\frac{1}{l}}+x^{1+\frac{1}{l}}\log ^{4}x.\end{eqnarray}$$
By combining (3.50), (3.51), (3.53), (3.54) and (3.59), for odd
$l$
and
$D^{1/2-{\it\delta}}\ll x\ll D^{1-{\it\delta}}$
, we have
$$\begin{eqnarray}\displaystyle M_{2,2}^{(1,l)} & = & \displaystyle ID^{1/2}l^{-1}K(1)K\left(\frac{1}{l}\right)x^{1+\frac{1}{l}}\nonumber\\ \displaystyle & & \displaystyle +\,O(l^{-1}D^{1/2}x^{1+\frac{1}{2l}}\log ^{2}x)+O(Dx^{1/2+\frac{1}{2l}})\nonumber\\ \displaystyle & & \displaystyle +\,O(x^{1+\frac{1}{l}}\log ^{4}x).\end{eqnarray}$$
By combining this and (3.46), for odd
$l\geqslant 3$
and
$D^{1/2-{\it\delta}}\ll x\ll D^{1-{\it\delta}}$
, we have
$$\begin{eqnarray}\displaystyle M_{2}^{(1,l)} & = & \displaystyle ID^{1/2}l^{-1}K(1)K\left(\frac{1}{l}\right)x^{1+\frac{1}{l}}\nonumber\\ \displaystyle & & \displaystyle +\,O(l^{-1}D^{1/2}x^{1+\frac{1}{2l}}\log ^{2}x)+O(lDx^{\frac{1}{l}}\log x)\nonumber\\ \displaystyle & & \displaystyle +\,O(Dx^{1/2+\frac{1}{2l}})+O(x^{1+\frac{1}{l}}\log ^{4}x).\end{eqnarray}$$
If
$l=1$
,
$D^{1/2-{\it\delta}}\ll x\ll D^{1-{\it\delta}}$
, by (3.45) and (3.60), we have
$$\begin{eqnarray}\displaystyle M_{2}^{(1,1)} & = & \displaystyle L(1)Dx\log x+IK(1)^{2}D^{1/2}x^{2}\nonumber\\ \displaystyle & & \displaystyle +\,O(Dx+D^{1/2}x^{3/2}\log ^{2}x+x^{2}\log ^{4}x).\end{eqnarray}$$
On the other hand, for odd
$l\geqslant 3$
and
$x=o(D^{1/2})$
, by (3.46) and (3.47), we have
$$\begin{eqnarray}M_{2}^{(1,l)}\ll lDx^{\frac{1}{l}}\log x+lx^{3/2(1+\frac{1}{l})}\log ^{2}x,\end{eqnarray}$$
and for
$l=1$
,
$x=o(D^{1/2})$
, by (3.45) and (3.47), we have
$$\begin{eqnarray}M_{2}^{(1,1)}=L(1)Dx\log x+O(Dx+x^{3}\log ^{2}x).\end{eqnarray}$$
(B) Next, we consider the case that
$l$
is even. In this case, we have
$$\begin{eqnarray}\displaystyle M_{2}^{(1,l)} & = & \displaystyle \mathop{\sum }_{p,q}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\mathop{\sum }_{d}\left(\frac{d}{p}\right)e^{-{\it\pi}d^{2}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle -\,\mathop{\sum }_{p,q}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\mathop{\sum }_{q|d}\left(\frac{d}{p}\right)e^{-{\it\pi}d^{2}/D^{2}}.\end{eqnarray}$$
Since
$$\begin{eqnarray}\mathop{\sum }_{d}\left(\frac{d}{p}\right)e^{-{\it\pi}d^{2}/D^{2}}\ll \sqrt{p}\log p,\end{eqnarray}$$
the first term on the right-hand side of (3.65) is
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p,q}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\mathop{\sum }_{d}\left(\frac{d}{p}\right)e^{-{\it\pi}d^{2}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad \ll \mathop{\sum }_{p}a\left(\frac{p}{x}\right)\sqrt{p}\log ^{2}p\mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\log q\nonumber\\ \displaystyle & & \displaystyle \quad \ll x^{3/2}\log x\cdot lx^{\frac{1}{l}}\nonumber\\ \displaystyle & & \displaystyle \quad \ll lx^{3/2+\frac{1}{l}}\log x.\end{eqnarray}$$
The second term on the right-hand side of (3.65) is
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p,q}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\mathop{\sum }_{q|d}\left(\frac{d}{p}\right)e^{-{\it\pi}d^{2}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{p}a\left(\frac{p}{x}\right)(\log p)\mathop{\sum }_{q}\left(\frac{q}{p}\right)a\left(\frac{q^{l}}{x}\right)(\log q)\mathop{\sum }_{d}\left(\frac{d}{p}\right)e^{-{\it\pi}q^{2}d^{2}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad \ll \mathop{\sum }_{p}a\left(\frac{p}{x}\right)(\log p)\mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)(\log q)\left|\mathop{\sum }_{d}\left(\frac{d}{p}\right)e^{-{\it\pi}q^{2}d^{2}/D^{2}}\right|.\end{eqnarray}$$
Now, since
$q$
satisfies
$q\leqslant (Bx)^{1/l}\ll D^{1-{\it\delta}}\ll D$
,
$D/q\gg 1$
holds. Therefore, by the Pólya–Vinogradov inequality, we have
$$\begin{eqnarray}\mathop{\sum }_{d}\left(\frac{d}{p}\right)e^{-{\it\pi}q^{2}d^{2}/D^{2}}\ll \sqrt{p}\log p.\end{eqnarray}$$
Therefore,
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{p,q}a\left(\frac{p}{x}\right)a\left(\frac{q^{l}}{x}\right)(\log p)(\log q)\mathop{\sum }_{q|d}\left(\frac{d}{p}\right)e^{-{\it\pi}d^{2}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad \ll \mathop{\sum }_{p}a\left(\frac{p}{x}\right)\sqrt{p}\log ^{2}p\mathop{\sum }_{q}a\left(\frac{q^{l}}{x}\right)\log q\nonumber\\ \displaystyle & & \displaystyle \quad \ll lx^{3/2+\frac{1}{l}}\log x.\end{eqnarray}$$
By combining (3.66) and (3.69), we obtain
$$\begin{eqnarray}M_{2}^{(1,l)}\ll lx^{3/2+\frac{1}{l}}\log x\end{eqnarray}$$
for even
$l$
.
Now, we have computed or evaluated
$M_{2}^{(1,l)}$
for all
$l$
. If
$x=o(D^{1/2})$
, by (3.63) and (3.70), we have
$$\begin{eqnarray}\mathop{\sum }_{l\geqslant 2}M_{2}^{(1,l)}\ll Dx^{1/3}\log x+x^{2}\log ^{2}x.\end{eqnarray}$$
By inserting (3.40), (3.64), (3.71) into (3.39), we obtain
$$\begin{eqnarray}M_{2}=L(1)Dx\log x+O(Dx+x^{3}\log ^{2}x).\end{eqnarray}$$
If
$D^{1/2-{\it\delta}}\ll x\ll D^{1-{\it\delta}}$
, by (3.61),
$$\begin{eqnarray}\mathop{\sum }_{l\geqslant 3,\text{odd}}M_{2}^{(1,l)}\ll D^{1/2}x^{4/3}+Dx^{2/3}+x^{4/3}\log ^{4}x.\end{eqnarray}$$
(It should be noticed that
$K(1/l)\ll l$
.) On the other hand, by (3.70),
$$\begin{eqnarray}\mathop{\sum }_{l\geqslant 2,\text{even}}M_{2}^{(1,l)}\ll x^{2}\log x.\end{eqnarray}$$
By inserting (3.40), (3.62), (3.73), (3.74) into (3.39), we obtain
$$\begin{eqnarray}\displaystyle M_{2} & = & \displaystyle L(1)Dx\log x+IK(1)^{2}D^{1/2}x^{2}\nonumber\\ \displaystyle & & \displaystyle +\,O(Dx+D^{1/2}x^{3/2}\log ^{2}x+x^{2}\log ^{4}x)\end{eqnarray}$$
for
$D^{1/2-{\it\delta}}\ll x\ll D^{1-{\it\delta}}$
.
3.5 Conclusion
Now, we have computed or evaluated all terms appearing on the right-hand side of (3.1). We fix
${\it\delta}>0$
sufficiently small. Then, we have the following.
(1) If
$1<x\ll D^{1/2-{\it\delta}}$
, by inserting (3.2), (3.3), (3.13), (3.14), (3.15), (3.17), (3.18), (3.31), (3.34) and (3.72) into (3.1), we obtain
$$\begin{eqnarray}\displaystyle & & \displaystyle f_{K}(x,D)\nonumber\\ \displaystyle & & \displaystyle \quad =L(1)Dx\log x-IK(1)^{2}D^{1/2}x^{2}-\frac{1}{2}K(1)^{2}x^{2}+a\left(\frac{1}{x}\right)^{2}D\log ^{2}D\nonumber\\ \displaystyle & & \displaystyle \quad \quad +\,a\left(\frac{1}{x}\right)O(xD^{1/2}\log D+x^{3/2}\log x\log D+Dx^{1/2}\log x\log D)\nonumber\\ \displaystyle & & \displaystyle \quad \quad +\,a\left(\frac{1}{x}\right)^{2}O(D\log D)+O(x^{2}D^{-1/2})+O(Dx+x^{3}\log ^{2}x)\nonumber\\ \displaystyle & & \displaystyle \quad \quad +\,O(D^{1/2}x^{3/2}\log ^{2}x)+O(\text{min}\{x^{2},x^{3}D^{-1/2}\})\nonumber\\ \displaystyle & & \displaystyle \quad \quad +\,O(\text{min}\{x^{2}D^{1/2},xD^{1/2}\log x\log D\})\nonumber\\ \displaystyle & & \displaystyle \quad \quad +\,O(\min \{xD\log x\log D,xD^{1/2}\log x\log D\nonumber\\ \displaystyle & & \displaystyle \quad \quad +\,x^{1/2}D\log x\log D(\log ^{2}D+\log ^{2}x)\})\nonumber\\ \displaystyle & & \displaystyle \quad \quad +\,O(\text{min}\{xD\log D,D\log x\log ^{2}D\})\nonumber\\ \displaystyle & & \displaystyle \quad \quad +\,O(\text{min}\{x^{2}D,D\log ^{2}x\log ^{2}D\}).\nonumber\end{eqnarray}$$
(2) If
$D^{1/2-{\it\delta}}\ll x\ll D^{1-{\it\delta}}$
, since
$a(1/x)=0$
, by inserting (3.2), (3.3), (3.13), (3.14), (3.15), (3.17), (3.18), (3.31), (3.34) and (3.75) into (3.1), we obtain
$$\begin{eqnarray}\displaystyle & & \displaystyle f_{K}(x,D)\nonumber\\ \displaystyle & & \displaystyle \quad =L(1)Dx\log x-\frac{1}{2}K(1)^{2}x^{2}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O(x^{2}D^{-1/2})+O(Dx+D^{1/2}x^{3/2}\log ^{2}x+x^{2}\log ^{4}x)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O(D^{1/2}x^{3/2}\log ^{2}x)+O(\text{min}\{x^{2},x^{3}D^{-1/2}\})\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O(\text{min}\{x^{2}D^{1/2},xD^{1/2}\log x\log D\})\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O(\min \{xD\log x\log D,xD^{1/2}\log x\log D\nonumber\\ \displaystyle & & \displaystyle \qquad +\,x^{1/2}D\log x\log D(\log ^{2}D+\log ^{2}x)\})\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O(\text{min}\{xD\log D,D\log x\log ^{2}D\})\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O(\text{min}\{x^{2}D,D\log ^{2}x\log ^{2}D\}).\nonumber\end{eqnarray}$$
By dividing both sides by
$xD\log D$
and putting
$x=D^{{\it\alpha}}$
(
${\it\alpha}>0$
), we obtain the following formulas.
(1) If
$0<{\it\alpha}\leqslant 1/2-{\it\delta}$
, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle F_{K}({\it\alpha},D)\nonumber\\ \displaystyle & & \displaystyle \quad =L(1){\it\alpha}+a(D^{-{\it\alpha}})^{2}D^{-{\it\alpha}}\log D+a(D^{-{\it\alpha}})\cdot O({\it\alpha}D^{-{\it\alpha}/2}\log D)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,a(D^{-{\it\alpha}})^{2}\cdot O(D^{-{\it\alpha}})+O(\text{min}\{1,{\it\alpha}D^{-{\it\alpha}}\log ^{2}D\})\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O(\min \{{\it\alpha}\log D,{\it\alpha}D^{-1/2}\log D+{\it\alpha}(1+{\it\alpha}^{2})D^{-{\it\alpha}/2}\log ^{3}D\})\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O(\text{min}\{D^{{\it\alpha}}(\log D)^{-1},{\it\alpha}^{2}D^{-{\it\alpha}}\log ^{3}D\})+o(1).\end{eqnarray}$$
(2) If
$1/2-{\it\delta}<{\it\alpha}\leqslant 1-{\it\delta}$
, we have
$$\begin{eqnarray}F_{K}({\it\alpha},D)=L(1){\it\alpha}+o(1).\end{eqnarray}$$
Here, the implied constants depend only on
$K(s)$
and
${\it\delta}>0$
. However, the identity (3.76) is still valid for
$1/2-{\it\delta}<{\it\alpha}\leqslant 1-{\it\delta}$
, since the terms except for the first term on the right-hand side of (3.76) are all
$o(1)$
if
${\it\alpha}$
is large. Hence, we obtain (1.5).◻
4 Applications of Theorem 1.1
We extend
$F_{K}({\it\alpha},D)$
to the whole of
${\it\alpha}\in \mathbf{R}$
by
$F_{K}({\it\alpha},D):=F_{K}(-{\it\alpha},D)$
for
${\it\alpha}<0$
. Then, the identity (1.5) holds for
$0<|{\it\alpha}|\leqslant 1-{\it\delta}$
, by replacing
${\it\alpha}$
on the right-hand side with
$|{\it\alpha}|$
. To investigate the low-lying zeros of quadratic
$L$
-functions, we consider the integral of
$F_{K}({\it\alpha},D)$
multiplied by some bounded function. First, we prove that in this integral, the contribution of the error terms in (1.5) is small.
Lemma 4.1. We have
$$\begin{eqnarray}\displaystyle & \displaystyle \int _{-1}^{1}|{\it\alpha}|D^{-|{\it\alpha}|/2}\log D\,d{\it\alpha}\ll \frac{1}{\log D}, & \displaystyle\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle \int _{-1}^{1}D^{-|{\it\alpha}|}\,d{\it\alpha}\ll \frac{1}{\log D}, & \displaystyle\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle \int _{-1}^{1}\text{min}\{1,|{\it\alpha}|D^{-|{\it\alpha}|}\log ^{2}D\}\,d{\it\alpha}\ll \frac{1}{\sqrt{\log D}}, & \displaystyle\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle \int _{-1}^{1}\text{min}\{D^{|{\it\alpha}|}(\log D)^{-1},{\it\alpha}^{2}D^{-|{\it\alpha}|}\log ^{3}D\}\,d{\it\alpha}\ll \frac{(\log \log D)^{2}}{\log D}, & \displaystyle\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle \int _{-1}^{1}\min \{|{\it\alpha}|\log D,|{\it\alpha}|D^{-1/2}\log D+|{\it\alpha}|(1+{\it\alpha}^{2})D^{-|{\it\alpha}|/2}\log ^{3}D\}\,d{\it\alpha} & \displaystyle \nonumber\\ \displaystyle & \displaystyle \hspace{36.0pt}\ll (\log D)^{-1/3}. & \displaystyle\end{eqnarray}$$
Proof. The estimates (4.1), (4.2) follow from direct computations of integrals. The estimate (4.3) follows from
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{-1}^{1}\text{min}\{1,|{\it\alpha}|D^{-|{\it\alpha}|}\log ^{2}D\}\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad \ll \int _{0}^{1/\sqrt{\log D}}1\,d{\it\alpha}+\int _{\frac{1}{\sqrt{\log D}}}^{1}{\it\alpha}D^{-{\it\alpha}}\log ^{2}D\,d{\it\alpha}\nonumber\end{eqnarray}$$
and
$$\begin{eqnarray}\displaystyle & \displaystyle \int _{0}^{1/\sqrt{\log D}}1\,d{\it\alpha}\ll \frac{1}{\sqrt{\log D}}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \int _{\frac{1}{\sqrt{\log D}}}^{1}{\it\alpha}D^{-{\it\alpha}}\log ^{2}D\,d{\it\alpha}\ll D^{-1/\sqrt{\log D}}\log ^{1/2}D. & \displaystyle \nonumber\end{eqnarray}$$
The estimate (4.4) follows from
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{-1}^{1}\text{min}\{D^{|{\it\alpha}|}(\log D)^{-1},{\it\alpha}^{2}D^{-|{\it\alpha}|}\log ^{3}D\}\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad \ll \int _{0}^{\log \log D/\log D}D^{{\it\alpha}}(\log D)^{-1}\,d{\it\alpha}+(\log ^{3}D)\int _{\frac{\log \log D}{\log D}}^{1}{\it\alpha}^{2}D^{-{\it\alpha}}\,d{\it\alpha}\nonumber\end{eqnarray}$$
and
$$\begin{eqnarray}\displaystyle \int _{0}^{\log \log D/\log D}D^{{\it\alpha}}(\log D)^{-1}\,d{\it\alpha} & \ll & \displaystyle \frac{\log \log D}{\log D},\nonumber\\ \displaystyle (\log ^{3}D)\int _{\frac{\log \log D}{\log D}}^{1}{\it\alpha}^{2}D^{-{\it\alpha}}\,d{\it\alpha} & \ll & \displaystyle \frac{(\log \log D)^{2}}{\log D}.\nonumber\end{eqnarray}$$
Finally, the left-hand side of (4.5) is at most
$$\begin{eqnarray}\displaystyle & & \displaystyle 2\int _{0}^{(\log D)^{-2/3}}{\it\alpha}\log D\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad +\,2\int _{(\log D)^{-2/3}}^{1}\{{\it\alpha}D^{-1/2}\log D+{\it\alpha}(1+{\it\alpha}^{2})D^{-{\it\alpha}/2}\log ^{3}D\}\,d{\it\alpha},\nonumber\end{eqnarray}$$
and by direct computations, we easily find that these integrals are at most
$O((\log D)^{-1/3})$
.◻
Next, we mention the
$L$
-functions associated to Kronecker symbols. If
$d\not \equiv 3$
$(\text{mod}\;4)$
,
${\it\chi}_{d}=(d/\cdot )$
becomes a Dirichlet character modulo
$4|d|$
or
$|d|$
. In this case, we denote the conductor of
${\it\chi}_{d}$
by
$d^{\ast }$
. If
$d\equiv 3$
(
$\text{mod}\;4$
), by the reciprocity law of Kronecker symbols, we find that
$$\begin{eqnarray}{\it\chi}_{d}(p^{k})=\left\{\begin{array}{@{}ll@{}}\displaystyle \left(\frac{p}{d}\right)^{k}\quad & (p\equiv 1\;(\text{mod}\;4)\;\text{or}\;p=2)\\ \displaystyle (-1)^{k}\left(\frac{p}{d}\right)^{k}\quad & (p\equiv 3\;(\text{mod}\;4))\end{array}\right.\end{eqnarray}$$
for primes
$p$
. Hence, the
$L$
-function associated to
${\it\chi}_{d}$
is expressed by
$$\begin{eqnarray}L(s,{\it\chi}_{d})=\frac{1}{1-\left(\frac{2}{d}\right)2^{-s}}\mathop{\prod }_{p\geqslant 3}\frac{1}{1-{\it\eta}_{4}(p)\left(\frac{p}{d}\right)p^{-s}},\end{eqnarray}$$
where
${\it\eta}_{4}$
is the nonprincipal character of modulo 4. In this case, we denote the conductor of
${\it\eta}_{4}(\cdot )(\cdot /d)$
by
$d^{\ast }$
. Let
$N({\it\chi}_{d},T)$
denote the number of zeros of
$L(s,{\it\chi}_{d})$
in the rectangle
$0<{\it\sigma}<1$
,
$-T\leqslant t\leqslant T$
. Then, it is well known that
$$\begin{eqnarray}N({\it\chi}_{d},T)=\frac{T}{{\it\pi}}\log \frac{d^{\ast }T}{2{\it\pi}e}+O\left(\frac{\log (d^{\ast }T)}{\log \log (d^{\ast }T+3)}\right)\end{eqnarray}$$
holds uniformly for
$d^{\ast }T>1$
(for example, see [Reference Selberg17]). Then, by partial integration, we have
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{{\it\rho}\in Z_{d}}|K({\it\rho})|^{2} & = & \displaystyle \int _{0}^{\infty }\left|K\left(\frac{1}{2}+it\right)\right|^{2}\,dN({\it\chi}_{d},t)\nonumber\\ \displaystyle & = & \displaystyle \frac{\log d^{\ast }}{{\it\pi}}\int _{0}^{\infty }\left|K\left(\frac{1}{2}+it\right)\right|^{2}\,dt+O_{K}\left(\frac{\log d^{\ast }}{\log \log d^{\ast }}\right).\end{eqnarray}$$
We define the constants
$A_{+}^{\ast }$
,
$A_{-}^{\ast }$
by
$$\begin{eqnarray}\begin{array}{@{}c@{}}\displaystyle A_{-}^{\ast }=\liminf _{D\rightarrow \infty }\frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log d^{\ast },\\ \displaystyle A_{+}^{\ast }=\limsup _{D\rightarrow \infty }\frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log d^{\ast }.\end{array}\end{eqnarray}$$
It should be noticed that since
$\log d^{\ast }\leqslant \log d+O(1)$
, we have
$A_{+}^{\ast }\leqslant 1$
.
Corollary 4.2. Assuming the GRH, for any
$C>0$
,
$0\leqslant {\it\mu}<2{\it\pi}$
,
${\it\epsilon}>0$
, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\#\left\{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}\;\;|\text{Im}({\it\rho}_{i})|\leqslant C\;(i=1,2),\vphantom{\frac{{\it\mu}}{\log D}}\right.\nonumber\\ \displaystyle & & \displaystyle \hspace{175.20007pt}\left.0<|{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{{\it\mu}}{\log D}\right\}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \frac{1}{\sin ^{4}1}\left\{\left(\frac{4}{9}C+{\it\epsilon}\right)g({\it\mu})-B_{-}^{\ast }C+{\it\epsilon}\right\}.\end{eqnarray}$$
Here,
$$\begin{eqnarray}g({\it\mu})=\left(\frac{\frac{{\it\mu}}{2}}{\sin \frac{{\it\mu}}{2}}\right)^{2},\qquad B_{-}^{\ast }=\frac{1}{3}A_{-}^{\ast },\end{eqnarray}$$
and
$A_{-}^{\ast }$
is the constant given by (4.7).
Proof. We fix
${\it\lambda}$
such that
$1/2<{\it\lambda}\leqslant 1-{\it\delta}$
, and put
$$\begin{eqnarray}r(u)=\left(\frac{\sin {\it\pi}{\it\lambda}u}{{\it\pi}{\it\lambda}u}\right)^{2}.\end{eqnarray}$$
Then, its Fourier transform is given by
$$\begin{eqnarray}\hat{r}({\it\alpha})={\it\lambda}^{-2}\text{max}\{{\it\lambda}-|{\it\alpha}|,0\}.\end{eqnarray}$$
In particular,
$\hat{r}({\it\alpha})$
is bounded and has a support in
$[-1+{\it\delta},1-{\it\delta}]$
. Therefore, by (1.5) and Lemma 4.1, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \hspace{-5.0pt}\int _{-\infty }^{\infty }F_{K}({\it\alpha},D)\hat{r}({\it\alpha})\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \hspace{25.20007pt}=L(1)\int _{-\infty }^{\infty }|{\it\alpha}|\hat{r}({\it\alpha})\,d{\it\alpha}+(\log D)\int _{-\infty }^{\infty }a(D^{-|{\it\alpha}|})^{2}D^{-|{\it\alpha}|}\hat{r}({\it\alpha})\,d{\it\alpha}+o(1).\end{eqnarray}$$
The implied constant is dependent only on
$K(s)$
and
${\it\lambda}$
. By a direct computation, the first integral on the right-hand side of (4.9) is
$$\begin{eqnarray}\displaystyle \int _{-\infty }^{\infty }|{\it\alpha}|\hat{r}({\it\alpha})\,d{\it\alpha}=\frac{{\it\lambda}}{3}. & & \displaystyle\end{eqnarray}$$
On the other hand, the second integral of the right hand side of (4.9) is
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{-\infty }^{\infty }a(D^{-|{\it\alpha}|})^{2}D^{-|{\it\alpha}|}\hat{r}({\it\alpha})\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad =2\int _{0}^{\infty }a(D^{-{\it\alpha}})^{2}D^{-{\it\alpha}}\hat{r}({\it\alpha})\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad =2{\it\lambda}^{-2}\int _{0}^{{\it\lambda}}({\it\lambda}-{\it\alpha})a(D^{-{\it\alpha}})^{2}D^{-{\it\alpha}}\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad =2{\it\lambda}^{-1}\int _{0}^{{\it\lambda}}a(D^{-{\it\alpha}})^{2}D^{-{\it\alpha}}\,d{\it\alpha}-2{\it\lambda}^{-2}\int _{0}^{{\it\lambda}}a(D^{-{\it\alpha}})^{2}{\it\alpha}D^{-{\it\alpha}}\,d{\it\alpha}.\end{eqnarray}$$
By the change of parameters
$t=D^{-{\it\alpha}}$
, the first integral on the right-hand side of (4.11) is
$$\begin{eqnarray}\displaystyle \int _{0}^{{\it\lambda}}a(D^{-{\it\alpha}})^{2}D^{-{\it\alpha}}\,d{\it\alpha} & = & \displaystyle \int _{0}^{\infty }a(D^{-{\it\alpha}})^{2}D^{-{\it\alpha}}\,d{\it\alpha}\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{\log D}\int _{0}^{1}a(t)^{2}\,dt\nonumber\\ \displaystyle & = & \displaystyle \frac{M}{\log D},\end{eqnarray}$$
where
$$\begin{eqnarray}M=M_{K}:=\int _{0}^{1}a(t)^{2}\,dt.\end{eqnarray}$$
On the other hand, since
$a(D^{-{\it\alpha}})^{2}$
is bounded, we have
$$\begin{eqnarray}\displaystyle \int _{0}^{{\it\lambda}}a(D^{-{\it\alpha}})^{2}{\it\alpha}D^{-{\it\alpha}}d{\it\alpha}\ll \int _{0}^{\infty }{\it\alpha}D^{-{\it\alpha}}\,d{\it\alpha}\ll \frac{1}{\log ^{2}D}. & & \displaystyle\end{eqnarray}$$
By inserting (4.12), (4.14) into (4.11), we have
$$\begin{eqnarray}\displaystyle \int _{-\infty }^{\infty }a(D^{-|{\it\alpha}|})^{2}D^{-|{\it\alpha}|}\hat{r}({\it\alpha})\,d{\it\alpha}=\frac{2M}{{\it\lambda}\log D}+O\left(\frac{1}{\log ^{2}D}\right). & & \displaystyle\end{eqnarray}$$
By inserting (4.10), (4.15) into (4.9), we obtain
$$\begin{eqnarray}\int _{-\infty }^{\infty }F_{K}({\it\alpha},D)\hat{r}({\it\alpha})\,d{\it\alpha}=\frac{{\it\lambda}}{3}L(1)+\frac{2}{{\it\lambda}}M+o(1).\end{eqnarray}$$
For
$C>0$
, we take
$$\begin{eqnarray}K(s)=K_{C}(s)=C^{2}\left(\frac{e^{(s-(1/2))/C}-e^{-(s-(1/2))/C}}{2s-1}\right)^{2}.\end{eqnarray}$$
Then, since
$K(1/2+it)=C^{2}(t^{-1}\sin (t/C))^{2}$
,
$K(s)$
is real on the line
$\text{Re}(s)=1/2$
and satisfies
$K(1/2+it_{1})\overline{K(1/2+it_{2})}\geqslant 0$
for all
$t_{1},t_{2}\in \mathbf{R}$
. Moreover, by an easy computation, its Mellin inverse transform
$a(x)=a_{C}(x)$
, defined in (1.2), is given by
$$\begin{eqnarray}a(x)=\left\{\begin{array}{@{}ll@{}}0\quad & \displaystyle (x<e^{-2/C}\;\text{or}\;x\geqslant e^{2/C})\\ \displaystyle \frac{C^{2}}{2}x^{-1/2}\left(\frac{1}{C}+\frac{1}{2}\log x\right)\quad & (e^{-2/C}\leqslant x<1)\\ \displaystyle \frac{C^{2}}{2}x^{-1/2}\left(\frac{1}{C}-\frac{1}{2}\log x\right)\quad & (1\leqslant x<e^{2/C}).\end{array}\right.\end{eqnarray}$$
This function surely satisfies the conditions given in Section 1. By the definition of
$F_{K}({\it\alpha},D)$
, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{-\infty }^{\infty }F_{K}({\it\alpha},D)\hat{r}({\it\alpha})\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}}K({\it\rho}_{1})\overline{K({\it\rho}_{2})}\int _{-\infty }^{\infty }D^{i{\it\alpha}({\it\gamma}_{1}-{\it\gamma}_{2})}\hat{r}({\it\alpha})\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}}K({\it\rho}_{1})\overline{K({\it\rho}_{2})}r\left(\frac{({\it\gamma}_{1}-{\it\gamma}_{2})\log D}{2{\it\pi}}\right)\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}}K({\it\rho}_{1})\overline{K({\it\rho}_{2})}\left(\frac{\sin \left(\frac{({\it\gamma}_{1}-{\it\gamma}_{2}){\it\lambda}\log D}{2}\right)}{\frac{({\it\gamma}_{1}-{\it\gamma}_{2}){\it\lambda}\log D}{2}}\right)^{2}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
We fix a real number
${\it\mu}$
satisfying
$0\leqslant {\it\mu}<2{\it\pi}$
. Since the function
$f(x):=((\sin x)/x)^{2}$
is even and decreasing in
$[0,{\it\pi}]$
, the inequality
$$\begin{eqnarray}f\left(\frac{({\it\gamma}_{1}-{\it\gamma}_{2}){\it\lambda}\log D}{2}\right)\geqslant f\left(\frac{{\it\mu}}{2}\right)\end{eqnarray}$$
holds if
${\it\gamma}_{1}$
,
${\it\gamma}_{2}$
satisfy
$|{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant {\it\mu}/({\it\lambda}\log D)$
. Therefore, by (4.17), we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{-\infty }^{\infty }F_{K}({\it\alpha},D)\hat{r}({\it\alpha})\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant \frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{\substack{ {\it\rho}_{1},{\it\rho}_{2}\in Z_{d} \\ |{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{{\it\mu}}{{\it\lambda}\log D}}}K({\it\rho}_{1})\overline{K({\it\rho}_{2})}f\left(\frac{({\it\gamma}_{1}-{\it\gamma}_{2}){\it\lambda}\log D}{2}\right)\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant \frac{f\left(\frac{{\it\mu}}{2}\right)}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{\substack{ {\it\rho}_{1},{\it\rho}_{2}\in Z_{d} \\ |{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{{\it\mu}}{{\it\lambda}\log D}}}K({\it\rho}_{1})\overline{K({\it\rho}_{2})}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
Now, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\substack{ {\it\rho}_{1},{\it\rho}_{2}\in Z_{d} \\ |{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{{\it\mu}}{{\it\lambda}\log D}}}K({\it\rho}_{1})\overline{K({\it\rho}_{2})}\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant \mathop{\sum }_{\substack{ {\it\rho}_{1},{\it\rho}_{2}\in Z_{d} \\ 0<|{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{{\it\mu}}{{\it\lambda}\log D} \\ |\text{Im}({\it\rho}_{1})|,|\text{Im}({\it\rho}_{2})|\leqslant C}}K({\it\rho}_{1})\overline{K({\it\rho}_{2})}+\mathop{\sum }_{{\it\rho}\in Z_{d}}|K({\it\rho})|^{2}.\end{eqnarray}$$
By (4.6), the second term on the right-hand side of (4.19) is given by
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{{\it\rho}\in Z_{d}}|K({\it\rho})|^{2} & = & \displaystyle \frac{\log d^{\ast }}{{\it\pi}}C^{4}\int _{0}^{\infty }\left(\frac{\sin \frac{t}{C}}{t}\right)^{4}\,dt+O_{C}\left(\frac{\log d^{\ast }}{\log \log d^{\ast }}\right)\nonumber\\ \displaystyle & = & \displaystyle \frac{C\log d^{\ast }}{{\it\pi}}\int _{0}^{\infty }\left(\frac{\sin t}{t}\right)^{4}\,dt+O_{C}\left(\frac{\log d^{\ast }}{\log \log d^{\ast }}\right)\nonumber\\ \displaystyle & = & \displaystyle \frac{C}{3}\log d^{\ast }+O_{C}\left(\frac{\log d^{\ast }}{\log \log d^{\ast }}\right).\end{eqnarray}$$
On the other hand, since
$$\begin{eqnarray}\min _{|y|\leqslant C}\left|K\left(\frac{1}{2}+iy\right)\right|^{2}=\sin ^{4}1,\end{eqnarray}$$
the first term on the right-hand side of (4.19) is
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\substack{ {\it\rho}_{1},{\it\rho}_{2}\in Z_{d} \\ 0<|{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{{\it\mu}}{{\it\lambda}\log D} \\ |\text{Im}({\it\rho}_{1})|,|\text{Im}({\it\rho}_{2})|\leqslant C}}K({\it\rho}_{1})\overline{K({\it\rho}_{2})}\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant (\sin ^{4}1)\#\left\{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}\;\;|\text{Im}({\it\rho}_{i})|\leqslant C\;(i=1,2),0<|{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{{\it\mu}}{{\it\lambda}\log D}\right\}\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant (\sin ^{4}1)\#\left\{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}\;\;|\text{Im}({\it\rho}_{i})|\leqslant C\;(i=1,2),0<|{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{{\it\mu}}{\log D}\right\}.\nonumber\end{eqnarray}$$
Hence, by (4.18), we obtain
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{-\infty }^{\infty }F_{K}({\it\alpha},D)\hat{r}({\it\alpha})\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant \frac{f\left(\frac{{\it\mu}}{2}\right)}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \left\{(\sin ^{4}1)\#\left\{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}\;\;|\text{Im}({\it\rho}_{i})|\leqslant C\;(i=1,2),\vphantom{\frac{{\it\mu}}{\log D}}\right.\right.\nonumber\\ \displaystyle & & \displaystyle \left.\left.\quad \quad \quad \quad \quad \quad 0<|{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{{\it\mu}}{\log D}\right\}+\frac{C}{3}\log d^{\ast }+O_{C}\left(\frac{\log d^{\ast }}{\log \log d^{\ast }}\right)\right\}.\end{eqnarray}$$
Combining (4.16) and (4.21), we obtain
$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\#\left\{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}\;\;|\text{Im}({\it\rho}_{i})|\leqslant C\;(i=1,2),\vphantom{\frac{{\it\mu}}{\log D}}\right.\nonumber\\ \displaystyle & & \displaystyle \left.\qquad \qquad \qquad \qquad \qquad \quad 0<|{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{{\it\mu}}{\log D}\right\}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \frac{1}{f\left(\frac{{\it\mu}}{2}\right)\sin ^{4}1}\left\{\vphantom{-\,\frac{f\left(\frac{{\it\mu}}{2}\right)}{D\log D}\cdot \frac{C}{3}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log d^{\ast }+o(1)}\frac{{\it\lambda}}{3}L(1)+\frac{2}{{\it\lambda}}M\right.\nonumber\\ \displaystyle & & \displaystyle \left.\qquad \qquad \qquad \quad \quad -\,\frac{f\left(\frac{{\it\mu}}{2}\right)}{D\log D}\cdot \frac{C}{3}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log d^{\ast }+o(1)\right\}.\end{eqnarray}$$
By direct computation, we find that
$$\begin{eqnarray}L(1)=\int _{0}^{\infty }a(x)^{2}\,dx=\frac{C}{3},\qquad M=\int _{0}^{1}a(x)^{2}\,dx=\frac{C}{6}.\end{eqnarray}$$
Moreover, by our assumption,
$$\begin{eqnarray}\frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log d^{\ast }\geqslant A_{-}^{\ast }.\end{eqnarray}$$
By inserting these results into (4.22) and putting
${\it\lambda}=1-{\it\delta}$
with sufficiently small
${\it\delta}>0$
, we obtain (4.8).◻
Next, we give a certain lower bound for the rate of simple zeros of quadratic
$L$
-functions.
Corollary 4.3. We assume the GRH. For any
${\it\epsilon}>0$
, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{\substack{ {\it\rho}\in Z_{d} \\ \text{simple}}}K({\it\rho})^{2}\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant \left(2B_{-}^{\ast }-\frac{4}{9}-{\it\epsilon}\right)(B_{+}^{\ast })^{-1}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{{\it\rho}\in Z_{d}}K({\it\rho})^{2}\end{eqnarray}$$
when
$D>1$
is sufficiently large. Here,
$$\begin{eqnarray}K(s)=K_{1}(s)=\left(\frac{e^{s-\frac{1}{2}}-e^{-s+\frac{1}{2}}}{2s-1}\right)^{2},\qquad B_{\pm }^{\ast }=\frac{1}{3}A_{\pm }^{\ast },\end{eqnarray}$$
and
$A_{\pm }^{\ast }$
are the constants defined by (4.7).
Proof. In the proof of Corollary 4.2, we showed that
$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}}K({\it\rho}_{1})\overline{K({\it\rho}_{2})}\left(\frac{\sin \left(\frac{({\it\gamma}_{1}-{\it\gamma}_{2}){\it\lambda}\log D}{2}\right)}{\frac{({\it\gamma}_{1}-{\it\gamma}_{2}){\it\lambda}\log D}{2}}\right)^{2}\nonumber\\ \displaystyle & & \displaystyle \qquad \quad =\int _{-\infty }^{\infty }F_{K}({\it\alpha},D)\hat{r}({\it\alpha})\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \qquad \quad =\frac{{\it\lambda}}{3}L(1)+\frac{2}{{\it\lambda}}M+o(1)\end{eqnarray}$$
holds for
${\it\lambda}=1-{\it\delta}$
. Moreover, in this case, we have
$$\begin{eqnarray}L(1)=\int _{0}^{\infty }a(x)^{2}\,dx=\frac{1}{3},\qquad M=\int _{0}^{1}a(x)^{2}\,dx=\frac{1}{6},\end{eqnarray}$$
since
$C=1$
. Therefore,
$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}}K({\it\rho}_{1})\overline{K({\it\rho}_{2})}\left(\frac{\sin \left(\frac{({\it\gamma}_{1}-{\it\gamma}_{2}){\it\lambda}\log D}{2}\right)}{\frac{({\it\gamma}_{1}-{\it\gamma}_{2}){\it\lambda}\log D}{2}}\right)^{2}\nonumber\\ \displaystyle & & \displaystyle \qquad \quad =\frac{{\it\lambda}}{9}+\frac{1}{3{\it\lambda}}+o(1).\end{eqnarray}$$
Let
$m_{{\it\rho},d}$
be the multiplicity of the zero of
$L(s,{\it\chi}_{d})$
at
${\it\rho}=1/2+i{\it\gamma}$
. Then,
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{\substack{ {\it\rho}\in Z_{d} \\ \text{simple}}}K({\it\rho})^{2} & {\geqslant} & \displaystyle \mathop{\sum }_{{\it\rho}\in Z_{d}}(2-m_{{\it\rho},d})K({\it\rho})^{2}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle 2\mathop{\sum }_{{\it\rho}\in Z_{d}}K({\it\rho})^{2}-\mathop{\sum }_{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}}K({\it\rho}_{1})K({\it\rho}_{2})\left(\frac{\sin \left(\frac{({\it\gamma}_{1}-{\it\gamma}_{2}){\it\lambda}\log D}{2}\right)}{\frac{({\it\gamma}_{1}-{\it\gamma}_{2}){\it\lambda}\log D}{2}}\right)^{2}.\nonumber\end{eqnarray}$$
Hence, we obtain
$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{\substack{ {\it\rho}\in Z_{d} \\ \text{simple}}}K({\it\rho})^{2}\nonumber\\ \displaystyle & & \displaystyle \qquad \geqslant \frac{2}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{{\it\rho}\in Z_{d}}K({\it\rho})^{2}\nonumber\\ \displaystyle & & \displaystyle \quad -\,\frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}}K({\it\rho}_{1})K({\it\rho}_{2})\left(\frac{\sin \left(\frac{({\it\gamma}_{1}-{\it\gamma}_{2}){\it\lambda}\log D}{2}\right)}{\frac{({\it\gamma}_{1}-{\it\gamma}_{2}){\it\lambda}\log D}{2}}\right)^{2}.\end{eqnarray}$$
The second term on the right-hand side is given by (4.25), since
$K({\it\rho}_{2})$
is real. On the other hand, the first term is
$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{{\it\rho}\in Z_{d}}K({\it\rho})^{2}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{{\it\pi}}\int _{0}^{\infty }\left(\frac{\sin t}{t}\right)^{4}\,dt\cdot \frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\log d^{\ast }+o(1)\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant \frac{1}{3}A_{-}^{\ast }+o(1)\nonumber\\ \displaystyle & & \displaystyle \quad =B_{-}^{\ast }+o(1).\end{eqnarray}$$
Therefore, by inserting (4.25), (4.27) into (4.26) and taking
${\it\delta}>0$
sufficiently small, we obtain
$$\begin{eqnarray}\frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{\substack{ {\it\rho}\in Z_{d} \\ \text{simple}}}K({\it\rho})^{2}\geqslant 2B_{-}^{\ast }-\frac{4}{9}-{\it\epsilon}.\end{eqnarray}$$
By combining this and
$$\begin{eqnarray}\frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{{\it\rho}\in Z_{d}}K({\it\rho})^{2}\leqslant B_{+}^{\ast }+o(1),\end{eqnarray}$$
we obtain (4.23).◻
Corollary 4.4. We assume the GRH, and that all zeros of
$L(s,{\it\chi}_{d})$
are simple. For
$0<{\it\lambda}<1$
, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{\substack{ {\it\rho}_{1},{\it\rho}_{2}\in Z_{d} \\ 0<|{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{2{\it\pi}{\it\lambda}}{\log D}}}K({\it\rho}_{1})K({\it\rho}_{2})\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant \frac{2}{3}{\it\lambda}-\frac{2}{9}{\it\lambda}^{2}-\frac{\cos 2{\it\pi}{\it\lambda}}{6{\it\pi}^{2}}+\frac{\sin 2{\it\pi}{\it\lambda}}{12{\it\pi}^{3}{\it\lambda}}-B_{+}^{\ast }+o(1)\end{eqnarray}$$
as
$D\rightarrow \infty$
, where
$B_{+}^{\ast }=A_{+}^{\ast }/3$
and
$$\begin{eqnarray}K(s)=K_{1}(s)=\left(\frac{e^{s-(1/2)}-e^{-s+(1/2)}}{2s-1}\right)^{2}.\end{eqnarray}$$
In particular, if
${\it\lambda}>{\it\lambda}_{0}=0.6073$
, we have
$$\begin{eqnarray}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{\substack{ {\it\rho}_{1},{\it\rho}_{2}\in Z_{d} \\ 0<|{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{2{\it\pi}{\it\lambda}}{\log D}}}K({\it\rho}_{1})K({\it\rho}_{2})\gg D\log D\end{eqnarray}$$
as
$D\rightarrow \infty$
.
Proof. Instead of the function
$r(u)$
used in the proof of Corollary 4.2, we use the Selberg minorant
$$\begin{eqnarray}h(u)=\left(\frac{\sin {\it\pi}u}{{\it\pi}u}\right)^{2}\frac{1}{1-u^{2}}.\end{eqnarray}$$
This function is bounded and satisfies
$h(u)\leqslant 1$
,
$h(u)<0$
if
$|u|>1$
. The Fourier transform of
$h(u)$
is given by
$$\begin{eqnarray}{\hat{h}}({\it\alpha})=\left\{\begin{array}{@{}ll@{}}\displaystyle 1-|{\it\alpha}|+\frac{\sin 2{\it\pi}|{\it\alpha}|}{2{\it\pi}}\quad & (|{\it\alpha}|\leqslant 1)\\ 0\quad & (|{\it\alpha}|>1)\\ \quad \end{array}\right.\end{eqnarray}$$
(for example, see [Reference Goldston, Gonek, Özlük and Snyder5]). For
$0<{\it\lambda}<1$
, we give lower and upper bounds for the integral
$$\begin{eqnarray}\int _{-\infty }^{\infty }F_{K}({\it\alpha},D)\cdot {\it\lambda}{\hat{h}}({\it\lambda}{\it\alpha})\,d{\it\alpha}.\end{eqnarray}$$
First, since the integrant is nonnegative and
$1/{\it\lambda}>1$
, by (1.5), we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{-\infty }^{\infty }F_{K}({\it\alpha},D)\cdot {\it\lambda}{\hat{h}}({\it\lambda}{\it\alpha})\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad =\int _{-\frac{1}{{\it\lambda}}}^{\frac{1}{{\it\lambda}}}F_{K}({\it\alpha},D)\cdot {\it\lambda}{\hat{h}}({\it\lambda}{\it\alpha})\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant \int _{-1}^{1}F_{K}({\it\alpha},D)\cdot {\it\lambda}{\hat{h}}({\it\lambda}{\it\alpha})\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad ={\it\lambda}L(1)\int _{-1}^{1}|{\it\alpha}|\left\{1-|{\it\lambda}{\it\alpha}|+\frac{\sin 2{\it\pi}{\it\lambda}|{\it\alpha}|}{2{\it\pi}}\right\}\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \qquad \quad +\,{\it\lambda}\log D\int _{-1}^{1}a(D^{-|{\it\alpha}|})^{2}D^{-|{\it\alpha}|}\left\{1-|{\it\lambda}{\it\alpha}|+\frac{\sin 2{\it\pi}{\it\lambda}|{\it\alpha}|}{2{\it\pi}}\right\}\,d{\it\alpha}+o(1).\end{eqnarray}$$
The first term on the right-hand side of (4.30) is
$$\begin{eqnarray}\displaystyle & & \displaystyle {\it\lambda}L(1)\int _{-1}^{1}|{\it\alpha}|\left\{1-|{\it\lambda}{\it\alpha}|+\frac{\sin 2{\it\pi}{\it\lambda}|{\it\alpha}|}{2{\it\pi}}\right\}\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad =2{\it\lambda}L(1)\int _{0}^{1}\left({\it\alpha}-{\it\lambda}{\it\alpha}^{2}+\frac{{\it\alpha}\sin 2{\it\pi}{\it\lambda}{\it\alpha}}{2{\it\pi}}\right)d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad =2{\it\lambda}L(1)\left\{\frac{1}{2}-\frac{{\it\lambda}}{3}-\frac{\cos 2{\it\pi}{\it\lambda}}{4{\it\pi}^{2}{\it\lambda}}+\frac{\sin 2{\it\pi}{\it\lambda}}{8{\it\pi}^{3}{\it\lambda}^{2}}\right\}.\end{eqnarray}$$
Next, we compute the second term on the right-hand side of (4.30). Since
$$\begin{eqnarray}1-|{\it\lambda}{\it\alpha}|+\frac{\sin 2{\it\pi}{\it\lambda}|{\it\alpha}|}{2{\it\pi}}=1+O\left(\frac{1}{\log ^{3}D}\right)\end{eqnarray}$$
for
$|{\it\alpha}|\ll 1/(\log D)$
, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle {\it\lambda}\log D\int _{-1}^{1}a(D^{-|{\it\alpha}|})^{2}D^{-|{\it\alpha}|}\left\{1-|{\it\lambda}{\it\alpha}|+\frac{\sin 2{\it\pi}{\it\lambda}|{\it\alpha}|}{2{\it\pi}}\right\}\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad =2{\it\lambda}\log D\int _{0}^{\infty }a(D^{-{\it\alpha}})^{2}D^{-{\it\alpha}}\left\{1+O\left(\frac{1}{\log ^{3}D}\right)\right\}\,d{\it\alpha}.\nonumber\end{eqnarray}$$
By the change of parameters
$D^{-{\it\alpha}}=v$
, we have
$$\begin{eqnarray}\int _{0}^{\infty }a(D^{-{\it\alpha}})^{2}D^{-{\it\alpha}}\,d{\it\alpha}=\frac{1}{\log D}\int _{0}^{1}a(v)^{2}\,dv=\frac{M}{\log D},\end{eqnarray}$$
where
$M=\int _{0}^{1}a(v)^{2}\,dv$
. Therefore,
$$\begin{eqnarray}\displaystyle {\it\lambda}\log D\int _{-1}^{1}a(D^{-|{\it\alpha}|})^{2}D^{-|{\it\alpha}|}\left\{1-|{\it\lambda}{\it\alpha}|+\frac{\sin 2{\it\pi}{\it\lambda}|{\it\alpha}|}{2{\it\pi}}\right\}\,d{\it\alpha}=2M{\it\lambda}+o(1). & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
By inserting (4.31), (4.32) into (4.30), we obtain
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{-\infty }^{\infty }F_{K}({\it\alpha},D)\cdot {\it\lambda}{\hat{h}}({\it\lambda}{\it\alpha})\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant 2{\it\lambda}L(1)\left\{\frac{1}{2}-\frac{{\it\lambda}}{3}-\frac{\cos 2{\it\pi}{\it\lambda}}{4{\it\pi}^{2}{\it\lambda}}+\frac{\sin 2{\it\pi}{\it\lambda}}{8{\it\pi}^{3}{\it\lambda}^{2}}\right\}+2M{\it\lambda}+o(1)\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{2}{3}{\it\lambda}-\frac{2}{9}{\it\lambda}^{2}-\frac{\cos 2{\it\pi}{\it\lambda}}{6{\it\pi}^{2}}+\frac{\sin 2{\it\pi}{\it\lambda}}{12{\it\pi}^{3}{\it\lambda}}+o(1),\end{eqnarray}$$
since
$L(1)=1/3$
,
$M=1/6$
. On the other hand, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{-\infty }^{\infty }F_{K}({\it\alpha},D)\cdot {\it\lambda}{\hat{h}}({\it\lambda}{\it\alpha})\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \qquad \quad =\frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{{\it\rho}_{1},{\it\rho}_{2}\in Z_{d}}K({\it\rho}_{1})K({\it\rho}_{2})h\left(\frac{({\it\gamma}_{1}-{\it\gamma}_{2})\log D}{2{\it\pi}{\it\lambda}}\right).\end{eqnarray}$$
Now, since
$h(({\it\gamma}_{1}-{\it\gamma}_{2})\log D/(2{\it\pi}{\it\lambda}))$
is negative if
$|{\it\gamma}_{1}-{\it\gamma}_{2}|>(2{\it\pi}{\it\lambda})/\log D$
, by (4.34), we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{-\infty }^{\infty }F_{K}({\it\alpha},D)\cdot {\it\lambda}{\hat{h}}({\it\lambda}{\it\alpha})\,d{\it\alpha}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{\substack{ {\it\rho}_{1},{\it\rho}_{2}\in Z_{d} \\ |{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{2{\it\pi}{\it\lambda}}{\log D}}}K({\it\rho}_{1})K({\it\rho}_{2})h\left(\frac{({\it\gamma}_{1}-{\it\gamma}_{2})\log D}{2{\it\pi}{\it\lambda}}\right)\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{{\it\rho}\in Z_{d}}K({\it\rho})^{2}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{\substack{ {\it\rho}_{1},{\it\rho}_{2}\in Z_{d} \\ 0<|{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{2{\it\pi}{\it\lambda}}{\log D}}}K({\it\rho}_{1})K({\it\rho}_{2})h\left(\frac{({\it\gamma}_{1}-{\it\gamma}_{2})\log D}{2{\it\pi}{\it\lambda}}\right)\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant B_{+}^{\ast }+\frac{1}{D\log D}\mathop{\sum }_{d}e^{-{\it\pi}d^{2}/D^{2}}\mathop{\sum }_{\substack{ {\it\rho}_{1},{\it\rho}_{2}\in Z_{d} \\ 0<|{\it\gamma}_{1}-{\it\gamma}_{2}|\leqslant \frac{2{\it\pi}{\it\lambda}}{\log D}}}K({\it\rho}_{1})K({\it\rho}_{2})+o(1).\nonumber\end{eqnarray}$$
By combining (4.33) and the above, we obtain (4.28). Since
$B_{+}^{\ast }=A_{+}^{\ast }/3\leqslant 1/3$
, the right-hand side of (4.28) becomes positive if
${\it\lambda}>{\it\lambda}_{0}=0.6073$
. Therefore, we obtain (4.29).◻
Acknowledgments
The author would like to express his gratitude to the referees for giving him valuable comments and suggestions.
