# Turán inequalities for the plane partition function

Ken Ono, Sudhir Pujahari, and Larry Rolen  Department of Mathematics,
University of Virginia, Charlottesville, VA 22904<EMAIL_ADDRESS>School of Mathematical Sciences, National Institute of Science Education and
Research, Bhubaneswar, HBNI, P. O. Jatni, Khurda 752050, Odisha, India
<EMAIL_ADDRESS>Department of Mathematics, Vanderbilt University,
Nashville, TN 37240<EMAIL_ADDRESS>
###### Abstract.

Heim, Neuhauser and Tröger recently established some inequalities for
MacMahon’s plane partition function $\mathrm{PL}(n)$ that generalize known
results for Euler’s partition function $p(n)$. They also conjectured that
$\mathrm{PL}(n)$ is log-concave for all $n\geq 12.$ We prove this conjecture.
Moreover, for every $d\geq 1$, we prove their speculation that
$\mathrm{PL}(n)$ satisfies the degree $d$ Turán inequalities for sufficiently
large $n$. The case where $d=2$ is the case of log-concavity.

###### Key words and phrases:

Plane partition function, Log-concavity, Turán inequalities

###### 2020 Mathematics Subject Classification:

Primary 11P82,05A17; Secondary 05A20

The first author thanks the Thomas Jefferson Fund and the NSF (DMS-2002265 and
DMS-2055118) for their generous support, as well as the Kavli Institute grant
NSF PHY-1748958. This work was supported by a grant from the Simons Foundation
(853830, LR). The third author is also grateful for support from a 2021-2023
Dean’s Faculty Fellowship from Vanderbilt University.

##  1\. Introduction and Statement of Results

A partition of a non-negative integer $n$ is any non-increasing sequence of
positive integers that sum to $n$. Hardy and Ramanujan famously proved that
the partition function $p(n),$ which counts the number of integer partitions
of $n,$ satisfies the asymptotic formula

$p(n)\sim\frac{1}{4n\sqrt{3}}\cdot e^{\pi\sqrt{2n/3}}.$

In the ’70s, Nicolas [13] employed such asymptotics to prove111This result was
reproved in recent work by DeSalvo and Pak [6]. that $p(n)$ is log-concave for
$n>25,$ where a sequence of real numbers $\\{\alpha(0),\alpha(1),\dots\\}$ is
said to be log-concave at $n$ if

(1.1) $\alpha(n)^{2}\geq\alpha(n-1)\alpha(n+1).$

The condition of log-concavity for nonvanishing real sequences $\\{\alpha(n)$}
is the first example of the Turán inequalities, which can be conveniently
formulated in terms of Jensen polynomials. The Jensen polynomial of degree $d$
and shift $n$ is defined by

(1.2) $J_{\alpha}^{d,n}(X)\,:=\,\sum_{j=0}^{d}\left(\begin{matrix}d\\\
j\end{matrix}\right)\alpha(n+j)\,X^{j}.$

For degree $d=2$ and shift $n-1,$ the roots are

$\frac{-\alpha(n)\pm\sqrt{\alpha(n)^{2}-\alpha(n-1)\alpha(n+1)}}{\alpha(n+1)}.$

Therefore, $\alpha(n)$ is log-concave at $n$ if and only if the roots of
$J_{\alpha}^{2,n-1}(X)$ are real. Generalizing log-concavity, a real sequence
is said to satisfy the degree $d$ Turán inequalities at $n$ if
$J_{a}^{d,n-1}(X)$ is hyperbolic, where a polynomial is hyperbolic if all of
its roots are real.

There have been several recent works on the higher Turán inequalities for
$p(n).$ Chen, Jia and Wang [4] proved that $J_{p}^{3,n}(X)$ is hyperbolic for
$n\geq 94,$ which inspired them to conjecture, for every integer $d\geq 1,$
that there is an integer $N_{p}(d)$ for which $J_{p}^{d,n}(X)$ is hyperbolic
for $n\geq N_{p}(d)$. Griffin, Zagier and two of the authors recently proved
(see Theorem 5 of [8]) this conjecture for all degrees $d$. Furthermore,
recent work by Larson and Wagner [11] established the optimal values
$N_{p}(4)=206$ and $N_{p}(5)=381,$ as well as the effective bound
$N_{p}(d)\leq(3d)^{24d}(50d)^{3d^{2}}.$

We consider such questions for plane partitions (for background, see
references by Andrews [2] and Stanley [16]). A plane partition of size $n$ is
an array of non-negative integers $\pi:=(\pi_{i,j})$ for which
$|\pi|:=\sum_{i,j}\pi_{i,j}=n$, in which the rows and columns are weakly
decreasing. The figure below offers a 3d rendering of a plane partition.

Figure 1. Example of a plane partition

If $\mathrm{PL}(n)$ is the number of size $n$ plane partitions, then MacMahon
[12] proved that

(1.3)
$f(x)=\sum_{n=0}^{\infty}\mathrm{PL}(n)x^{n}:=\prod_{n=1}^{\infty}\frac{1}{(1-x^{n})^{n}}=1+x+3x^{2}+6x^{3}+13x^{4}+24x^{5}+48x^{6}+\dots.$

This function also appears prominently in physics in connection with the
enumeration of small black holes in string theory. Indeed, $f(x)$ is the
generating function (for example, see Appendix E of [5]) for the number of BPS
bound states between a $D6$ brane and $D0$ branes on $\mathbb{C}^{3}.$

Heim, Neuhauser, and Tröger [9] have undertaken a study of $\mathrm{PL}(n)$ in
analogy with the aforementioned results on $p(n)$. In addition to proving many
inequalities satisfied by $\mathrm{PL}(n)$, they pose two further conjectures.
They prove (see Theorem 1.2 of [9]) that $\mathrm{PL}(n)$ is log-concave for
sufficiently large $n$, and they pose the following explicit conjecture (see
Conjecture 1 of [9]).

######  Conjecture (Heim, Neuhauser, and Tröger).

The function $\mathrm{PL}(n)$ is log-concave for $n\geq 12.$

Here we resolve this problem.

######  Theorem 1.1.

The Heim-Neuhauser-Tröger Conjecture on the log-concavity of $\mathrm{PL}(n)$
is true.

Heim et al. also conjectured the direct analog of the Chen-Jia-Wang Conjecture
on the higher degree Turán inequalities. We prove this conjecture.

######  Theorem 1.2.

If $d$ is a positive integer, then $J_{\mathrm{PL}}^{d,n}(X)$ is hyperbolic
for all sufficiently large $n$.

######  Remark.

In his Ph.D thesis [15], Pandey will obtain an effective form of Theorem 1.2
that is a counterpart to the bound of $N_{p}(d)\leq(3d)^{24d}(50d)^{3d^{2}}$
established [11] by Larson and Wagner for $p(n).$

The proofs of Theorems 1.1 and 1.2 require a strong asymptotic formula for
$\mathrm{PL}(n).$ In the 1930s, Wright [17] adapted the “circle method” of
Hardy and Ramanujan to prove asymptotic formulas for $\mathrm{PL}(n).$ He
obtained such a formula for every positive integer $r$, where the implied
error terms are smaller with larger choices of $r$ for large $n$. The bulk of
this paper is devoted to the lengthy and delicate task of obtaining the first
asymptotics with explicitly bounded error terms.

To state these formulas, we require the two constants

(1.4) $A:=\zeta(3)\approx 1.202056\dots,\ \ \ \ {\text{\rm and}}\ \ \ \
c:=2\int_{0}^{\infty}\frac{y\log\,y}{e^{2\pi
y}-1}dy=\zeta^{\prime}(-1)\approx-0.16542\dots.$

Furthermore, for any pair of non-negative integers $s$ and $m$, we define
coefficients $c_{s,m}(n)$ by

$\frac{(1+y)^{2s+2m+\frac{13}{12}}}{(3+2y)^{(m+\frac{1}{2})}}=:\sum_{n=0}^{\infty}c_{s,m}(n)y^{n}.$

In terms of these coefficients, we define the important numbers

(1.5) $b_{s,m}:=c_{s,m}(2m).$

The asymptotic formulas we obtain are defined in terms of special numbers
$\beta_{0},\beta_{1},\dots.$ To define them, for every positive integer $s$ we
let

(1.6) $\alpha_{s}:=\frac{2\Gamma(2s+2)\zeta(2s)\zeta(2s+2)}{s(2\pi)^{4s+2}}.$

The real numbers $\beta_{s}$ are the Taylor coefficients of

(1.7)
$\exp\left(-\sum_{i=1}^{\infty}\alpha_{i}y^{i}\right)=:\sum_{n=0}^{\infty}\beta_{s}y^{s}.$

For each $r,$ we use the numbers $\beta_{0},\dots,\beta_{r+1}$ to derive the
following explicit asymptotic formula.

######  Theorem 1.3.

If $r\in\mathbb{Z}^{+},$ then for every integer $n\geq\max(n_{r},\ell_{r},87)$
(see (2.8-2.9)) we have

$\mathrm{PL}(n)=\frac{e^{c+3AN_{n}^{2}}}{2\pi}\sum_{s=0}^{r+1}\sum_{m=0}^{r+1}\frac{(-1)^{m}\beta_{s}b_{s,m}\Gamma\left(m+\frac{1}{2}\right)}{A^{m+\frac{1}{2}}N_{n}^{2s+2m+\frac{25}{12}}}+E_{r}^{{\text{\rm
maj}}}(n)+E^{\min}(n),$

where $|E_{r}^{{\text{\rm maj}}}(n)|\leq\widehat{E}_{r}^{{\text{\rm maj}}}(n)$
(see definition (2.36)), $N_{n}:=(\frac{n}{2A})^{\frac{1}{3}}$ and

$|E^{\min}(n)|\leq\exp\left(\left(3A-\frac{2}{5}\right)n^{2}/(2A)^{\frac{2}{3}}\right).$

######  Three remarks.

(1) The $s=m=0$ term in Theorem 1.3 gives the well-known asymptotic222It is
well-known that Wright has a typographical error where his asymptotic is off
by a factor $\sqrt{3}.$

$\mathrm{PL}(n)\sim\frac{(2^{25}A^{7})^{\frac{1}{36}}e^{c}}{\sqrt{12\pi}\cdot
n^{\frac{25}{36}}}\exp\left(\sqrt[3]{\frac{27An^{2}}{4}}\right).$

Theorem 1.3 has two explicit error terms. The error $E^{\min}(n),$ which is
independent of $r$ and arises from “minor arc” integrals, is exponentially
smaller as $(3A-\frac{2}{5})/(2A)^{\frac{2}{3}}\approx
1.79<\sqrt[3]{27A/4}\approx 2.01.$ The error term $E_{r}^{{\text{\rm
maj}}}(n)$, which arises from “major arc” integrals, only offers small power
savings in $n$ that improve with larger choices of $r$. In addition to the
expected complications required to make error terms explicit, the proof of
Theorem 1.1 is hampered by the small size of these power savings. This
annoying problem does not arise333This comment for $p(n)$ applies for the
Fourier coefficients of all non-positive weight weakly holomorphic modular
forms. where working with further terms in the circle method gives
exponentially improved error terms. Since the effective bounds must be
sufficient to reduce the conjecture to a finite range that can be handled by
computer, the task of proving Theorem 1.1 is a delicate balance between theory
and practicality. To prove Theorem 1.1, we use the case of $r=2$, where the
major arc power savings is on the order of $n^{-\frac{7}{3}},$ and our
theoretical bounds are sufficient to confirm the conjecture for all $n\geq
8820.$

(2) Almkvist [1] and Govindarajan and Prabhakar [7] have refined Wright’s
asymptotic formula in a different way. At the expense of requiring more
summands as a function of $n$ (i.e. $\sim\kappa\sqrt[3]{n}$ many summands),
their formulas give precise asymptotics. Given a positive integer $n$,
choosing $r\sim\kappa\sqrt[3]{n}$ in Theorem 1.3 gives similarly strong
asymptotics, with the added benefit that the error terms are explicitly
bounded.

(3) Wright’s main asymptotic formula is presented as a single sum, as opposed
to the double sum in $s$ and $m$ in Theorem 1.3. This double sum formulation
is the main device in Wright’s proof of his asymptotic formula. He makes a
further simplifcation to obtain a single sum expression. We do not take this
extra step as it would introduce further error.

######  Examples.

For all $n\geq 105$, the $r=1$ case of Theorem 1.3 implies (after some
calculation) that

$\begin{split}\mathrm{PL}(n)=e^{3\cdot
2^{-\frac{2}{3}}A^{\frac{1}{3}}n^{\frac{2}{3}}}n^{-\frac{25}{36}}&\Big{(}\frac{2^{\frac{25}{36}}e^{c}A^{\frac{7}{36}}}{\sqrt{12\pi}}-\frac{\sqrt{3}\cdot
2^{\frac{13}{36}}e^{c}{\left(3A+1385\right)}}{25920\,\sqrt{\pi}A^{\frac{5}{36}}}n^{-\frac{2}{3}}\\\
&-\frac{\sqrt{3}\cdot
2^{\frac{1}{36}}e^{c}{\left(1377A^{2}-370650A+12525625\right)}}{1567641600\sqrt{\pi}A^{\frac{17}{36}}}n^{-\frac{4}{3}}+E(n)\Big{)},\end{split}$

where $|E(n)|\leq 527n^{-\frac{5}{3}}.$ If $\widehat{\mathrm{PL}}_{1}(n)$
denotes this formula without the error $E(n),$ then we have

$E(n):=\left(\mathrm{PL}(n)-\widehat{\mathrm{PL}}_{1}(n)\right)\cdot
e^{-3\cdot 2^{-\frac{2}{3}}A^{\frac{1}{3}}n^{\frac{2}{3}}}n^{\frac{25}{36}}.$

Table 1 illustrates the observed strength of the power savings obtained by the
$\widehat{\mathrm{PL}}_{1}(n)$ estimate.

$n$ | $\mathrm{PL}(n)$ | $E(n)$ | $527n^{-\frac{5}{3}}$
---|---|---|---
$100$ | $5.92\ldots\times 10^{16}$ | $-1.18\ldots\times 10^{-7}$ | $0.24\ldots$
$200$ | $4.06\ldots\times 10^{27}$ | $-3.00\ldots\times 10^{-8}$ | $0.07\ldots$
$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$
$500$ | $2.91\ldots\times 10^{52}$ | $-4.87\ldots\times 10^{-9}$ | $0.01\ldots$

Table 1. Numerics for $r=1$ case of Theorem 1.3

For $n\geq 87$, explicit calculations444We leave the details of the proof of
the simpler $r=1$ case to the reader. with the $r=2$ case of Theorem 1.3 gives
(see (3.6))

$\mathrm{PL}(n)=\widehat{\mathrm{PL}}_{2}(n)+E_{2}(n),$

where $\widehat{\mathrm{PL}}_{2}(n)$ is defined by (3.3), and
$|E_{2}(n)|\leq\mathcal{E}_{2}(n):=227e^{3AN_{n}^{2}}n^{-\frac{109}{36}}+e^{\left(3A-\frac{2}{5}\right)N_{n}^{2}}.$
The bound $\mathcal{E}_{2}(n)$ is smaller than $\widehat{\mathrm{PL}}_{2}(n)$
for $n\geq 96.$ Table 2 below illustrates the strength of this estimate.

$n$ | $\widehat{\mathrm{PL}}_{2}(n)-\mathcal{E}_{2}(n)$ | $\mathrm{PL}(n)$ | $\widehat{\mathrm{PL}}_{2}(n)+\mathcal{E}_{2}(n)$
---|---|---|---
$100$ | $5.932\ldots\times 10^{15}$ | $5.920\ldots\times 10^{16}$ | $1.124\ldots\times 10^{17}$
$200$ | $3.706\ldots\times 10^{27}$ | $4.066\ldots\times 10^{27}$ | $4.426\ldots\times 10^{27}$
$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$
$500$ | $2.913\ldots\times 10^{52}$ | $2.915\ldots\times 10^{52}$ | $2.917\ldots\times 10^{52}$
$1000$ | $3.542\ldots\times 10^{84}$ | $3.542\ldots\times 10^{84}$ | $3.542\ldots\times 10^{84}$

Table 2. Numerics for $r=2$ case of Theorem 1.3

In Section 2 we prove Theorem 1.3 by modifying Wright’s implementation of the
circle method. As is common for most applications of the circle method, the
proof follows by considering integrals over “major” and “minor” arcs. Our
analysis of the major arc contributions is essentially a necessarily lengthy
and careful refinement of Wright’s original work. However, our analysis of the
minor arc contributions follows a completely different approach, which relies
on work of Zagier and a careful application of Euler-Maclaurin summation. In
Section 3 we deduce Theorem 1.1 and Theorem 1.2 from Theorem 1.3. Theorem 1.2
follows from recent work by Griffin, Zagier, and two of the authors in [8] on
Jensen polynomials for suitable arithmetic sequences.

## Acknowledgements

The authors thank Kathrin Bringmann, Will Craig, Michael Griffin, Bernhard
Heim, Greg Moore, Boris Pioline, and Wei-Lun Tsai for useful comments.
Finally, we are grateful to the anonymous referees for their careful reading
of the manuscript and their suggestions.

##  2\. Effective form of Wright’s asymptotic formulas

For convenience, we begin by outlining Wright’s strategy for obtaining
asymptotics for $\mathrm{PL}(n),$ which is a modification of the classical
“circle method.” For positive integers $n,$ we recall that
$N_{n}=(\frac{n}{2A})^{\frac{1}{3}},$ where $A=\zeta(3)$ as in (1.4), and we
consider the circle

(2.1) $C_{N_{n}}:=\left\\{x:|x|=e^{-\frac{1}{N_{n}}}\right\\}.$

Throughout, we let $\theta_{x}$ denote the principal value of $\arg(x),$ and
we divide $C_{N_{n}}$ into a “major arc” $C^{{}^{\prime}}_{N_{n}},$ consisting
of those $x$ with $|\theta_{x}|<\frac{1}{N_{n}},$ and the “minor arc”
$C_{N_{n}}^{{}^{\prime\prime}}$ which is its complement.

In terms of the generating function $f(x)$ in (1.3), Cauchy’s integral formula
immediately gives

(2.2) $\mathrm{PL}(n)=J(n)+E^{\min}(n),$

where

(2.3) $J(n):=\frac{1}{2\pi
i}\int_{C_{N_{n}}^{{}^{\prime}}}\frac{f(x)}{x^{n+1}}dx\ \ \ \ \ \ {\text{\rm
and}}\ \ \ \ \ \ E^{\min}(n):=\frac{1}{2\pi
i}\int_{C_{N_{n}}^{{}^{\prime\prime}}}\frac{f(x)}{x^{n+1}}dx$

(i.e. with the usual counterclockwise orientation). Wright analyzes $J(n)$ and
$E^{\min}(n)$ separately555Wright referred to $J(n)$ as $J_{1}(n)$ (resp.
$E^{\min}(n)$ as $J_{2}(n)$).. The asymptotics arise from the major arc piece
$J(n)$, and the minor arc $E^{\min}(n)$ piece is a small error term. To prove
Theorem 1.3, we improve on Wright’s analysis of $E^{\min}(n)$ with a
completely different argument, and we meticulously estimate $J(n)$ to obtain
explicit estimates.

###  2.1. Explicit bounds over the minor arcs

Here we bound $E^{\min}(n)$ using a different method from that of Wright.
Instead of working directly with $f(x)$, we use its logarithmic derivative,
which is the generating function for the sums of squares of divisors. Thanks
to this interpretation, we make connection with work of Zagier [18], and we
can then effectively bound $E^{\min}(n)$ using Euler-Maclaurin summation and
calculus.

######  Proposition 2.1.

For all $n\geq 87,$ we have
$|E^{\min}(n)|\leq\exp\left(\left(3A-\frac{2}{5}\right)\left(\frac{n}{2A}\right)^{\frac{2}{3}}\right).$

####  2.1.1. Lemmata for Proposition 2.1

The next lemma bounds $\log f(|x|)$ on the minor arcs.

######  Lemma 2.2.

If $x\in C_{N_{n}}^{{}^{\prime\prime}},$ then we have $\log f(|x|)\leq
A{N_{n}}^{2}+0.33N_{n}-0.5.$

###### Proof.

We begin by noting that

$\log f(|x|)=\sum_{m\geq 1}\frac{|x|^{m}}{m(1-|x|^{m})^{2}},$

which is known as the MacMahon function. We let $|x|=:q$ and $t:=1/N_{n},$ and
so we have $|x|=q=e^{-t}$. Taking the derivative, we obtain the Lambert series

$L(q):=q\frac{d}{dq}\sum_{m\geq 1}\frac{q^{m}}{m(1-q^{m})^{2}}=\sum_{m\geq
1}\frac{q^{m}(1+q^{m})}{(1-q^{m})^{3}}.$

Thanks to the elementary fact that
$X(1+X)/(1-X)^{3}=\sum_{k=1}^{\infty}k^{2}X^{k},$ we have that

$L(q)=g_{3}(q):=\sum_{m\geq 1}\frac{m^{2}q^{m}}{1-q^{m}}.$

As $q=e^{2\pi i\tau}$ with $\tau=\frac{it}{2\pi}$, we have
$q\frac{d}{dq}=\frac{1}{2\pi i}\frac{d}{d\tau}=-\frac{d}{dt},$ we have
$\frac{d}{dt}\log f(e^{-t})=-g_{3}(e^{-t}).$ Integrating this relation and
adding the correct constant, we obtain

$\log f(|x|)=\log f(e^{-t})=\int_{t}^{1}g_{3}(e^{-z})dz+\log f(e^{-1}).$

Since $\log f(e^{-1})\approx 1.036$, we can bound this by

(2.4) $|\log f(|x|)|=\log
f(|x|)\leq\left|\int_{t}^{1}g_{3}(e^{-z})dz\right|+1.04.$

Estimating the integral in this inequality is more involved. We make use of
Zagier’s work [18] on generating functions of arbitrary divisor power sums. He
considers (see p. 15 of [18]) the function $g_{3}(e^{-z})$ as $z\searrow 0$.
In the $k=3$ case of Example 3, he applies Proposition 3 of [18] with
$F(t):=\frac{t^{2}}{e^{t}-1}$ to obtain an asymptotic expansion for
$g_{3}(e^{-t})=\frac{1}{t^{2}}\sum_{m\geq 1}F(mt).$

As we need an estimate with explicitly bounded error, as opposed to an
asymptotic expansion, we dig into the proof of Proposition 3 of [18]. This
gives, for each $k\geq 1$, an exact formula for $g_{3}(e^{-t})$, where $k$
controls the number of terms in this asymptotic expansion, and leaves an
integral that must be analyzed. The $k=1$ case666This is $N=1$ in Zagier’s
paper. gives

$\sum_{m\geq
1}F(mt)=\frac{1}{t}\int_{0}^{\infty}F(z)dz+\frac{(-1)^{0}B_{1}F(0)t^{0}}{1!}+(-t)^{0}\int_{0}^{\infty}\frac{F^{\prime}(x)\overline{B_{1}}(x)}{1!}dx,$

where $B_{1}=-1/2$ is the Bernoulli number, and $\overline{B_{1}}(x)=x-\lfloor
x\rfloor-\frac{1}{2}$ is a periodization of the first Bernoulli polynomial
$B_{1}(x)=x-\frac{1}{2}$. Since $F(t)$ has a removable singularity at $t=0$
with Taylor expansion $F(t)=t-\frac{1}{2}t^{2}+\ldots,$ we have $F(0)=0.$
Moreover, Zagier computed that

$\int_{0}^{\infty}F(z)dz=(3-1)!\zeta(3)=2\zeta(3)=2A.$

Combining these observations, we obtain

$g_{3}(e^{-t})=\frac{2A}{t^{3}}-\frac{1}{t^{2}}\int_{0}^{\infty}F^{\prime}(x)(x-\lfloor
x\rfloor-1/2)dx.$

As $|x-\lfloor x\rfloor-1/2|\leq\frac{1}{2}$, we have

$|g_{3}(e^{-t})|\leq\frac{2A}{t^{3}}+\frac{1}{2t^{2}}\int_{0}^{\infty}|F^{\prime}(x)|dx=\frac{2A}{t^{3}}+\frac{1}{2t^{2}}\int_{0}^{\infty}\left|\frac{(xe^{x}-2e^{x}+2)x}{(e^{x}-1)^{2}}\right|dx.$

Since $\int_{0}^{\infty}|F^{\prime}(x)|dx\approx 0.6471$, we have
$|g_{3}(e^{-t})|\leq 2A/t^{3}+0.33/t^{2}.$ Therefore, (2.4) gives

$\displaystyle\log f(|x|)$
$\displaystyle\leq\int_{t}^{1}\left(\frac{2A}{z^{3}}+\frac{0.33}{z^{2}}\right)dz+1.04\leq\frac{A}{t^{2}}+\frac{0.33}{t}-0.5.$

Letting $t=1/N_{n}$ gives the lemma. ∎

The proof of Proposition 2.1 also requires the following convenient lower
bounds.

######  Lemma 2.3.

If $x\in C_{N_{n}}^{{}^{\prime\prime}},$ then we have

$\frac{|x|}{(1-|x|)^{2}}-\frac{|x|}{|1-x|^{2}}\geq\frac{N_{n}^{2}}{2}-\frac{1}{12}.$

###### Proof.

We note that

(2.5)
$\frac{|x|}{(1-|x|)^{2}}-\frac{|x|}{|1-x|^{2}}=\frac{|x|}{(1-|x|)^{2}}\left(1-\left(\frac{1-|x|}{|1-x|}\right)^{2}\right).$

We now estimate

$\frac{1-|x|}{|1-x|}=\frac{1-e^{-1/N_{n}}}{|1-x|}$

on $C_{N_{n}}^{\prime\prime}$. Recall that this is the arc of the circle of
radius $e^{-1/N_{n}}$ with angles ranging from $1/N_{n}$ to $2\pi-1/N_{n}$.
Geometrically, we have that the closest a point $x$ on this arc can get to the
point $(1,0)$ in the plane is when the angle is $1/N_{n}$ (or $2\pi-1/N_{n}$),
and so

$\frac{1-|x|}{|1-x|}\leq\frac{1-e^{-1/N_{n}}}{|1-e^{-1/N_{n}}e^{i/N_{n}}|}.$

We obtain a formula from the denominator using the Law of Cosines. Namely, if
we draw a triangle (see Figure 2) with sides consisting of the line from
$(0,0)$ to $x=e^{-1/N_{n}}e^{i/N_{n}}$ on the circle, the line from $(0,0)$ to
$(1,0)$, and the line connecting $x$ to $(1,0)$, then the unknown length
$|1-x|$ is the opposite side of the angle $1/N_{n}$ bounded by side lengths
$1$ and $e^{-1/N_{n}}$.

(1,0)$x=e^{-1/N}e^{i/N}$$\tiny{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\frac{1}{{N_{n}}}}}$

Figure 2. Triangle used to evaluate $|1-x|$

After letting $t=1/N_{n},$ these facts imply that

$|1-e^{-1/N_{n}}e^{i/N_{n}}|=\sqrt{1+e^{-2/N_{n}}-2e^{-1/N_{n}}\cos(1/N_{n})}=\sqrt{1+e^{-2t}-2e^{-t}\cos(t)}.$

Thus, we get that

$h(t):=\frac{1-e^{-1/N_{n}}}{|1-e^{-1/N_{n}}e^{i/N_{n}}|}=\frac{1-e^{-t}}{\sqrt{1+e^{-2t}-2e^{-t}\cos(t)}}=\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{48}t^{2}+\ldots.$

Since $t=1/N_{n}=(2A/n)^{\frac{1}{3}}$ and $n\geq 1$, we have that
$t\leq(2A)^{\frac{1}{3}}\leq 1.34$. On the interval $[0,1.34]$, the maximum
absolute value of $h^{\prime\prime}(t)$ is $\lim_{t\rightarrow
0}h^{\prime\prime}(t)=\frac{\sqrt{2}}{24}$. Thus, by Taylor’s Theorem,

$|h(t)|=\frac{1}{\sqrt{2}}+O_{\leq}\left(\frac{\sqrt{2}}{48}t^{2}\right),$

where $O_{\leq}(\cdot)$ means that the expression is bounded by $\cdot$ in
absolute value (i.e. the implied constant can be chosen to be 1 with ordinary
$O$-notation). Hence, for $x\in C_{N_{n}}^{\prime\prime}$, we have

(2.6)
$\frac{1-|x|}{|1-x|}\leq\frac{1}{\sqrt{2}}+O_{\leq}\left(\frac{\sqrt{2}}{48N_{n}^{2}}\right).$

Returning to (2.5), we estimate

$\frac{|x|}{(1-|x|)^{2}}=\frac{e^{-1/N_{n}}}{(1-e^{-1/N_{n}})^{2}}=\frac{e^{-t}}{(1-e^{-t})^{2}}=t^{-2}-\frac{1}{12}+\frac{1}{240}t^{2}+\ldots$

By a similar use of Taylor’s Theorem, we have the strict inequality

$\frac{e^{-t}}{(1-e^{-t})^{2}}>t^{-2}-\frac{1}{12}=N_{n}^{2}-\frac{1}{12}.$

Using this bound, (2.5) and (2.6) completes the proof as

$\displaystyle\frac{|x|}{(1-|x|)^{2}}-\frac{|x|}{|1-x|^{2}}$
$\displaystyle\geq\left(N_{n}^{2}-\frac{1}{12}\right)\left(1-\left(\frac{1}{\sqrt{2}}+O_{\leq}\left(\frac{\sqrt{2}}{48N_{n}^{2}}\right)\right)^{2}\right)\geq\frac{N_{n}^{2}}{2}-\frac{1}{12}.$

∎

####  2.1.2. Proof of Proposition 2.1

To estimate

$E^{\min}(n)=\frac{1}{2\pi
i}\int_{C_{N_{n}}^{{}^{\prime\prime}}}\frac{f(x)}{x^{n+1}}dx,$

we bound $|f(x)/x^{n+1}|$. Following Wright (see page 184 of [17]), we note
that

(2.7) $|\log f(x)|\leq\log
f(|x|)-\left(\frac{|x|}{(1-|x|)^{2}}-\frac{|x|}{|1-x|^{2}}\right).$

Lemma 2.2 proved that $\log f(|x|)\leq A{N_{n}}^{2}+0.33N_{n}-0.5.$
Furthermore, Lemma 2.3 establishes that
$\frac{|x|}{(1-|x|)^{2}}-\frac{|x|}{|1-x|^{2}}\geq\frac{N_{n}^{2}}{2}-\frac{1}{12}.$
Therefore, (2.7) gives

$\displaystyle|\log f(x)|\leq
AN_{n}^{2}+0.33N_{n}-0.5-\left(\frac{N_{n}^{2}}{2}-\frac{1}{12}\right)\leq(A-1/2)N_{n}^{2}+0.33N_{n},$

and so $|f(x)|\leq e^{(A-1/2)N_{n}^{2}+0.33N_{n}}.$ Thus, the integrand in
$E^{\min}(n)$ is bounded by

$\frac{|f(x)|}{|x|^{n+1}}\leq
e^{(A-1/2)N_{n}^{2}+0.33N_{n}+(n+1)/N_{n}}=e^{(3A-1/2)N_{n}^{2}+0.33N_{n}+1/N_{n}}.$

Since the integral defining $E^{\min}(n)$ is along a curve of length bounded
by the circumference of the whole circle of radius $e^{-1/N_{n}}$, which is
$2\pi e^{-\frac{1}{N}_{n}}$, we finally find that

$|E^{\min}(n)|\leq
e^{(3A-1/2)N_{n}^{2}+0.33N_{n}+1/N_{n}-1/N_{n}}=e^{(3A-1/2)N_{n}^{2}+0.33N_{n}}.$

The claimed inequality for $n\geq 87$ follows by analyzing this last
expression.

###  2.2. Explicit major arc formulas

The size of $\mathrm{PL}(n)$ is given by the major arc integral $J(n).$ To
reduce the complexity of error terms, for each positive integer $r$ we define
thresholds

(2.8) $n_{r}:=\min\left\\{n\geq 1\ :\
0.056\cdot\sum_{s=1}^{r+1}\left(\frac{s\cdot
A^{\frac{1}{3}}}{2^{\frac{7}{6}}n^{\frac{1}{3}}}\right)^{2s}\left(\frac{\pi^{2}n^{\frac{1}{3}}}{(2A)^{\frac{1}{3}}s}+2\right)<1\right\\}$

and

(2.9) $\ell_{r}:=\min\left\\{n\geq 1\ :\
2^{r+4}\pi^{3}\alpha_{r+2}N_{n}^{-2r-4}+5e^{-4.7N_{n}}<\frac{1}{2}\right\\}.$

######  Remark.

The thresholds $\ell_{r}$ and $n_{r}$ are simple to compute. For instance, we
have that $\ell_{j}=1$ for $j\leq 22$ (resp. $\ell_{j}=2$ for $23\leq j\leq
30$), and $n_{1}=1,$ $n_{2}=2,\dots,n_{5}=18.$

The following explicit major arc estimate is the main result of this
subsection.

######  Proposition 2.4.

If $r\in\mathbb{Z}^{+},$ then for every $n\geq\max(\ell_{r},n_{r},55)$ we have

$J(n)=\frac{e^{c+3AN_{n}^{2}}}{2\pi}\sum_{s=0}^{r+1}\sum_{m=0}^{r+1}\frac{(-1)^{m}\beta_{s}b_{s,m}\Gamma\left(m+\frac{1}{2}\right)}{A^{m+\frac{1}{2}}N_{n}^{2s+2m+\frac{25}{12}}}+E_{r}^{{\text{\rm
maj}}}(n),$

where $|E_{r}^{{\text{\rm maj}}}(n)|\leq\widehat{E}_{r}^{{\text{\rm maj}}}(n)$
(see (2.36)).

####  2.2.1. Lemmata for Proposition 2.4

For fixed $n,$ Wright sets

(2.10)
$z=\log\left(\frac{1}{x}\right)=\log\left|\frac{1}{x}\right|-i\vartheta=\log\left(e^{\frac{1}{N}_{n}}\right)-i\vartheta=\frac{1}{N_{n}}-i\vartheta=:\rho
e^{i\phi},$

and he defines

(2.11) $w:=\mathrm{Re}\left(\frac{\pi}{2z}\right)=\frac{\pi\cos\phi}{2\rho}.$

On the major arc $C_{N_{n}}^{\prime}$, we have $|\vartheta|<\frac{1}{N}_{n}$,
and so

(2.12)
$\rho=\sqrt{N_{n}^{-2}+\vartheta^{2}}\in\left[\frac{1}{N_{n}},\frac{\sqrt{2}}{N_{n}}\right].$

Furthermore, we have

(2.13) $|\phi|=|\arctan(\vartheta N_{n})|\leq\arctan(1)=\frac{\pi}{4}.$

Wright uses the basic integral identity (see (3.17) of [17])

$\int_{0}^{\infty}t\log(1-e^{-tz})dt=-\frac{A}{z^{2}},$

which implies

$-\log f(x)+\frac{A}{z^{2}}=-\log f(x)-\int_{0}^{\infty}t\log(1-e^{-tz})dt.$

Using the generating function for $f(x)$, this becomes

$\sum_{m\geq 1}m\log(1-e^{-mz})-\int_{0}^{\infty}t\log(1-e^{-tz})dt.$

Now since $\frac{1}{e^{2\pi it}-1}-\frac{1}{1-e^{-2\pi it}}=-1,$ this can be
written as

$-\log f(x)+\frac{A}{z^{2}}=\sum_{m\geq
1}m\log(1-e^{-mz})+\int_{\Gamma}\left(\frac{t\log(1-e^{-tz})}{e^{2\pi
it}-1}-\frac{t\log(1-e^{-tz})}{1-e^{-2\pi it}}\right)dt,$

where $\Gamma$ is the path from $0$ to $\infty$ which travels along the real
axis, apart from sufficiently small semicircles at the positive integers above
the real axis to avoid the poles of the integrand.

$1$$2$$3$$4$

Figure 3. The path of integration $\Gamma$

Letting $\Gamma^{\prime}$ be the reflection of the path $\Gamma$ across the
$y$-axis, using the Residue Theorem (note that the residue of
$\frac{t\log(1-e^{-tz})}{e^{2\pi it}-1}$ at $t=m\in\mathbb{Z}$ is
$m\log(1-e^{-mz})/(2\pi i)$, Wright expresses777We note that Wright’s notation
does not clearly indicate that this is an exact identity. (see (3.18) of [17])
this as

(2.14) $-\log
f(x)+\frac{A}{z^{2}}=\int_{\Gamma^{\prime}}\frac{t\log(1-e^{-tz})}{e^{2\pi
it}-1}dt-\int_{\Gamma}\frac{t\log(1-e^{-tz})}{1-e^{-2\pi it}}dt.$

To obtain an effective estimate for $f(x)$, we study these two integrals.

######  Lemma 2.5.

Assuming the notation and hypotheses above, we have

$\begin{split}\int_{\Gamma}\frac{t\log(1-e^{-tz})}{1-e^{-2\pi
it}}dt&=\int_{0}^{iw}\frac{t\log(1-e^{-tz})}{1-e^{-2\pi
it}}dt+O_{\leq}\left(35N_{n}^{2}e^{-\pi^{2}N_{n}}\right),\\\
\int_{\Gamma^{{}^{\prime}}}\frac{t\log\,(1-e^{-tz})}{e^{2\pi
it}-1}dt&=\int_{0}^{-iw}\frac{t\log(1-e^{-tz})}{e^{2\pi
it}-1}dt+O_{\leq}\left(N_{n}^{2}e^{-\pi^{2}N_{n}}\right).\end{split}$

###### Proof.

By Cauchy’s Theorem, Wright showed that the integral over $\Gamma$ (see p. 182
of [17]) is

(2.15) $\int_{\Gamma}\frac{t\log(1-e^{-tz})}{1-e^{-2\pi
it}}dt=\int_{0}^{iw}\frac{t\log(1-e^{-tz})}{1-e^{-2\pi
it}}dt+\int_{iw}^{iw+\infty}\frac{t\log(1-e^{-tz})}{1-e^{-2\pi it}}dt,$

where in the second integral on the right hand side the path is horizontal and
parallel to the $x$-axis. To estimate the absolute value of the integrand,
note that by the reverse triangle inequality (letting $x\in(0,\infty)$ so that
$iw+x$ denotes a typical point on the path of integration)

$\left|\frac{1}{1-e^{-2\pi i(iw+x)}}\right|\leq\left|\frac{1}{1-|e^{-2\pi
i(iw+x)}|}\right|=\left|\frac{1}{1-e^{\pi^{2}\cos\phi/\rho}}\right|.$

Using (2.12), this is bounded by $\left|1/(1-e^{\pi^{2}N_{n}})\right|\leq
1.007e^{-\pi^{2}N_{n}}.$ To derive this bound, we used the fact that
$N_{n}\geq N_{1}\approx 0.75$ is monotonically increasing in $n$ and
$e^{-\pi^{2}N_{1}}|1-e^{\pi^{2}N_{1}}|\approx 0.99936.$ Therefore, we have

$\left|\int_{iw}^{iw+\infty}\frac{t\log(1-e^{-tz})}{1-e^{-2\pi
it}}dt\right|\leq
1.007e^{-\pi^{2}N_{n}}\int_{iw}^{iw+\infty}\left|t\log(1-e^{-tz})\right|dt.$

Following Wright again, and making the change of variables $v=tz$, this is
bounded by

(2.16) $\displaystyle
1.007e^{-\pi^{2}N_{n}}\int_{L}\frac{\left|t\log(1-e^{-v})\right|}{|z|^{2}}dv\leq
1.007N_{n}^{2}e^{-\pi^{2}N_{n}}\int_{L}\left|v\log(1-e^{-v})\right|dv=:U(n),$

where $L$ is the ray from $v=\frac{\pi i}{2}e^{i\phi}\cos\phi$ that forms the
angle $\phi$ with the (positive) real axis.

We split the integral in (2.16) into two pieces. Wright also does this, but we
make a slightly different choice below to assist us in our goal of obtaining
effective estimates. Throughout, let $v_{t}:=iwz+te^{i\phi}$, so that
$L=\\{v_{t}\ :\ t\in[0,\infty)\\}.$ Noting that

(2.17) $wz=\frac{\pi\cos\phi}{2\rho}\cdot\rho e^{i\phi}=\frac{\pi\cos\phi
e^{i\phi}}{2},$

we first estimate the piece $|\log(1-e^{-v_{t}})|$ in the integrand of (2.16).
We use (4.5.6) of DLMF [14], which states that for complex arguments $y$, to
obtain

(2.18) $|\log(1+y)|\leq-\log(1-|y|),\quad\text{ when }|y|<1.$

We will break up the line $L$ into a compact piece $L_{1}$ and a remaining
piece $L_{2},$ where we can utilize this bound. To see where it applies, we
compute

(2.19)
$|e^{-v_{t}}|=e^{-\operatorname{Re}(v_{t})}=e^{w\operatorname{Im}(z)-\operatorname{Re}(te^{i\phi})}=e^{-w\vartheta-t\cos\phi}=e^{-\cos\phi\left(\frac{\pi\vartheta}{2\rho}+t\right)}.$

In turn, this is less than or equal to one if and only if
$\cos\phi\left(\frac{\pi\vartheta}{2\rho}+t\right)\geq 0.$ Using (2.12) and
(2.13), and using $\vartheta\geq-1/N$, we have

$\cos\phi\left(\frac{\pi\vartheta}{2\rho}+t\right)>\frac{\sqrt{2}}{2}\left(-\frac{\pi}{2\sqrt{2}}+t\right)=-\frac{\pi}{4}+\frac{t}{\sqrt{2}},$

which gives

(2.20) $|e^{-v_{t}}|\leq e^{\frac{\pi}{4}-\frac{t}{\sqrt{2}}}.$

Hence, if we let

$L_{1}:=\left\\{v_{t}\colon t\in[0,\pi/\sqrt{8}+1)\right\\},\quad
L_{2}:=\left\\{v_{t}\colon t\in[\pi/\sqrt{8}+1,\infty)\right\\},$

then we can use (2.18) to estimate the integrand on $L_{2}.$ Thanks to (2.20),
for $t>\pi/\sqrt{8}$ we have

$|v_{t}\log(1-e^{-v_{t}})|\leq-|v_{t}|\cdot\log\left(1-|e^{-v_{t}}|\right)\leq-|v_{t}|\log\left(1-e^{\frac{\pi}{4}-\frac{t}{\sqrt{2}}}\right).$

Noting that $|v_{t}|\leq|\cos\phi|\frac{\pi}{2}+t\leq\frac{\pi}{2}+t,$ and
using the bound $-\log(1-x)<x/(1-x)$ for $0\neq x<1$ on real-valued logarithms
from (4.5.2) of DLMF [14], we find

$\left|v_{t}\log(1-e^{-v_{t}})\right|\leq\frac{\left(\frac{\pi}{2}+t\right)\cdot
e^{\frac{\pi}{4}-\frac{t}{\sqrt{2}}}}{1-e^{\frac{\pi}{4}-\frac{t}{\sqrt{2}}}}.$

Since $L_{1}$ is compact, we then find the following estimate for (2.16) (note
that when we integrate on $L_{2}$, since we are integrating absolute values,
the change of variables in the differential goes away as it has absolute value
1):

$\displaystyle
U(n)=1.007N_{n}^{2}e^{-\pi^{2}N_{n}}\int_{L_{1}}\left|v\log(1-e^{-v})\right|dv+1.007N_{n}^{2}e^{-\pi^{2}N_{n}}{N_{n}^{2}}\int_{L_{2}}\left|v\log(1-e^{-v})\right|dv$
$\displaystyle\ \ \ \ \leq
1.007N_{n}^{2}e^{-\pi^{2}N_{n}}\left(\frac{\pi}{\sqrt{8}}+1\right)\cdot\max\left\\{|v_{t}|\cdot|\log(1-e^{-v_{t}})|\
\colon t\in[0,\pi/\sqrt{8}+1)\right\\}$ $\displaystyle\ \ \ \ \ \ \ \ \ \
+1.007N_{n}^{2}e^{-\pi^{2}N_{n}}\int_{\frac{\pi}{\sqrt{8}}+1}^{\infty}\frac{\left(\frac{\pi}{2}+t\right)\cdot
e^{\frac{\pi}{4}-\frac{t}{\sqrt{2}}}}{1-e^{\frac{\pi}{4}-\frac{t}{\sqrt{2}}}}dt$
$\displaystyle\ \ \ \ \leq
1.007N_{n}^{2}e^{-\pi^{2}N_{n}}\left(\frac{\pi}{\sqrt{8}}+1\right)\cdot\max\left\\{|v_{t}|\cdot|\log(1-e^{-v_{t}})|\colon
t\in[0,\pi/\sqrt{8}+1)\right\\}+4.8N_{n}^{2}e^{-\pi^{2}N_{n}}.$

Now we estimate $\log(1-e^{-v_{t}})$ on for $t\in[0,\pi/\sqrt{8}+1)$. We begin
by recalling that for complex $y$, the principal branch of the logarithm is
given by $\log(y)=\log(|y|)+i\arg(y).$ Thus, we have

(2.21) $|\log(1-e^{-v_{t}})\left|\leq|\log|1-e^{-v_{t}}|\right|+\pi.$

To bound the logarithm, we find the maximum and minimum on the interval
$t\in[0,\pi/\sqrt{8}+1)$. The only critical point of $|1-e^{-v_{t}}|$ is at
$t=\cosh(i\phi).$ Thus, the potential extrema of $|1-e^{-v_{t}}|$ are at $0$,
$\pi/\sqrt{8}$, and $\cosh(i\phi)$. That is, the maximum of
$|\log|1-e^{-v_{t}}||$ is bounded by

$\max\left\\{|1-e^{-v_{0}}|,\ |1-e^{-v_{\pi/\sqrt{8}+1}}|,\
|1-e^{-v_{\cosh(i\phi)}}|\right\\}.$

We evaluate these in turn. For instance, we have that
$|1-e^{-v_{0}}|=\left|1-e^{-\frac{\pi i\cos\phi e^{i\phi}}{2}}\right|.$ On the
interval $\phi\in[-\pi/4,\pi/4]$, we have $0.75\leq|1-e^{-v_{0}}|\leq 1.85$
and $0.9\leq|1-e^{-v_{\pi/\sqrt{8}+1}}|\leq 1.4.$ Similarly, we find
$0.7\leq\left|1-e^{-v_{\cosh(i\phi)}}\right|\leq 1.45.$ Combining these
observations, we have $\left|\log|1-e^{-v_{t}}|\right|\leq 0.615\ldots$ on
$[0,\pi/\sqrt{8}+1)$. Thus, by (2.21), we have $|\log(1-e^{-v_{t}})|\leq
3.76.$ To bound $|v_{t}|$ on this interval, using (2.17) and the triangle
inequality, we find that
$|v_{t}|\leq\frac{\pi\cos\phi}{2}+t\leq\frac{\pi}{2}+\frac{\pi}{\sqrt{8}}+1=3.68\ldots.$
Therefore, we conclude that

$\displaystyle
1.007N_{n}^{2}e^{-\pi^{2}N_{n}}\left(\frac{\pi}{\sqrt{8}}+1\right)\max\left\\{|v_{t}|\cdot|\log(1-e^{-v_{t}})\colon
t\in[0,\pi/\sqrt{8}+1)\right\\}\leq 30N_{n}^{2}e^{-\pi^{2}N_{n}}.$

As a consequence, we obtain

$1.007N_{n}^{2}e^{-\pi^{2}N_{n}}\int_{L}\left|v\log(1-e^{-v})\right|dv\leq
35N_{n}^{2}e^{-\pi^{2}N_{n}}.$

Returning to (2.15), we have shown that

(2.22) $\int_{\Gamma}\frac{t\log(1-e^{-tz})}{1-e^{-2\pi
it}}dt=\int_{0}^{iw}\frac{t\log(1-e^{-tz})}{1-e^{-2\pi
it}}dt+O_{\leq}\left(35N_{n}^{2}e^{-\pi^{2}N_{n}}\right).$

This bounds the second integral in (2.14), which is the first claim in the
lemma.

The second claim in the lemma follows by arguing as above (after suitable sign
changes in the integrand) using the path of integration along
$\Gamma^{\prime}.$ Namely, we get

(2.23) $\int_{\Gamma^{{}^{\prime}}}\frac{t\log\,(1-e^{-tz})}{e^{2\pi
it}-1}dt=\int_{0}^{-iw}\frac{t\log(1-e^{-tz})}{e^{2\pi
it}-1}dt+O_{\leq}\left(N_{n}^{2}e^{-\pi^{2}N_{n}}\right).$

∎

We require bounds for three integrals, two of which make use of the
$\alpha_{s}$ defined by (1.6).

######  Lemma 2.6.

Assuming the notation and hypotheses above, the following are true.

(1) If $n$ is a positive integer, then we have

$\mathcal{I}_{1}:=2\int_{0}^{w}\frac{y\log(yz)}{e^{2\pi y}-1}dy=c+\frac{\log
z}{12}+O_{\leq}\left(3e^{-4.7N_{n}}\right).$

(2) If $n\geq n_{r}$ is an integer, then we have

$\mathcal{I}_{2}:=-2\sum_{s=1}^{r+1}\frac{\zeta(2s)z^{2s}}{s(2\pi)^{2s}}\int_{0}^{w}\frac{y^{2s+1}}{e^{2\pi
y}-1}dy=-\sum_{s=1}^{r+1}\alpha_{s}z^{2s}+O_{\leq}\left(e^{-\frac{\pi^{2}N_{n}}{2}}\right).$

(3) If $n$ is a positive integer, then we have

$\mathcal{I}_{3}:=-2\sum_{s\geq
r+2}\frac{\zeta(2s)z^{2s}}{s(2\pi)^{2s}}\int_{0}^{w}\frac{y^{2s+1}}{e^{2\pi
y}-1}dy\leq 2^{r+4}\pi^{3}\alpha_{r+2}N_{n}^{-2r-4}.$

###### Proof.

We estimate these three quantities one-by-one. For $\mathcal{I}_{1},$ we have

$\mathcal{I}_{1}=2\int_{0}^{w}\frac{y\log(yz)}{e^{2\pi
y}-1}dy=2\int_{0}^{\infty}\frac{y\log(yz)}{e^{2\pi
y}-1}dy-2\int_{w}^{\infty}\frac{y\log(yz)}{e^{2\pi y}-1}dy=c+\frac{\log
z}{12}-2\int_{w}^{\infty}\frac{y\log(yz)}{e^{2\pi y}-1}dy.$

To evaluate the last integral, we first note that
$w=\frac{\pi\cos\phi}{2\rho}\geq\frac{\pi\sqrt{2}}{4\rho}\geq\frac{\pi
N_{n}}{4}\geq 0.58,$ where we used (2.13) and (2.12) and the fact that $n\geq
1$. By direct manipulation, for $y\geq 0.58$, we have

(2.24) $\frac{1}{e^{2\pi y}-1}\leq 1.1\cdot e^{-2\pi y}.$

Then on the interval $w\geq 0.58$, we find that

$\displaystyle\left|-2\int_{w}^{\infty}\frac{y\log(yz)}{e^{2\pi
y}-1}dy\right|\leq 2.2\left|\int_{w}^{\infty}y\log(yz)e^{-2\pi y}dy\right|$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ =\frac{2.2}{4\pi^{2}}\cdot e^{-2\pi
w}\left|1-e^{2\pi w}\operatorname{Ei}(-2\pi w)+(1+2\pi w)\log(wz)\right|,$

where $\operatorname{Ei}(x):=-\int_{-x}^{\infty}\frac{e^{-t}dt}{t}$.
Straightforward manipulation then gives

(2.25) $\left|-2\int_{w}^{\infty}\frac{y\log(yz)}{e^{2\pi y}-1}dy\right|\leq
e^{-2\pi w}\left(1+0.056\cdot(1+2\pi w)(|\log\rho|+\pi)\right).$

If $n\geq 7$, then $\sqrt{2}/N_{n}<1$, and so $w\geq\pi N_{n}/4.$ Therefore,
(2.25) is bounded from above by

$e^{-\frac{\pi^{2}N_{n}}{2}}\left(1+0.056\cdot\left(1+\frac{\pi^{2}N_{n}}{2}\right)(\log
N_{n}+\pi)\right).$

It is straightforward to show that this is less than or equal to
$3e^{-4.7N_{n}}.$ By direct computation for $1\leq n\leq 6$ using (2.25), we
find that this bound holds in general, and so we have

(2.26) $\mathcal{I}_{1}=c+\frac{\log
z}{12}+O_{\leq}\left(3e^{-4.7N_{n}}\right).$

Using (2.24) and the integral representation of $\zeta(s),$ we can manipulate
$\mathcal{I}_{2}$ to obtain

$\displaystyle\mathcal{I}_{2}$
$\displaystyle=-2\sum_{s=1}^{r+1}\frac{\zeta(2s)z^{2s}}{s(2\pi)^{2s}}\int_{0}^{\infty}\frac{y^{2s+1}}{e^{2\pi
y}-1}dy+2\sum_{s=1}^{r+1}\frac{\zeta(2s)z^{2s}}{s(2\pi)^{2s}}\int_{w}^{\infty}\frac{y^{2s+1}}{e^{2\pi
y}-1}dy$
$\displaystyle=-2\sum_{s=1}^{r+1}\frac{\zeta(2s)z^{2s}}{s(2\pi)^{2s}}\int_{0}^{\infty}\frac{y^{2s+1}}{e^{2\pi
y}-1}dy+O_{\leq}\left(2.2\cdot\sum_{s=1}^{r+1}\frac{\zeta(2s)z^{2s}}{s(2\pi)^{2s}}\int_{w}^{\infty}y^{2s+1}e^{-2\pi
y}dy\right)$
$\displaystyle=-2\sum_{s=1}^{r+1}\frac{\zeta(2s)z^{2s}}{s(2\pi)^{2s}}\left(\frac{\Gamma(2+2s)\zeta(2+2s)}{(2\pi)^{2+2s}}\right)+O_{\leq}\left(2.2\cdot\sum_{s=1}^{r+1}\frac{\zeta(2s)z^{2s}}{s(2\pi)^{2s}}\left(\frac{\Gamma(2+2s;2\pi
w)}{(2\pi)^{2+2s}}\right)\right),$

where $\Gamma(a;x):=\int_{x}^{\infty}t^{a-1}e^{-t}dt$ is the incomplete Gamma
function. Using (1.6), this is

$\mathcal{I}_{2}=-\sum_{s=1}^{r+1}\alpha_{s}z^{2s}+O_{\leq}\left(2.2\sum_{s=1}^{r+1}\frac{\zeta(2s)z^{2s}}{s(2\pi)^{2s}}\left(\frac{\Gamma(2+2s;2\pi
w)}{(2\pi)^{2+2s}}\right)\right).$

To estimate this, we note that the proof of Lemma 2.2 of [3] shows, for $a>2,$
that

$\Gamma(a;x)\leq\frac{(x+b_{a})^{a}-x^{a}}{ab_{a}}e^{-x},$

where $b_{a}:=\Gamma(a+1)^{\frac{1}{a-1}}$. Combined with the Bernoulli number
formula for $\zeta(s)$ at positive even integers, this gives

$\left|2.2\sum_{s=1}^{r+1}\frac{\zeta(2s)z^{2s}}{s(2\pi)^{2s}}\left(\frac{\Gamma(2+2s;2\pi
w)}{(2\pi)^{2+2s}}\right)\right|\leq\frac{1.1e^{-2\pi
w}}{4\pi^{2}}\sum_{s=1}^{r+1}\frac{\rho^{2s}B_{2s}}{s(2\pi)^{2s}(2s)!}\left(2\pi
w+\left[(2s+1)!\right]^{\frac{1}{2s+1}}\right)^{2s+1}.$

A simple manipulation bounds this by

$\frac{1.1\cdot e^{-2\pi
w}}{4\pi^{2}}\sum_{s=1}^{r+1}\frac{z^{2s}B_{2s}}{s(2\pi)^{2s}(2s)!}\left(2\pi
w+2s\right)^{2s+1}.$

Using the Bernoulli number upper bound from (24.9.8) of [14], recalling that
$w\geq\pi N_{n}/4$, and noting that
$w=\frac{\pi\cos\phi}{2\rho}<\frac{\pi}{2\rho}<\frac{\pi N_{n}}{2},$ this is
bounded by

$\displaystyle 0.056\cdot$ $\displaystyle e^{-2\pi
w}\sum_{s=1}^{r+1}\frac{\rho^{2s}}{s(2\pi)^{4s}}(2\pi w+2s)^{2s+1}=0.056\cdot
e^{-2\pi
w}\sum_{s=1}^{r+1}\rho^{2s}\left(\frac{1}{2\pi}+\frac{s}{2\pi^{2}}\right)^{2s}\left(\frac{2\pi
w}{s}+2\right)$ $\displaystyle\leq 0.056\cdot
e^{-\frac{\pi^{2}N_{n}}{2}}\sum_{s=1}^{r+1}\left(\frac{s\sqrt{2}}{4N_{n}}\right)^{2s}\cdot\left(\frac{\pi^{2}N_{n}}{s}+2\right)=n_{r}\cdot
e^{-\frac{\pi^{2}N_{n}}{2}}.$

Therefore, if $n\geq n_{r}$, then we obtain the claimed inequality for
$\mathcal{I}_{2}.$

Finally, we turn to the bound for $\mathcal{I}_{3}$, which we recall is

$\mathcal{I}_{3}:=-2\sum_{s=r+2}^{\infty}\frac{\zeta(2s)z^{2s}}{s(2\pi)^{2s}}\int_{0}^{w}\frac{y^{2s+1}}{e^{2\pi
y}-1}dy.$

As $\zeta(2r+4)\geq\zeta(2s)$ for all $s\geq r+2,$ we have

$|\mathcal{I}_{3}|\leq\frac{\zeta(2r+4)}{r+2}\frac{\rho^{2r+4}}{(2\pi)^{2r+4}}\int_{0}^{w}\frac{y^{2r+5}}{e^{2\pi
y}-1}\sum_{s=0}^{\infty}\left(\frac{y\rho}{2\pi}\right)^{2s}dy.$

Therefore, using (1.6) we find that

$\displaystyle|\mathcal{I}_{3}|$
$\displaystyle\leq\frac{\zeta(2r+4)}{r+2}\cdot\frac{\rho^{2r+4}}{(2\pi)^{2r+4}}\int_{0}^{w}\frac{y^{2r+5}}{(e^{2\pi
y}-1)\left(1-(\frac{y\rho}{2\pi})^{2}\right)}dy$
$\displaystyle<\frac{\zeta(2r+4)}{r+2}\cdot\frac{\rho^{2r+4}}{(2\pi)^{2r+4}}\int_{0}^{\infty}\frac{y^{2r+5}}{e^{2\pi
y}-1}dy$
$\displaystyle<\frac{\Gamma(2r+6)\zeta(2r+4)\zeta(2r+6)\rho^{2r+4}}{(r+2)(2\pi)^{2r+7}}=4\pi^{3}\alpha_{r+2}\rho^{2r+4}.$

Since, $\rho\leq\frac{\sqrt{2}}{N_{n}},$ we obtain the claimed inequality for
$\mathcal{I}_{3}.$ ∎

####  2.2.2. Proof of Proposition 2.4

We begin by recalling (2.14), which asserts that

$-\log
f(x)+\frac{A}{z^{2}}=\int_{\Gamma^{\prime}}\frac{t\log(1-e^{-tz})}{e^{2\pi
it}-1}dt-\int_{\Gamma}\frac{t\log(1-e^{-tz})}{1-e^{-2\pi it}}dt.$

By combining the two integrals as a single integral, Lemma 2.5 gives

(2.27)
$\log\left(f(x)\right)=\frac{A}{z^{2}}+\int_{0}^{w}\frac{y\log\left(2\sin\left(\frac{yz}{2}\right)\right)}{e^{2\pi
y-1}}dy+O_{\leq}\left(36N_{n}^{2}e^{-\pi^{2}N_{n}}\right).$

To use this formula, we employ the identity $\sin\tau=\tau\prod_{m\geq
1}\left(1-\frac{\tau^{2}}{m^{2}\pi^{2}}\right),$ which implies that

$\log(\sin(\tau))=\log\tau-\sum_{s\geq 1}\sum_{m\geq 1}\frac{\tau^{2s}}{s\cdot
m^{2s}\pi^{2s}}=\log\tau-\sum_{s\geq 1}\frac{\zeta(2s)\tau^{2s}}{s\pi^{2s}}.$

Hence, for every $r\geq 1,$ the integral in (2.27) satisfies888Wright refers
to $\mathcal{I}_{1}$, $\mathcal{I}_{2}$, and $\mathcal{I}_{3}$ as $I_{4}$,
$S_{1}$, and $S_{2}$, respectively, on page 183 of [17].

$\displaystyle\mathcal{I}_{1}+\mathcal{I}_{2}+\mathcal{I}_{3}=\int_{0}^{w}\frac{y\log\left(2\sin\left(\frac{yz}{2}\right)\right)}{e^{2\pi
y-1}}dy$ $\displaystyle\ \ \ =2\int_{0}^{w}\frac{y\log(yz)}{e^{2\pi
y}-1}dy-2\sum_{s=1}^{r+1}\frac{\zeta(2s)z^{2s}}{s(2\pi)^{2s}}\int_{0}^{w}\frac{y^{2s+1}}{e^{2\pi
y}-1}dy-2\sum_{s\geq
r+2}\frac{\zeta(2s)z^{2s}}{s(2\pi)^{2s}}\int_{0}^{w}\frac{y^{2s+1}}{e^{2\pi
y}-1}dy.$

Thanks to Lemma 2.6, for $n\geq n_{r},$ (2.27) gives999This is an explicit
form of the first equation on p. 184 of [17]. We correct a typographical error
where the sum on $s$ accidentally starts with $s=0$ instead of $s=1.$

(2.28) $\displaystyle\log f(x)=\frac{A}{z^{2}}+c+\frac{\log
z}{12}-\sum_{s=1}^{r+1}\alpha_{s}z^{2s}+O_{\leq}\left(2^{r+4}\pi^{3}\alpha_{r+2}N_{n}^{-2r-4}+5e^{-4.7N_{n}}\right).$

We now make use of a complex-analytic version for the remainder terms of the
Taylor series of $f(x),$ which is required as our estimates make use of the
expressions involving $\beta_{s}$ as opposed to $\alpha_{s}$. Specifically,
the $\beta_{s}$ were defined in (1.7) to be the initial $r+2$ Taylor
coefficients of

$g_{r}(z):=e^{-\sum_{s=1}^{r+1}\alpha_{s}z^{s}}=\sum_{s=0}^{r+1}\beta_{s}z^{s}+R_{r}(z),$

where $R_{r}(z)$ is the remainder. For convenience, we assume that $n\geq 55$,
which guarantees that $|z|=\rho\leq\sqrt{2}N_{n}^{-1}\leq\frac{1}{2}.$ The
standard complex Taylor series remainder estimate (for example, see Theorem
B.21 of [10] with $R=1$) gives

$|R_{r}(z)|\leq\frac{\max_{|z|=1}(|g_{r}(z)|)\cdot|z|^{r+2}}{1-|z|}\leq
C_{r}|z|^{r+2},$

where

(2.29)
$C_{r}:=2\max_{|z|=1}\left(\left|e^{-\sum_{s=1}^{r+1}\alpha_{s}z^{s}}\right|\right).$

By replacing $z$ with $z^{2}$ (this is allowed since we demanded that $|z|<1$,
and so $|z^{2}|<1$ is still in the range of validity for the remainder
estimate), we obtain

$g_{r}(z^{2})=e^{-\sum_{s=1}^{r+1}\alpha_{s}z^{2s}}=\sum_{s=0}^{r+1}\beta_{s}z^{2s}+O_{\leq}\left(C_{r}|z|^{2r+4}\right).$

Therefore, by exponentiating (2.28), for $n\geq 55$ we obtain

$f(x)=e^{c}z^{\frac{1}{12}}e^{\frac{A}{z^{2}}}\left(\sum_{s=0}^{r+1}\beta_{s}z^{2s}+O_{\leq}\left(C_{r}|z|^{2r+4}\right)\right)\cdot
O_{\leq}\left(\exp(2^{r+4}\pi^{3}\alpha_{r+2}N_{n}^{-2r-4}+5e^{-4.7N_{n}})\right).$

To address the error term on the far right above, we assume that
$n\geq\ell_{r},$ which by (2.9) gives

$2^{r+4}\pi^{3}\alpha_{r+2}N_{n}^{-2r-4}+5e^{-4.7N_{n}}\leq\frac{1}{2}<1\
\text{ and }\
\frac{1}{1-(2^{r+4}\pi^{3}\alpha_{r+2}N_{n}^{-2r-4}+5e^{-4.7N_{n}})}\leq 2.$

Thanks to (4.5.11) of [14], which states that $e^{x}<1+x/(1-x)$ for $x<1$,
this gives

$\begin{split}f(x)&=e^{c}z^{\frac{1}{12}}e^{\frac{A}{z^{2}}}\left(\sum_{s=0}^{r+1}\beta_{s}z^{2s}+O_{\leq}\left(C_{r}|z|^{2r+4}\right)\right)\cdot
O_{\leq}\left(1+2^{r+5}\pi^{3}\alpha_{r+2}N_{n}^{-2r-4}+10e^{-4.7N_{n}}\right)\\\
&=M(x)+\mathcal{X}_{r}(n)+\mathcal{Y}_{r}(n),\end{split}$

where we let
$M(x):=e^{c+\frac{A}{z^{2}}}\sum_{s=0}^{r+1}\beta_{s}z^{2s+\frac{1}{12}},$ and
where

(2.30)
$\mathcal{X}_{r}(n):=e^{c+AN_{n}^{2}}2^{r+\frac{49}{24}}C_{r}N_{n}^{-2r-\frac{49}{12}}$

and

(2.31)
$\mathcal{Y}_{r}(n):=\left|e^{c+AN_{n}^{2}}\left(2^{r+5}\pi^{3}\alpha_{r+2}N_{n}^{-2r-4}+10e^{-4.7N_{n}}\right)\left(2^{r+\frac{49}{24}}C_{r}N_{n}^{-2r-\frac{49}{12}}+\sum_{s=0}^{r+1}2^{s+\frac{1}{24}}\beta_{s}N_{n}^{-2s-\frac{1}{12}}\right)\right|.$

This encodes the compilation of error on $C_{N_{n}}^{\prime}$ using the facts
that $\left|e^{\frac{A}{z^{2}}}\right|\leq
e^{\left|\frac{A}{\rho^{2}}\right|}\leq e^{AN_{n}^{2}}$ and
$|z|=\rho\leq\sqrt{2}N_{n}^{-1}$. Now, recalling that $n=2AN_{n}^{3}$, we
obtain

$\displaystyle J(n)$ $\displaystyle=\frac{1}{2\pi
i}\int_{\frac{1-i}{N_{n}}}^{\frac{1+i}{N_{n}}}f(e^{-z})e^{2AN_{n}^{3}z}dz=\frac{1}{2\pi
i}\int_{\frac{1-i}{N_{n}}}^{\frac{1+i}{N_{n}}}(M(x)+\mathcal{X}_{r}(n)+\mathcal{Y}_{r}(n))e^{2AN_{n}^{3}z}dz$
$\displaystyle=\frac{e^{c}}{2\pi
i}\int_{\frac{1-i}{N_{n}}}^{\frac{1+i}{N_{n}}}\left(\sum_{s=0}^{r+1}\beta_{s}z^{2s+\frac{1}{12}}\right)e^{\frac{A}{z^{2}}+2AN_{n}^{3}z}dz+O_{\leq}\left(\frac{(\mathcal{X}_{r}(n)+\mathcal{Y}_{r}(n))\cdot
e^{2AN_{n}^{2}}}{N_{n}\pi}\right),$

where we used that on the path of integration,
$|e^{2AN_{n}^{3}z}|=e^{2AN_{n}^{2}},$ and that the length is $2N_{n}^{-1}$. We
now let $v=N_{n}z,$ and introduce Wright’s

$P_{s}:=\frac{1}{2\pi
i}\int_{1-i}^{1+i}v^{2s+\frac{1}{12}}\exp\left(AN_{n}^{2}\left(2v+\frac{1}{v^{2}}\right)\right)dv,$

to find that

(2.32) $\displaystyle J(n)$
$\displaystyle=e^{c}\cdot\sum_{s=0}^{r+1}\frac{\beta_{s}P_{s}}{N_{n}^{2s+\frac{13}{12}}}+O_{\leq}\left(\frac{(\mathcal{X}_{r}(n)+\mathcal{Y}_{r}(n))\cdot
e^{2AN_{n}^{2}}}{N_{n}\pi}\right).$

To complete the proof we require an explicit version of Wright’s expansion of
$P_{s}$, that he obtained via the method of steepest descent. We follow Wright
in this regard. Using his notation, we first let $\mathcal{C}$ be the plane
curve defined by the equation $(x^{2}+y^{2})^{2}=x$, together with the
labelled points $E=(2^{-2/3},2^{-2/3})$, $D=(2^{-2/3},-2^{-2/3})$ on
$\mathcal{C}$ and the points $O=(0,0)$, $G=(1,1)$, and $F=(1,-1)$ in the
plane. This is illustrated in the following figure.

Figure 4. The curve $\mathcal{C}$ and the points $E,D,O,G,F$.

Wright noted that if $\xi_{s}(v):=(2\pi
i)^{-1}v^{2s+1/12}\exp(AN_{n}^{2}(2v+1/v^{2}))$, then since $|v|\leq\sqrt{2}$,
$x\leq 1$ on $OG$ and $OF$, then we have

$|\xi_{s}(v)|=\frac{|v|^{2s+\frac{1}{2}}}{2\pi}e^{2AN_{n}^{2}x}\leq\frac{2^{s+\frac{1}{4}}e^{2AN_{n}^{2}}}{2\pi},\quad(v\in
OF\cup OG),$

and he cleverly showed101010This corrects a typo of Wright’s concerning the
path of integration of the fourth integral on the right hand side. Wright
accidentally wrote $\int_{O}^{F}$ instead of $\int_{O}^{D}$. (see p. 186 of
[17]), in terms of the arc lengths in the diagram, that

(2.33) $\displaystyle P_{s}$
$\displaystyle=\int_{\mathcal{C}}\xi_{s}(v)dv+O_{\leq}\left(\left|\int_{D}^{F}\xi_{s}(v)dv\right|+\left|\int_{G}^{E}\xi_{s}(v)dv\right|+\left|\int_{E}^{O}\xi_{s}(v)dv\right|+\left|\int_{O}^{D}\xi_{s}(v)dv\right|\right)$
$\displaystyle=\int_{\mathcal{C}}\xi_{s}(v)dv+O_{\leq}\left(\frac{2^{s+\frac{1}{4}}e^{2AN_{n}^{2}}}{2\pi}\left(\mathrm{length}(DF)+\mathrm{length}(GE)+\mathrm{length}(EO)+\mathrm{length}(OD)\right)\right)$
$\displaystyle=\int_{\mathcal{C}}\xi_{s}(v)dv+O_{\leq}\left(0.64\cdot
2^{s}\cdot e^{2AN_{n}^{2}}\right).$

Making the change of variables $t^{2}=3-2v-v^{-2}$ to estimate the integral
$\int_{\mathcal{C}}\xi_{s}(v)dv,$ we have

$\int_{\mathcal{C}}\xi_{s}(v)dv=e^{3AN_{n}^{2}}\int_{\mathbb{R}}\chi_{s}(t)e^{-AN_{n}^{2}t^{2}}dt,$

where

$\chi_{s}(t)=\frac{v^{2s+\frac{37}{12}}\cdot t}{2\pi
i(1-v^{3})}=\frac{v^{2s+\frac{25}{12}}\sqrt{2v+1}}{2\pi(v^{2}+v+1)}$

is a smooth function on $\mathcal{C}.$ Expanding $\chi_{s}(t)=\sum_{m\geq
0}a_{s,m}t^{m}$, we can estimate the Taylor remainder for all $t\in\mathbb{R}$
by

$\chi_{s}(t)=\sum_{m=0}^{2r+3}a_{s,m}t^{m}+O_{\leq}\left(D_{r}\cdot|t|^{2r+4}\right),$

where

(2.34)
$D_{r}:=\frac{1}{(2r+4)!}\cdot\max\left\\{\max\left\\{\left|\chi_{s}^{(2r+4)}(t)\right|\right\\}_{t\in\mathbb{R}}\right\\}_{s=0}^{r+1}.$

We note that $D_{r}$ is explicitly computable, as $\chi_{s}(t)$ is a smooth
function on the compact curve $\mathcal{C}.$ Thus, we find that111111This
corrects a typo in Wright’s work, where he drops the factor $e^{3AN_{n}^{2}}$
temporarily in the final displayed equations of page 187.

$\displaystyle\int_{\mathcal{C}}\xi_{s}(v)dv$
$\displaystyle=e^{3AN_{n}^{2}}\sum_{m=0}^{2r+3}\int_{\mathbb{R}}a_{s,m}t^{m}e^{-AN_{n}^{2}t^{2}}dt+O_{\leq}\left(D_{r}\cdot
e^{3AN_{n}^{2}}\int_{\mathbb{R}}|t|^{2r+4}e^{-AN_{n}^{2}t^{2}}dt\right)$
$\displaystyle=e^{3AN_{n}^{2}}\sum_{m=0}^{2r+3}\int_{\mathbb{R}}a_{s,m}t^{m}e^{-AN_{n}^{2}t^{2}}dt+O_{\leq}\left(D_{r}\cdot\Gamma\left(r+\frac{5}{2}\right)(AN_{n}^{2})^{-\frac{5}{2}-r}e^{3AN_{n}^{2}}\right).$

Plugging into (2.33) gives

$\displaystyle
P_{s}=e^{3AN_{n}^{2}}\sum_{m=0}^{r+1}\frac{a_{s,2m}\Gamma\left(m+\frac{1}{2}\right)}{(AN_{n}^{2})^{m+\frac{1}{2}}}+O_{\leq}\left(D_{r}\cdot\Gamma\left(r+\frac{5}{2}\right)(AN_{n}^{2})^{-\frac{5}{2}-r}e^{3AN_{n}^{2}}+0.64\cdot
2^{s}e^{2AN_{n}^{2}}\right).$

Finally, Wright proved that $a_{s,2m}=(-1)^{m}b_{s,m}/2\pi,$ for $s\leq r+1,$
and so we obtain

$P_{s}=e^{3AN_{n}^{2}}\sum_{m=0}^{r+1}\frac{(-1)^{m}b_{s,m}\Gamma\left(m+\frac{1}{2}\right)}{2\pi\cdot(AN_{n}^{2})^{m+\frac{1}{2}}}+O_{\leq}\left(D_{r}\cdot\Gamma\left(r+\frac{5}{2}\right)(AN_{n}^{2})^{-\frac{5}{2}-r}e^{3AN_{n}^{2}}+0.64\cdot
2^{r+1}e^{2AN_{n}^{2}}\right).$

Plugging this into (2.32) gives

(2.35) $\displaystyle J(n)$
$\displaystyle=\frac{e^{c+3AN_{n}^{2}}}{2\pi}\sum_{s=0}^{r+1}\sum_{m=0}^{r+1}\frac{(-1)^{m}\beta_{s}b_{s,m}\Gamma\left(m+\frac{1}{2}\right)}{A^{m+\frac{1}{2}}N_{n}^{2s+2m+\frac{25}{12}}}+O_{\leq}\Bigg{(}\frac{(\mathcal{X}_{r}(n)+\mathcal{Y}_{r}(n))e^{2AN_{n}^{2}}}{N_{n}\pi}$
$\displaystyle+\left|e^{c}\left(D_{r}\cdot\Gamma\left(r+\frac{5}{2}\right)(AN_{n}^{2})^{-\frac{5}{2}-r}e^{3AN_{n}^{2}}+0.64\cdot
2^{r+1}e^{2AN_{n}^{2}}\right)\sum_{s=0}^{r+1}\beta_{s}N_{n}^{-2s-\frac{13}{12}}\right|\Bigg{)}.$

Therefore, the proof is complete by letting

(2.36) $\widehat{E}_{r}^{{\text{\rm
maj}}}(n):=\frac{(\mathcal{X}_{r}(n)+\mathcal{Y}_{r}(n))e^{2AN_{n}^{2}}}{N_{n}\pi}+|\mathcal{Z}_{r}(n)|,$

where

(2.37)
$\mathcal{Z}_{r}(n):=e^{c}\left(D_{r}\cdot\Gamma\left(r+\frac{5}{2}\right)(AN_{n}^{2})^{-\frac{5}{2}-r}e^{3AN_{n}^{2}}+0.64\cdot
2^{r+1}e^{2AN_{n}^{2}}\right)\sum_{s=0}^{r+1}\beta_{s}N_{n}^{-2s-\frac{13}{12}}.$

###  2.3. Proof of Theorem 1.3

Cauchy’s theorem (2.2) gives $\mathrm{PL}(n)=J(n)+E^{\min}(n).$ Therefore, the
theorem follows from Proposition 2.1 and Proposition 2.4.

##  3\. Proof of Theorems 1.1 and 1.2

Here we make use of Theorem 1.3 to prove Theorems 1.1 and 1.2.

###  3.1. Proof of Theorem 1.1

To apply Theorem 1.3 with a fixed $r,$ it is natural to express

(3.1)
$\mathrm{PL}(n)=\widehat{\mathrm{PL}}_{r}(n)+O_{\leq}\left(\mathcal{E}_{r}(n)\right),$

where $\widehat{\mathrm{PL}}_{r}(n)$ is the main term and $\mathcal{E}_{r}(n)$
is an explicit bound for the error. Then we can write

(3.2) $\begin{split}\mathrm{PL}(n)^{2}-&\mathrm{PL}(n-1)\mathrm{PL}(n+1)\\\ &\
\ \ \ \
\geq(\widehat{\mathrm{PL}}_{r}(n)-\mathcal{E}_{r}(n))^{2}-(\widehat{\mathrm{PL}}_{r}(n-1)+\mathcal{E}_{r}(n-1))\cdot(\widehat{\mathrm{PL}}_{r}(n+1)+\mathcal{E}_{r}(n+1)).\end{split}$

We apply Theorem 1.3 with $r=2.$ We first determine the $n$ to which it
applies. By (2.8), $n_{2}$ is the point beyond where the following expression
is guaranteed to be less than $1$:

$\frac{7\,{\left(243\cdot
2^{\frac{2}{3}}\pi^{2}n^{\frac{1}{3}}A^{\frac{5}{3}}+32\cdot
2^{\frac{1}{3}}\pi^{2}n^{\frac{5}{3}}A^{\frac{1}{3}}+64\pi^{2}nA+64\cdot
2^{\frac{2}{3}}n^{\frac{4}{3}}A^{\frac{2}{3}}+256\cdot
2^{\frac{1}{3}}n^{\frac{2}{3}}A^{\frac{4}{3}}+2916A^{2}\right)}}{32000n^{2}}.$

This expression is decreasing in $n.$ At $n=1$ it is $\approx 2.4$, and at
$n=2$ it is $\approx 0.8$, and so $n_{2}=2$. To compute $\ell_{2}$, we
determine when (2.9) is guaranteed to be less than $1/2$. This quantity is
bounded from above by $0.00005n^{-\frac{8}{3}}+5e^{-3.5n^{\frac{1}{3}}},$
which is decreasing in $n$ and is less than $0.16$ for $n=1$, and so we have
$\ell_{2}=1$. Therefore, $\max\\{\ell_{2},n_{2},87\\}=87$, and so Theorem 1.3
holds for $n\geq 87$.

The terms defining $\widehat{\mathrm{PL}}_{r}(n)$ come from the double sum in
Theorem 1.3, and can be organized by the powers of $n$ which are controlled by
$s+m$. To this end, we recall

$\mathrm{PL}(n)\approx\frac{e^{c+3AN_{n}^{2}}}{2\pi}\sum_{s=0}^{r+1}\sum_{m=0}^{r+1}\frac{(-1)^{m}\beta_{s}b_{s,m}\Gamma\left(m+\frac{1}{2}\right)}{A^{m+\frac{1}{2}}N_{n}^{2s+2m+\frac{25}{12}}}=:e^{3AN_{n}^{2}}n^{-\frac{25}{36}}\sum_{s=0}^{r+1}\sum_{m=0}^{r+1}f_{s,m}(n).$

The leading asymptotic is given by
$f_{0,0}(n)=2^{\frac{25}{36}}e^{c}A^{\frac{7}{36}}/\sqrt{12\pi}.$ The next
largest term, with a power saving of $n^{-\frac{2}{3}}$, is given by the terms
with $s+m=1$:

$f_{1,0}(n)+f_{0,1}(n)=-\frac{\sqrt{3}\cdot
2^{\frac{13}{36}}e^{c}{\left(3A+1385\right)}}{25920\,\sqrt{\pi}A^{\frac{5}{36}}}\cdot
n^{-\frac{2}{3}}.$

The terms with $s+m=2$ give

$f_{1,1}(n)+f_{2,0}(n)+f_{0,2}(n)=-\frac{\sqrt{3}\cdot
2^{\frac{1}{36}}e^{c}{\left(1377A^{2}-370650A+12525625\right)}}{1567641600\sqrt{\pi}A^{\frac{17}{36}}}\cdot
n^{-\frac{4}{3}}.$

The final term which contributes to our main estimate is the sum of terms with
$s+m=3$, giving

$\displaystyle f_{3,0}(n)$ $\displaystyle+f_{0,3}(n)+f_{2,1}(n)+f_{1,2}(n)$
$\displaystyle=-\frac{\sqrt{3}\cdot
2^{\frac{25}{36}}e^{c}{\left(609309A^{3}-90985275A^{2}+4957761375A+37576109375\right)}}{40633270272000\sqrt{\pi}A^{\frac{29}{36}}}\cdot
n^{-2}.$

These contributions together give our main term

(3.3)
$\displaystyle\widehat{\mathrm{PL}}_{2}(n):=e^{3AN_{n}^{2}}n^{-\frac{25}{36}}$
$\displaystyle(f_{0,0}(n)+f_{1,0}(n)+f_{0,1}(n)+f_{1,1}(n)+f_{2,0}(n)+f_{0,2}(n)$
$\displaystyle+$ $\displaystyle f_{3,0}(n)+f_{0,3}(n)+f_{2,1}(n)+f_{1,2}(n))$
$\displaystyle\approx$ $\displaystyle
e^{c+3AN_{n}^{2}}n^{-\frac{25}{36}}(0.23-0.056n^{-\frac{2}{3}}-0.006n^{-\frac{4}{3}}-0.001n^{-2}).$

The remaining terms with $s+m\geq 4$ are of equal or smaller magnitude than
the error which will come from $E^{\rm{maj}}_{2}(n)$ and $E^{\rm{min}}(n)$.
Thus, we will put these terms into our error estimate $\mathcal{E}_{2}(n)$. A
simple manipulation shows that for $n\geq 87$, we have

$f_{2,2}(n)+f_{3,1}(n)+f_{1,3}(n)+f_{3,2}(n)+f_{2,3}(n)+f_{3,3}(n)=O_{\leq}(10^{-5}n^{-\frac{7}{3}}).$

Thus, Theorem 1.3 implies that for $n\geq 87$, we have

(3.4)
$\mathrm{PL}(n)=\widehat{\mathrm{PL}}_{2}(n)+O_{\leq}\left(10^{-5}e^{3AN_{n}^{2}}n^{-\frac{109}{36}}+\widehat{E}_{2}^{\rm{maj}}(n)+E_{2}^{\rm{min}}(n)\right).$

We now aim to bound $\widehat{E}_{2}^{\rm{maj}}(n)$. This error is
$O(e^{3AN_{n}^{2}}n^{-\frac{109}{36}})$.121212At the top of page 189, Wright
makes a claim about the magnitude of this error term which is incorrect in the
power of $N_{n}$. This can be seen by comparing with the error estimate in the
final equation of page 188, which is correct and matches our power $-109/12$
here for $r=2$. Thus, we will compare our error terms to this expression. We
now turn to computing the relevant constants in turn. After computing
$\alpha(1)$, $\alpha(2)$, and $\alpha(3)$, we see that computing the constant
$C_{2}$ is equivalent to maximizing

$\left|e^{-\frac{e^{it}}{2880}-\frac{e^{2it}}{725760}-\frac{e^{3it}}{43545600}}\right|^{2}.$

The derivative of this function is

$\frac{1}{7257600}\,{\left(4\,\cos\left(t\right)^{2}+80\,\cos\left(t\right)+5039\right)}e^{\left(-\frac{1}{5443200}\,\cos\left(t\right)^{3}-\frac{1}{181440}\,\cos\left(t\right)^{2}-\frac{5039}{7257600}\,\cos\left(t\right)+\frac{1}{362880}\right)}\sin\left(t\right),$

which shows that there is a local maximum at $t=\pi$, which is the global
maximum. This directly gives $C_{2}=2\cdot e^{\frac{15061}{43545600}}\leq
2.0007$.

The computation of the constants $D_{r}$ is more involved. To compute $D_{r}$,
one recursively computes the derivatives $v^{(m)}(t)=v^{(m)}$ for
$m=1,\ldots,r+3$ by repeatedly differentiating the equation
$t^{2}=3-2v-v^{-2}$. Using the definition of $\chi_{s}(t)$ and the relation
$t=-i(v-1)\sqrt{2v+1}/v$ gives an expression for $\chi_{s}(t)$ as a function
of $v$. Here we need $D_{2}$, which requires the derivatives

$\displaystyle v^{\prime}$ $\displaystyle=tv^{3}/(v^{3}-1),$ $\displaystyle
v^{\prime\prime}$
$\displaystyle=v^{3}(1+3t^{2}v^{2}-2v^{3}+v^{6})/(v^{3}-1)^{3},$
$\displaystyle v^{(3)}$
$\displaystyle=3tv^{5}(3+5t^{2}v^{2}-6v^{3}+4t^{2}v^{5}+3v^{6})/(1-v^{3})^{5},$
$\displaystyle v^{(4)}$
$\displaystyle=3v^{5}(3+30t^{2}v^{2}-12v^{3}+\dots+20t^{4}v^{10}+24t^{2}v^{11}+3v^{12})/(1-v^{3})^{7},$
$\displaystyle v^{(5)}$
$\displaystyle=15tv^{7}(15+70t^{2}v^{2}-48v^{3}+\dots+24t^{4}v^{13}+40t^{2}v^{14}+12v^{15})/(1-v^{3})^{9},$
$\displaystyle v^{(6)}$
$\displaystyle=47v^{7}(5+105t^{2}v^{2}-26v^{3}+\dots+120t^{4}v^{19}+60t^{2}v^{20}+4v^{21})/(1-v^{3})^{11},$
$\displaystyle v^{(7)}$
$\displaystyle=315tv^{9}(35+315t^{2}v^{2}+\ldots+120t^{2}v^{23}+20v^{24})/(1-v^{3})^{13},$
$\displaystyle v^{(8)}$
$\displaystyle=315v^{9}(35+1260t^{2}v^{2}+\ldots+480t^{2}v^{29}+20v^{30})/(1-v^{3})^{15}.$

Differentiating and plugging into the definition of $\chi_{s}(t)$, we obtain
expressions such as

(3.5)
$\displaystyle\chi_{1}^{(6)}(t)=\frac{iv^{\frac{121}{12}}\sqrt{2v+1}}{859963392\pi(v^{2}+v+1)^{17}}\cdots(181387629768625+\ldots+2444688400v^{20}).$

Now that these are expressions of $v$, we can substitute $v=x\pm
i\sqrt{\sqrt{x}-x^{2}}$, which traces out the curve $\mathcal{C}$ as $x$
ranges from $0$ to $1$. That is, expressions such as the absolute value of
(3.5) can be evaluated for $x\in[0,1]$ numerically to obtain an estimate for
$D_{2}$. By symmetry, we only need to bound for $v=x+i\sqrt{\sqrt{x}-x^{2}}$
with $x\in[0,1]$. Performing this analysis, we find that $D_{2}\leq 5.3$.

These constants allow us to bound $\widehat{E}^{\rm{maj}}_{2}(n)$ by (2.36)
for $n\geq 87.$ A simple calculation gives

$\frac{\mathcal{X}_{2}(n)+\mathcal{Y}_{2}(n)}{N_{n}\pi}\cdot
e^{2AN_{n}^{2}}=O_{\leq}\left((127-10^{-5})e^{3AN_{n}^{2}}n^{-\frac{109}{36}}\right).$

Further simplification gives (see (2.37))
$\mathcal{Z}_{2}(n)=O_{\leq}\left(100e^{3AN_{n}^{2}}n^{-\frac{109}{36}}\right).$
Therefore, we find that

$\widehat{E}^{\rm{maj}}_{2}(n)=O_{\leq}\left((227-10^{-5})e^{3AN_{n}^{2}}n^{-\frac{109}{36}}\right).$

Since we have
$E^{\rm{min}}(n)=O_{\leq}\left(e^{\left(3A-\frac{2}{5}\right)N_{n}^{2}}\right),$
expression (3.4) yields

(3.6)
$\mathrm{PL}(n)=\widehat{\mathrm{PL}}_{2}(n)+O_{\leq}\left(227e^{3AN_{n}^{2}}n^{-\frac{109}{36}}+e^{\left(3A-\frac{2}{5}\right)N_{n}^{2}}\right).$

In other words, we can let
$\mathcal{E}_{2}(n):=227e^{3AN_{n}^{2}}n^{-\frac{109}{36}}+e^{\left(3A-\frac{2}{5}\right)N_{n}^{2}}.$
A simple calculation with (3.2) and (3.3) establishes that $\mathrm{PL}(n)$ is
log-concave for all $n\geq 8820.$ This completes the proof as log-concavity
has been confirmed on a computer for all $12\leq n\leq 10^{5}$ by Heim et al.
[9].

###  3.2. Proof of Theorem 1.2

The proof of Theorem 1.2 makes use of Theorem 1.3 and recent work by Griffin,
Zagier, and two of the authors of [8] on Jensen polynomials of suitable
sequences of real numbers. The main observation is that suitable real
sequences have Jensen polynomials that can be modeled by the Hermite
polynomials $H_{d}(X)$, which are orthogonal polynomials for the measure
$\mu(X)=e^{-X^{2}/4},$ and are given by the generating function

${\sum_{d=0}^{\infty}H_{d}(X)\,\frac{t^{d}}{d!}\;=\;e^{-t^{2}+Xt}}\;=\;1\
+X\,t\,+\,(X^{2}-2)\,\frac{t^{2}}{2!}\,+\,(X^{3}-6X)\,\frac{t^{3}}{3!}\,+\,\cdots$

######  Theorem 3.1 (Theorems 3 and 6 of [8]).

Let $\\{\alpha(n)\\},\\{A(n)\\},$ and $\\{\delta(n)\\}$ be sequences of
positive real numbers, with $\delta(n)$ tending to $0.$ For an integer $d\geq
3,$ suppose that there are real numbers $g_{3}(n),g_{4}(n),\dots g_{d}(n),$
for which

(3.7)
${\log\Bigl{(}{\frac{\alpha(n+j)}{\alpha(n)}\Bigr{)}\,\;=\;\,A(n)j-\delta(n)^{2}j^{2}+\sum_{i=3}^{d}g_{i}(n)j^{i}+\text{\rm
o}\bigl{(}\delta(n)^{d}\bigr{)}}}\qquad\text{\rm as $n\to\infty$},$

with $g_{i}(n)=o(\delta(n)^{i})$ for each $3\leq i\leq d.$ Then we have

(3.8)
$\lim_{n\to\infty}\biggl{(}\frac{\delta(n)^{-d}}{\alpha(n)}\,J_{\alpha}^{d,n}\Bigl{(}\frac{\delta(n)\,X\,-\,1}{\exp(A(n))}\Bigr{)}\biggr{)}\;=\;H_{d}(X).$

######  Remark.

Theorem 3.1 holds for $d\in\\{1,2\\}.$ In these cases there simply are no
numbers $g_{i}(n)$. In fact, one can obtain Theorem 1.2 for $d=2$ using this
result, which is essentially the method employed by Heim et. al. in [9]
without the formalism of Jensen polynomials.

Since the Hermite polynomials are hyperbolic, and since this property of a
polynomial with real coefficients is invariant under small deformation, we
have the following key consequence.

######  Corollary.

Assuming the hypotheses in Theorem 3.1, we have that the Jensen polynomials
$J_{\alpha}^{d,n}(X)$ are hyperbolic for all but finitely many values $n.$

###### Proof of Theorem 1.2.

Theorem 1.1 implies that $J_{\mathrm{PL}}^{2,n}(X)$ is hyperbolic for all
$n\geq 12.$ Therefore, without loss of generality we may assume that $d\geq
3$. To complete the proof, we apply Theorem 3.1.

Theorem 1.3 gives constants $\kappa>0$ and $\nu_{0},\nu_{1},\dots$ for which
$\mathrm{PL}(n)$ has an asymptotic expansion to all orders of $1/n$ of the
form

$\mathrm{PL}(n)\sim\exp\left(\sqrt[3]{\kappa
n^{2}}\right)n^{-\frac{25}{36}}\cdot\left(\nu_{0}-\sum_{m=1}^{\infty}\frac{\nu_{m}}{n^{\frac{2m}{3}}}\right).$

We choose $\nu_{0}$ so that the leading factor is a product of an exponential
with a power of $n.$ Using $\log(1-X)=-X-X^{2}/2-X^{3}/3-X^{4}/4-\dots,$ we
can rewrite this expression as

$\mathrm{PL}(n)\sim\exp(\sqrt[3]{\kappa
n^{2}})n^{-\frac{25}{36}}\cdot\exp\left(c_{0}+\frac{c_{1}}{n^{2/3}}+\frac{c_{2}}{n^{4/3}}+\frac{c_{3}}{n^{2}}+\dots\right),$

which in turn gives

$\log\left(\frac{\mathrm{PL}(n+j)}{\mathrm{PL}(n)}\right)\sim\sqrt[3]{\kappa}\sum_{i=1}^{\infty}\left(\begin{matrix}2/3\\\
i\end{matrix}\right)\frac{j^{i}}{n^{i-\frac{2}{3}}}-\frac{25}{36}\sum_{i=1}^{\infty}\frac{(-1)^{i-1}j^{i}}{in^{i}}+\sum_{s,t\geq
1}c_{t}\left(\begin{matrix}-t\\\
s\end{matrix}\right)\frac{j^{s}}{n^{2(s+t)/3}}.$

As $\left(\begin{matrix}2/3\\\ 1\end{matrix}\right)=2/3>0$ and
$\left(\begin{matrix}2/3\\\ 2\end{matrix}\right)=-1/9<0,$ Theorem 3.1 applies,
and so its corollary proves the theorem. ∎

## References

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and positron with the Fermi distribution

$\displaystyle n_{e}=\frac{1}{{\rm
e}^{(E-\mu)/T}+1}\,,\qquad\bar{n}_{e}=\frac{1}{{\rm e}^{(E+\mu)/T}+1}\,.$
(B.4)

The electromagnetic polarization tensor of plasmon interacting with electrons
and positrons to the leading order is given by the thermal integral

$\displaystyle\Pi^{\mu\nu}(k,T)$ $\displaystyle=16\pi\alpha\int\frac{{\rm
d}^{3}p}{(2\pi)^{3}}\frac{1}{2E}\left(n_{e}(E,T)+\bar{n}_{e}(E,T)\right)$
$\displaystyle\times\frac{p\cdot
k\left(p^{\mu}k^{\nu}+k^{\mu}p^{\nu}\right)-k^{2}p^{\mu}p^{\nu}-(p\cdot
k)^{2}g^{\mu\nu}}{(p\cdot k)^{2}-\left(k^{2}\right)^{2}/4}\,,$ (B.5)

where $p=(E,\vec{p})$ is the momentum of electrons and positrons and
$k=(\omega,\vec{k})$ is the plasmon momentum. The transverse and longitudinal
polarization functions are

$\displaystyle\Pi_{T}(\omega,k)$
$\displaystyle=\frac{1}{2}\left(\delta^{ij}-\hat{k}^{i}\hat{k}^{j}\right)\Pi^{ij}(\omega,\vec{k})\,,$
(B.6) $\displaystyle\Pi_{L}(\omega,k)$
$\displaystyle=\Pi^{00}(\omega,\vec{k})\,,$ (B.7)

where $\hat{k}$ is the normalized momentum vector. The effective propagator
can then be constructed in the Coulomb gauge ($\partial^{i}A_{i}=0$), and the
nonzero components are

$\displaystyle D_{T}^{ij}(\omega,\vec{k})$
$\displaystyle=\frac{1}{\omega^{2}-k^{2}-\Pi_{T}(\omega,k)}\left(\delta^{ij}-\hat{k}^{i}\hat{k}^{j}\right)\,,$
(B.8) $\displaystyle D_{L}(\omega,\vec{k})$
$\displaystyle=\frac{1}{k^{2}-\Pi_{L}(\omega,k)}\,,$ (B.9)

with the frequencies of the longitudinal and transverse modes defined by the
poles of effective propagators Eqs. (B.8) and (B.9)

$\displaystyle\omega_{T}(k)^{2}$
$\displaystyle=k^{2}+\Pi_{T}\left(\omega_{T}(k),k\right)\,,$ (B.10)
$\displaystyle\omega_{L}(k)^{2}$
$\displaystyle=\frac{\omega_{L}(k)^{2}}{k^{2}}\Pi_{L}\left(\omega_{L}(k),k\right)\,.$
(B.11)

The dressed polarization vectors are given by

$\tilde{\epsilon}_{L}^{\mu}(k)=\frac{\omega_{L}(k)}{k}\sqrt{Z_{L}(k)}\big{(}1,\vec{0}\big{)}^{\mu}\,,\qquad\tilde{\epsilon_{\pm}}^{\mu}(k)=\sqrt{Z_{T}(k)}\big{(}0,\vec{\epsilon}_{\pm}\big{)}^{\mu}\,,$
(B.12)

where the residue functions are

$\displaystyle Z_{L}(k)$
$\displaystyle=\left.\frac{k^{2}}{\omega^{2}}\left[-\frac{\partial\Pi_{L}}{\partial\omega^{2}}\left(\omega,k\right)\right]^{-1}\right|_{\omega\to\omega_{L}}\,,$
(B.13) $\displaystyle Z_{T}(k)$
$\displaystyle=\left.\left[1-\frac{\partial\Pi_{T}}{\partial\omega^{2}}\left(\omega,k\right)\right]^{-1}\right|_{\omega\to\omega_{T}}\,.$
(B.14)

The general dispersion relations valid at all temperatures and densities up to
first order in the electromagnetic fine structure constant $\alpha$ are given
by [101, 43],

$\displaystyle\omega_{T}^{2}$
$\displaystyle=k^{2}+\omega_{p}^{2}\frac{3\omega_{T}^{2}}{2v_{\star}^{2}k^{2}}\Big{[}1-\frac{\omega_{T}^{2}-v_{\star}^{2}k^{2}}{\omega_{T}^{2}}\frac{\omega_{T}}{2v_{\star}k}\ln\big{(}\frac{\omega_{T}+v_{\star}k}{\omega_{T}-v_{\star}k}\big{)}\Big{]}\,,\qquad
0\leq k<\infty$ (B.15) $\displaystyle\omega_{L}^{2}$
$\displaystyle=\omega_{p}^{2}\frac{3\omega_{L}^{2}}{v_{\star}^{2}k^{2}}\Big{[}\frac{\omega_{L}}{2v_{\star}k}\ln\big{(}\frac{\omega_{L}+v_{\star}k}{\omega_{L}-v_{\star}k}\big{)}-1\Big{]}\,,\qquad
0\leq k<k_{{\rm max}}$ (B.16)

where $k_{\mathrm{max}}$ is the maximum momentum of which the longitudinal
photon lies inside the light cone

$\displaystyle k_{{\rm max}}$
$\displaystyle=\frac{4\alpha}{\pi}\int_{0}^{\infty}{\rm
d}p\frac{p^{2}}{E}\Big{[}\frac{1}{v}\ln\big{(}\frac{1+v}{1-v}\big{)}-1\Big{]}[n_{e}(E)+\bar{n}_{e}(E)]$
$\displaystyle=\omega_{p}\sqrt{\frac{3}{v_{\star}^{2}}\Big{[}\frac{1}{2v_{\star}}\ln\big{(}\frac{1+v_{\star}}{1-v_{\star}}\big{)}-1}\Big{]}\,,$
(B.17)

where $v=p/E$ is the velocity of electrons and positrons.

The coupling to the electromagnetic current then gets renormalized with

$\tilde{\epsilon}_{T}^{\mu}=\sqrt{Z_{T}}\epsilon_{T}^{\mu}\,,\qquad\tilde{\epsilon}_{L}^{\mu}=\sqrt{Z_{L}}\epsilon_{L}^{\mu}\,,$
(B.18)

where

$\displaystyle Z_{T}(k)$
$\displaystyle=\frac{2\omega_{T}^{2}(\omega_{T}^{2}-v_{\star}^{2}k^{2})}{3\omega_{p}^{2}\omega_{T}^{2}+(\omega_{T}^{2}+k^{2})(\omega_{T}^{2}-v_{\star}^{2}k^{2})-2\omega_{T}^{2}(\omega_{T}^{2}-k^{2})}\,,$
(B.19) $\displaystyle Z_{L}(k)$
$\displaystyle=\frac{2(\omega_{L}^{2}-v_{\star}^{2}k^{2})}{3\omega_{p}^{2}-(\omega_{L}^{2}-v_{\star}^{2}k^{2})}\,.$
(B.20)

In this paper we will mostly deal with the relativistic limit, where we have
$T\gg m_{e}$, $v_{\star}=1$, $k_{{\rm max}}\to\infty$.In this case the
dispersion relations reduce to

$\displaystyle\omega_{T}^{2}$
$\displaystyle=k^{2}+\omega_{p}^{2}\frac{3\omega_{T}^{2}}{2k^{2}}\Big{[}1-\frac{\omega_{T}^{2}-k^{2}}{\omega_{T}^{2}}\frac{\omega_{T}}{2k}\ln\big{(}\frac{\omega_{T}+k}{\omega_{T}-k}\big{)}\Big{]}\,,\qquad
0\leq k<\infty$ (B.21) $\displaystyle\omega_{L}^{2}$
$\displaystyle=\omega_{p}^{2}\frac{3\omega_{L}^{2}}{k^{2}}\Big{[}\frac{\omega_{L}}{2k}\ln\big{(}\frac{\omega_{L}+k}{\omega_{L}-k}\big{)}-1\Big{]}\,,\qquad
0\leq k<\infty$ (B.22)

with the plasma frequency simplified to be

$\omega_{p,{\rm
rel}}^{2}=\frac{4\alpha}{3\pi}\big{(}\mu^{2}+\frac{\pi^{2}T^{2}}{3}\big{)}\,.$
(B.23)

The renormalized factors $Z_{T}$ and $Z_{L}$, given by Eqs. (B.19) and (B.20),
are also reduced with $v_{\star}=1$. In the relativistic limit when $T$ is
large, the number of net electrons is much smaller than the total number of
electron and positron pairs and thus the chemical potential $\mu$ can be
neglected. Hence, from Eq. (B.23) we see a relation $\omega_{p,{\rm rel}}\sim
0.1T$ in the relativistic limit. Recall the photon effective mass
$m^{2}(k,T)=\omega^{2}(k,T)-k^{2}$, the millicharge dark matter with mass $m$
can only be produced by the plasmon if the temperature is greater than $\sim
20\times m$. In this paper we focus on the dark matter mass ranging from 50
keV to 1 GeV, corresponding to a temperature $T>1$ MeV, which is safe to take
the relativistic limit (for the dark matter mass near 50 keV, see discussion
in [43], the relativistic limit is also valid).

We are interested in the plasmon decaying to the millicharge dark matter via
the channel $\gamma^{*}\rightarrow\chi\bar{\chi}$, and the thermally averaged
decay widths are given by

$\displaystyle\langle\Gamma\rangle_{\gamma_{L}^{*}\rightarrow\chi\bar{\chi}}n_{\gamma_{L}^{*}}$
$\displaystyle=\frac{Q_{\epsilon}}{(2\pi)^{3}}\int k^{2}{\rm
d}k\frac{Z_{L}(k)\omega_{L}(k)\left(m_{L}(k)^{2}+2m_{\chi}^{2}\right)\sqrt{m_{L}(k)^{2}\left(m_{L}(k)^{2}-4m_{\chi}^{2}\right)}}{3m_{L}(k)^{4}\left(e^{\omega_{L}(k)/T}-1\right)}\,,$
(B.24)
$\displaystyle\langle\Gamma\rangle_{\gamma_{T}^{*}\rightarrow\chi\bar{\chi}}n_{\gamma_{T}^{*}}$
$\displaystyle=\frac{4Q_{\epsilon}}{(2\pi)^{3}}\int k^{2}{\rm
d}k\frac{Z_{T}(k)\left(m_{T}(k)^{2}-m_{\chi}^{2}\right)\sqrt{m_{T}(k)^{2}\left(m_{T}(k)^{2}-4m_{\chi}^{2}\right)}}{3\omega_{T}(k)m_{T}(k)^{2}\left(e^{\omega_{T}(k)/T}-1\right)}\,,$
(B.25)

in which $Q_{\epsilon}$ is the millicharge given in Eq. (49).

##  Appendix C Relevant scattering cross-sections

The cross-sections for all relevant interactions are summarized in this
Appendix.

##### 1\. $\chi\bar{\chi}\rightarrow\gamma,\gamma^{\prime},Z\rightarrow
i\bar{i}$

$\displaystyle\sigma_{\chi\bar{\chi}\rightarrow
i\bar{i}}(s)=\frac{g_{X}^{2}Q_{\chi}^{2}}{48\pi}\sqrt{\frac{s-4m_{i}^{2}}{s-4m_{\chi}^{2}}}\left(1+\frac{2m_{\chi}^{2}}{s}\right)\times$
(C.1)
$\displaystyle\bigg{\\{}\frac{\mathcal{R}_{13}^{2}\left[a_{i}^{2}\left(s-4m_{i}^{2}\right)+v_{i}^{2}\left(2m_{i}^{2}+s\right)\right]}{\left(s-M_{Z}^{2}\right)^{2}+M_{Z}^{2}\Gamma_{Z}^{2}}+\frac{\mathcal{R}_{11}^{2}\left[a_{i}^{\prime
2}\left(s-4m_{i}^{2}\right)+v_{i}^{\prime
2}\left(2m_{i}^{2}+s\right)\right]}{\left(s-M_{\gamma^{\prime}}^{2}\right)^{2}+M_{\gamma^{\prime}}^{2}\Gamma_{\gamma^{\prime}}^{2}}$
$\displaystyle+\frac{2\mathcal{R}_{11}\mathcal{R}_{13}\left[a_{i}a_{i}^{\prime}\left(s-4m_{i}^{2}\right)+v_{i}v_{i}^{\prime}\left(2m_{i}^{2}+s\right)\right]}{\left[\left(s-M_{Z}^{2}\right)^{2}+M_{Z}^{2}\Gamma_{Z}^{2}\right]\left[\left(s-M_{\gamma^{\prime}}^{2}\right)^{2}+M_{\gamma^{\prime}}^{2}\Gamma_{\gamma^{\prime}}^{2}\right]}G$
$\displaystyle+\frac{4g_{2}s_{\theta}Q_{i}v_{i}\mathcal{R}_{13}\mathcal{R}_{12}\left(2m_{i}^{2}+s\right)\left(s-M_{Z}^{2}\right)}{s\left[\left(s-M_{Z}^{2}\right)^{2}+M_{Z}^{2}\Gamma_{Z}^{2}\right]}$
$\displaystyle+\frac{4g_{2}s_{\theta}Q_{i}v_{i}^{\prime}\mathcal{R}_{12}\mathcal{R}_{11}\left(2m_{i}^{2}+s\right)\left(s-M_{\gamma^{\prime}}^{2}\right)}{s\left[\left(s-M_{\gamma^{\prime}}^{2}\right)^{2}+M_{\gamma^{\prime}}^{2}\Gamma_{\gamma^{\prime}}^{2}\right]}$
$\displaystyle+\frac{4(g_{2}s_{\theta}Q_{i}\mathcal{R}_{12})^{2}\left(2m_{i}^{2}+s\right)}{s^{2}}\bigg{\\}}\,,$

with

$G=\left[\Gamma_{Z}M_{Z}\Gamma_{\gamma^{\prime}}M_{\gamma^{\prime}}+\left(s-M_{Z}^{2}\right)\left(s-M_{\gamma^{\prime}}^{2}\right)\right]\,.$
(C.2)

##### 2\. $\chi\bar{\chi}\rightarrow\gamma\gamma$

$\sigma_{\chi\bar{\chi}\rightarrow\gamma\gamma}(s)=\frac{g_{X}^{4}Q_{\chi}^{4}\mathcal{R}_{12}^{4}}{8\pi
s\left(s-4m_{\chi}^{2}\right)}\left\\{-\frac{\sqrt{s-4m_{\chi}^{2}}}{\sqrt{s}}\left(s+4m_{\chi}^{2}\right)+\frac{\log
A}{s}\left(s^{2}+4m_{\chi}^{2}s-8m_{\chi}^{4}\right)\right\\}\,,$ (C.3)

with

$A=\frac{\sqrt{s}+\sqrt{s-4m_{\chi}^{2}}}{\sqrt{s}-\sqrt{s-4m_{\chi}^{2}}}\,.$
(C.4)

##### 3\. $\gamma^{\prime}\rightarrow i\bar{i}$

$\Gamma_{\gamma^{\prime}\rightarrow
i\bar{i}}=\frac{M_{\gamma^{\prime}}}{48\pi}\sqrt{1-\frac{4m_{i}^{2}}{M_{\gamma^{\prime}}^{2}}}\left[v_{i}^{\prime
2}\left(1+\frac{2m_{i}^{2}}{M_{\gamma^{\prime}}^{2}}\right)+a_{i}^{\prime
2}\left(1-\frac{4m_{i}^{2}}{M_{\gamma^{\prime}}^{2}}\right)\right]\,.$ (C.5)

##### 4\. $i\bar{i}\rightarrow\gamma^{\prime}$

$\displaystyle\sigma_{i\bar{i}\rightarrow\gamma^{\prime}}(s)$
$\displaystyle=\sigma^{0}_{i\bar{i}\rightarrow\gamma^{\prime}}(s)\delta(s-M_{\gamma^{\prime}}^{2})$
$\displaystyle=\frac{\pi
M_{\gamma^{\prime}}^{2}}{4s}\sqrt{\frac{s}{s-4m_{i}^{2}}}\left[v_{i}^{\prime
2}\left(1+\frac{2m_{i}^{2}}{M_{\gamma^{\prime}}^{2}}\right)+a_{i}^{\prime
2}\left(1-\frac{4m_{i}^{2}}{M_{\gamma^{\prime}}^{2}}\right)\right]\delta(s-M_{\gamma^{\prime}}^{2})\,.$
(C.6)

##### 5\. $i\bar{i}\to\gamma\gamma^{\prime}$, $i\gamma\to i\gamma^{\prime}$

The squared scattering matrices summed over initial and final spins are given
by

$\displaystyle\sum_{\mathrm{spins}}\left|M\right|_{i\bar{i}\to\gamma\gamma^{\prime}}^{2}=$
$\displaystyle
2e^{2}Q_{i}^{2}\left\\{\mathcal{K}_{1}+\mathcal{K}_{2}\frac{1}{t_{i}u_{i}}+\mathcal{K}_{3}\left(\frac{1}{t_{i}^{2}}+\frac{1}{u_{i}^{2}}\right)+\mathcal{K}_{4}\left(\frac{1}{t_{i}}+\frac{1}{u_{i}}\right)+\mathcal{K}_{5}\left(\frac{u_{i}}{t_{i}}+\frac{t_{i}}{u_{i}}\right)\right\\}\,,$
(C.7) $\displaystyle\sum_{\mathrm{spins}}\left|M\right|_{i\gamma\to
i\gamma^{\prime}}^{2}=$ $\displaystyle
2e^{2}Q_{i}^{2}\left\\{\mathcal{K}_{1}+\mathcal{K}_{2}\frac{1}{s_{i}u_{i}}+\mathcal{K}_{3}\left(\frac{1}{s_{i}^{2}}+\frac{1}{u_{i}^{2}}\right)+\mathcal{K}_{4}\left(\frac{1}{s_{i}}+\frac{1}{u_{i}}\right)+\mathcal{K}_{5}\left(\frac{u_{i}}{s_{i}}+\frac{s_{i}}{u_{i}}\right)\right\\}\,,$
(C.8)

where $s_{i}=s-m_{i}^{2},t_{i}=t-m_{i}^{2},u_{i}=u-m_{i}^{2},$ and

$\displaystyle\mathcal{K}_{1}$ $\displaystyle=4a_{i}^{\prime
2}m_{i}^{2}/M_{\gamma^{\prime}}^{2}\,,$ $\displaystyle\mathcal{K}_{2}$
$\displaystyle=2\left[a_{i}^{\prime
2}\left(8m_{i}^{4}-6m_{i}^{2}M_{\gamma^{\prime}}^{2}+M_{\gamma^{\prime}}^{4}\right)+v_{i}^{\prime
2}\left(M_{\gamma^{\prime}}^{4}-4m_{i}^{4}\right)\right]\,,$
$\displaystyle\mathcal{K}_{3}$ $\displaystyle=-2m_{i}^{2}\left[a_{i}^{\prime
2}\left(M_{\gamma^{\prime}}^{2}-4m_{i}^{2}\right)+v_{i}^{\prime
2}\left(2m_{i}^{2}+M_{\gamma^{\prime}}^{2}\right)\right]\,,$
$\displaystyle\mathcal{K}_{4}$ $\displaystyle=-2\left[a_{i}^{\prime
2}\left(M_{\gamma^{\prime}}^{2}-4m_{i}^{2}\right)+v_{i}^{\prime
2}\left(2m_{i}^{2}+M_{\gamma^{\prime}}^{2}\right)\right]\,,$
$\displaystyle\mathcal{K}_{5}$ $\displaystyle=a_{i}^{\prime
2}\left(2m_{i}^{2}/M_{\gamma^{\prime}}^{2}-1\right)+v_{i}^{\prime 2}\,.$ (C.9)

The corresponding cross-sections are

$\displaystyle\sigma_{i\bar{i}\to\gamma\gamma^{\prime}}\left(s\right)=$
$\displaystyle\frac{e^{2}Q_{i}^{2}\left(s-M_{\gamma^{\prime}}^{2}\right)}{32\pi
s\sqrt{s\left(s-4m_{i}^{2}\right)}}\left\\{\left(\mathcal{K}_{1}-2\mathcal{K}_{5}\right)+\mathcal{K}_{3}\frac{2s}{\left(s-M_{\gamma^{\prime}}^{2}\right)^{2}m_{i}^{2}}\right.$
$\displaystyle\left.\quad\quad\quad\quad+\frac{2\sqrt{s}\left[\mathcal{K}_{2}-\mathcal{K}_{4}\left(s-M_{\gamma^{\prime}}^{2}\right)+\mathcal{K}_{5}\left(s-M_{\gamma^{\prime}}^{2}\right)^{2}\right]}{\left(s-M_{\gamma^{\prime}}^{2}\right)^{2}\sqrt{s\left(s-4m_{i}^{2}\right)}}\log
B\right\\}\,,$ (C.10) $\displaystyle\sigma_{i\gamma\to
i\gamma^{\prime}}\left(s\right)=$ $\displaystyle\frac{-e^{2}Q_{i}^{2}D}{64\pi
s\left(s-m_{i}^{2}\right)}\left\\{2\mathcal{K}_{1}-\frac{C\mathcal{K}_{5}}{s}+\frac{2\mathcal{K}_{4}}{s-m_{i}^{2}}+\frac{2\mathcal{K}_{3}}{\left(s-m_{i}^{2}\right)^{2}}\left(1-\frac{4s^{2}}{C^{2}-D^{2}}\right)\right.$
$\displaystyle\left.-\frac{2}{D\left(s-m_{i}^{2}\right)^{2}}\left[\mathcal{K}_{2}+\mathcal{K}_{4}\left(s-m_{i}^{2}\right)+\mathcal{K}_{5}\left(s-m_{i}^{2}\right)^{2}\right]\log\left(\frac{C+D}{C-D}\right)\right\\}\,,$
(C.11) $\displaystyle\sigma_{\gamma\gamma^{\prime}\to
i\bar{i}}\left(s\right)=$
$\displaystyle\frac{4s\left(s-4m_{i}^{2}\right)}{6\left(s-M_{\gamma^{\prime}}^{2}\right)^{2}}\cdot\sigma_{i\bar{i}\to\gamma\gamma^{\prime}}\left(s\right)\,,$
(C.12) $\displaystyle\sigma_{i\gamma^{\prime}\to i\gamma}\left(s\right)=$
$\displaystyle\frac{4\left(s-m_{i}^{2}\right)^{2}}{6D^{2}}\cdot\sigma_{i\gamma\to
i\gamma^{\prime}}\left(s\right)\,,$ (C.13)

with

$\displaystyle B=$
$\displaystyle\left(\sqrt{s}+\sqrt{s-4m_{i}^{2}}\right)/\left(\sqrt{s}-\sqrt{s-4m_{i}^{2}}\right)\,,$
$\displaystyle C=$ $\displaystyle m_{i}^{2}-M_{\gamma^{\prime}}^{2}+s\,,$
$\displaystyle D=$
$\displaystyle\sqrt{s^{2}-2\left(m_{i}^{2}+M_{\gamma^{\prime}}^{2}\right)s+\left(m_{i}^{2}-M_{\gamma^{\prime}}^{2}\right)^{2}}\,.$
(C.14)

##### 6\. $\chi\bar{\chi}\rightarrow\gamma^{\prime}\gamma^{\prime}$

$\displaystyle\sigma_{\chi\bar{\chi}\rightarrow\gamma^{\prime}\gamma^{\prime}}(s)$
$\displaystyle=\frac{g_{X}^{4}Q_{\chi}^{4}\mathcal{R}_{11}^{4}}{8\pi
s\left(s-4m_{\chi}^{2}\right)}\Bigg{\\{}-\frac{\sqrt{\left(s-4M_{\gamma^{\prime}}^{2}\right)\left(s-4m_{\chi}^{2}\right)}}{M_{\gamma^{\prime}}^{4}+m_{\chi}^{2}\left(s-4M_{\gamma^{\prime}}^{2}\right)}\left[2M_{\gamma^{\prime}}^{4}+m_{\chi}^{2}\left(s+4m_{\chi}^{2}\right)\right]$
$\displaystyle+\frac{\log
E}{s-2M_{\gamma^{\prime}}^{2}}\left(s^{2}+4m_{\chi}^{2}s+4M_{\gamma^{\prime}}^{4}-8m_{\chi}^{4}-8m_{\chi}^{2}M_{\gamma^{\prime}}^{2}\right)\Bigg{\\}}\,,$
(C.15)

with

$E=\frac{s-2M_{\gamma^{\prime}}^{2}+\sqrt{\left(s-4M_{\gamma^{\prime}}^{2}\right)\left(s-4m_{\chi}^{2}\right)}}{s-2M_{\gamma^{\prime}}^{2}-\sqrt{\left(s-4M_{\gamma^{\prime}}^{2}\right)\left(s-4m_{\chi}^{2}\right)}}\,.$
(C.16)

##### 7\. $\gamma^{\prime}\gamma^{\prime}\rightarrow\chi\bar{\chi}$

$9\left(s-4M_{\gamma^{\prime}}^{2}\right)\sigma_{\gamma^{\prime}\gamma^{\prime}\rightarrow\chi\bar{\chi}}(s)=8\left(s-4m_{\chi}^{2}\right)\sigma_{\chi\bar{\chi}\rightarrow\gamma^{\prime}\gamma^{\prime}}(s)\,.$
(C.17)

##### 8\. $\gamma^{\prime}\rightarrow\chi\bar{\chi}$

$\Gamma_{\gamma^{\prime}\rightarrow\chi\bar{\chi}}=\frac{g_{X}^{2}Q_{\chi}^{2}\mathcal{R}_{11}^{2}M_{\gamma^{\prime}}}{12\pi}\sqrt{1-\frac{4m_{\chi}^{2}}{M_{\gamma^{\prime}}^{2}}}\left(1+\frac{2m_{\chi}^{2}}{M_{\gamma^{\prime}}^{2}}\right)\,.$
(C.18)

##### 9\. $\chi\bar{\chi}\rightarrow\gamma^{\prime}$

$\displaystyle\sigma_{\chi\bar{\chi}\rightarrow\gamma^{\prime}}(s)$
$\displaystyle=\sigma^{0}_{\chi\bar{\chi}\rightarrow\gamma^{\prime}}(s)\delta(s-M_{\gamma^{\prime}}^{2})$
$\displaystyle=\frac{\pi
g_{X}^{2}Q_{\chi}^{2}\mathcal{R}_{11}^{2}M_{\gamma^{\prime}}^{2}}{s}\sqrt{\frac{s}{s-4m_{\chi}^{2}}}\left(1+\frac{2m_{\chi}^{2}}{M_{\gamma^{\prime}}^{2}}\right)\delta(s-M_{\gamma^{\prime}}^{2})\,.$
(C.19)

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