# New derivation of Time-Independent Perturbation Theory

A. N. Kvinikhidze<EMAIL_ADDRESS>Andrea Razmadze Mathematical
Institute of Tbilisi State University, 6, Tamarashvili Str., 0186 Tbilisi,
Georgia  College of Science and Engineering, Flinders University, Bedford
Park, SA 5042, Australia  B. Blankleider<EMAIL_ADDRESS>College of Science and Engineering, Flinders University, Bedford Park, SA
5042, Australia

###### Abstract

We propose a new derivation of Time-Independent Perturbation Theory (PT) that
has a fundamental advantage over the usual derivations presented in textbooks
on Quantum Mechanics (QM): it is simpler and much shorter. As such, it can
provide an easier and quicker way for students to learn PT, than afforded by
current methods. In spite of that, our approach does not require the
potentials to be energy independent or the inverse free Green function
$G_{0}^{-1}(E)$ to be a linear function of energy $E$, as is the case in QM,
and can be applied directly to various extensions of QM including Relativistic
QM, the Bethe-Salpeter equation, and all kinds of quasipotential approaches in
Quantum Field Theory.

##  I Introduction

In Quantum Mechanics (QM), almost none of the physically interesting problems
can be solved exactly. This makes Time-Independent Perturbation Theory (PT) a
fundamental part of QM, as it allows one to approximate, to any level of
accuracy, the energies and wave functions of any potential $K$ that differs
not too greatly from a potential $K_{0}$ whose energies and wave function are
known. For this reason, PT features as a major topic in all textbooks on QM.
What is noteworthy, is that the approach used to derive the expressions of PT
is standard: one starts with the Schrödinger equation for potential $K$
expressed as $K_{0}$ plus a “perturbation” $K_{1}$ (thus $K=K_{0}+K_{1}$), one
then introduces a small parameter $\lambda$ into $K_{1}$ (thus
$K=K_{0}+\lambda K_{1}$), and then one expands all $\lambda$-dependent terms
into a power series in $\lambda$. By utilising the orthonormality and
completeness of eigenstates, the coefficients of the power series can then be
determined by various means. At its core, this approach relies on the
potentials $K_{0}$ and $K_{1}$ being energy independent, thereby resulting in
a complete set of orthonormal eigenstates that are essential for the
derivation.

In this paper we would like to introduce a different way of deriving PT that
has a significant advantage over the standard one just described: it is
simpler and much shorter. Moreover, our formulation remains valid for
extensions of QM, like Relativistic QM (see [1] and reference therein), the
Bethe-Salpeter (BS) equation [2, 3], and all kinds of quasipotential
approaches in Quantum Field Theory [4, 5, 6, *Gross:1982nz, *Gross:1982ny],
where the potentials can be energy dependent and where the inverse free Green
function $G_{0}^{-1}(E)$ is not restricted to be the operator $E-H_{0}$ as in
QM, but can be any function of energy $E$.

##  II New derivation

###  II.1 Preliminaries

Textbooks on QM derive PT in a variety of ways, however they all share the
same overall approach (see for example [9, 10, 11, 12, 13, 14, 15]). For
completeness, we provide a quick summary of this standard approach in Appendix
A. Here we describe an alternative approach. The method is much simpler than
the one we previously proposed [16].

We consider a system subject to a potential $K=K_{0}+K_{1}$ where $K_{1}$ is
considered to be a perturbation to a potential $K_{0}$ for which the bound
state problem has been completely solved. Otherwise we make no assumptions
about the energy dependence of $K_{0}$, $K_{1}$, or of the Green function
$G_{0}(E)$ describing the free propagation of particles in this system. In
particular, we would like to find the solution of the bound state equation for
the perturbed system,

$[G^{-1}_{0}(E_{n})-K_{0}-K_{1}]|\psi_{n}\rangle=0,$ (1)

when the unperturbed system has a bound state described by the bound state
equation

$[G^{-1}_{0}({\cal E}_{n})-K_{0}]|\phi_{n}\rangle=0$ (2)

where not only ${\cal E}_{n}$ and $|\phi_{n}\rangle$ are known, but so is the
unperturbed Green function $G_{u}$ defined as the solution of the
inhomogeneous equation,

$G_{u}=G_{0}+G_{0}K_{0}G_{u}.$ (3)

In the case of QM, this statement of the PT problem is equivalent to the one
given in the standard approach described in Appendix A. Indeed, Eq. (1) is
just a rearranged version of Eq. (21), and Eq. (2) is just a rearranged
version of Eq. (18), and knowing all the solutions of the unperturbed
Schrödinger equation, Eq. (18), as is assumed in the standard approach,
implies that $G_{u}$ is known, since in that case,

$G_{u}(E)=\sum_{m}\frac{|\phi_{m}\rangle\langle\phi_{m}|}{E-{\cal E}_{m}},$
(4)

which follows from the definition of $G_{u}$ and the completeness relation Eq.
(20b). However, in the general case applicable to various extensions of QM,
our statement of the PT problem is less restrictive, it allows $K_{0}$ and
$K_{1}$ to be energy dependent and is applicable to any functional form of
$G^{-1}_{0}(E)$. Of course in the general case (for example in the BS
approach), Eq. (4) does not apply and $G_{u}(E)$ would need to be found by
solving Eq. (3).

Being defined by Eq. (3), $G_{u}$ is an interacting Green functions driven by
potential $K_{0}$. A general feature of interacting Green functions is that
they exhibit poles at energies corresponding to bound states or resonances
produced by their driving potentials, with the residue consisting of an outer
product of their corresponding bound state wave functions. Thus, if the
unperturbed potential $K_{0}$ gives rise to a bound state at energy ${\cal
E}_{n}$, one can express $G_{u}(E)$ as

$G_{u}(E)=\frac{|\phi_{n}\rangle\langle\phi_{n}|}{E-{\cal
E}_{n}}+G_{u}^{b}(E)$ (5)

where $|\phi_{n}\rangle$ is the corresponding bound state wave function and
$G_{u}^{b}(E)$ is a “background” term. Note that Eq. (5) implies the
normalisation condition $\langle\phi_{n}|\partial(G^{-1}_{0}-K_{0})/\partial
E|\phi_{n}\rangle=1$ which in the case of QM, reduces to
$\langle\phi_{n}|\phi_{n}\rangle=1$. In writing Eq. (5) we have assumed the
simplest case of no degeneracy for energy ${\cal E}_{n}$, but our presentation
is easily extended to the case of degeneracy simply by replacing the outer
product $|\phi_{n}\rangle\langle\phi_{n}|$ with ths sum
$\sum_{j}|\phi^{j}_{n}\rangle\langle\phi^{j}_{n}|$ where
$|\phi^{j}_{n}\rangle$ is the $j$’th degenerate state. In QM, Eq. (5) follows
from Eq. (4) and is a consequence of completeness, while in Quantum Field
Theory it is usually related the vacuum expectation of the time-ordered
product of fields, $G_{u}(E)=\langle
0|T\psi\psi\bar{\psi}\bar{\psi}|0\rangle$.

###  II.2 Derivation

Using Eq. (5) as representative of the general case where the potentials can
be energy dependent, the derivation of the perturbed bound state energy
$E_{n}$ and corresponding bound state function $|\psi_{n}\rangle$, is
remarkably short. The key idea is to use Eq. (3) and Eq. (5) to write Eq. (1)
in the form111$G^{b}\equiv[1-G^{b}_{u}(E_{n})K_{1}(E_{n})]^{-1}$ should be
understood as the solution of the equation
$G^{b}=G^{b}_{u}+G^{b}_{u}K_{1}G^{b}$ or as the infinite series of Eq. (10).

$|\psi_{n}\rangle=\frac{[1-G^{b}_{u}(E_{n})K_{1}(E_{n})]^{-1}|\phi_{n}\rangle\langle\phi_{n}|}{E_{n}-{\cal
E}_{n}}K_{1}(E_{n})|\psi_{n}\rangle,$ (6)

which is a bound state equation with a separable kernel, and which can
therefore be solved algebraically. Indeed, the shortness of the derivation is
due directly to the trick of reducing the PT problem to that of an equation
with a separable kernel, a trick that does not appear to have been noticed
previously. In particular, Eq. (6) can be used to write the equation for the
scalar combination $\langle\phi_{n}|K_{1}|\psi_{n}\rangle$,

$\langle\phi_{n}|K_{1}|\psi_{n}\rangle=\frac{\langle\phi_{n}|K_{1}(1-G^{b}_{u}K_{1})^{-1}|\phi_{n}\rangle}{E_{n}-{\cal
E}_{n}}\langle\phi_{n}|K_{1}|\psi_{n}\rangle.$ (7)

This is an algebraic equation for $\langle\phi_{n}|K_{1}|\psi_{n}\rangle$
which has a nontrivial solution only if

$E_{n}={\cal
E}_{n}+\langle\phi_{n}|K_{1}(E_{n})[1-G^{b}_{u}(E_{n})K_{1}(E_{n})]^{-1}|\phi_{n}\rangle.$
(8)

To solve Eq. (6) for $|\psi_{n}\rangle$ unambiguously, a normalisation
condition is needed, which we choose to correspond to the solution

$|\psi_{n}\rangle=[1-G^{b}_{u}(E_{n})K_{1}(E_{n})]^{-1}|\phi_{n}\rangle,$ (9)

as this choice simplifies the PT expressions. There is no problem to
renormalise $|\psi_{n}\rangle$ in order to construct amplitudes for scattering
processes where the bound states are involved [16]. Note also that in QM, Eq.
(9) implies $\langle\phi_{n}|\psi_{n}\rangle=1$, in agreement with the
normalisation chosen in standard derivations of PT [13, 14].

The simplicity and the shortness of the derivation of Eq. (8) and Eq. (9)
constitute the main results of this paper. They provide exact expressions for
$E_{n}$ and $|\psi_{n}\rangle$, albeit as functions of the yet undetermined
exact energy $E_{n}$. As such, it is possible that they could be solved
numerically by iteration, starting with ${\cal E}_{n}$ as the first estimate
of $E_{n}$. However, they can also be used to generate $E_{n}$ and
$|\psi_{n}\rangle$ at each order of $K_{1}$ in perturbation theory, by
introducing the expansion parameter $\lambda$ via the replacement
$K_{1}\rightarrow\lambda K_{1}$ and expanding Eq. (8) and Eq. (9) in powers of
$\lambda$. To pursue this procedure it may be convenient to expand Eq. (8) and
Eq. (9) in powers of $K_{1}$ beforehand, using

$[1-G^{b}_{u}(E_{n})K_{1}(E_{n})]^{-1}=\sum_{s=0}[G^{b}_{u}(E_{n})K_{1}(E_{n})]^{s}.$
(10)

##  III Discussion

We have presented a new derivation of PT that applies not only to QM, but also
to various extensions of QM for which the unperturbed potential $K_{0}$ and
the perturbation potential $K_{1}$ can be energy dependent, and for which the
inverse free Green function $G_{0}^{-1}(E)$ can be a nonlinear function of
$E$. However, the main feature of the new approach is its simplicity and the
shortness of its derivation, resulting in the expressions Eq. (8) and Eq. (9)
for the perturbed energy and wavefunction $E_{n}$ and $|\psi_{n}\rangle$,
respectively.

Although Eq. (8) and Eq. (9) were derived for the case where the only
difference between the perturbed and unperturbed systems is the presence or
absence of the perturbation potential $K_{1}$, our approach enables the use of
PT for a yet more general case, where in addition to $K_{1}$, the perturbed
and unperturbed systems differ by having different free Green functions.
Denoting by $G_{1}(E)$ and $G_{0}(E)$ the free Green functions of the
perturbed and unperturbed systems, respectively, it is clear that in the
derivation of Sec. II, all that needs to be done to describe this more general
case, is to replace Eq. (1) with

$\displaystyle[G^{-1}_{1}(E_{n})-K_{0}-K_{1}]|\psi_{n}\rangle=0,$ (11)

which can of course be written as

$\displaystyle[G^{-1}_{0}(E_{n})-K_{0}-\tilde{K}_{1}]|\psi_{n}\rangle=0,$ (12)

where $\tilde{K}_{1}=K_{1}+\Delta$ and

$\Delta=G^{-1}_{0}(E_{n})-G^{-1}_{1}(E_{n})$ (13)

is an energy-dependent addition to the perturbation potential $K_{1}$. Thus,
to obtain a PT for this more general case, all one needs to do is to use
$\tilde{K}_{1}$ instead of $K_{1}$ in Eq. (8) and Eq. (9). Such a PT would be
useful, for example, to describe a relativistic perturbation where the inverse
free Green function $G^{-1}_{1}(E)$ of the perturbed system is a non-linear
function of $E$, while the completely solved unperturbed system uses the
standard linear inverse free Green function $G^{-1}_{0}(E)=E-H_{0}$ of
nonrelativistic QM.

It is noteworthy that in the case where $K_{0}$ is energy independent, so that
the unperturbed Green function $G_{u}$ is given by Eq. (4), but where
$K_{1}(E)$ is still energy dependent, our perturbed wave function can be
written in the form given by Sakurai and Napolitano for QM (see Eq. (5.34) of
Ref. [14]),

$|\psi_{n}\rangle=[1-G^{b}_{u}({\cal E}_{n})\left(K_{1}(E_{n})-E_{n}+{\cal
E}_{n}\right)]^{-1}|\phi_{n}\rangle.$ (14)

The distinctive feature of this expression is that the argument of
$G^{b}_{u}({\cal E}_{n})$ is the unperturbed energy ${\cal E}_{n}$, not
$E_{n}$. Equation (14) is derived in Appendix B. In this case the perturbed
energy, Eq. (8), can similarly be written as

$E_{n}={\cal E}_{n}+\langle\phi|K_{1}(E_{n})\left[1-G^{b}_{u}({\cal
E}_{n})\left(K_{1}(E_{n})-E_{n}+{\cal E}_{n}\right)\right]^{-1}|\phi\rangle.$
(15)

To conclude, we note that the efficiency of our approach can be demonstrated
explicitly for the case of QM for which

$G^{b}_{u}(E)=\sum_{m\neq n}\frac{|\phi_{m}\rangle\langle\phi_{m}|}{E-{\cal
E}_{m}}.$ (16)

Substituting this expression into Eq. (10), the first few term of Eq. (8) and
Eq. (9) become

$\displaystyle E_{n}$ $\displaystyle={\cal
E}_{n}+\langle\phi_{n}|K_{1}|\phi_{n}\rangle+\sum_{m\neq
n}\frac{|\langle\phi_{m}|K_{1}|\phi_{n}\rangle|^{2}}{E_{n}-{\cal
E}_{m}}+\dots$ (17a) $\displaystyle|\psi_{n}\rangle$
$\displaystyle=|\phi_{n}\rangle+\sum_{m\neq
n}\frac{\langle\phi_{m}|K_{1}|\phi_{n}\rangle}{E_{n}-{\cal
E}_{m}}|\phi_{m}\rangle$ $\displaystyle+\sum_{k\neq n}\sum_{m\neq
n}\frac{\langle\phi_{m}|K_{1}|\phi_{k}\rangle\langle\phi_{k}|K_{1}|\phi_{n}\rangle}{(E_{n}-{\cal
E}_{m})(E_{n}-{\cal E}_{k})}|\phi_{m}\rangle+\dots$ (17b)

which immediately gives the perturbation series Eq. (26) of the standard
approach. This constitutes a much easier and shorter derivation of PT, and its
main results, than that of the standard approach used in all current textbooks
on QM. As such, our derivation may present students of QM a simpler and
quicker way to learn PT.

###### Acknowledgements.

A.N.K. was supported by the Shota Rustaveli National Science Foundation (Grant
No. FR17-354).

##  Appendix A Standard approach to PT in QM

In QM, the standard way of framing the task of PT is as follows (see for
example Griffiths and Schroeter [13]). Suppose one knows all the solutions of
the Schrödinger equation

$H_{u}|\phi_{n}\rangle={\cal E}_{n}|\phi_{n}\rangle$ (18)

where $H_{u}$ is some “unperturbed” Hamiltonian

$H_{u}=H_{0}+K_{0},$ (19)

$H_{0}$ being the kinetic energy operator and $K_{0}$ being the potential. In
almost all presentations it is assumed that the energy levels are discrete, so
that $n$ is an integer and the solutions $|\phi_{n}\rangle$ satisfy the
orthonormality and completeness conditions

$\displaystyle\langle\phi_{n}|\phi_{m}\rangle$ $\displaystyle=\delta_{nm},$
(20a) $\displaystyle\sum_{m}|\phi_{m}\rangle\langle\phi_{m}|$
$\displaystyle=I.$ (20b)

For simplicity, we shall also make this assumption while recognising that in
many practical applications one also needs to include scattering states [17].
Now suppose a perturbation is introduced into the system by adding a potential
$K_{1}$. The problem then is to determine the solutions of

$H|\psi_{n}\rangle=E_{n}|\psi_{n}\rangle$ (21)

where $H$ is the Hamiltonian with the perturbation included:

$H=H_{u}+K_{1}.$ (22)

The standard way to solve this problem is to introduce a parameter $\lambda$
such that

$H(\lambda)=H_{u}+\lambda K_{1}.$ (23)

One then solves the Schrödinger equation

$(H_{u}+\lambda
K_{1})|\psi_{n}(\lambda)\rangle=E_{n}(\lambda)|\psi_{n}(\lambda)\rangle$ (24)

by making a power expansion in $\lambda$,

$\displaystyle|\psi_{n}(\lambda)\rangle$
$\displaystyle=|\phi_{n}\rangle+\lambda|\psi^{1}_{n}\rangle+\lambda^{2}|\psi^{2}_{n}\rangle+\dots,$
(25a) $\displaystyle E_{n}(\lambda)$ $\displaystyle={\cal E}_{n}+\lambda
E_{n}^{1}+\lambda^{2}E_{n}^{2}+\dots.$ (25b)

The coefficients of these power series are then determined by various means.
After a substantial amount of algebra, and assuming there is no degeneracy,
one obtains the perturbation series [13, 14] (shown explicitly up to second
order in $K_{1}$)

$\displaystyle E_{n}$ $\displaystyle={\cal
E}_{n}+\langle\phi_{n}|K_{1}|\phi_{n}\rangle+\sum_{m\neq
n}\frac{|\langle\phi_{m}|K_{1}|\phi_{n}\rangle|^{2}}{{\cal E}_{n}-{\cal
E}_{m}}+\dots$ (26a) $\displaystyle|\psi_{n}\rangle$
$\displaystyle=|\phi_{n}\rangle+\sum_{m\neq
n}\frac{\langle\phi_{m}|K_{1}|\phi_{n}\rangle}{{\cal E}_{n}-{\cal
E}_{m}}|\phi_{m}\rangle-\sum_{m\neq
n}\frac{\langle\phi_{m}|K_{1}|\phi_{n}\rangle\langle\phi_{n}|K_{1}|\phi_{n}\rangle}{({\cal
E}_{n}-{\cal E}_{m})^{2}}|\phi_{m}\rangle$ $\displaystyle+\sum_{k\neq
n}\sum_{m\neq
n}\frac{\langle\phi_{m}|K_{1}|\phi_{k}\rangle\langle\phi_{k}|K_{1}|\phi_{n}\rangle}{({\cal
E}_{n}-{\cal E}_{m})({\cal E}_{n}-{\cal E}_{k})}|\phi_{m}\rangle+\dots$ (26b)

##  Appendix B Derivation of Eq. (14)

Assuming that $K_{0}$ (but not $K_{1}$) is energy independent, the unperturbed
Green function $G_{u}(E)$ is given by Eq. (4) and its background part,
$G^{b}_{u}(E)$, is given by Eq. (16). Defining

$\Delta_{n}=E_{n}-{\cal E}_{n},$ (27)

one can write down the Taylor expansion of $G^{b}_{u}(E_{n})$ around the point
${\cal E}_{n}$ as

$G^{b}_{u}(E_{n})=G^{b}_{u}({\cal E}_{n})-\Delta_{n}[G^{b}_{u}({\cal
E}_{n})]^{2}+\Delta^{2}_{n}[G^{b}_{u}({\cal E}_{n})]^{3}+\dots$ (28)

Writing $K_{1}\equiv K_{1}(E_{n})$, Eq. (9) then becomes

$\displaystyle|\psi_{n}\rangle$
$\displaystyle=[1-G^{b}_{u}(E_{n})K_{1}]^{-1}|\phi_{n}\rangle$
$\displaystyle=[1+G^{b}_{u}(E_{n})K_{1}+G^{b}_{u}(E_{n})K_{1}G^{b}_{u}(E_{n})K_{1}+\dots]|\phi_{n}\rangle$
$\displaystyle=[1+G^{b}_{u}({\cal E}_{n})K_{1}-\Delta_{n}[G^{b}_{u}({\cal
E}_{n})]^{2}K_{1}+G^{b}_{u}({\cal E}_{n})K_{1}G^{b}_{u}({\cal
E}_{n})K_{1}+\dots]|\phi_{n}\rangle$ $\displaystyle=[1+G^{b}_{u}({\cal
E}_{n})K_{1}+G^{b}_{u}({\cal E}_{n})(K_{1}-\Delta_{n})G^{b}_{u}({\cal
E}_{n})K_{1}+\dots]|\phi_{n}\rangle$ $\displaystyle=[1-G^{b}_{u}({\cal
E}_{n})(K_{1}{-\Delta_{n}})+G^{b}_{u}({\cal
E}_{n})(K_{1}-\Delta_{n})G^{b}_{u}({\cal
E}_{n})(K_{1}{-\Delta_{n}})+\dots]|\phi_{n}\rangle$
$\displaystyle=[1-G^{b}_{u}({\cal
E}_{n})(K_{1}-\Delta_{n})]^{-1}|\phi_{n}\rangle,$ (29)

where we have used the fact that $G^{b}_{u}({\cal E}_{n})|\phi_{n}\rangle=0$.

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# Variance of Squarefull Numbers in Short Intervals

Tsz Ho Chan

###### Abstract

In this paper, we study the variance of the number of squarefull numbers in
short intervals. As a result, we are able to prove that, for any
$0<\theta<1/2$, almost all short intervals $(x,x+x^{1/2+\theta}]$ contain
about $\frac{\zeta(3/2)}{2\zeta(3)}x^{\theta}$ squarefull numbers.

##  1 Introduction and Main Results

A number $n$ is squarefree if $p^{2}\not|\;n$ for every $p|n$, i.e. all the
exponents in its prime factorization are exactly one. In contrast, a number
$n$ is squarefull if $p^{2}|n$ for every prime $p|n$, i.e. all the exponents
in its prime factorization are at least two. It is well-known that every
squarefull number $n$ can be factored uniquely as $n=a^{2}b^{3}$ for some
integer $a$ and squarefree number $b$. Let $Q(x)$ be the number of squarefull
numbers which are less than or equal to $x$. Then

$Q(x)=\sum_{a^{2}b^{3}\leq x}\mu^{2}(b)$ (1)

where $\mu(n)$ is the Möbius function. Bateman and Grosswald [1] proved that

$Q(x)=\frac{\zeta(3/2)}{\zeta(3)}x^{1/2}+\frac{\zeta(2/3)}{\zeta(2)}x^{1/3}+O\bigl{(}x^{1/6}e^{-C_{1}(\log
x)^{4/7}(\log\log x)^{-3/7}}\bigr{)}$ (2)

for some absolute constant $C_{1}>0$ where $\zeta(s)$ is the Riemann zeta
function. We are interested in squarefull numbers in short intervals, namely

$Q(x+y)-Q(x)\;\;\text{ with }\;\;y=o(x).$

Bateman and Grosswald’s result (2) gives

$Q\bigl{(}x+x^{1/2+\theta}\bigr{)}-Q(x)\sim\frac{\zeta(3/2)}{2\zeta(3)}x^{\theta}$
(3)

for $1/6=0.1666...\leq\theta<1/2$. Subsequently, exponential sum, elementary,
and differencing methods were used to expand the range of validity for (3).
The current best result was obtained by Trifonov [7], showing that the
asymptotic formula (3) is true for $19/154=0.1233...<\theta<1/2$ (please refer
to the same paper for earlier developments and relevant references).

Inspired by the recent breakthrough work of Gorodetsky, Matomäki, Radizwiłł
and Rogers [3] on variance of squarefree numbers in short intervals and
arithmetic progressions, we are interested in studying the variance of
squarefull numbers in short intervals:

$\frac{1}{X}\int_{X}^{2X}\Big{|}Q(x+y)-Q(x)-\frac{\zeta(3/2)}{\zeta(3)}(\sqrt{x+y}-\sqrt{x})\Big{|}^{2}dx.$

Our main result is the following.

###### Theorem 1

Given $\epsilon>0$, $X>1$ and $H<X^{1/2-\epsilon}$. With $y=2\sqrt{x}H+H^{2}$,

$\frac{1}{X}\int_{X}^{2X}\Big{|}Q(x+y)-Q(x)-\frac{\zeta(3/2)}{\zeta(3)}H\Big{|}^{2}dx\ll_{\epsilon}H^{1.799}X^{2\epsilon}.$

Note: $\sqrt{x+y}-\sqrt{x}=H$ (which explains our choice of $y$).

Define

$S:=\Bigl{\\{}X\leq x\leq
2X:\Big{|}Q(x+y)-Q(x)-\frac{\zeta(3/2)}{\zeta(3)}H\Big{|}>H^{0.8995}X^{1.001\epsilon}\Bigr{\\}}.$

Then Theorem 1 implies
$|S|\cdot(H^{0.8995}X^{1.001\epsilon})^{2}\ll_{\epsilon}X\cdot
H^{1.799}X^{2\epsilon}$ where $|S|$ stands for the measure of $S$. Hence,
$|S|\ll_{\epsilon}X^{1-0.002\epsilon}$ which gives the following.

###### Corollary 1

Given $\epsilon>0$, $X>1$ and $X^{10\epsilon}<H<X^{1/2-\epsilon}$. With
$y=2\sqrt{x}H+H^{2}$,

$Q(x+y)-Q(x)=\frac{\zeta(3/2)}{\zeta(3)}H+O_{\epsilon}(H^{0.8995}X^{1.001\epsilon})$

for almost all $x\in[X,2X]$. Note: $H>X^{10\epsilon}$ is to ensure that the
error term is smaller than the main term.

Since $\epsilon$ can be arbitrarily small, Corollary 1 immediately implies

###### Corollary 2

For any $0<\theta<1/2$, the asymptotic formula (3) is true for almost all
$x\in[X,2X]$.

In both corollaries, almost all means all $x\in[X,2X]$ apart from a set of
measure $o(X)$. Hence, we confirm the expected count (3) of squarefull numbers
in short intervals almost surely. We did not minimize the exponents $1.799$
and $0.8995$ with our method as we believe that Theorem 1 is far from the
truth. One future direction of research is to lower these exponents to
something below $1$. Based on Propositions 2 and 3 below, we suspect the
following.

###### Conjecture 1

Given $\epsilon>0$, $X>1$ and $H<X^{1/2-\epsilon}$. With $y=2\sqrt{x}H+H^{2}$,

$\frac{1}{X}\int_{X}^{2X}\Big{|}Q(x+y)-Q(x)-\frac{\zeta(3/2)}{\zeta(3)}H\Big{|}^{2}dx\sim\frac{\zeta(4/3)}{\zeta(2)}\int_{0}^{\infty}\Bigl{(}\frac{\sin\pi
y}{\pi y}\Bigr{)}^{2}y^{1/3}dy\cdot H^{2/3}.$

Another direction of inquiry is to study the variance of squarefull numbers
over arithmetic progressions. However, the method of Gorodetsky, Matomäki,
Radizwiłł and Rogers [3] cannot be applied directly as they made use of
Poisson summation which does not translate easily into the “non-linear”
situation involving squarefull numbers.

Recently, Matomäki and Radizwiłł [6] made an important discovery on
multiplicative function $f(n)$ with $-1\leq f(n)\leq 1$ by showing that the
local average of $f$ on (almost all) short interval $[x,x+h]$ is close to its
global average between $[X,2X]$. It would be interesting to see if similar
behavior holds for multiplicative functions supported on squarefull numbers or
other sparse sequences.

Our proof of Theorem 1 uses different strategies based on various sizes of
$H$. For large $H$, one can simply apply Bateman and Grosswald’s asymptotic
formula (2). For medium $H$, we apply the following proposition which is based
on earlier work of Huxley and Trifonov [5] in studying squarefull numbers in
short intervals.

###### Proposition 1

For $X\leq x\leq 2X$ and $X^{0.14697}<H\leq X^{0.18529}$, we have

$Q(x+y)-Q(x)=\frac{\zeta(3/2)}{\zeta(3)}H+O\bigl{(}H^{0.8995}\bigr{)}$

when $y=2\sqrt{x}H+H^{2}$.

For small $H$, we establish the following two propositions by means of smooth
weights as in [3].

###### Proposition 2

Given a parameter $\lambda>0$. Then, for $H\leq X^{\frac{1}{6(1+2\lambda)}}$
and $y=2\sqrt{x}H+H^{2}$,

$\displaystyle\frac{1}{X}\int_{X}^{2X}$
$\displaystyle\Big{|}Q(x+y)-Q(x)-\frac{\zeta(3/2)}{\zeta(3)}H\Big{|}^{2}dx$
$\displaystyle=$
$\displaystyle(1+O_{\epsilon}(X^{-\epsilon/2}))2H^{2}\sum_{b\leq
X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3}}\sum_{n\geq
1}S\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}^{2}+O_{\epsilon}\Bigl{(}(H^{3\lambda}+H^{2-3\lambda}+H^{0.2468})X^{2\epsilon}\Bigr{)}$

###### Proposition 3

Given $\epsilon>0$ and $\lambda>0$. For $1\leq H\leq X^{1/2}$,

$2H^{2}\sum_{b\leq X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3}}\sum_{n\geq
1}S\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}^{2}=H^{2/3}\frac{\zeta(4/3)}{\zeta(2)}\int_{0}^{\infty}S(y)^{2}y^{1/3}dy+O_{\epsilon}\Bigl{(}\frac{H^{1+\lambda/2+\epsilon/4}}{X^{1/6}}+H^{2/3-\epsilon/6}\Bigr{)}$

where $S(x)=\frac{\sin\pi x}{\pi x}$.

This paper is organized as follows. First, we will give explicit construction
of certain smooth weights and include some needed lemmas. Then we will prove
Propositions 1, 2 and 3. At the end, we will combine everything together to
derive Theorem 1.

Notation. We use $[x]$ and $\|x\|$ to denote the integer part of $x$ and the
distance between $x$ and the nearest integer respectively. The symbols
$f(x)=O(g(x))$, $f(x)\ll g(x)$ and $g(x)\gg f(x)$ are equivalent to
$|f(x)|\leq Cg(x)$ for some constant $C>0$. $f(x)\asymp g(x)$ means that
$C_{1}f(x)\leq g(x)\leq C_{2}f(x)$ for some constants $0<C_{1}<C_{2}$.
$f(x)\sim g(x)$ means that $\lim_{x\rightarrow\infty}f(x)/g(x)=1$. Finally
$f(x)=O_{\lambda}(g(x))$, $f(x)\ll_{\lambda}g(x)$, $g(x)\gg_{\lambda}f(x)$ or
$g(x)\asymp_{\lambda}f(x)$ mean that the implicit constant may depend on
$\lambda$.

##  2 Construction of smooth functions and some lemmas

Let

$u(x):=\left\\{\begin{array}[]{ll}1,&\text{ if }x\geq 0,\\\ 0,&\text{ if
}x<0,\end{array}\right.\;\;\text{ and
}\;\;r_{L}(x):=\left\\{\begin{array}[]{ll}\frac{1}{L},&\text{ if }0\leq x\leq
L,\\\ 0,&\text{ otherwise}\end{array}\right.$

for some parameter $0<L\leq X$. Recall the definition of the convolution of
two integrable real-valued functions $f$ and $g$ as

$(f*g)(x):=\int_{-\infty}^{\infty}f(x-u)g(u)du.$

We define $u_{k}(x)$ recursively as follows:

$u_{0}(x)=u(x),\;\;\text{ and }\;\;u_{k+1}(x)=(u_{k}*r_{L})(x)\text{ for
integer }k\geq 1.$

One can show that $0\leq u_{k}(x)\leq 1$ on $[0,\infty)$ and $u_{k}(x)=1$ when
$x\in[kL,\infty)$. Also, one can prove inductively that $u_{k}(x)$ is $k-1$
times continuously differentiable for $k\geq 2$ and satisfies the derivative
bound $u_{k}^{(i)}(x)\ll_{k}\frac{1}{L^{i}}$ for $1\leq i\leq k-1$ by using
$(f*g)^{\prime}=f^{\prime}*g$ repeatedly. In fact, one can derive an explicit
formula for $u_{k+1}(x)$ via Laplace transform $\mathcal{L}$ and its
properties as follows:

$\displaystyle\mathcal{L}\\{u_{k+1}(x)\\}=$
$\displaystyle\mathcal{L}\\{u(x)\\}\cdot\mathcal{L}\\{r_{L}(x)\\}^{k}=\frac{1}{s}\cdot\Bigl{[}\frac{1}{L}\cdot\frac{1-e^{-Ls}}{s}\Bigr{]}^{k}=\frac{1}{L^{k}}\cdot\frac{(1-e^{-Ls})^{k}}{s^{k+1}}$
$\displaystyle=$
$\displaystyle\frac{1}{k!L^{k}}\sum_{i=0}^{k}(-1)^{i}\binom{k}{i}\frac{k!e^{-iLs}}{s^{k+1}},$
$\displaystyle u_{k+1}(x)=$
$\displaystyle\mathcal{L}^{-1}\Bigl{\\{}\frac{1}{k!L^{k}}\sum_{i=0}^{k}(-1)^{i}\binom{k}{i}\frac{k!e^{-iLs}}{s^{k+1}}\Bigr{\\}}=\frac{1}{k!L^{k}}\sum_{i=0}^{k}(-1)^{i}\binom{k}{i}u(x-iL)(t-iL)^{k}.$

For $0<L<X/(2k)$, we define the smooth weight functions

$\sigma_{k,X,L}^{-}(x):=\frac{1}{2}[u_{k}(x-X)+u_{k}(2X-x)]\;\;\text{ and
}\;\;\sigma_{k,X,L}^{+}(x):=\frac{1}{2}[u_{k}(x-X+kL)+u_{k}(2(X+kL)-x)].$

One can verify that $0\leq\sigma_{k,X,L}^{-}(x)\leq 1$ and
$0\leq\sigma_{k,X,L}^{+}(x)\leq 1$ are supported on $[X,2X]$ and
$[X-kL,2X+kL]$ respectively. Moreover, $\sigma_{k,X,L}^{-}(x)=1$ for
$x\in[X+kL,2X-kL]$ and $\sigma_{k,X,L}^{+}(x)=1$ for $x\in[X,2X]$. One also
has the derivative conditions:

$\frac{d^{i}\sigma_{k,X,L}^{\pm}(x)}{dx^{i}}\ll_{k}\frac{1}{L^{i}}\text{ for
}1\leq i\leq k-1.$

We will need the following lemma to show that oscillatory integral of certain
smooth function is small.

###### Lemma 1

Suppose $K>1$ is some large integer. Let $H\geq 1$, $W,Q,L,R>0$ satisfying
$QR/\sqrt{H}>1$ and $RL>1$. Suppose that $w(t)$ is a smooth function with
support on $[\alpha,\beta]$ satisfying

$w^{(j)}(t)\ll_{j}WL^{-j}\;\;\text{ for }j=0,1,2,...,K.$

Suppose $h(t)$ is a smooth function on $[\alpha,\beta]$ such that
$|h^{\prime}(t)|\geq R$ and

$h^{(j)}(t)\ll_{j}HQ^{-j},\;\;\text{ for }j=2,3,...,2K-1.$

Then the integral

$I=\int_{-\infty}^{\infty}w(t)e^{ih(t)}dt$

satisfies

$I\ll_{K}(\beta-\alpha)W[(QR/\sqrt{H})^{-K}+(RL)^{-K}].$

Proof: This is basically Lemma 8.1 in [2] but we give the proof here for
completeness. Note that we do not require $w$ to be infinitely differentiable
or in Schwartz class. Since $w$ has compact support and $h^{\prime}$ is non-
zero, repeated integration by parts gives

$\displaystyle I=$
$\displaystyle\int_{-\infty}^{\infty}w(t)d\frac{e^{ih(t)}}{ih^{\prime}(t)}=\int_{-\infty}^{\infty}\frac{w^{\prime}(t)}{ih^{\prime}(t)}e^{ih(t)}dt=\int_{-\infty}^{\infty}\frac{w^{\prime}(t)}{ih^{\prime}(t)}d\frac{e^{ih(t)}}{ih^{\prime}(t)}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}\frac{1}{ih^{\prime}(t)}\Bigl{(}\frac{w^{\prime\prime}(t)}{ih^{\prime}(t)}-\frac{w^{\prime}(t)h^{\prime\prime}(t)}{ih^{\prime}(t)^{2}}\Bigr{)}e^{ih(t)}dt$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}\Bigl{(}-\frac{w^{\prime\prime}(t)}{h^{\prime}(t)^{2}}+\frac{w^{\prime}(t)h^{\prime\prime}(t)}{h^{\prime}(t)^{3}}\Bigr{)}d\frac{e^{ih(t)}}{ih^{\prime}(t)}$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}\frac{1}{ih^{\prime}(t)}\Bigl{(}-\frac{w^{\prime\prime\prime}(t)}{h^{\prime}(t)^{2}}+\frac{2w^{\prime\prime}(t)h^{\prime\prime}(t)}{h^{\prime}(t)^{3}}+\frac{w^{\prime\prime}(t)h^{\prime\prime}(t)}{h^{\prime}(t)^{3}}+\frac{w^{\prime}(t)h^{\prime\prime\prime}(t)}{h^{\prime}(t)^{3}}-\frac{3w^{\prime}(t)h^{\prime\prime}(t)^{2}}{h^{\prime}(t)^{4}}\Bigr{)}e^{ih(t)}dt$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}\Bigl{(}-\frac{w^{\prime\prime\prime}(t)}{ih^{\prime}(t)^{3}}+\frac{3w^{\prime\prime}(t)h^{\prime\prime}(t)}{ih^{\prime}(t)^{4}}+\frac{w^{\prime}(t)h^{\prime\prime\prime}(t)}{ih^{\prime}(t)^{4}}-\frac{3w^{\prime}(t)h^{\prime\prime}(t)^{2}}{ih^{\prime}(t)^{5}}\Bigr{)}d\frac{e^{ih(t)}}{ih^{\prime}(t)}$
$\displaystyle\cdots$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}\sum_{\nu=K}^{2K-1}\sum_{\mu=1}^{\nu}\frac{w^{(\mu)}(t)}{h^{\prime}(t)^{\nu}}\sum_{2\gamma_{2}+3\gamma_{3}+\cdots+\nu\gamma_{\nu}=\nu-\mu}c_{\nu,\mu,\gamma_{2},\gamma_{3},\cdots,\gamma_{\nu}}h^{(2)}(t)^{\gamma_{2}}h^{(3)}(t)^{\gamma_{3}}\cdots
h^{(\nu)}(t)^{\gamma_{\nu}}e^{ih(t)}dt$

for some absolute coefficients
$c_{\nu,\mu,\gamma_{2},\gamma_{3},\cdots,\gamma_{\nu}}\in\mathbb{C}$. Hence,

$|I|\ll_{K}(\beta-\alpha)W\sum_{\nu=K}^{2K-1}R^{-\nu}\sum_{\mu=1}^{\nu}L^{-\mu}\frac{H^{(\nu-\mu)/2}}{Q^{\nu-\mu}}$

which gives the lemma.

We also need the following three lemmas to count squarefull numbers in short
intervals for medium size $H$.

###### Lemma 2

Given $\epsilon>0$. Let $x>1$, $\epsilon>0$,
$\sqrt{x}\leq\epsilon_{0}^{3}y\leq\epsilon_{0}^{5}x$, and
$y=x^{1/2+\theta}=2\sqrt{x}H+H^{2}$. Then

$Q(x+y)-Q(x)=\frac{\zeta(3/2)}{2\zeta(3)}x^{\theta}+O\bigl{(}\epsilon_{0}x^{\theta}\bigr{)}+O(R_{1}+R_{2})$

where $R_{1}$ is the number of pairs of positive integers $(m,k)$ with
$\epsilon_{0}x^{\theta}<m\leq x^{1/5}$ and $x<m^{2}k^{3}\leq x+y$ while
$R_{2}$ is the number of pairs of positive integers $(m,k)$ with
$\epsilon_{0}x^{\theta}<k\leq x^{1/5}$ and $x<m^{2}k^{3}\leq x+y$.

Proof: This is Lemma 5 in [5].

###### Lemma 3

Let $F(u)$ be a real function with three continuous derivatives for $1\leq
u\leq 2$, with $F^{\prime\prime}(u)$ and $F^{(3)}(u)$ non-zero. Let $A$ be a
positive integer and $N$ be a real number with $N\geq A$, and let

$\Delta=\frac{N}{A^{2}}\min_{1\leq a\leq 2}|F^{\prime\prime}(a)|.$

Suppose that $\Delta<1$. Then for

$0\leq\delta\leq\frac{1}{2}\sqrt{\Delta},$

the number of integers $a\in\\{A,A+1,...,2A\\}$ for which there exists an
integer $n$ with

$\Big{|}n-NF\Bigl{(}\frac{a}{A}\Bigr{)}\Big{|}\leq\delta$

is

$\displaystyle O\Bigl{(}1+\Delta^{3/10}A^{9/10}(\log A)^{1/2}+\Delta^{4/11}$
$\displaystyle A^{10/11}(\log A)^{5/11}+\delta^{1/8}\Delta^{3/8}A(\log
A)^{5/8}$ $\displaystyle+\delta^{1/7}\Delta^{1/7}A^{6/7}(\log
A)^{5/7}+\delta^{2/5}\Delta^{1/5}A\log A+\delta A\Bigr{)}.$

Proof: This is Theorem 2 in [5] with slightly different notation. Even though
$N$ was stated to be an integer in [5], one can check that it also works with
real number $N\geq A$.

###### Lemma 4

Let $F(u)$ be a real funcion with continuous non-zero derivatives up to the
fourth order for $1\leq u\leq 2$. Let $A$ be a positive integer and $N$ be a
positive real number and let $\delta\geq 0$. If

$\delta<c\min\Bigl{(}\frac{N}{A},\frac{N}{A^{2}}+\frac{1}{N}\Bigr{)}$

where $c$ is a positive constant constructed from the derivatives of $F(u)$,
then the number of integers $a\in\\{A,A+1,...,2A\\}$ for which there exists an
integer $n$ with

$\Big{|}n-NF\Bigl{(}\frac{a}{A}\Bigr{)}\Big{|}\leq\delta$

is

$O(A^{1/2}N^{1/6}+\delta^{1/4}A^{1/2}N^{1/4}+\delta^{1/3}A^{2/3}N^{1/6}+\delta
A).$

Proof: This is the $r=2$ case of Theorem 6 in [4] with slightly different
notation.

##  3 Proof of Proposition 1

We follow the proof of Theorem 2 in [5] closely with a substantially smaller
choice of $\epsilon_{0}=H^{-0.1005}$. When $X\leq x\leq 2X$,
$X^{0.14697}<H\leq X^{0.18529}$, one can check that the condition
$\sqrt{x}\leq\epsilon_{0}^{3}y\leq\epsilon_{0}^{5}x$ in Lemma 2 is satisfied
as $y=2\sqrt{x}H+H^{2}=x^{1/2+\theta}$. Hence, we can apply Lemma 2 to count
$Q(x+y)-Q(x)$. Note that the main term and first error term in Lemma 2 give

$\frac{\zeta(3/2)}{2\zeta(3)}\Bigl{(}2H+\frac{H^{2}}{\sqrt{x}}\Bigr{)}+O(H^{0.8995})=\frac{\zeta(3/2)}{\zeta(3)}H+O(H^{0.8995})$

by our choice of $\epsilon_{0}$ and $H\leq X^{0.18529}$.

To estimate $R_{1}$ in Lemma 2 (which was wrongly attributed as $R_{2}$ in
[5]), we divide $\epsilon_{0}(2H+H^{2}/\sqrt{x})<a\leq x^{1/5}$ into dyadic
intervals $(A,2A]$ with $H^{0.8995}\ll A\leq x^{1/5}/2\leq X^{1/5}$. We want
to count the number of integer pairs $(a,b)$ such that
$\frac{x^{1/3}}{a^{2/3}}<b\leq\frac{(x+y)^{1/3}}{a^{2/3}}$ which implies
$\|\frac{x^{1/3}}{a^{2/3}}\|\leq\frac{(x+y)^{1/3}}{a^{2/3}}-\frac{x^{1/3}}{a^{2/3}}\leq\frac{y}{3x^{2/3}A^{2/3}}$.
Set

$F(u)=\frac{1}{u^{2/3}},\;\;N=\frac{x^{1/3}}{A^{2/3}}\geq
A,\;\;\Delta=\frac{10x^{1/3}}{9\cdot
2^{8/3}A^{8/3}},\;\;\delta=\frac{H}{x^{1/6}A^{2/3}}.$

One can check that $\delta\leq\sqrt{\Delta}/2$ as $A\leq X^{1/5}$ and $H\leq
X^{0.18529}$. Hence, Lemma 3 gives a bound of

$O\Bigl{(}\Bigl{(}X^{1/10}A^{1/10}+\frac{X^{4/33}}{A^{2/33}}+\frac{H^{1/8}X^{5/48}}{A^{1/12}}+H^{1/7}X^{1/42}A^{8/21}+H^{2/5}A^{1/5}+\frac{HA^{1/3}}{X^{1/6}}\Bigr{)}\log
X\Bigr{)}.$

Summing over dyadic intervals, we have $R_{1}\ll H^{0.8995}$ within our range
of $H$.

To estimate of $R_{2}$ in Lemma 2 (which was wrongly attributed as $R_{1}$ in
[5]), we divide $\epsilon_{0}(2H+H^{2}/\sqrt{x})<a\leq x^{1/5}$ into dyadic
intervals $(A,2A]$ with $H^{0.8995}\ll A\leq x^{1/5}/2\leq X^{1/5}$. We want
to count the number of integer pairs $(b,a)$ such that
$\frac{x^{1/2}}{a^{3/2}}<b\leq\frac{(x+y)^{1/2}}{a^{3/2}}$ which implies
$\|\frac{x^{1/2}}{a^{3/2}}\|\leq\frac{(x+y)^{1/2}}{a^{3/2}}-\frac{x^{1/2}}{a^{3/2}}\leq\frac{y}{2x^{1/2}A^{3/2}}$.
Set

$F(u)=\frac{1}{u^{3/2}},\;\;N=\frac{x^{1/2}}{A^{3/2}}\geq
A,\;\;\Delta=\frac{15x^{1/2}}{4\cdot
2^{7/2}A^{7/2}},\;\;\delta=\frac{2H}{A^{3/2}}.$

One can check that $\delta\leq\sqrt{\Delta}/2$ as $A\leq X^{1/5}$ and $H\leq
X^{0.18529}$. Hence, Lemma 3 gives a bound of

$O\Bigl{(}\Bigl{(}\frac{X^{3/20}}{A^{3/20}}+\frac{X^{2/11}}{A^{4/11}}+\frac{H^{1/8}X^{3/16}}{A^{1/2}}+H^{1/7}X^{1/14}A^{1/7}+\frac{H^{2/5}X^{1/10}}{A^{3/10}}+\frac{H}{A^{1/2}}\Bigr{)}\log
X\Bigr{)}.$

Summing over dyadic intervals with $X^{0.15735}\leq A\leq X^{1/5}$ gives
$O(H^{0.8995})$ such integer pairs within our range of $H$. When
$A<X^{0.15735}$ and $H\leq X^{0.18529}$, one can verify that the condition on
$\delta$ in Lemma 4 is satisfied as

$\delta=\frac{H}{A^{3/2}}<\frac{1}{X^{0.00001}}\cdot\frac{X^{1/2}}{A^{7/2}}\;\;\text{
and
}\;\;\frac{X^{1/2}}{A^{7/2}}\leq\min\Bigl{(}\frac{X^{1/2}}{A^{5/2}},\frac{X^{1/2}}{A^{7/2}}+\frac{A^{3/2}}{X^{1/2}}\Bigr{)}\ll\min\Bigl{(}\frac{N}{A},\frac{N}{A^{2}}+\frac{1}{N}\Bigr{)}.$

Hence, Lemma 4 gives a bound of

$O\Bigl{(}X^{1/12}A^{1/4}+\frac{H^{1/4}X^{1/8}}{A^{1/4}}+\frac{H^{1/3}X^{1/12}}{A^{1/12}}+\frac{H}{A^{1/2}}\Bigr{)}.$

Summing over dyadic intervals with $H^{0.8995}\ll A<X^{0.15735}$ gives
$O(H^{0.8995})$ such integer pairs within our range of $H$. Hence, $R_{2}\ll
H^{0.8995}$ and we have the proposition by combining all of the above.

##  4 Proof of Proposition 2

By (1), we can rewrite

$\frac{1}{X}\int_{X}^{2X}\Big{|}Q(x+y)-Q(x)-\frac{\zeta(3/2)}{\zeta(3)}H\Big{|}^{2}dx=\frac{1}{X}\int_{X}^{2X}\Big{|}\sum_{x<a^{2}b^{3}\leq
x+y}\mu^{2}(b)-\frac{\zeta(3/2)}{\zeta(3)}H\Big{|}^{2}dx$ (4)

where $y=2\sqrt{x}H+H^{2}$ with $H\leq X^{\frac{1}{6(1+2\lambda)}}$. Observe
that

$\sum_{b=1}^{\infty}\frac{\mu^{2}(b)}{b^{s}}=\prod_{p}\Bigl{(}1+\frac{1}{p^{s}}\Bigr{)}=\prod_{p}\frac{(1-\frac{1}{p^{2s}})}{(1-\frac{1}{p^{s}})}=\frac{\zeta(s)}{\zeta(2s)}$
(5)

for $\Re{s}>1$. With some parameter $\lambda>0$, we will consider those
squarefull numbers with $b\leq(2X)^{1/3}/H^{\lambda}$ and
$b>(2X)^{1/3}/H^{\lambda}$ separately. The left hand side of (4) can be
rewritten as

$\displaystyle\frac{1}{X}\int_{X}^{2X}$
$\displaystyle\bigg{|}\mathop{\sum_{x<a^{2}b^{3}\leq x+y}}_{b\leq
X^{1/3}/H^{\lambda}}\mu^{2}(b)-H\sum_{b\leq
X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3/2}}$
$\displaystyle+\mathop{\sum_{x<a^{2}b^{3}\leq
x+y}}_{b>X^{1/3}/H^{\lambda}}\mu^{2}(b)-H\sum_{b>X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3/2}}\bigg{|}^{2}dx$
$\displaystyle=$ $\displaystyle I_{1}+I_{2}+O(I_{1}^{1/2}I_{2}^{1/2})$ (6)

by Cauchy-Schwarz inequality where

$I_{1}:=\frac{1}{X}\int_{X}^{2X}\bigg{|}\mathop{\sum_{x<a^{2}b^{3}\leq
x+y}}_{b\leq X^{1/3}/H^{\lambda}}\mu^{2}(b)-H\sum_{b\leq
X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3/2}}\bigg{|}^{2}dx$ (7)

and

$I_{2}:=\frac{1}{X}\int_{X}^{2X}\bigg{|}\mathop{\sum_{x<a^{2}b^{3}\leq
x+y}}_{b>X^{1/3}/H^{\lambda}}\mu^{2}(b)-H\sum_{b>X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3/2}}\bigg{|}^{2}dx.$
(8)

Observe that, among squarefull numbers $x<a^{2}b^{3}\leq x+y$ with
$b>X^{1/3}/H^{\lambda}$, the corresponding $a$’s must be distinct. For
otherwise, we would have $x<a^{2}b_{1}^{3}<a^{2}b_{2}^{3}\leq x+y$ which
implies

$2.1\sqrt{2X}H\geq 2\sqrt{x}H+H^{2}=y>a^{2}b_{2}^{3}-a^{2}b_{1}^{3}\geq
a^{2}(b_{1}+1)^{3}-a^{2}b_{1}^{3}>3a^{2}b_{1}^{2}>\frac{3X^{2/3}}{H^{2\lambda}}.$

This contradicts with $H\leq X^{\frac{1}{6(1+2\lambda)}}$. Thus,

$\mathop{\sum_{x<a^{2}b^{3}\leq
x+y}}_{b>X^{1/3}/H^{\lambda}}\mu^{2}(b)\leq\sqrt{\frac{x+y}{(\frac{X^{1/3}}{H^{\lambda}})^{3}}}\leq
2H^{3\lambda/2}\;\;\text{ and
}\;\;H\sum_{b>X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3/2}}\ll\frac{H^{1+\lambda/2}}{X^{1/6}}\leq
1$ (9)

as $H\leq X^{\frac{1}{6(1+2\lambda)}}$, $y=2\sqrt{x}H+H^{2}$ and $X\leq x\leq
2X$. Putting (9) into (8), we have

$I_{2}\ll H^{3\lambda}.$ (10)

We are going to study a smoothed version of $I_{1}$ first. Let
$L=X^{1-\epsilon/2}$ and $K$ be some sufficiently large integer (in terms of
$\epsilon$). Consider

$\frac{1}{X}\int_{-\infty}^{\infty}\sigma_{K,X,L}^{\pm}(x)\bigg{|}\mathop{\sum_{x<a^{2}b^{3}\leq
x+y}}_{b\leq X^{1/3}/H^{\lambda}}\mu^{2}(b)-H\sum_{b\leq
X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3/2}}\bigg{|}^{2}dx$ (11)

We have

$\displaystyle\mathop{\sum_{x<a^{2}b^{3}\leq x+y}}_{b\leq
X^{1/3}/H^{\lambda}}\mu^{2}(b)=$ $\displaystyle\sum_{b\leq
X^{1/3}/H^{\lambda}}\mu^{2}(b)\sum_{\sqrt{x/b^{3}}<a\leq\sqrt{(x+y)/b^{3}}}1$
$\displaystyle=$ $\displaystyle\sum_{b\leq
X^{1/3}/H^{\lambda}}\mu^{2}(b)\Bigl{[}\sqrt{\frac{x+y}{b^{3}}}-\sqrt{\frac{x}{b^{3}}}-\psi\Bigl{(}\sqrt{\frac{x+y}{b^{3}}}\Bigr{)}+\psi\Bigl{(}\sqrt{\frac{x}{b^{3}}}\Bigr{)}\Bigr{]}$
$\displaystyle=$ $\displaystyle H\sum_{b\leq
X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3/2}}-\sum_{b\leq
X^{1/3}/H^{\lambda}}\mu^{2}(b)\Bigl{[}\psi\Bigl{(}\sqrt{\frac{x+y}{b^{3}}}\Bigr{)}-\psi\Bigl{(}\sqrt{\frac{x}{b^{3}}}\Bigr{)}\Bigr{]}$
(12)

where $\psi(u)=u-[u]-1/2$. For $\psi(u)$, we apply the Fourier expansion

$\psi(u)=-\frac{1}{2\pi i}\sum_{0<|n|\leq
N}\frac{1}{n}e(nu)+O\Bigl{(}\min\bigl{(}1,\frac{1}{N\|u\|}\bigr{)}\Bigr{)}.$
(13)

We take $N=X^{10}$ and put (13) into (4). The arising error term is
$O(1/X^{4})$ unless $\|\sqrt{(x+y)/b^{3}}\|<X^{-5}$ or
$\|\sqrt{x/b^{3}}\|<X^{-5}$. When $\|\sqrt{(x+y)/b^{3}}\|<X^{-5}$ or
$\|\sqrt{x/b^{3}}\|<X^{-5}$, we have $|\sqrt{(x+y)/b^{3}}-a_{1}|<X^{-5}$ or
$|\sqrt{x/b^{3}}-a_{2}|<X^{-5}$ for some integers $a_{1},a_{2}$ which implies

$\Big{|}\frac{x+y}{b^{3}}-a_{1}^{2}\Bigr{|},\;\Big{|}\frac{x}{b^{3}}-a_{2}^{2}\Bigr{|}\ll\frac{1}{X^{4.5}}\;\;\text{
or }\;\;|x+y-a_{1}^{2}b^{3}|,\;|x-a_{2}^{2}b^{3}|\ll\frac{1}{X^{3.5}}$ (14)

as $b\leq X^{1/3}/H^{\lambda}$. Let $E(x)$ be the error in the second sum in
(4) coming from the error term of (13) when $\|\sqrt{(x+y)/b^{3}}\|<X^{-5}$ or
$\|\sqrt{x/b^{3}}\|<X^{-5}$ for some $b\leq X^{1/3}/H^{\lambda}$. As there are
$O(\sqrt{X})$ squarefull numbers in the interval $[X,3X]$ by (2),

$E(x)\leq\frac{X^{1/3}}{H^{\lambda}}\;\;\text{ and }\;\;E(x)\text{ is
supported on }O(\sqrt{X})\text{ intervals of length
}O\Bigl{(}\frac{1}{X^{3.5}}\Bigr{)}\text{ each}$ (15)

by (14). Thus, we can replace (11) by

$\displaystyle\frac{1}{4\pi^{2}X}$
$\displaystyle\int_{-\infty}^{\infty}\sigma_{K,X,L}^{\pm}(x)\bigg{|}\sum_{b\leq
X^{1/3}/H^{\lambda}}\mu^{2}(b)\sum_{0<|n|\leq
N}\frac{1}{n}e\Bigl{(}\frac{n\sqrt{x}}{b^{3/2}}\Bigl{)}\Bigl{[}1-e\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}\Bigr{]}+E(x)+O\Bigl{(}\frac{1}{X^{4}}\Bigr{)}\bigg{|}^{2}dx$
$\displaystyle=$
$\displaystyle\frac{1}{X}\Bigl{[}J+O\Bigl{(}\frac{J^{1/2}}{X^{7/6}}+\frac{1}{X^{7/3}}\Bigr{)}\Bigr{]}$
(16)

by Cauchy-Schwarz inequality and
$\int_{-\infty}^{\infty}\sigma_{K,X,L}^{\pm}(x)|E(x)+O(1/X^{4})|^{2}dx\ll
1/X^{7/3}$ by (15). Here

$J:=\frac{1}{4\pi^{2}}\int_{-\infty}^{\infty}\sigma_{K,X,L}^{\pm}(x)\bigg{|}\sum_{b\leq
X^{1/3}/H^{\lambda}}\mu^{2}(b)\sum_{0<|n|\leq
N}\frac{1}{n}e\Bigl{(}\frac{n\sqrt{x}}{b^{3/2}}\Bigl{)}\Bigl{[}1-e\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}\Bigr{]}\bigg{|}^{2}dx.$
(17)

Expanding things out,

$\displaystyle J=$ $\displaystyle\frac{1}{4\pi^{2}}\sum_{b_{1},b_{2}\leq
X^{1/3}/H^{\lambda}}\sum_{0<|n_{1}|,|n_{2}|\leq
N}\mu^{2}(b_{1})\mu^{2}(b_{2})\frac{1}{n_{1}n_{2}}$
$\displaystyle\times\Bigl{[}1-e\Bigl{(}\frac{n_{1}H}{b_{1}^{3/2}}\Bigr{)}\Bigr{]}\overline{\Bigl{[}1-e\Bigl{(}\frac{n_{2}H}{b_{2}^{3/2}}\Bigr{)}\Bigr{]}}\int_{-\infty}^{\infty}\sigma_{K,X,L}^{\pm}(x)e\Bigl{(}x^{1/2}\Bigl{(}\frac{n_{1}}{b_{1}^{3/2}}-\frac{n_{2}}{b_{2}^{3/2}}\Bigr{)}\Bigr{)}dx.$
(18)

By Lemma 1 with $H=X^{\epsilon}$, $W=1$, $Q=X$, $L=X^{1-\epsilon/2}$ and
$R=1/X^{1-\epsilon}$, the contribution from those terms with
$|\frac{n_{1}}{b_{1}^{3/2}}-\frac{n_{2}}{b_{2}^{3/2}}|>\frac{1}{X^{1/2-\epsilon}}$
is

$\ll X^{-K\epsilon/2}\ll\frac{1}{X^{5}}$ (19)

by picking $K=[20/\epsilon]$ for example. Hence, we may restrict the sum in
(4) to those integers for which

$\bigg{|}\frac{n_{1}}{b_{1}^{3/2}}-\frac{n_{2}}{b_{2}^{3/2}}\bigg{|}\leq\frac{1}{X^{1/2-\epsilon}}.$
(20)

We separate those $(n_{1},n_{2},b_{1},b_{2})$ for which
$\frac{n_{1}}{b_{1}^{3/2}}=\frac{n_{2}}{b_{2}^{3/2}}$ (i.e.
$n_{1}^{2}b_{2}^{3}=n_{2}^{2}b_{1}^{3}$) and those for which
$\frac{n_{1}}{b_{1}^{3/2}}\neq\frac{n_{2}}{b_{2}^{3/2}}$. In the first case,
we must have $b_{1}=b_{2}=b$ and $n_{1}=n_{2}=n$ which contributes

$\displaystyle\frac{\overline{\sigma_{K,X,L}^{\pm}}}{4\pi^{2}}\sum_{b\leq
X^{1/3}/H^{\lambda}}\mu^{2}(b)\sum_{n\neq
0}\frac{1}{n^{2}}\Big{|}1-e\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}\Big{|}^{2}+O\Bigl{(}\frac{1}{X^{5}}\Bigr{)}$

where the error term comes from adding $|n|>N=X^{10}$ and

$\overline{\sigma_{K,X,L}^{\pm}}=\int_{-\infty}^{\infty}\sigma_{K,X,L}^{\pm}(x)dx=X+O_{\epsilon}(X^{1-\epsilon/2}).$
(21)

As

$\Big{|}1-e\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}\Big{|}=2\Big{|}\sin\Bigl{(}\frac{n\pi
H}{b^{3/2}}\Bigr{)}\Big{|},$

the diagonal terms ($\frac{n_{1}}{b_{1}^{3/2}}=\frac{n_{2}}{b_{2}^{3/2}}$)
from (4) contribute

$(1+O(X^{-\epsilon/2}))H^{2}\sum_{b\leq
X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3}}\sum_{n\neq
0}S\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}^{2}+O\Bigl{(}\frac{1}{X^{5}}\Bigr{)}$
(22)

It remains to show that the contribution from those terms with
$n_{1}^{2}b_{2}^{3}\neq n_{2}^{2}b_{1}^{3}$ is also acceptable. Splitting
$n_{i}$ and $b_{i}$ dyadically, we need to bound

$\min\Bigl{(}\frac{1}{N_{1}},\frac{H}{B_{1}^{3/2}}\Bigr{)}\min\Bigl{(}\frac{1}{N_{2}},\frac{H}{B_{2}^{3/2}}\Bigr{)}\\#\Bigl{\\{}(b_{1},b_{2},n_{1},n_{2}):b_{i}\sim
B_{i},n_{i}\sim
N_{i},0<\Big{|}\frac{n_{1}}{b_{1}^{3/2}}-\frac{n_{2}}{b_{2}^{3/2}}\Big{|}\leq\frac{1}{X^{1/2-\epsilon}}\Bigr{\\}}$
(23)

for any $B_{1},B_{2}\leq X^{1/3}/H^{\lambda}$ and $N_{1},N_{2}\leq N$. For
$\lambda>\epsilon$, we have
$\frac{B_{1}^{3/2}}{X^{1/2-\epsilon}},\frac{B_{2}^{3/2}}{X^{1/2-\epsilon}}\ll\frac{X^{\epsilon}}{H^{3/2\lambda}}<0.1$.
So, we can assume that $N_{1}\geq\frac{10B_{1}^{3/2}}{X^{1/2-\epsilon}}$ and
$N_{2}\geq\frac{10B_{2}^{3/2}}{X^{1/2-\epsilon}}$. Thus,

$\Big{|}\frac{n_{1}}{b_{1}^{3/2}}-\frac{n_{2}}{b_{2}^{3/2}}\Big{|}\leq\frac{1}{X^{1/2-\epsilon}}$
(24)

has no solution unless $N_{1}B_{2}^{3/2}\asymp N_{2}B_{1}^{3/2}$. Also, (24)
implies

$|n_{1}^{2}b_{2}^{3}-n_{2}^{2}b_{1}^{3}|\ll
N_{1}B_{2}^{3/2}\frac{B_{1}^{3/2}B_{2}^{3/2}}{X^{1/2-\epsilon}}.$ (25)

Case 1: When
$\frac{B_{1}^{3/2}B_{2}^{3/2}}{X^{1/2-\epsilon}}>(N_{1}^{1/2}N_{2}^{1/2}B_{1}^{3/4}B_{2}^{3/4})^{0.1234}$,
Trifonov’s result on (3) tells us that the number of such pairs of squarefull
numbers $n_{1}^{2}b_{2}^{3}$ and $n_{2}^{2}b_{1}^{3}$ is bounded by
$O\bigl{(}(\frac{B_{1}^{3/2}B_{2}^{3/2}}{X^{1/2-\epsilon}})^{2}\bigr{)}$.
Hence, the contribution to (23) is

$\ll\sum_{B_{1},B_{2}\leq
X^{1/3}/H^{\lambda}}\frac{H}{B_{1}^{3/2}}\frac{H}{B_{2}^{3/2}}\frac{B_{1}^{3}B_{2}^{3}}{X^{1-2\epsilon}}\ll_{\epsilon}H^{2-3\lambda}X^{2\epsilon}.$
(26)

Case 2: When
$\frac{B_{1}^{3/2}B_{2}^{3/2}}{X^{1/2-\epsilon}}\leq(N_{1}^{1/2}N_{2}^{1/2}B_{1}^{3/4}B_{2}^{3/4})^{0.1234}$,
Trifonov’s result on (3) tells us that the number of such pairs of squarefull
numbers $n_{1}^{2}b_{2}^{3}$ and $n_{2}^{2}b_{1}^{3}$ is bounded by
$O\bigl{(}(N_{1}^{1/2}N_{2}^{1/2}B_{1}^{3/4}B_{2}^{3/4})^{0.2468}\bigr{)}$.
Hence, the contribution to (23) is

$\ll\sum_{B_{1},B_{2}\leq
X^{1/3}/H^{\lambda}}\frac{1}{N_{1}^{0.8766}}\frac{H^{0.1234}}{B_{1}^{0.1851}}\frac{1}{N_{2}^{0.8766}}\frac{H^{0.1234}}{B_{2}^{0.1851}}N_{1}^{0.1234}N_{2}^{0.1234}B_{1}^{0.1851}B_{2}^{0.1851}\ll
H^{0.2468}\log^{2}X.$ (27)

Combining (17), (4), (19), (22), (23), (26) and (27), we have

$J=(1+O(X^{-\epsilon/2}))2H^{2}\sum_{b\leq
X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3}}\sum_{n\geq
1}S\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}^{2}+O_{\epsilon}\Bigl{(}H^{2-3\lambda}X^{2\epsilon}+H^{0.2468}\log^{2}X\Bigr{)}.$
(28)

From Proposition 3, we know that $J\asymp_{\epsilon}H^{2/3}$. Since the
characteristic function of the interval $[X,2X]$ is between
$\sigma_{K,X,L}^{-}$ and $\sigma_{K,X,L}^{+}$, Proposition 2 follows from (4),
(4), (7), (8), (10), (11), (4) and (28).

##  5 Proof of Proposition 3

In this section, we shall focus on finding an asymptotic formula for the
diagonal contribution:

$2H^{2}\sum_{b\leq X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3}}\sum_{n\geq
1}S\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}^{2}\;\;\text{ where
}\;\;S(x)=\frac{\sin\pi x}{\pi x}.$ (29)

As in [3], we consider a smoother sum first. Fix $\epsilon>0$ and $K_{0}>0$.
Suppose that $w:\mathbb{R}\rightarrow\mathbb{C}$ satisfies

$|w^{(k)}(y)|\leq K_{0}\frac{H^{l\epsilon/4}}{(1+|y|)^{l}}$ (30)

for all $k,l\in\\{0,1,2,3,4\\}$. We claim that

$2H^{2}\sum_{b\leq X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3}}\sum_{n\geq
1}\Big{|}w\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}\Big{|}^{2}=H^{2/3}\frac{\zeta(4/3)}{\zeta(2)}\int_{0}^{\infty}|w(y)|^{2}y^{1/3}dy+O_{\epsilon}\Bigl{(}\frac{H^{1+\lambda/2+\epsilon/4}}{X^{1/6}}+H^{1/3}\Bigr{)}.$
(31)

Proof of claim: This is very similar to Lemma 8 in [3]. Firstly, we complete
the sum over all $b$. By (30) for $k=0$ and $l=0,1$, we have

$\sum_{n\geq 1}|w(n/\nu)|^{2}\ll\sum_{1\leq n\leq\nu
H^{\epsilon/4}}1+\sum_{n>\nu
H^{\epsilon/4}}\frac{H^{\epsilon/2}\nu^{2}}{n^{2}}\ll_{\epsilon}\nu
H^{\epsilon/4}$

for any $\nu>0$. Hence,

$2H^{2}\sum_{b>X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3}}\sum_{n\geq
1}\Big{|}w\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}\Big{|}^{2}\ll
H^{1+\epsilon/4}\sum_{b>X^{1/3}/H^{\lambda}}\frac{1}{b^{3/2}}\ll_{\epsilon}\frac{H^{1+\lambda/2+\epsilon/4}}{X^{1/6}}.$

Thus, the left hand side of (31) is

$2H^{2}\sum_{b\geq 1}\frac{\mu^{2}(b)}{b^{3}}\sum_{n\geq
1}\Big{|}w\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}\Big{|}^{2}+O_{\epsilon}\Bigl{(}\frac{H^{1+\lambda/2+\epsilon/4}}{X^{1/6}}\Bigr{)}.$
(32)

Secondly, we use contour integration to simplify (32). Define
$g(x):=|w(e^{x})|^{2}e^{x}$, which is smooth and decays exponentially as
$|x|\rightarrow\infty$ by (30). Its Fourier transform

$\hat{g}(\xi)=\int_{-\infty}^{\infty}|w(e^{x})|^{2}e^{x}e^{-2\pi
ix\xi}dx=\int_{0}^{\infty}|w(y)|^{2}y^{-2\pi i\xi}dy,$

and one can show that (i) $\hat{g}(\xi)$ is entire and (ii)
$\hat{g}(\xi)=O(H^{2\epsilon}/(|\xi|+1)^{3})$ uniformly for
$|\Im(\xi)|<1/(2\pi)$ by (30) and standard partial summation argument. Then
Fourier inversion gives, for $r>0$,

$|w(r)|^{2}=r^{-1}\frac{1}{2\pi i}\int_{(c)}r^{s}\hat{g}\Bigl{(}\frac{s}{2\pi
i}\Bigr{)}ds,$

where the integral is over $\Re(s)=c$ with $-1<c<1$. Hence, taking $c=-1/4$,

$2H^{2}\sum_{b\geq 1}\frac{\mu^{2}(b)}{b^{3}}\sum_{n\geq
1}\Big{|}w\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}\Big{|}^{2}=\frac{H}{i\pi}\sum_{b\geq
1}\frac{\mu^{2}(b)}{b^{3/2}}\sum_{n\geq
1}\frac{1}{n}\int_{(-1/4)}H^{s}n^{s}b^{-3s/2}\hat{g}\Bigl{(}\frac{s}{2\pi
i}\Bigr{)}ds.$

Due to absolute convergence of the series, one can take the sums inside the
integral and the above becomes

$\frac{H}{i\pi}\int_{(-1/4)}H^{s}\zeta(1-s)\frac{\zeta(\frac{3}{2}(1+s))}{\zeta(3(1+s))}\hat{g}\Bigl{(}\frac{s}{2\pi
i}\Bigr{)}ds$

by (5). Because of the bounds on $\hat{g}$ and $\zeta(1/2+it)\ll
t^{1/6}\log^{2}t$, we can push the contour integral above to the left to an
integral over $\Re(s)=-2/3$. Picking out a residue at $x=-1/3$ from the
singularity of $\zeta(\frac{3}{2}(1+s))$, the above simplifies to

$H^{2/3}\frac{\zeta(4/3)}{\zeta(2)}\hat{g}\Bigl{(}-\frac{1}{6\pi
i}\Bigr{)}+O(H^{1/3})=H^{2/3}\frac{\zeta(4/3)}{\zeta(4)}\int_{0}^{\infty}|w(y)|^{2}y^{1/3}dy+O(H^{1/3}).$

This and (32) finish the proof of the claim.

To deal with (29), we introduce the function

$w(y):=S(y)h\Bigl{(}\frac{y}{H^{\epsilon/4}}\Bigr{)}$

where $h$ is a smooth bump function such that $h(x)=1$ for $|x|\leq 1$ and
$h(x)=0$ for $|x|\geq 2$ (for example, one can take
$h(x)=\sigma_{10,2,0.1}^{+}(x-3)$). Then one can check that $w(y)$ satisfies
the hypothesis (30). For such $w$,

$\displaystyle 2H^{2}\sum_{b\leq X^{1/3}/H^{\lambda}}$
$\displaystyle\frac{\mu^{2}(b)}{b^{3}}\sum_{n\geq
1}\biggl{(}S\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}^{2}-w\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}^{2}\biggr{)}\ll
H^{2}\mathop{\sum_{b\leq X^{1/3}/H^{\lambda},n\geq 1}}_{nH^{1-\epsilon/4}\geq
b^{3/2}}\frac{1}{b^{3}}\frac{1}{(nH/b^{3/2})^{2}}$
$\displaystyle\ll\sum_{b\leq H^{2/3-\epsilon/6}}\sum_{n\geq
1}\frac{1}{n^{2}}+\sum_{b>H^{2/3-\epsilon/6}}\sum_{n\geq
b^{3/2}/H^{1-\epsilon/4}}\frac{1}{n^{2}}\ll_{\epsilon}H^{2/3-\epsilon/6}.$
(33)

On the other hand,

$\int_{0}^{\infty}|S(y)|^{2}y^{1/3}dy-\int_{0}^{\infty}|w(y)|^{2}y^{1/3}dy\ll\int_{H^{\epsilon/4}}^{\infty}y^{-5/3}dy\ll_{\epsilon}H^{-\epsilon/6}.$
(34)

Combining (31), (5) and (34), we have

$2H^{2}\sum_{b\leq X^{1/3}/H^{\lambda}}\frac{\mu^{2}(b)}{b^{3}}\sum_{n\geq
1}\Big{|}S\Bigl{(}\frac{nH}{b^{3/2}}\Bigr{)}\Big{|}^{2}=H^{2/3}\frac{\zeta(4/3)}{\zeta(2)}\int_{0}^{\infty}|S(y)|^{2}y^{1/3}dy+O_{\epsilon}\Bigl{(}\frac{H^{1+\lambda/2+\epsilon/4}}{X^{1/6}}+H^{2/3-\epsilon/6}\Bigr{)}$

which gives Proposition 3.

##  6 Proof of Theorem 1

From Propositions 2 and 3, we have, for $H\leq X^{\frac{1}{6(1+2\lambda)}}$,

$\frac{1}{X}\int_{X}^{2X}\Big{|}Q(x+y)-Q(x)-\frac{\zeta(3/2)}{\zeta(3)}H\Big{|}^{2}dx\ll_{\epsilon}(H^{3\lambda}+H^{2-3\lambda}+H^{0.2468})X^{2\epsilon}+H^{2/3}+\frac{H^{1+\lambda/2+\epsilon/4}}{X^{1/6}}.$

By picking $\lambda=0.067$, the above yields

$\frac{1}{X}\int_{X}^{2X}\Big{|}Q(x+y)-Q(x)-\frac{\zeta(3/2)}{\zeta(3)}H\Big{|}^{2}dx\ll_{\epsilon}H^{1.799}X^{2\epsilon}$
(35)

for $H\leq X^{0.14697}$. For $X^{0.14697}<H\leq X^{0.18529}$, we apply
Proposition 1 and get

$\frac{1}{X}\int_{X}^{2X}\Big{|}Q(x+y)-Q(x)-\frac{\zeta(3/2)}{\zeta(3)}H\Big{|}^{2}dx\ll
H^{1.799}.$ (36)

For $X^{0.18529}<H<X^{1/2-\epsilon}$, we simply apply (2) and get

$\displaystyle\frac{1}{X}\int_{X}^{2X}\Big{|}Q(x+y)-Q(x)-\frac{\zeta(3/2)}{\zeta(3)}H\Big{|}^{2}dx\ll$
$\displaystyle\frac{1}{X}\int_{X}^{2X}\Big{|}\frac{\zeta(2/3)}{\zeta(2)}(\sqrt[3]{x+y}-\sqrt[3]{x})+x^{1/6}\Big{|}^{2}dx$
$\displaystyle\ll$
$\displaystyle\frac{H^{2}}{X^{1/3}}+X^{1/3}\ll_{\epsilon}H^{1.799}.$ (37)

Combining (35), (36) and (6), we have Theorem 1.

## References

  * [1] P.T. Bateman and E. Grosswald, On a theorem of Erdős and Szekeres, Illinois J. Math. 2 (1958), 88–98.
  * [2] V. Blomer, R. Khan and M. Young, Distribution of Mass of Hecke eigenforms, Duke Math. J. 162 (2013) no. 14, 2609–2644.
  * [3] O. Gorodetsky, K. Matomäki, M. Radziwiłł and B. Rogers, On the variance of squarefree integers in short intervals and arithmetic progressions, Geom. Funct. Anal. 31 (2021), 111–149.
  * [4] M. Filaseta and O. Trifonov, The distribution of fractional parts with applications to gap results in number theory, Proc. London Math. Soc. (2) 3 (1996), 241–278.
  * [5] M.N. Huxley and O. Trifonov, The square-full numbers in an interval, Math. Proc. Camb. Phil. Soc. 119 (1996), 201–208.
  * [6] K. Matomäki and M. Radziwiłł, Multiplicative functions in short intervals, Annals of Math. 183 (2016), 1015–1056.
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Mathematics Department
Kennesaw State University
Marietta, GA 30060
<EMAIL_ADDRESS>


Zhihang Xu, Shaofei Zhang, Xi Wang, Jiajun Zhang, Wenning Wei, Lei He, Sheng
Zhao

# MuLanTTS: The Microsoft Speech Synthesis System for
Blizzard Challenge 2023

###### Abstract

In this paper, we present MuLanTTS, the Microsoft end-to-end neural text-to-
speech (TTS) system designed for the Blizzard Challenge 2023. About 50 hours
of audiobook corpus for French TTS as hub task and another 2 hours of speaker
adaptation as spoke task are released to build synthesized voices for
different test purposes including sentences, paragraphs, homographs, lists,
etc. Building upon DelightfulTTS, we adopt contextual and emotion encoders to
adapt the audiobook data to enrich beyond sentences for long-form prosody and
dialogue expressiveness. Regarding the recording quality, we also apply
denoise algorithms and long audio processing for both corpora. For the hub
task, only the 50-hour single speaker data is used for building the TTS
system, while for the spoke task, a multi-speaker source model is used for
target speaker fine-tuning. MuLanTTS achieves mean scores of quality
assessment 4.3 and 4.5 in the respective tasks, statistically comparable with
natural speech while keeping good similarity according to similarity
assessment. The excellent quality and similarity in this year's new and dense
statistical evaluation show the effectiveness of our proposed system in both
tasks.

Index Terms: Text-to-Speech, Blizzard Challenge 2023, French TTS, Audiobook,
Contextual Encoder, Emotion Encoder, Conformer

##  1 Introduction

The Blizzard Challenge aims to advance the technologies in TTS [1, 2, 3] by
comparing and understanding different approaches. It has been organized
annually since 2005 [4], with the basic task being to build high-quality TTS
systems based on the speech database provided by the organizers. Participants
use their developed systems to synthesize audio from the given test set, and
the generated audio samples are used to evaluate the performance of different
systems through subjective listening tests and objective metrics.

End-to-end neural TTS has achieved significant naturalness quality
improvements. Tacotron [1, 5] and TransformerTTS [6] introduce an attention-
based encoder-decoder acoustic model to predict mel-spectrograms from a
character sequence, and a WaveNet model [7] to synthesize waveforms from mel-
spectrograms. These autoregressive models suffer from robustness issues and
slow inference speed. RobuTrans [8] attempted to solve stability problems like
repeated phones by using linguistic features and phone duration with an
autoregressive Transformer. However, the inference is still slow due to the
autoregressive nature of the model. FastSpeech [2] shifted towards non-
autoregressive TTS by using a feed-forward transformer in the encoder and
decoder for parallel generation and a duration predictor for phoneme duration.
With the introduction of the Conformer [9] architecture and rich variance
adapter and prosody style [10, 11] into TTS systems, it could achieve human-
like quality in a specific speech corpus.

Compared with cascaded acoustic and vocoder models, single-stage end-to-end
TTS models as new paradigms have been proposed recently. These include neural
codec-based methods such as DelightfulTTS2 [10], VALLE [12], and Tortoise-TTS
[13], latent variable-based methods like VITS [14] and NaturalSpeech [15], and
pre-trained vector-based methods such as VQTTS [16], SpearTTS [17], and
UniCATS [18]. These models aim to reduce the training and inference mismatch
and capture more detailed intermediate speech representations other than
human-designed features like mel-spectrogram. With appropriate adversarial
criteria and training strategy, the single-stage pure end-to-end model has
room to achieve even higher fidelity.

With the fast-paced advancements in TTS, complex applications like paragraph
TTS for long-form text synthesis with expressiveness and contextual variations
have gained attention. Unlike single-sentence synthesis, it demands diverse
prosody and smooth transitions. Contextual TTS addresses traditional TTS
limitations by using extra contextual information for a more precise and
expressive portrayal of the intended message. Recent progress include
attention mechanisms [19, 20], conditional style prediction [21, 22], and the
integration of pre-trained NLP models into TTS systems [23, 22, 21].

According to this year's Blizzard Challenge with two tasks on 50 hours of
audiobook synthesis and 2 hours of spoke speech adaptation, we design the
model from two aspects: 1) incorporating rich text-level contextual and
emotion encoders within additional text corpus, to capture the subtle nuances
and variations that are characteristic of human speech in context variation.
2) building a multi-speaker source model to adapt to the spoke speaker. In
addition, we have incorporated advanced speech enhancement algorithms to
bolster the clarity and quality of the synthesized speech. These sophisticated
algorithms are designed to minimize noise and other audio disruptions, which
would benefit the perception of naturalness while noise may be amplified
during training. Furthermore, our data preparation process has played a
significant role in the transcription segmentation of long audios like
audiobook data.

Our system demonstrates superior performance in both tasks. In the MOS test,
the mean scores are 4.3 for FH1 and 4.5 for FS1, both of which outperformed
other TTS systems and are close to or matched the recording scores. In the
Similarity MOS test, though FH1 task is quite hard due to the large data
variation, our system still achieves the relatively highest mean score of 3.4
statistically comparing recordings. Meanwhile, in FS1, our System F obtains a
mean SMOS score of 4.3 which surpasses the recording's score a little. This
demonstrates the performance of our system in enhancing perceived quality and
naturalness. For audio samples, please visit our website111Audio samples:
https://cognitivespeech.github.io/mulantts.

##  2 System Structure

###  2.1 System Overview

As shown in Figure 1, Our system employs the conventional TTS strategy, which
comprises a frontend, an acoustic model, and a vocoder. The input text is
firstly normalized and converted into phone sequences using the TTS frontend,
and the resulting phone sequences are then fed into the acoustic model to
generate a 16k mel-spectrogram. To improve the pronunciation of poly-phone and
liaison phenomena in French, we utilize a BERT [24] model to make predictions
in a multitask fashion [25]. Additionally, to model long context, text
emotion, and narration-or-dialogue variance, we utilize BERT-based contextual
encoder [23] and emotion encoder to enhance the acoustic model's ability to
generate more expressive speech222All pretrained BERTs used in liaison,
polyphone, and contextual modeling are from https://huggingface.co/dbmdz/bert-
base-french-europeana-cased. Finally, the predicted mel-spectrogram is fed
into a high-fidelity HiFi-GAN [26] to generate a 24kHz waveform.

Figure 1: Our MuLanTTS system converts phonemes into a 16k mel-spectrogram
using a Conformal-based model, extracts long and emotional text details with
BERT-based encoders, and then applies HiFi-GAN to transform the mel-
spectrogram into a 24kHz waveform.

###  2.2 Liaison and Polyphone

####  2.2.1 Liaison

Liaison is a French pronunciation rule that occurs between a word that ends in
a consonant and one that starts with a vowel. It’s the practice of taking the
last letter of one word and linking it to the beginning of the next word and
primarily focuses on the final phonemes of 'n,' 't,' and 'z'. It usually is a
major issue for French phonetizers. Pontes and Furui [27] introduce a model of
liaison from data using a decision tree learning algorithm.

Unlike previous work, we utilize a BERT model to predict the liaison behavior
of target words. We fine-tune the BERT model using liaison data labeled by a
language expert, which encompasses approximately 1600 sentences and 1800
instances of liaisons. The result is a binary classification that predicts
liaison occurrence. Given the intricate relationship between liaison behavior
and part-of-speech (POS) in French, we concurrently train the model for the
POS task using data from TreeBank333https://universaldependencies.org/, which
consists of about 14000 sentences.

####  2.2.2 Polyphone

There are numerous homophones, which is attributed to the intricate
relationship between pronunciation and spelling in the French language. To
achieve improved French TTS pronunciation, a polyphone model is trained based
on BERT. Therefore, our final model encompasses multiple tasks, including POS
tagging, polyphone disambiguation, and liaison prediction. For most polyphone
words' training data, it's labeled by OpenAI text-davinci-003 model [28] with
employing appropriate prompts, each polyphone word is labeled with about 1000
sentences.

###  2.3 Contextual Modeling

####  2.3.1 Contextual Encoder

Given the same sentence with a different context, the prosody of the generated
speech would be different. Therefore, we adopt a text-based contextual encoder
[23] to model the contextual information to enhance the prosody coherence and
expressiveness of long-form content reading. Given a paragraph text, for each
sentence centered in a predefined number of sentences of context, our
contextual encoder extracts two kinds of contextual representations: token-
based and sentence-based representation. These two features are up-sampled to
phoneme-level and then added to the conformer encoder input as shown in Figure
1.

Specifically, we use a BERT and statistical feature extractor to extract the
token-level semantic embedding and syntactic information, which is derived as
a token-based contextual representation. For the sentence-based contextual
representation, we extract each sentence's BERT embedding (the average of BERT
token's embedding) for the predefined range of context in the paragraph and
assemble them into a matrix. The matrix is to stack sentence embeddings in the
order of sentences in the paragraph. The sentence embedding matrix will be
then encoded by a GRU module to generate a state vector, which contains both
historical and future information. With the above design of a hierarchical
contextual encoder, the Conformer encoder broadens the horizon of the current
phoneme to paragraph scope, which will enhance prosody coherence and
expressiveness.

####  2.3.2 Emotion Encoder

Generally, when humans read a passage, their emotional state could change
according to the context. Usually, calm is used for narration content while
dialogue content would convey more emotional states such as joy, anger, and
sorrow according to the context. In order to detect and express the emotion of
dialogue, we adopt a context-aware emotion encoder to capture the emotional
states.

In particular, we fine-tune a pre-trained BERT with a downstream emotion
classification task. To consider the contextual information, we use the
context of the dialogue (add up to 256 tokens), as BERT input and use the
average of token embedding of the current dialogue sentence in context as
sentence embedding (CSE) to build the emotion classification model, which
consists of a dense layer and cross-entropy loss.

Our training data comes from French novel texts found on the web, which we
labeled with an emotion tag using OpenAI’s text-davinci-003 model with
suitable prompts. In the TTS training process, we use the CSE as an implicit
emotion representation, replacing the emotion tag. This is because the
implicit embedding is smoother and includes emotional intensity. After
processing through a linear layer, we combine this emotion embedding with the
output from the Conformer encoder, as depicted in Figure 1. For narration
contents, which have no emotion, the CSE is zero-initialized.

###  2.4 Conformer based Acoustic Model

The Acoustic model we used follows the backbone in Blizzard Challenge 2021,
which is a phone-level duration-based fastspeech2 liked acoustic model but
changes the encoder and decoder module into the latest Conformer module.
Besides, we add some variance information to help handle the one-to-many
problem, which includes speaker ID, narration-or-dialogue ID, phone-level
pitch, utterance-level Global Style Token(GST) [29], and phone-level prosody
embedding. For simplicity, we've consolidated the phone-level pitch, GST, and
phone-level prosody embedding into a variance adaptor[3], as depicted in
Figure1.

The Conformer [9], a type of Transformer, blends elements from convolutional
neural networks (CNNs) and Transformer components. It effectively learns both
global and local interactions through a mix of self-attention and convolution,
a strategy found useful in TTS tasks.

The phone-level pitches are determined by averaging the ground truth frame-
level pitch values, according to the duration of the phone sequences.

For prosody at both utterance and phoneme levels, we employ two reference
encoders to extract values from reference mel during training, and two
separate predictors to estimate values during inference. The utterance-level
prosody vector is derived from a GST-based reference encoder, followed by a
GRU and style token layer. The GST predictor at the utterance level uses a GRU
and a bottleneck layer to predict the prosody vector.

Phoneme-level prosody vectors are derived from a Conformer-based reference
encoder. These vectors are predicted using the phoneme encoder's outputs. The
predictor for phoneme-level prosody, made up of two GRU layers, uses both text
encoder outputs and the utterance-level prosody vector for hierarchical
predictions.

##  3 Task Description, Data Processing, and Training Strategy

###  3.1 Task Description

Blizzard Challenge 2023 has 2 task tracks.

  * •

FH1: 50 hours paired text-speech data with 22.05 kHz sample rate from a female
native speaker of French as hub task.

  * •

FS1: 2 hours paired text-speech data with 22.05 kHz sample rate from a second
female native speaker of French as spoke task.

We build the FH1 TTS only on the given 50 hours of data and utilize a French
multi-speaker source model to adapt to spoke speaker for task FS1. The multi-
speaker source model for French is developed utilizing an internal corpus that
comprises over 100 hours of clear French data. This corpus includes
contributions from 53 speakers, with a balanced representation of genders,
including 25 females and 28 males, all exhibiting a general reading speech
style.

###  3.2 Data Processing

####  3.2.1 Raw data segmentation

The raw data provided by the organizer is divided into audiobook chapters. To
process this data, we use the provided timestamps to divide the lengthy audio
into smaller segments. To maintain the long and continuous context, we
concatenate short sentences using comma and quotation punctuation, ensuring
that the audio length falls within an approximate range of 5 to 20 seconds.
Furthermore, concatenation would be terminated if a sentence encountered an
EOS (End-Of-Sentence) mark or reached the end of a chapter.

####  3.2.2 Narration/Dialogue segmentation

Through a simple study of French dialogue format, we design several rules to
detect narration and dialogue in the text as follows:

  * •

$R_{1}$: Within the paragraph, the content enclosed in quotation marks is
considered as dialogue, but if the content is too short, that might be some
modal particles, etc., which are treated as the narration in our rules.

  * •

$R_{2}$: Paragraphs starting with ¬ and ending until the current paragraph are
considered as dialogue.

####  3.2.3 Post-process for model training

The French input text is converted into a phoneme sequence via sentence
separation, text normalization(TN), IPA-
based444https://www.internationalphoneticalphabet.org Grapheme-to-Phoneme(G2P)
model, and a French lexicon. We down-sample the audio used to 16 kHz to train
the acoustic model. As the speech data contains noise, we utilize a speech
enhancement tool [30] to reduce it, which we find advantageous for both
acoustic model and vocoder training, resulting in cleaner and more natural
synthesized speech. Following this, mel-spectrograms are extracted using a
short-time Fourier transform (STFT) with a frame size of 50 ms, a frame hop of
12.5 ms, and a Hann window function. For vocoder training, we adhere to the
strategy set out in the Blizzard Challenge 2021. We utilize the 24k audios and
16k mel-spectrograms to train the HiFi-GAN [26] using all the training data
mentioned in Section 3.1, and subsequently, we separately fine-tune the model
for each track. Lastly, to match the sample rate of the original recording, we
down-sample the audio from 24kHz to 22.05kHz.

###  3.3 Training Strategy

The loss function used in the MuLanTTS TTS system is shown in Equation 1

$L=L_{gst}+L_{phone}+L_{pitch}+L_{dur}+L_{mel}+L_{ssim}$ (1)

The function is multifaceted and includes several key components. For
instance, $L_{gst}$ represents the L1 loss between the predicted global style
token for an entire utterance and the token extracted from the recording using
a reference encoder. Similarly, $L_{phone}$ calculates the L1 loss between the
predicted phoneme-level prosody vectors and the vectors that are extracted
from the recording using a phoneme-level reference encoder. Furthermore, to
calculate the L1 loss between the predicted and actual values for pitch and
duration, we use $L_{pitch}$ and $L_{dur}$, respectively. The actual pitch
values are obtained using the internal pitch extractor tool, while the HMM
force alignment is employed to extract the actual duration values. $L_{mel}$
is the sum of mel-spectrogram L1 loss between the predicted and ground-truth
mel-spectrogram in each Conformer block. We also follow the Structural
Similarity measure to compare the predicted and ground-truth mel-spectrograms
in the final Conformer block, resulting in $L_{ssim}$.

The Conformer module includes 6 Conformer blocks, with an attention dimension
of 512 and a hidden size of 2048 for the convolutional feed-forward module. We
use the Ranger optimizer for training, with parameters set at $\alpha=6$,
$k=6$, and $\epsilon=1e-5$. The initial learning rate is $1e-3$, with a warm-
up period of 4000 steps and exponential decay. The training process is carried
out on 8 NVIDIA V100 GPUs.

##  4 Results

The Blizzard Challenge 2023 evaluation includes four tests: a naturalness test
using the Mean Opinion Score (MOS) and Multiple Stimuli with Hidden Reference
and Anchor (MUSHRA); a similarity test via the Similarity Mean Opinion Score
(SMOS); and two intelligibility tests, scored through the Word Error Rate
(WER) and a homograph test (HOMOS). In MOS, listeners rate the naturalness of
a single sample from 1 to 5. In SMOS, they rate similarity to a reference
voice from 1 to 5. The intelligibility tests involve listeners typing what
they hear from one-time-only audios. The audios, both synthesized and natural,
are rated by three types of listeners: paid, online volunteers, and speech
experts

###  4.1 Naturalness Test

####  4.1.1 FH1-MOS-MUSHRA

(a) Boxplot of MOS in track FH1.

(b) Boxplot of MUSHRA scores of Top-3 submitted system in track FH1.

Figure 2: MOS and MUSHRA scores for naturalness test in FH1. Our system is
identified as F with the recording (A) and FastSpeech baseline (BF).

Naturalness test evaluation results in FH1 task of all systems are shown in
Figure 2(a) and 2(b). The MOS result of our system shows a better statistical
score distribution than all the other TTS systems similar to recording, and we
have a close MOS compared to recording, 4.3(F) vs 4.4(recording). Although
there is a gap between the MUSHRA score of 0.75(F) and the recording score of
0.94(A), our system is also verified as the leading naturalness compared to
all other synthesized systems.

####  4.1.2 FS1-MOS-MUSHRA

(a) Boxplot of MOS in track FS1.

(b) Boxplot of MUSHRA scores of Top-4 submitted system in track FS1.

Figure 3: MOS and MUSHRA scores for naturalness test in FH1. Our system is
identified as F with the recording (A) and FastSpeech baseline (BF).

The naturalness test evaluation results for the FS1 task for all systems are
depicted in Figure 3(a) and 3(b). Our system achieved a higher MOS than all
other TTS systems among the synthesized systems, and our MOS is comparable to
that of the recording. In fact, our system is even more stable than the
recording, with a score of 4.5(F) versus 4.5(recording) and std 0.70(F) versus
0.77(A). According to mean MUSHRA score 0.76(F) and the recording score of
0.95(A), it also demonstrates our system gains the highest score than other
top submitted TTS systems.

###  4.2 Similarity Test

####  4.2.1 FH1-SMOS

Figure 4 displays the MOS for similarity evaluation results of all systems, as
rated by all listeners. Our System F attains the highest MOS of 3.4, which is
the same as the recording. However, this is not a particularly high similarity
score. We hypothesize that this could be due to the large timbre variance in
the FH1 track, which may be affected by changes in acoustic environments and
channels across different chapters and sections of audiobook data. The
similarity test may yield low scores if the reference audio samples differ
from the training data.

Figure 4: Boxplot of SMOS in track FH1. Our System is identified as F.

####  4.2.2 FS1-SMOS

Figure 5 illustrates the SMOS evaluation results across all systems. Our
System F attains a score of 4.3, which is statistically on par with System Q
and significantly higher than recording A (4.0).

Figure 5: Boxplot of SMOS in track FS1. Our System is identified as F.

###  4.3 Intelligibility of Sentences on SUS

Figure 6: Word error rates for SUS test.

The intelligibility test results for the SUS set, as evaluated by paid
listeners and native speakers, are depicted in Figure 6. The SUS set [31] was
manually generated utilizing specific grammar structures for this test. In
this evaluation, our proposed System F achieves the second-best mean WER
(0.101), slightly trailing System J (0.095).

###  4.4 Intelligibility of Sentences on Homographs

Figure 7: Accuracy for Homographs test.

Figure 7 shows our system F's HOMOS results, achieving nearly 80% accuracy,
which is above average. However, over 60% of the homographs have not been
addressed by our system. We plan to improve our TTS performance on these
homographs for better accuracy in the future.

##  5 Conclusions

This paper introduces Microsoft's neural TTS system, MuLanTTS, developed for
the Blizzard Challenge 2023 to create an audiobook speaker TTS and another
spoke speaker adaptation. Built on DelightfulTTS, it uses contextual and
emotion encoders to boost prosody and dialogue expressiveness. Denoising
algorithms and long audio processing also enhance recording quality.
MuLanTTS's performance, indicated by mean quality scores of 4.3 and 4.5, is
comparable to natural speech and maintains good similarity. The superior
performance of our system in this year's rigorous statistical evaluation
underscores the effectiveness of the proposed system in both tasks. It is also
noticed that the paragraph test samples haven't been evaluated yet in the
evaluation and there are still rooms for complex content synthesis in a long
context and highly expressive dialogue synthesis to improve for future work.

##  6 Acknowledgement

The authors appreciate Huaming Wang, Kun Cong, Hanye Xu, and Min Tang for
their valuable model on speech enhancement, and Gang Wang, Songze Wu, and
Zheng Niu for their support and discussions on French language processing and
model building. At last, we appreciate the organizers' great effort in the
dataset preparation, various evaluations, and detailed statistical analysis.

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