A computation identical to <cit.> gives

\begin{equation}\label{eqDZ1}
\lim_{X\to\infty}
\frac{\CS_\Sigma\bigl(Z_K^{(1)}(X),X\bigr)}{\CS_\Sigma(1,X)}
\end{equation}

To compute the $1$-level density, we need to compute the asymptotics
of $\CS_\Sigma(Z_K^{(2)}(X),X)$. We write

\begin{equation}\label{eqLLZZ2}
\begin{array}{rcl}
\displaystyle\CS_\Sigma(Z_K^{(2)}(X),X)&=&
\displaystyle-\frac{2}{\L_X}
\CS_\Sigma\Bigl(\sum_{n=1}^\infty\frac{\theta_K(n)\Lambda(n)}{\sqrt{n}}
\widehat{\Phi}\Bigl(\frac{\log n}{\L_X}\Bigr),X\Bigr)
\\[.2in]&=&
\displaystyle-\frac{2}{\L_X}
\sum_{p,m}\frac{\log p}{p^{m/2}}
\widehat{\Phi}\Bigl(\frac{m\log p}{\L_X}\Bigr)
\CS_\Sigma\bigl(\theta_K(p^m),X\bigr).
\end{array}
\end{equation}

We now have the following result estimating the ratios

Let $p$ be a prime number, and let $X\ge 1$ be a real number. Then,
for integers $m\geq 3$, we have

\begin{equation}\label{eqllzb}
\begin{array}{rcl}
\displaystyle \frac{\CS_\Sigma(\theta_K(p),X)}{\CS_\Sigma(1,X)}&\ll_\epsilon&
\displaystyle \frac{1}{p}+\frac{1}{p^{1/3}X^{1/6}}+\frac{p^{1/3}}{X^{1/3-\epsilon}};
\\[.2in]
\displaystyle \frac{\CS_\Sigma(\theta_K(p^2),X)}{\CS_\Sigma(1,X)}-1&\ll_\epsilon&
\displaystyle \frac{1}{p^2}+\frac{1}{X^{1/6}}+
\frac{p^{2/3}}{X^{1/3-\epsilon}};
\\[.2in]
\displaystyle \frac{\CS_\Sigma(\theta_K(p^m),X)}{\CS_\Sigma(1,X)}&\ll&
\displaystyle 1.
\end{array}
\end{equation}

From Lemma <ref> we have that $\theta_K(p)=\lambda_K(p)$.
The left-hand side of the first line of (<ref>) can be computed
from Theorem <ref>, yielding

\begin{equation*}
\frac{\CS_\Sigma(\theta_K(p),X)}{\CS_\Sigma(1,X)}\ll_\epsilon
\cA^\max(\lambda_p \chi_\Sigma)+X^{-1/6}\cC^\max(\lambda_p \chi_\Sigma)
\end{equation*}

The required bound then follows from the first part of Proposition
<ref>, together with the definitions of $\cA^\max$ and

The proof of the second inequality is identical: we merely use Theorem
<ref> instead of Theorem <ref> and the
third part of Proposition <ref> instead of the first.
Finally, Lemma <ref> states that $|\theta_K(p^m)|\leq 2$,
from which the third inequality follows immediately.

We are now ready to prove the main result of this section.

From (<ref>) and Proposition <ref>, we obtain

\begin{equation*}
\begin{array}{rcl}
\displaystyle-\frac{\CS_\Sigma(Z_K^{(2)}(X),X)}{\CS_\Sigma(1,X)}&=&
\displaystyle\frac{2}{\L_X}
\sum_{p}\frac{\log p}{p}
\widehat{\Phi}\Bigl(\frac{2\log p}{\L_X}\Bigr)
\frac{\CS_\Sigma(\theta_K(p^2),X)}{\CS_\Sigma(1,X)}+O\biggl(\frac{1}{\log X}
\sum_{\substack{p^m\ll X^a\\m\neq 2}}\frac{\log p}{p^{m/2}}
\frac{\CS_\Sigma(\theta_K(p^m),X)}{\CS_\Sigma(1,X)}\biggr)
\\[.25in]&=&
\displaystyle\frac{2}{\L_X}
\sum_{p}\frac{\log p}{p}\widehat{\Phi}\Bigl(\frac{2\log p}{\L_X}\Bigr)
+O_\epsilon\Bigl(\frac{1}{\log X}+X^{\frac{a-1}{6}}
\\[.25in]&&+
\displaystyle O_\epsilon\Bigl(\frac{1}{\log X}+X^{-\frac{1}{6}+\epsilon}
+O\Bigl(\frac{1}{\log X}\Bigr),
\end{array}
\end{equation*}

where the three error terms respectively arise from the three
estimates of Proposition <ref>. Assuming that
$a<\frac{2}{5}$, and using the above computation in conjunction
with (<ref>), gives

\begin{equation*}
\begin{array}{rcl}
\displaystyle
\lim_{X\to \infty} \cD(\FF_\Sigma(X),\Phi)&=&
\displaystyle\widehat{\Phi}(0)-\lim_{X\to \infty}
\displaystyle\frac{2}{\L_X}
\sum_{p}\frac{\log p}{p}\widehat{\Phi}\Bigl(\frac{2\log p}{\L_X}\Bigr)
\\[.2in]&=&
\displaystyle\widehat{\Phi}(0)-\frac12\int_{-1}^{1}\widehat{\Phi}(t)dt,
\end{array}
\end{equation*}

where the final equality follows from the prime number
theorem. This concludes the proof of Theorem <ref>.

§ MAIN TERM FOR THE AVERAGE CENTRAL VALUES

Let $\Sigma=(\Sigma_v)$ be a finite set of local
specifications.  Without loss of generality we assume that
$\Sigma_\infty$ is a singleton set, which is to say that either the
cubic fields prescribed by $\Sigma_\infty=\{\R\times \R \times \R\}$
are all totally real, or the cubic fields prescribed by $\Sigma_\infty
= \{\R\times \C\}$ are all complex.  We also assume (by adding a prime
if necessary) that there exists a prime $p$
such that $\Sigma_{p}=\{\Q_{p^3}\}$. Let $\FF_\Sigma$ denote the
family of cubic fields $K$ prescribed by the set $\Sigma$ of specifications,
namely such that for each place $v$ we have $K
\otimes_\Q \Q_v\in\Sigma_v$.

We let $V(\Z)(\Sigma)$ denote the set of elements $f\in V(\Z)$ such
that $\chi_\Sigma(f)=1$ and such that $\Delta(f)>0$ if
$\Sigma_\infty = \{\R\times \R\times \R\}$ (resp.  $\Delta(f)<0$ if
$\Sigma_\infty=\{\R\times \C\}$).  For each prime $p$,
let $\W_p(\Sigma)$ denote the set of elements in $V(\Z)(\Sigma)$ that
are nonmaximal at $p$. If $q$ is a squarefree positive integer, we set
$\W_q(\Sigma)=\cap_{p\mid q}\W_p(\Sigma)$. In particular

Thanks to the condition $\Sigma_{p}=\{\Q_{p^3}\}$, we have that every form $f\in V(\Z)(\Sigma)$
is irreducible. This implies that the set  $\overline{V(\Z)(\Sigma)^{\max}}$ of $\GL_2(\Z)$-orbits
parametrizes under the Delone–Faddeev correspondence
the family
$\FF_\Sigma$ of cubic fields prescribed by the finite set $\Sigma$ of specifications.

Let $\Psi:\R_{>0} \to \C$ be a smooth function of compact support with
$\int \Psi=1$. Note that since $\Stab(f)$ is nontrivial in our families
only when $f$ corresponds to a $C_3$-order, there are only $O(X^{1/2})$
such forms with discriminant bounded by $X$, so the contribution from
elements with nontrivial stabilizer is negligible. For the rest of
this paper, we automatically assume that every sum of binary cubic
forms $f$ is weighted by $1/|\Stab(f)|$. For a real number $X\ge 1$,
the inclusion-exclusion principle in conjunction with
Proposition <ref> yields:

\begin{equation}\label{eqnvzsieve}
\sum_{K\in \FF_\Sigma}L(\tfrac 12,\rho_K)\Psi\Bigl(\frac{|\Delta(K)|}{X}\Bigr)=
2\sum_{q\geq 1}\mu(q)\sum_{f\in
\overline{\W_q(\Sigma)}} S(f)
\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr) + O_{\epsilon}\bigl(X^{\frac34 -\delta + \epsilon}\bigr),
\end{equation}

$A_\Sigma(X)$, smoothed first moment of $L(\tfrac12,\rho_K)$
where $S(f)$ was defined in (<ref>) to be

\begin{equation}\label{eqL12f}
\end{equation}

with $V^\pm$ as in Proposition <ref> and where the sign is $+$
if $\Sigma_\infty=\{\R\times \R\times \R\}$ and $-$ if $\Sigma_\infty
= \{\R\times \C\}$.
The identity holds because for a maximal irreducible binary
cubic form $f\in V(\Z)^{\irr,\max}$ corresponding to the ring of
integers of a cubic field $K_f$, we have $2S(f)=L(\tfrac12,\rho_{K_f})$ by
Corollary <ref> and Proposition <ref>.

In this section, we will prove two results. First, we will prove an
upper bound on $A_\Sigma(X)$, which improves on the pointwise bound
coming from summing the best known upper bounds on $|L(\tfrac12,\rho_K)|$
over the associated fields $K$.  Second, assuming a sufficiently
strong upper bound on $|L(\tfrac12,\rho_K)|$, we obtain asymptotics for

§.§ Asymptotics for the terms with $q<Q$

For $Q\in \R_{\ge 1}$ to be chosen later, we split the right-hand side of
(<ref>) into two parts,

\begin{equation*}
\sum_{q<Q} \quad \text{and} \quad  \sum_{q\ge Q}.
\end{equation*}

This section is concerned with the first part:

\begin{equation}\label{eqlongest}
\displaystyle 2
\sum_{q<Q}\mu(q)
\sum_{f\in \overline{\W_q(\Sigma)}}
\sum_{n=1}^\infty
\frac{\lambda_n(f)}{n^{1/2}}
\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)
V^\pm \Bigl(\frac{n}{\sqrt{|\Delta(f)|}}\Bigr).
\end{equation}

It will be convenient for us to set some notation surrounding the
smooth functions above and their Mellin transforms.  For any
positive real number $y\in \R_{> 0}$, let $\H_y:\R_{>0}\to \C$
denote the compactly supported function

\begin{equation}\label{eqimpfunc}
\H_y(t):=
\Psi(t) \cdot
\end{equation}

The relevance of $\H_y(t)$ is that we have the equality

\begin{equation*}
\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)
\end{equation*}

$\H_y$, compactly supported function on $\R_{>0}$

$($i$)$ There exists a constant $C>0$ depending only on $\Psi$
such that for every $\epsilon \in [-1,1]$ and $y\in \R_{>0}$,

\begin{equation*}
E_\infty(\widetilde{\cH_y};\epsilon)= \int^\infty_{-\infty}
|\widetilde{\mathcal H_y}(-\epsilon + ir)| (1+|r|)^{2+4\epsilon} dr \le C.
\end{equation*}

$($ii$)$ There exists a constant $C>0$ depending only on $\Psi$ such that for every $y\in
\R_{>0}$,
$|\widetilde{\mathcal H_y}(\frac 56)| \le C$.

We have by definition (<ref>),

\[
\widetilde{V^\pm}(s) = G(s) \frac{\gamma^\pm(\tfrac12+s)}{\gamma^\pm(\tfrac12)\cdot s}.
\]

We deduce that the Mellin transform of $t\mapsto V^\pm(\frac{y}{\sqrt{t}})$ is equal to

\[
2 y^{2s} \widetilde{V^\pm}(-2s) = -y^{2s} G(-2s) \frac{\gamma^\pm(\tfrac12-2s)}{\gamma^\pm(\tfrac12)\cdot s}.
\]

Since $\mathcal H_y$ is the product of the two functions $\Psi$ and  $t\mapsto
V^\pm(\frac{y}{\sqrt{t}})$, its Mellin transform is the convolution of the Mellin transforms of
the respective functions:

\[
\widetilde{\mathcal H_y}(\sigma + ir) =
2 \int_{\Re(u)=\eta}
\widetilde \Psi(\sigma+ir + u)
y^{-2u} \widetilde{V^\pm}(2u) \frac{du}{2\pi i},
\]

where $0 <\eta < \tfrac 12$ is fixed.
We deduce the following inequality:

\[
|\widetilde{\mathcal H_y}(\sigma + ir)| \le
\frac{y^{-2\eta}}{\pi}   \int^\infty_{-\infty}
|\widetilde \Psi(\sigma +ir +\eta  + i\tau )| \cdot |\widetilde{V^\pm}(2\eta + 2i\tau )| d\tau.
\]

We shall use this inequality for $y\in [1,+\infty)$, in which case $y^{-2\eta}\le 1$.
We also shift the contour further to $\Re(u)=-\eta$, picking up a
simple pole at $u=0$, to obtain the inequality:

\[
|\widetilde{\mathcal H_y}(\sigma + ir)| \le
\frac{y^{2\eta}}{\pi}   \int^\infty_{-\infty}
|\widetilde \Psi(\sigma +ir - \eta  + i\tau )| \cdot |\widetilde{V^\pm}(-2\eta + 2i\tau )| d\tau + |\widetilde \Psi(\sigma +
ir)|\cdot |G(0)|.
\]

We shall use this inequality for the other interval $y\in (0,1]$, in which case $y^{2\eta}\le

Assertion (ii) follows immediately by inserting $\sigma=\tfrac 56$ and $r=0$.
Assertion (i) follows by inserting $\sigma=-\epsilon$ and integrating over $r$ because
$E_\infty(\widetilde{\cH_y};\epsilon)$ for $y\ge 1$ is bounded by

\[
\frac{1}{\pi}   \int^\infty_{-\infty}
|\widetilde \Psi(-\epsilon+\eta +ir)| (1+|r|)^{2+4\epsilon} dr
\cdot \int^\infty_{-\infty} |\widetilde{V^\pm}(2\eta + 2i\tau )| (1+|\tau|)^{2+4\epsilon} d\tau \le C,
\]

where $C$ depends only on $\Psi$. The estimate for $y\le 1$ is similar.

We are now ready to prove the main result of this subsection.

For every $\epsilon>0$ and $Q,X\ge 1$, the sum (<ref>) is
asymptotic to

\[
C_\Sigma \cdot X \cdot \bigl(\log X + \widetilde \Psi'(1)\bigr)
C'_{\Sigma} \cdot X
O_{\epsilon,\Sigma,\Psi}\Bigl( \frac{X^{1+\epsilon}}{Q} + X^{11/12+\epsilon} +  Q^{2+\epsilon}X^{3/4+\epsilon}\Bigr),
\]

where $C_\Sigma>0$ and $C'_\Sigma\in \R$ depend only on the finite set
$\Sigma$ of local specifications.
$C_\Sigma,C'_\Sigma$, main terms for the first moment

Since $V^\pm$ is a function rapidly decaying at infinity, we may truncate the $n$-sum in
the definition of $S(f)$ to $n < X^{1/2+\epsilon}$ with negligible
error term.  To estimate (<ref>) we switch order of
summation and consider

\[
\displaystyle 2\sum_{n< X^{1/2+\epsilon}}
\sum_{q<Q}\mu(q)
\sum_{f\in \overline{\W_q(\Sigma)}}
\frac{\lambda_n(f)}{n^{1/2}}
\H_{\frac{n}{\sqrt{X}}}\Bigl(\frac{|\Delta(f)|}{X}\Bigr).
\]

We use Corollary <ref>, to estimate the inner sum over $f$:

\begin{multline}\label{qlessQ}
\sum_{q<Q}\mu(q)
\left (\alpha^\pm \cA^{(q)}_{[n,r_\Sigma]}(\lambda_n \chi_\Sigma)
\cdot
\widetilde{H_{\frac{n}{\sqrt{X}}}}
\gamma^\pm
\cC^{(q)}_{[n,r_\Sigma]}(\lambda_n \chi_\Sigma)
\cdot \widetilde{H_{\frac{n}{\sqrt{X}}}}
(\frac 56) \cdot
\right )
\\
+\displaystyle O_{\epsilon,\Sigma,\Psi}\Bigl(\sum_{n< X^{1/2+\epsilon}}\frac{|g(\frac{n}{\sqrt{X}})|}{\sqrt{n}}
\sum_{q<Q}
\cdot
\Bigr),
\end{multline}

\begin{equation}\label{eqgXn}
g(y):=\widetilde{\H_y}(1) = \int_0^\infty \H_y(t)dt.
\end{equation}

$g(y)$, equal to $\widetilde{H_y}(1)$
The error term above is seen to be bounded by
$O_{\epsilon,\Sigma,\Psi}(Q^{2+\epsilon}X^{3/4+\epsilon})$ thanks to
Lemma <ref>.

Next, we bound the secondary term in (<ref>). Since $r_\Sigma$ is fixed, the
contribution to $\cC^{(q)}_{[n,r_\Sigma]}(\lambda_n \chi_\Sigma)$ from primes $p\mid r_\Sigma$ is bounded.
Therefore, we consider without further mention in the remainder of this paragraph only
the primes $p\nmid r_\Sigma$.
The contribution to $\cC^{(q)}_{[n,r_\Sigma]}(\lambda_n \chi_\Sigma)$ from
primes $p\mid q$ and $p\nmid n$ is given in <cit.> and <cit.> to be
$O(p^{-5/3})$. (Note that our quantity $\cC^{(p)}_1(1)$ defined in <ref>
corresponds to the quantity denoted $\cC_{p^2}(\Phi_p,1)$ in [29].)
The contribution to $\cC^{(q)}_{[n,r_\Sigma]}(\lambda_n \chi_\Sigma)$ from primes $p\mid
(q,n)$ is estimated
from <cit.> to also be $O(p^{-5/3})$. (If $a=(1^21_*)$, then $\mathcal
C_{p^2}(a,1)\asymp p^{1/3}$, and the cardinality of the orbit
a$ inside $V(\Z/p^2\Z)$ is equal to $p^4(p^2-1)$ by <cit.>, which yields $p^{1/3}
p^6 / p^8 = p^{-5/3}$, whereas the other nonmaximal types $a=(1^3_*)$,
$(1^3_{**})$, $(0)$ have a smaller contribution.)
The contribution to $\cC^{(q)}_{[n,r_\Sigma]}(\lambda_n \chi_\Sigma)$ from
primes $p\parallel n$
and $p\nmid q$ is computed from (<ref>) and Table <ref> to be
$O(p^{-1/3})$ (see also Lemma <ref>).
The contribution to $\cC^{(q)}_{[n,r_\Sigma]}(\lambda_n \chi_\Sigma)$ from primes $p^2\mid n$ and
$p\nmid q$ is bounded by $O(1)$.
The contribution to $\cC^{(q)}_{[n,r_\Sigma]}(\lambda_n \chi_\Sigma)$ from primes $p\nmid nq$ is
a convergent infinite product that is uniformly bounded. Therefore, letting $n_1=\prod_{p||n}
secondary term
in (<ref>) is

\begin{equation*}
\ll_{\epsilon,\Sigma,\Psi} X^{5/6} \cdot \sum_{n< X^{1/2+\epsilon}}\frac{1}{\sqrt{n}}\sum_{q<
Q}\frac{(n,q)^{1/3}}{q^{5/3-\epsilon}n_1^{1/3-\epsilon}}\ll_{\epsilon,\Sigma,\Psi} X^{11/12+\epsilon},
\end{equation*}

which is sufficiently small.

Finally, from Lemma <ref>, we see that the first term in (<ref>)
is equal to

\begin{equation*}
2\alpha^\pm \cdot X \cdot \sum_{n<X^{1/2+\epsilon}}\frac{g(\frac{n}{\sqrt
X})}{\sqrt{n}}\cA^\max(\lambda_n \chi_\Sigma)
+O_{\epsilon,\Sigma,\Psi}\Bigl(\sum_{n< X^{1/2+\epsilon}}\sum_{q=Q}^\infty
\frac{X}{\rad(n)^{3/2}q^{2-\epsilon}}\Bigr).
\end{equation*}

The result now follows with the values of the constants being

\begin{equation}\label{defCSigma}
T_\Sigma(s),\quad C'_\Sigma:= 2\alpha^\pm C',
\end{equation}

as follows from
Proposition <ref> below.

§.§ Computing the leading constants

We compute the constants $C_\Sigma,C'_\Sigma$ arising in Proposition
<ref>. We begin with the
following lemma.

The Mellin transform of the function $g$ in (<ref>) is

\begin{equation*}
\widetilde{g}(s)=\widetilde{\Psi}(1+s/2)\frac{G(s)}{s}
\frac{\gamma^\pm(1/2+s)}{\gamma^\pm(1/2)},
\end{equation*}

where $G$ is as in (<ref>). In particular, $\widetilde{g}(s)$ is
meromorphic on the half-plane $\Re(s)>-1/2$ with only a simple pole at $s=0$.

Unwinding definitions (<ref>) and (<ref>), we see that

\begin{equation*}
\begin{array}{rcl}
\displaystyle\widetilde{g}(s)&=&
\displaystyle\int_{0}^\infty \Psi(t)\int_{0}^\infty
\\[.2in]&=&
\displaystyle \int_{0}^\infty t^{s/2+1}
\Psi(t)\frac{dt}{t}\int_{0}^\infty V^\pm(u)u^s
\frac{du}{u}
\\[.2in]&=&
\displaystyle \widetilde{\Psi}(1+s/2)\widetilde{V^\pm}(s).
\end{array}
\end{equation*}

The lemma follows from the expression (<ref>) for

$T_\Sigma(s)$, Dirichlet series of $t_\Sigma(n)$
Define the Dirichlet series

\begin{equation}\label{defTSigma}
T_\Sigma(s):=\sum_{n=1}^\infty\frac{t_\Sigma(n) }{n^s},\end{equation}

$t_\Sigma(n)$, average of $\lambda_K(n)$ over $K$ in $\FF_\Sigma$
$t_\Sigma(n)=\cA^{\max}(\lambda_n\chi_\Sigma)$ is the average of $\lambda_K(n)$ over $K$ in
$\FF_\Sigma$ (note that this is actually a finite average, since the
value of $\lambda_K(n)$ is determined by the splitting type of $K$ at
the primes dividing $n$.)

The Dirichlet series $T_\Sigma(s)$ has a meromorphic continuation to
the half-plane $\Re(s)>1/3$ with a simple pole at $s=\frac
12$. Moreover, this simple pole has a positive residue.

For every integer $n\ge 1$, we have

\begin{equation*}
t_\Sigma(n)=\prod_{p^k\parallel n}\sum_{\sigma}
\frac{\lambda_{p^k}(\sigma)}{\# \mathcal O_\sigma},
\end{equation*}

where $\mathcal O_\sigma \subset V(\F_p)$ is the $\GL_2(\F_p)$-orbit attached to $\sigma$, and  $\sigma$
ranges over all
splitting types that are
compatible with $\Sigma_p$. The quantity $t_\Sigma(n)$ is clearly
multiplicative, and so $T_\Sigma(s)$ has an Euler product decomposition

\begin{equation*}
\end{equation*}

If $p\neq 3$ and there is no specification $\Sigma_p$ at $p$, then
Proposition <ref> asserts that
$t_\Sigma(p)=\frac{(p-1)(p^2-1)}{p^4}$ and that
$t_\Sigma(p^2)=\frac{(p^2-1)^2}{p^4}$.  Therefore, the
series $T_\Sigma(s)\zeta(2s)^{-1}$ converges absolutely for

It follows that the residue at $s=\tfrac 12$ is given by the following
convergent product

\[
\Res_{s=\frac12} T_\Sigma(s) = \frac{1}{2} \prod_p (1-p^{-1}) \sum^\infty_{k=0}
\frac{t_\Sigma(p^k)}{p^{k/2}}.
\]

We claim that each factor in the product is positive:

\[
\sum^\infty_{k=0}
\frac{t_\Sigma(p^k)}{p^{k/2}} >0
\quad
\text{for every prime $p$.}
\]

Indeed, $\lambda_{p^m}(f)$ is only negative if $\sigma_p(f)=(3)$ and
$m\equiv 1\pmod{3}$, in which case $\lambda_{p^m}(f)=-1$.
Therefore, the minimum possible value of
$ \sum^\infty_{k=0}\frac{t_\Sigma(p^k)}{p^{k/2}}$ occurs when $\Sigma_p=\{(3)\}$. In
this case

\begin{equation}\label{eqpostest}
\begin{array}{rcl}
\displaystyle \sum^\infty_{k=0}
\frac{t_\Sigma(p^k)}{p^{k/2}}&=&
\displaystyle \sum_{k\equiv 0\pmod 3}\frac{1}{p^{k/2}}-
\sum_{k\equiv 1\pmod{3}}\frac{1}{p^{k/2}},
\end{array}
\end{equation}

which is clearly positive since the $n$th term of the sum on the left
is greater than the $n$th term of the sum of the right.

As $X\to \infty$, we have the asymptotic

\[
\displaystyle\sum_{n=1}^\infty \frac{t_\Sigma(n)}{\sqrt{n}}  g(\frac{n}{\sqrt X})
\bigl(\log X + \widetilde \Psi'(1)\bigr) +
+O_{\epsilon,\Sigma,\Psi}\bigl(X^{-\frac1{12} + \frac \epsilon 2}
\bigr),
\]

\[
sT_\Sigma(\tfrac 12+s)
\frac{\gamma^\pm\bigl(\frac12+s\bigr)}{\gamma^\pm(1/2)}.
\]

From Lemma <ref>, we obtain

\begin{equation}\label{eqconstantint}
\begin{array}{rcl}
\displaystyle\sum_{n=1}^\infty
\frac{t_\Sigma(n)}{\sqrt{n}}
g(\frac{n}{\sqrt X})&=&\displaystyle
\frac{1}{2\pi i}\int_{\Re(s)=2}
T_\Sigma(\tfrac 12+s)\widetilde{g}(s) X^{s/2}ds\\[.2in]
\displaystyle
\frac{1}{2\pi i}\int_{\Re(s)=2}s T_\Sigma(\tfrac 12+s)\widetilde{\Psi}(1+s/2)
\\[.2in]
\displaystyle
\frac{1}{2\pi i}\int_{\Re(s)=2} J(s)X^{s/2}\frac{ds}{s^2},
\end{array}
\end{equation}

where the above equation serves as a definition of $J(s)$.

Since $\widetilde{\Psi}(1)=G(0)=1$, it follows
that $J(s)$ is holomorphic in $\Re(s)>-\frac16$, and
$J(0)=\Res_{s=\frac12}T_\Sigma(s)$. Expanding in Tayor series, we write

\begin{equation*}
J(s)X^{s/2}=J(0)+\Bigl(\frac{J(0)\log X}{2}+J'(0)\Bigr)s+\cdots
\end{equation*}

Shifting the integral to $\Re(s)=-\tfrac16 +\epsilon$ for some
$0<\epsilon<\frac16$, we therefore obtain

\begin{equation*}
\displaystyle\sum_{n=1}^\infty \frac{t_\Sigma(n)}{\sqrt{n}}  g(\frac{n}{\sqrt X})
=\frac{1}{2}\Res_{s=1/2}T_\Sigma(s)\cdot \log X+J'(0)+O_{\epsilon,\Sigma,\Psi}\bigl(X^{-\frac {1}{12}+\frac \epsilon 2}
\bigr).
\end{equation*}

Calculating $J'(0)$, we obtain, using that $G(s)$ is even:

\[
=\frac{1}{2}\Res_{s=\frac 12}T_\Sigma(s)\cdot  \widetilde \Psi'(1) +
\]

This concludes the proof of the proposition.

§.§ Upper bound for the first moment

In this subsection we investigate pointwise bounds for the tail of the
sieve when $q\ge Q$.

For every $Q,X\ge 1$ and $\epsilon >0$,

\[
\sum\limits_{q\ge Q}
\sum_{\substack{f\in \overline{\W_q^\irr}\\
|\Delta(f)| < X}}
\Bigl(\frac{X^{5/4-\delta+\epsilon}}{Q^{3/2-2\delta}}\Bigr),
\]

for $\delta=\frac{1}{128}$ as in
Theorem <ref>.

Let $f\in V(\Z)^\irr$ be an irreducible binary cubic form, and denote the
field of fractions of the ring associated to $f$ by $K_f$.

Note that for $f\in \mathcal W_q$ with $|\Delta(f)| < X$, we have
$|\Delta(K_f)|< X/q^2$, and recall from Proposition <ref> that

\begin{equation*}
\#\bigl\{f\in
\overline{\W_q}:\ |\Delta(f)|<X\bigr\}\ll_\epsilon\frac{X}{q^{2-\epsilon}}.
\end{equation*}

Therefore, we deduce from (<ref>) the estimate

\[
\sum\limits_{q\ge Q}
\sum_{\substack{f\in \overline{\W_q^\irr}\\
|\Delta(f)| < X}}
|S(f)|  \ll_\epsilon
\sum_{q\geq Q} (X/q^2)^{\theta+\epsilon}\cdot X/q^{2-\epsilon},
\]

where we recall that $\theta=1/4-\delta$. The result follows.

Optimizing, we pick $Q=X^{\frac{1-2\delta}{7-4\delta}}$ in
(<ref>). We have now established the following by combining
the two Propositions <ref> and <ref>.

For every $X\ge 1$ and $\epsilon >0$,

\begin{equation}\label{eqLfavgupbound}
\ll_{\epsilon,\Sigma,\Psi}
\end{equation}

\[
\frac{29-28\delta}{28-16\delta} = \frac{921}{892} = 1.0325\ldots
\]

for the best known value of $\delta=\frac{1}{128}$ of Theorem <ref>.

The exponent is smaller than $5/4-\delta = \tfrac{159}{128} = 1.2421875$,
thus (<ref>)
an improvement on the exponent arising from summing the pointwise bound on
$|L(\tfrac 12,\rho_K)|$ over cubic fields $K$ with discriminant bounded by

§ CONDITIONAL COMPUTATION OF THE FIRST MOMENT OF
$L(\TFRAC 12,\RHO_K)$

In this section, we shall compute the first moment of $L(\tfrac 12,\rho_K)$
assuming one of two hypotheses. More precisely, we prove the following

Assume one of the following two hypotheses:

(S) Strong Subconvexity: For every $K\in\FF_\Sigma$, we
have $|L(\frac12,\rho_K)|\ll |\Delta(K)|^{\frac{1}{6}-\vartheta}$
for some $\vartheta>0$.

(N) Nonnegativity: For every $K\in\FF_\Sigma$, we
have $L(\frac12,\rho_K)\geq 0$.

Then we have for small enough $\epsilon>0$,

\begin{equation*}
\sum_{K\in\FF_\Sigma}L\bigl(\tfrac12,\rho_K\bigr)
\Psi\Bigl (\frac{|\Delta(K)|}{X}\Bigr)
C_\Sigma \cdot X \cdot \bigl(\log X + \widetilde \Psi'(1) \bigr)+
C'_\Sigma \cdot X +
\end{equation*}

where $C_\Sigma$ and $C_\Sigma'$ are the constants arising in
Proposition <ref>.

Compared to Section <ref>, the proof is significantly more difficult, and will
several new inputs.  Indeed, recall that we have

\begin{equation}\label{eqincexcrep}
A_\Sigma(X)= 2 \sum_{q\ge 1}\mu(q)
\sum_{f\in \overline{\W_q(\Sigma)}}
\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr).
\end{equation}

Pick a small $\kappa_\downarrow>0$. Proposition <ref> provided
an estimate for the above sum with $q$ in the range
$q\in [X^{1/8-\kappa_\downarrow},X^{1/8+\kappa_\uparrow}]$, border range of the sieve
$q\geq X^{1/8+\kappa_\uparrow}$, large range of the sieve

For $q\geq X^{1/8-\kappa_\downarrow}$, our approach is to approximate the
smoothed sum of $S(f)$ with a smoothed sum of $D(\tfrac 12,f)$.  We do this
by breaking up these $q$ into two ranges: the “large range” and the
“border range”. Namely, we pick a small $\kappa_\uparrow>0$. Then the range
$q\geq X^{1/8+\kappa_\uparrow}$ is the large range while the range
$[X^{1/8-\kappa_\downarrow},X^{1/8+\kappa_\uparrow}]$ is the border range.  For $q$ in
both of these ranges we want to prove

\begin{equation}\label{eqSDCloseIntro9}
\sum_{f\in\overline{\W_q(\Sigma)}}S(f)\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)
\approx
\sum_{f\in\overline{\W_q(\Sigma)}}D(\tfrac12,f)\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr).
\end{equation}

On average over $f\in\overline{\W_q(\Sigma)}$, this is an unbalanced approximation of the central
value $D(\tfrac12,f)$ by the Dirichlet sum $S(f)$ of the coefficients $\lambda_n(f)$.

In <ref>, we establish (<ref>) with $q$ in the large
range, which is straightforward. The bulk of the section is devoted to proving
(<ref>) in the border range. This is proved in
<ref> and <ref> using the
unbalanced approximate functional equation of Proposition <ref>.
The crux of the proof is to estimate the average of the coefficients $e_k(f)$ of the
unbalanced Euler
factors $E_p(s,f)$ over the forms $f\in \overline{\W_q(\Sigma)}$.
Finally, in
<ref>, we compute the average
of $D(\tfrac 12,f)$ (assuming either nonnegativity or strong subconvexity of $L(\tfrac
thereby obtaining the average of $S(f)$ and finishing the proof of Theorem

§.§ Estimates for the large range

We begin by estimating $S(f)$ for integral binary cubic forms with large index.

For every integral binary cubic form $f\in V(\Z)^{\rm irr}$ and every
$\epsilon>0$, we have

\begin{equation*}
S(f)=D(\tfrac 12,f)+O_\epsilon\Bigl(\frac{|\Delta(f)|^{1/4+\epsilon}}{\ind(f)}\Bigr).
\end{equation*}

By definition, we have

\begin{equation*}
S(f)=\frac{1}{2\pi i}\int_{\Re(s)=2}D(\tfrac12+s,f)\widetilde{V^\pm}(s)
\end{equation*}

Shifting to the line $s=-1/2+\epsilon$, we pick up the pole of
$\widetilde{V^\pm}(s)$ at $0$ (with residue $1$), to obtain

\begin{equation*}
\begin{array}{rcl}
\displaystyle  S(f)&=&\displaystyle D(\tfrac 12,f)+
\frac{1}{2\pi i}\int_{\Re(s)=-1/2+\epsilon}D(\tfrac12+s,f)\widetilde{V^\pm}(s)
\\[.2in]&=&\displaystyle
\end{array}
\end{equation*}

where the final estimate follows since $D(s,f)$ is within
$|\Delta(f)|^\epsilon$ of $L(s,\rho_K)$ for $\Re(s)$ close to $0$. The
lemma now follows since $\Delta(f)=\ind(f)^2\Delta(K)$.

Adding up the above estimate for $f\in \overline{\W_q(\Sigma)}$, we
immediately obtain the following result.

For every square-free $q$, and $X\ge 1$, we have

\begin{equation*}
\sum_{f\in \overline{\W_q(\Sigma)}}S(f)\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)
=\sum_{f\in \overline{\W_q(\Sigma)}}D(\tfrac 12,f)\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)
\end{equation*}

The proposition follows from Lemma <ref> and the tail estimate in
Proposition <ref>.

An immediate consequence of the previous result is the following
estimate for $q$ in the large range.

For every small $\kappa_\uparrow>0$, square-free $q>X^{1/8+\kappa_\uparrow}$ and $X\ge 1$, we have

\begin{equation*}
\sum_{f\in \overline{\W_q(\Sigma)}}\bigl(S(f)-D(\tfrac 12,f)\bigr)
\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)\ll_{\epsilon,\kappa_\uparrow,\Sigma,\Psi}
\frac{X^{1-2\kappa_\uparrow+\epsilon}}{q}.
\end{equation*}

§.§ Preparations and strategy for the border range

In this subsection, we shall introduce spaces, notation, and
some preliminary results that will be useful subsequently in handling
the border range. One of the key tools in comparing $S(f)$ and
$D(\tfrac 12,f)$ is the unbalanced approximate functional equation of
Proposition <ref>. To apply this result, it is not possible
to only work with the information that forms $f \in \W_q(\Sigma)$ are nonmaximal at
primes dividing $q$. It is
necessary to work with the additional information of the index of $f$, including at
primes not dividing $q$.

To this end,
for a positive (not necessarily squarefree) integer $b$, let
$\U_{b}(\Sigma)$ denote the set of binary cubic forms $f\in
V(\Z)(\Sigma)$ such that $\ind(f)=b$.  $\U_b$, set of cubic
forms $f$ with $\ind(f)=b$  Note the inclusion $\U_b(\Sigma) \subset
\W_{\rad(b)}(\Sigma)$, and in fact we have

\begin{equation*}
\W_q(\Sigma) =\bigsqcup_{m\geq 1} \U_{mq}(\Sigma),
\end{equation*}

where the union is disjoint and $q$ is square-free. Let
$\overline{\U_{b}(\Sigma)}$ denote the set of $\GL_2(\Z)$-orbits on

$\Y_{b,r}$, subset of cubic forms $f\in \W_r$ with $b\parallel \ind(f)$
Let $b$ be a positive integer, and let $r$ be a positive squarefree
integer such that $(b,r)=1$. Finally, we define the set $\Y_{b,r}(\Sigma)$ to
be the subset of elements in $\W_{r}(\Sigma)$ whose index at primes $p$ dividing
$b$ is exactly $p^{v_p(b)}$. As usual we let
$\overline{\Y_{b,r}(\Sigma)}$ denote the set of $\GL_2(\Z)$-orbits on
$\Y_{b,r}(\Sigma)$. The significance of these subsets $\Y_{b,r}(\Sigma)$
is the following disjoint union

\[
\Y_{b,r}(\Sigma) = \bigsqcup_{(b,s)=1} \U_{brs}(\Sigma),
\]

hence for any function $\phi:\overline{\U_b(\Sigma)}\to\C$, we have

\begin{equation*}
\sum_{f\in \overline{\U_b(\Sigma)}}
\phi(f)\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)
=\sum_{(b,r)=1}\mu(r)\sum_{f\in \overline{\Y_{b,r}(\Sigma)}}
\phi(f)\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr).
\end{equation*}

Recall that the border range is what we are calling
$q\in[X^{1/8-\kappa_\downarrow},X^{1/8+\kappa_\uparrow}]$, where $\kappa_\downarrow,\kappa_{\uparrow}$
positive constants that can eventually be taken to be arbitrarily
small. We next estimate the sum of $S(f)-D(\tfrac 12,f)$ over $f$ in
$\overline{\U_{mq}(\Sigma)}$, where $m$ is somewhat large.

We begin by bounding the number of elements in $\overline{\U_{mq}(\Sigma)}\subset \overline{\U^{\rm
that have discriminant less than $X$.

For every positive integer $m$ and square-free $q$,
write $mq=m_1q_1$, where $m_1$ is powerful,
$(m_1,q_1)=1$, and $q_1$ is squarefree. Then for every $X\ge 1$,

\begin{equation}\label{eqUunif}
\overline{\U^{\rm irr}_{mq}}:|\Delta(f)|<X\}|\ll_\epsilon \frac{X^{1+\epsilon}}{m_1^{5/3}q_1^2}.
\end{equation}

The multiplicative constant depends only on $\epsilon$ (it is independent of $m,q,X$).

Elements $f$ in the left-hand side of (<ref>) are in bijection
with rings $R_f$ that have index $mq=m_1q_1$ in the maximal orders
$\O_{K_f}$ of their fields of fractions $K_f$. It follows that the
discriminants of these fields $K_f$ are less than $X/(m_1^2q_1^2)$. It
follows that the total number of such fields that can arise is bounded
by $O(X/m_1^2q_1^2)$.

To estimate the total number of rings $R_f$ that can arise, it
suffices to estimate the number of such rings $R_f$ within a single
$K_f$. This can be done prime by prime, for each prime dividing the
index $m_1q_1$. Let $p$ be a prime dividing $q_1$. Since $q_1$ is
squarefree, it follows that the index of $R_f$, at the prime $p$, is
$p$. Given the index $p$ overorder $R$ of $R_f$, it follows from
Proposition <ref>, that the number of index $p$ suborders
of $R$ is bounded by $3$.

For primes dividing $m_1$, this procedure is more complicated since
there can be many more subrings with prime power index. However, this
question is completely answered by work of Shintani [27]
and Datskovsky–Wright [11] (see <cit.>), who give
an explicit formula for the counting function of suborders $R$ of a fixed
cubic field $K$, which we state as Proposition <ref>.  They
show that the number of suborders of index $p^k$, for $k\geq 1$ is the
$k$th Dirichlet coefficient of the $p$th Euler factor of

\begin{equation*}
\frac{\zeta_K(s)}{\zeta_K(2s)}\zeta_\Q(3s)\zeta_\Q(3s-1).
\end{equation*}

We verify the lemma for a prime $p$ having splitting type $(12)$ in
$K$. For such a prime, the $p$th Euler factor of the above Dirichlet
series is:

\begin{equation*}
(1+p^{-2s})\Bigl(\sum_{k=0}^\infty p^{-3ks}\Bigr)
\Bigl(\sum_{k=0}^\infty p^{k-3ks}\Bigr).
\end{equation*}

It is thus clear that the $k$th Dirichlet coefficient is bounded by
$O(p^{k/3})$. Therefore, the number of possible suborders of index
$p^k$ is bounded by $O(p^{k/3})$. Other splitting types can be handled
in indentical fashion, giving the same bound.

Putting this together, it follows that the number of suborders of $K$,
having index $q_1m_1$ is bounded by $O(q_1^\epsilon m_1^{1/3})$. Multiplying
this quantity by $X/(m_1q_1)^2$ yields the result.

For $X\ge 1$, square-free $q$, and small enough $\eta>0$

\begin{equation*}
\sum_{m> X^\eta}\sum_{f\in \overline{\U_{mq}(\Sigma)}}
\bigl( S(f) - D(\tfrac 12,f) \bigr) \Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)=
\end{equation*}

From Lemma <ref>, it follows that for
$f\in\U_{mq}(\Sigma)$ with $|\Delta(f)|\asymp X$, we have
$S(f)-D(\tfrac 12,f)=O(\frac{X^{1/4+\epsilon}}{mq})$.  We write $mq$ as
$m_1q_1$, where $q_1$ is squarefree with $(q_1,m_1)=1$, and $m_1$ is
powerful.  We now have

\begin{equation*}
\begin{array}{rcl}
\displaystyle\sum_{m> X^\eta} \sum_{f\in \overline{\U_{mq}(\Sigma)}}
\bigl(S(f)-D(\tfrac 12,f)\bigr)
\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr) \ll_{\epsilon,\Sigma,\Psi}
\displaystyle\sum_{m>X^\eta}\frac{X^{1/4+\epsilon}}{mq}\cdot
\frac{X^{1+\epsilon}}{m_1^{5/3}q_1^2},
\end{array}
\end{equation*}

where the final estimate follows from Lemma <ref>.

We then have the following corollary.

Let $X\ge 1$, squarefree $q>X^{1/8-\kappa_\downarrow}$, and $\eta>0$ be
such that $\eta-2\kappa_\downarrow>0$. Then we have

\begin{equation*}
\sum_{m> X^\eta}\sum_{f\in \overline{\U_{mq}(\Sigma)}}
\bigl( S(f) - D(\tfrac 12,f) \bigr) \Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)=
\end{equation*}

Furthermore, $\kappa_\downarrow$ and hence $\eta$ can be taken to
be arbitrarily small. Therefore, a consequence of the above lemma is
that when $q$ is in the border range, sums over $\overline{\U_{mq}(\Sigma)}$ only have to
be considered for $m$ less than arbitrarily small powers of $X$.

Let $q\in [X^{1/8-\kappa_\downarrow},X^{1/8+\kappa_\uparrow}]$ be fixed for the rest of this
subsection. For
integer $m$, we write $mq=m_1q_1$, where $m_1$ is powerful,
$(m_1,q_1)=1$, and $q_1$ is squarefree. Note that since $m$ will be
taken to be very small ($\ll X^\eta$), $q_1$ will be quite close in
size to $q$. We restate Proposition <ref> for convenience:
for $f \in \overline{\U_{m_1q_1}(\Sigma)}$, we have

\begin{equation}\label{eqinex2}
S(f)=D(\tfrac 12,f)-
\sum_{k=1}^\infty
\frac{e_k(f)k^{1/2}}{q_1\rad(m_1)}
\sum_{n=1}^\infty
\frac{\lambda_n(f)}{n^{1/2}}
\left(
\frac{m_1^2 kn}{\rad(m_1)^2 |\Delta(f)|^{1/2}}
\right),
\end{equation}

where $e_k(f)k^{1/2}$ is the $k$th Dirichlet coefficient of the

\begin{equation*}
\sum_{k=1}^\infty \frac{e_k(f)k^{1/2}}{k^s}
\end{equation*}

Our next and final goal of this subsection is to perform a switching
trick, analogous to Theorem <ref>, in which our sums over
$\overline{\U_{m_1q_1}(\Sigma)}$ are replaced with sums over
$\overline{\U_{m_1}(\Sigma)}$.  We thus need to understand how the
quantity $e_k(f)$ behaves under such a switch.  The next lemma does
just that: more precisely, if $f$ is nonmaximal and switches to the
pair $(g,\alpha)$ with prime index $p$, then the next lemma determines
$e_k(f)$ in terms of $(g,\alpha)$.

As recalled in Proposition <ref>, the proof of <cit.> implies
that there is a
bijection between the zeros in $\P^1(\F_p)$ of the reduction modulo
$p$ of $g(x,y)$ and the set of cubic rings that are index-$p$ subrings
of $R_g$. Thus, $f$ corresponds uniquely to a pair $(g,\alpha)$, where
$\alpha\in \P^1(\F_p)$ is a root of $g(x,y)$ modulo $p$. Then the
following lemma determines $E_p(s,f)$ given this pair $(g,\alpha)$.

Let $g\in V(\Z)$ be a binary cubic form that is maximal at $p$. Let
$\alpha\in\P^1(\F_p)$ be a root of the reduction of $g$ modulo
$p$. Let $f\in V(\Z)$ be a binary cubic form corresponding to the index-$p$
subring of $R_g$ associated to the pair $(g,\alpha)$. Then $E_p(s,f)$,
and hence $e_k(f)$ for every $k$, is determined by the pair
$(g,\alpha)$.  More precisely, we have

(a) If $\sigma_p(g)=(111)$, then $\sigma_p(f)=(1^21)$ and

(b) If $\sigma_p(g)=(12)$, then $\sigma_p(f)=(1^21)$ and

(c) If $\sigma_p(g)=(1^21)$ and $\alpha$ is the single
root, then $\sigma_p(f)=(1^21)$ and $E_p(s,f)=1$;

(d) If $\sigma_p(g)=(1^21)$ and $\alpha$ is the double
root, then $\sigma_p(f)=(1^3)$ and $E_p(s,f)=1-p^{-s}$;

(e) If $\sigma_p(g)=(1^3)$, then $\sigma_p(f)=(1^3)$ and

The procedure to compute $f(x,y)$ given the pair $(g,\alpha)$ is as
follows: use the action of $\GL_2(\Z)$ to move $\alpha$ to the point
$[1:0]\in\P^1(\F_p)$. This yields the binary cubic form
$ax^3+bx^2y+cxy^2+dy^3$, where $p\mid a$. Moreover, since $g$ is
maximal at $p$, we see that $p\mid b$ implies that $p^2\nmid a$. Then
$f(x,y)$ can be taken to be $(a/p)x^3+bx^2y+pcxy^2+p^2dy^3$. Running
this procedure for the different splitting types of $g$ immediately
shows that the corresponding $f$ has the splitting type listed in the

For example, if $g$ has splitting type $(111)$ or $(12)$, then we may
bring one of the single roots (using a $\GL_2(\Z)$-transformation) to
infinity. Then we may write $g(x,y)=pax^3+bx^2y+cxy^2+dy^3$, where
$p\nmid b$ since $g$ is unramified.  Then the procedure gives
$f(x,y)=ax^3+bx^2y+pcxy^2+p^2dy^3$. Since $p\nmid b$, the splitting
type of $f(x,y)$ is $(1^21)$ as claimed. The other cases are similar,
and we omit them.

Finally, $e_{\ell,\beta}(f)$ is determined for $p\neq \ell|\ind(f)$ and all $\beta\ge 0$ as follows from
Lemma <ref>.

The final result of this subsection is to determine what happens to
the quantity $e_k(f)\lambda_n(f)$ after the switch.

Let $m_1$ and $q_1$ be positive integers, where $m_1$ is powerful,
$(m_1,q_1)=1$, and $q_1$ is squarefree. Let $k$ be a positive integer divisible only
by primes dividing $m_1q_1$. Let $n$ be a positive integer
and write $n=n_1\ell$ where $(\ell,m_1q_1)=1$ and $n_1$ is divisible only
by primes dividing $m_1q_1$. Then we have

\begin{equation*}
\sum_{f\in \overline{\U_{m_1q_1}(\Sigma)}}e_k(f)\lambda_n(f)\Psi(|\Delta(f)|)=
\sum_{g\in \overline{\U_{m_1}(\Sigma)}}c_{q_1}(g)d_{m_1}(g)
\lambda_\ell(g)\Psi(q_1^2|\Delta(g)|),
\end{equation*}

where $c_{q_1}$ and $d_{m_1}$ are congruence functions on $V(\Z)$
defined modulo $q_1$ and $m_1^3$, respectively. Furthermore, we have
$c_{q_1}(g)\ll_\epsilon q_1^\epsilon$ and
$d_{m_1}(g)\ll_\epsilon m_1^\epsilon$ uniformly for every $g\in V(\Z)$.

As in Section <ref>, we will write sums over $\overline{\U_{m_1q_1}(\Sigma)}$ in terms
of sums over $\overline{\U_{m_1}(\Sigma)}$. In this case, we have the simple

\begin{equation*}
\U_{m_1q_1}(\Sigma) \leftrightarrow \bigl\{(g,\alpha):g\in\U_{m_1}(\Sigma),
\alpha\in\Z/q_1\Z,g(\alpha)\equiv 0\pmod{q_1}\bigr\},
\end{equation*}

which follows by an argument similar to that of Lemma <ref>.

Since the functions $e_k$ and $\lambda_n$ are multiplicative, we may write

\begin{equation*}
e_k(f)=e_{k_1}(f)e_{k_2}(f);\quad \lambda_n(f)=\lambda_{n_1}(f)\lambda_{\ell}(f),
\end{equation*}

where $k_1$ is only divisible by primes dividing $q_1$, and $k_2$ is
only divisible by primes dividing $m_1$. Let $f\in
\overline{\U_{m_1q_1}(\Sigma)}$ correspond to the pair $(g,\alpha)$
with $g\in \overline{\U_{m_1}(\Sigma)}$.  It follows that
$\lambda_{n_1}(f)=0$ if $\alpha$ corresponds to a double root of $g$
modulo some prime $p\mid (q_1,n_1)$.  Otherwise, $\lambda_{n_1}(f)=1$.
Clearly, we have $\lambda_n(f)=\lambda_\ell(g)$, and

We have seen in Lemma <ref> that the value $e_{k_1}(f)$
depends only on the splitting type of $g$ modulo all the primes
dividing $q_1$, and on whether $\alpha$ is a single or double root
modulo all the primes dividing $q_1$.  It is clear that $e_{k_2}(g)$
is defined via congruence conditions modulo $m_1^3$. The first claim
of the lemma now follows from the bijection.

The bounds in the second claim of the lemma are immediate since
$\lambda_{n_1}$, $e_{k_1}$ and $e_{k_2}$, each are bounded by $\ll_\epsilon n_1^\epsilon$, $\ll_\epsilon
k_1^\epsilon$ and $\ll_\epsilon k_2^\epsilon$, respectively
Proposition <ref> and the examples just before Proposition
<ref> for
the claims regarding $e_{k_i}$).

§.§ Estimates for the border range

In this subsection, we assume that our integers $q$ lie in the border
range $[X^{1/8-\kappa_\downarrow},X^{1/8+\kappa_\uparrow}]$ with small enough
Our goal is to bound

\begin{equation*}
\sum_{f\in \overline{\W_{q}(\Sigma)}}
\bigl( S(f) - D(\tfrac 12,f) \bigr) \Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr),
\end{equation*}

for $q$ in this range. Recall that we have a disjoint union

\begin{equation*}
\overline{\W_q(\Sigma)}=\bigsqcup_{m\geq 1}\overline{\U_{mq}(\Sigma)},
\end{equation*}

and that we will be summing $S(f)-D(\tfrac 12,f)$ over $\overline{\U_{mq}(\Sigma)}$ (and then
summing over $m$) rather than simply summing over $\overline{\W_q(\Sigma)}$. From Lemma
<ref>, it follows that we may restrict the sum to
$m\leq X^\eta$, where $\eta$ may be taken to be arbitrarily small.
All multiplicative constants are understood to depend on the initial
choices of $\kappa_\downarrow,\kappa_\uparrow,\eta>0$.

We write $mq=m_1q_1$, where $m_1$ is powerful, $(m_1,q_1)=1$ and $q_1$
is squarefree. Note then that $m_1\leq m^2\leq X^{2\eta}$, and thus
$q_1\geq q/m\geq X^{1/8-\eta-\kappa_\downarrow}$.  We begin by fixing $k$ and $n$
in (<ref>), and bounding the sum over $f\in \overline{\U_{m_1q_1}(\Sigma)}$.

For every small enough $\kappa_1>0$, the following estimate holds.
Let $m_1$, $q_1$, $k$, and $n$ be positive integers and $X\ge
1$. Assume that $m_1$ is powerful, $(m_1,q_1)=1$, and $q_1$ is
squarefree. Write $n=n_1\ell_1$ where $(\ell_1,m_1q_1)=1$ and $n_1$ is
divisible only by primes dividing $m_1q_1$. Denote the radical of
$\ell_1$ by $\ell$. Then

\begin{equation*}
\sum_{f\in \overline{\U_{m_1q_1}(\Sigma)}}e_k(f)
\lambda_n(f)
\Bigr)
\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)\ll_{\epsilon,\Sigma,\Psi}
\end{equation*}

\begin{equation*}
H(n,m_1,q_1;X)=\frac{X}{q_1^2m_1^{5/3}\ell} +
\frac{X^{5/6+\kappa_1/3}}{q_1^{5/3}\ell^{1/3}}+
\ell q_1^2 m_1^{12} X^{9\kappa_1} + \frac{X^{1-\kappa_1}}{q_1^2 m_1^{5/3}}.
\end{equation*}

Applying the preceding Lemma <ref>,
we obtain

\begin{equation*}
\sum_{f\in \overline{\U_{m_1q_1}(\Sigma)}}e_k(f)
\lambda_n(f)V\Bigl(\frac{nkm_1^2}{\rad(m_1)^2\sqrt{|\Delta(f)|}}
\Bigr)\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)=
\sum_{f\in\overline{\U_{m_1}(\Sigma)}}c_{q_1}(f)d_{m_1}(f)\lambda_{\ell_1}(f)
\Psi_1\Bigl(\frac{q_1^2|\Delta(f)|}{X}\Bigr),
\end{equation*}

where $c_{q_1}$ is defined modulo $q_1$, $d_{m_1}$ is defined modulo
$m_1^3$, and $\Psi_1=\H_{\frac{nkm_1^2}{\sqrt{X}\rad(m_1)^2}}$. Recall
that in Corollary <ref>, we bound
$E_\infty(\widetilde{\Psi_1};-\epsilon)$ by an absolute constant.  For
brevity in this proof, we will write $\ll$ as a shorthand for

We perform an inclusion-exclusion principle to write the sum over
$\overline{\U_{m_1}(\Sigma)}$ in terms of sums over
$\overline{\Y_{m_1, r}(\Sigma)}$. This yields

\begin{equation*}
\sum_{f\in\overline{\U_{m_1}(\Sigma)}}c_{q_1}(f)d_{m_1}(f)\lambda_{\ell_1}(f)
\Psi_1\Bigl(\frac{q_1^2|\Delta(f)|}{X}\Bigr)=
\sum_{(m_1,r)=1}\mu( r )\sum_{f\in\overline{\Y_{m_1, r }(\Sigma)}}
\Psi_1\Bigl(\frac{q_1^2|\Delta(f)|}{X}\Bigr).
\end{equation*}

We split up the above sum into two sums, corresponding to the ranges
$ r < B$ and $ r \geq B$, for some $B>1$.

We estimate each summand in the range $ r <B$ using Theorem
<ref>, and each summand in the range $ r \geq B$ using Lemma
<ref>, to respectively obtain

\begin{equation*}
\begin{array}{rcl}
\displaystyle\sum_{f\in\overline{\Y_{m_1, r }(\Sigma)}}
\Psi_1\Bigl(\frac{q_1^2|\Delta(f)|}{X}\Bigr)
\displaystyle\frac{X^{1+\epsilon}(\ell, r )}{q_1^{2}m_1^{5/3} r ^2\ell}+
\frac{X^{5/6+\epsilon}(\ell,r)}{q_1^{5/3}r^{5/3}\ell^{1/3}}+\ell q_1^2 m_1^{12} r ^{8}X^\epsilon
\\[.2in]&\ll&
\displaystyle\frac{X^{1+\epsilon}}{q_1^{2}m_1^{5/3} r \ell}+
\frac{X^{5/6+\epsilon}}{q_1^{5/3}r^{2/3}\ell^{1/3}}+\ell q_1^2 m_1^{12} r ^{8}X^\epsilon;
\\[.2in]
\displaystyle\sum_{f\in\overline{\Y_{m_1, r }(\Sigma)}}
\Psi_1\Bigl(\frac{q_1^2|\Delta(f)|}{X}\Bigr)
\displaystyle\frac{X^{1+\epsilon}}{q_1^2m_1^{5/3} r ^2}.
\end{array}
\end{equation*}

The second bound is simply an application of the tail estimate of
Lemma <ref>. The first bound is more complicated, and
we explain how it is derived. Summing over $\overline{\Y_{m_1, r
}(\Sigma)}$ can be replaced by summing a function $\phi\chi_\Sigma$
over $\overline{V(\Z)}$, where $\phi$ is defined modulo $m_1^2r^2$ and
$\chi_\Sigma$ is the indicator function defined in <ref>
before Corollary <ref>. In the above equation, we are
therefore summing a function defined over $r^2m_1^3q_1\ell r_\Sigma$
(here, we also use Lemma <ref>).  Moreover $q_1$ is
squarefree, and the function defined modulo $\ell$ is
$\lambda_{\ell_1}$. Therefore, the error term with applying Theorem
<ref> is bounded by $\ll \ell q_1^2m_1^{12}r^8X^\epsilon$.

We now estimate the first and second main terms. The density of the
first main term follows from the uniformity estimates and the bound
$\cA_{\ell_1}(\lambda_{\ell_1}) \ll\frac{1}{\ell}$ from
Lemma <ref>. The second main term computation follows
similarly using the bound $\cC_{\ell_1}(\lambda_{\ell_1})\ll
\frac{1}{\ell^{1/3}}$ from Lemma <ref>.

Adding the above bounds over the appropriate ranges of $ r $ yields

\[
\sum_{f\in \overline{\U_{m_1q_1}(\Sigma)}}e_k(f)
\lambda_n(f)V\Bigl(\frac{nkm_1^2}{\rad(m_1)\sqrt{|\Delta(f)|}}
\Bigr)\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)
\ll \frac{X^{1+\epsilon} \log B}{q_1^2m_1^{5/3}\ell} +
\frac{X^{5/6+\epsilon} B^{1/3}}{q_1^{5/3}\ell^{1/3}}+
\ell q_1^2 m_1^{12} B^9X^{\epsilon} + \frac{X^{1+\epsilon}}{q_1^2 m_1^{5/3} B}.
\]

Choosing $B=X^{\kappa_1}$ concludes the proof of the proposition.

Let notation be as in the beginning of this section. We have

\begin{equation*}
\begin{array}{rcl}
&&\displaystyle\sum_{f\in\overline{\W_q(\Sigma)}}\bigl(D(\tfrac 12,f-S(f)\bigr)
\Psi\Bigl(\frac{\Delta(f)}{X}\Bigr)\\[.2in]&=&
\displaystyle\sum_{f\in\overline{\W_q(\Sigma)}}\Psi\Bigl(\frac{\Delta(f)}{X}\Bigr)
\sum_{k=1}^\infty\frac{e_k(f)k^{1/2}}{\rad(\ind(f))}\sum_{n=1}^\infty
\frac{\lambda_n(f)}{n^{1/2}}V^{{\rm sgn}(\Delta(f))}\Bigl(
\frac{\ind(f)^2kn}{\rad(\ind(f))^2|\Delta(f)|^{1/2}}\Bigr)
\\[.2in]&=&
\displaystyle\sum_{m=1}^{\infty}\sum_{f\in\overline{\U_{mq}(\Sigma)}}
\Psi\Bigl(\frac{\Delta(f)}{X}\Bigr)
\sideset{}{^\flat}\sum_{k\geq 1}
\frac{e_k(f)k^{1/2}}{q_1\rad(m_1)}\sum_{n=1}^\infty
\frac{\lambda_n(f)}{n^{1/2}}V^{{\rm sgn}(\Delta(f))}\Bigl(
\frac{m_1^2kn}{\rad(m_1)^2|\Delta(f)|^{1/2}}\Bigr)
\\[.2in]&=&
\displaystyle\sum_{m=1}^{X^\eta}\sum_{f\in\overline{\U_{mq}(\Sigma)}}
\Psi\Bigl(\frac{\Delta(f)}{X}\Bigr)
\sideset{}{^\flat}\sum_{k\geq 1}\frac{e_k(f)k^{1/2}}{q_1\rad(m_1)}
\sum_{n\le \frac{X^{1/2+\epsilon}}{k}}
\frac{\lambda_n(f)}{n^{1/2}}V^{{\rm sgn}(\Delta(f))}\Bigl(
\frac{m_1^2kn}{\rad(m_1)^2|\Delta(f)|^{1/2}}\Bigr)
\\[.2in]
\end{array}
\end{equation*}

where the final estimate follows from Corollary <ref>, and the rapid decay
of $V^{\pm}$ to truncate the $n$-sum, and where the $\flat$ above indicates that the sum
over $k$ is supported on
multiples of $q_1$ and ranges only over integers whose prime factors
are all divisors of $mq$.

Next, we truncate the sum over $k$ as follows.

For every small enough $\kappa_2>0$, $X\ge 1$, and $q_1,m_1$ as above (i.e., satisfying $m_1\leq
X^{2\eta}$ and $q_1 \ge X^{1/8-\eta-\kappa_{\downarrow}}$), we have

\begin{equation}\label{eqkupbd}
\displaystyle\sum_{m=1}^{X^\eta}\sum_{\substack{f\in\overline{\U^{\rm irr}_{mq}}\\ |\Delta(f)|<X}}
\sideset{}{^\flat}\sum_{k > q_1^2X^{2\eta+3\kappa_2}}\frac{|e_k(f)|k^{1/2}}{q_1\rad(m_1)}
\sum_{n\le \frac{X^{1/2+\epsilon}}{k}}
\frac{|\lambda_n(f)|}{n^{1/2}}
\ll_{\epsilon,\kappa_2}
\displaystyle\frac{X^{1-\kappa_2+4\eta+2\kappa_\downarrow+\epsilon}}{q}.
\end{equation}

The integers $k$ that arise range over products of powers of primes
dividing $mq$. Write $k=k_1k_2$ where $(k_1,k_2)=1$, $v_p(k_1)\le 2$
and $v_p(k_2)\ge 3$ for every prime $p|k$. We have $k_1 \le q_1^2
\rad(m_1)^2$ since $q_1$ is square-free.  Hence $k>q_1^2
X^{2\eta+3\kappa_2}$ implies $k_2 > X^{3\kappa_2}$.  It follows from
Proposition <ref> that

\[
e_{k}(f) \ll_\epsilon \frac{\rad(k_2)^2}{k_2} X^\epsilon
\le k_2^{-1/3} X^{\epsilon} < X^{-\kappa_2 + \epsilon}.
\]

Hence the sum over $k> q_1^2 X^{2\eta+3\kappa_2}$ is bounded by

\[
\frac{X^{-\kappa_2+\epsilon}}{q_1\rad(m_1)} X^{1/4+\epsilon}
= \frac{X^{1/4-\kappa_2+2\epsilon}}{q_1\rad(m_1)}.
\]

We already know from Lemma <ref> that

\begin{equation*}
\sum_{\substack{f\in\overline{\U^{\rm irr}_{mq}}\\ |\Delta(f)|<X}} 1
\ll_\epsilon  \frac{X^{1+\epsilon}}{m_1^{5/3}q_1^2}.
\end{equation*}

Therefore, the left-hand side of (<ref>) is bounded by

\begin{equation*}
\ll_{\epsilon,\kappa_2}
X^{5/4 -\kappa_2 + 3\epsilon} \cdot
\sum^{X^\eta}_{m=1}
\frac{1}{m_1^{5/3}\rad(m_1) q_1^3}
\ll_{\epsilon,\kappa_2}
\frac{X^{5/4-\kappa_2+\eta+3\epsilon}}{q_1^3}\le
\frac{X^{1-\kappa_2+3\eta+2\kappa_\downarrow+3\epsilon}}{q_1},
\end{equation*}

which is sufficient because $q_1\ge q/m$ and $m\le X^\eta$.

We input Proposition <ref>, which bounds the
sum over $f$, and obtain with Corollary <ref> and (<ref>):

\begin{equation}\label{eqafterfsum}
\begin{array}{rcl}
\displaystyle\sum_{f\in\overline{\W_q(\Sigma)}}\bigl(D(\tfrac 12,f)-S(f)\bigr)
\Psi\Bigl(\frac{\Delta(f)}{X}\Bigr)&\ll_{\epsilon,\kappa_2,\Sigma,\Psi}&
\displaystyle\sum_{m_1=1}^{X^{2\eta}}
\sideset{}{^\flat}\sum_{k\leq q_1^2X^{2\eta+3\kappa_2}}
\sum_{n\le \frac{X^{1/2+\epsilon}}{k}}
\frac{k^{1/2}}{n^{1/2}q_1\rad(m_1)}
X^{\epsilon} H(n,m_1,q_1;X)
\\[.2in]
&&\displaystyle +\frac{X^{1-\eta+2\kappa_\downarrow+\epsilon}}{q}
\end{array}
\end{equation}

In our next result, we estimate the top line of (<ref>):

For every square-free $q\in
[X^{1/8-\kappa_\downarrow},X^{1/8+\kappa_\uparrow}]$ and $X\ge 1$, we have

\begin{equation*}
\sum_{m_1=1}^{X^{2\eta}}
\sideset{}{^\flat}\sum_{k\leq q_1^2X^{2\eta+3\kappa_2}}
\sum_{n\le \frac{X^{1/2+\epsilon}}{k}}
\frac{k^{1/2}}{n^{1/2}q_1\rad(m_1)}
H(n,m_1,q_1;X)\ll_{\epsilon,\kappa_1,\kappa_2} H(q;X),
\end{equation*}

where $H(q;X)$ is the sum of the final terms in Equations
(<ref>), (<ref>), (<ref>), and (<ref>).

We take each term in $H(n,m_1,q_1;X)$ by turn, and sum it over $n$,
then $k$, and then $m_1$. In this proof we shall write $\ll$ as a
shorthand for $\ll_{\epsilon,\kappa_1,\kappa_2}$.  Recall that we had
written $n=n_1\ell_1$, where $n_1$ is only divisible by primes
dividing $mq$, while $(\ell_1, mq)=1$, and that we denoted the radical
of $\ell_1$ by $\ell$. We begin with the first term: in this case, the
sum over $\ell$ (and also the sum over $m_1$) converges and so we

\begin{equation}\label{eqest1}
\frac{X}{q_1^2}\sum_{m_1=1}^{X^{2\eta}}
\sideset{}{^\flat}\sum_{k\leq q_1^2X^{2\eta+3\kappa_2}}
\sum_{n_1\le \frac{X^{1/2+\epsilon}}{k}}
\sum_{\ell\le \frac{X^{1/2+\epsilon}}{kn_1}}
\frac{k^{1/2}}{n_1^{1/2}\ell^{1/2}q_1\rad(m_1)}
\frac{1}{m_1^{5/3}\ell}\ll \frac{X^{7/8+3\eta+(3/2)\kappa_2+\kappa_\downarrow+\epsilon}}{q}.
\end{equation}

For the second term, we write

\begin{equation}\label{eqest2}
\frac{X^{5/6+\kappa_1/3}}{q_1^{5/3}}
\sum_{m_1=1}^{X^{2\eta}}
\sideset{}{^\flat}\sum_{k\leq q_1^2X^{2\eta+3\kappa_2}}
\sum_{n_1\le \frac{X^{1/2+\epsilon}}{k}}
\sum_{\ell\le \frac{X^{1/2+\epsilon}}{kn_1}}
\frac{k^{1/2}}{n_1^{1/2}\ell^{1/2}q_1\rad(m_1)}
\frac{1}{\ell^{1/3}}\ll \frac{X^{11/12+(8/3)\eta+\kappa_1/3+\kappa_2+\epsilon}}{q^2}.
\end{equation}

To estimate the third term, we write

\begin{equation}\label{eqest3}
\begin{array}{rcl}
\displaystyle q_1^2X^{9\kappa_1}
\sum_{m_1=1}^{X^{2\eta}}
\sideset{}{^\flat}\sum_{k\leq q_1^2X^{2\eta+3\kappa_2}}
\sum_{n\le \frac{X^{1/2+\epsilon}}{k}}
\frac{k^{1/2}}{n^{1/2}q_1\rad(m_1)}
\ell m_1^{12}&\ll&
\displaystyle q_1X^{3/4+9\kappa_1+26\eta+\epsilon}
\sideset{}{^\flat}\sum_{k\leq q_1^2X^{2\eta+3\kappa_2}}\frac{1}{k}
\\[.2in]&\ll&\displaystyle
\frac{X^{7/8+9\kappa_1+26\eta+\kappa_\uparrow+\epsilon}}{q},
\end{array}
\end{equation}

where the final estimate follows because non-zero values of $k$ are
all multiples of the squarefree $q_1$: see Proposition
<ref>. Finally, we have

\begin{equation}\label{eqest4}
\frac{X^{1-\kappa_1}}{q_1^2}
\sum_{m_1=1}^{X^{2\eta}}
\sideset{}{^\flat}\sum_{k\leq q_1^2X^{2\eta+3\kappa_2}}
\sum_{n\le \frac{X^{1/2+\epsilon}}{k}}
\frac{k^{1/2}}{n^{1/2}q_1\rad(m_1)}
\frac{1}{m_1^{5/3}}\ll \frac{X^{1-\kappa_1+3\eta+2\kappa_\downarrow+\epsilon}}{q}.
\end{equation}

This concludes the proof of Proposition <ref>.

We are now ready to prove the main result of this subsection.

There exist positive constants $\kappa_\uparrow,\kappa_\downarrow,\kappa_3$ such that
the following holds. For every $X \ge 1$ and every squarefree $q\in
[X^{1/8-\kappa_\downarrow},X^{1/8+\kappa_\uparrow}]$, we have

\begin{equation*}
\sum_{f\in \overline{\W_q(\Sigma)}}
\Bigl(S(f)-D(\tfrac 12,f)\Bigr)
\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)=
\end{equation*}

We apply (<ref>) and then apply Proposition
<ref>. It is only necessary to ensure that the exponent of
$X$ is less than $1$ for each of the 6 different error terms. This is
easily done. First, we temporarily pick any positive $\kappa_\uparrow$
and $\kappa_\downarrow$. Next we pick $\eta>2\kappa_\downarrow$. Then
we pick $\kappa_1>3\eta+2\kappa_\downarrow$ and
$\kappa_2>4\eta+2\kappa_\downarrow$.  This takes care
of (<ref>) and of the last two terms of (<ref>).

Finally, to ensure that the exponents of $X$ in the final terms of
(<ref>), (<ref>), and (<ref>) are less than $1$,
we simply divide our constants
$\kappa_\uparrow,\kappa_\downarrow,\eta,\kappa_1,\kappa_2$ by the same
sufficiently large number.

We now put together our results for the border range and the large

There exists an absolute constant $\varkappa>0$ such that for every
$X\ge 1$ and every squarefree $q\ge X^{1/8-\varkappa}$, we have

\begin{equation*}
\sum_{f\in\overline{\W_q(\Sigma)}}
\Bigl(
S(f) -  D(\tfrac 12,f)
\Bigr)
\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)
O_{\Sigma,\Psi} \Bigl(\frac{X^{1-\varkappa}}{q}\Bigr).
\end{equation*}

We combine Corollary <ref> and Proposition <ref>, where we
= \min(\kappa_{\uparrow},\kappa_3)$.

There exists an absolute constant $\mu>0$ such that for every
$X\ge 1$, we have

\begin{equation}\label{eqStoD}
\sum_{\substack{q \;{\rm squarefree}\\
q\geq X^{1/8-\mu}}}
\left|
\sum_{f\in\overline{\W_q(\Sigma)}}
\Bigl(
S(f) -  D(\tfrac 12,f)
\Bigr)
\Psi\Bigl(\frac{|\Delta(f)|}{X}\Bigr)
\right|
O_{\Sigma,\Psi} \bigl(X^{1-\mu}\bigr).
\end{equation}

Adding up the above result for $q\geq X^{1/8-\varkappa}$, we note that
$\{f\in \overline{\W_q(\Sigma)}:|\Delta(f)|<X\}$ is empty for $q\ge X^{1/2}$ because $\Delta(f) = \ind(f)^2
\Delta(K_f) \ge q^2 \Delta(K_f) \ge q^2$ for $f\in \W_q$.

Choosing $X^{\kappa_1}=\frac{X^{1/10}}{\ell^{1/10}q_1^{2/5}m_1^{7/5}}$
in the proof of Proposition <ref> yields the equality of the
last two terms and the following value

\begin{equation*}
H(n,m_1,q_1;X)=\frac{X}{q_1^2m_1^{5/3}\ell} +
\frac{\ell^{1/10} X^{9/10}}{q_1^{8/5}m_1^{3/5}} +
\frac{X^{5/6}}{q_1^{5/3}\ell^{1/3}}.
\end{equation*}

Then the remainders (<ref>) and (<ref>) are equal, and
we may further choose $\kappa_\uparrow$ so that they are also equal to
the remainder in Corollary <ref> which yields a specific
choice of the constant $\kappa_\uparrow$.  In the end, an admissible set of
values of the constants is as follows: $ \kappa_\downarrow =
\tfrac{1}{3000}$, $\kappa_{\uparrow}=\tfrac{1}{300}$,
$\eta=\tfrac{1}{1000}$ $\kappa_1=\tfrac{1}{100}$,
$\kappa_2=\tfrac{1}{200}, \varkappa = \tfrac{1}{3000}$.  To verify the
admissibility of these numerical values, it suffices to insert them in
each of the remainder terms of Proposition <ref>,
Corollary <ref>, Corollary <ref>,
(<ref>), (<ref>), (<ref>), (<ref>), and

§.§ Counting suborders

In this subsection we prove Theorem <ref> by conditionally

\begin{equation*}
\sum_{q>X^{1/8-\varkappa}}\sum_{f\in\overline{\W_q(\Sigma)}}S(f).
\end{equation*}

Note that by Corollary <ref>, we may replace $S(f)$ in the
above sum by $D(\tfrac 12,f)$. The advantage of using $D(\tfrac 12,f)$ over
$S(f)$ is that the values of $D(\tfrac 12,f)$ for binary cubic forms $f$
corresponding to suborders of a fixed cubic field $K$ can be
simultaneously controlled in terms of $L(\tfrac 12,\rho_K)$. To this
end, we start by recalling the following result, due to works of
Shintani [27] and Datskovsky–Wright [11] (see
<cit.>), giving an explicit formula for the counting
function of suborders $R$ of a fixed cubic field $K$.

Let $K$ be a cubic field with ring of integers $\O_K$. For an order
$R\subset\O_K$, let $\ind(R)$ denote the index of $R$ in $\O_K$. Then

\begin{equation*}
\sum_{R\subset
\O_K}\frac{1}{\ind(R)^s}=\frac{\zeta_K(s)}{\zeta_K(2s)}\zeta_\Q(3s)\zeta_\Q(3s-1).
\end{equation*}

We thus obtain the following corollary regarding the number $N_K(Z)$
of orders of $\O_K$ with index less than $Z$ for a cubic field $K$.

For every $\epsilon>0$, $Z\ge 1$ and cubic field $K$, we have

\begin{equation*}
\begin{array}{rcl}
N_K(Z)\ll_\epsilon Z^{1+\epsilon}|\Delta(K)|^\epsilon.
\end{array}
\end{equation*}

The implied constant is independent of $K$ and $Z$.

This follows from Perron's formula integrating along the
vertical line $\Re(s)=1+\epsilon$.

The above result can be used to give a very useful bound on the sum of
$D(\tfrac 12,f)$, over $f\in\overline{\W_q(\Sigma)}$ for $q$ greater than some
positive $Q$.

For every $Q,X\ge 1$ and $\epsilon>0$,

\begin{equation}\label{eqsumK}
\sum_{q\ge Q} \sum_{
\substack{f\in \overline{\W_q(\Sigma)}\\ |\Delta(f)|< X}}
\ll_{\epsilon,\Sigma}
X^{\frac12 + \epsilon} \sum_{2^{\mathbb N}\ni Y\le X/Q^2}
\sum_{\substack{K\in \FF_\Sigma \\
Y\leq|\Delta(K)|<2Y}}|L(\tfrac 12,\rho_K)|.
\end{equation}

Consider a real number $Y$ with $Y\ll X/Q^2$ and a cubic field
$K$ such that $Y\leq |\Delta(K)|<2Y$. Then the number of binary cubic
forms $f\in \cup_{q\ge Q} \overline{W_q(\Sigma)}$ such that $|\Delta(f)|<X$ and $K_f = K$ is bounded by

\[
N_K\Bigl(\frac{X^{\frac12}}{Y^{\frac12}}\Bigr) = O_\epsilon\bigl(X^{1/2+\epsilon}/Y^{1/2}\bigr),
\]

using Corollary <ref>.

Summing over all $K$ in the discrimant range $Y\leq|\Delta(K)|< 2Y$,
and then summing over $Y\in 2^{\mathbb N}$ such that the dyadic ranges
$[Y,2Y)$ cover (more than) the interval $[1,X/Q^2]$, we capture the
sum over $f\in \overline{\W_q(\Sigma)}$, for all $q>Q$, such that
$|\Delta(f)|< X$.

Recall from (<ref>) that we have $D(\tfrac12,f) =
L(\tfrac12,\rho_{K_f}) E(\tfrac12,f)$ and $E(\tfrac12,f)
=\prod_{p|\Delta(f)} (1+O(p^{-\frac12})) = |\Delta(f)|^{o(1)}$, which
concludes the proof of the lemma.

The above lemma yields the following consequence, which clarifies how
nonnegativity is used by us.

For every cubic field $K\in\FF_\Sigma$, assume that $L(\tfrac12,\rho_K)\geq
0$. Then for $Q,X\geq 1$, we have

\begin{equation}\label{eqcornonnegbound}
\sum_{q\geq Q}
\sum_{\substack{f\in\overline{\W_q(\Sigma)}\\|\Delta(f)|<X}} D(\tfrac12,f)
\ll_{\epsilon,\Sigma} X^{29/28+\epsilon}Q^{-15/14}.
\end{equation}

First note that the assumption $L(\tfrac12,\rho_K)\geq 0$ for all cubic
fields $K$ implies that $D(\tfrac12,f)\geq 0$ for all irreducible integral
binary cubic forms. Thus, we may apply the previous lemma to estimate
the left-hand side of (<ref>).

From Theorem <ref> (using a smooth function which
dominates the characteristic function of $[1,2]$), we obtain

\begin{equation*}
\sum_{\substack{K\in \FF_\Sigma \\ Y\leq|\Delta(K)|<2Y}}|L(\tfrac
12,\rho_K)|\ll_{\epsilon,\Sigma} Y^{\frac{29-28\delta}{28-16\delta}+\epsilon},
\end{equation*}

for $\delta = 1/128$. Even the bound with $\delta=0$ in conjunction
with (<ref>), yields the result.

We are now ready to prove Theorem <ref>.

Proof assuming strong subconvexity: The hypothesis (S) would imply
that the central value in the right-hand side of (<ref>) is bounded by $Y^{\frac 16
-\vartheta}$.  Hence the bound in (<ref>) becomes $X^{\frac
12 + \epsilon} (X/Q^2)^{\frac{2}{3}-\vartheta}$.  We pick
$Q=X^{\frac{1}{8}-\kappa_{\downarrow}}$ with
$\epsilon,\kappa_\downarrow >0$ sufficiently small such that $\tfrac12
+ \epsilon + (\tfrac34 + 2 \kappa_\downarrow)(\tfrac 23 - \vartheta)<1$.
Proposition <ref>, together with
Corollary <ref> and Corollary <ref>, now yield
the result.

Proof assuming nonnegativity: We pick $Q=X^{1/8-\varkappa}$,
with $\varkappa$ as in Theorem <ref>. It follows that we have

\begin{equation*}
\sum_{q\geq Q}\sum_{\substack{f\in \overline{\W_q(\Sigma)}}}S(f) \Psi\Bigl (\frac{|\Delta(K)|}{X}\Bigr)=
\sum_{q\geq Q}\sum_{\substack{f\in \overline{\W_q(\Sigma)}}}D(\tfrac12,f) \Psi\Bigl (\frac{|\Delta(K)|}{X}\Bigr)
\end{equation*}

Since we are assuming hypothesis (N), Corollary <ref> asserts
that we have

\begin{equation*}
\sum_{q\geq Q}
\sum_{\substack{f\in \overline{\W_q(\Sigma)}\\ |\Delta(f)|<X}}
\end{equation*}

which is sufficiently small. The result now follows from Proposition

§ PROOFS OF THEOREMS <REF> AND <REF>

In addition to the quantity $A_\Sigma(X)$, that we defined in
(<ref>), we also define
$MA_\Sigma(X)$, sum of $\lvert L(\tfrac12,\rho_K) \rvert $ for $K\in \FF_\Sigma(X)$

\begin{equation*}
\begin{array}{rcl}
\sum_{\substack{K\in \FF_\Sigma\\ X\le |\Delta(K)|<2X}}
|L(\tfrac 12,\rho_K)|;
\\[.2in]\displaystyle
\sum_{\substack{K\in \FF_\Sigma\\ X/2 \le |\Delta(K)|< 3X \\L(\tfrac 12,\rho_K)\ge
L(\tfrac 12,\rho_K).
\end{array}
\end{equation*}

The letter M stands for maximal and the letter P for positive.
$PA_\Sigma(X)$, sum of $L(\tfrac12,\rho_K)\ge 0$ for $K\in \FF_\Sigma(X)$

For every $\epsilon>0$ and $X\ge 1$, we have the asymptotic inequality

\[
MA_\Sigma(X) \le 2 PA_\Sigma(X) +
O_{\epsilon,\Sigma}\bigl(X^{\frac{29-28\delta}{28-16\delta} +\epsilon}\bigr).
\]

We let $\Psi_1:\R_{>0}\to [0,1]$ be a smooth function compactly supported on the
interval $[\frac12,3]$ such that $\Psi_1(t)=1$ for $t\in [1,2]$.
We have an inequality followed by a basic identity

\begin{equation}\label{eqbasicidentity}
\begin{aligned}
MA_\Sigma(X) &\le
\sum_{\substack{K\in \FF_\Sigma}}
|L(\tfrac 12,\rho_K)| \Psi_1\Bigl(\frac{|\Delta(K)|}{X}\Bigr)
\\
\sum_{\substack{K\in \FF_\Sigma\\ L(\tfrac 12,\rho_K)\ge 0}}
L(\tfrac 12,\rho_K) \Psi_1\Bigl(\frac{|\Delta(K)|}{X}\Bigr)
\sum_{\substack{K\in \FF_\Sigma}}
L(\tfrac 12,\rho_K) \Psi_1\Bigl(\frac{|\Delta(K)|}{X}\Bigr),
\end{aligned}
\end{equation}

which follows from $|x| = 2\max(x,0) - x$ for every $x\in \R$.
The first sum is $\le 2 PA_\Sigma(X)$. (Note that
in the respective definitions of $MA_\Sigma(X)$ and $PA_\Sigma(X)$, the discriminant range
has increased from  $X\le |\Delta(K)|<2X$ to $X/2 \le |\Delta(K)|< 3X$ for this purpose).
The second sum is equal to $A_\Sigma(X)$ for which we have established the
estimate (<ref>). This concludes the proof.

We finally arrive at the proof of our main result of this paper.
In Section <ref>, we have estimated the terms $q<Q$ of the first moment
$A_\Sigma(X)$. In Section <ref>, we have estimated for the other terms
$q\ge Q$ the difference $S(f) - D(\tfrac12,f)$. The conclusion of all these results is
summarized in the
following which was stated in the introduction as Theorem <ref>:

There is an absolute constant $\mu>0$ such that the following holds.
For every $0<\nu\le \mu$, $\epsilon>0$, and $X\ge 1$,

\begin{equation}\label{eqASignewerror}
A_\Sigma(X)-C_\Sigma\cdot X \bigl(\log X +  \widetilde \Psi'(1) \bigr)  - C'_\Sigma \cdot X
\ll_{\epsilon,\nu,\Sigma,\Psi}
X^{1/2+\epsilon} \cdot \sum_{ 2^{\mathbb N} \ni Y\le
\end{equation}

where the sum over $Y$ is dyadic, namely $Y\in 2^{\mathbb N}$ is constrained to be a
of $2$.

The result will follow from Proposition <ref>, Corollary
<ref> and (<ref>).
It follows from Proposition <ref> that

\[
A_{\Sigma}(X) -
C_\Sigma\cdot X \bigl(\log X +  \widetilde \Psi'(1) \bigr)  - C'_\Sigma \cdot X
\ll_{\epsilon, \Sigma,\Psi}
\frac{X^{1+\epsilon}}{Q}+X^{\frac{11}{12}+\epsilon}
Q^{2+\epsilon} X^{\frac34+\epsilon} +
\sum_{q\ge Q}
\left|
\sum_{f\in \overline{W_q(\Sigma)}} S(f) \Psi\left(
\frac{|\Delta(f)|}{X}
\right)
\right|.
\]

Let $a>0$ be sufficiently small such that $\Psi(t)=0$ whenever $a^2t\ge 1$.
Choose $Q= a^{-1} X^{1/8-  \nu/ 2}$.
Using that $Q\gg_\Psi X^{1/8-\mu}$, we can apply Corollary <ref> to obtain the bound

\[
\sum_{q\ge Q}
\left|
\sum_{f\in \overline{W_q(\Sigma)}}
\Bigl(
S(f) -  D(\tfrac 12,f)
\Bigr)
\Psi\left(
\frac{|\Delta(f)|}{X}
\right)
\right|
\ll_{\Sigma,\Psi} X^{1-\mu} \le X^{1-\nu}.
\]

The estimate (<ref>) yields

\[
\begin{aligned}
\sum_{q\ge Q}
\sum_{f\in \overline{W_q(\Sigma)}}
\left|
D(\tfrac 12,f)
\right|
\Psi\left(
\frac{|\Delta(f)|}{X}
\right)
\sum_{q\ge Q}
\sum_{\substack{f\in \overline{W_q(\Sigma)}\\ |\Delta(f)|<X/a^2}}
\left|
D(\tfrac 12,f)
\right|
\\
X^{\frac12 + \epsilon}
\cdot \sum_{ 2^{\mathbb N} \ni Y\le
\frac{MA_\Sigma(Y)}{Y^{1/2}}.
\end{aligned}
\]

It remains to observe that $(X/a^2)/Q^2 = X^{3/4 + \nu}$ to conclude the proof.

We are now ready to prove our main Theorem <ref>.
Recall that the qualitative version in Theorem <ref> follows from Theorem

Recall that $C_\Sigma>0$ in Proposition <ref>.
We distinguish two cases depending on the size of the sum of $MA_\Sigma(Y)$
in the right-hand
side of (<ref>).

In the first case, if the right-hand
side of (<ref>) is $< X$, then we have
$A_\Sigma(X)\sim C_\Sigma \cdot X\cdot \log X$. In combination with
Theorem <ref>, we obtain that $\gg_{\epsilon,\Sigma}
X^{\frac{3}{4}+\delta-\epsilon}$ cubic fields
$K\in\FF_\Sigma$ with $|\Delta(K)|<X$ satisfy $L(\tfrac 12,\rho_K)>0$.

\begin{equation}\label{firstcase}
\delta_\Sigma(X) \ge \frac{3}{4} + \delta - \epsilon - O_\epsilon\bigl(\frac{1}{\log X}\bigr),
\end{equation}

which is sufficient to imply Theorem <ref> in that case.

Assume in the second case that the right-hand side of (<ref>) is $\ge
X$, namely

\begin{equation*}
\sum_{2^{\mathbb N} \ni Y\le X^{3/4+\nu}}\frac{MA_\Sigma(Y)}{Y^{1/2}}\ge
\end{equation*}

This implies that there exists $Y\in 2^{\mathbb N}$ with $Y\leq X^{3/4+\nu}$ such that
$MA_\Sigma(Y)\ge X^{1/2-\epsilon}Y^{1/2}$.
It follows from Proposition <ref> that

\begin{equation*}
2 PA_\Sigma(Y)\ge X^{1/2-\epsilon}Y^{1/2}
+O_{\epsilon,\Sigma}\bigl(Y^{\frac{29-28\delta}{28-16\delta} +\epsilon}\bigr).
\end{equation*}

Since $Y \leq X^{3/4+\nu}$, the error term is negligible. (The convexity bound
suffices for this). We deduce in the second case:

\begin{equation}\label{eqYineq}
2PA_\Sigma(Y)\ge X^{1/2-\epsilon}Y^{1/2}.
\end{equation}

Theorem <ref> and (<ref>) imply that $\gg_{\epsilon}
X^{1/2-\epsilon}Y^{1/4+\delta-\epsilon}$ cubic fields
$K\in\FF_\Sigma$ with $|\Delta(K)|<Y$ satisfy the inequality
$L(\tfrac 12,\rho_K)>0$.

\begin{equation}\label{secondcase}
\delta_\Sigma(Y) \ge \frac{\log X}{2 \log Y} + \bigl(\frac 14+\delta -\epsilon\bigr)
-O_\epsilon\bigl(\frac{1}{\log Y}\bigr).
\end{equation}

Since (<ref>) implies that $Y\to \infty$ we deduce

\begin{equation*}
\limsup_{X\to \infty} \delta_\Sigma(X) \ge \frac 34 + \delta,
\end{equation*}

because the inequality is satisfied either in the first case by $\delta_\Sigma(X)$
in (<ref>) or in the
second case by $\delta_\Sigma(Y)$ in (<ref>), since $\frac{\log X}{\log Y} \ge \frac{4}{3+4\nu}\ge

To conclude a lower bound on the $\liminf$, we need a lower bound on $Y$ in the second case.
Theorem <ref> implies $PA_\Sigma(Y) = O_\epsilon(Y^{\frac 54 -\delta+\epsilon})$.
Together with (<ref>), this yields the following lower bound:

\begin{equation}\label{lowerY}
Y\gg_\epsilon X^{\frac{2}{3-4\delta}-\epsilon}.
\end{equation}

This implies

\begin{equation}\label{deltaSigmage}
\delta_\Sigma(X) \ge \frac 12 + \bigl(\frac 14 +\delta\bigr)\cdot \frac{2}{3-4\delta} - \epsilon -
O_\epsilon\bigl(\frac{1}{\log X}\bigr).
\end{equation}

The first two terms of (<ref>) simplify to $\frac{2}{3-4\delta}$, hence

\[
\liminf_{X\to \infty} \delta_\Sigma(X) \ge \frac{2}{3-4\delta}.
\]

This concludes the proof of Theorem <ref>.

The same argument implies an Omega result $MA_\Sigma(X) =\Omega_\Sigma(X^{1-o(1)})$ as
\infty$. Namely, there is a sequence $X_k\to \infty$ such that for every $\epsilon >0$, $MA_\Sigma(X
_k)/X_k^{1-\epsilon}\to \infty$.

For completeness, we also record the following lower bound for
the first moment:

For every $\epsilon>0$ and $X\ge 1$,

\[
X^{\frac{5-4\delta}{6-8\delta}-\epsilon} \ll_{\epsilon,\Sigma}
\sum_{K\in \FF_\Sigma(X)} \bigl|L\bigl(\tfrac12,\rho_K\bigr)\bigr|.
\]

The lower bound (<ref>) for $Y$ implies the lower bound in
Proposition <ref>
as follows:

\[
\sum_{K\in \FF_\Sigma(X)}
|L(\tfrac 12,\rho_K)| \ge
\sum_{K\in \FF_\Sigma(Y)}
|L(\tfrac 12,\rho_K)|
\gg_{\epsilon,\Sigma} X^{\frac12 -\epsilon} Y^{\frac12},
\]

and $\tfrac12 + \tfrac{1}{3-4\delta} = \tfrac{5-4\delta}{6-8\delta}$.

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E-mail address, Department of Mathematics, University of Toronto, Toronto, ON, M5S 2E4,
E-mail address, Department of Mathematical Sciences, Chalmers University of Technology and the University of

[0ex]0ex0exGothenburg, SE-412 96 Gothenburg, Sweden
E-mail address, Department of Mathematics, Cornell University, Ithaca, NY 14853, USA