# Bounds on the Möbius-signed partition numbers

Taylor Daniels  Dept. of Mathematics, Purdue Univ., 150 N University St, W
Lafayette, IN 47907

###### Abstract.

For $n\in\mathbb{N}$ let $\Pi[n]$ denote the set of partitions of $n$, i.e.,
the set of positive integer tuples $(x_{1},x_{2},\ldots,x_{k})$ such that
$x_{1}\geq x_{2}\geq\cdots\geq x_{k}$ and $x_{1}+x_{2}+\cdots+x_{k}=n$. Fixing
$f:\mathbb{N}\to\\{0,\pm 1\\}$, for $\pi=(x_{1},x_{2},\ldots,x_{k})\in\Pi[n]$
let $f(\pi):=f(x_{1})f(x_{2})\cdots f(x_{k})$. In this way we define the
signed partition numbers

$p(n,f)=\sum_{\pi\in\Pi[n]}f(\pi).$

Following work of Vaughan and Gafni on partitions into primes and prime
powers, we derive asymptotic formulae for quantities $p(n,\mu)$ and
$p(n,\lambda)$, where $\mu$ and $\lambda$ denote the Möbius and Liouville
functions from prime number theory, respectively. In addition we discuss how
quantities $p(n,f)$ generalize the classical notion of restricted partitions.

###### 2020 Mathematics Subject Classification:

Primary: 11P55, 11P82, 11M26.
_Keywords and phrases_ : partitions, Möbius function, Liouville function.

##  1\. Introduction

For $n\in\mathbb{N}$ let $\Pi[n]$ denote the set of _partitions_ of $n$, i.e.,
the set of positive integer tuples $(x_{1},x_{2},\ldots,x_{k})$ such that
$x_{1}\geq x_{2}\geq\ldots\geq x_{k}$ and $x_{1}+x_{2}+\cdots+x_{k}=n$. For
fixed $f:\mathbb{N}\to\\{0,\pm 1\\}$, for $n\in\mathbb{N}$ and
$\pi=(x_{1},x_{2},\ldots,x_{k})\in\Pi[n]$ let

$f(\pi):=f(x_{1})f(x_{2})\cdots f(x_{k}).$ (1.1)

With this we define for $n\in\mathbb{N}$ the _signed partition numbers_
$p(n,f)$ via

$p(n,f)=\sum_{\pi\in\Pi[n]}f(\pi).$ (1.2)

Definition (1.2) generalizes several classical partition-related quantities.
For example, with the constant function $1$ the $p(n,1)$ are the _ordinary
partition numbers_ $p(n)$, and with the indicator function $1_{A}$ for some
$A\subset\mathbb{N}$ the $p(n,1_{A})$ are the _$A$ -restricted partition
numbers_ $p_{A}(n)$, which we write as $p(n,A)$.

Many sequences of restricted partition numbers are classical and well studied
(see, e.g., [andrews1976partitions, erdos1942elementary]). Despite this, the
past ten years have seen a number of new publications on the asymptotics of
some restricted partition numbers, e.g., [berndt2018partitions,
das2022partitions, gafni2016power, vaughan2015squares].

Our two sequences $(p(n,f))_{\mathbb{N}}$ of interest are those using (for
$f$) the Möbius $\mu$ and Liouville $\lambda$ functions from multiplicative
number theory. If $n\in\mathbb{N}$ has prime factorization
$p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{r}^{a_{r}}$ with distinct primes $p_{i}$
and all $a_{i}\geq 1$, then

$\lambda(n)=(-1)^{a_{1}+\cdots+a_{r}}\qquad\text{and}\qquad\mu(n)=\begin{cases}(-1)^{r}&\text{if
all $a_{i}=1$,}\\\ 0&\text{otherwise.}\end{cases}$

We note that $\mu(n)=\lambda(n)$ on the support of $\mu$ and $\mu^{2}$, i.e.,
on the set of _squarefree_ $n$.

We first state simple asymptotic results on the quantities $p(n,\mu)$ and
$p(n,\lambda)$ which are immediate corollaries of our main results.

######  Theorem 1.1.

For all $\varepsilon>0$, as $n\to\infty$ one has

$\displaystyle
p(n,\mu)=O\big{(}e^{(1+\varepsilon)\sqrt{n}}\big{)}\qquad\text{and}\qquad
p(n,\lambda)=O\big{(}e^{(1+\varepsilon)\sqrt{\zeta(2)n}}\big{)},$ (1.3)

where $\zeta(2)=\pi^{2}/6$. In addition, for $n=2k$ with $k\in\mathbb{N}$, as
$k\to\infty$ one has

$\log p(2k,\mu)\sim\sqrt{2k}\qquad\text{and}\qquad\log
p(2k,\lambda)\sim\sqrt{\zeta(2)2k},$ (1.4)

where the relation $a_{m}\sim b_{m}$ indicates that
$\lim_{m\to\infty}a_{m}/b_{m}=1$.

The generating function $\Phi(z)=\Phi(z,f)$ for the sequence
$(p(n,f))_{\mathbb{N}}$ is defined via

$\Phi(z)=\prod_{n=1}^{\infty}(1-f(n)z^{n})^{-1}=1+\sum_{n=1}^{\infty}p(n,f)z^{n}\qquad(|z|<1),$
(1.5)

and $\Psi(z)=\Psi(z,f)$ is defined via

$\Psi(z)=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{f^{k}(n)}{k}z^{nk}\qquad(|z|<1)$
(1.6)

so that

$\Phi(z)=\exp\Psi(z).$ (1.7)

By comparison to the functions $\Phi(z,1)$ and $\Psi(z,1)$, one sees that
$\Phi(z,f)$ and $\Psi(z,f)$ are analytic for $|z|<1$, so by Cauchy’s theorem
one has

$p(n,f)=\frac{1}{2\pi
i}\int_{|z|=\rho}\Phi(z)z^{-n-1}\,dz=\rho^{-n}\int_{0}^{1}\Phi(\rho
e(\alpha))e(-n\alpha)\,d\alpha$ (1.8)

for all $0<\rho<1$, where $e(\alpha)=\exp(2\pi i\alpha)$. Using integrals
(1.8), we define the quantities $p(x,f)$ for general $x>0$ by using $x$ in
place of $n$ therein.

When $f$ is nonnegative and $f(1)=1$, the integrals (1.8) are dominated by the
effect of the singularity of $\Phi(z)$ at $z=+1$. Thus, for a given $x$ it is
natural to select $\rho=\rho(x)$ in (1.8) so as to minimize the quantity
$\rho^{-x}\Phi(\rho)$, which is equivalent to minimizing
$\Psi(\rho)-x\log\rho$. For these $f$ the quantity $\Psi(\rho)-x\log\rho$
grows to $+\infty$ as $\rho$ approaches $0$ and $1$, it follows that the
minimum of said quantity occurs when

$\rho\Psi^{\prime}(\rho)=x.$ (1.9)

Moreover, such a $\rho$ satisfying (1.9) is necessarily unique, as the
nonnegativity of $f$ implies that $\rho\Psi^{\prime}(\rho)$ is strictly
increasing on $(0,1)$. Equation (1.9) is the _saddle-point equation_ common to
many analytic-combinatorial problems.

When $f=\mu$ or $f=\lambda$, the integrals (1.8) for $p(x,\mu)$ and
$p(x,\lambda)$ are both dominated by the effects of _two_ singularities,
namely those at $z=+1$ and $z=-1$. We are thus motivated to (individually)
minimize the quantities $\Psi(\rho)-x\log\rho$ and $\Psi(-\rho)-x\log\rho$,
leading to the equations

$\displaystyle\rho\Psi^{\prime}(\rho)$ $\displaystyle=x,$ (sp)
$\displaystyle(-\rho)\Psi^{\prime}(-\rho)$ $\displaystyle=x,$ (sp$*$)

where $(-\rho)\Psi^{\prime}(-\rho)$ is $z\Psi^{\prime}(z)$ evaluated at
$z=-\rho$. Because $\mu$ and $\lambda$ take both positive and negative values,
it is no longer clear that solutions to equations (sp) and (sp$*$) even exist,
much less unique ones. This is in fact the case for all sufficiently large
$x$, as proved in section 8.

For nonnegative integer $j$ let $\Psi_{(j)}(z):=(z\partial_{z})^{j}\Psi(z)$.
To preempt any confusion, we note that $(-1)^{j}\Psi_{(j)}(-\rho)$ and
$\Psi_{(j)}(-\rho)$ are uniquely valued regardless of their interpretations
vis-à-vis the chain-rule. When $x$ is sufficiently large and $\rho\in(0,1)$ is
a solution to one of the equations (sp) and (sp$*$) with
$\Psi(z)=\Psi(z,\rho)$, we have the following relations between $x$, $\rho$,
and certain derivatives of $\Psi(\pm\rho)$.

######  Theorem 1.2.

Let $A>0$ and $\Psi(z)=\Psi(z,\mu)$. For all sufficiently large $x$, if
$\rho=\rho(x)$ satisfies at least one of the equations (sp) and (sp$*$), then

$\displaystyle-x\log\rho$
$\displaystyle=\tfrac{1}{2}\sqrt{x}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right],$ $\displaystyle\Psi(\pm\rho)$
$\displaystyle=\tfrac{1}{2}\sqrt{x}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right].$

In general, for these $x$ and $\rho$ and for $j\geq 1$, one has

$\displaystyle(\pm 1)^{j}\Psi^{(j)}(\pm\rho)$ $\displaystyle=j!\hskip
0.80002pt2^{j-1}x^{\frac{j+1}{2}}\mathopen{}\mathclose{{}\left[1+O_{j}((\log
x)^{-A})}\right],$ $\displaystyle\Psi_{(j)}(\pm\rho)$ $\displaystyle=j!\hskip
0.80002pt2^{j-1}x^{\frac{j+1}{2}}\mathopen{}\mathclose{{}\left[1+O_{j}((\log
x)^{-A})}\right].$

Our next theorem gives a formula for $p(x,\mu)$ modeled after Vaughan’s
asymptotic formula for $p(n,\mathbb{P})$, the number of partitions into prime
numbers [vaughan2008number]*Thm. 2. In particular, with $\Phi(z)$ and
$\Psi(z)$ denoting $\Phi(z,\mathbb{P})$ and $\Psi(z,\mathbb{P})$,
respectively, Vaughan establishes that as $n\to\infty$ one has

$p(n,\mathbb{P})=\frac{\rho^{-n}\Phi(\rho)}{\sqrt{2\pi\Psi_{(2)}(\rho)}}\big{[}1+O(n^{-1/5})\big{]},$
(1.10)

where $\rho\in(0,1)$ is the unique solution to the equation
$\rho\Psi^{\prime}(\rho)=n$.

######  Theorem 1.3.

Let $\Phi(z)=\Phi(z,\mu)$ and $\Psi(z)=\Psi(z,\mu)$, and for sufficiently
large $x$ let $\rho$ and $\rho_{*}$ denote the unique solutions to (sp) and
(sp$*$), respectively. As $x\to\infty$, one has

$p(x,\mu)=\frac{\rho^{-x}\Phi(\rho)}{\sqrt{2\pi\Psi_{(2)}(\rho)}}\big{[}1+O(x^{-1/5})\big{]}+(-1)^{-x}\frac{\rho_{*}^{-x}\Phi(-\rho_{*})}{\sqrt{2\pi\Psi_{(2)}(-\rho_{*})}}\big{[}1+O(x^{-1/5})\big{]},$
(1.11)

where $(-1)^{-x}=\exp(-\pi ix)$.

For the case of $\Phi(z,\lambda)$ and $\Psi(z,\lambda)$ we have results
similar to those above. Due to its ubiquity in our discussions of
$\Psi(z,\lambda)$, it is convenient to let

$\mathfrak{z}=\zeta(2)=\tfrac{\pi^{2}}{6}.$ (1.12)

######  Theorem 1.4.

Let $A>0$ and $\Psi(z)=\Psi(z,\lambda)$. For all sufficiently large $x$, if
$\rho=\rho(x)$ satisfies at least one of the equations (sp) and (sp$*$), then

$\displaystyle-x\log\rho$
$\displaystyle=\tfrac{1}{2}\sqrt{\mathfrak{z}x}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right],$ $\displaystyle\Psi(\pm\rho)$
$\displaystyle=\tfrac{1}{2}\sqrt{\mathfrak{z}x}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right].$

In general, for these $x$ and $\rho$ and for $j\geq 1$, one has

$\displaystyle(\pm 1)^{j}\Psi^{(j)}(\pm\rho)$ $\displaystyle=j!\hskip
0.80002pt2^{j-1}\mathfrak{z}^{\frac{1-j}{2}}x^{\frac{j+1}{2}}\mathopen{}\mathclose{{}\left[1+O_{j}((\log
x)^{-A})}\right],$ $\displaystyle\Psi_{(j)}(\pm\rho)$ $\displaystyle=j!\hskip
0.80002pt2^{j-1}\mathfrak{z}^{\frac{1-j}{2}}x^{\frac{j+1}{2}}\mathopen{}\mathclose{{}\left[1+O_{j}((\log
x)^{-A})}\right],$

######  Theorem 1.5.

Let $\Phi(z)=\Phi(z,\lambda)$ and $\Psi(z)=\Psi(z,\lambda)$, and for
sufficiently large $x$ let $\rho$ and $\rho_{*}$ denote the unique solutions
to (sp) and (sp$*$), respectively. As $x\to\infty$ one has

$p(x,\lambda)=\frac{\rho^{-x}\Phi(\rho)}{\sqrt{2\pi\Psi_{(2)}(\rho)}}\big{[}1+O(x^{-1/5})\big{]}+(-1)^{-x}\frac{\rho_{*}^{-x}\Phi(-\rho_{*})}{\sqrt{2\pi\Psi_{(2)}(-\rho_{*})}}\big{[}1+O(x^{-1/5})\big{]},$
(1.13)

where $(-1)^{-x}=\exp(-\pi ix)$.

We note that the results of Theorem 1.1 immediately follow from Theorems
1.2–1.5. Indeed, for the case of $p(n,\mu)$, using the relations in Theorem
1.2 one may deduce that the two fractions in equation (1.11) both grow like
$\exp(\sqrt{n}+o(\sqrt{n}))$, from which the bound (1.3) for $p(n,\mu)$
follows at once. Because $(-1)^{-x}=+1$ when $x=2k$ with $k\in\mathbb{N}$, the
asymptotic for $\log p(2k,\mu)$ in (1.4) similarly follows at once. The
statements concerning $p(n,\lambda)$ and $\log p(2k,\lambda)$ similarly follow
from Theorems 1.4 and 1.5.

We now discuss three classical results on certain sequences
$(p(n,A))_{\mathbb{N}}$ to help contextualize the relations of Theorem 1.1.
The most well known of these is the relation

$\log p(n,1)\sim\kappa\sqrt{n},$

where

$\kappa:=\pi\sqrt{2/3},$

which is an immediate corollary of Hardy and Ramanujan’s main results in
[hardy1918asymptotic].

Before stating the second result, we recall that $A\subset\mathbb{N}$ is said
to have _density_ $\delta_{A}$ if the ratio $|A\cap\\{1,2,\ldots,N\\}|N^{-1}$
tends to $\delta_{A}$ as $N\to\infty$. It is evident that
$\delta_{\mathbb{N}}=1$, and it is well known that the set of squarefree
numbers has density $6/\pi^{2}\approx 0.601$. The prime number theorem implies
that $\delta_{\mathbb{P}}=0$. A remarkable theorem of Erdős
[erdos1942elementary] states that: if $A\subset\mathbb{N}$ and
$\mathrm{gcd}(A)=1$, then $A$ has density $\delta>0$ if and only if

$\log p(n,A)\sim\kappa\sqrt{\delta n}.$ (1.14)

We remark that the proof of the “only if” statement in Erdős’ theorem is
omitted in [erdos1942elementary] since the proof of a similar result is given
there, and that a complete proof of both statements may be found in
[nathanson2008elementary]*Ch. 16. The third result, though originally due to
Roth and Szekeres [roth1954some], can be deduced as a corollary to Vaughan’s
formula (1.10). Namely, as $n\to\infty$ one has

$\log p(n,\mathbb{P})\sim\kappa\sqrt{\frac{2n}{\log n}}.$ (1.15)

Noting that $\kappa=2\sqrt{\zeta(2)}=2\sqrt{\mathfrak{z}}$, Theorem 1.1 thus
states that as $n\to\infty$ one has

$p(n,\mu)=O(e^{(1+\varepsilon)\sqrt{n}})\qquad\text{and}\qquad
p(n,\lambda)=O(e^{(\frac{1}{2}+\varepsilon)\kappa\sqrt{n}}),$

and that for $n=2k$ one has

$\log p(2k,\mu)\sim\sqrt{2k}\qquad\text{and}\qquad\log
p(2k,\lambda)\sim\tfrac{1}{2}\kappa\sqrt{2k}.$

Finally we comment on our use of the term “signed partition number”, since it
is apparent from (1.1) and (1.2) that the term “weighted partition number” is
an equally appropriate term. We avoid this latter descriptor due to its
ubiquity in the literature as a moniker for partition number sequences having
generating functions of the form

$\Phi(z)=\prod_{n=1}^{\infty}(1-z^{n})^{-f(n)}=1+\sum_{n=1}^{\infty}p_{f}(n)z^{n}$
(1.16)

for some nonnegative integer-valued $f(n)$. These $p_{f}(n)$ coincide with our
$p(n,f)$ if $f$ takes only the values $0$ and $1$, but given the different
roles $f(n)$ plays in (1.5) and (1.16), we favor the descriptor “signed” for
$p(n,f)$ rather than “weighted.”

### Acknowledgements.

The author would like thank Trevor Wooley for both his suggestion of these
research problems and his financial support, and would like to additionally
thank both Trevor Wooley and Ben McReynolds for their numerous, invaluable
suggestions and feedback in preparing this work. Thanks are also extended to
Nicolas Robles and Alexandru Zaharescu for many helpful conversations in
formulating this paper.

##  2\. Notation and structure

Expressions $O(g(x))$ with $g(x)>0$ are quantities bounded by some constant
multiple of $g(x)$. The relation $f(x)\ll g(x)$ indicates that $f(x)=O(g(x))$
for all sufficiently large $x>0$, and $f(x)\asymp g(x)$ indicates that
$f(x)\ll g(x)$ and $g(x)\ll f(x)$. Quantities of the form $X_{0}(A)$ indicate
some $X_{0}>0$ depending on $A$. Statements involving $\varepsilon$ are
understood to hold for all sufficiently small $\varepsilon>0$ unless noted
otherwise, and constants’ dependencies on $\varepsilon$ are generally
suppressed.

For real $\alpha$ let $e(\alpha)=\exp(2\pi i\alpha)$, let
$\|\alpha\|=\min_{n\in\mathbb{Z}}|\alpha-n|$, and let
$\\{\alpha\\}=\alpha-\lfloor\alpha\rfloor$, and for real $\alpha$ and
$f:\mathbb{N}\to\mathbb{C}$ let

$S_{f}(t,\alpha)=\sum_{n\leq t}f(n)e(n\alpha).$

Expressions $n\equiv a$ (mod $q$) are abbreviated as $n\equiv a\,(q)$. We
recall $\Psi_{(j)}(z):=(z\partial_{z})^{j}\Psi(z)$.

Let $1\leq Q\leq X$. If $0\leq a<q$ and $(a,q)=1$, we define the _arc_ about
$a/q$ as the set

$\mathfrak{M}(a/q)=\mathopen{}\mathclose{{}\left\\{\alpha\in[0,1):|\alpha-a/q|\leq
Q/(qX)}\right\\},$

noting that these sets are generally denoted $\mathfrak{M}(q,a)$ in the
literature. Among the sets $\mathfrak{M}(a/q)$, the _major arcs_ are those
$\mathfrak{M}(a/q)$ with $q\leq Q$, and $\mathfrak{M}(X,Q)$ denotes the union
of all major arcs. The connected components of
$[0,1)\setminus\mathfrak{M}(X,Q)$ are the _minor arcs_ , and their union is
denoted $\mathfrak{m}(X,Q)$.

Ultimately, only $\alpha$ lying in small subsets of the arcs
$\mathfrak{M}(0/1)$, $\mathfrak{M}(1/2)$, and $\mathfrak{M}(1/1)$
significantly contribute to the integral (1.8), with all other $\alpha$
contributing to an error term. Anticipating this, for $X,X_{*}>1$ we define

$\displaystyle\mathfrak{P}$ $\displaystyle=\\{\alpha\in[0,1):\|\alpha\|\leq
X^{-1}(\log X)^{-\frac{1}{4}}\\},$ $\displaystyle\mathfrak{P}_{*}$
$\displaystyle=\\{\alpha\in[0,1):|\alpha-\tfrac{1}{2}|\leq X_{*}^{-1}(\log
X_{*})^{-\frac{1}{4}}\\}.$

Following Gafni [gafni2021partitions] we term the sets $\mathfrak{P}$ and
$\mathfrak{P}_{*}$ the _principal arcs_ , though colloquially the terms
“super-major arcs” and “major-major arcs” have also been used.

Functions $\Phi(z,f)$, $\Psi(z,f)$, etc. are simply written as $\Phi(z)$,
$\Psi(z)$, etc. when the $f$ used is clear or when the discussions apply to
general $f:\mathbb{N}\to\\{0,\pm 1\\}$. For $\rho,\rho_{*}\in(0,1)$ it is
convenient to write

$X=\frac{1}{\log(1/\rho)}\qquad\text{and}\qquad
X_{*}=\frac{1}{\log(1/\rho_{*})}$ (2.1)

so that

$\rho=e^{-1/X}\qquad\text{and}\qquad\rho_{*}=e^{-1/X_{*}},$ (2.2)

and this convention is maintained throughout this paper.

### Structure.

This paper has three major components. Section 3 provides a general recipe for
our analyses, after which sections 4–7 consider the quantities $\Psi(\rho
e(\alpha),\mu)$ when $\alpha$ is in the minor, major, and principal arcs and
$\rho$ is sufficiently close to $1$ (rather, when $X$ is sufficiently large).
In section 8 we examine the implicit relationships between $\rho$, $X$, and
$x$ when $x$ is large and $\rho=\rho(x)$ is defined as the (unique) solution
to the equation (sp). Similar analyses are made for $\rho_{*}$ and $X_{*}$
with $\rho_{*}$ defined as the solution to (sp$*$), concluding the proof of
Theorem 1.2.

Using our results on $\Psi(z,\mu)$, sections 9 and 10 establish that only
those $\alpha$ in the principal arcs significantly contribute to the integral
(1.8) for $p(x,\mu)$. Following this, section 11 treats these principal-arc
integrals to ultimately establish Theorem 1.3.

Finally, section 12 considers the quantities $p(x,\lambda)$. Because the
Liouville $\lambda$ function is far easier to handle than the Möbius $\mu$
function, the techniques of sections 3–8 allow for a rapid analysis of
$\Psi(z,\lambda)$ and a proof of Theorem 1.4. Due to the overwhelming
similarities of the relations of Theorems 1.2 and 1.4, the arguments of
sections 9–11 that prove Theorem 1.3 allow for an almost-immediate proof of
Theorem 1.5.

### A note on $\rho_{*}$ and $X_{*}$.

In sections 3–7 our results involve only $\rho$ and $X$, and not $\rho_{*}$
and $X_{*}$. Indeed, until section 8 our results do not require that $\rho$
(or $\rho_{*}$) be a solution to equation (sp) (or (sp$*$)), so there is no
need to discuss two different $\rho,\rho_{*}\in(0,1)$ until then. Moreover, in
section 8 we find that when $\rho$ and $\rho_{*}$ are the (unique) solutions
to (sp) and (sp$*$), respectively, one has $X\sim X_{*}\sim 2\sqrt{x}$, making
the results of sections 3–7 equally applicable if one uses $\rho_{*}$ and
$X_{*}$ in place of $\rho$ and $X$, respectively.

##  3\. A template for our analyses

Our analyses of $\Psi(z,\mu)$ and $\Psi(z,\lambda)$ for $z=\rho e(\alpha)$
with $\alpha$ in the sets $\mathfrak{M}(X,Q)$ and $\mathfrak{m}(X,Q)$ more or
less follow the same sequence of steps. This section outlines these steps and
summarizes them in Lemma 3.4.

By comparison with $\Psi(z,1)$, the double series for $\Psi(z,f)$ in equation
(1.6) converges absolutely for $|z|<1$. Thus, we exchange the sums in (1.6)
and define

$F_{k}(z)=\sum_{n=1}^{\infty}f^{k}(n)z^{n}$

so that

$\Psi(z)=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{f^{k}(n)}{k}z^{nk}=\sum_{k=1}^{\infty}\frac{1}{k}F_{k}(z^{k}).$

A cornerstone of our analyses is the observation that if $f(n)$ only takes
values $0$ or $\pm 1$, then the functions $F_{k}(z)$ are only unique up to the
parity of $k$. Indeed, as $f^{k}(n)=f(n)$ for odd $k$ and
$f^{k}(n)=f^{2}(n)=|f(n)|$ for even $k$, we see that
$F_{k}(z)=F_{k\,\,(\mathrm{mod}\,2)}(z)$ for all $k$. We thus let

$\displaystyle F_{0}(z)$
$\displaystyle=\sum_{n=1}^{\infty}f^{2}(n)z^{n},\qquad$ $\displaystyle
F_{1}(z)=\sum_{n=1}^{\infty}f(n)z^{n},$ (3.1) $\displaystyle\Psi_{0}(z)$
$\displaystyle=\sum_{k\equiv 0\,(2)}\frac{1}{k}F_{0}(z^{k}),\qquad$
$\displaystyle\Psi_{1}(z)=\sum_{k\equiv 1\,(2)}\frac{1}{k}F_{1}(z^{k}),$ (3.2)

so that

$\Psi(z)=\Psi_{0}(z)+\Psi_{1}(z).$

######  Remark 3.1.

We write $F_{0}(z)$ instead of $F_{2}(z)$ because the sums
$\sum_{n}f^{2}(n)z^{n}$ provide the main asymptotic terms in our analyses of
$p(n,\mu)$ and $p(n,\lambda)$.

For $j=0,1$ we split the functions $\Psi_{j}(z)$ into leading sums and tail
sums, namely

$\Psi_{j}(z)=\sum_{\begin{subarray}{c}k\leq K\\\ k\equiv
j\,(2)\end{subarray}}\frac{1}{k}F_{j}(z^{k})+\sum_{\begin{subarray}{c}k>K\\\
k\equiv j\,(2)\end{subarray}}\frac{1}{k}F_{j}(z^{k}),$ (3.3)

where $K>1$ is a real parameter of our choosing. The tail sums in equation
(3.3) are always part of our error terms, so we record the following lemma
bounding these tail sums independently of $j$, and even independently of $f$.

######  Lemma 3.2.

Let $f:\mathbb{N}\to\\{0,\pm 1\\}$. For real $\alpha$ and large $X$ and $K$
one has that

$\sum_{k>K}\frac{1}{k}F_{k}(\rho^{k}e(k\alpha))\ll\frac{X}{K}.$

###### Proof.

For fixed $k$ we see that

$F_{k}(\rho^{k}e(k\alpha))=\sum_{n=1}^{\infty}f^{k}(n)\rho^{nk}e(nk\alpha)\ll\sum_{n=1}^{\infty}\rho^{nk}=\frac{\rho^{k}}{1-\rho^{k}}.$

Using the relation $\rho=e^{-1/X}$, it follows that

$\sum_{k>K}\frac{1}{k}F_{k}(\rho^{k}e(k\alpha))\ll\sum_{k>K}\frac{1}{k}\,\frac{e^{-k/X}}{1-e^{-k/X}}=\sum_{k>K}\frac{1/k}{e^{k/X}-1}\ll\sum_{k>K}\frac{X}{k^{2}},$

and the result follows. ∎

######  Remark 3.3.

It is clear that Lemma 3.2 extends to all bounded $f:\mathbb{N}\to\mathbb{C}$.

The analyses of $\Psi_{0}(z)$ and $\Psi_{1}(z)$ are thus reduced to analyses
of $F_{0}(z^{k})$ and $F_{1}(z^{k})$ for $k\leq K$. In the following
discussion let $j\in\\{0,1\\}$ and let $X$ be sufficiently large.

Suppose that for $z$ in some arbitrary $\Omega\subset\\{|z|<1\\}$ there exist
functions $\mathcal{X}_{0}(z)$ and $\mathcal{X}_{1}(z)$ such that: For
$z\in\Omega$ and $k\leq K$ one has

$F_{j}(z^{k})=\mathcal{X}_{j}(z)/k.$ (3.4)

Then

$\sum_{\begin{subarray}{c}k\leq K\\\ k\equiv
j\,(2)\end{subarray}}\frac{\mathcal{X}_{j}(z)}{k^{2}}=c_{j}\mathcal{X}_{j}(z)+O\mathopen{}\mathclose{{}\left(\frac{\mathcal{X}_{j}(z)}{K}}\right)\qquad(z\in\Omega),$

where $c_{0}=\tfrac{1}{4}\zeta(2)$ and $c_{1}=\tfrac{3}{4}\zeta(2)$, and it
follows from Lemma 3.2 that

$\Psi_{j}(z)=c_{j}\mathcal{X}_{j}(z)+O\mathopen{}\mathclose{{}\left(\frac{\mathcal{X}_{j}(z)}{K}}\right)+O\mathopen{}\mathclose{{}\left(\frac{X}{K}}\right)\qquad(z\in\Omega).$
(3.5)

If instead of (3.4) one has $F_{j}(z^{k})\ll\mathcal{X}_{j}(z)/k$ for
$z\in\Omega$, then one concludes that

$\Psi_{j}(z)\ll\mathcal{X}_{j}(z)+X/K\qquad(z\in\Omega)$

in place of (3.5). The above discussions are summarized in the following
lemma.

######  Lemma 3.4.

Let $f:\mathbb{N}\to\\{0,\pm 1\\}$, and for $j\in\\{0,1\\}$ let $F_{j}(z)$ and
$\Psi_{j}(z)$ be defined as in (3.1) and (3.2), so that

$\Psi(z)=\sum_{k\equiv 0\,(2)}\frac{1}{k}F_{0}(z^{k})+\sum_{k\equiv
1\,(2)}\frac{1}{k}F_{1}(z^{k})=\Psi_{0}(z)+\Psi_{1}(z)\qquad(|z|<1).$

Fixing $j\in\\{0,1\\}$, if for $z\in\Omega$ and $1\leq k\leq K$ one has
$F_{j}(z^{k})=\mathcal{X}_{j}(z)/k$, then

$\Psi_{j}(z)=c_{j}\mathcal{X}_{j}(z)+O\mathopen{}\mathclose{{}\left(\frac{\mathcal{X}_{j}(z)}{K}}\right)+O\mathopen{}\mathclose{{}\left(\frac{X}{K}}\right)\qquad(z\in\Omega),$
(3.6)

where $c_{0}=\tfrac{1}{4}\zeta(2)$ and $c_{1}=\tfrac{3}{4}\zeta(2)$. If
instead of the relation $F_{j}(z^{k})=\mathcal{X}_{j}(z)/k$ one has that
$F_{j}(z^{k})\ll\mathcal{X}_{j}(z)/k$, then in place of equation (3.6) one
concludes that

$\Psi_{j}(z)\ll\mathcal{X}_{j}(z)+X/K\qquad(z\in\Omega).$

##  4\. The minor arcs

We now begin our specific analyses of $\Psi(z,\mu)$, so $\Psi(z)$ always
indicates $\Psi(z,\mu)$ unless otherwise stated. As $\Psi_{1}(z)$ is easier to
estimate than $\Psi_{0}(z)$, we analyze $\Psi_{1}(z)$ first.

######  Lemma 4.1.

Fix $A>0$. For $X>X_{0}(A)$ and $\alpha\in\mathfrak{m}(X,Q)$ one has

$\Psi_{1}(\rho e(\alpha))\ll X(\log X)^{-A}.$

###### Proof.

With $f=\mu$, the function $F_{1}(z)$ of (3.1) is
$\sum_{n=1}^{\infty}\mu(n)z^{n}$, whereby

$\Psi_{1}(\rho e(\alpha))=\sum_{k\equiv
1\,(2)}\frac{1}{k}F_{1}(\rho^{k}e(k\alpha))=\sum_{k\equiv
1\,(2)}\frac{1}{k}\sum_{n=1}^{\infty}\mu(n)e^{-nk/X}e(nk\alpha).$

Let $k\leq X^{1/2}$. Integrating by parts, we find that

$F_{1}(\rho^{k}e(k\alpha))=\frac{k}{X}\int_{0}^{\infty}e^{-tk/X}S_{\mu}(t,k\alpha)\,dt.$
(4.1)

When $t\geq X^{3/4}/k$, we bound $S_{\mu}(t,k\alpha)$ using Davenport’s
inequality [davenport1937some]:

$S_{\mu}(t,\theta)\ll t(\log t)^{-A}\qquad\text{for $t\geq 3$, uniformly in
$\theta$.}$ (4.2)

Trivially bounding $S_{\mu}(t,k\alpha)\ll t$ when $t<X^{3/4}/k$, it follows
that integral (4.1) is

$\displaystyle\ll\frac{k}{X}\mathopen{}\mathclose{{}\left[\int_{0}^{X^{3/4}k^{-1}}e^{-tk/X}t\,dt+\int_{X^{3/4}k^{-1}}^{\infty}e^{-tk/X}t(\log
t)^{-A}\,dt}\right]$
$\displaystyle\ll\frac{X}{k}\mathopen{}\mathclose{{}\left[\int_{0}^{X^{-1/4}}e^{-u}u\,du+\log^{-A}(X^{3/4}k^{-1})\int_{X^{-1/4}}^{\infty}e^{-u}u\,du}\right],$
(4.3)

and we note that the assumption that $k\leq X^{1/2}$ ensures that our use of
inequality (4.2) remains valid as $X\to\infty$. We also note the change of
$k/X$ to $X/k$ in expression (4.3). As $\int_{0}^{a}e^{-u}u\,du\ll a$ and
$\int_{a}^{\infty}e^{-u}u\,du\ll 1$ when $0<a<1$, we deduce that

$F_{k}(\rho^{k}e(k\alpha))\ll\frac{X}{k}\Big{[}X^{-\frac{1}{4}}+\log^{-A}(X^{\frac{3}{4}}k^{-1})\Big{]}\ll\frac{X/k}{(\log
X)^{A}}\qquad(k\leq X^{1/2}).$

Applying Lemma 3.4 with $K=X^{1/2}$ we find that

$\Psi_{1}(\rho e(\alpha))\ll X(\log X)^{-A}+X/X^{1/2},$

and the result follows. ∎

######  Remark 4.2.

The reader may notice that the proof of Lemma 4.1 did not utilize the
assumption that $\alpha\in\mathfrak{m}(X,Q)$. Indeed, because Davenport’s
inequality (4.2) is uniform in the variable $\theta$ (which became $k\alpha$
above), we immediately deduce the following corollary.

######  Corollary 4.3.

For fixed $A>0$, real $\alpha$, and all $X>X_{0}(A)$, one has that

$\Psi_{1}(\rho e(\alpha))\ll X(\log X)^{-A}.$

The analysis of $\Psi_{0}(\rho e(\alpha))$ is much more involved than that of
$\Psi_{1}(\rho e(\alpha))$, as we require a suitable replacement for
Davenport’s inequality (4.2) for the sums $S_{\mu^{2}}(t,\theta)$. Our
starting point is a theorem of Brüdern and Perelli that is built on previous
work of Brüdern, Granville, Perelli, Vaughan, and Wooley
[brudern1998exponential].

######  Lemma 4.4 ([brudern1999exponential], Thm. 4).

If $1\leq Q\leq X^{3/7}$, then

$\sup_{\theta\in\mathfrak{m}(X,Q)}\mathopen{}\mathclose{{}\left|S_{\mu^{2}}(X,\theta)}\right|\ll
X^{1+\varepsilon}Q^{-1}.$ (4.4)

To motivate how Lemma 4.4 is used to establish our substitute for inequality
(4.2) (and ultimately our bound on $\Psi_{0}$), we discuss a technical item to
be addressed. Namely, to apply Lemma 3.4 we need that
$F_{0}(\rho^{k}e(k\alpha))\ll\mathcal{X}_{0}(\rho^{k}e(k\alpha))/k$ for $k\leq
K$, with $\mathcal{X}_{0}$ not depending on $k$. However, the multiples
$k\alpha$ may not all be contained in $\mathfrak{m}(X,Q)$ as is necessary to
apply the bound (4.4). In fact, it may be that some $k\alpha>1$; however, as
$S_{f}(t,\theta)=S_{f}(t,\\{\theta\\})$ for any $f$, it suffices to consider
the fractional parts $\\{k\alpha\\}$ of the $k\alpha$.

Suppose for the moment that $K$, $\mathcal{X}$, and $\mathcal{Q}$ are
quantities such that $\\{k\alpha\\}\in\mathfrak{m}(\mathcal{X},\mathcal{Q})$
for all $\alpha\in\mathfrak{m}(X,Q)$ and $k\leq K$. In this case we have (for
a general $f$) that

$\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}(X,Q)\\\ k\leq
K\end{subarray}}|S_{f}(t,k\alpha)|=\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}(X,Q)\\\
k\leq
K\end{subarray}}|S_{f}(t,\\{k\alpha\\})|\leq\sup_{\theta\in\mathfrak{m}(\mathcal{X},\mathcal{Q})}|S_{f}(t,\theta)|.$
(4.5)

Next, considering the definitions of the sets
$\mathfrak{M}(\mathcal{X},\mathcal{Q})$ and
$\mathfrak{m}(\mathcal{X},\mathcal{Q})$, one sees that if $t\geq\mathcal{X}$,
then
$\mathfrak{m}(\mathcal{X},\mathcal{Q})\subset\mathfrak{m}(t,\mathcal{Q})$.
From (4.5) then, for all $t\geq\mathcal{X}$ we have that

$\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}(X,Q)\\\ k\leq
K\end{subarray}}|S_{f}(t,k\alpha)|\leq\sup_{\theta\in\mathfrak{m}(\mathcal{X},\mathcal{Q})}|S_{f}(t,\theta)|\leq\sup_{\theta\in\mathfrak{m}(t,\mathcal{Q})}|S_{f}(t,\theta)|.$
(4.6)

Thus we need only ensure that we may apply Lemma 4.4 to the final expression
in (4.6) with $f=\mu^{2}$. These ideas are summarized in the following lemma.

######  Lemma 4.5.

Suppose that $1\leq Q\leq X$, that $1\leq\mathcal{Q}\leq\mathcal{X}^{3/7}$,
that $K>0$, and that: For all $\alpha\in\mathfrak{m}(X,Q)$ and $k\leq K$ one
has $\\{k\alpha\\}\in\mathfrak{m}(\mathcal{X},\mathcal{Q})$. Then for
$t\geq\mathcal{X}$ one has

$\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}(X,Q)\\\ k\leq
K\end{subarray}}|S_{\mu^{2}}(t,k\alpha)|\leq\sup_{\theta\in\mathfrak{m}(t,\mathcal{Q})}|S_{\mu^{2}}(t,\theta)|\ll
t^{1+\varepsilon}\mathcal{Q}^{-1}.$ (4.7)

###### Proof.

Immediate by inequality (4.6) and Lemma 4.4. ∎

We now consider how to select $K$, $\mathcal{X}$, and $\mathcal{Q}$. The
simple choices $\mathcal{X}=X$ and $\mathcal{Q}=Q$ yield the estimate
$\Psi(\rho e(\alpha))\ll X^{1+\varepsilon}$, which is insufficient since
$\Psi(\rho e(\alpha))$ is comparable to $X$ when $\alpha\in\mathfrak{M}(X,Q)$.
Examining Lemma 4.4 and the proof of Lemma 4.1, one may deduce that
_decreasing_ $\mathcal{X}$ has a far greater effect on the resulting estimates
of $\Psi_{0}(\rho e(\alpha))$ than does _increasing_ $\mathcal{Q}$, so we
prioritize decreasing $\mathcal{X}$. The quantities $\mathcal{X}$ and
$\mathcal{Q}$ are linked through the set
$\mathfrak{m}(\mathcal{X},\mathcal{Q})$ however, so we must consider an
“exchange” between them. The mechanism of this exchange is shown in the
following lemma.

######  Lemma 4.6.

Fix $\kappa\in(0,1)$, let $1\leq Q\leq X$, let $\alpha\in\mathfrak{m}(X,Q)$,
and let $1\leq k\leq Q^{\kappa}$. If $(a_{k},q_{k})=1$ and

$\mathopen{}\mathclose{{}\left|q_{k}k\alpha-a_{k}}\right|\leq Q^{1-\kappa}/X,$
(4.8)

then $q_{k}>Q^{1-\kappa}$. In particular one has
$\\{k\alpha\\}\in\mathfrak{m}(X,Q^{1-\kappa})$. Finally, if $0<\nu<1-\kappa$
then

$\mathfrak{m}(X,Q^{1-\kappa})\subset\mathfrak{m}(XQ^{-\nu},Q^{1-\kappa-\nu}).$
(4.9)

###### Proof.

We begin by recalling that $\mathfrak{m}(X,Q)$ consists of those
$\beta\in[0,1)$ such that: if $0\leq b<r$, $(b,r)=1$, and $|r\beta-b|\leq
Q/X$, then it holds that $r>Q$.

By Dirichlet’s approximation theorem there exist $q_{k}\in\mathbb{N}$ and
$a_{k}\in\mathbb{Z}$ with $q_{k}\leq X/Q$, $(a_{k},q_{k})=1$, and
$|q_{k}k\alpha-a_{k}|\leq Q/X$. By writing
$a_{k}/(kq_{k})=a_{k}^{\prime}/(k^{\prime}q_{k})$ with
$(a_{k}^{\prime},k^{\prime}q_{k})=1$ and $k^{\prime}\leq k$, we deduce that

$\mathopen{}\mathclose{{}\left|\alpha-\frac{a_{k}^{\prime}}{k^{\prime}q_{k}}}\right|=\mathopen{}\mathclose{{}\left|\alpha-\frac{a_{k}}{kq_{k}}}\right|\leq\frac{Q}{kq_{k}X}\leq\frac{Q}{k^{\prime}q_{k}X}.$

As $(a_{k}^{\prime},k^{\prime}q_{k})=1$ and $\alpha\in\mathfrak{m}(X,Q)$, we
thus have $k^{\prime}q_{k}>Q$ and $q_{k}>Q/k^{\prime}\geq Q^{1-\kappa}$. Again
using Dirichlet’s approximation theorem, there exist $0\leq b_{k}<r_{k}$ with
$(b_{k},r_{k})=1$ and $|r_{k}\\{k\alpha\\}-b_{k}|\leq Q/X$. Setting
$a_{k}:=b_{k}+r_{k}\lfloor k\alpha\rfloor$, we have that $(a_{k},r_{k})=1$ and
$|r_{k}k\alpha-a_{k}|\leq Q/X$, whereby $r_{k}>Q^{1-\kappa}$ and
$\\{k\alpha\\}\in\mathfrak{m}(X,Q^{1-\kappa})$.

Now let $0<\nu<1-\kappa$ and $\beta\in\mathfrak{m}(X,Q^{1-\kappa})$, and
suppose that $(b,r)=1$ and

$|r\beta-b|\leq Q^{1-\kappa-\nu}/(XQ^{-\nu}).$ (4.10)

Of course $|r\beta-b|\leq Q^{1-\kappa}/X$ then, whereby $r>Q^{1-\kappa}\geq
Q^{1-\kappa-\nu}$, yielding the result. ∎

We now establish the desired substitute for Davenport’s inequality (4.2).

######  Lemma 4.7.

Let $X$ be sufficiently large, and let $\delta,\kappa,\nu\in(0,1)$ be such
that $\nu<1-\kappa$ and $\delta(1-\kappa-\nu)\leq\tfrac{3}{7}(1-\delta\nu)$.
Let $Q=X^{\delta}$. For all $x\geq XQ^{-\nu}$ one has

$\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}(X,Q)\\\ k\leq
Q^{\kappa}\end{subarray}}\mathopen{}\mathclose{{}\left|S_{\mu^{2}}(x,k\alpha)}\right|\ll
x^{1+\varepsilon}Q^{\kappa+\nu-1}.$ (4.11)

###### Proof.

By Lemma 4.6, we have
$\\{k\alpha\\}\in\mathfrak{m}(XQ^{-\nu},Q^{1-\kappa-\nu})$ for all
$\alpha\in\mathfrak{m}(X,Q)$ and $k\leq Q^{\kappa}$. As
$Q^{1-\kappa-\nu}\leq(XQ^{-\nu})^{3/7}$ by our assumptions, we apply Lemma 4.5
with $\mathcal{X}=XQ^{-\nu}$, $\mathcal{Q}=Q^{1-\kappa-\nu}$, and
$K=Q^{\kappa}$ to conclude that

$\sup_{\begin{subarray}{c}\alpha\in\mathfrak{m}(X,Q)\\\ k\leq
Q^{\kappa}\end{subarray}}\mathopen{}\mathclose{{}\left|S_{\mu^{2}}(t,k\alpha)}\right|\ll
t^{1+\varepsilon}(Q^{1-\kappa-\nu})^{-1}\qquad(t\geq XQ^{-\nu}).$ (4.12)

∎

At last we estimate $\Psi_{0}(\rho e(\alpha))$, and thereby $\Psi(\rho
e(\alpha))$, for all $\alpha\in\mathfrak{m}(X,Q)$.

######  Proposition 4.8.

Let $A>0$ be fixed and let $Q=X^{2/5}$. For all $X>X_{0}(A)$ and all
$\alpha\in\mathfrak{m}(X,Q)$ one has

$\Psi(\rho e(\alpha))\ll X(\log X)^{-A}.$

###### Proof.

Since $\Psi_{1}(\rho e(\alpha))\ll X(\log X)^{-A}$ by Lemma 4.1, it is more
than sufficient to show that $\Psi_{0}(\rho e(\alpha))\ll X^{9/10}$. Recalling
from (3.1) that $F_{0}(z)=\sum_{n=1}^{\infty}\mu^{2}(n)z^{n}$, we sum by parts
to find that

$F_{0}(\rho^{k}e(k\alpha))=\sum_{n=1}^{\infty}\mu^{2}(n)\rho^{nk}e(nk\alpha)=\frac{k}{X}\int_{0}^{\infty}e^{-tk/X}S_{\mu^{2}}(t,k\alpha)\,dt.$

For the sake of exposition we let $\delta,\kappa,\nu\in(0,1)$ be unknowns such
that $\nu<1-\kappa$ and $\delta(1-\kappa-\nu)\leq\tfrac{3}{7}(1-\delta\nu)$,
let $Q=X^{\delta}$, and let $k\leq Q^{\kappa}$. Bounding
$S_{\mu^{2}}(t,k\alpha)\ll t$ when $t\leq XQ^{-\nu}$ and using (4.12) when
$t>XQ^{-\nu}$, we have

$\displaystyle F_{0}(\rho^{k}e(k\alpha))$
$\displaystyle\ll\frac{k}{X}\mathopen{}\mathclose{{}\left[\int_{0}^{XQ^{-\nu}}e^{-tk/X}t\,dt+Q^{\kappa+\nu-1}\int_{XQ^{-\nu}}^{\infty}e^{-tk/X}t^{1+\varepsilon}\,dt}\right]$
$\displaystyle=\frac{X}{k}\mathopen{}\mathclose{{}\left[\int_{0}^{kQ^{-\nu}}e^{-u}u\,du+Q^{\kappa+\nu-1}\mathopen{}\mathclose{{}\left(\frac{X}{k}}\right)^{\varepsilon}\int_{kQ^{-\nu}}^{\infty}e^{-u}u\,du}\right]$
$\displaystyle\ll\frac{X}{k}\Big{[}1-\mathopen{}\mathclose{{}\left(1+kQ^{-\nu}}\right)\exp\mathopen{}\mathclose{{}\left(-kQ^{-\nu}}\right)+Q^{\kappa+\nu-1}X^{\varepsilon}\Big{]}.$

As $k\leq Q^{\kappa}$ and the function $1-(1+t)e^{-t}$ is increasing in $t$,
this is

$\ll\frac{X}{k}\Big{[}1-(1+Q^{\kappa-\nu})\exp(-Q^{\kappa-\nu})+Q^{\kappa+\nu-1}X^{\varepsilon}\Big{]}.$
(4.13)

If, in addition to our previous assumptions, we suppose that $\kappa-\nu<0$,
then the inequality $1-(1+t)e^{-t}\ll t^{2}$ (for small $t$) and the fact that
$Q$ may be assumed to be large (since $Q=X^{\delta}$ and $X$ is large)
together imply that (4.13) is

$\ll\frac{X}{k}\Big{[}Q^{2(\kappa-\nu)}+Q^{\kappa+\nu-1}X^{\varepsilon}\Big{]}=\frac{X}{k}\Big{[}X^{-2\delta(\nu-\kappa)}+X^{-\delta(1-\kappa-\nu)+\varepsilon}\Big{]},$
(4.14)

and taking $\varepsilon<\tfrac{1}{5}\delta(1-\kappa-\nu)$ we deduce that

$F_{0}(\rho^{k}e(k\alpha))\ll\frac{1}{k}\mathopen{}\mathclose{{}\left[X^{1-2\delta(\nu-\kappa)}+X^{1-\frac{4}{5}\delta(1-\kappa-\nu)}}\right].$
(4.15)

Applying Lemma 3.4 using (4.15) and $K=Q^{\kappa}=X^{\delta\kappa}$, it
follows that

$\Psi_{0}(\rho e(\alpha))\ll
X^{1-2\delta(\nu-\kappa)}+X^{1-\frac{4}{5}\delta(1-\kappa-\nu)}+X^{1-\delta\kappa}.$

Setting $\delta=\tfrac{2}{5}$, $\kappa=\tfrac{1}{4}$, and $\nu=\tfrac{2}{5}$,
we see that these satisfy the conditions

$\nu<1-\kappa,\quad\kappa<\nu,\quad\text{and}\quad\delta(1-\kappa-\nu)\leq\tfrac{3}{7}(1-\delta\nu)$

and yield the bound

$\Psi_{0}(\rho e(\alpha))\ll
X^{\frac{17}{25}}+X^{\frac{111}{125}}+X^{\frac{9}{10}}\ll X^{\frac{9}{10}},$

and the proposition follows. ∎

##  5\. The major arcs

We now analyze $\Psi(\rho e(\alpha))$ for $\alpha\in\mathfrak{M}(X,Q)$, which
includes when $\alpha$ in $\mathfrak{P}$ or $\mathfrak{P}_{*}$. Although the
choice $Q=X^{2/5}$ mentioned in the proof of Proposition 4.8 is ultimately
used, we again treat $Q$ as a free parameter until this is choice is
necessary.

In truth, we need not further analyze $\Psi_{1}(\rho e(\alpha))$ since
Corollary 4.3 provides a satisfactory estimate already. While our analysis of
$\Psi_{0}(\rho e(\alpha))$ on the minor arcs requires estimates of the
exponential sums $S_{\mu^{2}}(t,\theta)$, our analysis on the major arcs
requires results on sums of $\mu^{2}(n)$ as $n$ runs over arithmetic
progressions.

We recall that $\mu(n)\neq 0$ if and only if $p^{2}\nmid n$ for all prime $p$
and that such $n$ said to be squarefree or _quadratfrei_. For positive $q$ and
$r$, not necessarily coprime, we define

$\mathcal{Q}(t;q,r)=\sum_{\begin{subarray}{c}n\leq t\\\ n\equiv
r\,(q)\end{subarray}}\mu^{2}(n),$

which is simply the number of squarefree $n\leq t$ that are congruent to $r$
modulo $q$. The following function determines the asymptotic behavior of
$\mathcal{Q}(t;q,r)$, so it is convenient to define it now.

######  Definition 5.1.

If $(q,r)$ is squarefree set

$g(q,r):=\sum_{\begin{subarray}{c}n=1\\\ (n^{2},q)\hskip 0.27995pt|\hskip
0.27995ptr\end{subarray}}^{\infty}\frac{\mu(n)(n^{2},q)}{n^{2}q},$ (5.1)

and set $g(q,r):=0$ otherwise.

We note that (5.1) immediately implies that $g(q,r)=g(q,(q,r))$. Although the
following result is well known, we include a proof here for completeness.

######  Lemma 5.2.

For all positive integers $q$ and $r$ and all sufficiently large $t$, one has

$\mathcal{Q}(t;q,r)=g(q,r)t+O(t^{1/2})$

with implicit constant independent of $q$ and $r$.

###### Proof.

Writing $n=ab^{2}$ with $a$ squarefree and using the well known identity

$\mu^{2}(n)=\sum_{d^{2}\mid n}\mu(d),$

we rearrange terms in the sum $\mathcal{Q}(t;q,r)$ to find that

$\mathcal{Q}(t;q,r)=\sum_{b\leq\sqrt{t}}\mu(b)\bigg{(}\sum_{\begin{subarray}{c}a\leq
t/b^{2}\\\ ab^{2}\equiv r\,(q)\end{subarray}}1\bigg{)}=\sum_{d\hskip
0.27995pt|\hskip 0.27995ptr}\sum_{\begin{subarray}{c}b\leq\sqrt{t}\\\
(b^{2},q)=d\end{subarray}}\mu(b)\bigg{(}\sum_{\begin{subarray}{c}a\leq
t/b^{2}\\\ ab^{2}\equiv r\,(q)\end{subarray}}1\bigg{)}.$

For any $b$ let $d=(b^{2},q)$, let $b^{2}=\beta d$, let $q=\theta d$, and let
$r=\rho d$. Then $ab^{2}\equiv r\,(q)$ if and only if
$a\beta\equiv\rho\,(\theta)$ and, defining $\beta^{\prime}$ so that
$\beta^{\prime}\beta\equiv 1\,(\theta)$ (valid since $(\beta,\theta)=1$), we
have that $ab^{2}\equiv r\,(q)$ if and only if
$a\equiv\beta^{\prime}\rho\,(\theta)$. It follows that

$\mathcal{Q}(t;q,r)=\sum_{d\hskip 0.27995pt|\hskip
0.27995ptr}\sum_{\begin{subarray}{c}b\leq\sqrt{t}\\\
(b^{2},q)=d\end{subarray}}\mu(b)\bigg{(}\sum_{\begin{subarray}{c}a\leq
t/b^{2}\\\
a\equiv\beta^{\prime}\rho\,(\theta)\end{subarray}}1\bigg{)}=\sum_{d\hskip
0.27995pt|\hskip 0.27995ptr}\sum_{\begin{subarray}{c}b\leq\sqrt{t}\\\
(b^{2},q)=d\end{subarray}}\mu(b)\mathopen{}\mathclose{{}\left(\frac{t}{b^{2}\theta}+O\mathopen{}\mathclose{{}\left(1}\right)}\right).$

Writing $1/(b^{2}\theta)=d/(b^{2}q)$ we deduce that

$\mathcal{Q}(t;q,r)=t\sum_{\begin{subarray}{c}b\leq\sqrt{t}\\\ (b^{2},q)\hskip
0.27995pt|\hskip
0.27995ptr\end{subarray}}\frac{\mu(b)(b^{2},q)}{b^{2}q}+O(t^{1/2}),$ (5.2)

noting the implicit constant to be independent of $q$ and $r$. Extending the
sum over $b$ in equation (5.2) to sum over all $b\geq 1$ at a cost of
$O(t^{-1/2})$, we conclude that

$\mathcal{Q}(t;q,r)=t\bigg{(}\sum_{\begin{subarray}{c}n=1\\\ (n^{2},q)\hskip
0.27995pt|\hskip
0.27995ptr\end{subarray}}^{\infty}\frac{\mu(n)(n^{2},q)}{n^{2}q}\bigg{)}+O\mathopen{}\mathclose{{}\left(t^{1/2}}\right)$

as claimed, and the implicit constant is again independent of $q$ and $r$. ∎

We now record a lemma on an exponential sum involving the constants $g(q,r)$.

######  Definition 5.3.

Let $G(q)$ be the multiplicative function defined on prime powers by

$G(p^{i})=\begin{cases}-(p^{2}-1)^{-1}&i=1,2,\\\
0&\text{otherwise.}\end{cases}$

With this $G(q)$, as shown in [vaughan2005variance]*Lem. 2.5, for all $q$ one
has

$\sum_{r=1}^{q}e(r/q)g(q,r)=\zeta(2)^{-1}G(q).$ (5.3)

We now derive the primary asymptotic formula for our major arcs.

######  Proposition 5.4.

Fix $A>0$, let $Q=X^{2/5}$, and suppose that $\alpha=a/q+\beta$ with $0\leq
a<q\leq Q$, $(a,q)=1$, and $|\beta|\leq Q/(qX)$. There exists a real constant
$V(q)$ such that, as $X\to\infty$, one has

$\Psi(\rho e(\alpha))=V(q)\mathopen{}\mathclose{{}\left(\frac{X}{1-2\pi
iX\beta}}\right)+O(X(\log X)^{-A}).$ (5.4)

Moreover, one has $V(1)=V(2)=1/4$, $V(4)=0$, and $|V(q)|\leq 1/5$ for all
other $q$. Finally, the implicit constant in formula (5.4) is independent of
$q$ and $a$.

######  Remark 5.5.

The constants $V(q)$ have explicit formula

$V(q)=\zeta(2)^{-1}\sum_{d\hskip 0.27995pt\mid\hskip
0.27995ptq}G\mathopen{}\mathclose{{}\left(\frac{q}{d}}\right)\sum_{\begin{subarray}{c}k\equiv
0\,(2)\\\ (k,q)=d\end{subarray}}\frac{1}{k^{2}}$

as seen in the following proof. This formula is omitted from the above
proposition to simplify its statement and because only those values of $V(q)$
listed therein are relevant for our analysis of $p(x,\mu)$.

###### Proof of Proposition 5.4..

Let $\tau:=X^{-1}(1-2\pi iX\beta)$ so that $\rho e(\alpha)=e(a/q)e^{-\tau}$.
As $\Psi_{1}(\rho e(\alpha))\ll X(\log X)^{-A}$ for $X>X_{0}(A)$, it suffices
to show that

$\Psi_{0}(\rho e(\alpha))=V(q)\tau^{-1}+O(X^{\frac{9}{10}}).$ (5.5)

For all $k\in\mathbb{N}$ set

$q_{k}:=q/(q,k)\qquad\text{and}\qquad a_{k}:=ak/(q,k).$

Recalling definition (3.1) we have
$F_{0}(z)=\sum_{n=1}^{\infty}\mu^{2}(n)z^{n}$, and grouping summands into
residue classes modulo $q_{k}$ we have

$F_{0}(\rho^{k}e(k\alpha))=\sum_{n=1}^{\infty}\mu^{2}(n)e\mathopen{}\mathclose{{}\left(\frac{nka}{q}}\right)e^{-nk\tau}=\sum_{r=1}^{q_{k}}e\mathopen{}\mathclose{{}\left(\frac{a_{k}r}{q_{k}}}\right)\sum_{n\equiv
r\,(q_{k})}\mu^{2}(n)e^{-nk\tau}.$ (5.6)

Summing by parts and applying Lemma 5.2, we find that

$\sum_{n\equiv
r\,(q_{k})}\mu^{2}(n)e^{-nk\tau}=k\tau\int_{0}^{\infty}e^{-tk\tau}\mathcal{Q}(t;q_{k},r)\,dt=k\tau\int_{0}^{\infty}e^{-tk\tau}(tg(q_{k},r)+O(t^{\frac{1}{2}}))\,dt.$
(5.7)

As $|\tau|\ll X^{-1}(1+X|\beta|)\ll Q/(qX)$, the error term here is

$k\tau\int_{0}^{\infty}e^{tk\tau}O(t^{1/2})\ll\frac{kQ}{qX}\int_{0}^{\infty}e^{-tk/X}t^{1/2}\,dt\ll\frac{kQ}{qX}\mathopen{}\mathclose{{}\left(\frac{X}{k}}\right)^{3/2}=\frac{Q}{q}\mathopen{}\mathclose{{}\left(\frac{X}{k}}\right)^{1/2},$
(5.8)

and we remark that it is in equation (5.7) and inequality (5.8) that the
independence (from $q_{k}$ and $r$) of the constant in Lemma 5.2 is used.

To consider the remaining integral in equation (5.7), we first rewrite

$g(q_{k},r)\int_{0}^{\infty}e^{-tk\tau}tk\tau\,dt=(k\tau)^{-1}g(q_{k},r)\int_{0}^{\infty}e^{-tk\tau}tk\tau\,d(tk\tau).$
(5.9)

Observing that $|\arg k\tau|=|\arg(1-2\pi iX\beta)|<\frac{\pi}{2}$ for all
$|\beta|\leq Q/qX$, we recognize the integral in (5.9) to be equal to
$\Gamma(2)$ (see, e.g., [montgomery2007multiplicative]*App. C) and conclude
that

$g(q_{k},r)\int_{0}^{\infty}e^{-tk\tau}tk\tau\,dt=\frac{g(q_{k},r)}{k\tau}.$
(5.10)

Returning to equation (5.6) and using (5.8) and (5.10), we find that

$\sum_{r=1}^{q_{k}}e\mathopen{}\mathclose{{}\left(\frac{a_{k}r}{q_{k}}}\right)\sum_{n\equiv
r\,(q_{k})}\mu^{2}(n)e^{-nk\tau}=\frac{1}{k\tau}\sum_{r=1}^{q_{k}}e\mathopen{}\mathclose{{}\left(\frac{a_{k}r}{q_{k}}}\right)g(q_{k},r)+O\mathopen{}\mathclose{{}\left(\frac{X^{1/2}Q}{k^{1/2}}}\right).$
(5.11)

As $(a_{k},q_{k})=1$, we define $a_{k}^{\prime}$ so that
$a_{k}a_{k}^{\prime}\equiv 1\,(q_{k})$ and recall from (5.1) that
$g(q,r)=g(q,(q,r))$ to deduce that

$g(q_{k},r)=g(q_{k},(q_{k},r))=g(q_{k},(q_{k},ra_{k}^{\prime}))=g(q_{k},ra_{k}^{\prime}).$
(5.12)

By equation (5.12) one has

$\sum_{r=1}^{q_{k}}e(a_{k}r/q_{k})g(q_{k},r)=\sum_{r=1}^{q_{k}}e(r/q_{k})g(q_{k},ra_{k}^{\prime})=\sum_{r=1}^{q_{k}}e(r/q_{k})g(q_{k},r),$
(5.13)

which by equation (5.3) is equal to $\zeta(2)^{-1}G(q_{k})$.

Combining equations (5.7), (5.11), and (5.13) we conclude that

$F_{0}(\rho^{k}e(k\alpha))=\frac{G(q_{k})}{\tau
k\zeta(2)}+O\mathopen{}\mathclose{{}\left(\frac{X^{1/2}Q}{k^{1/2}}}\right),$
(5.14)

but we note that Lemma 3.4 cannot be applied here, as the quantity
$\mathcal{X}_{0}(z)$ indicated by (5.14) is dependent on $k$. Thus considering
the definition of $\Psi_{0}(z)$ directly, it follows from (5.14) that

$\Psi_{0}(\rho e(\alpha))=\sum_{k\equiv
0\,(2)}\frac{1}{k}F_{0}(\rho^{k}e(k\alpha))=(\tau\zeta(2))^{-1}\sum_{k\equiv
0\,(2)}\frac{G(q_{k})}{k^{2}}+O(X^{1/2}Q),$ (5.15)

and we are motivated to define

$V(q)=\zeta(2)^{-1}\sum_{k\equiv 0\,(2)}\frac{G(q_{k})}{k^{2}}.$ (5.16)

The quantity $V(q)$ is clearly real-valued, and since the sum (5.16) converges
absolutely (note that $|G|<1$), we set $d:=(q,k)$ and rearrange terms to see
that

$V(q)=\zeta(2)^{-1}\sum_{d\hskip 0.27995pt\mid\hskip
0.27995ptq}G\mathopen{}\mathclose{{}\left(\frac{q}{d}}\right)\sum_{\begin{subarray}{c}k\equiv
0\,(2)\\\ (k,q)=d\end{subarray}}\frac{1}{k^{2}}.$ (5.17)

Using formula (5.17), the remaining properties of $V(q)$ can be verified
directly—albeit tediously—so their demonstrations are relegated to the
appendix.

So that the $O(X^{1/2}Q)$ term in equation (5.15) is $o(X)$, it suffices to
take $Q=X^{2/5}$, and equation (5.5) (and thereby (5.4)) is established. We
finally note that the implicit constants throughout our proof of equation
(5.5) are independent of $q$ and $a$, and therefore the implicit constant in
equation (5.4) is independent of $q$ and $a$ as well. ∎

##  6\. The principal arcs I

We recall that the principal arcs are

$\displaystyle\mathfrak{P}$ $\displaystyle=\\{\alpha\in[0,1):\|\alpha\|\leq
X^{-1}(\log X)^{-1/4}\\},$ $\displaystyle\mathfrak{P}_{*}$
$\displaystyle=\\{\alpha\in[0,1):|\alpha-\tfrac{1}{2}|\leq X_{*}^{-1}(\log
X_{*})^{-1/4}\\}.$

For brevity, let

$\eta:=X^{-1}(\log X)^{-1/4}\qquad\text{and}\qquad\eta_{*}:=X_{*}^{-1}(\log
X_{*})^{-1/4}.$ (6.1)

Because $\Psi(re(\alpha))=\Psi(re(\alpha-1))$ for all $\alpha$, it is
convenient to identify the sets

$\\{e(\alpha):\alpha\in\mathfrak{P}\\}\qquad\text{and}\qquad\\{e(\beta):\text{$\beta\in\mathbb{R}$
and $|\beta|\leq\eta$}\\}$ (6.2)

and consider $\Psi(\rho e(\alpha))$ with $\alpha\in\mathfrak{P}$ as $\Psi(\rho
e(\beta))$ with $|\beta|\leq\eta$. As $e(1/2)=-1$, we may similarly identify
the sets

$\\{e(\alpha):\alpha\in\mathfrak{P}_{*}\\}\qquad\text{and}\qquad\\{-e(\beta):\text{$\beta\in\mathbb{R}$
and $|\beta|\leq\eta_{*}$}\\}$ (6.3)

and consider $\Psi(\rho_{*}e(\alpha))$ with $\alpha\in\mathfrak{P}_{*}$ as
$\Psi(-\rho_{*}e(\beta))$ with $|\beta|\leq\eta_{*}$.

Despite the above definitions’ in terms of $X_{*}$, as discussed in section 2
we do not presently distinguish $X$ and $X_{*}$ because this and the following
section do not involve solutions to the saddle-point equations (sp) and
(sp$*$). Considering this with (6.2) and (6.3) then, we see that our results
on $\Psi(\pm\rho e(\beta))$ with $|\beta|\leq\eta$ sufficiently cover the
analyses of $\Psi(\rho e(\alpha))$ with $\alpha$ in the principal arcs.

We now introduce some convenient notation for the analyses of $\Psi(\rho
e(\alpha))$ when $\alpha$ is in the principal arcs. Because the discussions
below are applicable to general $f:\mathbb{N}\to\\{0,\pm 1\\}$, we temporarily
return to discussion of a general $\Psi(z)=\Psi(z,f)$.

For complex $s$ we write $s=\sigma+it$, and $\sigma$ and $t$ always represent
the real and imaginary parts of some $s\in\mathbb{C}$, respectively. Integrals
$\frac{1}{2\pi i}\int$ are abbreviated using dashed integrals
$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int$.
Let $\mathfrak{D}[f;s]$ denote the Dirichlet series
$\sum_{n=1}^{\infty}f(n)n^{-s}$ and let $\sigma_{0}=\sigma_{0}(f)$ and
$\sigma_{1}=\sigma_{1}(f)$ denote the abcissae of absolute convergence for
$\mathfrak{D}[f^{2};s]$ and $\mathfrak{D}[f;s]$, respectively. For simplicity
all $\sigma_{0}$ and $\sigma_{1}$ below are assumed to be positive.

We recall from the definitions (3.1) and (3.2) that

$\Psi(z)=\Psi_{0}(z)+\Psi_{1}(z)=\sum_{\begin{subarray}{c}k\equiv
0\,(2)\end{subarray}}\sum_{n=1}^{\infty}\frac{f^{2}(n)}{k}z^{nk}+\sum_{k\equiv
1\,(2)}\sum_{n=1}^{\infty}\frac{f(n)}{k}z^{nk}.$

For real $\beta$ and $z=\rho e(\beta)$ we again set $\tau:=\frac{1}{X}(1-2\pi
iX\beta)$ so that $z=e^{-\tau}$. Using the well known Cahen-Mellin formula

$e^{-w}=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\sigma-i\infty}^{\sigma+i\infty}\Gamma(s)z^{-s}\,ds\qquad(\operatorname{Re}w>0,\,\,\sigma>0),$

we express $\Psi_{0}(z)$ and $\Psi_{1}(z)$ as sums of integrals and
interchange the summations and integrations using the series’ absolute
convergence to derive the formulae

$\displaystyle\Psi_{0}(z)=\Psi_{0}(e^{-\tau})$
$\displaystyle=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{1+\sigma_{0}-i\infty}^{1+\sigma_{0}+i\infty}2^{-s-1}\zeta(s+1)\mathfrak{D}[f^{2};s]\Gamma(s)\tau^{-s}\,ds,$
(6.4) $\displaystyle\Psi_{1}(z)=\Psi_{1}(e^{-\tau})$
$\displaystyle=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{1+\sigma_{1}-i\infty}^{1+\sigma_{1}+i\infty}(1-2^{-s-1})\zeta(s+1)\mathfrak{D}[f;s]\Gamma(s)\tau^{-s}\,ds.$
(6.5)

We note that if $f$ is nonnegative the distinction between $\Psi_{0}$ and
$\Psi_{1}$ is unnecessary, and that the addition of (6.4) and (6.5) yields
formula (6.2.6) of [andrews1976partitions] as expected. In general, recalling
that $\Psi_{(j)}(z):=(z\partial_{z})^{j}\Psi(z)$, for $j\geq 0$ one has

$\Psi_{(j)}(z)=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}k^{j-1}n^{j}f^{k}(n)z^{nk},$
(6.6)

yielding (after a change of variable of $s$ to $s+j$) the general formulae

$\displaystyle(\Psi_{0})_{(j)}(z)$
$\displaystyle=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{1+\sigma_{0}-i\infty}^{1+\sigma_{0}+i\infty}2^{-s-1}\zeta(s+1)\mathfrak{D}[f^{2};s]\Gamma(s+j)\tau^{-s-j}\,ds,$
(6.7) $\displaystyle(\Psi_{1})_{(j)}(z)$
$\displaystyle=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{1+\sigma_{1}-i\infty}^{1+\sigma_{1}+i\infty}(1-2^{-s-1})\zeta(s+1)\mathfrak{D}[f;s]\Gamma(s+j)\tau^{-s-j}\,ds.$
(6.8)

When $z=-\rho e(\beta)=-e^{-\tau}$ our derivations require only small changes.
As $(-z)^{nk}=z^{nk}$ when $k$ is even, one sees from equation (6.6) that
$\Psi_{0}(z)$ and the derivatives $(\Psi_{0})_{(j)}(z)$ are even functions,
whereby (6.7) may be reused for $(\Psi_{0})_{(j)}(-z)$. When $k$ is odd the
equality $(-1)^{nk}=(-1)^{n}$ shows that

$(\Psi_{1})_{(j)}(-z)=\sum_{k\equiv
1\,(2)}\sum_{n=1}^{\infty}(-1)^{n}n^{-j}k^{-j-1}f(n)z^{nk},$ (6.9)

and one derives a formula like (6.8) by replacing $\mathfrak{D}[f(n);s]$ with
$\mathfrak{D}[(-1)^{n}f(n);s]$.

##  7\. The principal arcs II

We now return to the specific study of $\Psi(z,\mu)$, again abbreviating it to
$\Psi(z)$. In addition we maintain the definitions $\eta=X^{-1}(\log
X)^{-1/4}$ and $\tau:=X^{-1}(1-2\pi iX\beta)$ so that $\pm\rho e(\beta)=\pm
e^{-\tau}$. The goal of this section is the proof of the following
proposition.

######  Proposition 7.1.

Fix $A>0$. For all $j\geq 0$, all $X>X_{0}(A)$, and all $|\beta|\leq\eta$ one
has

$\Psi_{(j)}(\pm\rho
e(\beta))=\frac{j!}{4}\mathopen{}\mathclose{{}\left(\frac{X}{1-2\pi
iX\beta}}\right)^{j+1}+O_{j}(X^{j+1}(\log X)^{-A})$ (7.1)

and

$(\pm 1)^{j}\Psi_{(j)}(\pm\rho
e(\beta))=\frac{j!}{4}\mathopen{}\mathclose{{}\left(\frac{X}{1-2\pi
iX\beta}}\right)^{j+1}+O_{j}(X^{j+1}(\log X)^{-A}).$ (7.2)

We record two useful, classical properties of $\zeta(s)$ and $\Gamma(s)$.
First, there exist universal positive constants $c$ and $t_{0}$ such that
$\zeta(s)\neq 0$ when

$|t|>t_{0}\qquad\text{and}\qquad\sigma\geq 1-\frac{c}{\log|t|},$ (7.3)

and on this region one has [montgomery2007multiplicative]*Thm. 6.6 and 6.7

$|\zeta(\sigma+it)|^{-1}\ll\log{t}\ll t^{\varepsilon}.$ (7.4)

Second, one has [paris2001asymptotics]*ineq. (2.1.19)

$|\Gamma(s)|\ll|s|^{\sigma-\frac{1}{2}}\exp\mathopen{}\mathclose{{}\left(-\tfrac{1}{2}\pi|t|+\tfrac{1}{6}|s|^{-1}}\right)\qquad(\sigma>0).$
(7.5)

######  Lemma 7.2.

Fix $A>0$. For all $j\geq 0$, all $X>X_{0}(A)$, and all $|\beta|\leq\eta$, one
has

$(\Psi_{1})_{(j)}(\rho e(\beta))\ll_{j}X^{j+1}(\log X)^{-A}.$

###### Proof.

As $\rho e(\beta)=e^{-\tau}$ and $\mathfrak{D}[\mu;s]=1/\zeta(s)$, we have
from (6.8) that

$(\Psi_{1})_{(j)}(\rho
e(\beta))=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{2-i\infty}^{2+i\infty}(1-2^{-s-1})\frac{\zeta(s+1)}{\zeta(s)}\Gamma(s+j)\tau^{-s-j}\,ds.$
(7.6)

Letting $T$ be a sufficiently large parameter of our choosing, we first
truncate integral (7.6) at height $T$ with error term

$\ll\int_{T}^{\infty}\mathopen{}\mathclose{{}\left|\frac{\Gamma(2+j+it)\tau^{-j-2-it}}{\zeta(2+it)}}\right|\,dt.$
(7.7)

Having assumed that $|\beta|\leq\eta=X^{-1}(\log X)^{-1/4}$, for sufficiently
large $X$ one has

$|\arg\tau|=|\arg(1-2\pi iX\beta)|\leq|\arctan(2\pi(\log
X)^{-1/4})|\leq\frac{\pi}{4},$

where $\arctan(x)$ has values in $(-\tfrac{\pi}{2},\tfrac{\pi}{2})$, and since
$|\tau|^{-1}\leq X$ this implies that

$|\tau^{-2-j-it}|=|\tau|^{-j-2}e^{t\arg\tau}\leq X^{j+2}e^{\pi
t/4}\qquad(t>0).$ (7.8)

Using inequalities (7.4), (7.5), and (7.8), it follows that integral (7.7) is

$\ll_{j}X^{j+2}\int_{T}^{\infty}t^{j+3/2+\varepsilon}e^{-\pi
t/4}\,dt\ll_{j}X^{j+2}e^{-\pi T/5},$

whereby

$(\Psi_{1})_{(j)}(\rho
e(\beta))=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{2-iT}^{2+iT}(1-2^{-s-1})\frac{\zeta(s+1)}{\zeta(s)}\Gamma(s+j)\tau^{-s-j}\,ds+O_{j}(X^{j+2}e^{-\pi
T/5})$ (7.9)

Let $\sigma_{*}:=1-c/\log T$, where $c$ is the constant of the zero free
region (7.3). Using Cauchy’s theorem, we see that the integral in (7.9) is

$=\mathopen{}\mathclose{{}\left[\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{2-iT}^{\sigma_{*}-iT}+\,\,\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\sigma_{*}-iT}^{\sigma_{*}+iT}+\,\,\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\sigma_{*}+iT}^{2+iT}}\right](1-2^{-s-1})\frac{\zeta(s+1)}{\zeta(s)}\Gamma(s+j)\tau^{-s-j}\,ds.$

Again using (7.4), (7.5), and (7.8), we see that the integral from
$\sigma_{*}+iT$ to $2+iT$ here is

$\ll\int_{\sigma_{*}}^{2}\mathopen{}\mathclose{{}\left|\frac{\Gamma(j+\sigma+iT)\tau^{-j-\sigma-
iT}}{\zeta(\sigma+iT)}}\right|\,d\sigma\ll_{j}X^{j+2}T^{j+3/2+\varepsilon}e^{-\pi
T/4}\ll_{j}X^{j+2}e^{-\pi T/5},$ (7.10)

and similarly for the integral form $2-iT$ to $\sigma_{*}-iT$. Using similar
arguments for the integral from $\sigma_{*}-iT$ to $\sigma_{*}+iT$, we find
that

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\sigma_{*}-iT}^{\sigma_{*}+iT}(1-2^{-s-1})\frac{\zeta(s+1)}{\zeta(s)}\Gamma(s+j)\tau^{-s-j}\,ds\ll_{j}X^{j+1-c/\log
T}\log T.$ (7.11)

Combining equation (7.9) and inequalities (7.10) and (7.11), we deduce that

$(\Psi_{1})_{(j)}(\rho
e(\beta))\ll_{j}X^{j+1}\mathopen{}\mathclose{{}\left(Xe^{-\pi
T/5}+X^{-c/\log{T}}\log{T}}\right).$ (7.12)

Now letting $T=\exp(\sqrt{\log X})$, we have that

$X\exp\mathopen{}\mathclose{{}\left(-\frac{\pi}{5}T}\right)=\exp\mathopen{}\mathclose{{}\left(\log
X-\frac{\pi}{5}e^{\sqrt{\log
X}}}\right)\ll\exp\mathopen{}\mathclose{{}\left(-\frac{\pi}{6}e^{\sqrt{\log
X}}}\right)$

and

$X^{-c/\log{T}}\log{T}=\mathopen{}\mathclose{{}\left|\frac{X}{1-2\pi
iX\beta}}\right|^{-c/\sqrt{\log X}}\sqrt{\log X}\ll e^{-c\sqrt{\log
X}}\sqrt{\log X},$

and the result follows from inequality (7.12) since

$\exp\mathopen{}\mathclose{{}\left(-\frac{\pi}{6}e^{\sqrt{\log
X}}}\right)+e^{-c\sqrt{\log X}}\sqrt{\log X}\ll(\log X)^{-A}.$

∎

The analysis of $(\Psi_{0})_{(j)}$ is largely similar to that of
$(\Psi_{1})_{(j)}$. As $\mathfrak{D}[\mu^{2};s]=\zeta(s)/\zeta(2s)$, equation
(6.7) becomes

$(\Psi_{0})_{(j)}(\rho
e(\beta))=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{2-i\infty}^{2+i\infty}2^{-s-1}\zeta(s+1)\frac{\zeta(s)}{\zeta(2s)}\Gamma(s+j)\tau^{-s-j}\,ds,$
(7.13)

and we see that the integrand here has a simple pole at $s=1$ with residue

$2^{-2}\frac{\zeta(2)}{\zeta(2)}\Gamma(j+1)\tau^{-j-1}=\frac{j!}{4}\mathopen{}\mathclose{{}\left(\frac{X}{1-2\pi
iX\beta}}\right)^{j+1}.$ (7.14)

Under the same assumptions as in Lemma 7.2, we account for this residue and
shift the line of integration in (7.13) left to
$\sigma=\tfrac{1}{2}+\varepsilon$. We then bound

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\frac{1}{2}+\varepsilon-i\infty}^{\frac{1}{2}+\varepsilon+i\infty}2^{-s-1}\zeta(s+1)\frac{\zeta(s)}{\zeta(2s)}\Gamma(s+j)\tau^{-s-j}\,ds\ll_{j}X^{j+\frac{1}{2}+\varepsilon}$

using arguments like in the proof of Lemma 7.2 and thereby deduce that

$(\Psi_{0})_{(j)}(\rho
e(\beta))=\frac{j!}{4}\mathopen{}\mathclose{{}\left(\frac{X}{1-2\pi
iX\beta}}\right)^{j+1}+O_{j}(X^{j+\frac{1}{2}+\varepsilon}).$ (7.15)

Equation (7.1) then follows at once by combining Lemmata 7.2 and equation
(7.15).

We now consider $\Psi_{(j)}(-\rho e(\beta))$ for $|\beta|\leq\eta$, again
recalling that $\eta=X^{-1}(\log X)^{-1/4}$. First, equation (6.9) shows that

$(\Psi_{1})_{(j)}(-\rho e(\beta))=(\Psi_{1})_{(j)}(-e^{-\tau})=\sum_{k\equiv
1\,(2)}\sum_{n=1}^{\infty}(-1)^{n}\mu(n)n^{j}k^{j-1}e^{-nk\tau},$

and we now require a closed formulae for
$\mathfrak{D}[(-1)^{n}\mu(n);s]=\sum_{n=1}^{\infty}(-1)^{n}\mu(n)n^{-s}$. If
$2\mid k$ then $\mu(2k)=0$, and if $2\nmid k$ then $\mu(2k)=-\mu(k)$, whereby

$\sum_{k=1}^{\infty}\frac{\mu(2k)}{(2k)^{s}}=-2^{-s}\sum_{2\,\nmid\,k}\frac{\mu(k)}{k^{s}}=-2^{-s}\prod_{\begin{subarray}{c}p\text{
prime}\\\
p>2\end{subarray}}(1-p^{-s})=-\mathopen{}\mathclose{{}\left(\frac{2^{-s}}{1-2^{-s}}}\right)\frac{1}{\zeta(s)}.$

We similarly see that $\sum_{n\equiv
1\,(2)}(-1)^{n}\mu(n)n^{-s}=-[(1-2^{-s})\zeta(s)]^{-1}$, and we deduce that

$\mathfrak{D}[(-1)^{n}\mu(n);s]=-\mathopen{}\mathclose{{}\left(\frac{1+2^{-s}}{1-2^{-s}}}\right)\frac{1}{\zeta(s)}.$
(7.16)

The only significant difference between $\mathfrak{D}[\mu;s]$ and
$\mathfrak{D}[(-1)^{n}\mu(n);s]$ is the addition of simple poles at points
$s=2\pi ik/\log 2$ for $k\in\mathbb{Z}$. Because the arguments used in proving
Lemma 7.2 only involve $s$ with $\sigma\geq 1-c/\log T$, the analysis of
$(\Psi_{1})_{(j)}(-\rho e(\beta))$ is identical to that of
$(\Psi_{1})_{(j)}(+\rho e(\beta))$, yielding the bound

$(\Psi_{1})_{(j)}(-\rho e(\beta))\ll X(\log X)^{-A},$

mutatis mutandis. Recalling the remarks at the end of section 6, the functions
$\Psi_{0}(z)$ and $(\Psi_{0})_{(j)}(z)$ are even, so by equation (7.15) we
immediately conclude that

$(\Psi_{0})_{(j)}(-\rho
e(\beta))=\frac{j!}{4}\mathopen{}\mathclose{{}\left(\frac{X}{1-2\pi
iX\beta}}\right)^{j+1}+O_{j}(X^{j+\frac{1}{2}+\varepsilon}),$

and formula (7.1) for $z=-\rho e(\beta)$ follows.

We now turn our attention to equation (7.2), following the arguments used in
[vaughan2008number]*Lem. 2.3. Arguing by induction, we find that for all
$j\geq 1$ there exist integers $s(j,k)$ for which

$z^{j}\Psi^{(j)}(z)=\sum_{k=1}^{j}s(j,k)\Psi_{(j)}(z).$ (7.17)

Moreover, because $s(j,j)=1$ for all $j$, the leading term in the sum (7.17)
is $\Psi_{(j)}(z)$. Using the assumption that $|\beta|\leq X^{-1}(\log
X)^{-1/4}$, we observe that

$\rho e(\beta)=(1+O(1/X))(1+O(\beta))=1+O(1/X),$

whereby $(\pm\rho e(\beta))^{j}=(\pm 1)^{j}+O_{j}(X^{-1})$. Using this with
equations (7.1) and (7.17), equation (7.2) is established for $z=\pm\rho
e(\beta)$ and the proof of Proposition 7.1 is complete.

######  Remark 7.3.

We note that the $s(j,k)$ in (7.17) are the _Stirling numbers of the second
kind_ from elementary combinatorics. One also has a kind of dual relation to
(7.17), in that for $j\in\mathbb{N}$ one has

$\Psi_{(j)}(z)=\sum_{k=1}^{j}S(j,k)z^{j}\Psi^{(j)}(z),$ (7.18)

where the $S(j,k)$ are the _Stirling numbers of the first kind_. Equations
(7.17) and (7.18) may both be proved using induction and the recursive
relations satisfied by the $S(j,k)$ and $s(j,k)$. See, e.g.,
[graham1994concrete]*Exer. 6.13.

##  8\. Proof of Theorem 1.2

We now examine how $\rho$ and $\rho_{*}$ (and thereby $X$ and $X_{*}$) should
depend on $x$, recalling that for sufficiently large $x$ we want $\rho$ and
$\rho_{*}$ to be solutions to the equations

$\Psi_{(1)}(\rho)=\rho\Psi^{\prime}(\rho)=x\qquad\text{and}\qquad\Psi_{(1)}(-\rho)=(-\rho)\Psi^{\prime}(-\rho)=x,$

i.e., the equations (sp) and (sp$*$), respectively.

######  Lemma 8.1.

Let $\Psi(z)=\Psi(z,\mu)$. For all $x>x_{0}$ there exist unique $\rho=\rho(x)$
and $\rho_{*}=\rho_{*}(x)$ in $(0,1)$ satisfying equations (sp) and (sp$*$),
respectively. Moreover, as $x\to\infty$ one has $\rho(x)\to 1$ and
$\rho_{*}(x)\to 1$.

###### Proof.

Setting $g(\rho):=\rho\Psi^{\prime}(\rho)$, it is evident that $g(\rho)$ is
continuous on $[0,1)$ and that $g(0)=0$. Because Proposition 7.1 shows that
$\lim_{\rho\to 1^{-}}g(\rho)=\infty$, a solution $\rho(x)\in(0,1)$ for (sp) is
guaranteed by the intermediate value theorem. Moreover, since $g(\rho)$ is
bounded on intervals $[0,\rho_{0}]\subset[0,1)$ it is necessary that
$\rho(x)\to 1$ as $x\to\infty$.

We have $g^{\prime}(\rho)=\Psi^{\prime}(\rho)+\rho\Psi^{\prime\prime}(\rho)$,
and equation (7.2) implies that $\lim_{\rho\to 1^{-}}g^{\prime}(\rho)=\infty$,
so there necessarily exists $\rho_{0}\in(0,1)$ such that $g^{\prime}(\rho)>0$
for $\rho>\rho_{0}$. It follows that if $x>x_{0}:=\sup_{[0,\rho_{0}]}g(\rho)$,
then $\rho(x)>\rho_{0}$ and the solution $\rho(x)$ to equation (sp) must be
unique since $g(\rho)$ is monotonic on $(\rho_{0},1)$.

The arguments concerning equation (sp$*$) and $\rho_{*}$ are identical. ∎

######  Remark 8.2.

Because $\rho=e^{-1/X}$ and $\rho(x)\to 1$ as $x\to\infty$, the statements “as
$x\to\infty$”, “as $X\to\infty$”, and “as $\rho\to 1$” are equivalent, whereby
asymptotic statements such as $X\sim 2\sqrt{x}$ may be made precise. For
simplicity we omit further references to this technicality, and the
dependencies of $\rho$ and $X$ on $x$ is understood in future discussions. The
same is done for $\rho_{*}$, $X_{*}$, and $x$.

Fix $A>0$. By Proposition 7.1, for $X>X_{0}(A)$ one has

$x=\tfrac{1}{4}X^{2}\mathopen{}\mathclose{{}\left[1+O((\log X)^{-A})}\right],$
(8.1)

whereby $\log x=2\log X+O\mathopen{}\mathclose{{}\left(1}\right)$, and in
particular $\log x\asymp\log X$. This implies that the term $O((\log X)^{-A})$
in equation (8.1) may be replaced with $O((\log x)^{-A})$, so that

$X=2\sqrt{x}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right]^{-\frac{1}{2}}=2\sqrt{x}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right]$ (8.2)

and

$-x\log\rho=\frac{x}{X}=\tfrac{1}{2}\sqrt{x}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right]^{-1}=\tfrac{1}{2}\sqrt{x}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right].$ (8.3)

Recalling from Proposition 7.1 that $\Psi_{(j)}(\rho)=\tfrac{1}{4}j!\hskip
0.80002ptX^{j+1}+O(X^{j+1}(\log X)^{-A})$, for $j\geq 1$ we use equation (8.2)
to deduce that

$\displaystyle\Psi_{(j)}(\rho)$ $\displaystyle=2^{j-1}j!\hskip
0.80002ptx^{\frac{j+1}{2}}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right]^{j-1}=2^{j-1}j!\hskip
0.80002ptx^{\frac{j+1}{2}}\mathopen{}\mathclose{{}\left[1+O_{j}((\log
x)^{-A})}\right].$ (8.4)

Just as equation (7.2) is proved using equation (7.17), it follows from (8.4)
that

$(\pm 1)^{j}\Psi^{(j)}(\rho)=2^{j-1}j!\hskip
0.80002ptx^{\frac{j+1}{2}}\mathopen{}\mathclose{{}\left[1+O_{j}((\log
X)^{-A})}\right],$ (8.5)

and we have established the relations of Theorem 1.2 for $\rho$ satisfying
$\rho\Psi^{\prime}(\rho)=x$.

Now considering the case of $\rho_{*}$ and $X_{*}$, again by Proposition 7.1
we have

$x=\tfrac{1}{4}X_{*}^{2}\mathopen{}\mathclose{{}\left[1+O((\log{X_{*}})^{-A})}\right].$
(8.6)

Considering the identical forms of equations (8.1) and (8.6), the analogues of
equations (8.2)–(8.5) for $-x\log(\rho_{*})$, $\Psi_{(j)}(-\rho_{*})$, etc.
follow mutatis mutandis, completing the proof of Theorem 1.2.

##  9\. Arc transference and $*$-arcs

Having considered $\Psi(\rho e(\alpha))$ for different $\alpha\in[0,1)$ and
for $\rho$ and $\rho_{*}$ as functions of $x$, we now return to the integral
formula for $p(x,\mu)$, namely the integral

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{|z|=r}\Phi(z)z^{-x-1}\,dz=r^{-x}\int_{0}^{1}\exp\Psi(\rho
e(\alpha))e(-x\alpha)\,d\alpha\qquad(0<r<1),$ (9.1)

and consider the different contributions made to (9.1) as $\alpha$ runs over
$[0,1)$. As mentioned in section 2, ultimately only those $\alpha$ in small
neighborhoods of $0$, $1$, and $1/2$ significantly contribute to this
integral.

In what follows let $x>0$ be sufficiently large. In analyzing integral (9.1)
when $\alpha$ is near $0$ and $1$, control of our error terms requires that we
take the radius $r$ of integration to be the solution $\rho$ of (sp). On the
other hand, controlling our error terms in (9.1) for $\alpha$ near $1/2$
similarly requires that $r$ be the solution $\rho_{*}$ of (sp$*$). As we
expect for almost all $x$ that $\rho(x)$ and $\rho_{*}(x)$ are not equal, we
must make a “compromise” by modifying the original circle $|z|=r$ of
integration in (9.1).

As $\Phi(z)$ is analytic for $|z|<1$, we split the interval $[0,1)$ into the
sets

$\mathscr{A}:=[0,\tfrac{1}{4})\cup[\tfrac{3}{4},1)\qquad\text{and}\qquad\mathscr{A}_{*}:=[\tfrac{1}{4},\tfrac{3}{4})$
(9.2)

and change the path of integration on the left side of (9.1) to be the
positively oriented path consisting of the two semicircles

$\rho e(\mathscr{A}):=\\{\rho
e(\alpha):\alpha\in\mathscr{A}\\}\qquad\text{and}\qquad\rho_{*}e(\mathscr{A}_{*}):=\\{\rho_{*}e(\alpha):\alpha\in\mathscr{A}_{*}\\}$

and the two connecting line segments $[i\rho,i\rho_{*}]$ and
$[-i\rho_{*},-i\rho]$, as shown in Figure 9.1.

$\rho_{*}$$\rho$$1$$-1$$i$$-i$$\rho
e(\mathscr{A})$$\rho_{*}e(\mathscr{A}_{*})$ Figure 9.1. The new contour of
integration. We note that the figure draws $\rho_{*}>\rho$ for illustrative
purposes only—it is possible that $\rho_{*}\leq\rho$.

Using this new contour of integration in (9.1) we have

$p(x,\mu)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\rho
e(\mathscr{A})}\Phi(z)z^{-x-1}\,dz+\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\rho_{*}e(\mathscr{A}_{*})}\Phi(z)z^{-x-1}\,dz+\mathfrak{T}(\rho,\rho_{*}),$
(9.3)

where

$\mathfrak{T}(\rho,\rho_{*}):=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{i\rho}^{i\rho_{*}}\Phi(z)z^{-n-1}\,dz+\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{-i\rho_{*}}^{-i\rho}\Phi(z)z^{-n-1}\,dz,$
(9.4)

and we note that

$\mathfrak{T}(\rho,\rho_{*})=2\operatorname{Re}\mathopen{}\mathclose{{}\left[\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{i\rho}^{i\rho_{*}}\Phi(z)z^{-n-1}\,dz}\right]=2\operatorname{Re}\mathopen{}\mathclose{{}\left[\frac{i^{-n-1}}{2\pi}\int_{\rho}^{\rho_{*}}\Phi(it)t^{-n-1}\,dt}\right].$

######  Remark 9.1.

Before proceeding further we must address a technical issue introduced by this
“arc transference” in equation (9.3). Because $\rho(x)$ and $\rho_{*}(x)$ are
unlikely to be equal, so then are $X$ and $X_{*}$ unlikely to be equal. Thus,
the division of $[0,1)$ into major, minor, and principal arcs using $X$ and
$Q$ (which is a power of $X$) is now problematic. In particular, we must
define the “dual” $*$-arcs using $X_{*}$ and $Q_{*}$ via

$\mathfrak{M}_{*}(a/q):=\\{\alpha\in[0,1):|\alpha-a/q|\leq Q_{*}/(qX_{*})\\}$

and consider $*$-versions of the major and minor arcs, say
$\mathfrak{M}(X_{*},Q_{*})$ and $\mathfrak{m}(X_{*},Q_{*})$, noting that we
have already defined $\mathfrak{P}_{*}$.

Fortunately, this technicality does not create too much trouble. In
particular, recalling the note at the end of section 2, the proofs of
Propositions 4.8, 5.4, and 7.1 do not require that $\rho$ be a solution to one
of the equations (sp) and (sp$*$), said propositions hold by identical proofs
if one replaces $\rho$, $X$, and $Q$ with $\rho_{*}$, $X_{*}$, and $Q_{*}$,
respectively. That this replacement is valid is perhaps made intuitively clear
by recalling that, per Theorem 1.2, both $X\sim 2\sqrt{x}$ and $X_{*}\sim
2\sqrt{x}$.

We now establish that this “arc transference” made in equation (9.3) only adds
a small error term to our formula for $p(x,\mu)$; that is, small relative to
the main terms of (9.3), which are of order $\exp(\sqrt{x})$. For convenience
we recall from Theorem 1.2 and the previous section that

$\displaystyle-x\log\rho$
$\displaystyle=\tfrac{1}{2}\sqrt{x}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right],\qquad X$
$\displaystyle=2\sqrt{x}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right],$ (9.5) $\displaystyle-x\log\rho_{*}$
$\displaystyle=\tfrac{1}{2}\sqrt{x}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right],\qquad X_{*}$
$\displaystyle=2\sqrt{x}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right].$ (9.6)

######  Lemma 9.2.

For $x>x_{0}$ let $\rho$ and $\rho_{*}$ be the unique solutions to equations
(sp) and (sp$*$), respectively. For all $\varepsilon>0$, as $x\to\infty$ one
has

$\mathfrak{T}(\rho,\rho_{*})\ll\exp((\tfrac{1}{2}+\varepsilon)\sqrt{x}).$

###### Proof.

Fix $A,\varepsilon>0$ and let $x>x_{0}(A)$. It is evident that
$\mathfrak{T}(\rho,\rho_{*})=-\mathfrak{T}(\rho_{*},\rho)$, so there is no
loss of generality in assuming that $\rho\leq\rho_{*}$. First, using (9.5) and
(9.6) we find that

$\displaystyle\begin{aligned}
&\int_{\rho}^{\rho_{*}}\frac{dt}{t^{x+1}}=\frac{\rho_{*}^{-x}-\rho^{-x}}{x}=\frac{1}{x}\mathopen{}\mathclose{{}\left[e^{\frac{1}{2}\sqrt{x}+o(\sqrt{x})}-e^{\frac{1}{2}\sqrt{x}+o(\sqrt{x})}}\right]\ll
e^{(\frac{1}{2}+\varepsilon)\sqrt{x}}.\end{aligned}$ (9.7)

For $\rho\leq t\leq\rho_{*}$ let $t=\exp(-1/X_{t})$, and assume that $x$ (and
therefore $\rho$) is large enough that $X_{t}(\log X_{t})^{-A}$ is increasing
in $t$ for $t\geq\rho$. Because $V(4)=0$ in formula (5.4), for sufficiently
large $x$ it holds that

$\operatorname{Re}\Psi(it)\ll X_{t}(\log X_{t})^{-A}\ll X_{*}(\log
X_{*})^{-A}\ll\sqrt{x}(\log x)^{-A}\leq\varepsilon\sqrt{x}\qquad$ (9.8)

for $\rho\leq t\leq\rho_{*}$. Using inequalities (9.7) and (9.8) we have

$\mathopen{}\mathclose{{}\left|\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{i\rho}^{i\rho_{*}}\Phi(z)z^{-x-1}\,dz}\right|\ll
e^{\varepsilon\sqrt{x}}\int_{\rho}^{\rho_{*}}\frac{dt}{t^{x+1}}\ll
e^{(\frac{1}{2}+2\varepsilon)\sqrt{x}},$

and the result follows by repeating our arguments with $\varepsilon/2$ in
place of $\varepsilon$. ∎

######  Corollary 9.3.

For $x>x_{0}$ let $\rho$ and $\rho_{*}$ be the unique solutions to equations
(sp) and (sp$*$), respectively. For all $\varepsilon>0$, as $x\to\infty$ one
has

$\displaystyle p(x,\mu)$ $\displaystyle=\rho^{-x}\int_{\mathscr{A}}\Phi(\rho
e(\alpha))e(-x\alpha)\,d\alpha+\rho_{*}^{-x}\int_{\mathscr{A}_{*}}\Phi(\rho_{*}e(\alpha))e(-x\alpha)\,d\alpha+O\big{(}e^{(\frac{1}{2}+\varepsilon)\sqrt{x}}\big{)}.$

##  10\. The contributions of nonprincipal arcs

We are now ready to bound the contributions of all nonprincipal arcs to the
integrals

$\rho^{-x}\int_{\mathscr{A}}\Phi(\rho
e(\alpha))e(-x\alpha)\,d\alpha\qquad\text{and}\qquad\rho_{*}^{-x}\int_{\mathscr{A}_{*}}\Phi(\rho_{*}e(\alpha))e(-x\alpha)\,d\alpha,$

recalling that

$\mathscr{A}:=[0,\tfrac{1}{4})\cup[\tfrac{3}{4},1)\qquad\text{and}\qquad\mathscr{A}_{*}:=[\tfrac{1}{4},\tfrac{3}{4})$

and that

$\displaystyle\mathfrak{P}$ $\displaystyle=\\{\alpha\in[0,1):\|\alpha\|\leq
X^{-1}(\log X)^{-1/4}\\},$ $\displaystyle\mathfrak{P}_{*}$
$\displaystyle=\\{\alpha\in[0,1):|\alpha-\tfrac{1}{2}|\leq X_{*}^{-1}(\log
X_{*})^{-1/4}\\}.$

Our primary tools toward this end are the following inequalities on
$\operatorname{Re}\Psi(z)$.

######  Lemma 10.1.

Let $X>X_{0}$. For all $\alpha\in\mathscr{A}\setminus\mathfrak{P}$ one has

$\operatorname{Re}\Psi(\rho
e(\alpha))\leq\mathopen{}\mathclose{{}\left(1-\frac{1}{\log
X}}\right)\Psi(\rho),$ (10.1)

and for all $\alpha\in\mathscr{A}_{*}\setminus\mathfrak{P}_{*}$ one has

$\operatorname{Re}\Psi(\rho_{*}e(\alpha))\leq\mathopen{}\mathclose{{}\left(1-\frac{1}{\log
X}}\right)\Psi(-\rho_{*}).$ (10.2)

######  Remark 10.2.

In the following proof we require that $A>1$ in contrast to our usual
allowance that $A>0$. This restriction does affect any of our previous results
since said results allowed $A>0$ to be arbitrary.

###### Proof.

Fix $A>1$, let $X>X_{0}(A)$, and let $Q=X^{2/5}$. Suppose that
$\alpha\in\mathscr{A}$ can be written as $\alpha=a/q+\beta$ with $0\leq
a<q\leq Q$, $(a,q)=1$, and $|\beta|\leq Q/(qX)$. For real $\beta$ we define

$\Delta=(1+4\pi^{2}X^{2}\beta^{2})^{-\frac{1}{2}}\qquad\text{and}\qquad\varphi=\arctan(2\pi
X\beta),$

using the branch of $\arctan$ with values in $(-\frac{\pi}{2},\frac{\pi}{2})$,
so that $(1-2\pi iX\beta)^{-1}=\Delta e^{i\varphi}$ and Proposition 5.4 states
that

$\Psi(\rho e(\alpha))=V(q)X\Delta e^{i\varphi}+O(X(\log X)^{-A}).$ (10.3)

As $V(1)=\tfrac{1}{4}$, we have

$\Psi(\rho)=\tfrac{1}{4}X+O(X(\log X)^{-A}).$ (10.4)

Since $\Delta\leq 1$ and $|V(q)|\leq\tfrac{1}{5}$ when $q>2$, it follows at
once from (10.3) and (10.4) that inequality (10.1) holds for all $\alpha$ in
sets $\mathscr{A}\cap\mathfrak{M}(a/q)$ with $3\leq q\leq Q$. Recalling from
Proposition 4.8 that for $\alpha\in\mathscr{A}\cap\mathfrak{m}(X,Q)$ one has
$\Psi(\rho e(\alpha))\ll X(\log X)^{-A}$ already, then a fortiori inequality
(10.1) holds for these $\alpha$ as well.

It thus remains to consider only those $\alpha$ in the major arcs
$\mathfrak{M}(\frac{0}{1})$ and $\mathfrak{M}(\frac{1}{1})$ (since
$\mathfrak{M}(\frac{1}{2})$ is not included in $\mathscr{A}$) that are _not_
in the principal arc $\mathfrak{P}$. Considering the case of
$\mathfrak{M}(\frac{0}{1})$, let $\alpha=\beta$ with $X^{-1}(\log
X)^{-1/4}<\beta\leq Q/X$. Here we find that

$\Delta\leq\big{(}1+4\pi^{2}(\log
X)^{-\frac{1}{2}}\big{)}^{-\frac{1}{2}}=1-2\pi^{2}(\log
X)^{-\frac{1}{2}}+O((\log X)^{-1})\leq 1-(\log X)^{-\frac{1}{2}}$

for sufficiently large $X$, whereby equation (10.3) shows that

$\operatorname{Re}\Psi(\rho e(\beta))\leq\tfrac{1}{4}(1-(\log
X)^{-\frac{1}{2}})X+O(X(\log X)^{-A}).$ (10.5)

Since the main term in $(1-(\log X)^{-1})\Psi(\rho)$ is $\tfrac{1}{4}(1-(\log
X)^{-1})X$ and we have $A>1$, we may “spend” the extra saved $(\log X)^{-1/2}$
in (10.5) to overcome any differences between the error terms in (10.4) and
(10.5) and thereby conclude that inequality (10.1) holds for all
$\beta\in\mathfrak{M}(\tfrac{0}{1})\setminus\mathfrak{P}$.

A similar argument shows that inequality (10.1) holds for
$\alpha\in\mathfrak{M}(\tfrac{1}{1})\setminus\mathfrak{P}$, so the lemma’s
first assertion is established. Since $V(2)$ is also $\tfrac{1}{4}$,
Proposition 5.4 similarly implies that

$\Psi(-\rho_{*})=\tfrac{1}{4}X_{*}+O(X_{*}(\log X_{*})^{-A}).$

Using this in place of (10.4), identical arguments to those above show that
inequality (10.2) holds for
$\alpha\in\mathscr{A}_{*}\setminus\mathfrak{P}_{*}$, and the proof of the
lemma is completed. ∎

Using the results of Lemma 10.1 we now take our first major step toward
establishing the asymptotic formula (1.11) for $p(x,\mu)$.

######  Proposition 10.3.

For all $x>x_{0}$ let $\rho$ and $\rho_{*}$ be the unique solutions to
equations (sp) and (sp$*$), respectively. For fixed $B>0$, as $x\to\infty$ one
has

$\displaystyle p(x,\mu)$ $\displaystyle=\rho^{-x}\int_{\mathfrak{P}}\Phi(\rho
e(\alpha))e(-x\alpha)\,d\alpha+\rho_{*}^{-x}\int_{\mathfrak{P}_{*}}\Phi(\rho_{*}e(\alpha))e(-x\alpha)\,d\alpha$
(10.6)
$\displaystyle\qquad+O(\rho^{-x}\Phi(\rho)x^{-B})+O(\rho_{*}^{-x}\Phi(-\rho_{*})x^{-B})+O(e^{(\frac{1}{2}+\varepsilon)\sqrt{x}}).$

###### Proof.

Fix $B>0$ and let $x>x_{0}(B)$. When
$\alpha\in\mathscr{A}\setminus\mathfrak{P}$ Lemma 10.1 implies that

$\Phi(\rho e(\alpha))=\exp\Psi(\rho
e(\alpha))\ll\exp\mathopen{}\mathclose{{}\left(\Psi(\rho)-\frac{\Psi(\rho)}{\log
X}}\right),$ (10.7)

and since $\Psi(\rho)\asymp X\asymp x^{1/4}$ and $\log X\asymp\log x$ by
(9.5), we have

$\exp\mathopen{}\mathclose{{}\left(-\frac{\Psi(\rho)}{\log
X}}\right)\ll\exp\mathopen{}\mathclose{{}\left(-\frac{x^{1/2}}{\log
x}}\right)\ll\exp(-x^{1/4})\ll x^{-B}.$ (10.8)

By (10.7) and (10.8) then, we have

$\rho^{-x}\int_{\mathscr{A}\setminus\mathfrak{P}}\Phi(\rho
e(\alpha))e(-x\alpha)\,d\alpha\ll\rho^{-x}e^{\Psi(\rho)}x^{-B}=\rho^{-x}\Phi(\rho)x^{-B}$

and similarly for the integral over $\mathscr{A}_{*}\setminus\mathfrak{P}_{*}$
(by way of inequality (10.2)), and the result follows at once from Corollary
9.3. ∎

##  11\. The proof of Theorem 1.3

To establish our asymptotic formula for $p(x,\mu)$ in Theorem 1.3, it only
remains to treat the principal-arc integrals in equation (10.6). We recall
that we identify

$\\{e(\alpha):\alpha\in\mathfrak{P}\\}\qquad\text{and}\qquad\\{e(\alpha):\alpha\in\mathfrak{P}_{*}\\}$

with

$\\{e(\beta):\text{$\beta\in\mathbb{R}$ and
$|\beta|\leq\eta$}\\}\qquad\text{and}\qquad\\{-e(\beta):\text{$\beta\in\mathbb{R}$
and $|\beta|\leq\eta_{*}$}\\},$

respectively, where

$\eta=X^{-1}(\log X)^{-1/4}\qquad\text{and}\qquad\eta_{*}=X_{*}^{-1}(\log
X_{*})^{-1/4}.$

When $f$ is nonnegative it is clear that $|\Psi(z,f)|\leq\Psi(|z|,f)$ for
$|z|<1$, and similarly for all derivatives of $\Psi(z,f)$. We recover a
similar bound for $\Psi(z,\mu)$ in the following lemma.

######  Lemma 11.1.

Fix $A>0$. There exists a constant $X_{A}>0$ such that for all $0\leq j\leq
3$, $X>X_{A}$, and $|\beta|\leq\eta$ one has that

$|\Psi_{(j)}(\rho e(\beta))|\leq\sqrt{2}\Psi_{(j)}(\rho).$ (11.1)

###### Proof.

For $0\leq j\leq 3$ let $c_{j}$ denote the implicit constants in equation
(7.1). By said equation, for $X>X_{0}(A)$ we have

$\frac{|\Psi_{(j)}(\rho e(\beta))|}{\Psi_{(j)}(\rho)}=|1-2\pi
iX\beta|^{-j-1}\mathopen{}\mathclose{{}\left|\frac{1+O_{j}(\log^{-A}X)}{1+O_{j}(\log^{-A}X)}}\right|\leq\frac{1+c_{j}\log^{-A}X}{1-c_{j}\log^{-A}X}.$

The latter fraction here is at most $\sqrt{2}$ for $X>X_{0}(j,A)$, and
inequality (11.1) follows by setting $X_{A}:=\max_{0\leq j\leq 3}X_{0}(j,A)$.
∎

We first consider the integral in (10.6) over $\mathfrak{P}$ using arguments
from [vaughan2008number]*sec. 4.

######  Proposition 11.2.

For $x>x_{0}$ let $\rho$ be the unique solution to equation (sp). As
$x\to\infty$, one has

$\rho^{-x}\int_{\mathfrak{P}}\Phi(\rho
e(\alpha))e(-x\alpha)\,d\alpha=\frac{\rho^{-x}\Phi(\rho)}{\sqrt{2\pi\Psi_{(2)}(\rho)}}\mathopen{}\mathclose{{}\left[1+O(x^{-1/5})}\right]$
(11.2)

######  Remark 11.3.

In the following proof, quantities $c_{1},c_{2},\ldots$ are positive
constants.

###### Proof.

For $|\beta|\leq\eta$ let $\Psi(\rho e(\beta))=R(\beta)+iI(\beta)$ for real-
valued functions $R(\beta)$ and $I(\beta)$, and consider the third-order
Taylor expansion with remainder:

$\displaystyle R(\beta)$ $\displaystyle=R(0)+\beta
R^{\prime}(0)+\tfrac{1}{2}\beta^{2}R^{\prime\prime}(0)+\tfrac{1}{6}\beta^{3}R^{\prime\prime\prime}(c_{R}\beta),$
$\displaystyle I(\beta)$ $\displaystyle=I(0)+\beta
I^{\prime}(0)+\tfrac{1}{2}\beta^{2}I^{\prime\prime}(0)+\tfrac{1}{6}\beta^{3}I^{\prime\prime\prime}(c_{I}\beta),$

where $c_{R}$ and $c_{I}$ are in $(0,1)$ and depend on $\beta$. One may easily
see that

$R^{(j)}(\beta)+iI^{(j)}(\beta)=\partial_{\beta}^{\,j}\big{[}\Psi(\rho
e(\beta))\big{]}=(2\pi i)^{j}\Psi_{(j)}(\rho e(\beta))$ (11.3)

for all $j\geq 1$, and in particular that

$\displaystyle R^{\prime}(\beta)+iI^{\prime}(\beta)$ $\displaystyle=2\pi
i\Psi_{(1)}(\rho e(\beta)),$ (11.4a) $\displaystyle
R^{\prime\prime}(\beta)+iI^{\prime\prime}(\beta)$
$\displaystyle=-4\pi^{2}\Psi_{(2)}(\rho e(\beta)),$ (11.4b) $\displaystyle
R^{\prime\prime\prime}(\beta)+iI^{\prime\prime\prime}(\beta)$
$\displaystyle=-8\pi^{3}i\Psi_{(3)}(\rho e(\beta)).$ (11.4c)

Applying Lemma 11.1 to equation (11.4c), we have

$|R^{\prime\prime\prime}(\beta)|\leq 8\pi^{3}\sqrt{2}\Psi_{(3)}(\rho)$

and likewise for $|I^{\prime\prime\prime}(\beta)|$. With
$c_{0}:=8\pi^{3}\sqrt{2}\Psi_{(3)}(\rho)$, we implicitly define $w(\beta)$ via

$\beta^{3}\big{[}R^{\prime\prime\prime}(c_{R}\beta)+iI^{\prime\prime\prime}(c_{I}\beta)\big{]}=2c_{0}|\beta|^{3}w(\beta)$
(11.5)

and see that $|w(\beta)|\leq 1$. Combining equations (11.3)–(11.5), we deduce
that

$\Psi(\rho e(\beta))=\Psi(\rho)+2\pi
i\beta\Psi_{(1)}(\rho)-2\pi^{2}\beta^{2}\Psi_{(2)}(\rho)+\tfrac{8}{3}\pi^{3}w(\beta)|\beta|^{3}\Psi_{(3)}(\rho).$
(11.6)

Recalling that $x=\Psi_{(1)}(\rho)$ we have $e(-x\beta)=\exp(-2\pi
i\beta\Psi_{(1)}(\rho))$, and we thus write

$\rho^{-x}\int_{\mathfrak{P}}\exp\big{[}\Psi(\rho
e(\alpha))\big{]}e(-x\alpha)\,d\alpha=\rho^{-x}\int_{-\eta}^{\eta}\exp\Theta(\rho,\beta)\,d\beta,$
(11.7)

where

$\Theta(\rho,\beta):=\Psi(\rho)-2\beta^{2}\pi^{2}\Psi_{(2)}(\rho)+\tfrac{8}{3}\pi^{3}w(\beta)|\beta|^{3}\Psi_{(3)}(\rho).$
(11.8)

By Theorem 1.2 we have $\Psi_{(2)}(\rho)\gg X^{3}$ and $\Psi_{(3)}(\rho)\ll
X^{4}$, so that when $X$ is sufficiently large and
$|\beta|\leq\eta=X^{-1}(\log X)^{-1/4}$ we have

$\operatorname{Re}\Theta(\rho,\beta)\leq\Psi(\rho)-1\pi^{2}\beta^{2}\Psi_{(2)}(\rho).$

For those $\beta$ such that $|\beta|\geq X^{-3/2}(\log X)^{3/2}$ as well, we
further deduce that

$\operatorname{Re}\Theta(\rho,\beta)\leq\Psi(\rho)-c_{1}(\log X)^{3}.$ (11.9)

Fixing $B>1$ and setting

$\delta:=X^{-3/2}(\log X)^{3/2},$ (11.10)

we use (11.9) and repeat the arguments used for Proposition 10.3 to conclude
that

$\rho^{-x}\int_{\delta\leq|\beta|\leq\eta}\exp\Theta(\rho,\beta)\,d\beta\ll\rho^{-x}\Phi(\rho)x^{-B}.$
(11.11)

It thus remains to consider (11.2) when $|\beta|<\delta$. For these $\beta$,
Theorem 1.2 implies that

$\displaystyle\tfrac{8}{3}\pi^{3}|\beta|^{3}w(\beta)\Psi_{(3)}(\rho)\ll
X^{-9/2}(\log X)^{9/2}X^{4}\sim
c_{2}x^{-1/4}\mathopen{}\mathclose{{}\left[(\log
x)^{9/2}+O\mathopen{}\mathclose{{}\left(1}\right)}\right]\ll x^{-1/5},$

and we deduce that

$\rho^{-x}\int_{-\delta}^{\delta}\exp\Theta(\rho,\beta)\,d\beta=\rho^{-x}e^{\Psi(\rho)}\mathopen{}\mathclose{{}\left[1+O(x^{-\frac{1}{5}})}\right]\int_{-\delta}^{\delta}\exp\mathopen{}\mathclose{{}\left(-2\pi^{2}\Psi_{(2)}(\rho)\beta^{2}}\right)\,d\beta.$
(11.12)

We have

$\int_{-\delta}^{\delta}\exp\mathopen{}\mathclose{{}\left(-2\pi^{2}\Psi_{(2)}(\rho)\beta^{2}}\right)\,d\beta=\frac{1}{\sqrt{2\pi\Psi_{(2)}(\rho)}}+O\mathopen{}\mathclose{{}\left(\frac{\exp(-\delta^{2}\Psi_{(2)}(\rho))}{\delta\Psi_{(2)}(\rho)}}\right),$

and since $\Psi_{(2)}(\rho)\asymp X^{3}\asymp x^{3/2}$ our error term here is

$\ll\frac{\exp(-c_{3}(\log X)^{3})}{X^{3/2}(\log X)^{3/2}}\ll\exp(-c_{4}(\log
x)^{3})\ll x^{-B}.$

In total then, we find that

$\rho^{-x}\int_{-\delta}^{\delta}\exp\Theta(\rho,\beta)\,d\beta=\rho^{-x}\Phi(\rho)\mathopen{}\mathclose{{}\left[1+O(x^{-1/5})}\right]\mathopen{}\mathclose{{}\left[\frac{1}{\sqrt{2\pi\Psi_{(2)}(\rho)}}+O(x^{-B})}\right].$

Using the assumption that $B>1$ we deduce that

$\rho^{-x}\int_{-\delta}^{\delta}\exp\Theta(\rho,\beta)\,d\beta=\frac{\rho^{-x}\Phi(\rho)}{\sqrt{2\pi\Psi_{(2)}(\rho)}}\mathopen{}\mathclose{{}\left[1+O(x^{-1/5})}\right],$
(11.13)

and the result follows from adding equations (11.11) and (11.13), again due to
the facts that $\Psi_{(2)}(\rho)\asymp x^{3/2}$ and $B>1$. ∎

As is par for the course, the analysis of $\Psi(\rho_{*}e(\alpha))$ for
$\alpha$ in $\mathfrak{P}_{*}$ is largely identical to the analysis for
$\alpha$ in $\mathfrak{P}$. Thus, let $\rho_{*}$ be the unique solution to
(sp$*$), and for $\alpha\in\mathfrak{P}_{*}$ write $\alpha=\tfrac{1}{2}+\beta$
and $\rho_{*}e(\alpha)=-\rho_{*}e(\beta)$ with
$|\beta|\leq\eta_{*}=X_{*}^{-1}(\log X_{*})^{-1/4}$.

Like in Lemma 11.1, we easily deduce that for fixed $A>0$ there is some
$X_{A}>0$ such that: For $0\leq j\leq 3$, $X_{*}>X_{A}$, and
$\alpha\in\mathfrak{P}_{*}$ one has

$|\Psi_{(j)}(\rho_{*}e(\alpha))|=|\Psi_{(j)}(-\rho_{*}e(\beta))|\leq\sqrt{2}\Psi_{(j)}(-\rho_{*}).$

As $x=\Psi_{(1)}(-\rho_{*})$ now, this time we have

$e(-x\alpha)=e(-\tfrac{1}{2}x)e(-x\beta)=e(-\tfrac{1}{2}x)\exp(-2\pi
i\beta\Psi_{(1)}(-\rho_{*})).$

Letting $\Psi(-\rho_{*}e(\beta))=R(\beta)+iI(\beta)$ for some new $R$ and $I$,
we again compute that

$\Psi(-\rho_{*}e(\beta))=\Psi(-\rho_{*})+2\pi
i\beta\Psi_{(1)}(-\rho_{*})-2\pi^{2}\beta^{2}\Psi_{(2)}(-\rho_{*})+\tfrac{8}{3}\pi^{3}w_{*}(\beta)|\beta|^{3}\Psi_{(3)}(-\rho_{*}),$

cf. (11.6), where $w_{*}(\beta)$ is defined as $w(\beta)$ was in equation
(11.5). It follows that

$\rho_{*}^{-x}\int_{\mathfrak{P}_{*}}\exp\\!\big{[}\Psi(\rho_{*}e(\alpha))\big{]}e(-x\alpha)\,d\alpha=e(-\tfrac{1}{2}x)\rho_{*}^{-x}\int_{-\eta_{*}}^{\eta_{*}}\exp\Theta_{*}(\rho_{*},\beta)\,d\beta,$
(11.14)

where

$\Theta_{*}(\rho_{*},\beta):=\Psi(-\rho_{*})-2\pi^{2}\beta^{2}\Psi_{(2)}(-\rho_{*})+\tfrac{8}{3}\pi^{3}w_{*}(\beta)|\beta|^{3}\Psi_{(3)}(-\rho_{*}).$
(11.15)

Considering the overwhelming similarities between equations (11.14) and
(11.15) and equations (11.7) and (11.8), respectively, and the symmetric
relations for $\Psi_{(j)}(\pm\rho)$ in Theorem 1.2, we simply repeat the proof
of Proposition 11.2 to conclude that

$\rho_{*}^{-x}\int_{\mathfrak{P}_{*}}\Phi(\rho_{*}e(\alpha))e(-x\alpha)\,d\alpha=e(-\tfrac{1}{2}x)\frac{\rho_{*}^{-x}\Phi(-\rho_{*})}{\sqrt{2\pi\Psi_{(2)}(-\rho_{*})}}\mathopen{}\mathclose{{}\left[1+O(x^{-1/5})}\right].$
(11.16)

######  Remark 11.4.

To maintain consistency between the formulae for $p(x,f)$ and $p(n,f)$, in the
remainder of this paper we write $e(-\tfrac{1}{2}x)=(-1)^{-x}$ and understand
all future occurrences of $(-1)^{-x}$ to use this identification.

Combining Propositions 10.3 and 11.2 and equation (11.16), as $x\to\infty$ one
has

$\displaystyle p(x,\mu)$
$\displaystyle=\frac{\rho^{-x}\Phi(\rho)}{\sqrt{2\pi\Psi_{(2)}(\rho)}}\mathopen{}\mathclose{{}\left[1+O(x^{-1/5})}\right]+(-1)^{-x}\frac{\rho_{*}^{-x}\Phi(-\rho_{*})}{\sqrt{2\pi\Psi_{(2)}(-\rho_{*})}}\mathopen{}\mathclose{{}\left[1+O(x^{-1/5})}\right]$
(11.17) $\displaystyle\qquad+O(e^{(\frac{1}{2}+\varepsilon)\sqrt{x}}),$

which is nearly equation (1.11). Using Theorem 1.2 one finds that

$\frac{\rho^{-x}\Phi(\rho)}{\sqrt{2\pi\Psi_{(2)}(\rho)}}=\exp\\!\Big{(}\Psi(\rho)-x\log(\rho)-\tfrac{1}{2}\log(2\pi\Psi_{(2)}(\rho))\Big{)}=\exp\mathopen{}\mathclose{{}\left(\sqrt{x}+o(\sqrt{x})}\right),$

whereby we simplify equation (11.17) by writing

$\frac{\rho^{-x}\Phi(\rho)}{\sqrt{2\pi\Psi_{(2)}(\rho)}}\mathopen{}\mathclose{{}\left[1+O\big{(}x^{-1/5}\big{)}}\right]+O\big{(}e^{(\frac{1}{2}+\varepsilon)\sqrt{x}}\big{)}=\frac{\rho^{-x}\Phi(\rho)}{\sqrt{2\pi\Psi_{(2)}(\rho)}}\mathopen{}\mathclose{{}\left[1+O\big{(}x^{-1/5}\big{)}}\right]$

to finally yield the formula

$p(x,\mu)=\frac{\rho^{-x}\Phi(\rho)}{\sqrt{2\pi\Psi_{(2)}(\rho)}}\mathopen{}\mathclose{{}\left[1+O(x^{-1/5})}\right]+(-1)^{-x}\frac{\rho_{*}^{-x}\Phi(-\rho_{*})}{\sqrt{2\pi\Psi_{(2)}(-\rho_{*})}}\mathopen{}\mathclose{{}\left[1+O(x^{-1/5})}\right],$
(11.18)

and the proof of Theorem 1.3 is complete.

##  12\. Formulae for $p(x,\lambda)$

Given the functions’ strong connection, results for one of Möbius’ $\mu$ and
Liouville’s $\lambda$ functions are often easily translated into results for
the other, and our formulae for $p(x,\mu)$ and $p(x,\lambda)$ conform to this
pattern. Many of the proofs of our lemmata for $p(x,\lambda)$ are nearly
identical to the corresponding ones for $p(x,\mu)$, and indeed often easier
since $\lambda^{2}(n)=1$ for all $n$, so our analyses for $p(x,\lambda)$ in
this section are notably briefer than those of the previous sections.

While the sets $\mathfrak{M}(X,Q)$, $\mathfrak{m}(X,Q)$, $\mathfrak{P}$, and
$\mathfrak{P}_{*}$ are all defined as in section 2, our choice of $Q$ for
$p(x,\lambda)$ is not the same as it is for $p(x,\mu)$, so we start anew with
an unknown $Q$ such that $1\leq Q\leq X$. All other notations are also
maintained, but we emphasize that $\Phi(z)$ and $\Psi(z)$ now denote
$\Phi(z,\lambda)$ and $\Psi(z,\lambda)$, respectively. In particular, the
equations (sp) and (sp$*$) are still written as $\Psi_{(1)}(\rho)=x$ and
$\Psi_{(1)}(-\rho)=x$. As done in our analysis of $\Psi(\rho e(\alpha),\mu)$,
in analyzing $\Psi(\rho e(\alpha),\lambda)$ we do not assume $\rho$ to be a
solution to (sp) or (sp$*$) until necessary.

Finally we note that the functions $F_{0}(z)$ and $F_{1}(z)$ of (3.1) are now
given by the formulae

$F_{0}(z)=\sum_{n=1}^{\infty}z^{n}\qquad\text{and}\qquad
F_{1}(z)=\sum_{n=1}^{\infty}\lambda(n)z^{n}.$

For the analysis of $\Psi(\rho e(\alpha))$ for $\alpha\in\mathfrak{m}(X,Q)$,
Davenport’s inequality (4.2) on $S_{\mu}(t,\theta)$ is replaced by a similar
bound on $S_{\lambda}(t,\theta)$ due to Bateman and Chowla.

######  Lemma 12.1 ([bateman1963some]*Lem. 1).

Fix $A>0$. For $t>t_{0}$ one has $S_{\lambda}(t,\theta)\ll t(\log t)^{-A}$
uniformly for $\theta\in\mathbb{R}$.

###### Proof.

By the well known identity $\lambda(n)=\sum_{d^{2}\hskip 0.27995pt|\hskip
0.27995ptn}\mu(n/d^{2})$, one has

$\sum_{n\leq t}\lambda(n)e(n\theta)=\sum_{d\leq\sqrt{t}}\,\sum_{m\leq
t/d^{2}}\mu(m)e(md^{2}\theta)=\sum_{d\leq\sqrt{t}}S_{\mu}(td^{-2},d^{2}\theta).$

Using Davenport’s inequality (4.2) when $d\leq t^{1/4}$ and trivially bounding

$\mathopen{}\mathclose{{}\left|S_{\mu}(td^{-2},d^{2}\theta)}\right|\leq\sum_{t^{1/4}<d\leq\sqrt{t}}td^{-2}\leq
t\sum_{d>t^{1/4}}d^{-2}\ll t\cdot t^{-1/4},$

the result follows. ∎

Because Lemma 12.1 provides a perfect substitute for Davenport’s inequality
(4.2), we simply repeat the proof of Lemma 4.1 to immediately recover the
estimate

$\Psi_{1}(\rho e(\alpha))\ll X(\log X)^{-A}\qquad(\alpha\in\mathbb{R}).$
(12.1)

Our analysis of $\Psi_{0}(\rho e(\alpha))$ is greatly expedited by the well
known inequality

$S_{\lambda^{2}}(t,\theta)=\sum_{n\leq
t}e(n\theta)\ll\min\\{t,\|\theta\|^{-1}\\},$ (12.2)

and we remark that (12.2) serves as the $\Psi(z,\lambda)$-analogue of Lemma
4.4.

######  Lemma 12.2.

Fix $A>0$ and let $Q=X^{1/3}$. For $X>X_{0}(A)$ and
$\alpha\in\mathfrak{m}(X,Q)$ one has

$\Psi(\rho e(\alpha))\ll X(\log X)^{-A}.$

###### Proof.

By inequality (12.1) it suffices to show that $\Psi_{0}(\rho e(\alpha))\ll
X(\log X)^{-A}$.

For the sake of exposition we begin with parameters $\delta,\kappa\in(0,1/2)$
and let $Q=X^{\delta}$, let $K=Q^{\kappa}$, and let $k\leq K$. By Lemma 4.6 we
have $\\{k\alpha\\}\in\mathfrak{m}(X,Q^{1-\kappa})$ for all
$\alpha\in\mathfrak{m}(X,Q)$ and all $k\leq Q^{\kappa}$, which implies for
said $\alpha$ and $k$ that $\|k\alpha\|>Q^{1-\kappa}/X$, i.e., that

$\|k\alpha\|^{-1}<XQ^{\kappa-1}.$ (12.3)

As $F_{0}(z)=\sum_{n=1}^{\infty}z^{n}$, by partial summation we have

$F_{0}(\rho^{k}e(k\alpha))=\sum_{n=1}^{\infty}e^{nk/X}e(k\alpha)=\frac{k}{X}\int_{0}^{\infty}e^{-tk/X}S_{\lambda^{2}}(t,k\alpha)\,dt.$

Bounding $S_{\lambda^{2}}(t,k\alpha)\ll t$ when $t\leq XQ^{\kappa-1}$ and
$S_{\lambda^{2}}(t,k\alpha)\ll\|k\alpha\|^{-1}$ otherwise, it follows that

$\displaystyle
F_{0}(\rho^{k}e(k\alpha))\ll\frac{X}{k}\int_{0}^{kQ^{\kappa-1}}e^{-u}u\,du+\|k\alpha\|^{-1}\int_{kQ^{\kappa-1}}^{\infty}e^{-u}\,du.$
(12.4)

Integrating by parts and applying (12.3), the right side of (12.4) is seen to
be

$\ll\frac{X}{k}\big{(}1-(1+kQ^{\kappa-1})e^{-kQ^{\kappa-1}}\big{)}+XQ^{\kappa-1}e^{-kQ^{\kappa-1}}=\frac{X}{k}(1-e^{-kQ^{\kappa-1}}).$

As $kQ^{\kappa-1}\leq Q^{2\kappa-1}$ and $2\kappa-1<0$, we bound $1-e^{-u}\leq
u$ for $0\leq u\leq 1$ to deduce that

$F_{0}(\rho^{k}e(k\alpha))\ll\frac{X}{k}(1-e^{-X^{\delta(2\kappa-1)}})\leq\frac{X}{k}X^{\delta(2\kappa-1)}.$

Using $K=Q^{\kappa}=X^{\delta\kappa}$ we apply Lemma 3.4 to conclude that

$\Psi_{0}(\rho e(\alpha))\ll X^{1+\delta(2\kappa-1)}+X^{1-\delta\kappa},$

and letting $\delta=\kappa=\tfrac{1}{3}$ yields the bound $\Psi_{0}(\rho
e(\alpha))\ll X^{\frac{9}{10}}$, which is more than needed. ∎

We now consider $\Psi(\rho e(\alpha))$ with $\alpha$ in $\mathfrak{M}(X,Q)$,
recalling the definition

$\mathfrak{z}=\zeta(2)=\tfrac{\pi^{2}}{6}.$

######  Lemma 12.3.

Fix $A>0$, let $X>X_{0}(A)$, and let $Q=X^{1/3}$. Suppose that
$\alpha=a/q+\beta$ with $0\leq a<q\leq Q$, $(a,q)=1$, and $|\beta|\leq Q/qX$.
Then one has

$\Psi(\rho
e(\alpha))=W(q)\mathopen{}\mathclose{{}\left(\frac{\mathfrak{z}X}{1-2\pi
iX\beta}}\right)+O(X(\log X)^{-A}),$

where

$W(q):=\begin{cases}\tfrac{1}{4}q^{-2}&2\nmid q,\\\ q^{-2}&2\mid
q.\end{cases}$

###### Proof.

As $\Psi_{1}(\rho e(\alpha))\ll X(\log X)^{-A}$ per (12.1), it remains only to
analyze $\Psi_{0}(\rho e(\alpha))$. Again let $\tau=X^{-1}(1-2\pi iX\beta)$ so
that $\rho e(\alpha)=e(a/q)e^{-\tau}$, and for all $k$ set $q_{k}:=q/(q,k)$
and $a_{k}:=ak/(q,k)$. Grouping summands modulo $q_{k}$, we have

$F_{0}(\rho^{k}e(k\alpha))=\sum_{n=1}^{\infty}e\mathopen{}\mathclose{{}\left(\frac{nka}{q}}\right)\rho^{nk}e(nk\beta)=\sum_{\ell=1}^{q_{k}}e\mathopen{}\mathclose{{}\left(\frac{a_{k}\ell}{q_{k}}}\right)\sum_{n\equiv\ell\,(q_{k})}e^{-nk\tau}.$
(12.5)

Summing by parts we find that $\sum_{n\equiv\ell\,(q_{k})}e^{-nk\tau}$ is
equal to

$\displaystyle
k\tau\int_{0}^{\infty}e^{-xk\tau}\mathopen{}\mathclose{{}\left(\frac{x}{q_{k}}+O(1)}\right)\,dx=\frac{1}{q_{k}}\int_{0}^{\infty}e^{-xk\tau}xk\tau\,dx+O\mathopen{}\mathclose{{}\left(k|\tau|\int_{0}^{\infty}e^{-xk/X}\,dx}\right),$

and arguing as in the derivation of (5.10) we deduce that

$\sum_{n\equiv\ell\,(q_{k})}e^{-nk\tau}=\frac{1}{kq_{k}\tau}+O(|1-2\pi
iX\beta|)=\frac{1}{kq_{k}\tau}+O\mathopen{}\mathclose{{}\left(\frac{Q}{q}}\right).$
(12.6)

Combining (12.5) and (12.6), it follows that

$F_{0}(\rho^{k}e(k\alpha))=\frac{1}{kq_{k}\tau}\sum_{\ell=1}^{q_{k}}\mathopen{}\mathclose{{}\left(\frac{\ell
a_{k}}{q_{k}}}\right)+O(Q).$

Since $(a_{k},q_{k})=1$, the sum here is equal to $q_{k}$ when $q_{k}=1$
(i.e., when $q\hskip 0.39993pt|\hskip 0.39993ptk$) and equal to 0 otherwise,
whereby

$\sum_{\begin{subarray}{c}k\leq K\\\ 2\hskip 0.27995pt\mid\hskip
0.27995ptk\end{subarray}}\frac{1}{k}F_{0}(\rho^{k}e(k\alpha))=\tau^{-1}\sum_{\begin{subarray}{c}k\leq
K\\\ 2\hskip 0.27995pt\mid\hskip 0.27995ptk\\\ q\hskip 0.27995pt\mid\hskip
0.27995ptk\end{subarray}}\frac{1}{k^{2}}+O\mathopen{}\mathclose{{}\left(Q\log
K}\right).$

Conditioning on the parity of $q$ and noting that $|\tau^{-1}|\leq X$, we find
that

$\sum_{\begin{subarray}{c}k\leq K\\\ 2\hskip 0.27995pt\mid\hskip
0.27995ptk\end{subarray}}\frac{1}{k}F_{0}(\rho^{k}e(k\alpha))=\mathfrak{z}W(q)\tau^{-1}+O\mathopen{}\mathclose{{}\left(\frac{X}{K}}\right)+O(Q\log
K),$

and by applying Lemma 3.2 it follows that

$\Psi_{0}(\rho
e(\alpha))=\mathfrak{z}W(q)\tau^{-1}+O(X/K)+O\mathopen{}\mathclose{{}\left(Q\log
K}\right).$

The result then follows upon substituting $Q=X^{1/3}$ and $K=Q^{1/3}=X^{1/9}$,
in agreement with the parameters $\delta$ and $\kappa$ used in proving Lemma
12.2. ∎

######  Remark 12.4.

Considering the constants $W(q)$ we find that

$W(1)=W(2)=\tfrac{1}{4},\quad W(4)=\tfrac{1}{16},\quad\text{and}\quad
W(q)\leq\tfrac{1}{36}\quad\text{otherwise}.$

Recalling from Proposition 5.4 that

$V(1)=V(2)=\tfrac{1}{4},\quad
V(4)=0,\quad\text{and}\quad|V(q)|\leq\tfrac{1}{5}\quad\text{otherwise,}$

we see that $W(q)$ and $V(q)$ behave similarly as functions of $q$, excepting
when $q=4$.

As done for our analysis of $\Psi(\rho e(\alpha),\mu)$ for $\alpha$ in the
principal arcs, we again let

$\eta=X^{-1}(\log X)^{-1/4}\qquad\text{and}\qquad\eta_{*}=X_{*}^{-1}(\log
X_{*})^{-1/4}$

and identify the sets

$\\{e(\alpha):\alpha\in\mathfrak{P}\\}\qquad\text{and}\qquad\\{e(\alpha):\alpha\in\mathfrak{P}_{*}\\}$

with

$\\{e(\beta):\text{$\beta\in\mathbb{R}$ and
$|\beta|\leq\eta$}\\}\qquad\text{and}\qquad\\{-e(\beta):\text{$\beta\in\mathbb{R}$
and $|\beta|\leq\eta_{*}$}\\},$

respectively. As such we have the following analogue of Proposition 7.1.

######  Lemma 12.5.

Fix $A>0$. For all $X>X_{0}(A)$, $j\geq 0$, and $|\beta|\leq\eta$, one has

$\displaystyle\Psi_{(j)}(\pm\rho e(\beta))$ $\displaystyle=j!\hskip
0.80002pt\frac{\mathfrak{z}}{4}\mathopen{}\mathclose{{}\left(\frac{X}{1-2\pi
iX\beta}}\right)^{j+1}+O_{j}(X^{j+1}(\log X)^{-A}),$ (12.7) $\displaystyle(\pm
1)^{j}\Psi^{(j)}(\pm\rho e(\beta))$ $\displaystyle=j!\hskip
0.80002pt\frac{\mathfrak{z}}{4}\mathopen{}\mathclose{{}\left(\frac{X}{1-2\pi
iX\beta}}\right)^{j+1}+O_{j}(X^{j+1}(\log X)^{-A}).$ (12.8)

###### Proof.

The proof is nearly identical to the proof of Proposition 7.1 following the
requisite substitutions. In particular, the integrands now involve

$\displaystyle\mathfrak{D}[\lambda;s]$
$\displaystyle=\frac{\zeta(2s)}{\zeta(s)}\qquad\text{and}\qquad\mathfrak{D}[\lambda^{2};s]=\zeta(s)$

in place of $\mathfrak{D}[\mu;s]$ and $\mathfrak{D}[\mu^{2};s]$, respectively.
Our main terms in (12.7) and (12.8) now come from a residue of

$2^{-2}\zeta(2)\Gamma(j+1)\tau^{-j-1}=j!\hskip
0.80002pt\frac{\mathfrak{z}}{4}\mathopen{}\mathclose{{}\left(\frac{X}{1-2\pi
iX\beta}}\right)^{j+1},$

cf. expression (7.14). When $z=-\rho e(\beta)$, similar to formula (7.16) we
find that

$\mathfrak{D}[(-1)^{n}\lambda(n);s]=-(1+2^{1-s})\frac{\zeta(2s)}{\zeta(s)},$

and again the arguments used in proving Proposition 7.1 yield the result. ∎

Using Lemma 12.3 in place of Proposition 5.4, the arguments of section 8 for
Lemma 8.1 and Theorem 1.2 similarly yield the uniqueness of the solutions
$\rho$ and $\rho_{*}$ of (sp) and (sp$*$), respectively, as well as the
relations of Theorem 1.4. For convenience, again using $\mathfrak{z}=\zeta(2)$
we record here the particular relations

$\displaystyle-x\log\rho$
$\displaystyle=\tfrac{1}{2}\sqrt{\mathfrak{z}x}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right],\qquad X$
$\displaystyle=2\sqrt{x/\mathfrak{z}}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right],$ (12.9) $\displaystyle-x\log\rho_{*}$
$\displaystyle=\tfrac{1}{2}\sqrt{\mathfrak{z}x}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right],\qquad X_{*}$
$\displaystyle=2\sqrt{x/\mathfrak{z}}\mathopen{}\mathclose{{}\left[1+O((\log
x)^{-A})}\right].$ (12.10)

As done in section 9 we again use an “arc transference” to write

$p(x,\lambda)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\rho
e(\mathscr{A})}\Phi(z)z^{-x-1}\,dz+\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\rho_{*}e(\mathscr{A}_{*})}\Phi(z)z^{-x-1}\,dz+\mathfrak{T}(\rho,\rho_{*}),$
(12.11)

where again

$\mathscr{A}:=[0,\tfrac{1}{4})\cup[\tfrac{3}{4},1),\qquad\mathscr{A}_{*}:=[\tfrac{1}{4},\tfrac{3}{4}),$

and

$\mathfrak{T}(\rho,\rho_{*}):=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{i\rho}^{i\rho_{*}}\Phi(z)z^{-n-1}\,dz+\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{-i\rho_{*}}^{-i\rho}\Phi(z)z^{-n-1}\,dz.$

Once again, the term $\mathfrak{T}(\rho,\rho_{*})$ only contributes a small
error term to our formula (12.11).

######  Lemma 12.6 (cf. Lem. 9.2).

For $x>x_{0}$ let $\rho$ and $\rho_{*}$ be the unique solutions to equations
(sp) and (sp$*$), respectively. For all $\varepsilon>0$, as $x\to\infty$ one
has

$\mathfrak{T}(\rho,\rho_{*})\ll\exp\mathopen{}\mathclose{{}\left((\tfrac{3}{4}+\varepsilon)\sqrt{\mathfrak{z}x}}\right).$

###### Proof.

Fixing $A,\varepsilon>0$ and letting $x>x_{0}(A)$, the relations (12.9) and
(12.10) show that

$\int_{\rho}^{\rho_{*}}\frac{dt}{t^{x+1}}=\frac{1}{x}\mathopen{}\mathclose{{}\left[e^{\frac{1}{2}\sqrt{\mathfrak{z}x}+o(\sqrt{x})}-e^{\frac{1}{2}\sqrt{\mathfrak{z}x}+o(\sqrt{x})}}\right]\ll
e^{(\frac{1}{2}+\varepsilon)\sqrt{\mathfrak{z}x}}.$ (12.12)

For $\rho\leq t\leq\rho_{*}$ define $X_{t}$ via $t=\exp(-1/X_{t})$, and assume
$x$ (and $\rho$) to be large enough that $X_{t}(\log X_{t})^{-A}$ is
increasing as a function of $t$ for $t\geq\rho$. As $W(4)=\tfrac{1}{16}$ in
Lemma 12.3, for $\rho\leq t\leq\rho_{*}$ and sufficiently large $\rho$,
$X_{*}$, and $x$, we have

$\operatorname{Re}\Psi(it)=\frac{\mathfrak{z}}{16}X_{t}+O(X_{t}(\log
X_{t})^{A})\leq\frac{\mathfrak{z}}{16}X_{*}+O(X_{*}(\log
X_{*})^{-A})\leq\frac{\mathfrak{z}}{8}X_{*},$ (12.13)

and, using equation (12.10), we further have

$\frac{\mathfrak{z}}{8}X_{*}=\tfrac{1}{4}\sqrt{\mathfrak{z}x}+O(\sqrt{x}(\log
x)^{-A})\leq(\tfrac{1}{4}+\varepsilon)\sqrt{\mathfrak{z}x}.$ (12.14)

The result then follows from (12.12)–(12.14), akin to the proof of Lemma 9.2.
∎

Due to the overwhelming similarities between Theorems 1.2 and 1.4 and between
Lemmata 9.2 and 12.6, and the parallels drawn in Remark 12.4, the remaining
arguments of sections 9–11 are naturally ported to analogous arguments
concerning $p(x,\lambda)$, $\Phi(z,\lambda)$, etc.. As such, there is little
to be gained by repeating said arguments, so we merely state the salient
results concerning $p(x,\lambda)$ and reference their respective
$p(x,\mu)$-analogues.

######  Lemma 12.7 (cf. Lem. 10.1 and Prop. 10.3).

For $x>x_{0}$ let $\rho$ and $\rho_{*}$ be the unique solutions to equations
(sp) and (sp$*$), respectively. For fixed $B>0$, as $x\to\infty$ one has

$\displaystyle p(x,\lambda)$
$\displaystyle=\rho^{-x}\int_{\mathfrak{P}}\Phi(\rho
e(\alpha))e(-x\alpha)\,d\alpha+\rho_{*}^{-x}\int_{\mathfrak{P}_{*}}\Phi(\rho
e(\alpha))e(-x\alpha)\,d\alpha$
$\displaystyle\qquad+O(\rho^{-x}\Phi(\rho)x^{-B})+(\rho_{*}^{-x}\Phi(-\rho_{*})x^{-B})+O(e^{(\frac{3}{4}+\varepsilon)\sqrt{\mathfrak{z}x}}).$

######  Lemma 12.8 (cf. Prop. 11.2 and eq. (11.16)).

For all $x>x_{0}$ let $\rho$ and $\rho_{*}$ be the unique solutions to
equations (sp) and (sp$*$), respectively. As $x\to\infty$, one has

$\displaystyle\rho^{-x}\int_{\mathfrak{P}}\Phi(\rho
e(\alpha))e(-x\alpha)\,d\alpha$
$\displaystyle=\frac{\rho^{-x}\Phi(\rho)}{\sqrt{2\pi\Psi_{(2)}(\rho)}}\mathopen{}\mathclose{{}\left[1+O(x^{-1/5})}\right],$
$\displaystyle\rho_{*}^{-x}\int_{\mathfrak{P}_{*}}\Phi(\rho_{*}e(\alpha))e(-x\alpha)\,d\alpha$
$\displaystyle=(-1)^{-x}\frac{\rho_{*}^{-x}\Phi(-\rho_{*})}{\sqrt{2\pi\Psi_{(2)}(-\rho_{*})}}\mathopen{}\mathclose{{}\left[1+O(x^{-1/5})}\right].$

where $(-1)^{-x}=\exp(-\pi ix)$.

Combining Lemmata 12.7 and 12.8 and repeating the arguments used in deriving
equation (11.18) from equation (11.17), Theorem 1.5 is similarly established.

## Appendix

Here we provide the details verifying that the constants $V(q)$ of Proposition
5.4 satisfy the relations $V(1)=V(2)=\tfrac{1}{4}$, $V(4)=0$, and
$|V(q)|\leq\tfrac{1}{5}$ for all other $q$. We recall that $G(q)$ is the
multiplicative function defined on prime powers by

$G(p^{k})=\begin{cases}-(p^{2}-1)^{-1}&k=1,2,\\\ 0&k>2,\end{cases}$

and that $V(q)$ is given by

$V(q)=\frac{1}{\zeta(2)}\sum_{d\hskip 0.27995pt\mid\hskip
0.27995ptq}G\mathopen{}\mathclose{{}\left(\frac{q}{d}}\right)\sum_{\begin{subarray}{c}k\equiv
0\,(2)\\\ (k,q)=d\end{subarray}}\frac{1}{k^{2}}.$

The values for $1$, $2$, and $4$ are immediately computed, while the bound
$|V(q)|\leq\tfrac{1}{5}$ requires more slightly more work. First, for odd
primes $p$ we compute that

$\displaystyle\zeta(2)V(p)$
$\displaystyle=G(p)\sum_{\begin{subarray}{c}k\equiv 0\,(2)\\\
(k,p)=1\end{subarray}}\frac{1}{k^{2}}+G(1)\sum_{\begin{subarray}{c}k\equiv
0\,(2)\\\ (k,p)=p\end{subarray}}\frac{1}{k^{2}}$
$\displaystyle=G(p)\mathopen{}\mathclose{{}\left(\sum_{k=1}^{\infty}\frac{1}{k^{2}}-\sum_{p\hskip
0.27995pt\mid\hskip
0.27995ptk}\frac{1}{k^{2}}}\right)+\frac{\zeta(2)}{4p^{2}}$
$\displaystyle=\zeta(2)\mathopen{}\mathclose{{}\left[G(p)\frac{p^{2}-1}{4p^{2}}+\frac{1}{4p^{2}}}\right]=0,$

so that $V(p)=0$ for all odd primes $p$. Next, using the notation $2^{a}\|k$
to indicate that $2^{a}\mid k$ and $2^{a+1}\nmid k$, if $n\geq 3$ then

$\displaystyle\zeta(2)V(2^{n})$
$\displaystyle=G(4)\sum_{2^{n-2}\,\|\,k}\frac{1}{k^{2}}+G(2)\sum_{2^{n-1}\,\|\,k}\frac{1}{k^{2}}+G(1)\sum_{2^{n}\hskip
0.27995pt\mid\hskip 0.27995ptk}\frac{1}{k^{2}}$
$\displaystyle\leq\mathopen{}\mathclose{{}\left[\frac{1/3}{(2^{n-2})^{2}}+\frac{1/3}{(2^{n-1})^{2}}+\frac{1}{(2^{n})^{2}}}\right]\zeta(2)$
$\displaystyle=\frac{\zeta(2)}{2^{2n}}\cdot\frac{23}{3},$

and it follows that $V(2^{n})\leq\tfrac{23}{192}\approx 0.12$ for $n\geq 3$.
For all other $q$ we first crudely bound

$|V(q)|\leq\sum_{d\,|\,q}\frac{|G(q/d)|}{d^{2}}=\frac{1}{q^{2}}\sum_{d\,|\,q}|G(d)|d^{2}.$
(A.1)

The latter sum in equation (A.1) is a multiplicative function of $q$. Writing
$q=p_{1}^{a_{1}}\ldots p_{r}^{a_{r}}$, said sum is a product of factors either
of the form

$\frac{2p_{i}^{2}-1}{p_{i}^{2}(p_{i}^{2}-1)}\quad\text{or}\quad\frac{p_{i}^{4}+2p_{i}^{2}-1}{p_{i}^{2a_{i}}(p_{i}^{2}-1)},$
(A.2)

where the factor is of the first form when $a_{i}=1$ and of the second form
when $a_{i}>1$. The second factor in (A.2) is bounded above by

$\frac{(p_{i}^{2}+1)^{2}}{p_{i}^{4}(p^{2}-1)},$

and direct inspection shows that the majorant in (A.1) is largest when $q=6$
(among those $q$ not equal to 1, any prime $p$, or a power of $2$), yielding a
bound of $\tfrac{7}{12}\cdot\tfrac{17}{72}\approx 0.138<\tfrac{1}{5}$.

## References