Q_{1}^{\prime\prime}m_{1}},$

by (4.29). Note that

$X={PZ\over Q_{1}^{\prime\prime}m_{1}},$

where $Z$ is defined as in (4.26). It is at this point that we apply Lemma
4.16, with $R=Q/Q_{1}$, $S=V/Q_{1}$, and $W=[m,Q_{1}]=q_{1}$. Note that

$(U,X)=K(EQ_{1},X)=K\frac{1}{mQ_{1}^{\prime\prime}}(EQ_{1}mQ_{1}^{\prime\prime},PZ)=K\frac{1}{q_{1}}(Eq_{1}Q_{1},Z).$

Here $K$ is an absolute constant, not the same in each occurrence. Then

$(U,(U,X)W)=K(EQ_{1},EQ_{1}q_{1},Z)\ll(EQ_{1},Z).$

Now applying Lemma 4.16 gives:

$\displaystyle\Bigg{[}\cdot\Bigg{]}$ $\displaystyle\ll_{\varepsilon}$
$\displaystyle Q^{\varepsilon}\left({V(EQ_{1},Z)\over
Q{EQ_{1}}q_{1}}+{V^{2}\over Q^{2}}+\frac{Q_{1}c}{Q}+{V|X|\over
Q^{2}c}\right).$

Returning to the original summation, we have:

$\displaystyle\hskip-36.135pt\left|\sum_{Q\leq q\leq
Q+V}\sum_{r(q)}^{\prime}\mathcal{S}_{q}(r,k,\ell;\gamma)\overline{\mathcal{S}_{q}(r,k^{\prime},\ell^{\prime};\gamma^{\prime})}\right|$
$\displaystyle\ll_{\varepsilon}$ $\displaystyle Q^{\varepsilon}\sum_{E\mid
BcP^{2}\atop q_{1}=BcP^{2}/E}{\bf 1}_{\\{d\ell\equiv ak\equiv
d^{\prime}\ell^{\prime}\equiv
a^{\prime}k^{\prime}(\operatorname{mod}(q_{1},c))\\}}\sum_{Q_{1}\mid
q_{1}\atop{p\mid Q_{1}\Longrightarrow(p^{\infty},q_{1})\mid
Q_{1}\atop{Q_{1}^{\prime}=Q_{1}^{2}\atop{m=q_{1}/Q_{1},\
(m,Q_{1})=1,\atop(E,q_{1}/Q_{1})=1}}}}q_{1}\mathcal{N}_{m}{m\over\phi(m)}$
$\displaystyle\times\left(\sum_{r_{1}(Q_{1})}^{\prime}{\bf 1}_{\\{-\ell\equiv
Pr_{1}(Ba+Dc)(\operatorname{mod}Q_{1})\atop-k\equiv
Pr_{1}(Ac+Bd)(\operatorname{mod}Q_{1})\\}}{\bf 1}_{\\{-\ell^{\prime}\equiv
Pr_{1}(Ba^{\prime}+Dc)(\operatorname{mod}Q_{1})\atop-k^{\prime}\equiv
Pr_{1}(Ac+Bd^{\prime})(\operatorname{mod}Q_{1})\\}}\right)\left[{V(EQ_{1},Z)\over
QEQ_{1}q_{1}}+{V^{2}\over Q^{2}}+\frac{Q_{1}c}{Q}+{V|X|\over Q^{2}c}\right].$

Note that the $r_{1}(\operatorname{mod}Q_{1})$ summation is exactly
$\mathcal{N}_{Q_{1}}$, and since $(m,Q_{1})=1$, we have that
$\mathcal{N}_{m}\cdot\mathcal{N}_{Q_{1}}=\mathcal{N}_{q_{1}}$. Now we have,
crudely, that

$\displaystyle\hskip-72.26999pt\left|\sum_{Q\leq q\leq
Q+V}\sum_{r(q)}^{\prime}\mathcal{S}_{q}(r,k,\ell;\gamma)\overline{\mathcal{S}_{q}(r,k^{\prime},\ell^{\prime};\gamma^{\prime})}\right|$
$\displaystyle\ll_{\varepsilon}$ $\displaystyle Q^{\varepsilon}\sum_{E\mid
BcP^{2}\atop q_{1}=BcP^{2}/E}{\bf 1}_{\\{d\ell\equiv ak\equiv
d^{\prime}\ell^{\prime}\equiv
a^{\prime}k^{\prime}(\operatorname{mod}(q_{1},c))\\}}\sum_{Q_{1}\mid
q_{1}\atop{p\mid Q_{1}\Longrightarrow(p^{\infty},q_{1})\mid
Q_{1}\atop(E,q_{1}/Q_{1})=1}}\mathcal{N}_{q_{1}}$
$\displaystyle\times\left[{V(EQ_{1},Z)\over QEQ_{1}}+{V^{2}c\over
Q^{2}}+\frac{c^{3}}{Q}+{V|Z|\over Q^{2}}\right],$

from which is the claim. ∎

Next we analyze the size of $\mathcal{N}_{m}$ in (4.27).

######  Lemma 4.33.

Let $m\mid BcP^{2}$. Then

$\mathcal{N}_{m}\ll 1.$ (4.34)

###### Proof.

Our goal is to count the number of $r$ in $(\mathbb{Z}/m)^{\times}$ satisfying
$-\ell\equiv Pr(Ba+Dc)$, $-k\equiv Pr(Ac+Bd),$ and also $-\ell^{\prime}\equiv
Pr(Ba^{\prime}+Dc),$ $-k^{\prime}\equiv Pr(Ac+Bd^{\prime})$. Let

$m_{1}:=(m,c),$

so that $m_{1}\mid c$, and $m\ll m_{1}$ (since $B$ and $P$ are fixed).
Reducing the moduli mod $m_{1}$ gives the equations:

$-\bar{a}\ell\equiv rPB,\ -\bar{d}k\equiv rPB,\
-\bar{a}^{\prime}\ell^{\prime}\equiv rPB,\ -\bar{d}^{\prime}k^{\prime}\equiv
rPB(\operatorname{mod}m_{1}).$

Here we used that $(a,c)=1$ since
$\gamma\in\operatorname{SL}_{2}(\mathbb{Z})$, etc. There are clearly a bounded
number of solutions in $r$ to the above, which gives the claim. ∎

And lastly, we analyze the number of elements in
$\operatorname{SL}_{2}(\mathbb{Z})$ with a given value of $c$ and satisfying a
congruence in $Z$ in (4.26).

######  Lemma 4.35.

Let $\gamma\in\operatorname{SL}_{2}(\mathbb{Z})\cap B_{T}$ be given with
$\gamma_{c}=c$, and fix a divisor $Z_{1}\mid c$ with $c\ll T$. Also fix
$k,\ell,k^{\prime},\ell^{\prime}$ with $k\ell=k^{\prime}\ell^{\prime}$. Then
the number of $\gamma^{\prime}\in\operatorname{SL}_{2}(\mathbb{Z})$ with
$\gamma_{c}^{\prime}=c$ and $|a^{\prime}|,|b^{\prime}|,|d^{\prime}|\asymp T$
and satisfying $Z\equiv 0(\operatorname{mod}Z_{1})$ is bounded by

$\ll{T\over Z_{1}}(Z_{1},\ell d-ka),$ (4.36)

as $T\to\infty$.

###### Proof.

In the variables $a^{\prime},d^{\prime}$, we have the pair of equations:
$a^{\prime}d^{\prime}\equiv 1(\operatorname{mod}Z_{1})$ and

$Z=(Ac\ell+Bak+Bd\ell+Dck)-(Ac\ell^{\prime}+Ba^{\prime}k^{\prime}+Bd^{\prime}\ell^{\prime}+Dck^{\prime})\equiv
0(Z_{1}),$

or

$Bk^{\prime}a^{\prime 2}-(Bak+Bd\ell)a^{\prime}+B\ell^{\prime}\equiv
0(Z_{1}).$

Let $B_{1}:=(B,Z_{1})$ and set $Z_{2}:=Z_{1}/B_{1}$ and $B_{2}:=B/B_{1}$, with
$(B_{2},Z_{2})=1$. Dividing through by $B_{1}$, the equation reduces to

$k^{\prime}a^{\prime 2}-(ak+d\ell)a^{\prime}+\ell^{\prime}\equiv 0(Z_{2}).$

Let

$\tilde{Z}:=(k^{\prime},\ell^{\prime},Z_{2}).$

Since $k\ell=k^{\prime}\ell^{\prime}$, we have that $\tilde{Z}^{2}\mid k\ell$.
Working locally, suppose that $\tilde{Z}\mid\ell$. Then reducing mod
$\tilde{Z}$ gives the equation

$-(ak)a^{\prime}\equiv 0(\tilde{Z}).$

But $(a,\tilde{Z})=(a,c)=1$ and $a^{\prime}$ is also coprime to $\tilde{Z}$,
which implies that $k\equiv 0(\tilde{Z})$. Therefore there are no solutions
unless both $\tilde{Z}\mid k$ and $\tilde{Z}\mid\ell$. In this case, we can
divide the whole equation by $\tilde{Z}$. Set $k_{1}:=k/\tilde{Z},\ \dots,\
\ell_{1}^{\prime}:=\ell^{\prime}/\tilde{Z}$ and $Z_{3}:=Z_{2}/\tilde{Z}$. Then
the equation becomes

$k_{1}^{\prime}a^{\prime
2}-(ak_{1}+d\ell_{1})a^{\prime}+\ell_{1}^{\prime}\equiv 0(Z_{3}),$

with $(k_{1}^{\prime},\ell_{1}^{\prime},Z_{3})=1$. Working locally, we may
assume that $(k_{1}^{\prime},Z_{3})=1$. Now we can simply solve the equation.
Assuming for simplicity that $Z_{3}$ is odd (with minor modifications
otherwise), we have that

$(a^{\prime}-\bar{2}\bar{k}_{1}^{\prime}(ak_{1}+d\ell_{1}))^{2}\equiv-\bar{k}_{1}^{\prime}\ell_{1}^{\prime}+\bar{4}\bar{k}_{1}^{\prime
2}(ak_{1}+d\ell_{1})^{2}\equiv(\bar{2}\bar{k}_{1}^{\prime}(ak_{1}-d\ell_{1}))^{2}(\operatorname{mod}Z_{3}),$

where we again used that $k\ell=k^{\prime}\ell^{\prime}$ and $ad\equiv
1(\operatorname{mod}Z_{3})$. The equation is now a difference of squares, so

$(a^{\prime}-\bar{k}_{1}^{\prime}\ell_{1}d)(a^{\prime}-\bar{k}_{1}^{\prime}k_{1}a)\equiv
0(\operatorname{mod}Z_{3}).$

Again working locally, suppose that $Z_{3}=p^{U}$. Then for $V+W=U$, we get a
solution

$a^{\prime}\equiv\bar{k}_{1}^{\prime}\ell_{1}d(\operatorname{mod}p^{V})\text{
and }a^{\prime}\equiv\bar{k}_{1}^{\prime}k_{1}a(\operatorname{mod}p^{W}).$

Assume WLOG that $V\leq W$. Then for there to be any solutions, it must be the
case that $\ell_{1}d-k_{1}a\equiv 0(\operatorname{mod}p^{V})$, and if this is
the case, then there are $p^{U-W}=p^{V}$ solutions for $a^{\prime}$. Let

$Z_{4}:=(Z_{3},\ell_{1}d-k_{1}a).$

By the above discussion, the number of solutions for
$a^{\prime}(\operatorname{mod}Z_{3})$ is at most

$\min(Z_{4},\sqrt{Z_{3}}).$

So once the value of $a^{\prime}$ mod $Z_{3}$ is fixed, the total number of
such $a^{\prime}\asymp T$ is

$\ll\frac{T}{Z_{3}}\ll{T\tilde{Z}\over
Z_{2}}\ll{T(k^{\prime},\ell^{\prime},Z_{1})\over Z_{1}}.$

Thus we can bound the total number of values of $a^{\prime}$ by

$Z_{4}T(k^{\prime},\ell^{\prime},Z_{1})/Z_{1}\ll
T(Z_{3},\ell_{1}d-k_{1}a)(k^{\prime},\ell^{\prime},Z_{1})/Z_{1}\ll
T(Z_{1},\ell d-ka)/Z_{1}.$

With $a^{\prime}$ and $c$ fixed, the number of $d^{\prime},b^{\prime}$ is $\ll
1$, since they are all of order $T$ and $a^{\prime}d^{\prime}-b^{\prime}c=1$.
This completes the proof. ∎

##  5\. Major Arc Analysis

######  Theorem 5.1.

There is an $\eta>0$ and a set $\mathscr{E}\subset\mathbb{Z}$ of “exceptional”
$n$, of zero density,

$\frac{1}{N}\\#(\mathscr{E}\cap[-N,N])=O(N^{-\eta}),$

such that, for $n\notin\mathscr{E}$, $n\asymp N$, we have that

$\mathcal{M}_{N}(n)\ \gg\ \mathfrak{S}(n){\widehat{\mathcal{R}_{N}}(0)\over
N}+O\left({\widehat{\mathcal{R}_{N}}(0)\over N}N^{-\eta}\right),$

as $N\to\infty$, where, for admissible $n\notin\mathscr{E}$ and for any
$\varepsilon>0$, the “singular series” $\mathfrak{S}(n)$ satisfies

$\mathfrak{S}(n)\gg_{\varepsilon}|n|^{-\varepsilon}.$

The implied constants are absolute.

###### Proof.

We begin with (2.17):

$\displaystyle\mathcal{M}_{N}(n)$ $\displaystyle=$
$\displaystyle\int_{0}^{1}\mathfrak{M}(\theta)\widehat{\mathcal{R}_{N}}(\theta)e(-n\theta)d\theta$
$\displaystyle=$
$\displaystyle\int_{0}^{1}\sum_{q<Q_{0}}\sum_{r(q)}^{\prime}\sum_{m\in\mathbb{Z}}\psi((\beta+m)\tfrac{N}{K_{0}})\widehat{\mathcal{R}_{N}}(\tfrac{r}{q}+\beta)e(-n(\tfrac{r}{q}+\beta))d\beta$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}}\sum_{q<Q_{0}}\sum_{r(q)}^{\prime}\widehat{\mathcal{R}_{N}}(\tfrac{r}{q}+\beta)e(-n(\tfrac{r}{q}+\beta))\psi(\beta\tfrac{N}{K_{0}})d\beta.$

We have that

$\mathcal{R}_{N}(\tfrac{r}{q}+\beta)=\sum_{x,y\in\mathbb{Z}}\Upsilon\left({x\over
X}\right)\Upsilon\left({y\over
X}\right)\sum_{\gamma_{0}\in\Gamma(q)\backslash\Gamma}e_{q}(r\mathfrak{f}_{\gamma_{0}}(x,y))\left[\sum_{\gamma\in\mathscr{F}_{T}\atop\gamma\equiv\gamma_{0}(\operatorname{mod}q)}e(\beta\mathfrak{f}_{\gamma}(x,y))\right].$

For the bracketed term, we apply Lemma 3.4, together with
$|\beta|<K_{0}/N<1/X^{2}$.

$\displaystyle\mathcal{R}_{N}(\tfrac{r}{q}+\beta)$ $\displaystyle=$
$\displaystyle\sum_{x,y\in\mathbb{Z}}\Upsilon\left({x\over
X}\right)\Upsilon\left({y\over
X}\right)\sum_{\gamma_{0}\in\Gamma(q)\backslash\Gamma}e_{q}(r\mathfrak{f}_{\gamma_{0}}(x,y))$
$\displaystyle\left[\frac{1}{[\Gamma:\Gamma(q)]}\sum_{\gamma\in\mathscr{F}_{T}}e(\beta\mathfrak{f}_{\gamma}(x,y))+O(|\mathscr{F}_{T}|N^{-\Theta})\right].$

Inserting this into $\mathcal{M}_{N}$ gives:

$\displaystyle\mathcal{M}_{N}(n)$ $\displaystyle=$
$\displaystyle\sum_{x,y\in\mathbb{Z}}\Upsilon\left({x\over
X}\right)\Upsilon\left({y\over
X}\right)\left[\sum_{q<Q_{0}}\frac{1}{[\Gamma:\Gamma(q)]}\sum_{\gamma_{0}\in\Gamma(q)\backslash\Gamma}\sum_{r(q)}^{\prime}e_{q}(r(\mathfrak{f}_{\gamma_{0}}(x,y)-n))\right]$
$\displaystyle\times\left[\sum_{\gamma\in\mathscr{F}_{T}}\int_{\mathbb{R}}\psi(\beta\tfrac{N}{K_{0}})e(\beta(\mathfrak{f}_{\gamma}(x,y)-n))d\beta\right]$
$\displaystyle+O\left({\widehat{\mathcal{R}_{N}}(0)\over
N}N^{-\Theta}{Q_{0}^{5}K_{0}}\right),$

where we have split into modular and archimedean components. The proof then
follows on applying Theorem 3.39 and Theorem 3.42 to the modular component,
and Lemma 3.5 to the archimedean part, together with the choice of parameters
in (2.15). ∎

##  6\. Minor Arc Analysis

Our goal is to estimate $\mathcal{E}_{N}$ in $\ell^{2}$, or what is the same
(by Parseval), to bound

$\|\mathcal{E}_{N}\|^{2}\ =\ \|\widehat{\mathcal{E}_{N}}\|^{2}\ =\
\int_{0}^{1}|1-\mathfrak{M}(\theta)|^{2}|\widehat{\mathcal{R}_{N}}(\theta)|^{2}d\theta.$

The main result of this section is the following.

######  Theorem 6.1.

There exists some $\eta>0$ so that, as $N\to\infty$,

$\|\mathcal{E}_{N}\|^{2}\ll{|\widehat{\mathcal{R}_{N}}(0)|^{2}\over
N}N^{-\eta}.$

A standard argument concludes the main Theorem 1.13 (and hence Theorem 1.8)
from Theorem 5.1 and Theorem 6.1. We begin the analysis as follows. In
Dirichlet’s approximation theorem, we choose the level

$M\ :=\ TX,$

so that for every $\theta\in[0,1]$, there is a $q<M$ and $(r,q)=1$ so that
$\theta=\frac{r}{q}+\beta$ with

$\left|\theta-\frac{r}{q}\right|=|\beta|<\frac{1}{qM}.$

Now we decompose the circle into dyadic regions of the form

$W_{Q}\ :=\ \left\\{\theta=\frac{r}{q}+\beta\ :\ q\asymp
Q,(r,q)=1,|\beta|\ll\frac{1}{QM}\right\\},$

so that

$\|\mathcal{E}_{N}\|^{2}\ \ll\
\sum_{Q<M\atop{dyadic}}\int_{W_{Q}}|1-\mathfrak{M}(\theta)|^{2}|\widehat{\mathcal{R}_{N}}(\theta)|^{2}d\theta.$

This decomposes further into three ranges, according to whether $Q$ satisfies:
$Q<Q_{0}$, or $Q_{0}<Q<X/Y$, or $X/Y<Q<XT=M$. Here we have set

$Y:=T^{(2\delta-1)/10}=N^{y},\qquad y>0,$ (6.2)

to be a small power of $N$.

On the latter two ranges, the weight $|1-\mathfrak{M}(\theta)|^{2}$ is exactly
$1$. To keep track, we define the integrals:

$\mathcal{I}_{Q_{0},K_{0}}\ :=\ \int_{\theta=\frac{r}{q}+\beta\atop
q<Q_{0},|\beta|<K_{0}/N}\left|\beta{N\over
K_{0}}\right|^{2}|\widehat{\mathcal{R}_{N}}(\theta)|^{2}d\theta,$
$\mathcal{I}_{Q_{0}}\ :=\ \int_{\theta=\frac{r}{q}+\beta\atop
q<Q_{0},K_{0}/N\leq|\beta|<1/(qM)}|\widehat{\mathcal{R}_{N}}(\theta)|^{2}d\theta,$
$\mathcal{I}_{Q}\ :=\ \int_{\theta=r/q+\beta\atop Q\leq
q<2Q,(r,q)=1,|\beta|<1/(QM)}\left|\widehat{\mathcal{R}_{N}}(\theta)\right|^{2}d\theta.$
(6.3)

###  6.1. Preliminaries

We first estimate $\widehat{\mathcal{R}_{N}}(\theta)$ for
$\theta=\frac{r}{q}+\beta$ as follows. We begin by decomposing $x$ and $y$
according to their residue classes mod $q$ and applying Poisson summation in
$x$ and $y$ gives:

$\displaystyle\mathcal{R}_{N}(\theta)$ $\displaystyle=$
$\displaystyle\sum_{\gamma\in\mathscr{F}_{T}}\sum_{x,y\in\mathbb{Z}}\Upsilon\left({x\over
X}\right)\Upsilon\left({y\over
X}\right)e((\frac{r}{q}+\beta)\mathfrak{f}_{\gamma}(x,y))$ (6.4)
$\displaystyle=$ $\displaystyle
X^{2}\sum_{\gamma\in\mathscr{F}_{T}}\sum_{k,\ell\in\mathbb{Z}}\mathcal{S}_{q}(r,k,\ell;\gamma)\mathcal{J}_{X}(\beta,k,\ell,q;\gamma),$

where $\mathcal{S}_{q}$ is given in (4.1) and

$\mathcal{J}_{X}(\beta,k,\ell,q;\gamma)\ :=\
\iint_{x,y\in\mathbb{R}}\Upsilon\left({x}\right)\Upsilon\left({y}\right)e(\beta\mathfrak{f}_{\gamma}(xX,yX)-\tfrac{X}{q}(kx+\ell
y))dxdy.$

######  Lemma 6.5.

Suppose that $q<X/Y$. Then for any $L<\infty$, we have that

$|\mathcal{J}_{X}(\beta,k,\ell,q;\gamma)|\ \ll_{L}\ X^{-L},$ (6.6)

unless $|k|,|\ell|\ll 1$, in which case, we have:

$|\mathcal{J}_{X}(\beta,k,\ell,q;\gamma)|\ll\min(1,\frac{1}{N|\beta|}).$ (6.7)

Alternatively, if $X/Y\leq q$, then we have the same arbitrary cancellation
(6.6), unless $|k|,|\ell|\ll Y\frac{q}{X}$ (in which case, we only need the
trivial bound).

###### Proof.

The phase of $\mathcal{J}_{X}$ can be written as $e(g)$ where

$g(x,y)=\beta\mathfrak{f}_{\gamma}(xX,yX)-\frac{X}{q}(kx+\ell y).$

Inputting (2.8), the partial derivatives of $g$ are

$\partial_{x}g(x,y)=\frac{X}{q}\left(\beta
qBc_{\gamma}P^{2}Xy-k\right)+O(|\beta|TX)$ (6.8)

and similarly

$\partial_{y}g(x,y)=\frac{X}{q}\left(\beta
qBc_{\gamma}P^{2}Xx-\ell\right)+O(|\beta|TX).$ (6.9)

First consider the case that $q<X/Y$. Recalling that $|\beta|\ll 1/(QTX)$,
$B,P,x,y\asymp 1$, and $c_{y}\asymp T$, we have that $\nabla g\neq 0$ unless

$|k|,|\ell|\ll|\beta|qTX\ll 1.$

Outside of this range, we apply non-stationary phase, giving (6.6).

Now suppose that the pair $(k,\ell)$ is such that $\nabla g$ does vanish at
some point $p$ in the support of $\Upsilon\times\Upsilon$. In this case, we
apply stationary phase to show that

$|\mathcal{J}_{X}|\ll\min(1,\Delta_{p}^{-1/2}),$

where $\Delta_{p}$ is absolute determinant of the Hessian of $g$ at $p$. Since

$\Delta_{p}=|\det(\partial_{i,j}(g))|=(|\beta|Bc_{\gamma}P^{2}X^{2})^{2}\asymp(|\beta|TX^{2})^{2},$

we have that

$\Delta_{p}^{-1/2}\ll\frac{1}{TX^{2}|\beta|}=\frac{1}{N|\beta|},$

which gives (6.7). (In fact, since the form is bilinear in the variables, it
is possible to evaluate the integrals explicitly, though this is not needed
here.)

In the case that $X/Y\leq q$, we can only apply non-stationary phase if the
phase is actually growing, which is the case if
$\max(|k|,|\ell|)>Y\frac{q}{X}$. This completes the proof. ∎

###  6.2. Minor Arcs I: Case $q<Q_{0}$

######  Proposition 6.10.

Assume that $q<X$. Then

$\left|\widehat{\mathcal{R}_{N}}\left(\frac{r}{q}+\beta\right)\right|\ll{X^{2}|\mathscr{F}_{T}|\over
N|\beta|}.$ (6.11)

###### Proof.

Inserting Lemma 6.5 gives

$\widehat{\mathcal{R}_{N}}\left(\frac{r}{q}+\beta\right)\ll
X^{2}\sum_{\gamma\in\mathscr{F}_{T}}\sum_{k,\ell\ll
1}|\mathcal{S}_{q}(r,k,\ell;\gamma)|\frac{1}{N|\beta|},$

which gives the result on trivially estimating $|\mathcal{S}_{q}|\leq 1$. ∎

######  Corollary 6.12.

We have:

$\mathcal{I}_{Q_{0},K_{0}}\ \ll\ {|\widehat{\mathcal{R}_{N}}(0)|^{2}\over
N}{Q_{0}^{2}\over K_{0}},$

and

$\mathcal{I}_{Q_{0}}\ \ll\ {|\widehat{\mathcal{R}_{N}}(0)|^{2}\over
N}{Q_{0}^{2}\over K_{0}}.$

###### Proof.

Inserting the $L^{\infty}$ bound (6.11) gives:

$\displaystyle\mathcal{I}_{Q_{0},K_{0}}$ $\displaystyle\ll$
$\displaystyle\int_{\theta=\frac{r}{q}+\beta\atop
q<Q_{0},|\beta|<K_{0}/N}\left|\beta{N\over
K_{0}}\right|^{2}{X^{4}|\mathscr{F}_{T}|^{2}\over N^{2}|\beta|^{2}}d\theta\
\ll\ {|\widehat{\mathcal{R}_{N}}(0)|^{2}\over K_{0}^{2}}Q_{0}^{2}{K_{0}\over
N},$

giving the claim. Similarly,

$\displaystyle\mathcal{I}_{Q_{0}}$ $\displaystyle\ll$
$\displaystyle\int_{\theta=\frac{r}{q}+\beta\atop
q<Q_{0},K_{0}/N<|\beta|}\left|{X^{2}|\mathscr{F}_{T}|\over
N|\beta|}\right|^{2}d\theta\ll{|\widehat{\mathcal{R}_{N}}(0)|^{2}\over
N^{2}}{Q_{0}^{2}N\over K_{0}},$

as claimed. ∎

The choice of parameters (2.14) ensures that these are power savings, as
required in Theorem 6.1.

###  6.3. Minor Arcs II: Case $Q_{0}\leq Q<X/Y$

Next take the intermediate range, where $Q_{0}\leq Q<X/Y$. We need to estimate
(6.3). Inserting (6.4) and opening the square gives

$\displaystyle\mathcal{I}_{Q}$ $\displaystyle=$
$\displaystyle\int_{\theta=r/q+\beta\atop Q\leq
q<2Q,(r,q)=1,|\beta|<1/(QM)}\left|X^{2}\sum_{\gamma\in\mathscr{F}_{T}}\sum_{k,\ell\in\mathbb{Z}}\mathcal{S}_{q}(r,k,\ell;\gamma)\mathcal{J}_{X}(\beta,k,\ell,q;\gamma)\right|^{2}d\theta$
$\displaystyle=$ $\displaystyle
X^{4}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}}\sum_{k,\ell,k^{\prime},\ell^{\prime}\in\mathbb{Z}}\sum_{q\asymp
Q}\left[\sum_{r(q)}^{\prime}\mathcal{S}_{q}(r,k,\ell;\gamma)\overline{\mathcal{S}_{q}(r,k^{\prime},\ell^{\prime};\gamma^{\prime})}\right]$
$\displaystyle\hskip
36.135pt\int_{|\beta|<1/(QM)}\mathcal{J}_{X}(\beta,k,\ell,q;\gamma)\overline{\mathcal{J}_{X}(\beta,k^{\prime},\ell^{\prime},q;\gamma^{\prime})}d\beta.$

We apply Lemma 6.5 to get

$\mathcal{I}_{Q}\ \ll\ {X^{4}\over
N}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}}\sum_{k,\ell,k^{\prime},\ell^{\prime}\ll
1}\sum_{q\asymp
Q}\left|\sum_{r(q)}^{\prime}\mathcal{S}_{q}(r,k,\ell;\gamma)\overline{\mathcal{S}_{q}(r,k^{\prime},\ell^{\prime};\gamma^{\prime})}\right|.$
(6.13)

Now we introduce a parameter

$H:=Q_{0}^{\eta_{0}/4},$ (6.14)

where $\eta_{0}$ is the constant in (3.8), and decompose

$\mathcal{I}_{Q}\ \leq\ \mathcal{I}_{Q}^{(<)}+\mathcal{I}_{Q}^{(\geq)},$

according to whether $\gcd(c,c^{\prime})<H$ or $\gcd(c,c^{\prime})\geq H$. We
first deal with the large gcd.

######  Proposition 6.15.

There exists some $\eta>0$ so that:

$\mathcal{I}_{Q}^{(\geq)}\ll{|\widehat{\mathcal{R}_{N}}(0)|^{2}\over
N}N^{-\eta}.$

###### Proof.

Let $h\geq H$ be the gcd of $c$ and $c^{\prime}$. Then applying Corollary 4.8
(and notation therein) to (6.13), we have

$\displaystyle\mathcal{I}_{Q}^{(\geq)}$ $\displaystyle\ll$
$\displaystyle{X^{4}\over N}\sum_{\gamma\in\mathscr{F}_{T}}\sum_{h\mid
c_{\gamma}\atop h\geq H}\sum_{q_{1}\mid BcP^{2}\atop q_{1}\ll
Q}\sum_{\gamma^{\prime}\in\mathscr{F}_{T}\atop c^{\prime}\equiv
0(h)}\sum_{q_{1}^{\prime}\mid Bc^{\prime}P^{2}\atop q_{1}^{\prime}\ll
Q}\sum_{q\asymp Q\atop q\equiv
0((q_{1},q_{1}^{\prime}))}{(q_{1},q_{1}^{\prime})\over q},$

where in the last sum, we weakened the condition that $q$ is divisible by both
$q_{1}$ and $q_{1}^{\prime}$ to just being divisible by their gcd. Now
estimating divisor sums and applying Nullstellensatz (3.8) in the
$\gamma^{\prime}$ summation gives

$\displaystyle\mathcal{I}_{Q}^{(\geq)}$ $\displaystyle\ll_{\varepsilon}$
$\displaystyle N^{\varepsilon}{X^{4}\over
N}H^{-\eta_{0}}|\mathscr{F}_{T}|^{2},$

from which the claim follows. ∎

Next we handle the small gcd.

######  Proposition 6.16.

There exists some $\eta>0$ so that:

$\mathcal{I}_{Q}^{(<)}\ll{|\widehat{\mathcal{R}_{N}}(0)|^{2}\over
N}N^{-\eta}.$

###### Proof.

We begin with the observation that $(c,c^{\prime})<H$ implies that $c\neq
c^{\prime}$, since $c\asymp T$ and $H=o(T)$. We apply Lemma 4.12, estimate
$\gcd\left(q(q_{1},q_{1}^{\prime}),J\right)$ by
$\gcd\left((q_{2},q_{2}^{\prime}),J\right){q(q_{1},q_{1}^{\prime})\over(q_{2},q_{2}^{\prime})},$
and apply (4.6), giving

$\displaystyle\mathcal{I}_{Q}^{(<)}$ $\displaystyle\ll_{\varepsilon}$
$\displaystyle{X^{4}\over
N}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop(c,c^{\prime})<H}\sum_{q\asymp
Q}q^{-1+\varepsilon}{1\over(q_{2},q_{2}^{\prime})^{1/4}}(q_{1}q_{1}^{\prime})^{1/2}(q_{1},q_{1}^{\prime})^{1/2}$
$\displaystyle\hskip
7.22743pt\left(\gcd\left((q_{2},q_{2}^{\prime}),(c-c^{\prime})(B^{2}-\Delta
cc^{\prime})\right)\right)^{1/4}.$

Since $q=q_{1}q_{2}=q_{1}^{\prime}q_{2}^{\prime}$, we have that
$(q_{2},q_{2}^{\prime})=q/[q_{1},q_{1}^{\prime}]$. We crudely estimate

$\mathcal{I}_{Q}^{(<)}\ \ll_{\varepsilon}\ N^{\varepsilon}{X^{4}\over
N}\sum_{q\asymp
Q}q^{-5/4}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop{c\neq
c^{\prime}\atop{q_{1}=(BcP^{2},q)\atop
q_{1}^{\prime}=(Bc^{\prime}P^{2},q)}}}(q_{1}q_{1}^{\prime})\left(\gcd\left(q,BP^{2}(c-c^{\prime})(B^{2}-\Delta
cc^{\prime})\right)\right)^{1/4}.$ (6.17)

Let $h=(q_{1},q_{1}^{\prime})$, then $h\mid BP^{2}(c,c^{\prime})$, and
$h\ll\min(Q,H)$. Then we can write $q_{1}=hg$, $q_{1}^{\prime}=hg^{\prime}$,
with $(g,g^{\prime})=1$. Also note that
$h\mid(q,BP^{2}(c-c^{\prime})(B^{2}-\Delta cc^{\prime}))$, so we can write

$(q,BP^{2}(c-c^{\prime})(B^{2}-\Delta cc^{\prime}))=h\tilde{g}.$

Then it follows that

$(\tilde{g},g)\ll(BP^{2}(c-c^{\prime})/h,g)\cdot(BP^{2}(B^{2}-\Delta
cc^{\prime}),g).$

The first gcd on the right hand side above is $1$, since a factor of both
should have been included in $h$. Since $g\mid BP^{2}\Delta cc^{\prime}$, the
second gcd is equal to $(BP^{2}B^{2},g)\ll 1$. A similar argument shows that

$(\tilde{g},g^{\prime})\ll 1,$

and hence we have the bound

$[hg,hg^{\prime},h\tilde{g}]\gg gg^{\prime}\tilde{g}$

on their least common multiple. (Since $h$ may be small, it will not help us
in this estimate.)

Similarly, we note that

$(\tilde{g},c)\ll(BP^{2}(c-c^{\prime}),c)\cdot(B^{2}-\Delta cc^{\prime},c).$

The second gcd is again bounded, while the first is bounded by $H,$ by the
definition of $\mathcal{I}_{Q}^{(<)}$.

Thus we can estimate:

$\displaystyle\mathcal{I}_{Q}^{(<)}$ $\displaystyle\ll_{\varepsilon}$
$\displaystyle N^{\varepsilon}{X^{4}\over
N}Q^{-5/4}\sum_{\gamma\in\mathscr{F}_{T}}\sum_{\gamma^{\prime}\in\mathscr{F}_{T}\atop{c\neq
c^{\prime},(c,c^{\prime})<H}}\sum_{h\mid BP^{2}(c,c^{\prime})\atop h\ll
Q}\sum_{g\mid BP^{2}c\atop g\ll Q}\sum_{g^{\prime}\mid BP^{2}c^{\prime}\atop
g^{\prime}\ll Q}$ $\displaystyle\times\sum_{\tilde{g}\mid
BP^{2}(c-c^{\prime})(B^{2}-\Delta cc^{\prime})\atop gg^{\prime}\tilde{g}\ll
Q,\ (\tilde{g},c)\ll
H}(hghg^{\prime})\left(h\tilde{g}\right)^{1/4}\sum_{q\asymp Q\atop q\equiv
0([hg,hg^{\prime},h\tilde{g}])}1$ $\displaystyle\ll_{\varepsilon}$
$\displaystyle N^{\varepsilon}{X^{4}\over
N}Q^{-5/4}H^{9/4}\sum_{\gamma\in\mathscr{F}_{T}}\sum_{\tilde{g}\ll
Q\atop(\tilde{g},c)\ll
H}{Q\over\tilde{g}^{3/4}}\sum_{\gamma^{\prime}\in\mathscr{F}_{T}\atop
BP^{2}(c-c^{\prime})(B^{2}-\Delta cc^{\prime})\equiv 0(\tilde{g})}1.$

For fixed $c$, let
$\mathcal{R}=\mathcal{R}(\tilde{g})\subset\mathbb{Z}/\tilde{g}\mathbb{Z}$
denote the set of roots mod $\tilde{g}$ of the polynomial
$BP^{2}(c-x)(B^{2}-\Delta cx)$. If $\Delta=0$, then this is a linear
polynomial with leading coefficient $BP^{2}$, so $\\#\mathcal{R}\ll 1$. If
$\Delta\neq 0$, then this is a reducible quadratic polynomial with leading
coefficient $BP^{2}\Delta c$; since $(\tilde{g},c)\ll H$, it follows that
$\\#\mathcal{R}\ll_{\varepsilon}H^{1+\varepsilon}$. Finally, we have that

$\displaystyle\mathcal{I}_{Q}^{(<)}$ $\displaystyle\ll_{\varepsilon}$
$\displaystyle N^{\varepsilon}{X^{4}\over
N}Q^{-5/4}H^{9/4}\sum_{\gamma\in\mathscr{F}_{T}}\sum_{\tilde{g}\ll
Q\atop(\tilde{g},c)\ll
H}{Q\over\tilde{g}^{3/4}}\sum_{\alpha\in\mathcal{R}(\tilde{g})}\sum_{\gamma^{\prime}\in\mathscr{F}_{T}\atop
c^{\prime}\equiv\alpha(\tilde{g})}1.$

We apply Nullstellensatz (3.8) in the last summation to obtain:

$\displaystyle\mathcal{I}_{Q}^{(<)}$ $\displaystyle\ll_{\varepsilon}$
$\displaystyle N^{\varepsilon}{X^{4}\over
N}Q^{-5/4}H^{9/4}\sum_{\gamma\in\mathscr{F}_{T}}\sum_{\tilde{g}\ll
Q}{Q\over\tilde{g}^{3/4}}H{1\over\tilde{g}^{\eta_{0}}}|\mathscr{F}_{T}|$
$\displaystyle\ll_{\varepsilon}$ $\displaystyle N^{\varepsilon}{X^{4}\over
N}Q^{-1/4}H^{13/4}Q^{1/4-\eta_{0}}|\mathscr{F}_{T}|^{2}.$

The choice of the parameter $H$ in (6.14) ensures that we have saved a power
of $Q\geq Q_{0}$. The claim follows since $Q_{0}$ is a power of $N$ (by
(2.13)). ∎

These two propositions establish Theorem 6.1 in the intermediate range of $Q$.

###  6.4. Minor Arcs III: Case $X/Y\leq Q<M$

In this largest range, we return to the exact evaluation:

$\displaystyle\mathcal{I}_{Q}$ $\displaystyle=$ $\displaystyle
X^{4}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}}\sum_{k,\ell,k^{\prime},\ell^{\prime}\in\mathbb{Z}}\sum_{q\asymp
Q}\left[\sum_{r(q)}^{\prime}\mathcal{S}_{q}(r,k,\ell;\gamma)\overline{\mathcal{S}_{q}(r,k^{\prime},\ell^{\prime};\gamma^{\prime})}\right]$
$\displaystyle\hskip
36.135pt\int_{|\beta|<1/(QM)}\mathcal{J}_{X}(\beta,k,\ell,q;\gamma)\overline{\mathcal{J}_{X}(\beta,k^{\prime},\ell^{\prime},q;\gamma^{\prime})}d\beta.$

Now we break

$\mathcal{I}_{Q}\leq\mathcal{I}_{Q}^{=}+\mathcal{I}_{Q}^{\neq}$

depending on whether $c=c^{\prime}$ or not. We first handle the latter case.

######  Proposition 6.18.

There is an $\eta>0$ so that

$\displaystyle\mathcal{I}_{Q}^{\neq}$ $\displaystyle\ll$
$\displaystyle{|\widehat{\mathcal{R}}_{N}(0)|^{2}\over N}N^{-\eta},$

as $N\to\infty$.

###### Proof.

To begin, we can use the last part of Lemma 6.5 to show that:

$\mathcal{I}_{Q}^{\neq}\ll{X^{4}\over
QM}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}}\sum_{|k|,|\ell|,|k^{\prime}|,|\ell^{\prime}|\ll{Yq\over
X}}\ \sum_{q\asymp
Q}\left|\sum_{r(q)}^{\prime}\mathcal{S}_{q}(r,k,\ell;\gamma)\overline{\mathcal{S}_{q}(r,k^{\prime},\ell^{\prime};\gamma^{\prime})}\right|\
+\ O(N^{-100}).$ (6.19)

We will omit this last term henceforth. In the case $c\neq c^{\prime}$, we
estimate using Lemma 4.12, crudely (e.g., $q_{1}\ll T$, etc) giving:

$\displaystyle\mathcal{I}_{Q}^{\neq}$ $\displaystyle\ll$
$\displaystyle{X^{4}\over QM}\sum_{q\asymp
Q}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop
q_{1}=(BcP^{2},q),q_{1}^{\prime}=(Bc^{\prime}P^{2},q)}$
$\displaystyle\times\sum_{|k|,|\ell|,|k^{\prime}|,|\ell^{\prime}|\ll{Yq\over
X}}\
q^{-5/4+\varepsilon}(q_{1}q_{1}^{\prime})^{1/2}(q_{1},q_{1}^{\prime})^{1/4}\gcd\left(q(q_{1},q_{1}^{\prime}),J,K\right)^{1/4}$
$\displaystyle\ll$ $\displaystyle{N^{\varepsilon}X^{4}T^{3/2}\over
Q^{9/4}M}\sum_{q\asymp
Q}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}}\sum_{|k|,|\ell|,|k^{\prime}|,|\ell^{\prime}|\ll{Yq\over
X}}\ \gcd\left(q,J,K\right)^{1/4}.$

Recall from Lemma 4.4 that $J$ does not depend on
$k,k^{\prime},\ell,\ell^{\prime}$ but $K$ does. Then

$\mathcal{I}_{Q}^{\neq}\ll{Y^{4}X^{4}T^{3/2}\over Q^{9/4}M}\left[{Q^{4}\over
X^{4}}+1\right]\sum_{q\asymp
Q}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop c\neq
c^{\prime}}\gcd\left(q,J\right)^{1/4}.$

By (4.6), we have that:

$\gcd\left(q,J\right)\ll T^{2}\gcd\left((q_{2},q_{2}^{\prime}),J\right)\ll
T^{2}\gcd\left((q_{2},q_{2}^{\prime}),(c-c^{\prime})(B^{2}-\Delta
cc^{\prime})\right)\ll T^{5},$

since $B^{2}-\Delta cc^{\prime}$ is never zero. Then

$\displaystyle\mathcal{I}_{Q}^{\neq}$ $\displaystyle\ll$
$\displaystyle{N^{\varepsilon}X^{4}T^{3/2}\over Q^{9/4}M}\left[{Q^{4}\over
X^{4}}+1\right]Q|\mathscr{F}_{T}|^{2}T^{5/4}\ll
Y^{4}\left[{T^{3/2}}M^{7/4}+{X^{4}T^{3/2}\over
X^{5/4}M}\right]|\mathscr{F}_{T}|^{2}T^{5/4}$ $\displaystyle\ll$
$\displaystyle Y^{4}|\mathscr{F}_{T}|^{2}X^{2}{T^{5}\over X^{1/4}}\ll
Y^{4}{|\widehat{\mathcal{R}_{N}(0)}|^{2}\over N}{T^{6}\over X^{1/4}}.$

The claim again follows due to the large power of $X$ savings (relative to the
small loss of powers of $T$ and $Y$); see (2.9) and (6.2). ∎

Next we analyze the case that $c=c^{\prime}$. At this stage, we decompose
$\mathcal{I}_{Q}^{=}$ further according to whether
$k\ell=k^{\prime}\ell^{\prime}$ or not,

$\mathcal{I}_{Q}^{=}\ll\mathcal{I}_{Q}^{=,=}+\mathcal{I}_{Q}^{=,\neq}.$

We first analyze the case that $k\ell\neq k^{\prime}\ell^{\prime}$.

######  Proposition 6.20.

There is an $\eta>0$ so that

$\displaystyle\mathcal{I}_{Q}^{=,\neq}$ $\displaystyle\ll$
$\displaystyle{|\widehat{\mathcal{R}}_{N}(0)|^{2}\over N}N^{-\eta},$

as $N\to\infty$.

###### Proof.

We again apply Lemma 6.5 as in (6.19). In this case, we then apply Lemma 4.14,
which gives:

$\displaystyle\mathcal{I}_{Q}^{=,\neq}$ $\displaystyle\ll$
$\displaystyle{X^{4}\over
QM}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop
c=c^{\prime}}\sum_{|k|,|\ell|,|k^{\prime}|,|\ell^{\prime}|\ll{Yq\over X}\atop
k\ell\neq k^{\prime}\ell^{\prime}}\ \sum_{q\asymp
Q}\left[q^{-5/4+\varepsilon}q_{1}^{2}\gcd\left(q,\ell^{\prime}k^{\prime}-\ell
k\right)^{1/4}\right].$

The $\gcd$ is bounded by $|k\ell|\ll(Yq/X)^{2}$, giving:

$\displaystyle\mathcal{I}_{Q}^{=,\neq}$ $\displaystyle\ll$
$\displaystyle{Y^{1/2}X^{4}\over
QM}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop
c=c^{\prime}}\sum_{|k|,|\ell|,|k^{\prime}|,|\ell^{\prime}|\ll{Yq\over X}\atop
k\ell\neq k^{\prime}\ell^{\prime}}\ \sum_{q\asymp
Q}\left[q^{-5/4+\varepsilon}q_{1}^{2}(q/X)^{1/2}\right]$ $\displaystyle\ll$
$\displaystyle Y^{5}{X^{4}\over QM}|\mathscr{F}_{T}|^{2}\left[{Q^{4}\over
X^{4}}+1\right]QQ^{-5/4}T^{2}Q^{1/2}X^{-1/2}$ $\displaystyle\ll$
$\displaystyle Y^{5}{|\widehat{\mathcal{R}_{N}}(0)|^{2}\over N}{T^{6}\over
X^{1/4}}.$

The claim again follows due to the large power of $X$ savings. ∎

The last case is when $c=c^{\prime}$ and $k\ell=k^{\prime}\ell^{\prime}$; here
we will fight not for powers of $Q$ but powers of the much smaller parameter
$T$. We can save a factor of $T^{2\delta-1}$ from the fact that $c=c^{\prime}$
(and hence there are only $T$ values for $\gamma^{\prime}$, not
$T^{2\delta}\asymp|\mathscr{F}_{T}|$). But this is insufficient for a power
gain in the end. So new ideas are needed.

######  Proposition 6.21.

There is an $\eta>0$ so that

$\displaystyle\mathcal{I}_{Q}^{=,=}$ $\displaystyle\ll$
$\displaystyle{|\widehat{\mathcal{R}}_{N}(0)|^{2}\over N}N^{-\eta},$

as $N\to\infty$.

Before beginning the proof, we return to the original formulation:

$\displaystyle\mathcal{I}_{Q}^{=,=}$ $\displaystyle=$ $\displaystyle
X^{4}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop
c=c^{\prime}}\sum_{k,\ell,k^{\prime},\ell^{\prime}\in\mathbb{Z}\atop
k\ell=k^{\prime}\ell^{\prime}}\sum_{q\asymp
Q}\left[\sum_{r(q)}^{\prime}\mathcal{S}_{q}(r,k,\ell;\gamma)\overline{\mathcal{S}_{q}(r,k^{\prime},\ell^{\prime};\gamma^{\prime})}\right]$
$\displaystyle\hskip
36.135pt\int_{|\beta|<1/(QM)}\mathcal{J}_{X}(\beta,k,\ell,q;\gamma)\overline{\mathcal{J}_{X}(\beta,k^{\prime},\ell^{\prime},q;\gamma^{\prime})}d\beta.$

We apply the last part of Lemma 6.5 to truncate the
$k,\ell,k^{\prime},\ell^{\prime}$ range:

$\displaystyle\mathcal{I}_{Q}^{=,=}$ $\displaystyle=$ $\displaystyle
X^{4}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop
c=c^{\prime}}\sum_{|k|,|\ell|,|k^{\prime}|,|\ell^{\prime}|\ll{YQ\over X}\atop
k\ell=k^{\prime}\ell^{\prime}}\sum_{q\asymp
Q}\left[\sum_{r(q)}^{\prime}\mathcal{S}_{q}(r,k,\ell;\gamma)\overline{\mathcal{S}_{q}(r,k^{\prime},\ell^{\prime};\gamma^{\prime})}\right]$
$\displaystyle\hskip
36.135pt\int_{|\beta|<1/(QM)}\mathcal{J}_{X}(\beta,k,\ell,q;\gamma)\overline{\mathcal{J}_{X}(\beta,k^{\prime},\ell^{\prime},q;\gamma^{\prime})}d\beta.$

Over this range of $k,\ell,k^{\prime},\ell^{\prime}$, we need to level out the
$q$ dependence from the archimedean component $\mathcal{J}$. But the range of
$q\asymp Q$ is too long for this purpose, so we decompose the sum into $U$
intervals, where $U$ is a parameter chosen to be

$U=Q^{1/2}.$

Each interval is of length $Q/U=Q^{1/2}$, which is much larger than $T$. Let
$Q_{1}$ range over the starting points of these intervals. Then we have:

$\displaystyle\mathcal{I}_{Q}^{=,=}$ $\displaystyle=$ $\displaystyle
X^{4}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop
c=c^{\prime}}\sum_{|k|,|\ell|,|k^{\prime}|,|\ell^{\prime}|\ll{YQ\over X}\atop
k\ell=k^{\prime}\ell^{\prime}}\sum_{j\in\\{1,\dots,U\\}\atop
Q_{1}=Q+j\frac{Q}{U}}\sum_{Q_{1}\leq q\leq
Q_{1}+\frac{Q}{U}}\left[\sum_{r(q)}^{\prime}\mathcal{S}_{q}(r,k,\ell;\gamma)\overline{\mathcal{S}_{q}(r,k^{\prime},\ell^{\prime};\gamma^{\prime})}\right]$
$\displaystyle\hskip
36.135pt\int_{|\beta|<1/(QM)}\mathcal{J}_{X}(\beta,k,\ell,q;\gamma)\overline{\mathcal{J}_{X}(\beta,k^{\prime},\ell^{\prime},q;\gamma^{\prime})}d\beta.$

On each of these sub-intervals, we will replace $q$ in $\mathcal{J}$ by
$Q_{1}$, thereby freeing the $q$ variable for a purely modular analysis, as
follows.

######  Lemma 6.22.

For any $\varepsilon>0$, we have that:

$\displaystyle\mathcal{I}_{Q}^{=,=}$ $\displaystyle\ll_{\varepsilon}$
$\displaystyle\frac{1}{QM}X^{4}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop
c=c^{\prime}}\sum_{|k|,|\ell|,|k^{\prime}|,|\ell^{\prime}|\ll{YQ\over X}\atop
k\ell=k^{\prime}\ell^{\prime}}\sum_{j\in\\{1,\dots,U\\}\atop
Q_{1}=Q+j\frac{Q}{U}}\left|\sum_{Q_{1}\leq q\leq
Q_{1}+\frac{Q}{U}}\sum_{r(q)}^{\prime}\mathcal{S}_{q}(r,k,\ell;\gamma)\overline{\mathcal{S}_{q}(r,k^{\prime},\ell^{\prime};\gamma^{\prime})}\right|$
$\displaystyle\hskip 36.135pt\ +\
Y^{4}{|\widehat{\mathcal{R}}_{N}(0)|^{2}\over N}\frac{T}{U}.$

###### Proof.

Returning to the definition of $\mathcal{J}$, we see that

$\displaystyle\hskip-21.68121pt\left|\mathcal{J}_{X}(\beta,k,\ell,q;\gamma)-\mathcal{J}_{X}(\beta,k,\ell,Q_{1};\gamma)\right|\ll\frac{Y}{U},$

since $k,\ell\ll YQ/X$. We replace the two appearances of $q$ in $\mathcal{J}$
one at a time.

Each time, we apply Lemma 4.15 to the resulting difference, which is bounded
by

$\displaystyle\ll$ $\displaystyle Y^{2}\frac{1}{U}{X^{4}\over
QM}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop
c=c^{\prime}}\sum_{|k|,|\ell|,|k^{\prime}|,|\ell^{\prime}|\ll{Yq\over X}\atop
k\ell=k^{\prime}\ell^{\prime}}\ \sum_{q\asymp Q\atop
q_{1}=(c,q)}\left[{(c,q)^{2}\over q^{2}}\sum_{r(q)}^{\prime}{\bf
1}_{\\{-\ell\equiv Pr(Ba+Dc)(\operatorname{mod}q_{1})\atop-k\equiv
Pr(Ac+Bd)(\operatorname{mod}q_{1})\\}}\right]$ $\displaystyle\ll$
$\displaystyle Y^{2}\frac{1}{U}{X^{4}\over
Q^{3}M}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop
c=c^{\prime}}\sum_{q_{1}\mid c\atop q_{1}\ll Q}q_{1}^{2}\sum_{q\asymp Q\atop
q\equiv 0(\operatorname{mod}q_{1})}\sum_{|k|,|\ell|\ll{Yq\over X}}\
\sum_{r(q)}^{\prime}{\bf 1}_{\\{-\ell\equiv
PrBa(\operatorname{mod}q_{1})\atop-k\equiv PrBd(\operatorname{mod}q_{1})\\}},$

where we used that $c\equiv 0(q_{1})$. Since $(a,c)=(c,d)=1$, for given
$\ell$, the value of $r$ is determined up to constants mod $q_{1}$, so there
are $\ll q/q_{1}$ values of $r$ contributing. For each value of $(\ell,r)$, we
have that $k$ is determined mod $q_{1}$, but we won’t use this. In total, we
bound the difference by

$\displaystyle\ll$ $\displaystyle Y^{2}\frac{1}{U}{X^{4}\over
Q^{3}M}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop
c=c^{\prime}}\sum_{q_{1}\mid c\atop q_{1}\ll Q}q_{1}^{2}\sum_{q\asymp Q\atop
q\equiv 0(\operatorname{mod}q_{1})}{Q\over X}{Q\over q_{1}}{Q\over X}$
$\displaystyle\ll$ $\displaystyle Y^{2}\frac{1}{U}{X^{4}\over
Q^{3}M}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop
c=c^{\prime}}\sum_{q_{1}\mid c\atop q_{1}\ll Q}q_{1}^{2}{Q\over q_{1}}{Q\over
X}{Q\over q_{1}}{Q\over X}$ $\displaystyle\ll$ $\displaystyle
Y^{2}{|\widehat{\mathcal{R}}_{N}(0)|^{2}\over N}\frac{T}{U}.$

Finally, we estimate trivially that

$\int_{|\beta|<1/(QM)}\mathcal{J}_{X}(\beta,k,\ell,Q_{1};\gamma)\overline{\mathcal{J}_{X}(\beta,k^{\prime},\ell^{\prime},Q_{1};\gamma^{\prime})}d\beta\
\ll\ \frac{1}{QM},$

whence the claim follows, since $Q\geq X/Y$. ∎

Now we have leveled out the sum, and are in position to apply the crucial
Lemma 4.25.

###### Proof of Proposition 6.21.

Inserting Lemma 4.25 into Lemma 6.22 gives:

$\displaystyle\mathcal{I}_{Q}^{=,=}$ $\displaystyle\ll_{\varepsilon}$
$\displaystyle\frac{1}{QM}X^{4}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop
c=c^{\prime}}\sum_{|k|,|\ell|,|k^{\prime}|,|\ell^{\prime}|\ll{YQ\over X}\atop
k\ell=k^{\prime}\ell^{\prime}}U$ $\displaystyle\times
Q^{\varepsilon}\sum_{q_{1}\mid BcP^{2}\atop E=BcP^{2}/q_{1}}{\bf
1}_{\\{d\ell\equiv ak\equiv d^{\prime}\ell^{\prime}\equiv
a^{\prime}k^{\prime}(\operatorname{mod}(q_{1},c))\\}}\mathcal{N}_{q_{1}}\sum_{Q_{1}\mid
q_{1}\atop{p\mid Q_{1}\Longrightarrow(p^{\infty},q_{1})\mid
Q_{1}\atop(E,q_{1}/Q_{1})=1}}$ $\displaystyle\times\left[{(EQ_{1},Z)\over
UEQ_{1}}+{c\over U^{2}}+\frac{c^{3}}{Q}+{|Z|\over UQ}\right]$
$\displaystyle\hskip 36.135pt\ +\
N^{\varepsilon}{|\widehat{\mathcal{R}}_{N}(0)|^{2}\over N}\frac{T}{U},$

where $Z$ and $\mathcal{N}_{q_{1}}$ are as defined in the statement of Lemma
4.25.

We first handle the contribution from the latter three terms in (6.4)

$\displaystyle\frac{Q^{\varepsilon}}{QM}X^{4}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop
c=c^{\prime}}\sum_{|k|,|\ell|,|k^{\prime}|,|\ell^{\prime}|\ll{YQ\over X}\atop
k\ell=k^{\prime}\ell^{\prime}}\left[{c\over U}+\frac{c^{3}U}{Q}+{|Z|\over
Q}\right]$
$\displaystyle\ll\frac{N^{\varepsilon}Y^{2}}{QM}X^{4}|\mathscr{F}_{T}|^{2}\left(\frac{Q}{X}\right)^{2}\left[{T\over
U}+\frac{T^{3}U}{Q}+{T^{2}\over Q}\right]$ $\displaystyle\ll
N^{\varepsilon}Y^{2}{|\mathcal{R}_{N}(0)|^{2}\over N}\left[{T^{2}\over
U}+\frac{T^{4}U}{X}+{T^{3}\over X}\right],$

where we bounded $Z\ll TYQ/X\ll YT^{2}$ from (4.26) and used (4.34). With
$U=Q^{1/2}$, this is sufficient savings if $T$ is small enough relative to
$Q\geq X/Y$; these are all power savings. Only the first term of (6.4) remains
to be handled.

$\displaystyle\frac{Q^{\varepsilon}}{QM}X^{4}\sum_{\gamma,\gamma^{\prime}\in\mathscr{F}_{T}\atop
c=c^{\prime}}\sum_{|k|,|\ell|,|k^{\prime}|,|\ell^{\prime}|\ll{YQ\over X}\atop
k\ell=k^{\prime}\ell^{\prime}}\sum_{q_{1}\mid BcP^{2}\atop
E=BcP^{2}/q_{1}}{\bf 1}_{\\{d\ell\equiv ak\equiv d^{\prime}\ell^{\prime}\equiv
a^{\prime}k^{\prime}(\operatorname{mod}(q_{1},c))\\}}\sum_{Q_{1}\mid
q_{1}\atop{p\mid Q_{1}\Longrightarrow(p^{\infty},q_{1})\mid
Q_{1}\atop(E,q_{1}/Q_{1})=1}}{(EQ_{1},Z)\over EQ_{1}}$
$\displaystyle\ll\frac{Q^{\varepsilon}}{QM}X^{4}\sum_{\gamma\in\mathscr{F}_{T}}\sum_{|k|,|\ell|,|k^{\prime}|,|\ell^{\prime}|\ll{YQ\over
X}\atop{k\ell=k^{\prime}\ell^{\prime}}}\sum_{q_{1}\mid BcP^{2}\atop
E=BcP^{2}/q_{1}}{\bf 1}_{\\{d\ell\equiv ak\equiv d^{\prime}\ell^{\prime}\equiv
a^{\prime}k^{\prime}(\operatorname{mod}(q_{1},c))\\}}$ $\displaystyle\hskip
36.135pt\sum_{Q_{1}\mid q_{1}\atop{p\mid
Q_{1}\Longrightarrow(p^{\infty},q_{1})\mid
Q_{1}\atop(E,q_{1}/Q_{1})=1}}{1\over EQ_{1}}\sum_{Z_{1}\mid
EQ_{1}}Z_{1}\sum_{\gamma^{\prime}\in\mathscr{F}_{T}\atop c=c^{\prime}}{\bf
1}_{\\{Z(\gamma^{\prime})\equiv 0(\operatorname{mod}Z_{1})\\}}.$

Now we apply (4.36) to the last summation, expanding
$\gamma^{\prime}\in\mathscr{F}_{T}$ to all of
$\operatorname{SL}_{2}(\mathbb{Z})$ (recalling that in $\mathscr{F}_{T}$, all
entries are $\asymp T$). This gives

$\displaystyle\frac{Q^{\varepsilon}}{QM}X^{4}\sum_{\gamma\in\mathscr{F}_{T}}\sum_{|k|,|\ell|,|k^{\prime}|,|\ell^{\prime}|\ll{YQ\over
X}\atop{k\ell=k^{\prime}\ell^{\prime}}}\sum_{q_{1}\mid BcP^{2}\atop
E=BcP^{2}/q_{1}}{\bf 1}_{\\{d\ell\equiv ak\equiv d^{\prime}\ell^{\prime}\equiv
a^{\prime}k^{\prime}(\operatorname{mod}(q_{1},c))\\}}$ $\displaystyle\hskip
36.135pt\sum_{Q_{1}\mid q_{1}\atop{p\mid
Q_{1}\Longrightarrow(p^{\infty},q_{1})\mid
Q_{1}\atop(E,q_{1}/Q_{1})=1}}{1\over EQ_{1}}\sum_{Z_{1}\mid
EQ_{1}}T(Z_{1},\ell d-ka)$
$\displaystyle\ll\frac{Q^{\varepsilon}}{QM}TX^{4}\sum_{\gamma\in\mathscr{F}_{T}}\sum_{q_{1}\mid
BcP^{2}\atop E=BcP^{2}/q_{1}}\sum_{Q_{1}\mid q_{1}\atop{p\mid
Q_{1}\Longrightarrow(p^{\infty},q_{1})\mid
Q_{1}\atop(E,q_{1}/Q_{1})=1}}{1\over EQ_{1}}\sum_{Z_{1}\mid
EQ_{1}}\sum_{Z_{2}\mid Z_{1}}Z_{2}$ $\displaystyle\hskip
36.135pt\sum_{|k|,|\ell|\ll{YQ\over X}\atop{d\ell\equiv
ak(\operatorname{mod}(q_{1},c))\atop d\ell\equiv
ak(\operatorname{mod}Z_{2})}}\sum_{|k^{\prime}|,|\ell^{\prime}|\ll{YQ\over
X}\atop{k\ell=k^{\prime}\ell^{\prime}}}1.$

The $k^{\prime},\ell^{\prime}$ sum is a divisor sum. Note that
$(q_{1},c)\asymp q_{1}$. Replace the condition $d\ell\equiv
ak\operatorname{mod}(q_{1},c)$ by $d\ell\equiv
ak\operatorname{mod}(q_{1}/Q_{1},c)$. Note that $q_{1}/Q_{1}$ is coprime to
$EQ_{1}$. So with $k$ fixed, $\ell$ is restricted to a residue class mod
$(q_{1}/Q_{1},c)Z_{2}$. This gives

$\displaystyle\frac{Y^{4}}{QM}TX^{4}\sum_{\gamma\in\mathscr{F}_{T}}\sum_{q_{1}\mid
BcP^{2}\atop E=BcP^{2}/q_{1}}\sum_{Q_{1}\mid q_{1}\atop{p\mid
Q_{1}\Longrightarrow(p^{\infty},q_{1})\mid
Q_{1}\atop(E,q_{1}/Q_{1})=1}}{1\over EQ_{1}}\sum_{Z_{2}\mid EQ_{1}}Z_{2}$
$\displaystyle\hskip 36.135pt{Q\over X}\left({Q\over
X(q_{1}/Q_{1},c)Z_{2}}+1\right)$
$\displaystyle\ll{Y^{2}}X^{2}|\mathscr{F}_{T}|\ \ll\
{Y^{2}}{|\widehat{\mathcal{R}_{N}}(0)|^{2}\over N}{1\over T^{2\delta-1}}.$

This gives the claim, by the choice of $Y$ in (6.2). ∎

Theorem 6.1 has now been established in all ranges of $Q$, thus completing the
proof Theorem 1.8.

## References

  * [Bea65] A. F. Beardon.  The Hausdorff dimension of singular sets of properly discontinuous groups in $N$-dimensional space.  Bull. Amer. Math. Soc., 71:610–615, 1965.
  * [BGS11] J. Bourgain, A. Gamburd, and P. Sarnak.  Generalization of Selberg’s 3/16th theorem and affine sieve.  Acta Math, 207:255–290, 2011.
  * [BK10] J. Bourgain and A. Kontorovich.  On representations of integers in thin subgroups of SL$(2,{{\bf{Z}}})$.  GAFA, 20(5):1144–1174, 2010.
  * [BK14a] J. Bourgain and A. Kontorovich.  On Zaremba’s conjecture.  Annals Math., 180(1):137–196, 2014.
  * [BK14b] Jean Bourgain and Alex Kontorovich.  On the local-global conjecture for integral Apollonian gaskets.  Invent. Math., 196(3):589–650, 2014.
  * [BK15] Jean Bourgain and Alex Kontorovich.  The Affine Sieve Beyond Expansion I: Thin Hypotenuses.  Int. Math. Res. Not. IMRN, (19):9175–9205, 2015.
  * [BK18] Jean Bourgain and Alex Kontorovich.  Beyond expansion IV: Traces of thin semigroups.  Discrete Anal., pages Paper No. 6, 27, 2018.
  * [BKS10] J. Bourgain, A. Kontorovich, and P. Sarnak.  Sector estimates for hyperbolic isometries.  GAFA, 20(5):1175–1200, 2010.
  * [Bur62] D. A. Burgess.  On character sums and $L$-series.  Proc. London Math. Soc. (3), 12:193–206, 1962.
  * [BV12] Jean Bourgain and Péter P. Varjú.  Expansion in $SL_{d}({\bf Z}/q{\bf Z}),\,q$ arbitrary.  Invent. Math., 188(1):151–173, 2012.
  * [CCC80] S. Chowla, J. Cowles, and M. Cowles.  On the number of conjugacy classes in ${\rm SL}(2,\,{\bf Z})$.  J. Number Theory, 12(3):372–377, 1980.
  * [Gol74] Dorian M. Goldfeld.  A simple proof of Siegel’s theorem.  Proc. Nat. Acad. Sci. U.S.A., 71:1055, 1974.
  * [IK04] Henryk Iwaniec and Emmanuel Kowalski.  Analytic number theory, volume 53 of American Mathematical Society Colloquium Publications.  American Mathematical Society, Providence, RI, 2004.
  * [Kon13] Alex Kontorovich.  From Apollonius to Zaremba: local-global phenomena in thin orbits.  Bull. Amer. Math. Soc. (N.S.), 50(2):187–228, 2013.
  * [Kon14] Alex Kontorovich.  Levels of distribution and the affine sieve.  Ann. Fac. Sci. Toulouse Math. (6), 23(5):933–966, 2014.
  * [Kon16] Alex Kontorovich.  Applications of thin orbits.  In Dynamics and analytic number theory, volume 437 of London Math. Soc. Lecture Note Ser., pages 289–317. Cambridge Univ. Press, Cambridge, 2016.
  * [LP82] P.D. Lax and R.S. Phillips.  The asymptotic distribution of lattice points in Euclidean and non-Euclidean space.  Journal of Functional Analysis, 46:280–350, 1982.
  * [Mar96] J. Marklof.  On multiplicities in length spectra of arithmetic hyperbolic three-orbifolds.  Nonlinearity, 9(2):517–536, 1996.
  * [McM12] C. McMullen.  Dynamics of units and packing constants of ideals, 2012.  Online lecture notes, http://www.math.harvard.edu/~ctm/expositions/home/text/papers/cf/slides/slides.pdf.
  * [McM20] C. T. McMullen.  Billiards, heights and the arithmetic of non-arithmetic groups, 2020.  Preprint.
  * [MW83] D. W. Masser and G. Wüstholz.  Fields of large transcendence degree generated by values of elliptic functions.  Invent. Math., 72(3):407–464, 1983.
  * [Ogr21] Brooke Ogrodnik.  On the local-global conjecture for commutator traces, 2021.  Preprint (Rutgers University PhD Thesis), arXiv:2103.14594.
  * [Pat76] S.J. Patterson.  The limit set of a Fuchsian group.  Acta Mathematica, 136:241–273, 1976.
  * [Sul84] D. Sullivan.  Entropy, Hausdorff measures old and new, and limit sets of geometrically finite Kleinian groups.  Acta Math., 153(3-4):259–277, 1984.
  * [Zha20] Xin Zhang.  On representation of integers from thin subgroups of $SL(2,\mathbb{Z})$ with parabolics.  Int. Math. Res. Not. IMRN, (18):5611–5629, 2020.


# The mean number of divisors for rough, dense and practical numbers

Andreas Weingartner  Department of Mathematics, 351 West University Boulevard,
Southern Utah University, Cedar City, Utah 84720, USA<EMAIL_ADDRESS>Dedicated to Maurice Margenstern for his insightful conjectures on practical
numbers

(Date: April 14, 2021)

###### Abstract.

We give asymptotic estimates for the mean number of divisors of integers
without small prime factors, integers with bounded ratios of consecutive
divisors, and for practical numbers. In the last case, this confirms a
conjecture of Margenstern.

###### 2010 Mathematics Subject Classification:

11N25, 11N37

##  1\. Introduction

Let $\tau(n)$ be the number of positive divisors of $n$. We explore the
average size of $\tau(n)$ as $n$ runs through one of several integer
sequences. First, we look at the $y$-rough numbers, i.e. integers with no
prime factors $\leq y$. Let $P^{-}(n)$ denote the smallest prime factor of
$n\geq 2$ and put $P^{-}(1)=\infty$. Define the function $\xi(u)$ by
$\xi(u)=0$ for $u<1$ and

$u\xi(u)=2+2\int_{1}^{u-1}\xi(t)\,dt\qquad(u\geq 1).$ (1)

######  Theorem 1.

Uniformly for $x\geq 1$, $y\geq 2$, $u=\log x/\log y$,

$S(x,y):=\sum_{n\leq x\atop P^{-}(n)>y}\tau(n)=1+x(\log x)\prod_{p\leq
y}\left(1-\frac{1}{p}\right)^{2}\\\ +\frac{x}{\log
y}\left\\{\xi(u)-ue^{-2\gamma}-\left.\frac{2y}{x}\right|_{x\geq
y}+O\left(\frac{1}{\log y}\right)\right\\}$

and

$\xi(u)=(u+2)e^{-2\gamma}+O\left(u^{-u}\right)\qquad(u\geq 1).$

With its three main terms of magnitude $xu/\log y$, one of which is negative,
Theorem 1 is somewhat difficult to make sense of. However, this form is
preferable when we apply it to prove Theorems 2 and 3. The estimate for
$\xi(u)$ and Mertens’ formula allow us to simplify Theorem 1 as follows.

######  Corollary 1.

We have

$S(x,y)=x\log(xy^{2})\prod_{p\leq
y}\left(1-\frac{1}{p}\right)^{2}\left\\{1+O\left(\frac{1}{\log
x}+u^{-u}\right)\right\\}\quad(x\geq y\geq 2),$ $S(x,y)=\frac{x\xi(u)-2y}{\log
y}\left\\{1+O\left(\frac{1}{\log x}+e^{-\sqrt{\log
y}}\right)\right\\}\quad(x\geq 2y\geq 4).$

Dividing these estimates by that of Lemma 2 for $\Phi(x,y)$, the number of
$y$-rough integers $n\leq x$, we obtain the following results for the mean
value of $\tau(n)$. Throughout, $\omega(u)$ denotes Buchstab’s function.

######  Corollary 2.

We have

$\frac{S(x,y)}{\Phi(x,y)}=\log(xy^{2})\prod_{p\leq
y}\left(1-\frac{1}{p}\right)\left\\{1+O\left(\frac{1}{\log
x}+u^{-u}\right)\right\\}\quad(x\geq y\geq 2),$
$\frac{S(x,y)}{\Phi(x,y)}=\frac{\xi(u)}{\omega(u)}\left\\{1+O\left(\frac{1}{\log
x}+e^{-\sqrt{\log y}}\right)\right\\}\quad(x\geq 2y\geq 4).$

Corollary 2 shows that, for fixed $u$, the average of $\tau(n)$ over
$x^{1/u}$-rough integers $n\leq x$ is asymptotic to $\xi(u)/\omega(u)$, as
$x\to\infty$. The graphs of $\xi(u)$ and $\xi(u)/\omega(u)$ are shown in
Figure 1.

Figure 1. A graph of the function $\xi(u)$ (solid) and its asymptote
$(u+2)e^{-2\gamma}$ (dashed), and the ratio $\xi(u)/\omega(u)$ (dot-dashed)
and its asymptote $(u+2)e^{-\gamma}$ (dotted).

We will see in Section 3 that Theorem 1 follows without much difficulty from
Lemma 2. Our main motivation for including Theorem 1 is its role in the proofs
of the following results.

Next, we consider integers with bounded ratios of consecutive divisors. We say
that $n$ is $t$-dense if the ratios of consecutive divisors of $n$ do not
exceed $t$. Let $\mathcal{D}(x,t)$ denote the set of $t$-dense integers $n\leq
x$ and write $D(x,t)=|\mathcal{D}(x,t)|$. Let $\lambda(v)$ be given by
$\lambda(v)=0$ for $v<0$ and

$\lambda(v)=v-\int_{0}^{(v-1)/2}\lambda(u)\,\xi\left(\frac{v-u}{u+1}\right)\frac{du}{u+1}\quad(v\geq
0).$ (2)

######  Theorem 2.

Uniformly for $x\geq 1$, $t\geq 2$, $v=\log x/\log t$,

$T(x,t):=\sum_{n\in\mathcal{D}(x,t)}\tau(n)=x\alpha_{t}(\log
t)\lambda(v)+O(x),$

where $0<\alpha_{0}\leq\alpha_{t}=1+O(1/\log t)$ and

$\lambda(v)=\lambda_{0}(v+1)^{\delta}+\lambda_{1}(v+1)^{-1}+O((v+1)^{-1.962}),$
(3)
$\delta=0.7136125...,\quad\lambda_{0}=1.118192...,\quad\lambda_{1}=\frac{2}{3e^{-2\gamma}-2}=-1.897014...$

The constants $\delta$ and $\lambda_{0}$ are defined in Lemma 6.

Figure 2. The function $\lambda(v)$ (solid) and its approximation
$\lambda_{0}(v+1)^{\delta}+\lambda_{1}(v+1)^{-1}$ (dashed).

If one is only concerned with the order of magnitude of $T(x,t)$, Theorem 2
and the estimate $D(x,t)\asymp x/v$ of Saias [6, Thm. 1] imply the following
simple formulas.

######  Corollary 3.

Uniformly for $x\geq t\geq 2$, $v=\log x/\log t$,

$T(x,t)\asymp xv^{\delta}\log t,\qquad\frac{T(x,t)}{D(x,t)}\asymp
v^{\delta}\log x.$

Corollary 3 shows how the average size of $\tau(n)$ varies from $\asymp(\log
x)^{1+\delta}=(\log x)^{1.713...}$ in the case of $2$-dense integers, to
$\asymp\log x$ when $t=x$ and $\mathcal{D}(x,t)$ contains all natural numbers
up to $x$. As in the case of natural numbers, the typical size of $\tau(n)$ is
much smaller than the average size. Theorem 1 of [17] shows that almost all
$n\in\mathcal{D}(x,t)$ satisfy

$\tau(n)^{1+o(1)}=v^{C\log 2}(\log t)^{\log 2},\quad
C=(1-e^{-\gamma})^{-1}=2.280...,$

which in the case of $2$-dense integers is $\asymp(\log x)^{C\log 2}=(\log
x)^{1.580...}$.

For more precise information, we combine the estimates for $T(x,t)$,
$\lambda(v)$ and $\alpha_{t}$ in Theorem 2, to obtain the following three
asymptotic estimates, which reveal the behavior of $T(x,y)$ in each of the
three cases: (i) fixed $t$, (ii) $t\to\infty$, $v\to\infty$ and (iii) fixed
$v$.

######  Corollary 4.

Uniformly for $x\geq t\geq 2$, $v=\log x/\log t$,

$T(x,t)=a_{t}x(\log xt)^{\delta}\left\\{1+O\left(\frac{1}{v^{\delta}\log
t}+\frac{1}{v^{1+\delta}}\right)\right\\},$

where $a_{t}=\lambda_{0}\alpha_{t}(\log t)^{1-\delta}$,

$T(x,t)=\lambda_{0}x(v+1)^{\delta}(\log t)\left\\{1+O\left(\frac{1}{\log
t}+\frac{1}{v^{1+\delta}}\right)\right\\},$ $T(x,t)=x(\log
t)\lambda(v)\left\\{1+O\left(\frac{1}{\log t}\right)\right\\}.$

With the corresponding three estimates for $D(x,t)$ in [13, Corollaries
1.1-1.3], we obtain the following formulas for the mean value of $\tau(n)$
over $t$-dense integers $n\leq x$.

######  Corollary 5.

Uniformly for $x\geq t\geq 2$, $v=\log x/\log t$,

$\frac{T(x,t)}{D(x,t)}=b_{t}(\log
xt)^{1+\delta}\left\\{1+O\left(\frac{1}{v^{\delta}\log
t}+\frac{1}{v^{1+\delta}}\right)\right\\},$

where $b_{t}=\lambda_{0}(1-e^{-\gamma})\beta_{t}(\log t)^{-\delta}$,
$0<\beta_{0}\leq\beta_{t}=1+O(1/\log t)$,

$\frac{T(x,t)}{D(x,t)}=\lambda_{0}(1-e^{-\gamma})(v+1)^{\delta}\log(xt)\left\\{1+O\left(\frac{1}{\log
t}+\frac{1}{v^{1+\delta}}\right)\right\\},$ $\frac{T(x,t)}{D(x,t)}=(\log
t)\frac{\lambda(v)}{d(v)}\left\\{1+O\left(\frac{1}{\log t}\right)\right\\},$

where the function $d(v)$ is as in [13].

The $t$-dense integers are a special case of a family of integer sequences
that arise as follows. Let $\theta$ be an arithmetic function. Let
$\mathcal{B}=\mathcal{B}_{\theta}$ be the set of positive integers containing
$n=1$ and all those $n\geq 2$ with prime factorization
$n=p_{1}^{\alpha_{1}}\cdots p_{k}^{\alpha_{k}}$, $p_{1}<p_{2}<\ldots<p_{k}$,
which satisfy

$p_{i}\leq\theta\big{(}p_{1}^{\alpha_{1}}\cdots
p_{i-1}^{\alpha_{i-1}}\big{)}\qquad(1\leq i\leq k).$ (4)

We write $\mathcal{B}(x)=\mathcal{B}\cap[1,x]$ and $B(x)=|\mathcal{B}(x)|$.
When $\theta(n)=nt$, then $\mathcal{B}$ is the set of $t$-dense integers [6,
10, 13]. If $\theta(n)=\sigma(n)+1$, where $\sigma(n)$ is the sum of the
positive divisors of $n$, then $\mathcal{B}$ is the set of practical numbers
[3, 6, 7, 8, 9, 10, 13], i.e. integers $n$ such that every $m\leq n$ can be
expressed as a sum of distinct positive divisors of $n$. By making some
adjustments to the proof of Theorem 2, we will obtain the following result.

######  Theorem 3.

Assume $\max(2,n)\leq\theta(n)\ll n\exp((\log n)^{a})$ for $n\geq 1$ , where
$a$ is any constant with $a<(1-\delta)/(2-\delta)=0.2226...$ Then

$T(x):=\sum_{n\in\mathcal{B}(x)}\tau(n)=\nu_{\theta}x(\log x)^{\delta}+O(x),$

where $\delta=0.7136125...$ is as in Lemma 6 and $\nu_{\theta}$ is a positive
constant.

In 1991, Margenstern [3] proposed seven conjectures about practical numbers,
together with empirical evidence to support them. Theorem 3 confirms
Conjecture 4 of [3], which says that $T(x)\sim\nu x(\log x)^{\delta}$ for some
constants $\nu,\delta\in(1/2,1)$. Although we have not determined the constant
factor $\nu$, empirical evidence suggests that $\nu=0.54...$ in the case of
practical numbers. Thirty years after the publication of Margenstern’s paper
[3], six of the seven conjectures have now essentially been resolved [4, 5,
13, 17], while the remaining one [3, Conjecture 2], which is about the count
of twin practical numbers, has seen significant progress [5, 17].

We take this opportunity to give an improved version of [13, Thm. 5.1], an
asymptotic estimate for the counting function $B(x)=|\mathcal{B}(x)|$. Theorem
4 has a sharper error term and allows for $\theta(n)$ to grow faster. Let

$l(x)=\exp\left(\frac{\log x}{\log_{2}x\log_{3}^{3}x}\right).$

######  Theorem 4.

Assume $\max(2,n)\leq\theta(n)\leq nf(n)$, where $f(x)\ll l(x)$, $f(x)$ is
non-decreasing and $(\log f(x))/\log x$ is eventually decreasing. Then

$B(x)=\frac{c_{\theta}x}{\log x}\bigl{\\{}1+O\left(E(x)\right)\bigr{\\}},$

where

$E(x):=\int_{x}^{\infty}\frac{\log f(y)}{y\log^{2}y}dy$

and the positive constant $c_{\theta}$ is given by

$c_{\theta}=\frac{1}{1-e^{-\gamma}}\sum_{n\in\mathcal{B}}\frac{1}{n}\Biggl{(}\sum_{p\leq\theta(n)}\frac{\log
p}{p-1}-\log n\Biggr{)}\prod_{p\leq\theta(n)}\left(1-\frac{1}{p}\right).$

For example, if $\theta(n)\ll n(\log n)^{A}$, where $A$ is an arbitrary
constant, then $E(x)\ll\log\log x/\log x$, while [13, Thm. 5.1] only allows
for $A<1$ and has a relative error term of $O(\log^{A}x/\log x)$ in that case.
Comparing Theorem 4 with [13, Thm. 5.1], we have essentially replaced $f(y)$
with $\log f(y)$ in the definition of the error term $E(x)$. The improvement
comes from a more careful treatment of the penultimate term in Lemma 2, when
this estimate is inserted into Lemma 4. If $B(x)\sim c_{\theta}x/\log x$, then
$\theta(n)$ cannot grow much faster than is allowed by Theorem 4, since
$\theta(n)\asymp n^{1+\varepsilon}$ implies $B(x)\sim c_{\theta}x/(\log
x)^{1-\eta}$ for some $\eta>0$, by [14, Thm. 3].

##  2\. The function $\xi(u)$

If we sort the integers $n\leq x$ according to their smallest prime factor
$p=P^{-}(n)$ and ignore higher powers of $p$, we expect that

$S(x,y)=\sum_{n\leq x\atop P^{-}(n)>y}\tau(n)\approx\sum_{\sqrt{x}<p\leq
x}2+\sum_{y<p\leq\sqrt{x}}2S(x/p,p).$

This suggests the definition (1), provided $S(x,y)\approx x\xi(u)/\log y$. We
have

$u\xi(u)=\begin{cases}0&u<1\\\ 2&1\leq u<2\\\ 4\log(u-1)+2\quad&2\leq
u<3\end{cases}$

and

$(u\xi(u))^{\prime}=2\xi(u-1)\qquad(u\neq 1,2).$ (5)

As with Buchstab’s function $\omega(u)$, $\xi(u)$ is $k$ times continuously
differentiable for $u>k+1$. By induction on $\lfloor u\rfloor$, (1) easily
implies the bounds

$\frac{u+2}{4}\leq\xi(u)\leq u+1\qquad(u\geq 1).$ (6)

Equation (5) shows that $\xi(u)$ belongs to a family of functions $f$ that
satisfy $uf^{\prime}(u)+af(u)+bf(u-1)=0$, where $a$ and $b$ are constants.
These equations were studied in great detail by Hildebrand and Tenenbaum [2].

######  Lemma 1.

We have

$\displaystyle\xi(u)=2\omega(u)$
$\displaystyle+\int_{0}^{u}\omega(t)\omega(u-t)dt\qquad$ $\displaystyle(u\geq
0),$ (7) $\displaystyle|\xi^{\prime\prime}(u)|$
$\displaystyle<\frac{2^{u}}{7\Gamma(u+1)}\qquad$ $\displaystyle(u\geq 3.5),$
(8) $\displaystyle|\xi^{\prime}(u)-e^{-2\gamma}|$
$\displaystyle<\frac{2^{u}}{7\Gamma(u+1)}\qquad$ $\displaystyle(u\geq 2.5),$
(9) $\displaystyle|\xi(u)-(u+2)e^{-2\gamma}|$
$\displaystyle<\frac{2^{u}}{7\Gamma(u+1)}\qquad$ $\displaystyle(u\geq 1.5),$
(10) $\displaystyle\xi(u)-(u+2)e^{-2\gamma}$
$\displaystyle\ll\left(\frac{e}{u\log 2u}\right)^{u}\qquad$
$\displaystyle(u\geq 1).$ (11)

###### Proof.

With equation (5), we find that the Laplace transform $\hat{\xi}(s)$ is given
by

$1+\hat{\xi}(s)=s^{-2}\exp\left(-2\gamma+2\int_{0}^{s}(1-e^{-t})t^{-1}dt\right)=(1+\hat{\omega}(s))^{2},$
(12)

by [11, Thm. III.6.5]. This implies (7).

For the next three estimates, we find exact formulas for
$\xi(u),\xi^{\prime}(u),\xi^{\prime\prime}(u)$ in terms of polylogarithms, on
the interval $[0,5]$, with the help of Mathematica. This shows that the claims
are valid for $u\leq 5$. For $u>3$, $\xi(u)$ is twice continuously
differentiable, so that (5) implies

$u\xi^{\prime\prime}(u)=2\xi^{\prime}(u-1)-2\xi^{\prime}(u)=-2\int_{u-1}^{u}\xi^{\prime\prime}(t)dt\qquad(u>4).$

Assuming that (8) is false, we define

$M:=\inf\\{u\geq 3.5:|\xi^{\prime\prime}(u)|\geq 2^{u}/(7\Gamma(u+1))\\},$

so that $M>5$. We have

$M|\xi^{\prime\prime}(M)|\leq
2\int_{M-1}^{M}|\xi^{\prime\prime}(t)|dt<2\int_{M-1}^{M}\frac{2^{t}}{7\Gamma(t+1)}dt<2\frac{2^{M-1}}{7\Gamma(M)},$

since $2^{t}/(7\Gamma(t+1))$ is decreasing for $t\geq 2$. Dividing by $M$, we
get $|\xi^{\prime\prime}(M)|<2^{M}/(7\Gamma(M+1)$, which is impossible in
light of the continuity of $\xi^{\prime\prime}(u)$ at $u=M>5$. This
establishes (8).

Since $\omega(u)-e^{-\gamma},\omega^{\prime}(u)\ll 1/\Gamma(u+1)$ and
$\int_{0}^{\infty}(\omega(u)-e^{-\gamma})du=e^{-\gamma}-1$, (7) implies
$\xi(u)-(u+2)e^{-2\gamma}=o(1)$ and $\xi^{\prime}(u)-e^{-2\gamma}=o(1)$ as
$u\to\infty$. The estimate (8) yields

$\left|\xi^{\prime}(u)-e^{-2\gamma}\right|=\left|-\int_{u}^{\infty}\xi^{\prime\prime}(t)dt\right|\leq\int_{u}^{\infty}\frac{2^{t}}{7\Gamma(t+1)}dt<\frac{2^{u}}{7\Gamma(u+1)},$

where the last inequality is valid for $u\geq 5$. Similarly, (9) yields

$\left|\xi(u)-(u+2)e^{-2\gamma}\right|=\left|\int_{u}^{\infty}(\xi^{\prime}(t)-e^{-2\gamma})dt\right|\leq\int_{u}^{\infty}\frac{2^{t}dt}{7\Gamma(t+1)}<\frac{2^{u}}{7\Gamma(u+1)},$

for $u\geq 5$.

The estimate (11) follows from [2, Thms. 1 and 2 ], which contains more
precise information about the error term. ∎

##  3\. Proof of Theorem 1

Define

$\Phi(x,y)=|\\{n\leq x:P^{-}(n)>y\\}|.$

In Lemma 5 of [17], we obtained the following estimate for $\Phi(x,y)$ by
combining several results from Section III.6 of Tenenbaum [11]. Having three
separate terms of size $x/\log y$ in our estimate of $\Phi(x,y)$, rather than
one composite term, simplifies our proof.

######  Lemma 2.

Uniformly for $x\geq 1$, $y\geq 2$, $u=\log x/\log y$,

$\begin{split}\Phi(x,y)&=1+x\prod_{p\leq
y}\left(1-\frac{1}{p}\right)+\frac{x}{\log
y}\\!\left\\{\omega(u)-e^{-\gamma}\\!-\left.\frac{y}{x}\right|_{x\geq
y}+O\left(\frac{e^{-u/3}}{\log y}\right)\\!\right\\}.\\\ \end{split}$

######  Lemma 3.

Uniformly for $x\geq 1$, $y\geq 2$, $u=\log x/\log y$,

$\sum_{n\leq x\atop P^{-}(n)>y}\frac{1}{n}=1+(\log x)\prod_{p\leq
y}\left(1-\frac{1}{p}\right)+\int_{0}^{u}(\omega(v)-e^{-\gamma})dv+O\left(\frac{1}{\log
y}\right).$

###### Proof.

This follows from Lemma 2 and partial summation. ∎

###### Proof of Theorem 1.

We write

$S(x,y):=\sum_{n\leq x\atop P^{-}(n)>y}\tau(n)=\sum_{n\leq x\atop
P^{-}(n)>y}\sum_{d|n}1=\sum_{d\leq x\atop P^{-}(d)>y}\Phi(x/d,y).$

With Lemma 2, we obtain

$\begin{split}S(x,y)=&O\left(\frac{x}{\log^{2}y}\right)+\Phi(x,y)+\Biggl{(}\sum_{d\leq
x\atop P^{-}(d)>y}\frac{1}{d}\Biggr{)}x\prod_{p\leq
y}\left(1-\frac{1}{p}\right)\\\ &+\frac{x}{\log y}\sum_{d\leq x\atop
P^{-}(d)>y}\frac{1}{d}\left\\{\omega\left(\frac{\log x/d}{\log
y}\right)-e^{-\gamma}-\left.\frac{y}{x/d}\right|_{d\leq
x/y}\right\\}.\end{split}$

We estimate $\Phi(x,y)$ with Lemma 2, and the first sum with Lemma 3. In the
second sum, the contribution from $d=1$ is

$\frac{x(\omega(u)-e^{-\gamma})-y|_{x\geq y}}{\log y},$

while the contribution from $1<d\leq x$ is

$\frac{x}{\log
y}\int_{0}^{u}\omega(v)(\omega(u-v)-e^{-\gamma})dv+O\left(\frac{x}{\log^{2}y}\right),$

by partial summation, Lemma 3 and integration by parts. Combining these
estimates with Mertens’ formula, we obtain Theorem 1. ∎

##  4\. Proof of Theorem 2

We assume that $\theta$ satisfies

$\theta:\mathbb{N}\to\mathbb{R}\cup\\{\infty\\},\quad\theta(1)\geq
2,\quad\theta(n)\geq P^{+}(n)\quad(n\geq 2).$ (13)

######  Lemma 4.

Assume (13) and let $f(n)$ be multiplicative. For $x\geq 0$ we have

$\sum_{m\leq x}f(m)=\sum_{n\in\mathcal{B}(x)}f(n)\Bigl{(}1+\sum_{2\leq r\leq
x/n\atop P^{-}(r)>\theta(n)}f(r)\Bigr{)}.$

###### Proof.

Each $m\geq 1$ factors uniquely as $m=nr$, with $n\in\mathcal{B}$ and
$P^{-}(r)>\theta(n)$ if $r>1$. ∎

######  Theorem 5.

Assume (13). We have

$\sum_{n\in\mathcal{B}(x)}\tau(n)=(1-L)x\log x+o(x\log x),$

where $0\leq L\leq 1$ and

$L=\sum_{n\in\mathcal{B}}\frac{\tau(n)}{n}\prod_{p\leq\theta(n)}\left(1-\frac{1}{p}\right)^{2}.$
(14)

Thus, $L=1$ if and only if $\sum_{n\in\mathcal{B}(x)}\tau(n)=o(x\log x)$.

###### Proof.

For every $N\leq x$, Lemma 4 yields

$\sum_{m\leq
x}\tau(m)\geq\sum_{n\in\mathcal{B}(N)}\tau(n)\Bigl{(}1+\sum_{2\leq r\leq
x/n\atop P^{-}(r)>\theta(n)}\tau(r)\Bigr{)}.$

We divide both sides by $x\log x$, fix $N$ and let $x\to\infty$. Theorem 1
implies

$1\geq\sum_{n\in\mathcal{B}(N)}\frac{\tau(n)}{n}\prod_{p\leq\theta(n)}\left(1-\frac{1}{p}\right)^{2}.$

This shows that the series in (14) converges to a value $L\in[0,1]$. Inserting
Theorem 1 into the last sum of Lemma 4, the convergence of the series (14)
implies

$x\log x=\sum_{n\in\mathcal{B}(x)}\tau(n)+Lx\log x+o(x\log x).$

∎

######  Corollary 6.

Assume (13). If $\max(2,n)\leq\theta(n)\ll n^{2}$ for $n\geq 1$, then

$x(\log x)^{0.58}\ll\sum_{n\in\mathcal{B}(x)}\tau(n)\ll x(\log x)^{0.99998}$

and $L=1$.

###### Proof.

Theorem 3 and Table 1 of [14] show that $\theta(n)\ll n^{2}$ implies $B(x)\ll
x(\log x)^{-0.4191}.$ We have

$\sum_{n\in\mathcal{B}(x)\atop\Omega(n)\leq 2.01\log_{2}x}\tau(n)\leq
B(x)2^{2.01\log_{2}x}\ll\frac{x(\log x)^{(2.01)\log 2}}{(\log x)^{0.4191}}\ll
x(\log x)^{0.98}$

and

$\sum_{n\in\mathcal{B}(x)\atop\Omega(n)\geq
2.01\log_{2}x}\tau(n)\leq\sum_{n\leq x\atop\Omega(n)\geq
2.01\log_{2}x}\tau(n)\ll x(\log x)^{0.99998},$

by Lemma 5 with $\alpha=2.01$. Theorem 5 shows that $L=1$.

The lower bound follows from [17, Cor. 3]. The exponent $0.58$ can be replaced
by any constant less than $C\log 2-1$, where $C=(1-e^{-\gamma})^{-1}$. ∎

######  Lemma 5 (Lemma 2 of [17]).

Let $\varepsilon>0$. For $2\leq\alpha\leq 4-\varepsilon$ we have

$\sum_{n\leq x\atop\Omega(n)\geq\alpha\log_{2}x}\tau(n)\ll x(\log
x)^{\alpha(\log 2-\log\alpha+1)-1}.$

###### Proof of Theorem 2.

We insert Theorem 1 into the last sum of Lemma 4. The contribution from the
error term of Theorem 1 is $O(x)$, since $L\leq 1$ by Theorem 5. The
contribution form the penultimate term of Theorem 1 is

$\ll\sum_{n\in\mathcal{D}(x,t)\atop n\leq\sqrt{x/t}}\tau(n)\frac{nt}{\log
nt}\ll\frac{\sqrt{xt}}{\log\sqrt{xt}}\sum_{n\leq\sqrt{x/t}}\tau(n)\ll x.$

For the remaining terms, we use the fact that $L=1$ for each $t\geq 2$, by
Corollary 6, and that $\xi(u)=0$ for $u<0$. With Mertens’ formula, we find
that, for $x\geq 1$,

$T(x,t)=x\sum_{n\in\mathcal{D}_{t}}\frac{\tau(n)}{n\log nt}\left\\{\frac{\log
x}{\log nt}e^{-2\gamma}-\xi\left(\frac{\log x/n}{\log
nt}\right)\right\\}+O(x),$

where $\mathcal{D}_{t}$ denotes the set of all $t$-dense integers. Since
$\xi(u)=(u+2)e^{-2\gamma}+O(e^{-u})$ and
$\xi^{\prime}(u)=e^{-2\gamma}+O(e^{-u})$ for $u>2$, partial summation yields

$T(x,t)=x\log
x\int_{1}^{\infty}\frac{T(y,t)e^{-2\gamma}}{y^{2}\log^{2}yt}dy-x\int_{1}^{x}\frac{T(y,t)}{y^{2}\log
yt}\,\xi\left(\frac{\log xt}{\log yt}-1\right)dy+O(x),$

for $x\geq 1$. With the change of variables

$x=t^{e^{z}-1},\quad y=t^{e^{u}-1},\quad G_{t}(z)=\frac{T(x,t)}{x},$

we obtain, for $z\geq 0$,

$G_{t}(z)=e^{-2\gamma}\hat{G}_{t}(1)(e^{z}-1)-\int_{0}^{z}G_{t}(u)h(z-u)du+E_{t}(z),$

where $h(z)=\xi(e^{z}-1)$ and $E_{t}(z)=O(1)$. We multiply by $e^{-sz}$, where
$\operatorname{Re}s>1$, and integrate over $z\geq 0$, to get the equation of
Laplace transforms,

$\hat{G}_{t}(s)=\frac{e^{-2\gamma}\hat{G}_{t}(1)}{s(s-1)}-\hat{G}_{t}(s)\hat{h}(s)+\hat{E}_{t}(s)\qquad(\operatorname{Re}s>1).$

Solving for $\hat{G}_{t}(s)$, we find that

$\hat{G}_{t}(s)=e^{-2\gamma}\hat{G}_{t}(1)\hat{\Lambda}(s)+\hat{\Lambda}(s)s(s-1)\hat{E}_{t}(s)\qquad(\operatorname{Re}s>1),$

where $\hat{\Lambda}(s)$ is as in Lemma 6. Since $\Lambda(0)=0$ and
$\Lambda^{\prime}(0)=1$,

$G_{t}(z)=e^{-2\gamma}\hat{G}_{t}(1)\Lambda(z)+\int_{0}^{z}\Lambda^{\prime\prime}(z-u)E_{t}(u)du-\int_{0}^{z}\Lambda^{\prime}(z-u)E_{t}(u)du+E_{t}(z),$

for $z\geq 0$. Now $E_{t}(z)=O(1)$, so Lemma 6 yields

$\begin{split}\int_{0}^{z}\Lambda^{\prime}(z-u)E_{t}(u)du&=\int_{0}^{z}\lambda_{0}\delta
e^{\delta(z-u)}E_{t}(u)du+O(1)\\\ &=\lambda_{0}\delta e^{\delta
z}\int_{0}^{\infty}e^{-\delta u}E_{t}(u)du+O(1)\\\
&=\lambda_{0}\delta\hat{E}_{t}(\delta)e^{\delta z}+O(1)\\\
&=\delta\hat{E}_{t}(\delta)\Lambda(z)+O(1).\end{split}$

Similarly,

$\int_{0}^{z}\Lambda^{\prime\prime}(z-u)E_{t}(u)du=\delta^{2}\hat{E}_{t}(\delta)\Lambda(z)+O(1).$

Since $L=1$, $e^{-2\gamma}\hat{G}_{t}(1)=\log t+O(1)$, by partial summation.
Thus,

$G_{t}(z)=\alpha_{t}(\log t)\Lambda(z)+O(1),$

where

$\alpha_{t}=\frac{e^{-2\gamma}\hat{G}_{t}(1)+\delta(\delta-1)\hat{E}_{t}(\delta)}{\log
t}=1+O\left(\frac{1}{\log t}\right).$

The claim $\alpha_{t}\geq\alpha_{0}>0$ follows from the lower bound in
Corollary 6. Together with Lemma 6, this completes the proof of Theorem 2. ∎

######  Lemma 6.

Let $\Lambda(z)=\lambda(e^{z}-1)$ and $h(z)=\xi(e^{z}-1)$. The Laplace
transform of $\Lambda$ is given by

$\hat{\Lambda}(s)=\frac{1}{s(s-1)(1+\hat{h}(s))},$ (15)

$\hat{\Lambda}(s)$ has a simple pole at $s=\delta=0.7136125...$ with residue
$\lambda_{0}=1.118192...$, a simple pole at $s=-1$ with residue

$\lambda_{1}=\frac{2}{3e^{-2\gamma}-2}=-1.897011...,$

and a pair of simple poles at $-1.962...\pm 11.57...i$ with corresponding
residues $-0.0078...\pm 0.0031...i$. There are no other singularities with
$\operatorname{Re}s\geq-3$. We have

$\Lambda(z)=\lambda_{0}e^{\delta z}+\lambda_{1}e^{-z}+O(e^{-1.962z}),$
$\Lambda^{\prime}(z)=\lambda_{0}\delta e^{\delta
z}-\lambda_{1}e^{-z}+O(e^{-1.962z}),$
$\Lambda^{\prime\prime}(z)=\lambda_{0}\delta^{2}\delta e^{\delta
z}+\lambda_{1}e^{-z}+O(e^{-1.962z}).$

The constant $\delta$ is the unique value of $s\in(0,1)$ such that

$0=1+\int_{0}^{\infty}\left(\xi(v)-(v+2)e^{-2\gamma}\right)\frac{dv}{(v+1)^{s+1}}+\frac{e^{-2\gamma}}{s}+\frac{e^{-2\gamma}}{s-1}.$

We have $\lambda_{0}=(\delta(\delta-1)I)^{-1}$, where

$I=-\int_{0}^{\infty}\left(\xi(v)-(v+2)e^{-2\gamma}\right)\frac{\log(v+1)}{(v+1)^{1+\delta}}dv-\frac{e^{-2\gamma}}{\delta^{2}}-\frac{e^{-2\gamma}}{(\delta-1)^{2}}.$

###### Proof.

Since $0\leq\xi(u)\leq u+1$ for $u\geq 0$, by (6), the definition (2) of
$\lambda(v)$ yields, by induction on $\lfloor v\rfloor$, the preliminary (and
very crude) estimate $|\lambda(v)|\leq(v+1)^{2}$, and hence $|\Lambda(z)|\leq
e^{2z}$. It follows that the Laplace transform
$\hat{\Lambda}(s)=\int_{0}^{\infty}\Lambda(z)e^{-zs}dz$ converges absolutely
for $\operatorname{Re}s>2$ and is analytic there. The relation (15) follows
from (2). We write

$g(s):=1+\hat{h}(s)=1+\int_{0}^{\infty}\left(\xi(v)-(v+2)e^{-2\gamma}\right)\frac{dv}{(v+1)^{s+1}}+\frac{e^{-2\gamma}}{s}+\frac{e^{-2\gamma}}{s-1}.$

Since $|\xi(v)-(v+2)e^{-2\gamma}|\ll v^{-v}$, by Lemma 1, the integral is an
entire function of $s$, which means that $g(s)$ is analytic everywhere except
for simple poles at $s=0$ and $s=1$. Thus, (15) extends $\hat{\Lambda}(s)$ to
a meromorphic function on the entire complex plane, whose poles are exactly
the zeros of $g(s)$.

Since $\xi(v)=0$ for $v<1$, integration by parts shows that

$g(s)=1+2^{1-s}\left(\frac{e^{-2\gamma}}{s-1}+\frac{1-e^{-2\gamma}}{s}\right)+\frac{1}{s}\int_{1}^{\infty}\left(\xi^{\prime}(v)-e^{-2\gamma}\right)\frac{dv}{(v+1)^{s}}.$
(16)

Thus, if $g(s)=0$, we must have

$|\tau|\leq
H(\sigma):=2^{1-\sigma}+\int_{1}^{\infty}\left|\xi^{\prime}(v)-e^{-2\gamma}\right|\frac{dv}{(v+1)^{\sigma}}.$

Let $P_{a}$ denote the finite set of poles of $\hat{\Lambda}(s)$ (i.e. zeros
of $g(s)$) with $\operatorname{Re}s\geq-a$. Since $\hat{\Lambda}(s)$ is
analytic for $\sigma>2$ and $H(\sigma)$ is decreasing, $P_{a}$ is a subset of
the finite rectangle $-a\leq\sigma\leq 2$, $|\tau|\leq H(-a)$.

From (2), $\lambda(v)$ is continuously differentiable for $v>0$,
$\lambda^{\prime}(v)$ is continuously differentiable for $v>0,v\neq 1$ and
$\lambda^{\prime\prime}(v)$ has a finite jump discontinuity at $v=1$. Let
$F(z)$ be any one of
$\Lambda(z),\Lambda^{\prime}(z),\Lambda^{\prime\prime}(z)$, and $\hat{F}(s)$
be its Laplace transform. We have

$F(z)=\frac{1}{2\pi i}\int_{3-i\infty}^{3+i\infty}\hat{F}(s)e^{zs}\,ds.$

Let $T=e^{z(a+3)}$. Since the result is trivial for bounded $z$, we may assume
that $z$ is sufficiently large such that $T>2H(-a)$. We have
$|\hat{h}(s)|\leq\frac{H(\sigma)}{|\tau|}$, so that $|\hat{h}(s)|\leq 1/2$
whenever $|\tau|\geq 2H(\sigma)$. Thus,

$\frac{1}{1+\hat{h}(s)}=1-\hat{h}(s)+O(\hat{h}(s)^{2})=1-2^{1-s}\left(\frac{e^{-2\gamma}}{s-1}+\frac{1-e^{-2\gamma}}{s}\right)+O_{\sigma}(\tau^{-2}),$

for $|\tau|\geq 2H(\sigma)$, since the integral in (16) is
$O_{\sigma}(\tau^{-1})$, as can be seen by applying integration by parts a
second time. With (15), it follows that the Laplace transforms of
$\Lambda(z),\Lambda^{\prime}(z)$ and $\Lambda^{\prime\prime}(z)$ satisfy

$\hat{\Lambda}(s)\ll_{a}\tau^{-2},\qquad\widehat{\Lambda^{\prime}}(s)=s\hat{\Lambda}(s)=\frac{1}{s-1}+O_{a}(\tau^{-2}),$
$\widehat{\Lambda^{\prime\prime}}(s)=s(s-1)\hat{\Lambda}(s)+s\hat{\Lambda}(s)-1=\frac{1}{s-1}-2^{1-s}\left(\frac{e^{-2\gamma}}{s-1}+\frac{1-e^{-2\gamma}}{s}\right)+O_{a}(\tau^{-2}),$

for $|\tau|\geq 2H(-a)$ and $-a\leq\sigma\leq 2$. We find that

$\int_{3+iT}^{3+i\infty}\hat{F}(s)e^{zs}\,ds=O\left(e^{-az}\right),$
$\int_{-a+iT}^{3+iT}\hat{F}(s)e^{zs}\,ds\ll_{a}\int_{-a}^{3}\frac{e^{z\sigma}}{T}d\sigma=O_{a}\left(e^{-az}\right),$

and

$\int_{-a+i2H(-a)}^{-a+iT}\hat{F}(s)e^{zs}\,ds=O_{a}\left(e^{-az}\right).$

For the remaining segment, we assume that $\hat{\Lambda}(s)$ has no poles with
$\sigma=a$. If it does, one just replaces $-a$ by $-a-\varepsilon$. We have

$\left|\int_{-a-i2H(-a)}^{-a+i2H(-a)}\hat{F}(s)e^{zs}\,ds\right|\leq
4H(-a)\max_{|\tau|\leq
2H(-a)}\left|\hat{F}(-a+i\tau)\right|e^{-az}=O_{a}\left(e^{-az}\right).$

The residue theorem now yields

$F(z)=\sum_{s_{k}\in
P_{a}}\operatorname{Res}\left(\hat{F}(s)e^{zs};s_{k}\right)+O_{a}\left(e^{-az}\right).$

We now determine the poles with $\sigma\geq-3$. To estimate $H(-3)$, we use
exact values of $\xi^{\prime}(v)$ on $[1,5]$, and estimate the tail with the
help of (9). This shows that for $\sigma\geq-3$, all zeros of $g(s)$ satisfy
$|\tau|\leq H(-3)<62$.

For real $s\in(0,1)$, we have $\lim_{s\to 0^{+}}g(s)=+\infty$ and $\lim_{s\to
1^{-}}g(s)=-\infty$, so $g(s)$ has at least one real zero in the interval
$(0,1)$. We will see in a moment that there is exactly one such zero, say at
$s=\delta\in(0,1)$. We have $g(-1)=0$, since (5) implies

$\frac{d}{dv}(v+1)\left(\xi(v+1)-(v+3)e^{-2\gamma}\right)=2\left(\xi(v)-(v+2)e^{-2\gamma}\right),$
(17)

for $v>0$, $v\neq 1$. Define

$g_{5}(s)=1+\int_{0}^{5}\left(\xi(v)-(v+2)e^{-2\gamma}\right)\frac{dv}{(v+1)^{s+1}}+\frac{e^{-2\gamma}}{s}+\frac{e^{-2\gamma}}{s-1}.$

By using exact formulas for $\xi(v)$ on the interval $[0,5]$, we can calculate
$g_{5}(s)$ with arbitrary precision.

Let $R$ be the rectangle defined by the lines $\text{Re}(s)=\pm 3$,
$\text{Im}(s)=\pm 62$. We evaluate the contour integral

$\frac{1}{2\pi i}\int_{R}\frac{g_{5}^{\prime}(s)}{g_{5}(s)}ds=2,$

by numerical integration. This shows that $g_{5}(s)$ has exactly four zeros
(and two poles) inside of $R$. We need to estimate the error

$|g(s)-g_{5}(s)|\leq\int_{5}^{\infty}\left|\xi(v)-(v+2)e^{-2\gamma}\right|\,\frac{dv}{(v+1)^{\sigma+1}}.$

Lemma 1 shows that

$\xi(v)-(v+2)e^{-2\gamma}=2r(v)+2e^{-\gamma}(v+1)r(v+1)+\int_{0}^{v}r(u)r(v-u)du,$

where $r(v)=\omega(v)-e^{-\gamma}$. With a table of zeros and relative extrema
of $\omega(u)-e^{-\gamma}$ on the interval $[5,10.3355]$ due to Cheer and
Goldston [1], and the estimate $|\omega(u)-e^{-\gamma}|<1/\Gamma(u+1)$ from
[12, Lemma 1], we find that

$|g(s)-g_{5}(s)|<0.0035\quad(\text{Re}(s)\geq-3).$

On the boundary of the rectangle $R$, we find that $|g_{5}(s)|>0.0051$ .
Rouché’s Theorem now shows that $g(s)$ also has exactly four zeros in the
rectangle $R$. Applying a similar argument to the small square $Q$, bounded by
the lines $\text{Re}(s)=-1.963,-1.961$, $\text{Im}(s)=11.574,11.576$, we find
that $g(s)$ has a zero inside $Q$. Similarly, we find that the real zero at
$\delta\in(0,1)$ satisfies $0.713611<\delta<0.713614$. Thus, the four zeros of
$g(s)$ with $\text{Re}(s)\geq-3$ are $\delta=0.71361...$, $-1$, $-1.96...\pm
11.57...i$. Replacing $g_{5}(s)$ by $g_{6}(s)$, where we use numerical
integration to evaluate $\xi(v)$ for $5<v\leq 6$, we find that
$\delta=0.7136125...$ and the complex zeros are at $-1.962...\pm 11.57...i$.

Since all poles $s_{k}$ with $\operatorname{Re}s_{k}\geq-3$ are simple, the
corresponding residues satisfy
$\lambda_{k}=(s_{k}(s_{k}-1)g^{\prime}(s_{k}))^{-1}$. In the case of
$\lambda_{1}$, integration by parts and (17) allow us to find the exact value.
∎

##  5\. Another equation for $\delta$

Besides the definition of $\delta$ in Lemma 6, $\delta$ satisfies another
equation, which allows us to independently confirm the numerical calculations
of Lemma 6. Let

$Q(s):=\int_{0}^{\infty}u^{s}\left(e^{2J(u)}-1\right)du\qquad(\operatorname{Re}s>1),$

where

$J(u):=\int_{u}^{\infty}\frac{e^{-t}}{t}dt.$

If we develop $e^{2J(u)}-1$ into the series $\sum_{k\geq 0}b_{k}u^{k-2}$, and
write

$e^{2J(u)}-1=\left(e^{2J(u)}-1-\sum_{k=0}^{K}b_{k}u^{k-2}\right)+\sum_{k=0}^{K}b_{k}u^{k-2},$

we find that the contribution from $u\in[0,1]$ to $Q(s)$, and hence $Q(s)$
itself, extends to a meromorphic function on all of $\mathbb{C}$, with a
simple pole at every $s=1-k$, $k\in\mathbb{N}\cup\\{0\\}$, for which
$b_{k}\neq 0$.

From the proof of Lemma 6, we know that $\hat{h}(s)$, the Laplace transform of
$h(z)=\xi(e^{z}-1)$, extends to a meromorphic function on all of $\mathbb{C}$,
whose only poles are at $s=0,1$. Moreover, the poles of $\hat{\Lambda}(s)$ are
precisely the zeros of $1+\hat{h}(s)$. With Lemma 7, we can locate these zeros
by studying $Q(s)$, without the need to evaluate $\xi(v)$.

######  Lemma 7.

For $s\in\mathbb{C}$ we have

$(s+1)Q(s)=2\Gamma(s+1)\left(1+\hat{h}(s)\right).$

###### Proof.

For $\operatorname{Re}s>1$, equation (12) implies

$\begin{split}Q(s)&=\int_{0}^{\infty}u^{s}\hat{\xi}(u)du=\int_{0}^{\infty}u^{s}\int_{0}^{\infty}\xi(v)e^{-uv}dv\,du\\\
&=\int_{0}^{\infty}\xi(v)\int_{0}^{\infty}u^{s}e^{-uv}du\,dv\\\
&=\int_{0}^{\infty}\xi(v)\int_{0}^{\infty}\left(\frac{w}{v}\right)^{s}e^{-w}\frac{dw}{v}\,dv\\\
&=\Gamma(s+1)\int_{0}^{\infty}v\xi(v)\frac{dv}{v^{s+2}}\\\
&=\frac{2\Gamma(s+1)}{s+1}\left(1+\int_{0}^{\infty}\xi(v)\frac{dv}{(v+1)^{s+1}}\right),\end{split}$

where the last equation follows from (5) and integration by parts. ∎

It follows from Lemma 7 that the constant $\delta$ in Lemma 6 and Theorem 2 is
the unique value of $s\in(0,1)$ which satisfies $Q(s)=0$.

######  Remark 1.

Let $\Omega(z)=\omega(e^{z}-1)$. The poles of the Laplace transform appearing
in [12, Lemma 5], in connection with the study of $D(x,t)$, are the zeros of
$(1+\hat{\Omega}(s-1))$. The same reasoning as in Lemma 7 shows that

$s\int_{0}^{\infty}u^{s-1}\left(e^{J(u)}-1\right)du=\Gamma(s)\left(1+\hat{\Omega}(s-1)\right).$

This provides an alternative for locating the poles of the Laplace tranform in
[12, Lemma 5], without the need to approximate $\omega(u)$.

##  6\. Proof of Theorem 3

######  Lemma 8.

Assume $\max(2,n)\leq\theta(n)\ll n\exp(\log n)^{a})$ for $n\geq 1$ , where
$a$ is any constant with $a<(1-\delta)/(2-\delta)=0.2226...$ Then $L=1$,

$T(x):=\sum_{n\in\mathcal{B}(x)}\tau(n)\ll x(\log x)^{\delta+a(1-\delta)}$

and

$\sum_{n\in\mathcal{B}}\frac{\tau(n)\log^{a}n}{n\log^{2}n}\ll 1.$

###### Proof.

Since $\theta(n)\leq An\exp((\log n)^{a})$ for some constant $A$,
$\mathcal{B}(x)\subset\mathcal{D}(x,t)$, where $t=A\exp((\log x)^{a})$.
Corollary 3 shows that

$\sum_{n\in\mathcal{B}(x)}\tau(n)\leq\sum_{n\in\mathcal{D}(x,t)}\tau(n)\ll
x(\log t)v^{\delta}\ll x(\log x)^{\delta+a(1-\delta)}.$

The first claim now follows from Theorem 5. Partial summation yields the last
claim. ∎

###### Proof of Theorem 3.

We follow the proof of Theorem 2, with some adjustments. We insert Theorem 1
into the last sum of Lemma 4. As before, the contribution from the error term
of Theorem 1 is $O(x)$, since $L\leq 1$ by Theorem 5. To estimate the
contribution form the penultimate term of Theorem 1, note that
$\log\theta(n)\asymp\log(n\theta(n))$. We write

$\sum_{n\in\mathcal{B}(x)\atop n\theta(n)\leq
x}\tau(n)\frac{\theta(n)}{\log\theta(n)}\asymp\sum_{k\geq
0}\frac{x}{2^{k}\log(x/2^{k})}\sum_{n\in\mathcal{B}(x)\atop n\theta(n)\in
I_{k}}\frac{\tau(n)}{n},$ (18)

where $I_{k}=(x/2^{k+1},x/2^{k}]$. The contribution from $k$ with
$2^{k}>\sqrt{x}$ is trivially $O(x)$, since $\sum_{n\leq
x}\tau(n)/n\ll\log^{2}x$. If $2^{k}\leq\sqrt{x}$, then
$\log(x/2^{k})\asymp\log x$. The conditions on $\theta$ mean that
$n\theta(n)\in I_{k}$ implies that $n\in(M_{k},N_{k}]$ where
$N_{k}=\sqrt{x2^{-k}}$ and $\log(N_{k}/M_{k})\ll(\log x)^{a}$. With
$b:=\delta+a(1-\delta)$, partial summation and Lemma 8 yield

$\sum_{n\in\mathcal{B}(x)\atop n\theta(n)\in
I_{k}}\frac{\tau(n)}{n}\ll\frac{T(N_{k})}{N_{k}}+\int_{M_{k}}^{N_{k}}\frac{T(y)}{y^{2}}dy\ll(\log
N_{k})^{b}+\frac{(\log N_{k})^{b+1}-(\log M_{k})^{b+1}}{b+1}.$

Writing $\log M_{k}=\log N_{k}-\log N_{k}/M_{k}=\log N_{k}+O(\log^{a}x)$, we
find that the last expression is $\ll(\log x)^{a+b}\leq\log x$. Inserting this
into (18), we conclude that the contribution from the penultimate term of
Theorem 1 is $O(x)$.

For the remaining terms, we use the fact that $L=1$, by Lemma 8, and that
$\xi(u)=0$ for $u<0$. With Mertens’ formula, we find that, for $x\geq 1$,

$T(x)=x\sum_{n\in\mathcal{B}}\frac{\tau(n)}{n\log\theta(n)}\left\\{\frac{\log
x}{\log\theta(n)}e^{-2\gamma}-\xi\left(\frac{\log
x/n}{\log\theta(n)}\right)\right\\}+O(x).$

We now replace each occurrence of $\theta(n)$ by $2n$. Since
$\xi(u)=(u+2)e^{-2\gamma}+O(e^{-u})$ and
$\xi^{\prime}(u)=e^{-2\gamma}+O(e^{-u})$, Lemma 8 shows that

$T(x)=x\sum_{n\in\mathcal{B}}\frac{\tau(n)}{n\log 2n}\left\\{\frac{\log
x}{\log 2n}e^{-2\gamma}-\xi\left(\frac{\log x/n}{\log
2n}\right)\right\\}+O(x).$

The rest of the proof is identical to that of Theorem 2, with $t=2$. The
conclusion is that $T(x)=x\alpha_{2}(\log 2)\lambda(v)+O(x)$. Theorem 3 now
follows from (3). ∎

##  7\. Proof of Theorem 4

The condition $\theta(n)\ll nl(n)$ implies that $B(x)\ll x/\log x$, by [5,
Prop. 1] (with $y=x,z=1$). This upper bound for $B(x)$ is used throughout this
proof. In Lemma 4, with $f(n)=1$, we estimate the last sum with Lemma 2. The
contribution from the error term of Lemma 2 is $\ll x/\log^{2}x$.

To estimate the contribution form the penultimate term of Lemma 2, note that
$\log\theta(n)\asymp\log(n\theta(n))$. We write

$\sum_{n\in\mathcal{B}(x)\atop n\theta(n)\leq
x}\frac{\theta(n)}{\log\theta(n)}\asymp\sum_{k\geq
0}\frac{x}{2^{k}\log(x/2^{k})}\sum_{n\in\mathcal{B}(x)\atop n\theta(n)\in
I_{k}}\frac{1}{n},$ (19)

where $I_{k}=(x/2^{k+1},x/2^{k}]$. The contribution from $k$ with
$2^{k}>\sqrt{x}$ is trivially $O(x/\log^{2}x)$, so we may assume
$2^{k}\leq\sqrt{x}$ and $\log(x/2^{k})\asymp\log x$. The conditions on
$\theta$ mean that $n\theta(n)\in I_{k}$ implies that $n\in(M_{k},N_{k}]$
where $N_{k}=\sqrt{x2^{-k}}$ and $\log(N_{k}/M_{k})\ll\log f(x)$. Partial
summation yields

$\sum_{n\in\mathcal{B}(x)\atop n\theta(n)\in
I_{k}}\frac{1}{n}\leq\frac{B(N_{k})}{N_{k}}+\int_{M_{k}}^{N_{k}}\frac{B(y)}{y^{2}}dy\ll\frac{1}{\log
N_{k}}+\log\log N_{k}-\log\log M_{k}.$

Writing $\log M_{k}=\log N_{k}-\log N_{k}/M_{k}=\log N_{k}+O(\log f(x))$, we
find that the last expression is $O((\log f(x))/\log x)$. Inserting this into
(19), we conclude that the contribution from the penultimate term of Lemma 2
is $O(x(\log f(x))/\log^{2}x)$.

For the contribution from the term $x\prod_{p\leq y}(1-1/p)$ in Lemma 2, we
have

$\sum_{n\in\mathcal{B}}\frac{1}{n}\prod_{p\leq\theta(n)}\left(1-\frac{1}{p}\right)=1,$

by [14, Thm. 1]. With Mertens’ formula and the fact that $\omega(u)=0$ for
$u<0$, we obtain

$B(x)=x\sum_{n\in\mathcal{B}}\frac{1}{n\log\theta(n)}\left\\{e^{-\gamma}-\omega\left(\frac{\log
x/n}{\log\theta(n)}\right)\right\\}+O\left(\frac{x\log
f(x)}{\log^{2}x}\right).$

We now replace each instance of $\theta(n)$ by $2n$, incurring an error of

$\ll\frac{x\log f(x)}{\log^{2}x}+x\int_{x}^{\infty}\frac{\log
f(y)}{y\log^{3}y}dy\ll\frac{x\log f(x)}{\log^{2}x},$

since $\log f(y)/\log y$ is eventually decreasing. Next, we use partial
summation on the last sum, with an error $\ll x/\log^{2}x$, to obtain

$B(x)=x\int_{1}^{\infty}\frac{1}{y\log
2y}\left\\{e^{-\gamma}-\omega\left(\frac{\log x/y}{\log
2y}\right)\right\\}+O\left(\frac{x\log f(x)}{\log^{2}x}\right).$

The rest of the proof is the same as that of [13, Thm. 5.1]. In the process of
inverting a Laplace transform, two additional error terms appear. They are

$\ll\frac{x}{\log^{3}x}\int_{1}^{x}\frac{\log f(y)}{y}dy+\frac{x}{\log
x}\int_{x}^{\infty}\frac{\log f(y)}{y\log^{2}y}dy\leq\frac{x\log
f(x)}{\log^{2}x}+\frac{xE(x)}{\log x}.$

Since $f(x)$ is non-decreasing, $f(x)\geq f(1)\geq 2$ and $E(x)\geq(\log
f(x))/\log x\geq\log 2/\log x$, so that all error terms are $O(xE(x)/\log x)$.

The proof of the formula for $c_{\theta}$ is identical to that of [15, Thm.
1], where this formula is derived under a more restrictive upper bound
condition on $\theta(n)$. This stronger condition was needed in [15], only
because [13, Thm. 5.1] required it to ensure that $B(x)\sim c_{\theta}x/\log
x$ holds.

## References

  * [1] A. Y. Cheer and D. A. Goldston, A differential delay equation arising from the sieve of Eratosthenes, Math. Comp. 55 (1990), 129–141.
  * [2] A. Hildebrand and G. Tenenbaum, On a class of differential-difference equations arising in number theory, J. Anal. Math. 61 (1993), 145–179.
  * [3] M. Margenstern, Les nombres pratiques: théorie, observations et conjectures, J. Number Theory 37 (1991), 1–36.
  * [4] G. Melfi, On two conjectures about practical numbers, J. Number Theory 56 (1996), 205–210.
  * [5] C. Pomerance and A. Weingartner, On primes and practical numbers, Ramanujan J. (2021). https://doi.org/10.1007/s11139-020-00354-y
  * [6] E. Saias, Entiers à diviseurs denses 1, J. Number Theory 62 (1997), 163–191.
  * [7] W. Sierpiński, Sur une propriété des nombres naturels, Ann. Mat. Pura Appl. (4) 39 (1955), 69–74.
  * [8] A. K. Srinivasan, Practical numbers, Current Sci. 17 (1948), 179–180.
  * [9] B. M. Stewart, Sums of distinct divisors, Amer. J. Math. 76 (1954), 779–785.
  * [10] G. Tenenbaum, Sur un problème de crible et ses applications, Ann. Sci. École Norm. Sup. (4) 19 (1986), 1–30.
  * [11] G. Tenenbaum, Introduction to Analytic and Probabilistic Number Theory, Cambridge Studies in Advanced Mathematics, Vol. 46, Cambridge Univ. Press, Cambridge, 1995.
  * [12] A. Weingartner, Integers with dense divisors 3, J. Number Theory 142 (2014), 211-222.
  * [13] A. Weingartner, Practical numbers and the distribution of divisors, Q. J. Math. 66 (2015), no. 2, 743–758.
  * [14] A. Weingartner, A sieve problem and its application, Mathematika 63 (2017), no. 1, 213–229.
  * [15] A. Weingartner, On the constant factor in several related asymptotic estimates, Math. Comp. 88 (2019), no. 318, 1883–1902.
  * [16] A. Weingartner, The constant factor in the asymptotic for practical numbers, Int. J. Number Theory, 16 (2020), no. 3, 629–638.
  * [17] A. Weingartner, The number of prime factors of integers with dense divisors, arXiv:2101.11585.