# On the Tong-type identity and the mean square of the error term for an
extended Selberg class

Xiaodong Cao, Yoshio Tanigawa and Wenguang Zhai

###### Abstract

In 1956, Tong established an asymptotic formula for the mean square of the
error term in the summatory function of the Piltz divisor function $d_{3}(n).$
The aim of this paper is to generalize Tong’s method to a class of Dirichlet
series that satisfy a functional equation. As an application, we can establish
the asymptotic formulas for the mean square of the error terms for a class of
functions in the well-known Selberg class. The Tong-type identity and formula
established in this paper can be viewed as an analogue of the well-known
Voronoï’s formula.

0002010 Mathematics Subject Classification:11M41,11N37000Key words and
phrases: Selberg class, Dirichlet series, functional equation, Tong-type
identity, Voronoï’s formula, mean square, error term, arithmetical function,
distribution function, cusp form, Maass form, Dedekind zeta function.000This
work is supported by the National Key Basic Research Program of China(Grant
No. 2013CB834201), the National Natural Science Foundation of China(Grant No.
11171344), the Natural Science Foundation of Beijing(Grant No. 1112010) and
the Fundamental Research Funds for the Central Universities in
China(2012YS01).

###### Contents

  1. 1 Introduction and main results
    1. 1.1 The Selberg class
    2. 1.2 Statements of main results
    3. 1.3 A remark on Tong’s method
    4. 1.4 Organization of this paper
  2. 2 A class of more general arithmetic functions
  3. 3 Some preliminary lemmas
  4. 4 The Tong-type identity for the error term
  5. 5 Large and small values of $E_{\varrho}(y)$ in short intervals
  6. 6 The truncated Tong-type formula for an integral involving the error term
    1. 6.1  Lemmas for generalized Bessel functions
    2. 6.2  Proof of Theorem 7
  7. 7 Some estimates for the weighted integral of the error term
  8. 8 The proof of Theorem 1
    1. 8.1 Tong’s formula of $E(y)$
    2. 8.2 Evaluation of $\int_{T}^{(1+\delta)T}K_{1}^{2}dy$
    3. 8.3 Upper bound of $\int_{T}^{(1+\delta)T}K_{2}^{2}dy$
    4. 8.4 Proof of Theorem 1
    5. 8.5 Proofs of Corollaries 1 and 2
  9. 9 Proofs of Theorem 2 and Theorem 3
  10. 10 Some applications of our main results
    1. 10.1 Some functions of degree $2$
    2. 10.2 Some functions of degree $3$
    3. 10.3 Some examples of degree $4$

##  1 Introduction and main results

Let $a(n)(n\geq 1)$ be a sequence of complex numbers, which is an arithmetic
function. One of the most basic goals of analytic number theory is to
establish the asymptotic formula for the summatory function of $a(n),$ as
accurate as possible. Especially it is important to study the properties of
the so-called error term of this asymptotic formula, such as the upper bound,
the moments, the sign changes and $\Omega$-results, etc.

One of the important tools in this area is Voronoï’s formula of the error term
when the sequence $a(n)(n\geq 1)$ satisfy good conditions, for example, they
are the coefficients of $L$-functions of any degree. When $a(n)\ (n\geq 1)$
are the coefficients of an $L$-function of degree two, Voronoï’s formula of
the corresponding error term is a very strong tool to study the properties of
the error term. A well-known example is the Dirichlet divisor problem. For a
survey of Voronoï’s formula, see for example, Steuding [47].

However, when $a(n)\ (n\geq 1)$ are the coefficients of an $L$-function of
degree $\geq 3$, Voronoï’s formula is not strong enough to get good results
for the properties of the error term. Especially, it fails to give an
asymptotic formula for the mean square of the error term even when the degree
is $3.$

In 1956, Tong [49] established an asymptotic formula for the mean square of
the error term in the summatory function of the Piltz divisor function
$d_{3}(n).$ This is the first result in this area for the case of degree $3$.
Tong’s main ingredient is to replace the error term by a corresponding
integral such that the difference between the error term and this integral is
very small on average.

The aim of this paper is to generalize Tong’s method to a class of Dirichlet
series that satisfy a functional equation. Especially as an application, we
establish the asymptotic formulas for the mean square of the error terms for a
class of functions in the well-known Selberg class.

###  1.1 The Selberg class

To study a summatory function of $a(n)$ and its error term, we consider the
Dirichlet series with coefficients $a(n)$ which satisfies a general functional
equation. There are many results in this direction, see for example,
Chandrasekharan and Narasimhan [3], Redmond [42], Hafner [12, 13], Ivić [20],
Meurman [36], Lau [33], Kanemitsu, Sankaranarayanan and Tanigawa [29],
Friedlander and Iwaniec [9]. We shall study such a problem for $L$-functions
in the so-called Selberg class.

The well-known Selberg class ${\cal S}$ (see for example [26, 44, 46])
consists of non-vanishing Dirichlet series

${\cal L}(s):=\sum_{n=1}^{\infty}\frac{a(n)}{n^{s}},$

which satisfies the following hypotheses:

I. Ramanujan’s conjecture: $a(n)\ll n^{\varepsilon}$ for any $\varepsilon>0$.
II. Analytic continuation: There exists a non-negative integer $m_{\cal L}$
such that $(s-1)^{m_{\cal L}}{\cal L}(s)$ is an entire function of finite
order.
III. Functional equation: ${\cal L}(s)$ satisfies a functional equation of
type

(1.1) $\Lambda_{{\cal L}}(s)=\omega\overline{\Lambda_{{\cal L}}(1-\bar{s})},$

where

(1.2) $\Lambda_{\cal L}(s):={\cal
L}(s)Q^{s}\prod_{j=1}^{L}\Gamma(\alpha_{j}s+\beta_{j}),$

and $Q>0,|\omega|=1$ and $\alpha_{j}>0,\beta_{j}\in\mathbb{C}$ with ${\rm
Re\,}\beta_{j}\geq 0$ for all $1\leq j\leq L$. The number
$d=2\sum_{j}\alpha_{j}$ is called the degree of ${\cal L}(s).$
IV. Euler product: ${\cal L}(s)$ satisfies

${\cal L}(s)=\prod_{p}\exp\left(\sum_{n\geq 1}\frac{b(p^{n})}{p^{ns}}\right)$

with suitable coefficients $b(p^{n})$ satisfying $b(p^{n})\ll p^{nc}$ for some
$c<1/2.$

Many well-known functions are contained in the Selberg class ${\cal S}$. We
recall some examples. The well-known Riemann zeta-function $\zeta(s)$ and
Dirichlet $L$-functions are functions in ${\cal S}$ of degree $1.$ The product
of any $\ell$ functions in ${\cal S}$ of degree $1$ is a function in ${\cal
S}$ of degree $\ell.$ So $\zeta^{2}(s)$ is a function in ${\cal S}$ of degree
$2$ and $\zeta^{3}(s)$ is of degree $3.$ Let $f(z)$ be a holomorphic cusp form
with respect to $SL_{2}({\mathbb{Z}})$. If $f(z)$ is an eigenform of all Hecke
operators, the automorphic $L$-function attached to $f(z)$ is a function in
${\cal S}$ of degree $2$. The Dedekind zeta-function over an algebraic number
field of degree $\kappa\geq 2$ is a function in ${\cal S}$ of degree $\kappa.$

The extended Selberg class ${\cal S}^{\\#}$ (see [26, 27, 28] for an
introduction) consists of all Dirichlet series $\sum_{n\geq 1}a(n)n^{-s}$
which satisfy the conditions $\mathrm{I}^{*}$, II and III, where
$\mathrm{I}^{*}$ means

$\mathrm{I}^{*}$: $\sum_{n\geq 1}a(n)n^{-s}$ is absolutely convergent for
$\sigma>1.$

Consider the sum

$A(y):={\sum_{n\leq y\,}}^{\prime}a(n).$

Define $Q(y)$ as the sum of the residues of the function ${\cal
L}(s)y^{s}s^{-1}$ and the error term $E(y)$ is defined by

$E(y):=A(y)-Q(y).$

Estimates for $E(x)$ and $\int_{0}^{T}|E(y)|^{2}dy$ were first studied by
Voronoï in 1904 and Cramér in 1922 for the special case of the Dirichlet
divisor problem respectively, see [52] and [6]. Since then this has been
generalized for larger classes of Dirichlet series. There are many results in
this direction, see for example, [4, 5, 12, 13, 33, 42, 43].

Roton [43] proved that if $L\in{\cal S}^{\\#}$ is a function of degree $d\geq
2.$ Then

$E(y)\ll y^{(d-1)/(d+1)+\varepsilon},$

which is a generalization of Landau’s classical result [32]. She also proved
that if

$\sum_{n\leq y}|a(n)|^{2}\ll y^{1+\varepsilon},$

then

$\displaystyle\int_{0}^{T}E^{2}(y)dy\ll\left\\{\begin{array}[]{ll}T^{2-1/d},&\mbox{if
$0<d<3,$}\\\ T^{3-4/d+\varepsilon},&\mbox{if $d\geq 3$.}\end{array}\right.$

###  1.2 Statements of main results

Suppose $0\leq\theta<1$ is a real number. Let ${\cal S}^{\theta}$ denote the
set of all Dirichlet series ${\cal L}(s)=\sum_{n\geq 1}a(n)n^{-s}$ which
satisfy II, III and

(1.4) ${\mathrm{I}}^{\prime}:\ \ \ |a(n)|\ll n^{\theta+\varepsilon},\ \ \
\sum_{n\leq y}|a(n)|^{2}\ll y^{1+\varepsilon}.$

In this paper we assume that $\sum_{j=1}^{L}\beta_{j}$ is real for the sake of
simplicity. Obviously ${\cal S}\subset{\cal S}^{\theta}\subset{\cal S}^{\\#}.$
Let ${\cal S}_{real}^{\theta}$ denote the set of all functions in ${\cal
S}^{\theta}$ with real coefficients.

Now suppose $a(n)(n\geq 1)$ is a sequence of real numbers such that its
corresponding Dirichlet series ${\cal L}(s)\in{\cal S}_{real}^{\theta}$ and is
of degree $d\geq 2.$ Without loss of generality, we suppose that $d$ is an
integer. Let $1/2\leq\sigma^{*}<1$ denote a fixed real number such that the
estimate

(1.5) $\int_{0}^{T}|{\cal L}(\sigma^{*}+it)|^{2}dt\ll T^{1+\varepsilon}$

holds for any $\varepsilon>0.$ We suppose that $\sigma^{*}$ satisfies the
condition

(1.6) $\sigma^{*}<(d+1)/2d,$

which plays an important role in our paper.

######  Theorem 1.

Suppose that $d\geq 2$ is a fixed integer, $0\leq\theta\leq 1/2-1/2d$ is a
real number. Suppose that ${\cal L}(s)\in{\cal S}_{real}^{\theta}$ is a
function of degree $d\geq 2$ such that (1.5) and (1.6) hold. Then we have

(1.7)
$\int_{1}^{T}E^{2}(y)dy=C_{d}T^{2-1/d}+O(T^{2-\frac{3-4\sigma^{*}}{2d(1-\sigma^{*})-1}+\varepsilon}),$

where $C_{d}>0$ is a positive constant.

######  Corollary 1.

Suppose $0\leq\theta\leq 1/4$ is a real number and ${\cal L}(s)\in{\cal
S}_{real}^{\theta}$ is of degree $2,$ then we have

(1.8)
$\int_{1}^{T}E^{2}(y)dy=C_{2}T^{3/2}+O_{\varepsilon}(T^{1+\varepsilon}).$

######  Remark 1.

When ${\cal L}(s)\in{\cal S}_{real}^{0}$ is of degree 2, the result obtained
by Voronoï’s formula is sometimes stronger. See for example, Meurman [36] or
Lau and Tsang [34] in the case of Dirichlet divisor problem. We note that the
error term in the asymptotic formula (1.8) is best possible when disregarding
$\varepsilon.$

######  Corollary 2.

Suppose $0\leq\theta\leq 1/3$ is a real number. Let ${\cal L}(s)\in{\cal
S}_{real}^{\theta}$ is a function of degree $3$ such that ${\cal L}(s)={\cal
L}_{1}(s){\cal L}_{2}(s),$ where ${\cal L}_{1}(s)\in{\cal S}_{real}^{\theta}$
is a function of degree $1$, and ${\cal L}_{2}(s)\in{\cal S}_{real}^{\theta}$
is a function of degree $2$. Then we have

(1.9)
$\int_{1}^{T}E^{2}(y)dy=C_{3}T^{5/3}+O_{\varepsilon}(T^{8/5+\varepsilon}).$

Furthermore if we assume that

(1.10) $\int_{0}^{T}|\mathcal{L}_{2}(1/2+it)|^{6}dt\ll T^{2+\varepsilon},$

then we have

(1.11)
$\int_{1}^{T}E^{2}(y)dy=C_{3}T^{5/3}+O_{\varepsilon}(T^{14/9+\varepsilon}).$

######  Remark 2.

The assumption (1.10) for degree $2$ $L$-functions is very natural. The first
result in this direction is the twelfth power moment of the Riemann zeta-
function $\zeta(s)$ (the sixth moment of $\zeta^{2}(s)$ over the critical
line) due to Heath-Brown [16]. Meurmann [38] proved the twelfth power moment
for any Dirichlet L-functions. From Heath-Brown [16] and Meurmann [38] we see
that (1.10) holds for Dedekind zeta functions of quadratic fields. From [25]
we know that (1.10) holds for the automorphic $L$-functions attached to
holomorphic cusp forms.

######  Remark 3.

Tong [49] first established Theorem 1 for the Piltz divisor problem of
dimension $d\geq 3,$ which becomes a true asymptotic formula when $d=3.$
Fomenko [10] followed Tong’s method to study the case of the Dedekind zeta-
function of cubic fields.

######  Theorem 2.

Suppose that $d\geq 2$ is an integer, $0\leq\theta\leq 1/2-1/2d$ is a real
number. Suppose that ${\cal L}(s)\in{\cal S}_{real}^{\theta}$ is a function of
degree $d$ such that (1.5) and (1.6) hold. Then the error term $E(t)$ has a
distribution function $f(\alpha)$ in the sense that, for any interval
$I\subset{\mathbb{R}}$ we have

$T^{-1}mes\\{t\in[1,T]:t^{-(d-1)/2d)}E(t)\in
I\\}\rightarrow\int_{I}f(\alpha)d\alpha$

as $T\rightarrow\infty.$ The function $f(\alpha)$ and its derivatives satisfy

$\frac{d^{k}}{d\alpha^{k}}f(\alpha)\ll_{A,k}(1+|\alpha|)^{-A}$

for $k=0,1,2,\cdots$ and $f(\alpha)$ can be extended to an entire function.

######  Corollary 3.

Suppose $0\leq\theta\leq 1/2-1/2d$ is a real number. Suppose that ${\cal
L}(s)\in{\cal S}_{real}^{\theta}$ is a function of degree $2,$ or is a
function of degree $3$ such that it can be written as a product of a function
of degree $1$ and a function of degree $2$, then Theorem 2 holds.

######  Theorem 3.

Suppose that $d\geq 2$ is an integer, $0\leq\theta\leq 1/2-1/2d$ is a real
number. Suppose that ${\cal L}(s)\in{\cal S}_{real}^{\theta}$ is a function of
degree $d$ such that (1.5) and (1.6) hold. Then for any real number $0\leq
u\leq 2,$ the mean value

$\lim_{T\rightarrow\infty}T^{-1-\frac{(d-1)u}{2d}}\int_{1}^{T}|E(t)|^{u}dt$

converges to a finite limit as $T$ tends to infinity.

######  Corollary 4.

Suppose $0\leq\theta\leq 1/2-1/2d$ is a real number. Suppose that ${\cal
L}(s)\in{\cal S}_{real}^{\theta}$ is a function of degree $2,$ or is a
function of degree $3$ such that it can be written as a product of a function
of degree $1$ and a function of degree $2$, Then Theorem 3 holds.

###  1.3 A remark on Tong’s method

The most important tool up to now to study the behaviour of $E(y)$ is
Voronoï’s formula, which is usually of the form

(1.12) $E(y)=c_{1}y^{\frac{d-1}{2d}}\sum_{n\leq
N}\frac{a(n)}{n^{\frac{d+1}{2d}}}\cos(c_{2}(ny)^{1/d}+c_{3})+O(y^{\frac{d-1}{d}+\varepsilon}N^{-\frac{1}{d}}),$

where $c_{1},c_{2},c_{3}$ are constants and $1\ll N\ll y.$ For a proof of the
above general formula, see for example Friedlander and Iwaniec [9]. Roughly
speaking, this formula gives a good approximation to the error term, which is
usually a finite exponential sum, plus a “permissible small” error.

When $d=2,$ the formula (1.12) is strong enough for us to study the properties
of $E(x).$ For example, its upper bound, power moments, sign changes, etc. A
good example is the Dirichlet divisor problem, see for example, the survey
paper Tsang [50].

However when $d\geq 3,$ the error term in (1.12) is always $\gg
y^{1-2/d+\varepsilon}$, which is much larger than the expected order
$y^{(d-1)/2d}$. So (1.12) can’t be used to study the asymptotic behaviour of
power moments of $E(y),$ even for the mean square.

In [48, 49], Tong developed a method to study the mean square of the error
term in the summatory function of the Piltz divisor function $d_{\ell}(n),$
which denotes the number of ways such that $n$ can be written as a product of
$\ell$ natural numbers. When $\ell=3,$ Tong’s method gives a true asymptotic
formula of the mean square of the error term $\Delta_{3}(y)$. Tong’s main
ingredient is to replace the error term by a corresponding integral such that
the difference between the error term and this integral is very small on
average.

The aim of this paper is to generalize Tong’s method to the general case.
Actually we shall generalize Tong’s method to a class of functions much more
general than ${\cal S}_{real}^{\theta}$. As applications, we establish the
asymptotic formula of the mean square of $E(y)$ for functions in ${\cal
S}_{real}^{\theta}$.

###  1.4 Organization of this paper

The organization of this paper is as follows. In Section 2 we shall introduce
a class of more general functions. In Section 3 we give some preliminary
lemmas about the so-called generalized Bessel functions. In Section 4 we shall
establish a Tong-type identity for the corresponding error term $E(y)$. As
some simple applications of the Tong’s identity, in Section 5 we give a lower
bound for integrals involving $E(y)$ and study the small values of $E(y).$ In
Section 6 we shall establish a truncated Tong-type formula for an integral
involving $E(y).$ In Section 7 we shall estimate some exponential integrals,
which are important in the proof. In Section 8 we shall prove Theorem 1. In
Section 9 we shall prove Theorem 2 and Theorem 3. In Section 10, we give some
examples.

##  2 A class of more general arithmetic functions

For future applications in mind (e.g. [2]), we shall derive a Tong-type
identity of the error term for a more general class of functions than that of
Selberg class. In this section, following basically to Chandrasekharan and
Narasimhan [3, 4] and Hafner [12], we introduce such a class of functions.

Definition.  Let $\\{a(n)\\}$ and $\\{b(n)\\}$ be two sequences of complex
numbers, not identically zero. Let $\\{\lambda_{n}\\}$ and $\\{\mu_{n}\\}$ be
two strictly increasing sequences of positive numbers tending to infinity.
Suppose that the series

$\displaystyle\varphi(s)=\sum_{n=1}^{\infty}a(n)\lambda_{n}^{-s},\quad\psi(s)=\sum_{n=1}^{\infty}b(n)\mu_{n}^{-s}$

converge in some half-plane and have abscissae of absolute convergence
$\sigma_{a}^{*}$ and $\sigma_{b}^{*}$, respectively. Define two gamma factors

(2.1)
$\displaystyle\Delta_{1}(s)=\prod_{j=1}^{N}\Gamma(\alpha_{j}s+\beta_{j})$ and
(2.2)
$\displaystyle\Delta_{2}(s)=\prod_{h=1}^{N^{\prime}}\Gamma(\alpha_{h}^{\prime}s+\beta_{h}^{\prime}),$

where $\alpha_{j}$ and $\alpha_{h}^{\prime}$ are positive real numbers and
$\beta_{j}$ and $\beta_{h}^{\prime}$ are complex numbers. We assume throughout
that

(2.3)
$\alpha:=\sum_{j=1}^{N}\alpha_{j}=\sum_{h=1}^{N^{\prime}}\alpha_{h}^{\prime}.$

Let $r$ be real. We say that $\varphi$ and $\psi$ satisfy the functional
equation

(2.4) $\displaystyle\Delta_{1}(s)\varphi(s)=\Delta_{2}(r-s)\psi(r-s)$

if there exists in the $s$-plane a domain $D$ that is the exterior of a
compact set $\mathfrak{S}$ (we call it a “singularity set”) and on which there
exists a holomorphic function $\chi(s)$ ($s=\sigma+it$, $\sigma$ and $t$ real)
such that

$\displaystyle\texttt{(i)}\quad\lim_{|t|\rightarrow\infty}\chi(\sigma+it)=0$

uniformly in every interval
$-\infty<\sigma_{1}\leq\sigma\leq\sigma_{2}<+\infty$, and

$\displaystyle\texttt{(ii)}\quad\chi(s)=\begin{cases}\Delta_{1}(s)\varphi(s)&\mbox{for
$\sigma>\sigma_{a}^{*}$,}\\\ \Delta_{2}(r-s)\psi(r-s)&\mbox{for
$\sigma<r-\sigma_{b}^{*}$.}\end{cases}.$

Finally suppose that both $\varphi(s)$ and $\psi(s)$ have only a finite number
of poles on the complex plane ${\mathbb{C}}.$

######  Remark 4.

When $\Delta_{1}(s)=\Delta_{2}(s),$ we get the class of functions defined in
Chandrasekharan and Narasimhan [3, 4]. Clearly any function in the extended
Selberg class ${\cal S}^{\\#}$ satisfies the functional equation of the form
(2.4).

######  Remark 5.

There are many other arithmetic functions in the above class but
$\Delta_{1}(s)\not=\Delta_{2}(s).$ One example is the well-known asymmetric
many-dimensional divisor problem. For a survey of the asymmetric many-
dimensional divisor problem, see Krätzel [31] or Ivić etc. [23]. A forthcoming
paper [2] in this direction is in preparation through the approach of this
paper.

From now on we always assume that $a(n),b(n)$ are real. We suppose that there
are infinitely many $n$ such that $|a(n)|\gg 1$ and the same holds for $b(n).$

For $y>0$ and a real number $\varrho$, we define the summatory function
$A_{\varrho}(y)$ of the arithmetical function $a(n)$ by

$\displaystyle
A_{\varrho}(y)=\frac{1}{\Gamma(\varrho+1)}{\sum_{\lambda_{n}\leq
y}}^{\prime}a(n)(y-\lambda_{n})^{\varrho},$

where the symbol $\prime$ indicates that the last term has to be halved if
$\varrho=0$ and $y=\lambda_{n}$. When $\varrho$ is negative, $A_{\varrho}(x)$
is defined only for those positive $y$ not equal to any $\lambda_{n}$.

Let $s_{0}=\sup\\{|s|:s\in\mathfrak{S}\\},$ where $\mathfrak{S}$ is the
”singularity set” in Definition, and
$t_{0}=\max\\{|\frac{\beta_{j}}{\alpha_{j}}|,|\frac{\beta_{h}^{\prime}}{\alpha_{h}^{\prime}}|:j=1,2,\ldots,N,h=1,2,\ldots,N^{\prime}\\}$.
Choose two constants $c>\max\\{\sigma_{a}^{*},\sigma_{b}^{*},s_{0},t_{0}\\}$
and $R>\max\\{s_{0},t_{0}\\}$. Choose the third constant $b\not\in\mathbb{Q}$
such that $r-b$ is not an integer and $b>\max\\{c,r\\}$. We choose
$a\leq\min\\{\sigma_{b}^{*}-\frac{2}{\alpha},\frac{r}{2}-\frac{1}{2\alpha}\\}$.
In fact $a$ will be chosen to be small so that the relevant integral converges
absolutely.

Let $\mathcal{C}$ be the rectangle with vertices $c\pm iR$ and $r-b\pm iR$,
taken in the counter-clockwise direction. For convenience, we shall use
$\mathcal{C}_{u,v}$ to denote the oriented polygonal path with vertices
$u-i\infty,u-iR$, $v-iR,v+iR,u+iR$, and $u+i\infty$ in this order.

We define the residual function or the main term

(2.5) $\displaystyle Q_{\varrho}(y)=\frac{1}{2\pi
i}\int_{\mathcal{C}}\frac{\Gamma(s)\varphi(s)y^{\varrho+s}}{\Gamma(s+\varrho+1)}ds.$

From our choices of $b,c$ and $R$, the path $\mathcal{C}$ encircles all of
$S$. By the residue theorem we have

(2.6) $Q_{\varrho}(y)=\sum_{s_{j}}y^{\varrho+s_{j}}P_{s_{j}}(\log y),$

where $s_{j}$ runs over all the poles of
$\Gamma(s)\varphi(s)y^{\varrho+s}/\Gamma(s+\varrho+1)$ inside $\mathcal{C}$,
$P_{s_{j}}(t)$ is a polynomial of $t$ such that its degree is the order of the
pole $s_{j}$ minus $1$.

We define the error term in the asymptotic formula for $A_{\varrho}(y)$ as

$E_{\varrho}(y)=A_{\varrho}(y)-Q_{\varrho}(y).$

As for $\psi(s)$ we define similarly that

$\displaystyle A_{\varrho}^{*}(y)$
$\displaystyle=\frac{1}{\Gamma(\varrho+1)}{\sum_{\mu_{n}\leq
y}}^{\prime}b(n)(y-\mu_{n})^{\varrho},$ $\displaystyle Q_{\varrho}^{*}(y)$
$\displaystyle=\frac{1}{2\pi
i}\int_{\mathcal{C}}\frac{\Gamma(s)\psi(s)y^{\varrho+s}}{\Gamma(s+\varrho+1)}ds,$
$\displaystyle E_{\varrho}^{*}(y)$
$\displaystyle=A_{\varrho}^{*}(y)-Q_{\varrho}^{*}(y).$

Similarly to (2.6) we have

$Q^{*}_{\varrho}(y)=\sum_{s_{j}}y^{\varrho+s_{j}}P_{s_{j}}^{*}(\log y),$

where $s_{j}$ runs over all the poles of
$\Gamma(s)\psi(s)y^{\varrho+s}/\Gamma(s+\varrho+1)$ inside $\mathcal{C}$,
$P_{s_{j}}^{*}(t)$ is a polynomial such that its degree is the order of the
pole $s_{j}$ minus $1$.

When $\varrho=0,$ for simplicity, we also use the following notation

$\displaystyle A(y)$ $\displaystyle=A_{0}(y),$ $\displaystyle Q(y)$
$\displaystyle=Q_{0}(y),$ $\displaystyle E(y)$ $\displaystyle=E_{0}(y),$
$\displaystyle A^{*}(y)$ $\displaystyle=A_{0}^{*}(y),$ $\displaystyle
Q^{*}(y)$ $\displaystyle=Q_{0}^{*}(y),$ $\displaystyle E^{*}(y)$
$\displaystyle=E_{0}^{*}(y).$

We suppose that $\sigma^{*}<\sigma_{b}^{*}$ is a real number such that the
estimate

(2.7) $\displaystyle\int_{-T}^{T}|\psi(\sigma^{*}+it)|^{2}dt\ll
T^{1+\varepsilon}$

holds. Since there are infinitely many $n$ such that $b(n)\gg 1,$ we have
$\sigma^{*}\geq 0.$ We also suppose that $\psi(s)$ doesn’t have any poles in
the region $\sigma\leq\sigma^{*}.$

##  3 Some preliminary lemmas

We shall prove a fundamental lemma of the asymptotic expansion of the integral
which we need later. First we recall the well-known Stirling’s formula for
gamma function.

######  Lemma 1 (Stirling’s formula).

Let $c$ be a constant. Then there exist some constants $\gamma_{j}$ and
$\gamma_{j}^{\prime}$ depending on $c$ such that for any positive integer $m$,

(3.1) $\displaystyle\log\Gamma(s+c)=\left(s+c-\frac{1}{2}\right)\log
s-s+\frac{1}{2}\log
2\pi+\sum_{j=1}^{m}\frac{\gamma_{j}}{s^{j}}+O\left(\frac{1}{|s|^{m+1}}\right)$

and

(3.2) $\displaystyle\Gamma(s+c)=\sqrt{2\pi}e^{\left(s+c-\frac{1}{2}\right)\log
s-s}\left(1+\sum_{j=1}^{m}\frac{\gamma_{j}^{\prime}}{s^{j}}+O\left(\frac{1}{|s|^{m+1}}\right)\right)$

uniformly for $|\arg s|<\pi-\delta$ for fixed $\delta>0$, as $|s|\to\infty$.

The constants $\gamma_{j}$ in (3.1) are given explicitly by

$\gamma_{j}=\frac{(-1)^{j-1}B_{j+1}(c)}{j(j+1)},$

where $B_{j}(x)$ is the Bernoulli polynomials of degree $j$. See for example,
Wang and Guo [54, p. 123 (5)] and see also A. Erdélyi et al. [8].

For two gamma factors $\Delta_{1}(s)$ and $\Delta_{2}(s)$ defined by (2.1) and
(2.2), respectively, we define

(3.3) $\displaystyle\mu$
$\displaystyle=\sum_{j=1}^{N}\left(\beta_{j}-\frac{1}{2}\right)+\frac{1}{2},$
$\displaystyle\mu^{\prime}$
$\displaystyle=\sum_{h=1}^{N^{\prime}}\left(\beta_{h}^{\prime}-\frac{1}{2}\right)+\frac{1}{2},$
(3.4) $\displaystyle\nu$
$\displaystyle=\sum_{j=1}^{N}\left(\beta_{j}-\frac{1}{2}\right)\log\alpha_{j},$
$\displaystyle\nu^{\prime}$
$\displaystyle=\sum_{h=1}^{N^{\prime}}\left(\beta_{h}^{\prime}-\frac{1}{2}\right)\log\alpha_{h}^{\prime},$
(3.5) $\displaystyle\tau$
$\displaystyle=\sum_{j=1}^{N}\alpha_{j}\log\alpha_{j},$
$\displaystyle\tau^{\prime}$
$\displaystyle=\sum_{h=1}^{N^{\prime}}\alpha_{h}^{\prime}\log\alpha_{h}^{\prime}.$

Furthermore we define

(3.6)
$\theta_{\varrho}=\frac{r}{2}-\frac{1}{4\alpha}+\varrho\left(1-\frac{1}{2\alpha}\right)+\frac{\mu^{\prime}-\mu}{2\alpha}$

and

$h=2\alpha e^{-\frac{\tau+\tau^{\prime}}{2\alpha}}.$

Let $\mathcal{C}_{a,b}$ be a curve defined in the previous section such that
all poles of $\Delta_{2}(s)$ lie in the left hand side of $\mathcal{C}_{a,b}$.
If $\beta_{j}$ and $\beta_{h}^{\prime}$ are real, this condition means that
$b$ is greater than the maximal pole of $\Delta_{2}(s)$.

######  Lemma 2.

We assume that

(3.7)
$\overline{\Delta_{1}(s)}=\Delta_{1}(\bar{s})\quad\mbox{and}\quad\overline{\Delta_{2}(s)}=\Delta_{2}(\bar{s}).$

Let $\omega$ be a real number and $M$ an integer. Let $a$ be a real number
such that

(3.8) $a<\frac{r}{2}-\frac{\mu^{\prime}-\mu+\omega+M+1}{2\alpha}.$

Let $\mathcal{D}$ be a domain such that

$\mathcal{D}=\mathbb{C}\backslash\\{s\in\mathbb{C}\,|\,{\rm
Re\,}s<b^{\prime},|{\rm Im\,}s|<R^{\prime}\\}$

with some constant $b^{\prime}$ and $R^{\prime}$. We suppose that
$\mathcal{D}\supset\mathcal{C}_{a,b}$, and furthermore if $\omega\neq 0$ we
assume that $b^{\prime}>0$. Suppose that a function $F(s)$ is regular in the
domain $\mathcal{D}$ and has an asymptotic expansion

(3.9) $F(s)\sim s^{M}\sum_{n=0}^{\infty}\frac{d_{n}}{s^{n}}\ \ \ (d_{0}\neq
0)$

with real coefficients $d_{n}$ as $|s|\to\infty$. Let $\mathcal{I}(x)$ be the
function defined by

(3.10)
$\displaystyle\mathcal{I}(x):=\int_{\mathcal{C}_{a,b}}\frac{\Delta_{2}(s)}{\Delta_{1}(r-s)}s^{\omega}F(s)x^{-s}ds.$

Then for any positive integer $m$ we have an asymptotic expansion

$\displaystyle\mathcal{I}(x)$
$\displaystyle=2iC\sum_{k=0}^{m}c_{k}x^{\frac{\tilde{M}-k+\frac{1}{2}}{2\alpha}}\cos\left(hx^{\frac{1}{2\alpha}}-a_{k}\pi\right)+O\left(x^{\frac{\tilde{M}-m-\frac{1}{2}}{2\alpha}}\right),$

where

(3.11) $\displaystyle\tilde{M}$ $\displaystyle=\mu^{\prime}-\mu-\alpha
r+\omega+M,$ $\displaystyle a_{k}$
$\displaystyle=\frac{1}{2}\left(M+\omega-k+\alpha
r+\mu+\mu^{\prime}-\frac{1}{2}\right),$ $\displaystyle C$
$\displaystyle=(2\pi)^{\frac{N^{\prime}-N}{2}}e^{\nu^{\prime}-\nu-\tau
r-\frac{(\tau+\tau^{\prime})(\tilde{M}+\frac{1}{2})}{2\alpha}}$

and $c_{k}$ are real constants which do not depend on $x$. In particular

$c_{0}=\frac{\sqrt{\pi}}{\sqrt{\alpha}}d_{0}$

and

$c_{1}=\frac{\sqrt{\pi}}{\sqrt{\alpha}}\left\\{d_{0}\left(-\frac{1}{2\alpha}\left(\frac{\tilde{M}^{2}}{2}-\frac{1}{24}\right)+\frac{1}{2}\left(\sum_{h=1}^{N^{\prime}}\frac{B_{2}(\beta_{h}^{\prime})}{\alpha_{h}^{\prime}}+\sum_{j=1}^{N}\frac{B_{2}(\alpha_{j}r+\beta_{j})}{\alpha_{j}}\right)\right)+d_{1}\right\\},$

where $B_{2}(x)$ is the Bernoulli polynomial of degree 2.

######  Remark 6.

This lemma is a generalization of Theorem 3 in Tong [48]. But Tong omitted the
proof of his Theorem 3.

###### Proof.

We note first that the integral (3.10) is convergent absolutely under the
condition (3.8).

We derive an asymptotic expansion of the integrand of (3.10). Here for
simplicity we use the symbol $\sim$ to denote that the right hand side of
$\sim$ is an asymptotic expansion of a function in the left hand side.

Consider $\Delta_{2}(s)$ first. By (3.2), there exist constants
$b_{h}^{\prime}$ such that

$\displaystyle\Delta_{2}(s)=\prod_{h=1}^{N^{\prime}}\Gamma(\alpha_{h}^{\prime}s+\beta_{h}^{\prime})\sim(2\pi)^{N^{\prime}/2}\exp(g_{2}(s))\left(1+\frac{b_{1}^{\prime}}{s}+\frac{b_{2}^{\prime}}{s^{2}}+\cdots\right),$

where $g_{2}(s)$ is given by

$\displaystyle g_{2}(s)$
$\displaystyle=\sum_{h=1}^{N^{\prime}}\left\\{\left(\alpha_{h}^{\prime}s+\beta_{h}^{\prime}-\frac{1}{2}\right)(\log
s+\log\alpha_{h}^{\prime})-\alpha_{h}^{\prime}s\right\\}$
$\displaystyle=\left(\alpha s+\mu^{\prime}-\frac{1}{2}\right)\log
s+(\tau^{\prime}-\alpha)s+\nu^{\prime}.$

Similarly we have, with some constants $b_{j}^{\prime\prime}$,

$\displaystyle\Delta_{1}(r-s)$
$\displaystyle=\prod_{j}\Gamma(-\alpha_{j}s+\alpha_{j}r+\beta_{j})$
$\displaystyle\sim(2\pi)^{N/2}\exp(g_{1}(s))\left(1+\frac{b_{1}^{\prime\prime}}{s}+\frac{b_{2}^{\prime\prime}}{s^{2}}+\cdots\right),$

where $g_{1}(s)$ is given by

$\displaystyle g_{1}(s)$
$\displaystyle=\sum_{j=1}^{N}\left\\{\left(-\alpha_{j}s+\alpha_{j}r+\beta_{j}-\frac{1}{2}\right)(\log(-s)+\log\alpha_{j})+\alpha_{j}s\right\\}$
$\displaystyle=\left(-\alpha s+\alpha
r+\mu-\frac{1}{2}\right)\log(-s)-(\tau-\alpha)s+\tau r+\nu.$

In particular

$b_{1}^{\prime}=\frac{1}{2}\sum_{h=1}^{N^{\prime}}\frac{B_{2}(\beta_{h}^{\prime})}{\alpha_{h}^{\prime}}\qquad\mbox{and}\qquad
b_{1}^{\prime\prime}=-\frac{1}{2}\sum_{j=1}^{N}\frac{B_{2}(\alpha_{j}r+\beta_{j})}{\alpha_{j}}.$

By using

$\log(-s)=\log s-\pi i\,{\rm sgn}(t)$

for non-zero and non-negative $s$, where $t={\rm Im\,}s$, we can see easily
that

$\displaystyle g(s):$ $\displaystyle=g_{2}(s)-g_{1}(s)$
$\displaystyle=(2\alpha s+\mu^{\prime}-\mu-\alpha r)\log s-\pi i\alpha s\,{\rm
sgn}(t)$ $\displaystyle\quad+\pi i\left(\alpha r+\mu-\frac{1}{2}\right){\rm
sgn}(t)+(\tau+\tau^{\prime}-2\alpha)s+\nu^{\prime}-\nu-\tau r.$

Combing these formulas we find that with some constants $p_{n}$,

$\displaystyle\frac{\Delta_{2}(s)}{\Delta_{1}(r-s)}s^{\omega}F(s)x^{-s}$
$\displaystyle\sim(2\pi)^{(N^{\prime}-N)/2}\exp(g(s))e^{-s\log
x}\left(1+\frac{b_{1}^{\prime}}{s}+\cdots\right)\left(1+\frac{b_{1}^{\prime\prime}}{s}+\cdots\right)^{-1}$
$\displaystyle\qquad\times
s^{\omega+M}\left(d_{0}+\frac{d_{1}}{s}+\cdots\right)$
$\displaystyle=(2\pi)^{(N^{\prime}-N)/2}\exp(g(s))e^{-s\log
x}s^{\omega+M}\sum_{n=0}^{\infty}\frac{p_{n}}{s^{n}}.$

By (3.7) and the assumption that $d_{n}$ are real, we have

$\overline{\int_{\mathcal{C}_{a,b}\cap\\{s|t>0\\}}\frac{\Delta_{2}(s)}{\Delta_{1}(r-s)}s^{\omega}F(s)x^{-s}ds}=-\int_{\mathcal{C}_{a,b}\cap\\{s|t<0\\}}\frac{\Delta_{2}(s)}{\Delta_{1}(r-s)}s^{\omega}F(s)x^{-s}ds,$

hence in order to evaluate $\mathcal{I}(x)$ it is enough to consider the case
$t>0$ and take the imaginary part. So we suppose that $t>0$. Let

$f(s)=2\alpha s\log s+(\tau+\tau^{\prime}-2\alpha)s-\pi i\alpha s-s\log x.$

Then

(3.12)
$\displaystyle\frac{\Delta_{2}(s)}{\Delta_{1}(r-s)}s^{\omega}F(s)x^{-s}\sim$
$\displaystyle C_{0}e^{\pi i(\alpha r+\mu-1/2)}s^{\mu^{\prime}-\mu-\alpha
r+\omega+M}\exp(f(s))\sum_{n=0}^{\infty}\frac{p_{n}}{s^{n}}$

where we put

$C_{0}=(2\pi)^{(N^{\prime}-N)/2}e^{\nu^{\prime}-\nu-\tau r}.$

Now let

$y:=xe^{-(\tau+\tau^{\prime})}.$

We define the new parameters $w$, $\xi$ and $\eta$ by

$\displaystyle s=y^{\frac{1}{2\alpha}}w$ $\displaystyle w=i(1+\xi),\ \
\mbox{$|\xi|$ small}$ $\displaystyle\eta^{2}=-2\alpha
i\left((1+\xi)\log(1+\xi)-\xi\right).$

The branch of $\eta$ will be taken as

$\xi=\xi(\eta)=\frac{i^{1/2}\eta}{\sqrt{\alpha}}+\frac{1}{6}\left(\frac{i^{1/2}\eta}{\sqrt{\alpha}}\right)^{2}-\frac{1}{72}\left(\frac{i^{1/2}\eta}{\sqrt{\alpha}}\right)^{3}+\cdots.$

Let $\delta$ be a small constant. We put

$w_{1}=u_{1}+iv_{1}=i(1+\xi(-\delta))\ \ w_{2}=u_{2}+iv_{2}=i(1+\xi(\delta)).$

Define the paths of integration by

$\displaystyle\mathcal{L}_{0}$ $\displaystyle=\left\\{s\ |\
s=y^{\frac{1}{2\alpha}}i(1+\xi(\eta)),-\delta\leq\eta\leq\delta\right\\}$
$\displaystyle\mathcal{L}_{1}$ $\displaystyle=\left\\{s\ |\
s=y^{\frac{1}{2\alpha}}(u_{1}+iv),0\leq v\leq v_{1}\right\\}$
$\displaystyle\mathcal{L}_{2}$ $\displaystyle=\left\\{s\ |\
s=y^{\frac{1}{2\alpha}}(u_{2}+iv),v\geq v_{2}\right\\}.$

By Cauchy’s theorem and the remark above,

$\mathcal{I}(x)=2i\sum_{j=0}^{2}{\rm
Im\,}\int_{\mathcal{L}_{j}}\frac{\Delta_{2}(s)}{\Delta_{1}(r-s)}s^{\omega}F(s)x^{-s}ds.$

First we consider the integral

$J_{0}=\int_{\mathcal{L}_{0}}\frac{\Delta_{2}(s)}{\Delta_{1}(r-s)}s^{\omega}F(s)x^{-s}ds.$

From the above choice of parameters, we have

$\displaystyle f(s)$ $\displaystyle=2\alpha s\log s-2\alpha s-\pi i\alpha
s-s\log y$ $\displaystyle=2\alpha
y^{\frac{1}{2\alpha}}i(1+\xi)\Bigl{(}\log(1+\xi)-1\Bigr{)}$
$\displaystyle=-2\alpha y^{\frac{1}{2\alpha}}i+2\alpha
y^{\frac{1}{2\alpha}}i\Bigl{(}(1+\xi)\log(1+\xi)-\xi\Bigr{)}.$

Therefore we have

$\displaystyle\frac{\Delta_{2}(s)}{\Delta_{1}(r-s)}s^{\omega}F(s)x^{-s}=C_{0}e^{-2\alpha
iy^{\frac{1}{2\alpha}}}e^{\pi i(\alpha
r+\mu-1/2)}\exp(-y^{\frac{1}{2\alpha}}\eta^{2})\sum_{n=0}^{\infty}p_{n}s^{\tilde{M}-n},$

where $\tilde{M}$ is defined by (3.11). In terms of the parameter $\eta$, we
have

$\displaystyle\sum_{n=0}^{\infty}p_{n}s^{\tilde{M}-n}=\sum_{n=0}^{\infty}p_{n}\left(iy^{\frac{1}{2\alpha}}(1+\xi)\right)^{\tilde{M}-n}$
$\displaystyle=\sum_{n=0}^{\infty}p_{n}(iy^{\frac{1}{2\alpha}})^{\tilde{M}-n}\left(1+(\tilde{M}-n)\xi+\frac{(\tilde{M}-n)(\tilde{M}-n-1)}{2}\xi^{2}+\cdots\right)$
$\displaystyle=\sum_{n=0}^{\infty}p_{n}(iy^{\frac{1}{2\alpha}})^{\tilde{M}-n}\left(1+(\tilde{M}-n)\frac{i^{\frac{1}{2}}\eta}{\sqrt{\alpha}}+\frac{(\tilde{M}-n)(\tilde{M}-n-\frac{2}{3})}{2}\left(\frac{i^{\frac{1}{2}}\eta}{\sqrt{\alpha}}\right)^{2}+\cdots\right)$

and

$ds=iy^{\frac{1}{2\alpha}}d\xi=iy^{\frac{1}{2\alpha}}\frac{i^{\frac{1}{2}}}{\sqrt{\alpha}}\left(1+\frac{i^{\frac{1}{2}}\eta}{3\sqrt{\alpha}}-\frac{i\eta^{2}}{24\alpha}+\cdots\right)d\eta,$

hence

$\displaystyle\left(\sum_{n=0}^{\infty}p_{n}s^{\tilde{M}-n}\right)ds=\sum_{n=0}^{\infty}p_{n}(iy^{\frac{1}{2\alpha}})^{\tilde{M}-n+1}i^{\frac{1}{2}}\sum_{j=0}^{\infty}q_{n,j}(i^{\frac{1}{2}}\eta)^{j}d\eta$

with some real constants $q_{n,j}$.

Now take the integral of $\eta$ over $[-\delta,\delta]$. The terms of odd
powers of $\eta$ vanish. By using the well known formula

$\int_{-\delta}^{\delta}e^{-a\eta^{2}}\eta^{2h}d\eta=a^{-h-1/2}\Gamma(h+1/2)+O_{\delta,h}\left(\frac{e^{-a\delta^{2}}}{a}\right)\
\ a>0,$

we find that

$\displaystyle\int_{\mathcal{L}_{0}}\left(\exp\left(-y^{\frac{1}{2\alpha}}\eta^{2}\right)\sum_{n=0}^{\infty}p_{n}s^{\tilde{M}-n}\right)ds$
$\displaystyle\sim\sum_{n=0}^{\infty}\sum_{h=0}^{\infty}p_{n}q_{n,2h}\left(iy^{\frac{1}{2\alpha}}\right)^{\tilde{M}-n+1}i^{\frac{1}{2}+h}\int_{-\delta}^{\delta}e^{-y^{\frac{1}{2\alpha}}\eta^{2}}\eta^{2h}d\eta$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{h=0}^{\infty}p_{n}q_{n,2h}\left(iy^{\frac{1}{2\alpha}}\right)^{\tilde{M}-n+1}i^{\frac{1}{2}+h}\left(y^{-\frac{h+\frac{1}{2}}{2\alpha}}\Gamma(h+1/2)+O\left(\frac{e^{-y^{\frac{1}{2\alpha}}\delta^{2}}}{y^{\frac{1}{2\alpha}}}\right)\right)$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{h=0}^{\infty}p_{n}q_{n,2h}\Gamma(h+1/2)i^{\tilde{M}-n+\frac{3}{2}+h}y^{\frac{\tilde{M}-n-h+\frac{1}{2}}{2\alpha}}+O\left(y^{\frac{\tilde{M}}{2\alpha}}e^{-y^{\frac{1}{2\alpha}}\delta^{2}}\right).$

Hence for any $m>0$, we get

$J_{0}=C_{0}e^{-2\alpha iy^{\frac{1}{2\alpha}}}e^{\pi i(\alpha
r+\mu-\frac{1}{2})}\sum_{k=0}^{m}c_{k}^{\prime}i^{\tilde{M}-k+\frac{3}{2}}y^{\frac{\tilde{M}-k+\frac{1}{2}}{2\alpha}}+O\left(y^{\frac{\tilde{M}-m-1/2}{2\alpha}}\right),$

where we put

$c_{k}^{\prime}=\sum_{n+h=k}p_{n}q_{n,2h}\Gamma(h+1/2)(-1)^{h}.$

Next we consider the integral on $\mathcal{L}_{2}$. Let

$J_{2}=\int_{\mathcal{L}_{2}}\frac{\Delta_{2}(s)}{\Delta_{1}(r-s)}s^{\omega}F(s)x^{-s}ds.$

From the definition of $\mathcal{L}_{2}$, we have

$\displaystyle|J_{2}|$
$\displaystyle\leq\int_{v_{2}}^{\infty}\left|\frac{\Delta_{2}(s)}{\Delta_{1}(r-s)}s^{\omega}F(s)x^{-s}\right|y^{\frac{1}{2\alpha}}dv$

with $s=y^{\frac{1}{2\alpha}}w$, $(w=u_{2}+iv)$. By using (3.12), the above
integrand is bounded as

$\ll x^{-u_{2}}y^{\frac{\tilde{M}+1}{2\alpha}}|w|^{\tilde{M}}|\exp(f(s))|.$

Since

$\displaystyle f(y^{\frac{1}{2\alpha}}w)$ $\displaystyle=2\alpha
y^{\frac{1}{2\alpha}}w\left(\frac{1}{2\alpha}\log y+\log w\right)-2\alpha
y^{\frac{1}{2\alpha}}w-\pi i\alpha
y^{\frac{1}{2\alpha}}w-y^{\frac{1}{2\alpha}}w\log y$ $\displaystyle=2\alpha
y^{\frac{1}{2\alpha}}\left(w\log w-w-\frac{\pi i}{2}w\right),$

we find that

$\displaystyle|J_{2}|$ $\displaystyle\ll
x^{-u_{2}}y^{\frac{\tilde{M}+1}{2\alpha}}\max_{v_{2}\leq
v<\infty}G(v)\int_{v_{2}}^{\infty}|w|^{-2}dv,$

where we put

$\displaystyle G(v)=v^{\tilde{M}+2}\exp\left(2\alpha y^{\frac{1}{2\alpha}}{\rm
Re\,}\left(w\log w-w-\frac{\pi i}{2}w\right)\right).$

We shall see that $G(v)$ is a decreasing function. Differentiate $G(v)$ with
respect to $v$, then we have

$\displaystyle\frac{d}{dv}G(v)=$ $\displaystyle
v^{\tilde{M}+1}\exp\left(2\alpha y^{\frac{1}{2\alpha}}{\rm Re\,}\left(w\log
w-w-\frac{\pi i}{2}w\right)\right)$
$\displaystyle\times\left(\tilde{M}+2-2\alpha y^{\frac{1}{2\alpha}}v\left(\arg
w-\frac{\pi}{2}\right)\right).$

Note that

$\lim_{v\to\infty}v\left(\arg w-\frac{\pi}{2}\right)=-u_{2}>0.$

Therefore for sufficiently large $x$ (hence for sufficiently large $y$),
$G(v)$ is a decreasing function. Thus $G(v)$ attains its maximal value at
$v=v_{2}$, namely, we have

$\displaystyle|J_{2}|\ll$ $\displaystyle
x^{-u_{2}}y^{\frac{\tilde{M}+1}{2\alpha}}\exp\left(2\alpha
y^{\frac{1}{2\alpha}}{\rm Re\,}\left(w_{2}\left(\log w_{2}-1-\frac{\pi
i}{2}\right)\right)\right).$

Let

$\delta_{2}=\arg w_{2}>\frac{\pi}{2}.$

Then

${\rm Re\,}\left(w_{2}\left(\log w_{2}-1-\frac{\pi
i}{2}\right)\right)=u_{2}(\log|w_{2}|-1)-v_{2}\left(\delta_{2}-\frac{\pi}{2}\right),$

which is a constant smaller than 0 (since $\delta$ is also small).
Consequently $J_{2}$ is exponentially decayed as $x\to\infty$.

Similarly it is shown that $J_{1}$ is also exponentially decayed as
$x\to\infty$.

Collecting these formulas, we get

$\displaystyle J:$
$\displaystyle=\int_{\mathcal{L}_{0}+\mathcal{L}_{1}+\mathcal{L}_{2}}\frac{\Delta_{2}(s)}{\Delta_{1}(r-s)}s^{\omega}F(s)x^{-s}ds$
$\displaystyle=C_{0}e^{-2\alpha iy^{\frac{1}{2\alpha}}}e^{\pi i(\alpha
r+\mu-\frac{1}{2})}\sum_{k=0}^{m}c_{k}^{\prime}i^{\tilde{M}-k+\frac{3}{2}}y^{\frac{\tilde{M}-k+\frac{1}{2}}{2\alpha}}+O\left(y^{\frac{\tilde{M}-m-1/2}{2\alpha}}\right)$

for any $m$. Noting that $\mathcal{I}(x)=J-\bar{J}$ and
$y=xe^{-(\tau+\tau^{\prime})}$, we conclude that

$\displaystyle\mathcal{I}(x)=2iC_{0}\sum_{k=0}^{m}c_{k}^{\prime}y^{\frac{\tilde{M}-k+\frac{1}{2}}{2\alpha}}\sin\left(-2\alpha
y^{\frac{1}{2\alpha}}+\left(\tilde{M}-k+2\alpha
r+2\mu+\frac{1}{2}\right)\frac{\pi}{2}\right)$
$\displaystyle\quad+O\left(y^{\frac{\tilde{M}-m-\frac{1}{2}}{2\alpha}}\right)$
$\displaystyle=2iC_{0}e^{-\frac{(\tau+\tau^{\prime})(\tilde{M}+\frac{1}{2})}{2\alpha}}\sum_{k=0}^{m}c_{k}x^{\frac{\tilde{M}-k+\frac{1}{2}}{2\alpha}}\cos\left(hx^{\frac{1}{2\alpha}}-\left(M+\omega-k+\alpha
r+\mu^{\prime}+\mu-\frac{1}{2}\right)\frac{\pi}{2}\right)$
$\displaystyle\quad+O\left(x^{\frac{\tilde{M}-m-\frac{1}{2}}{2\alpha}}\right),$

where we write

$c_{k}=c_{k}^{\prime}e^{\frac{(\tau+\tau^{\prime})}{2\alpha}k}.$

This completes the proof of Lemma 2. ∎

######  Lemma 3.

We assume (3.7) in Lemma 2. Let $F(s)$ be a function which satisfies the same
assumptions in Lemma 2. Define the function $\mathcal{J}(x)$ by

$\mathcal{J}(x)=\frac{1}{2\pi
i}\int_{\mathcal{C}_{a,b}}\frac{\Gamma(r-s)\Delta_{2}(s)}{\Gamma(r+\rho-s)\Delta_{1}(r-s)}F(s)x^{-s}ds,$

with real $\varrho$. Then there exist constants $a_{l}$ and $c_{l}$ such that
for any positive integer $m$, we have

(3.13) $\displaystyle\mathcal{J}(x)=$
$\displaystyle\sum_{l=0}^{m}a_{l}x^{\frac{\tilde{M}-l+1/2}{2\alpha}}\cos(hx^{\frac{1}{2\alpha}}+c_{l}\pi)+O(x^{\frac{\tilde{M}-m-1/2}{2\alpha}})+O(x^{-b}),$

where

$\tilde{M}=\mu^{\prime}-\mu-\alpha r-\varrho+M.$

If we take $b$ large, we have

(3.14) $\mathcal{J}(x)\ll x^{\frac{\mu^{\prime}-\mu-\alpha
r-\varrho+M+1/2}{2\alpha}}.$

###### Proof.

The right hand side except the last error term in (3.13) are obtained by
putting $\omega=-\rho$ in Lemma 2. When we deform the path of integration
$\mathcal{C}_{a,b}$ to the path
$\mathcal{L}_{0}\cup\mathcal{L}_{1}\cup\mathcal{L}_{2}$ and its complex
conjugate, it may pass across the poles of $\Gamma(r-s)$ greater than $b$, but
they are finite depending on $m$. Hence the contribution from these poles is
$O(x^{-b})$. The assertion (3.14) is obtained by taking $m=0$ in (3.13). ∎

We note that if $F(s)=P(s)/Q(s)$ is a rational function of $s$, we can take
$M$ as $M=\deg P-\deg Q$.

######  Lemma 4.

Let $x>0$ and $\varrho>\min\\{2\alpha a-r\alpha+\mu^{\prime}-\mu,-1\\}$.
Define the function $f_{\varrho}(x)$ by

$f_{\varrho}(x)=\frac{1}{2\pi
i}\int_{\mathcal{C}_{a,b}}G_{\varrho}(s)x^{r+\varrho-s}ds$

where

$\displaystyle
G_{\varrho}(s)=\frac{\Gamma(r-s)\Delta_{2}(s)}{\Gamma(r+\varrho+1-s)\Delta_{1}(r-s)}.$

Then

$\displaystyle\frac{d}{dx}f_{\varrho+1}(x)=f_{\varrho}(x)$

and for any non-negative integer $m$, we have an asymptotic expansion

(3.15) $\displaystyle f_{\varrho}(x)=$
$\displaystyle\sum_{l=0}^{m}\kappa_{l}x^{\theta_{\varrho}-\frac{l}{2\alpha}}\cos\left(hx^{\frac{1}{2\alpha}}+c_{l}\pi\right)+O\left(x^{\theta_{\varrho}-\frac{m+1}{2\alpha}}+x^{r+\varrho-b}\right),$

where

$\displaystyle
c_{l}=-\frac{1}{2}\left(\mu+\mu^{\prime}+r\alpha+\varrho+\frac{1}{2}\right)+\frac{l}{2}$

and $\kappa_{l}$ are constants. In particular,

$\kappa_{0}=(2\pi)^{\frac{N^{\prime}-N}{2}}\left(\frac{2\alpha}{h}\right)^{\varrho}\sqrt{\frac{2}{h\pi}}\,e^{\nu^{\prime}-\nu-\frac{(\tau^{\prime}+\tau)(\mu^{\prime}-\mu)}{2\alpha}+\frac{r}{2}(\tau^{\prime}-\tau)},$

and

$\kappa_{1}=\kappa_{0}\frac{2\alpha}{h}\left(-\frac{1}{2\alpha}\left(\frac{(\mu^{\prime}-\mu-\alpha
r-\varrho-1)^{2}}{2}-\frac{1}{24}\right)+B\right),$

where we put

$B=\frac{1}{2}\left(\sum_{h=1}^{N^{\prime}}\frac{B_{2}(\beta_{h}^{\prime})}{\alpha_{h}^{\prime}}+\sum_{j=1}^{N}\frac{B_{2}(\alpha_{j}r+\beta_{j})}{\\\
a_{j}}-B_{2}(r)+B_{2}(r+\varrho+1)\right).$

###### Proof.

(3.15) is obtained from Lemma 3. The other assertions are calculated
explicitly by using Lemma 2. ∎

##  4 The Tong-type identity for the error term

In this section we shall prove the Tong-type identity for the integral (or
multiple integral) of the error term $E_{\varrho}(y)$, which is a
generalization of Theorem 1 in Tong [48] for the error term in the Piltz
divisor problem. The main result of this section is

######  Theorem 4.

Let $x\geq 1$, $L\geq 0$. Suppose that

$\varrho+k>2\alpha\sigma_{b}^{*}-\alpha r-\frac{1}{2}+\mu^{\prime}-\mu.$

Then for $x+ky>0$ we have

$\displaystyle\int_{0}^{y}\cdots\int_{0}^{y}E_{\varrho}(x+y_{1}+\cdots+y_{k})dy_{1}\cdots
dy_{k}$ $\displaystyle=\sum_{\mu_{n}\leq
L}\frac{b(n)}{\mu_{n}^{r+\varrho}}\int_{0}^{y}\cdots\int_{0}^{y}f_{\varrho}\left((x+y_{1}+\cdots+y_{k})\mu_{n}\right)dy_{1}\cdots
dy_{k}$
$\displaystyle\quad+\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}\sum_{\mu_{n}>L}\frac{b(n)}{\mu_{n}^{r+\varrho+k}}f_{\varrho+k}\left((x+jy)\mu_{n}\right).$

######  Remark 7.

Cao [1] and Fomenko [11] used Tong’s idea in [48] to study the asymmetric
many-dimensional problem and mean value theorems for automorphic
$L$-functions, respectively.

###### Proof.

We assume first that $\varrho\geq 0$ and
$\varrho+k>\alpha(2c-r)+\mu^{\prime}-\mu$. Then by Perron’s formula (see, Ivić
[19, (A.8) and (A.4)]) we have

$\displaystyle A_{\varrho}(x)=\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\frac{\Gamma(s)\varphi(s)x^{s+\varrho}}{\Gamma(s+\varrho+1)}ds,$

where $c$ is the constant chosen in Section 2. By the residue theorem and
(2.5) we have

$\displaystyle E_{\varrho}(x)=\frac{1}{2\pi
i}\int_{\mathcal{C}_{c,r-b}}\frac{\Gamma(s)\varphi(s)x^{s+\varrho}}{\Gamma(s+\varrho+1)}ds,$
and $\displaystyle\int_{\lambda_{1}}^{x}E_{\varrho}(u)du=\frac{1}{2\pi
i}\int_{\mathcal{C}_{c,r-b}}\frac{\Gamma(s)\varphi(s)x^{s+\varrho+1}}{\Gamma(s+\varrho+2)}ds+c_{1}$

with some constant $c_{1}$. By induction we deduce

$\displaystyle\int_{0}^{y}E_{\varrho}(x+y_{1})dy_{1}=$
$\displaystyle\int_{\lambda_{1}}^{x+y}E_{\varrho}(u)du-\int_{\lambda_{1}}^{x}E_{\varrho}(u)du$
$\displaystyle=$ $\displaystyle\frac{1}{2\pi
i}\int_{\mathcal{C}_{c,r-b}}\frac{\Gamma(s)\varphi(s)\left((x+y)^{s+\varrho+1}-x^{s+\varrho+1}\right)}{\Gamma(s+\varrho+2)}ds,$

and

(4.1)
$\displaystyle\int_{0}^{y}\cdots\int_{0}^{y}E_{\varrho}(x+y_{1}+\cdots+y_{k})dy_{1}\cdots
dy_{k}$ $\displaystyle=\frac{1}{2\pi
i}\int_{0}^{y}\cdots\int_{0}^{y}dy_{2}\cdots
dy_{k}\int_{\mathcal{C}_{c,r-b}}\frac{\Gamma(s)\varphi(s)\left((x+y+\sum\limits_{j=2}^{k}y_{j})^{s+\varrho+1}-(x+\sum\limits_{j=2}^{k}y_{j})^{s+\varrho+1}\right)}{\Gamma(s+\varrho+2)}ds$
$\displaystyle=\frac{1}{2\pi
i}\int_{\mathcal{C}_{c,r-b}}\frac{\Gamma(s)\varphi(s)}{\Gamma(s+\varrho+2)}\int_{0}^{y}\cdots\int_{0}^{y}\left(\Bigl{(}x+y+\sum_{j=2}^{k}y_{j}\Bigr{)}^{s+\varrho+1}-\Bigl{(}x+\sum_{j=2}^{k}y_{j}\Bigr{)}^{s+\varrho+1}\right)dy_{2}\cdots
dy_{k}$ $\displaystyle=\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}\frac{1}{2\pi
i}\int_{\mathcal{C}_{c,r-b}}\frac{\Gamma(s)\varphi(s)(x+jy)^{s+\varrho+k}}{\Gamma(s+\varrho+k+1)}ds.$

Furthermore, under the condition $\varrho+k>\alpha(2c-r)+\mu^{\prime}-\mu$. we
can change the path of integration in (4.1) to $\mathcal{C}_{r-c,r-b}$, and by
using the functional equation along with a change of variable from $s$ to
$r-s$, we have

$\displaystyle\frac{1}{2\pi
i}\int_{\mathcal{C}_{c,r-b}}\frac{\Gamma(s)\varphi(s)u^{s+\varrho+k}}{\Gamma(s+\varrho+k+1)}ds=\frac{1}{2\pi
i}\int_{\mathcal{C}_{c,b}}G_{\varrho+k}(s)\psi(s)u^{r+\varrho+k-s}ds\quad(u>0).$

Since $b>c>\sigma_{b}^{*}$, $\psi(s)$ is expressed as a Dirichlet series.
Exchanging the order of summation and integration, we get

(4.2)
$\displaystyle\sum_{n=1}^{\infty}\frac{b(n)}{\mu_{n}^{r+\varrho+k}}\frac{1}{2\pi
i}\int_{\mathcal{C}_{c,b}}G_{\varrho+k}(s)(u\mu_{n})^{r+\varrho+k-s}ds=\sum_{n=1}^{\infty}\frac{b(n)}{\mu_{n}^{r+\varrho+k}}f_{\varrho+k}(u\mu_{n})$

for $a<c$. Since $f_{\varrho+k}(u\mu_{n})\ll(u\mu_{n})^{\theta_{\varrho+k}}$,
the series (4.2) converges absolutely if

(4.3) $\varrho+k>2\alpha\sigma_{b}^{\ast}-\alpha
r-\frac{1}{2}+\mu^{\prime}-\mu.$

Thus under this assumption, the exchange of the order of summation and
integration is justified and the equality

(4.4)
$\displaystyle\int_{0}^{y}\cdots\int_{0}^{y}E_{\varrho}\left(x+\sum_{j=1}^{k}y_{j}\right)dy_{1}\cdots
dy_{k}$
$\displaystyle\qquad=\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}\sum_{n=1}^{\infty}\frac{b(n)}{\mu_{n}^{r+\varrho+k}}f_{\varrho+k}\left((x+jy)\mu_{n}\right)$

is valid under (4.3).

Finally we note that if $\min\\{u,u+y\\}>0$, then for any non-negative integer
$j$ we have

(4.5)
$\displaystyle\int_{0}^{y}f_{\varrho+j}\left((u+y_{1})\mu_{n}\right)dy_{1}=\int_{0}^{y}dy_{1}\frac{1}{2\pi
i}\int_{\mathcal{C}_{a,b}}G_{\varrho+j}(s)\left((u+y_{1})\mu_{n}\right)^{r+\varrho+j-s}ds$
$\displaystyle=\frac{1}{2\pi
i\mu_{n}}\int_{\mathcal{C}_{a,b}}\frac{\Gamma(r-s)\Delta_{2}(s)}{\Gamma(r+\varrho+j+2-s)\Delta_{1}(r-s)}$
$\displaystyle\hskip
56.9055pt\times\left(\left((u+y_{1})\mu_{n}\right)^{r+\varrho+j+1-s}-(u\mu_{n})^{r+\varrho+j+1-s}\right)ds$
$\displaystyle=\frac{1}{\mu_{n}}\left(f_{\varrho+j+1}(u+y)-f_{\varrho+j+1}(u)\right).$

By induction we get

(4.6)
$\displaystyle\int_{0}^{y}\cdots\int_{0}^{y}f_{\varrho}\left(\Bigl{(}x+\sum_{j=1}^{k}y_{j}\Bigr{)}\mu_{n}\right)dy_{1}\cdots
dy_{k}=\frac{1}{\mu_{n}^{k}}\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}f_{\varrho+k}\left((x+jy)\mu_{n}\right).$

Theorem 4 follows from (4.4) and (4.6) at once. ∎

##  5 Large and small values of $E_{\varrho}(y)$ in short intervals

In this section we shall study the large and small values of the error term
$E_{\varrho}(y)$ for $y$ in some ”short” interval. We always suppose that
$\alpha>\frac{1}{2}$.

We first state some conditions.

  1. (A1)

$b(n)\ll_{\varepsilon}{\mu_{n}}^{\ell_{1}+\varepsilon}$ for some $\ell_{1}\geq
0$ and $b(n)\gg 1$ for infinitely many $n$.

  2. (A2)

$n^{\hbar}\ll\mu_{n}\ll n^{\hbar}$ for some $0<\hbar$, and
$\mu_{m}^{c}-\mu_{n}^{c}\gg m^{\hbar c}-n^{\hbar c}$ for
$0<c\leq\frac{1}{\hbar}$ and $m>n$.

  3. (A3)

$a(n)\ll_{\varepsilon}{\lambda_{n}}^{\ell+\varepsilon}$ for some $\ell\geq 0$
and $a(n)\gg 1$ for infinitely many $n$.

  4. (A4)

$a(n)\geq 0$ for all $n\in\mathbb{N}$ and $a(n)\gg 1$ for infinitely many $n$.

######  Theorem 5.

Let $\lambda\geq 1$ be a fixed real number. Assume that the conditions (A1)
and (A2) hold. Then there exists a constant $B>0$ such that for $x\geq x_{0}$
and $Bx^{1-\frac{1}{2\alpha}}\leq U\leq x$ we have

(5.1) $\displaystyle\int_{x}^{x+U}|E_{\varrho}(y)|^{\lambda}dy\gg
x^{\lambda\theta_{\varrho}}U,\quad\int_{1}^{x}|E_{\varrho}(y)|^{\lambda}dy\gg
x^{1+\lambda\theta_{\varrho}},$

where $\theta_{\varrho}$ is defined by (3.6). In particular

(5.2) $\displaystyle\int_{x}^{x+U}|E_{\varrho}(y)|dy\gg
x^{\theta_{\varrho}}U,\quad\int_{1}^{x}|E_{\varrho}(y)|dy\gg
x^{1+\theta_{\varrho}}.$

######  Theorem 6.

Under the conditions (A1) and (A2), there exist two positive constants $B$ and
$B^{\prime}$ such that for $x\geq x_{0}$, we have

$\max\limits_{x\leq y\leq x+Bx^{1-\frac{1}{2\alpha}}}\pm
E_{\varrho}(y)>B^{\prime}x^{\theta_{\varrho}}.$

Furthermore we have:

  1. (i)

If the conditions (A1), (A2) and (A3) hold, then there exists a point
$x^{*}\in[x,x+Bx^{1-\frac{1}{2\alpha}}]$ with $|E(x^{*})|\ll
x^{\ell+\varepsilon}$ and $E(y)$ is continuous at $x^{*}$.

  2. (ii)

If the conditions (A1), (A2) and (A4) hold, then for any $t$ with
$|t|<B^{\prime}x^{\theta_{0}}$ there exists at least a point
$x^{*}\in[x,x+Bx^{1-\frac{1}{2\alpha}}]$ such that $E(x^{*})=t$ and $E(y)$ is
continuous at $x^{*}$.

######  Remark 8.

Tong [48] first proved Theorem 6 for the error term in the summatory function
of the the general Piltz divisor problem $d_{\ell}(n)$ for any $\ell\geq 2$.
A. Ivić [20] studied the general case for $\Delta_{1}(s)=\Delta_{2}(s).$ Our
result here provides a new proof of Ivić’s result.

######  Remark 9.

When $\varrho>0$, $E_{\varrho}(y)$ is a continuous function of $y$. Then for
any $t$ with $|t|<B^{\prime}x^{\theta_{\varrho}}$ there exists at least a
point $x^{*}\in[x,x+Bx^{1-\frac{1}{2\alpha}}]$ such that $E(x^{*})=t$ and
$E(y)$ is continuous at $x^{*}$.

######  Remark 10.

For $E_{\varrho}^{*}(y),$ we have similar results. Suppose that
$\\{\lambda_{n}\\}(n\geq 1)$ satisfy

(A2*): $n^{\hbar^{*}}\ll\lambda_{n}\ll n^{\hbar^{*}}$ for some $0<\hbar^{*}$,
and $\mu_{m}^{c^{*}}-\mu_{n}^{c^{*}}\gg m^{\hbar^{*}c^{*}}-n^{\hbar^{*}c^{*}}$
for $0<c^{*}\leq\frac{1}{\hbar^{*}}$ and $m>n$.

So we have the following results:

  1. (1)

If (A3) and (A2*) hold, then Theorem 5 holds for $E_{\varrho}^{*}(y)$ with
$\theta_{\varrho}$ replaced by

$\theta_{\varrho}^{\prime}=:\frac{r}{2}-\frac{1}{4\alpha}+\varrho\left(1-\frac{1}{2\alpha}\right)+\frac{\mu-\mu^{\prime}}{2\alpha}.$

  2. (2)

If (A3) and (A2*) hold, there exists two positive constants $B$ and
$B^{\prime}$ such that for $x\geq x_{0}$, we have

$\max\limits_{x\leq y\leq x+Bx^{1-\frac{1}{2\alpha}}}\pm
E_{\varrho}^{*}(y)>B^{\prime}x^{\theta_{\varrho}^{\prime}}.$

  3. (3)

If the conditions (A3), (A2*) and (A1) hold, then there exists a point
$x^{*}\in[x,x+Bx^{1-\frac{1}{2\alpha}}]$ with $|E^{*}(x^{*})|\ll
x^{\ell_{1}+\varepsilon}$ and $E^{*}(y)$ is continuous at point $x^{*}$.

  4. (4)

If the conditions (A3), (A2*) hold, $b(n)\geq 0(n\geq 1)$ and $b(n)\gg 1$ for
infinitely many $n,$ then for any $t$ with
$|t|<B^{\prime}x^{\theta_{0}^{\prime}}$ there exists at least a point
$x^{*}\in[x,x+Bx^{1-\frac{1}{2\alpha}}]$ such that $E^{*}(x^{*})=t$ and
$E^{*}(y)$ is continuous at point $x^{*}$.

Proof of Theorem 5. We let $k$ be a fixed large integer such that
$v=r+\varrho+k-\theta_{\varrho+k}>\sigma_{b}^{*}+1/2$. Therefore the series

$g(t)=\sum_{n=1}^{\infty}\frac{b(n)}{\mu_{n}^{v}}\cos\left(h(t\mu_{n})^{\frac{1}{2\alpha}}+c_{0}\pi\right)$

is absolutely convergent. By Lemma 2 of Ivić [20], there exist two constants
$B^{\prime}>0,D^{\prime}>0$ such that for $x\geq x_{0}$ every interval
$[x,x+D^{\prime}x^{1-\frac{1}{2\alpha}}]$ contains two points $x_{1},x_{2}$
for which

$g(x_{1})>B^{\prime},\quad g(x_{2})<-B^{\prime}.$

Take $U^{\prime}=c^{\prime}x^{1-\frac{1}{2\alpha}}$, where $c^{\prime}$ be a
constant to be specified below, and $x^{*}=x_{j}$ for $j=1$ or 2. Applying
Theorem 4 with $L=0$, Lemma 4 and (4.5) we have

$\displaystyle\int_{0}^{U^{\prime}}dy\int_{0}^{y}\cdots\int_{0}^{y}E_{\varrho}(x^{*}+y_{1}+\cdots+y_{k})dy_{1}\cdots
dy_{k}$
$\displaystyle=\int_{0}^{U^{\prime}}\left(\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}\sum_{n=1}^{\infty}\frac{b(n)}{\mu_{n}^{r+\varrho+k}}f_{\varrho+k}\left((x^{*}+jy)\mu_{n}\right)\right)dy$
$\displaystyle=(-1)^{k}\kappa_{0}(\varrho+k)U^{\prime}g(x^{*})(x^{*})^{\theta_{\varrho+k}}+O\left(U^{\prime}x^{\theta_{\varrho+k}-\frac{1}{2\alpha}}\right)$
$\displaystyle\quad+\sum_{j=1}^{k}\frac{(-1)^{k-j}}{j}\binom{k}{j}\sum_{n=1}^{\infty}\frac{b(n)}{\mu_{n}^{r+\varrho+k+1}}\left(f_{k+\varrho+1}\left((x^{*}+jU^{\prime})\mu_{n}\right)-f_{k+\varrho+1}(x^{*}\mu_{n})\right).$

Using Lemma 4 again we obtain

(5.3)
$\displaystyle\left|\int_{0}^{U^{\prime}}dy\int_{0}^{y}\cdots\int_{0}^{y}E_{\varrho}(x^{*}+y_{1}+\cdots+y_{k})dy_{1}\cdots
dy_{k}\right|$
$\displaystyle\geq(x^{*})^{\theta_{\varrho+k+1}}\left(c^{\prime}\kappa_{0}(\varrho+k)B^{\prime}+O\left(\frac{U^{\prime}}{x}\right)\right.$
$\displaystyle\qquad\left.-\kappa_{0}(\varrho+k+1)2^{k+1}\psi_{1}(r+\varrho+k+1-\theta_{\varrho+k+1})\left(1+\frac{kU^{\prime}}{x}\right)^{\theta_{\varrho+k+1}}\right),$

where $\psi_{1}(s)=\sum_{n=1}^{\infty}\frac{|b(n)|}{\mu_{n}^{s}}$. Now we can
choose a constant $c^{\prime}$ such that

(5.4)
$c^{\prime}>\frac{\kappa_{0}(\varrho+k+1)2^{k+2}\psi_{1}(r+\varrho+k+1-\theta_{\varrho+k+1})}{\kappa_{0}(\varrho+k)B^{\prime}}.$

Combining (5.3) and (5.4) we have for some constant $B_{0}>0$ as $x>x_{0}$

(5.5) $\displaystyle B_{0}x^{\theta_{\varrho+k+1}}$
$\displaystyle\leq\left|\int_{0}^{U^{\prime}}dy\int_{0}^{y}\cdots\int_{0}^{y}E_{\varrho}(x^{*}+y_{1}+\cdots+y_{k})dy_{1}\cdots
dy_{k}\right|$
$\displaystyle\leq\int_{0}^{U^{\prime}}dy\int_{0}^{y}\cdots\int_{0}^{y}|E_{\varrho}(x^{*}+y_{1}+\cdots+y_{k})|dy_{1}\cdots
dy_{k}$
$\displaystyle\leq\int_{0}^{U^{\prime}}dy\int_{0}^{U}|E_{\varrho}(x+y_{1})|dy_{1}\int_{0}^{y}\cdots\int_{0}^{y}dy_{2}\cdots
dy_{k}$ $\displaystyle\leq
x^{k(1-\frac{1}{2\alpha})}\int_{x}^{x+U}|E_{\varrho}(y)|dy,$

here $U=(k+2)U^{\prime}$.This proves the first assertion in (5.2).

Next, we have

$\displaystyle\int_{1}^{x}|E_{\varrho}(y)|dy\geq\sum_{j=0}^{\frac{1}{2B}x^{\frac{1}{2\alpha}}}\int_{\frac{x}{2}+jBx^{1-\frac{1}{2\alpha}}}^{\frac{x}{2}+(j+1)Bx^{1-\frac{1}{2\alpha}}}|E_{\varrho}(y)|dy\gg
x^{\frac{1}{2\alpha}}\left(x^{\theta_{\varrho}}x^{1-\frac{1}{2\alpha}}\right)=x^{1+\theta_{\varrho}},$

and this completes the proof of the second assertion in (5.2).

By Hölder’s inequality and (5.2) we immediately get (5.1).

Proof of Theorem 6. Similarly to (5.5) we see that for $j=1,2$

$\displaystyle x^{\theta_{\varrho+k+1}}$
$\displaystyle\ll\max\pm\int_{0}^{U^{\prime}}dy\int_{0}^{y}\cdots\int_{0}^{y}E_{\varrho}(x_{j}+y_{1}+\cdots+y_{k})dy_{1}\cdots
dy_{k}$ $\displaystyle\ll\max_{x\leq y\leq x+U}\pm
E_{\varrho}(y)\times\int_{0}^{U^{\prime}}dy\int_{0}^{y}\cdots\int_{0}^{y}dy_{2}\cdots
dy_{k}.$

Hence for $x>x_{0}$

(5.6) $\max_{x\leq y\leq x+U}\pm E_{\varrho}(y)\gg x^{\theta_{\varrho}}.$

When $\varrho=0$, for every interval $[x,x+2U]$, it follows from (5.6) that
there exists two points $y^{\prime},y^{\prime\prime}\in[x,x+2U]$ such that

$\displaystyle E(y^{\prime})>0,E(y^{\prime\prime})<0,\quad
y^{\prime}<y^{\prime\prime}.$

If the condition (A4) holds, then $Q(y),$ which is the main term of the
summatory function $a(n),$ is of the form (2.6) with $\varrho=0$, always
positive and increasing. And the function $E(y)$ is a continuous and strictly
decreasing in each interval $(\lambda_{n},\lambda_{n+1})$. Moreover, $E(y)$
has a jump up to $a(n)/2$ at the point $\lambda_{n}\ \ (n=1,2,\cdots)$, i.e.
$E(\lambda_{n})=E(\lambda_{n}-0)+a(n)/2$. Hence there exists at least a zero
point $x^{*}$ of $E(x)$ in the interval $(y^{\prime},y^{\prime\prime})$ with
$E(x)$ is continuous at point $x^{*}$.

Otherwise, we may suppose that for any $x\in(y^{\prime},y^{\prime\prime})$
with $E(x)\neq 0$. Without loss of generality, we assume that for all
$\lambda_{n}\in(y^{\prime},y^{\prime\prime})$ with
$E(\lambda_{n}+0)E(\lambda_{n+1}-0)>0$. Hence we can find at least a
$\lambda_{n}\in(y^{\prime},y^{\prime\prime})$ such that
$E(\lambda_{n}-0)\times E(\lambda_{n}+0)<0$, in this time we can choose
$x^{*}=\frac{\lambda_{n}+\lambda_{n+1}}{2}$. This completes the proof of
Theorem 6.

##  6 The truncated Tong-type formula for an integral involving the error term

Suppose $x>10$ is a large parameter. For any integrable function $g(y)$, we
define

$\displaystyle\tilde{y}:=y+\frac{1}{x}(y_{1}+\cdots+y_{k}),$
$\displaystyle\int_{\mathbf{E}_{k}}g(\tilde{y})d\mathrm{Y}_{k}:=\int_{0}^{1}\cdots\int_{0}^{1}g(\tilde{y})dy_{1}\cdots
dy_{k}.$

The integral $\int_{\mathbf{E}_{k}}E(\tilde{y})d\mathrm{Y}_{k}$ is very
important in Tong’s theory. The aim of this section is to give its truncated
expression under some suitable conditions.

Recall the definition of $A^{*}(y).$ Let $\hat{Q}(y)$ denote the sum of
residues over all poles of the function $\psi(s)y^{s}s^{-1}$ except the pole
$s=0.$ And define the error term $\hat{E}(y)$ by

$\hat{E}(y):=A^{*}(y)-\hat{Q}(y).$

Since $\psi(s)$ doesn’t have poles when $\sigma\leq\sigma^{*}$ by assumption,
we have immediately that

$Q^{*}(y)=\hat{Q}(y)+\psi(0),\ \ E^{*}(y)=\hat{E}(y)-\psi(0).$

Let $y\geq 1$, $\frac{1}{2}\leq M\leq N<\infty$, $\lambda$ be a real number,
and define

(6.1) $\displaystyle I(\lambda,M,N,y)=2\pi
i\int_{M}^{N}u^{\lambda}\hat{E}(u)\exp\left(-ih(uy)^{\frac{1}{2\alpha}}\right)du.$

We assume in this section that

  1. (B1)

There exists a constant $1-\sigma_{b}^{*}\leq\omega_{0}<1$ such that
$|b(n)|\ll\mu_{n}^{\sigma_{b}^{\ast}-1+\omega_{0}}$.

  2. (B2)

$\sum_{\mu_{n}\leq y}|b(n)|\ll y^{\sigma_{b}^{\ast}}\log^{A}y,\ y\geq 2.$

  3. (B3)

There exists a constant $0\leq\omega_{1}\leq 1$ such that

$\int_{1}^{T}|\hat{E}(u)|du\ll T^{\sigma_{b}^{\ast}+\omega_{1}}.$

Throughout this section we let $k$ to be a fixed positive integer such that

(6.2) $k>\max\left\\{2\alpha\sigma_{b}^{*}-\alpha
r-\frac{1}{2}-\varrho+\mu^{\prime}-\mu,\alpha r\right\\}.$

Furthermore we assume that

(6.3)
$\displaystyle\sigma_{b}^{\ast}+\omega_{1}-r-\frac{1}{4\alpha}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha}-1<0.$

We note that (6.2) and (6.3) implies

(6.4)
$\sigma_{b}^{\ast}+\omega_{1}-\frac{r}{2}-\frac{1}{4\alpha}-1-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha}-\frac{k}{\alpha}<0.$

Let $1\leq x\leq y\leq(1+\delta)x$, $N=[x^{4\alpha-1-\varepsilon}]$ and
$J=\left[(4\alpha^{2}r+4\alpha)\varepsilon^{-1}\right]$, where $\delta>0$ is a
fixed small positive constant. Without loss of generality, we suppose
$N\not=\mu_{n}$ for any $n.$ Furthermore let

$\displaystyle\lambda_{0}:=\theta_{\varrho}+\frac{1}{2\alpha}-r-\varrho-1=-\frac{r}{2}+\frac{1}{4\alpha}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha}-1.$

From Theorem 6 we can find a real number $M$ in every subinterval
$[t,t+Bt^{1-\frac{1}{2a}}]$ in $[1,\sqrt{N}]$ such that $M\neq\mu_{n}$ for
$n=1,2,\ldots$ and $\hat{E}(M)=0$ if $b(n)\geq 0$ for all $n$ or
$\hat{E}(M)\ll M^{\sigma_{b}^{*}-1+\omega_{0}}$ otherwise.

The truncated Tong-type formula can be stated as

######  Theorem 7.

Assume the conditions (A2), (B1), (B2) and (B3) hold. Let $M\in[1,\sqrt{N}]$
be chosen as above. Then we have

(6.5)
$\displaystyle\int_{\mathbf{E}_{k}}E_{\varrho}(\tilde{y})d\mathrm{Y}_{k}=\sum_{j=1}^{6}R_{j}(y;M),$

where

(6.6) $\displaystyle R_{1}(y;M)=$
$\displaystyle\kappa_{0}y^{\theta_{\varrho}}\sum_{\mu_{n}\leq
M}\frac{b(n)}{\mu_{n}^{r+\varrho-\theta_{\varrho}}}\cos(h(y\mu_{n})^{\frac{1}{2\alpha}}+c_{0}\pi),$
(6.7) $\displaystyle R_{2}(y;M)=$ $\displaystyle
y^{\theta_{\varrho}+\frac{1}{2\alpha}}{\rm
Re\,}\left\\{c_{00}I\left(\lambda_{0},M,N,y\right)\right\\},$ (6.8)
$\displaystyle R_{3}(y;M)=$
$\displaystyle\mathop{{\sum_{l=0}^{J}\sum_{m=0}^{J}}}_{l+m\neq 0}{\rm
Re\,}\left\\{c_{lm}I\left(\lambda_{0}+\frac{l-m}{2\alpha},M,N,y\right)\right\\}x^{-l}y^{-l+\theta_{\varrho}+\frac{1}{2\alpha}+\frac{l-m}{2\alpha}},$
(6.9) $\displaystyle R_{4}(y;M)=$
$\displaystyle\sum_{j=0}^{k}\sum_{m=0}^{k}{\rm
Re\,}\left\\{c_{jm}^{\prime}I\left(\lambda_{0}-\frac{k+m}{2\alpha},N,\infty,y+\frac{j}{x}\right)\right\\}$
$\displaystyle\hskip 142.26378pt\times
x^{k}\left(y+\frac{j}{x}\right)^{k+\theta_{\varrho}+\frac{1}{2\alpha}-\frac{k+m}{2\alpha}},$
(6.10) $\displaystyle R_{5}(y;M)\ll$ $\displaystyle
x^{\theta_{\varrho}-\frac{1}{2\alpha}}M^{\max(\sigma_{b}^{\ast}-\frac{r}{2}-\frac{3}{4\alpha}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha},0)}\log^{A}M$
$\displaystyle+x^{-2+\theta_{\varrho}+\frac{1}{2\alpha}}M^{\max(\sigma_{b}^{\ast}-\frac{r}{2}+\frac{1}{4\alpha}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha},0)}\log^{A}M$
$\displaystyle+x^{\theta_{\varrho}-\frac{r}{2}}M^{\sigma_{b}^{\ast}+\omega_{1}-r-\frac{1}{4\alpha}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha}-1}$
$\displaystyle+x^{(4\alpha-1)(\sigma_{b}^{\ast}+\omega_{1})-2k+r-\varrho+2(\mu^{\prime}-\mu)+\frac{2k}{\alpha}-2\alpha
r-4\alpha},$ (6.11) $\displaystyle R_{6}(y;M)\ll$
$\displaystyle\begin{cases}0,&\mbox{if $b(n)\geq 0\ (n\geq 1)$,}\\\
x^{\theta_{\varrho}}M^{\sigma_{b}^{\ast}-1+\omega_{0}-\frac{r}{2}-\frac{1}{4\alpha}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha}},&\mbox{if
the condition (B1) holds.}\end{cases}$

Here $\kappa_{0},c_{0}$ and $c_{00},c_{lm},c_{jm}^{\prime}$ are certain real
and complex constants, respectively.

###  6.1  Lemmas for generalized Bessel functions

Before proving Theorem 7 we shall prepare some preliminary assertions for the
so-called generalized Bessel functions.

By induction one easily verifies ($r-s\notin-\mathbb{N}$)

(6.12)
$\displaystyle\int_{\mathbf{E}_{k}}\tilde{y}^{r-s}d\mathrm{Y}_{k}=x^{k}\sum_{l=0}^{k}(-1)^{k-l}\binom{k}{l}\frac{(y+lx^{-1})^{k+r-s}}{(1+r-s)(2+r-s)\cdots(k+r-s)}$

For $j\geq 0$, define

$\displaystyle f_{\varrho,j}(x)=\frac{1}{2\pi
i}\int_{\mathcal{C}_{a,b}}\frac{\Gamma(r-s)}{(r+\varrho-s)^{j}\Gamma(r+\varrho+1-s)}\frac{\Delta_{2}(s)}{\Delta_{1}(r-s)}x^{r+\varrho-s}ds$

and

$g_{\varrho,j}(u)=\int_{\mathbf{E}_{k}}f_{\varrho,j}(\tilde{y}u)d\mathrm{Y_{k}}.$

By definition $f_{\varrho}(x)=f_{\varrho,0}(x)$. We note that
$f_{\varrho,j}(x)$ converges absolutely if

$a<\frac{r}{2}+\frac{\varrho+j-\mu^{\prime}+\mu}{2\alpha}.$

######  Lemma 5.

Let $x\leq y\leq(1+\delta)x$, $b$ is a sufficiently large positive constant.
Then we have

(6.13) $\displaystyle f_{\varrho,j}(x)$ $\displaystyle\ll
x^{\theta_{\varrho}-\frac{j}{2\alpha}},$ and (6.14) $\displaystyle
g_{\varrho,j}(u)$ $\displaystyle\ll
x^{2k}(ux)^{\theta_{\varrho}-\frac{k+j}{2\alpha}}.$

###### Proof.

The first assertion (6.13) is obtained immediately by replacing $\varrho$ by
$\varrho+1$ and taking $M=-j$ in (3.14). ($M$ is the parameter in the
definition of $F(s),$ see (3.9).)

To prove (6.14), substituting the definitions of $f_{\varrho,j}(x)$ into
$g_{\varrho,j}(u)$ and using (6.12), we get

$\displaystyle g_{\varrho,j}(u)$
$\displaystyle=x^{k}\sum_{l=0}^{k}(-1)^{k-l}\binom{k}{l}\left(y+\frac{l}{x}\right)^{k}$
$\displaystyle\quad\times\frac{1}{2\pi
i}\int_{\mathcal{C}_{a,b}}\frac{\Gamma(r-s)}{(r+\varrho-s)^{j}\Gamma(r+\varrho+k+1-s)}\frac{\Delta_{2}(s)}{\Delta_{1}(r-s)}\left(\left(y+\frac{l}{x}\right)u\right)^{r+\varrho-s}ds.$

Replacing $\varrho$ by $\lambda=\varrho+k+1$ and taking $M=-j$ in (3.14), we
get

$\displaystyle g_{\varrho,j}(u)\ll
x^{2k}(ux)^{\theta_{\varrho}-\frac{k+j}{2\alpha}}.$

∎

We also need the asymptotic behaviour of the function $\Phi_{\varrho}(y)$
defined by

$\Phi_{\varrho}(y)=\frac{d}{du}\left(\frac{f_{\varrho}(yu)}{u^{r+\varrho}}\right).$

######  Lemma 6.

Let $l$ be a fixed positive integer and $x\leq y\leq(1+\delta)x$. Then there
are constants $b_{lm}^{\prime}$ and $b_{jm}^{\prime\prime}$ such that for any
positive integer $H$, we have

(6.15) $\displaystyle\frac{d^{l}}{dy^{l}}\Phi_{\varrho}(y)$
$\displaystyle=\sum_{m=0}^{H}{\rm
Re\,}\left\\{b_{lm}^{\prime}\frac{(uy)^{\theta_{\varrho}+\frac{1}{2\alpha}+\frac{l-m}{2\alpha}}}{y^{l}u^{r+\varrho+1}}e^{-ih(uy)^{\frac{1}{2\alpha}}}+O\left(\frac{(uy)^{\theta_{\varrho}+\frac{l-H}{2\alpha}}}{y^{l}u^{r+\varrho+1}}\right)\right\\}$
$\displaystyle\quad+O\left(y^{r+\varrho-b-l}u^{-1-b}\right),$ and (6.16)
$\displaystyle\int_{\mathbf{E}_{k}}\Phi_{\varrho}(\tilde{y})d\mathrm{Y_{k}}$
$\displaystyle=\sum_{j=0}^{k}\left\\{\sum_{m=0}^{H}{\rm
Re\,}\left(b_{jm}^{\prime\prime}\frac{x^{k}(y+\frac{j}{x})^{k+\theta_{\varrho}+\frac{1}{2\alpha}-\frac{k+m}{2\alpha}}}{u^{r+\varrho+1-\frac{1}{2\alpha}-\theta_{\varrho}+\frac{k+m}{2\alpha}}}e^{-ih((y+\frac{j}{x})u)^{\frac{1}{2\alpha}}}\right)\right.$
$\displaystyle\qquad\left.+O\left(\frac{x^{2k+\theta_{\varrho}+\frac{1}{2\alpha}-\frac{k+H+1}{2\alpha}}}{u^{r+\varrho+1-\frac{1}{2\alpha}-\theta_{\varrho}+\frac{k+H+1}{2\alpha}}}\right)\right\\}+O\left(x^{2k+r+\varrho-b}u^{-1-b}\right).$

###### Proof.

From the definition of $f_{\varrho}(yu)$, we have

(6.17) $\displaystyle\Phi_{\varrho}(y)=\frac{1}{2\pi
i}\int_{\mathcal{C}_{a,b}}\frac{\Gamma(r-s)\Delta_{2}(s)(-s)}{\Gamma(r+\varrho+1-s)\Delta_{1}(r-s)}y^{r+\varrho-s}u^{-s-1}ds,$

hence

$\displaystyle\frac{d^{l}}{dy^{l}}\Phi_{\varrho}(y)$
$\displaystyle=\frac{1}{y^{l}u^{r+\varrho+1}}\frac{(-1)^{l-1}}{2\pi
i}\int_{\mathcal{C}_{a,b}}\frac{\Gamma(r-s)\Delta_{2}(s)}{\Gamma(r+\varrho+1-s)\Delta_{2}(r-s)}$
$\displaystyle\quad\times
s(s-r-\varrho)(s-r-\varrho+1)\cdots(s-r-\varrho+l-1)(yu)^{r+\varrho-s}ds.$

Replacing $\varrho$ by $\varrho+1$ and taking $M=l+1$ in the formula (3.13),
we find that

$\displaystyle\frac{d^{l}}{dy^{l}}\Phi_{\varrho}(y)$
$\displaystyle\sim\frac{1}{y^{l}u^{r+\varrho+1}}\Bigl{\\{}\sum_{m=0}^{\infty}b_{lm}(yu)^{\theta_{\varrho}+\frac{1}{2\alpha}+\frac{l-m}{2\alpha}}\cos(h(yu)^{\frac{1}{2\alpha}}+c_{m}\pi)$
$\displaystyle\hskip 85.35826pt+O((yu)^{r+\varrho-b})\Bigr{\\}},$

which proves the assertion (6.15).

Next we prove (6.16). From (6.17) and (6.12) we have

(6.18)
$\displaystyle\int_{\mathbf{E}_{k}}\Phi_{\varrho}(\tilde{y})d\mathrm{Y_{k}}$
$\displaystyle=\frac{1}{2\pi
i}\int_{\mathcal{C}_{a,b}}\frac{\Gamma(r-s)\Delta_{2}(s)(-s)}{\Gamma(r+\varrho+1-s)\Delta_{1}(r-s)}u^{-s-1}\int_{\mathbf{E}_{k}}\tilde{y}^{r+\varrho-s}d\mathrm{Y_{k}}ds$
$\displaystyle=\frac{x^{k}}{u^{r+\varrho+1}}\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}\left(y+\frac{j}{k}\right)^{k}$
$\displaystyle\quad\times\frac{1}{2\pi
i}\int_{\mathcal{C}_{a,b}}\frac{\Gamma(r-s)\Delta_{2}(s)(-s)}{\Gamma(r+\varrho+k+1-s)\Delta_{1}(r-s)}\left(\left(y+\frac{j}{x}\right)u\right)^{r+\varrho-s}ds.$

Replacing $\varrho$ by $\varrho+k+1$ and taking $M=1$ in (3.13), we find that

$\displaystyle\frac{1}{2\pi
i}\int_{\mathcal{C}_{a,b}}\frac{\Gamma(r-s)\Delta_{2}(s)(-s)}{\Gamma(r+\varrho+k+1-s)\Delta_{1}(r-s)}\left(\left(y+\frac{j}{x}\right)u\right)^{r+\varrho-s}ds$
$\displaystyle\sim\sum_{m=0}^{\infty}b_{jm}^{\prime}\left(\left(y+\frac{j}{x}\right)u\right)^{\theta_{\varrho}+\frac{1}{2\alpha}-\frac{k+m}{2\alpha}}\cos\left(h\left(\left(y+\frac{j}{x}\right)u\right)^{\frac{1}{2\alpha}}+c_{m}\pi\right)$
$\displaystyle\quad+O\left((yu)^{r+\varrho-b}\right)$

with certain real constants $b_{jm}^{\prime}$. Substituting this expression to
(6.18), we get (6.16). ∎

###  6.2  Proof of Theorem 7

Now we turn to the proof of Theorem 7. Let $M\in[1,\sqrt{N}]$ be a real number
as is chosen before, namely. $M\neq\mu_{n}$ for $n=1,2,\ldots$ and
$\hat{E}(M)=0$ if $b(n)\geq 0$ for all $n$ or $\hat{E}(M)\ll
M^{\sigma_{b}^{*}-1+\omega_{0}}$ otherwise. We always suppose that there are
infinitely many $n$ for which $b(n)\gg 1.$

Recall the Tong-type identity:

$\displaystyle
I:=\int_{\mathbf{E}_{k}}E_{\varrho}\left(y+\frac{\sum_{j=1}^{k}y_{j}}{x}\right)d\mathrm{Y}_{k}=\sum_{n=1}^{\infty}\frac{b(n)}{\mu_{n}^{r+\varrho}}\int_{\mathbf{E}_{k}}f_{\varrho}\left(\left(y+\frac{\sum_{j=1}^{k}y_{j}}{x}\right)\mu_{n}\right)d\mathrm{Y}_{k}.$

In the right hand side we divide the sum over $n$ into two parts, namely

$I=\sum_{\mu_{n}\leq M}+\sum_{\mu_{n}>M}=:S_{1}(M)+S_{2}(M).$

We first treat $S_{1}(M)$. Let $F(\tilde{y})=f_{\varrho}(\tilde{y}u)$. By
Taylor’s formula we have

$F(\tilde{y})=F(y)+F^{\prime}(y_{1})(\tilde{y}-y)$

with some $y\leq y_{1}\leq\tilde{y}$. Then from the definition
$f_{\varrho}(x)$ and (3.15) of Lemma 4 we have an asymptotic expansion

(6.19) $\displaystyle F(y)$
$\displaystyle\sim\sum_{l=0}^{\infty}\kappa_{l}(yu)^{\theta_{\varrho}-\frac{l}{2\alpha}}\cos(h(yu)^{\frac{1}{2\alpha}}+c_{l}\pi)+O((yu)^{r+\varrho-b})$
and the upper bound $\displaystyle\ll(yu)^{\theta_{\varrho}}$

if we take $b$ large. Furthermore since
$F^{\prime}(y_{1})=uf_{\varrho-1}(y_{1}u)$, we have

$\displaystyle|F^{\prime}(y_{1})(\tilde{y}-y)|$
$\displaystyle\ll\frac{u}{x}(xu)^{\theta_{\varrho-1}}$
$\displaystyle=x^{-2}(xu)^{\theta_{\varrho}+\frac{1}{2\alpha}}.$

Take out the term corresponding to $l=0$ in (6.19), we get

$\displaystyle S_{1}(M)$ $\displaystyle=\sum_{\mu_{n}\leq
M}\frac{b(n)}{\mu_{n}^{r+\varrho}}\left\\{\kappa_{0}(y\mu_{n})^{\theta_{\varrho}}\cos(h(y\mu_{n})^{\frac{1}{2\alpha}}+c_{0}\pi)+O((y\mu_{n})^{\theta_{\varrho}-\frac{1}{2\alpha}})\right.$
$\displaystyle\hskip
56.9055pt\left.+O((y\mu_{n})^{r+\varrho-b})+O\left(x^{-2}(x\mu_{n})^{\theta_{\varrho}+\frac{1}{2\alpha}}\right)\right\\}$
$\displaystyle=:R_{1}(y;M)+V_{1}+V_{2}+V_{3}.$

The error term $V_{2}$ can be negligible by taking $b$ large. On the hand, by
(B2), we have

(6.20) $\displaystyle V_{1}$ $\displaystyle\ll
x^{\theta_{\varrho}-\frac{1}{2\alpha}}\sum_{\mu_{n}\leq
M}|b(n)|\mu_{n}^{-r-\varrho+\theta_{\varrho}-\frac{1}{2\alpha}}$
$\displaystyle\ll
x^{\theta_{\varrho}-\frac{1}{2\alpha}}M^{\max(\sigma_{b}^{\ast}-\frac{r}{2}-\frac{3}{4\alpha}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha},0)}\log^{A}M.$

Similarly we have

(6.21) $V_{3}\ll
x^{-2+\theta_{\varrho}+\frac{1}{2\alpha}}M^{\max(\sigma_{b}^{\ast}-\frac{r}{2}+\frac{1}{4\alpha}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha},0)}\log^{A}M.$

Since $M\neq\mu_{n}$ for all $n$, it is easy to see that if $K(u)$ is any
continuously differentiable function such that

$K(u)Q^{*}(u)\rightarrow 0,\int_{u}^{\infty}Q^{*}(t)K^{\prime}(t)dt\rightarrow
0\ (u\rightarrow\infty),$

then

$\int_{M}^{\infty}K(u)d\sum_{\mu_{n}\leq
u}b(n)=\int_{M}^{\infty}K(u)dA^{*}(u).$

So for $S_{2}(M)$ we can write

$\displaystyle S_{2}(M)$
$\displaystyle=\int_{M}^{\infty}\left(\frac{1}{u^{r+\varrho}}\int_{E_{k}}f_{\varrho}(\tilde{y}u)d\mathrm{Y_{k}}\right)d\sum_{\mu_{n}\leq
u}b(n)$
$\displaystyle=\int_{M}^{\infty}\left(\frac{1}{u^{r+\varrho}}\int_{E_{k}}f_{\varrho}(\tilde{y}u)d\mathrm{Y_{k}}\right)dA^{*}(u)$
$\displaystyle=\int_{M}^{\infty}\left(\frac{1}{u^{r+\varrho}}\int_{E_{k}}f_{\varrho}(\tilde{y}u)d\mathrm{Y_{k}}\right)d\hat{Q}(u)+\int_{M}^{\infty}\left(\frac{1}{u^{r+\varrho}}\int_{E_{k}}f_{\varrho}(\tilde{y}u)d\mathrm{Y_{k}}\right)d\hat{E}(u)$
$\displaystyle=\int_{M}^{\infty}\frac{\hat{Q}^{\prime}(u)}{u^{r+\varrho}}\int_{E_{k}}f_{\varrho}(\tilde{y}u)d\mathrm{Y_{k}}du+\left.\frac{\hat{E}(u)}{u^{r+\varrho}}\int_{E_{k}}f_{\varrho}(\tilde{y}u)d\mathrm{Y_{k}}\right|_{u=M}^{\infty}$
$\displaystyle\quad-\int_{M}^{\infty}\hat{E}(u)du\int_{\mathbf{E}_{k}}\frac{d}{du}\left(\frac{f_{\varrho}(\tilde{y}u)}{u^{r+\varrho}}\right)d\mathrm{Y}_{k}$
$\displaystyle=:S_{21}(M)+S_{22}(M)-S_{23}(M).$

We shall consider $S_{21}(M)$ first. From the definition of
$f_{\varrho,j}(x)$, one easily checks that

$\displaystyle\frac{d}{du}f_{\varrho,j+1}(yu)=\frac{1}{u}f_{\varrho,j}(yu)\quad(1\leq
j\leq k+1).$

Let $w_{j}(u)$ be the functions defined by

$w_{1}(u)=\hat{Q}^{\prime}(u)u^{1-r-\varrho}$

and

$w_{j+1}(u)=w_{j}^{\prime}(u)u\ (j\geq 1).$

Then by repeated integration by parts and the definition of $g_{\varrho,j}$ we
have

(6.22)
$\displaystyle\int_{M}^{M_{1}}\frac{\hat{Q}^{\prime}(u)}{u^{r+\varrho}}\int_{E_{k}}f_{\varrho}(\tilde{y}u)d\mathrm{Y_{k}}du$
$\displaystyle=\sum_{j=1}^{k+1}(-1)^{j+1}\,w_{j}(u)g_{\varrho,j}(u)\Big{|}_{M}^{M_{1}}$
$\displaystyle\quad+(-1)^{k+1}\int_{M}^{M_{1}}w_{k+1}^{\prime}(u)g_{\varrho,k+1}(u)du.$

Recall that $\hat{Q}(u)$ is the sum of residues of of all poles of
$\psi(s)x^{s}s^{-1}$ except $0$. Let $\gamma$ be a pole with order $\nu$ and
$\gamma_{0}$ the pole with the maximal real part. Then the residue at $\gamma$
has the form $\hat{Q}_{\gamma}(u)=u^{\gamma}P(\log u)$, where $P(u)$ is a
polynomial of degree $\nu-1$. Let $w_{j,\gamma}(u)$ be functions defined by
the same way for $\hat{Q}_{\gamma}(u)$. Then we see easily

$\displaystyle w_{1,\gamma}(u)$ $\displaystyle=u^{1-r-\varrho}(\gamma
u^{\gamma-1}P(\log u)+u^{\gamma}P^{\prime}(\log u)/u)$ $\displaystyle\ll
u^{{\rm Re\,}\gamma-r-\varrho}\log^{\nu-1}u.$

Similarly we have

$w_{j,\gamma}(u)\ll u^{{\rm Re\,}\gamma-r-\varrho}\log^{\nu-1}u$

for $j\geq 2$. Hence by (6.14), we have

$\displaystyle w_{j,\gamma}(M_{1})g_{\varrho,j}(M_{1})\ll M_{1}^{{\rm
Re\,}\gamma-r-\varrho}(\log
M_{1})^{\nu-1}\,x^{2k}(xM_{1})^{\theta_{\varrho}-\frac{k+j}{2\alpha}}.$

Since $k$ is chosen as (6.2), the exponent of $M_{1}$ is negative,
$w_{j,\gamma}(M_{1})g_{\varrho,j}(M_{1})$, as well as
$w_{j}(M_{1})g_{\varrho,j}(M_{1})$ tend to $0$ when $M_{1}\to\infty$.

Next we consider the term with $u=M$. Applying (6.13) in this case, we get

(6.23) $\displaystyle|w_{j,\gamma}(M)g_{\varrho,j}(M)|$
$\displaystyle\leq|w_{j,\gamma}(M)|\int_{E_{k}}|f_{\varrho,j}(\tilde{y}u)|d\mathrm{Y_{k}}$
$\displaystyle\ll M^{{\rm Re\,}\gamma-r-\varrho}(\log
M)^{\nu-1}\,(xM)^{\theta_{\varrho}-\frac{1}{2\alpha}}$
$\displaystyle=x^{\theta_{\varrho}-\frac{j}{2\alpha}}M^{{\rm
Re\,}\gamma-\frac{r}{2}-\frac{1+2j}{4\alpha}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha}}(\log
M)^{\nu-1}.$

For the last integral of (6.22) which corresponds to $\gamma$, (6.13) implies
that

(6.24)
$\displaystyle\int_{M}^{M_{1}}w_{k+1,\gamma}^{\prime}(u)g_{\varrho,k+1}(u)du$
$\displaystyle\ll\int_{M}^{M_{1}}u^{{\rm Re\,}\gamma-r-\varrho}(\log
u)^{\nu-1}\,(xu)^{\theta_{\varrho}-\frac{k+1}{2\alpha}}du$
$\displaystyle=x^{\theta_{\varrho}-\frac{k+1}{2\alpha}}M^{{\rm
Re\,}\gamma-\frac{r}{2}-\frac{1}{4\alpha}-\frac{\varrho}{2}+\frac{\mu^{\prime}-\mu}{2\alpha}-\frac{k+1}{2\alpha}}(\log
M)^{\nu-1}.$

From (6.23) and (6.24) we get the estimate from the pole $\gamma$. It is
easily seen that the term $j=1$ coming from $\gamma_{0}$ gives the maximal
estimate, namely we have

$S_{21}(M)\ll x^{\theta_{\varrho}-\frac{1}{2\alpha}}M^{{\rm
Re\,}\gamma_{0}-\frac{r}{2}-\frac{3}{4\alpha}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha}}(\log
M)^{A^{\prime}}$

with some integer $A^{\prime}$. Now we note that ${\rm
Re\,}\gamma_{0}\leq\sigma_{b}^{\ast}$, hence $S_{21}(M)$ is absorbed into the
right hand side of (6.20).

Next we consider $S_{22}(M)$. By the trivial bound $\hat{E}(x)\ll
x^{\sigma_{b}^{\ast}+\varepsilon}$ and (6.14), we see that

$\displaystyle\frac{\hat{E}(M_{1})}{M_{1}^{r+\varrho}}\int_{\mathbf{E}_{k}}f_{\varrho}(\tilde{y}M_{1})d\mathrm{Y_{k}}$
$\displaystyle\ll
M_{1}^{\sigma_{b}^{\ast}-r-\varrho+\varepsilon}x^{2k}(yM_{1})^{\theta_{\varrho}-\frac{k}{2\alpha}}$

The exponent of $M_{1}$ is negative by (6.2), hence this term vanishes when
$M_{1}\to\infty$.

Consider the term
$\frac{\hat{E}(M)}{M^{r+\varrho}}\int_{E_{k}}f_{\varrho}(\tilde{y}M)dY$. If we
assume that $b(n)\geq 0(n\geq 1)$, then from Theorem 6 (ii) we can find $M$
such that $\hat{E}(M)=0$, hence this term becomes zero. If we assume the
condition (B1), there exists $M$ such that $\hat{E}(M)\ll
M^{\sigma_{b}^{\ast}-1+\omega_{0}}$. Hence by (6.13)

$\displaystyle S_{22}(M)$ $\displaystyle\ll
x^{\theta_{\varrho}}M^{\sigma_{b}^{\ast}-1+\omega_{0}-(r+\varrho)+\theta_{\varrho}}$
$\displaystyle=x^{\theta_{\varrho}}M^{\sigma_{b}^{\ast}-1+\omega_{0}-\frac{r}{2}-\frac{1}{4\alpha}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha}}.$

We call the above estimate of $S_{22}(M)$ as $R_{6}(M)$.

Finally we consider $S_{23}(M)$. Divide the integral as

$\displaystyle
S_{23}(M)=\left(\int_{M}^{N}+\int_{N}^{\infty}\right)\hat{E}(u)\int_{\mathbf{E}_{k}}\frac{d}{du}\left(\frac{f_{\varrho}(\tilde{y}u)}{u^{r+\varrho}}\right)d\mathrm{Y_{k}}du=:S_{231}+S_{232}.$

To treat $S_{231}$, we substitute Taylor’s formula for
$\Phi_{\varrho}(\tilde{y})=\frac{d}{du}\left(\frac{f_{\varrho}(\tilde{y}u)}{u^{r+\varrho}}\right)$:

(6.25)
$\displaystyle\Phi_{\varrho}(\tilde{y})=\sum_{l=0}^{J}\left(\frac{d^{l}}{dy^{l}}\Phi_{\varrho}(y)\right)\frac{(\tilde{y}-y)^{l}}{l!}+\left.\frac{d^{J+1}}{dv^{J+1}}\Phi_{\varrho}(v)\right|_{v=v_{0}}\frac{(\tilde{y}-y)^{J+1}}{(J+1)!},$

where $v_{0}=y+\theta(\tilde{y}-y)$ with some $0\leq\theta\leq 1$. Let $V_{4}$
be the contribution in $S_{231}$ from the last error term of (6.25). From
(6.16) and $v_{0}\sim y\sim x$, we have

$\displaystyle V_{4}$
$\displaystyle=\int_{M}^{N}\hat{E}(u)du\int_{\mathbf{E}_{k}}\left.\frac{d^{J+1}}{dv^{J+1}}\Phi_{\varrho}(v)\right|_{v=v_{0}}\frac{(\tilde{y}-y)^{J+1}}{(J+1)!}d\mathrm{Y}_{k}$
$\displaystyle\ll\int_{M}^{N}|\hat{E}(u)|\frac{(xu)^{\theta_{\varrho}+\frac{1}{2\alpha}+\frac{J+1}{2\alpha}}}{x^{J+1}u^{r+\varrho+1}}\frac{1}{x^{J+1}}du$
$\displaystyle\ll\int_{M}^{N}|\hat{E}(u)|\frac{(xu)^{\theta_{\varrho}-\frac{r}{2}}}{u^{r+\varrho+1}}(xu)^{\frac{r}{2}+\frac{1}{2\alpha}}\left(\frac{(xu)^{\frac{1}{2\alpha}}}{x^{2}}\right)^{J+1}du.$

Since $u\leq N=[x^{4\alpha-1-\varepsilon}]$ and
$J=[(4\alpha^{2}r+4\alpha)\varepsilon^{-1}]$, it follows that

$\displaystyle(ux)^{\frac{r}{2}+\frac{1}{2\alpha}}\left(\frac{(ux)^{\frac{1}{2\alpha}}}{x^{2}}\right)^{J+1}$
$\displaystyle\leq(x^{4\alpha-1-\varepsilon}x)^{\frac{r}{2}+\frac{1}{2\alpha}}\left(\frac{(x^{4\alpha-1-\varepsilon}x)^{\frac{1}{2\alpha}}}{x^{2}}\right)^{J+1}$
$\displaystyle\leq x^{-\varepsilon(\frac{r}{2}+\frac{1}{2\alpha})}\leq 1.$

Therefore

$\displaystyle V_{4}$ $\displaystyle\ll
x^{\theta_{\varrho}-\frac{r}{2}}\int_{M}^{N}|\hat{E}(u)|u^{-r-\frac{1}{4\alpha}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha}-1}du.$

Now by the assumption (B3) we get the upper bound of $V_{4}$:

(6.26) $V_{4}\ll
x^{\theta_{\varrho}-\frac{r}{2}}M^{\sigma_{b}^{\ast}+\omega_{1}-r-\frac{1}{4a}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha}-1}.$

Next we treat the first $J$-terms of the Taylor expansion of
$\Phi_{\varrho}(\tilde{y})$. Their contribution to $S_{231}$ is

$\displaystyle\sum_{l=0}^{J}\frac{x^{-l}}{l!}\int_{M}^{N}\hat{E}(u)\frac{d^{l}}{dy^{l}}\Phi_{\varrho}(y)du\int_{\mathbf{E}_{k}}(y_{1}+\cdots+y_{k})^{l}d\mathrm{Y}_{k}$
$\displaystyle=\sum_{l=0}^{J}c_{l}^{*}x^{-l}\int_{M}^{N}\hat{E}(u)\frac{d^{l}}{dy^{l}}\Phi_{\varrho}(y)du$
$\displaystyle=\sum_{l=0}^{J}\sum_{m=0}^{J}{\rm
Re\,}\left\\{c_{l}^{*}b_{lm}^{\prime}\int_{M}^{N}\hat{E}(u)\frac{(uy)^{\theta_{\varrho}+\frac{1}{2\alpha}+\frac{l-m}{2\alpha}}}{(xy)^{l}u^{r+\varrho+1}}e^{-ih(uy)^{\frac{1}{2\alpha}}}du\right\\}$
$\displaystyle\quad+O\left(\sum_{l=0}^{J}\int_{M}^{N}|\hat{E}(u)|\frac{(ux)^{\theta_{\varrho}+\frac{l-J}{2\alpha}}}{x^{2l}u^{r+\varrho+1}}du\right)+O\left(\sum_{l=0}^{J}x^{-l+r+\varrho-b-1}\int_{M}^{N}|\hat{E}(u)|u^{-1-b}du\right).$

The last $O$-term can be removed by taking $b$ large. In the first $O$-term,
the integrand can be transformed into

$\displaystyle|\hat{E}(u)|\frac{(ux)^{\theta_{\varrho}-\frac{J}{2\alpha}}}{u^{r+\varrho+1}}\left(\frac{(ux)^{\frac{1}{2\alpha}}}{x^{2}}\right)^{l}.$

Since $\frac{(ux)^{\frac{1}{2\alpha}}}{x^{2}}\leq 1$, this $O$-term is bounded
by

$\ll\sum_{l=0}^{J}\int_{M}^{N}|\hat{E}(u)|\frac{(ux)^{\theta_{\varrho}-\frac{J}{2\alpha}}}{u^{r+\varrho+1}}du\ll
x^{\theta_{\varrho}-\frac{J}{2\alpha}}M^{\sigma_{b}^{\ast}+\omega_{1}-\frac{r}{2}-\frac{1}{4\alpha}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha}-\frac{J}{2\alpha}-1},$

which is contained in the right hand side of (6.26). Combining these formulas
we obtain that

$\displaystyle S_{231}$ $\displaystyle=\sum_{l=0}^{J}\sum_{m=0}^{J}{\rm
Re\,}\left\\{c_{l}^{*}b_{lm}^{\prime}\int_{M}^{N}\hat{E}(u)\frac{(uy)^{\theta_{\varrho}+\frac{1}{2\alpha}+\frac{l-m}{2\alpha}}}{(xy)^{l}u^{r+\varrho+1}}e^{-ih(uy)^{\frac{1}{2\alpha}}}du\right\\}$
$\displaystyle\quad+O\left(x^{\theta_{\varrho}-\frac{r}{2}}M^{\sigma_{b}^{\ast}+\omega_{1}-r-\frac{1}{4a}-\frac{\varrho}{2\alpha}+\frac{\mu^{\prime}-\mu}{2\alpha}-1}\right).$

As for the double sum over $l$ and $m$, let $R_{2}(y;M)$ denote the term
corresponding to $l=m=0$ and $R_{3}(y;M)$ the term corresponding to $l+m>0$.
With the notation of the function $I(\lambda,M,N,y)$, we can express these
terms as

$\displaystyle R_{2}(y;M)$
$\displaystyle=y^{\theta_{\varrho}+\frac{1}{2\alpha}}{\rm
Re\,}\left\\{c_{00}I\left(\theta_{\varrho}+\frac{1}{2\alpha}-r-\varrho-1,M,N,y\right)\right\\},$
$\displaystyle R_{3}(y;M)$
$\displaystyle=\mathop{{\sum_{l=0}^{J}\sum_{m=0}^{J}}}_{l+m\neq 0}{\rm
Re\,}\left\\{c_{lm}I\left(\theta_{\varrho}+\frac{1}{2\alpha}+\frac{l-m}{2\alpha}-r-\varrho-1,M,N,y\right)\right\\}x^{-l}y^{\theta_{\varrho}+\frac{1}{2\alpha}-l+\frac{l-m}{2\alpha}}.$

Finally we consider $S_{232}$. From (6.16) with $H=k$, we have

$\displaystyle\int_{\mathbf{E}_{k}}\Phi_{\varrho}(\tilde{y})d\mathrm{Y_{k}}$
$\displaystyle=\sum_{j=0}^{k}\left\\{\sum_{m=0}^{k}{\rm
Re\,}\left(b_{jm}^{\prime\prime}\frac{x^{k}(y+\frac{j}{x})^{k+\theta_{\varrho}+\frac{1}{2\alpha}-\frac{k+m}{2\alpha}}}{u^{r+\varrho+1-\frac{1}{2\alpha}-\theta_{\varrho}+\frac{k+m}{2\alpha}}}e^{-ih((y+\frac{j}{x})u)^{\frac{1}{2\alpha}}}\right)\right.$
$\displaystyle\qquad\left.+O\left(\frac{x^{2k+\theta_{\varrho}+\frac{1}{2\alpha}-\frac{2k+1}{2\alpha}}}{u^{r+\varrho+1-\frac{1}{2\alpha}-\theta_{\varrho}+\frac{2k+1}{2\alpha}}}\right)\right\\}+O\left(x^{2k+r+\varrho-b}u^{-1-b}\right).$

Substituting this expression to the definition of $S_{232}$ and noting (6.4),
the contribution from the second $O$-term becomes

$\displaystyle\ll\int_{N}^{\infty}|\hat{E}(u)|\frac{x^{2k+\theta_{\varrho}+\frac{k}{\alpha}}}{u^{\frac{r}{2}+\frac{1}{4\alpha}+1+\frac{\varrho}{2\alpha}-\frac{\mu^{\prime}-\mu}{2\alpha}+\frac{k}{\alpha}}}du$
$\displaystyle\ll
x^{2k+\theta_{\varrho}+\frac{k}{\alpha}}N^{\sigma_{b}^{\ast}+\omega_{1}-(\frac{r}{2}+\frac{1}{4\alpha}+1+\frac{\varrho}{2\alpha}-\frac{\mu^{\prime}-\mu}{2\alpha}+\frac{k}{\alpha})},$
$\displaystyle\ll
x^{(4\alpha-1)(\sigma_{b}^{\ast}+\omega_{1})-2k+r-\varrho+2(\mu^{\prime}-\mu)+\frac{2k}{\alpha}-2\alpha
r-4\alpha}.$

The contribution from the last $O$-term is also negligible by taking $b$
large. Combining these, we deduce that

(6.27) $\displaystyle S_{232}$ $\displaystyle=\sum_{j=0}^{k}\sum_{m=0}^{k}{\rm
Re\,}\left\\{c_{jm}^{\prime}\int_{N}^{\infty}\hat{E}(u)\frac{x^{k}(y+\frac{j}{x})^{k+\theta_{\varrho}+\frac{1}{2\alpha}-\frac{k+m}{2\alpha}}}{u^{r+\varrho+1-\frac{1}{2\alpha}-\theta_{\varrho}+\frac{k+m}{2\alpha}}}e^{-ih((y+\frac{j}{x})u)^{\frac{1}{2\alpha}}}du\right\\}$
$\displaystyle\quad+O\left(x^{(4\alpha-1)(\sigma_{b}^{\ast}+\omega_{1})-2k+r-\varrho+2(\mu^{\prime}-\mu)+\frac{2k}{\alpha}-2\alpha
r-4\alpha}\right).$

The term of sum over $j$ and $m$, which we denote by $R_{4}(y;M)$, can be
expressed as

$\displaystyle R_{4}(y;M)=$ $\displaystyle
x^{k}\sum_{j=0}^{k}\sum_{m=0}^{k}{\rm
Re\,}\left\\{c_{lm}^{\prime}\left(y+\frac{j}{x}\right)^{k+\theta_{\varrho}+\frac{1}{2\alpha}-\frac{k+m}{2\alpha}}\right.$
$\displaystyle\times\left.I\left(-r-\varrho-1+\frac{1}{2\alpha}+\theta_{\varrho}-\frac{k+m}{2\alpha},N,\infty,y+\frac{j}{x}\right)\right\\}$

We call the sum of remaining terms (6.20), (6.21), (6.26) and (6.27) as
$R_{5}(y;M)$. This completes the proof of Theorem 7. ∎

##  7 Some estimates for the weighted integral of the error term

In this section we shall give several estimates involving the exponential
integral $I(\lambda,M,N,y)$ defined by (6.1), which is closely related to the
mean square bound $\eqref{psi-condition}$ and important in this paper. For
convenience we use the notation
$\int_{(\sigma)}f(s)ds=\int_{\sigma-i\infty}^{\sigma+i\infty}f(s)ds$.

We first give the following preliminary lemma.

######  Lemma 7.

Let $x\geq 1,\frac{1}{2}\leq P\leq P_{1}<P_{2}\leq(1+\delta)P$. We have

(7.1) $\displaystyle\int_{P_{1}}^{P_{2}}u^{\lambda}e^{\pm i(t\log u-ux)}du\ll
x^{-\frac{1}{2}}P^{\lambda+\frac{1}{2}}(1+|t|)^{-\varepsilon}(Px)^{\varepsilon}.$

###### Proof.

Let $f(u)=t\log u-ux$. Clearly

$f^{\prime}(u)=\frac{t}{u}-x,\quad f^{\prime\prime}(u)=-\frac{t}{u^{2}}.$

First we suppose $(1-\delta)Px\leq t\leq(1+2\delta)Px$, we apply the second
mean value theorem and the second derivative test to get

(7.2) $\displaystyle\int_{P_{1}}^{P_{2}}u^{\lambda}e^{\pm i(t\log u-ux)}du\ll
P^{\lambda}\max_{P_{1}\leq P_{3}\leq P_{4}\leq
P_{2}}\left|\int_{P_{3}}^{P_{4}}e^{\pm i(t\log u-ux)}du\right|$
$\displaystyle\ll P^{\lambda}\max_{P_{1}\leq u\leq
P_{2}}\sqrt{\frac{u^{2}}{t}}\ll x^{-\frac{1}{2}}P^{\lambda+\frac{1}{2}},$

from which the assertion of the lemma follows in this case.

If $(1+2\delta)Px<t<\infty$ or $-\infty<t<(1-\delta)Px$, by the first
derivative test we get

(7.3) $\displaystyle\int_{P_{1}}^{P_{2}}u^{\lambda}e^{\pm i(t\log u-ux)}du\ll
P^{\lambda}\max_{P_{1}\leq P_{3}<P_{4}\leq
P_{2}}\left\\{\left|\frac{t}{P_{3}}-x\right|^{-1},\left|\frac{t}{P_{4}}-x\right|^{-1}\right\\}$
$\displaystyle\ll P^{\lambda}x^{-1}(1+|t|)^{-\varepsilon}(Px)^{\varepsilon}\
(\mbox{or}\ \ \ll P^{\lambda}x^{-1}).$

The assertion of the lemma can be checked easily by the above two cases.∎

######  Lemma 8.

Let $M<N\leq x^{A}$ ($A$ a fixed positive number), $w$ be a real number and
$0<\mu_{n}\leq\frac{M}{2}$. Then we have

(7.4)
$\displaystyle\int_{x}^{(1+\delta)x}I(\lambda,M,N,y)y^{w}\cos\left(h(\mu_{n}y)^{\frac{1}{2\alpha}}+c\right)dy\ll
x^{w+1-\frac{3}{4\alpha}+\varepsilon^{\prime}}\max_{M\leq P\leq
N}P^{\lambda+\sigma^{*}+1-\frac{3}{4\alpha}}.$

###### Proof.

From (2.7) we easily get (by using Cauchy’s inequality )

(7.5) $\displaystyle\int_{-T}^{T}\left|\psi(\sigma^{*}+it)\right|^{2}dt$
$\displaystyle\ll T^{1+\varepsilon},$
$\displaystyle\int_{-T}^{T}\left|\psi(\sigma^{*}+it)\right|dt$
$\displaystyle\ll\sqrt{2T\int_{-T}^{T}\left|\psi(\sigma^{*}+it)\right|^{2}dt}\ll
T^{1+\frac{\varepsilon}{2}},$

and (by using integration by parts)

(7.6) $\displaystyle\int_{0}^{\infty}\frac{\left|\psi(\sigma^{*}\pm
it)\right|}{(1+|t|)^{1+\varepsilon}}dt=\left.\frac{\int_{0}^{t}\left|\psi(\sigma^{*}\pm
iu)\right|du}{(1+|t|)^{1+\varepsilon}}\right|_{t=0}^{\infty}+(1+\varepsilon)\int_{0}^{\infty}\frac{\int_{0}^{t}\left|\psi(\sigma^{*}\pm
iu)\right|du}{(1+|t|)^{2+\varepsilon}}dt\ll 1.$

By Perron’s formula and the residue theorem we have

$\displaystyle\hat{E}(u)$
$\displaystyle=\lim_{T\rightarrow\infty}\frac{1}{2\pi
i}\int_{\sigma_{b}^{*}+\varepsilon-
iT}^{\sigma_{b}^{*}+\varepsilon+iT}\psi(s)\frac{u^{s}}{s}ds-\hat{Q}(u)$
$\displaystyle=\lim_{T\rightarrow\infty}\frac{1}{2\pi
i}\int_{\sigma^{*}-iT}^{\sigma^{*}+iT}\psi(s)\frac{u^{s}}{s}ds,$

since $\psi(s)s^{-1}\rightarrow 0$ uniformly in the strip $\sigma^{*}\leq{\rm
Re\,}s\leq\sigma_{b}^{*}$ as $t\rightarrow\pm\infty$ (also see page 357 in
Ivić [19]). Let $\frac{1}{2}\leq P<Q\leq(1+\delta)P$, then from (6.1)

(7.7) $\displaystyle I(\lambda,P,Q,y)$
$\displaystyle=\int_{P}^{Q}u^{\lambda}e^{-ih(uy)^{\frac{1}{2\alpha}}}du\lim_{T\rightarrow\infty}\int_{\sigma^{*}-iT}^{\sigma^{*}+iT}\psi(s)\frac{u^{s}}{s}ds$
$\displaystyle=\lim_{T\rightarrow\infty}\int_{\sigma^{*}-iT}^{\sigma^{*}+iT}\frac{\psi(s)}{s}ds\int_{P}^{Q}u^{\lambda+s}e^{-ih(uy)^{\frac{1}{2\alpha}}}du$
$\displaystyle=\int_{\sigma^{*}-i\infty}^{\sigma^{*}+i\infty}\frac{\psi(s)}{s}ds\int_{P}^{Q}u^{\lambda+\sigma^{*}+it}e^{-ih(uy)^{\frac{1}{2\alpha}}}du,\quad(s=\sigma^{*}+it),$

where the inversion of order of two integrations is justified by (7.1) and
(7.6).

Assume $0<\mu_{n}\leq\frac{P}{2}$, we get

(7.8) $\displaystyle\int_{x}^{(1+\delta)x}I(\lambda,P,Q,y)y^{w}e^{\pm
h(\mu_{n}y)^{\frac{1}{2\alpha}}}dy$
$\displaystyle=\int_{x}^{(1+\delta)x}y^{w}e^{\pm
ih(\mu_{n}y)^{\frac{1}{2\alpha}}}dy\int_{\sigma^{*}-i\infty}^{\sigma^{*}+i\infty}\frac{\psi(s)}{s}ds\int_{P}^{Q}u^{\lambda+\sigma^{*}+it}e^{-ih(uy)^{\frac{1}{2\alpha}}}du$
$\displaystyle=\int_{\sigma^{*}-i\infty}^{\sigma^{*}+i\infty}\frac{\psi(s)}{s}ds\int_{P}^{Q}\int_{x}^{(1+\delta)x}y^{w}u^{\lambda+\sigma^{*}+it}e^{i(\pm
h(\mu_{n}y)^{\frac{1}{2\alpha}}-h(uy)^{\frac{1}{2\alpha}})}dydu$
$\displaystyle=:\int_{\sigma^{*}-i\infty}^{\sigma^{*}+i\infty}\frac{\psi(s)}{s}I_{1}(s)ds,$

where the inversion of order of the integrations is justified by (7.1) and
(7.6). Set
$u^{\frac{1}{2\alpha}}=U,U_{0}=P^{\frac{1}{2\alpha}},U_{1}=Q^{\frac{1}{2\alpha}}$,
$hy^{\frac{1}{2\alpha}}=Y,X_{0}=hx^{\frac{1}{2\alpha}}$ and
$X_{1}=h\left((1+\delta)x\right)^{\frac{1}{2\alpha}}$, hence

(7.9) $\displaystyle
I_{1}(s)=\frac{(2\alpha)^{2}}{h^{2\alpha(w+1)}}\int_{U_{0}}^{U_{1}}U^{2\alpha(\lambda+1+\sigma^{*}+it)-1}dU\int_{X_{0}}^{X_{1}}Y^{2\alpha(w+1)-1}e^{i(\pm\mu_{n}^{\frac{1}{2\alpha}}-U)Y}dY.$

We write $\eta=2\alpha(w+1)-1$. Applying integration by parts we get

$\displaystyle\int_{X_{0}}^{X_{1}}Y^{\eta}e^{i(\pm\mu_{n}^{\frac{1}{2\alpha}}-U)Y}dY$
$\displaystyle=\frac{1}{(\pm\mu_{n}^{\frac{1}{2\alpha}}-U)i}\left(\left.Y^{\eta}e^{i(\pm\mu_{n}^{\frac{1}{2\alpha}}-U)Y}\right|_{X_{0}}^{X_{1}}-\eta\int_{X_{0}}^{X_{1}}Y^{\eta-1}e^{i(\pm\mu_{n}^{\frac{1}{2\alpha}}-U)Y}dY\right).$

Substituting the above expression into (7.9) we obtain

$\displaystyle I_{1}(s)\ll X_{0}^{\eta}\max_{X_{0}\leq Y\leq
X_{1}}\left|\int_{U_{0}}^{U_{1}}\frac{U^{2\alpha(\lambda+1+\sigma^{*})-1}}{\pm\mu_{n}^{\frac{1}{2\alpha}}-U}e^{i(2\alpha
t\log U-UY)}dU\right|.$

Now we apply the second mean value theorem and (7.1) to get

$\displaystyle I_{1}(s)$
$\displaystyle\ll\frac{X_{0}^{\eta}}{U_{0}\mp\mu_{n}^{\frac{1}{2\alpha}}}\max_{X_{0}\leq
Y\leq X_{1},U_{0}\leq U^{*}\leq
U_{1}}\left|\int_{U_{0}}^{U^{*}}U^{2\alpha(\lambda+1+\sigma^{*})-1}e^{i(2\alpha
t\log U-UY)}dU\right|$
$\displaystyle\ll\frac{X_{0}^{\eta}}{U_{0}}X_{0}^{-\frac{1}{2}}U_{0}^{2\alpha(\lambda+1+\sigma^{*})-\frac{1}{2}}(X_{0}U_{0})^{\varepsilon}(1+|t|)^{-\varepsilon}$
$\displaystyle\ll
x^{w+1-\frac{3}{4\alpha}}P^{\lambda+1+\sigma^{*}-\frac{3}{4\alpha}}(Px)^{\varepsilon}(1+|t|)^{-\varepsilon}.$

We substitute the above estimate into (7.8) and then use (7.6) to get

$\displaystyle\int_{x}^{(1+\delta)x}I(\lambda,P,Q,y)y^{w}e^{\pm
ih(\mu_{n}y)^{\frac{1}{2\alpha}}}dy=\int_{\sigma^{*}-i\infty}^{\sigma^{*}+i\infty}\frac{\psi(s)}{s}I_{1}(s)ds$
$\displaystyle\qquad\ll
x^{w+1-\frac{3}{4\alpha}}P^{\lambda+1+\sigma^{*}-\frac{3}{4\alpha}}(Px)^{\varepsilon}\int_{-\infty}^{\infty}\frac{|\psi(\sigma^{*}+it)|}{(1+|t|)^{1+\varepsilon}}dt$
$\displaystyle\qquad\ll
x^{w+1-\frac{3}{4\alpha}}P^{\lambda+1+\sigma^{*}-\frac{3}{4\alpha}}(Px)^{\varepsilon}.$

Now applying a simple splitting argument to the interval $[M,N]$ would suffice
to complete the proof of the lemma.∎

######  Lemma 9.

Let $2(\lambda+\sigma^{*})\neq-1$, $M<N\leq x^{A}(A\mbox{ a fixed positive
number})$ and $\delta>0$ with $(1+\delta)^{\frac{1}{\alpha}}-1<\frac{1}{4}$.
Then we have

(7.10)
$\displaystyle\int_{x}^{(1+\delta)x}\left|I(\lambda,M,N,y)\right|^{2}dy\ll
x^{1-\frac{1}{\alpha}+\varepsilon^{\prime}}\max_{M\leq P\leq
N}P^{2(\lambda+\sigma^{*}+1)-\frac{1}{\alpha}}.$

###### Proof.

From (7.7) we obtain

(7.11)
$\displaystyle\int_{x}^{(1+\delta)x}\left|I(\lambda,P,Q,y)\right|^{2}dy$
$\displaystyle=\int_{x}^{(1+\delta)x}dy\left(\int_{(\sigma^{*})}\frac{\psi(s)}{s}ds\int_{P}^{Q}u^{\lambda+\sigma^{*}}e^{i\bigl{(}t\log
u-h(uy)^{\frac{1}{2\alpha}}\bigr{)}}du\right)$ $\displaystyle\quad\
\times\left(\int_{(\sigma^{*})}\frac{\psi(\bar{s}_{1})}{\bar{s}_{1}}d\bar{s}_{1}\int_{P}^{Q}v^{\lambda+\sigma^{*}}e^{i\bigl{(}-t_{1}\log
v+h(vy)^{\frac{1}{2\alpha}}\bigr{)}}dv\right)$
$\displaystyle=\int_{(\sigma^{*})}\int_{(\sigma^{*})}\frac{\psi(s)\psi(\bar{s}_{1})}{s\bar{s}_{1}}dsd\bar{s}_{1}\times
I(s,s_{1}),$

where the inversion of order of the integrations is justified by (7.1) and
(7.6), and we use notation $s=\sigma^{*}+it$, $s_{1}=\sigma^{*}+it_{1}$ and

$\displaystyle
I(s,s_{1}):=\int_{x}^{(1+\delta)x}dy\int_{P}^{Q}u^{\lambda+\sigma^{*}}e^{i\bigl{(}t\log
u-h(uy)^{\frac{1}{2\alpha}}\bigr{)}}du\int_{P}^{Q}v^{\lambda+\sigma^{*}}e^{i\bigl{(}-t_{1}\log
v+h(vy)^{\frac{1}{2\alpha}}\bigr{)}}dv.$

Set
$h(uy)^{\frac{1}{2\alpha}}=U,h(vy)^{\frac{1}{2\alpha}}=V,P^{\frac{1}{2\alpha}}=U_{0},Q^{\frac{1}{2\alpha}}=U_{1},hy^{\frac{1}{2\alpha}}=Y,hx^{\frac{1}{2\alpha}}=X_{0},h((1+\delta)x)^{\frac{1}{2\alpha}}=X_{1}$,
$t_{2}=2\alpha t,t_{3}=2\alpha t_{1}$, $\mu=2\alpha(\lambda+\sigma^{*}+1)-1$
and $\beta=-4\alpha(\lambda+\sigma^{*})-2\alpha-1$. We have

(7.12) $\displaystyle I(s,s_{1})=$
$\displaystyle\frac{(2\alpha)^{3}}{h^{2\alpha}}\int_{X_{0}}^{X_{1}}Y^{\beta}e^{i(t_{3}-t_{2})\log
Y}G(Y;t_{2},t_{3})dY,$

where

$\displaystyle G(Y;t_{2},t_{3}):=$
$\displaystyle\int_{U_{0}Y}^{U_{1}Y}U^{\mu}e^{i(t_{2}\log
U-U)}dU\int_{U_{0}Y}^{U_{1}Y}V^{\mu}e^{i(-t_{3}\log V+V)}dV.$

Since $P<Q\leq(1+\delta)P$, so
$U_{0}X_{0}<U_{1}X_{1}\leq(1+\delta)^{\frac{1}{\alpha}}U_{0}X_{0}:=(1+\delta_{1})U_{0}X_{0}$.
Here note that $0<\delta_{1}=(1+\delta)^{\frac{1}{\alpha}}-1<\frac{1}{4}.$

To estimate $I(s,s_{1})$ we need to consider the following four cases.

(Case 1).
$t_{2},t_{3}\notin\left((1-2\delta_{1})U_{0}X_{0},(1+4\delta_{1})U_{0}X_{0}\right)$.
In this case applying (7.3) to estimate the two integrals in
$G(Y;t_{2},t_{3})$ we obtain

$\displaystyle G(Y;t_{2},t_{3})$ $\displaystyle\ll\max_{X_{0}\leq Y\leq
X_{1}}\left|\int_{U_{0}Y}^{U_{1}Y}U^{\mu}e^{i(t_{2}\log
U-U)}dU\right|\times\max_{X_{0}\leq Y\leq
X_{1}}\left|\int_{U_{0}Y}^{U_{1}Y}V^{\mu}e^{i(-t_{3}\log V+V)}dV\right|$
$\displaystyle\ll(U_{0}X_{0})^{2\mu}(U_{0}X_{0})^{2\varepsilon}(1+|t_{2}|)^{-\varepsilon}(1+|t_{3}|)^{-\varepsilon}.$

We substitute the above estimate into (7.12) to get

(7.13) $\displaystyle I(s,s_{1})\ll
U_{0}^{2\mu}X_{0}^{2\mu+\beta+1}(U_{0}X_{0})^{2\varepsilon}(1+|t_{2}|)^{-\varepsilon}(1+|t_{3}|)^{-\varepsilon}.$

(Case 2).
$t_{2}\in\left((1-2\delta_{1})U_{0}X_{0},(1+4\delta_{1})U_{0}X_{0}\right)$,
$t_{3}\notin\left((1-2\delta_{1})U_{0}X_{0},(1+4\delta_{1})U_{0}X_{0}\right)$.
If $|t_{2}-t_{3}|\leq\delta_{1}U_{0}X_{0}$, then
$t_{2}<(1-\delta_{1})U_{0}X_{0}$ or $t_{2}>(1+3\delta_{1})U_{0}X_{0}$. Clearly
the estimate (7.13) also holds.

Now we assume $|t_{2}-t_{3}|>\delta_{1}U_{0}X_{0}$. Applying integration by
parts to the variable $Y$ in (7.12) we get

(7.14)
$\displaystyle\frac{h^{2\alpha}}{(2\alpha)^{3}}I(s,s_{1})=\int_{X_{0}}^{X_{1}}Y^{\beta}e^{i(t_{3}-t_{2})\log
Y}G(Y;t_{2},t_{3})dY$
$\displaystyle=\frac{1}{\beta+1+(t_{3}-t_{2})i}\int_{X_{0}}^{X_{1}}G(Y;t_{2},t_{3})dY^{\beta+1+i(t_{3}-t_{2})}$
$\displaystyle=\frac{1}{\beta+1+(t_{3}-t_{2})i}\left(\sum_{j=0}^{1}(-1)^{j+1}X_{j}^{\beta+1}e^{i(t_{3}-t_{2})\log
X_{j}}G(X_{j};t_{2},t_{3})\right.$ $\displaystyle\
\left.+\sum_{j=0}^{1}(-1)^{j}U_{j}^{\mu+1+it_{2}}I_{1j}+\sum_{j=0}^{1}(-1)^{j}U_{j}^{\mu+1-it_{3}}I_{2j}\right),$

where

$\displaystyle I_{1j}:=\int_{X_{0}}^{X_{1}}Y^{\beta+\mu+1}e^{i(t_{3}\log
Y-U_{j}Y)}dY\int_{U_{0}Y}^{U_{1}Y}V^{\mu}e^{i(-t_{3}\log V+V)}dV$
$\displaystyle I_{2j}:=\int_{X_{0}}^{X_{1}}Y^{\beta+\mu+1}e^{i(-t_{2}\log
Y+U_{j}Y)}dY\int_{U_{0}Y}^{U_{1}Y}U^{\mu}e^{i(t_{2}\log U-U)}dU.$

Applying (7.2) to the first integral over $U$ in $G(X_{j};t_{2},t_{3})$, and
(7.3) to the second integral over $V$ in $G(X_{j};t_{2},t_{3})$ respectively,
we obtain

(7.15) $\displaystyle
G(X_{j};t_{2},t_{3})\ll(U_{0}X_{0})^{\mu+\frac{1}{2}}(U_{0}X_{0})^{\mu}\ll(U_{0}X_{0})^{2\mu+\frac{1}{2}},\quad(j=0,1).$

Applying (7.3) to $I_{1j}$ one has

(7.16) $\displaystyle I_{1j}$ $\displaystyle\ll
X_{0}^{\beta+\mu+2}\max_{X_{0}\leq Y\leq
X_{1}}\left|\int_{U_{0}Y}^{U_{1}Y}V^{\mu}e^{i(-t_{3}\log V+V)}dV\right|$
$\displaystyle\ll X_{0}^{\beta+\mu+2}(U_{0}X_{0})^{\mu}\ll
U_{0}^{\mu}X_{0}^{\beta+2\mu+2}.$

To treat $I_{2j}$ we define for $j=0,1$

$X_{2j}=\frac{t_{2}}{U_{j}}-\sqrt{\frac{X_{0}}{U_{j}}}\quad
X_{3j}=\frac{t_{2}}{U_{j}}+\sqrt{\frac{X_{0}}{U_{j}}}.$

First we consider the case when $X_{0}\leq X_{2j}<X_{3j}\leq X_{1}$, we divide
the interval $[X_{0},X_{1}]$ into three subintervals and write

(7.17) $\displaystyle
I_{2j}=\int_{X_{0}}^{X_{2j}}+\int_{X_{2j}}^{X_{3j}}+\int_{X_{3j}}^{X_{1}}:=I_{3j}+I_{4j}+I_{5j}.$

We let $\kappa=\beta+\mu+1$ and
$H(Y,t_{2}):=\int_{U_{0}Y}^{U_{1}Y}U^{\mu}e^{i(t_{2}\log U-U)}dU$. It follows
that from integration by parts

(7.18) $\displaystyle I_{3j}$
$\displaystyle=\int_{X_{0}}^{X_{2j}}Y^{\kappa}H(Y,t_{2})e^{i(-t_{2}\log
Y+U_{j}Y)}dY$
$\displaystyle=\int_{X_{0}}^{X_{2j}}\frac{Y^{\kappa+1}}{i(-t_{2}+U_{j}Y)}H(Y,t_{2})de^{i(-t_{2}\log
Y+U_{j}Y)}$ $\displaystyle=I_{6j}+I_{7j}-I_{8j},$

where***In Tong’s paper[49], page 521, he missed the term $I_{8j}$.

$\displaystyle I_{6j}:$
$\displaystyle=\left.\frac{Y^{\kappa+1}}{i(-t_{2}+U_{j}Y)}H(Y,t_{2})e^{i(-t_{2}\log
Y+U_{j}Y)}\right|_{X_{0}}^{X_{2j}},$ $\displaystyle I_{7j}:$
$\displaystyle=\sum_{l=0}^{1}(-1)^{l}U_{l}^{\mu+1+it_{2}}\int_{X_{0}}^{X_{2j}}\frac{Y^{\kappa+\mu+1}}{i(t_{2}-U_{j}Y)}e^{i(U_{j}-U_{l})Y}dY,$
$\displaystyle I_{8j}:$
$\displaystyle=\int_{X_{0}}^{X_{2j}}\left(\frac{(\kappa+1)Y^{\kappa}}{i(-t_{2}+U_{j}Y)}-\frac{Y^{\kappa+1}U_{j}}{i(-t_{2}+U_{j}Y)^{2}}\right)H(Y,t_{2})e^{i(-t_{2}\log
Y+U_{j}Y)}dY.$

By (7.2) we get

$\displaystyle I_{6j}$ $\displaystyle\ll X_{0}^{\kappa+1}\max_{X_{0}\leq Y\leq
X_{2j}}\frac{1}{|t_{2}-U_{j}Y|}\times(U_{0}X_{0})^{\mu+\frac{1}{2}}$
$\displaystyle\ll
X_{0}^{\kappa+1}\frac{1}{\sqrt{X_{0}U_{0}}}(U_{0}X_{0})^{\mu+\frac{1}{2}}\ll
U_{0}^{\mu}X_{0}^{\kappa+\mu+1}.$

We also have

$\displaystyle I_{7j}\ll
U_{0}^{\mu+1}\int_{X_{0}}^{X_{2j}}\frac{X_{0}^{\kappa+\mu+1}}{t_{2}-U_{j}Y}dY\ll
U_{0}^{\mu}X_{0}^{\kappa+\mu+1}\log(U_{0}X_{0}).$

Applying (7.2) to the integral $H(Y,t_{2})$ in $I_{8j}$ we obtain

$\displaystyle I_{8j}$
$\displaystyle\ll\int_{X_{0}}^{X_{2j}}\left(\frac{X_{0}^{\kappa}}{t_{2}-U_{j}Y}+\frac{X_{0}^{\kappa+1}U_{0}}{(t_{2}-U_{j}Y)^{2}}\right)(U_{0}X_{0})^{\mu+\frac{1}{2}}dY$
$\displaystyle\ll
U_{0}^{\mu+\frac{1}{2}}X_{0}^{\kappa+\mu+\frac{1}{2}}\int_{X_{0}}^{X_{2j}}\frac{1}{t_{2}-U_{j}Y}dY+U_{0}^{\mu+\frac{3}{2}}X_{0}^{\kappa+\mu+\frac{3}{2}}\int_{X_{0}}^{X_{2j}}\frac{1}{(t_{2}-U_{j}Y)^{2}}dY$
$\displaystyle\ll
U_{0}^{\mu-\frac{1}{2}}X_{0}^{\kappa+\mu+\frac{1}{2}}\log(U_{0}X_{0})+U_{0}^{\mu}X_{0}^{\kappa+\mu+1}\ll
U_{0}^{\mu}X_{0}^{\kappa+\mu+1}.$

Combining the above three estimates and (7.18) we have

(7.19) $\displaystyle I_{3j}\ll
U_{0}^{\mu}X_{0}^{\kappa+\mu+1}\log(U_{0}X_{0}).$

Similarly to the estimate for $I_{8j}$ one has

(7.20) $\displaystyle I_{4j}$
$\displaystyle\ll\int_{X_{2j}}^{X_{3j}}X_{0}^{\kappa}(U_{0}X_{0})^{\mu+\frac{1}{2}}dY\ll(X_{3j}-X_{2j})U_{0}^{\mu+\frac{1}{2}}X_{0}^{\kappa+\mu+\frac{1}{2}}$
$\displaystyle\ll\sqrt{\frac{X_{0}}{U_{0}}}U_{0}^{\mu+\frac{1}{2}}X_{0}^{\kappa+\mu+\frac{1}{2}}\ll
U_{0}^{\mu}X_{0}^{\kappa+\mu+1}.$

Similarly to the estimate for $I_{3j}$ we can prove

(7.21) $\displaystyle I_{5j}\ll
U_{0}^{\mu}X_{0}^{\kappa+\mu+1}\log(U_{0}X_{0}).$

Substituting (7.19), (7.20) and (7.21) into (7.17) we obtain when $X_{0}\leq
X_{2j}<X_{3j}\leq X_{1}$

$\displaystyle I_{2j}\ll U_{0}^{\mu}X_{0}^{\kappa+\mu+1}\log(U_{0}X_{0}).$

It remains to consider the other four cases: $X_{2j}<X_{3j}\leq X_{0}<X_{1}$,
$X_{2j}\leq X_{0}<X_{3j}\leq X_{1}$, $X_{0}\leq X_{2j}\leq X_{1}<X_{3j}$ and
$X_{0}\leq X_{1}\leq X_{2j}<X_{3j}$. By a very similar approach to the case
$X_{0}\leq X_{2j}<X_{3j}\leq X_{1}$, one may get the same estimate of $I_{2j}$
for the above four cases. For example, when $\leq X_{2j}\leq X_{0}<X_{3j}\leq
X_{1}$ we use decomposition

$I_{2j}=\int_{X_{0}}^{X_{1}}=\int_{X_{0}}^{X_{3j}}+\int_{X_{3j}}^{X_{1}}.$

Hence we always have

(7.22) $\displaystyle I_{2j}\ll
U_{0}^{\mu}X_{0}^{\beta+2\mu+2}\log(U_{0}X_{0}).$

Since $|t_{2}-t_{3}|>\delta_{1}U_{0}X_{0}$ and
$t_{2}\in\left((1-2\delta_{1})U_{0}X_{0},(1+4\delta_{1})U_{0}X_{0}\right)$, it
is easy to check

$|t_{2}-t_{3}|>(U_{0}X_{0})^{1-2\varepsilon}(1+|t_{2}|)^{-\varepsilon}(1+|t_{3}|)^{-\varepsilon}.$

Substituting (7.15), (7.16) and (7.22) into (7.14), and noting the above
estimate, in (Case 2) we finally obtain

(7.23) $\displaystyle I(s,s_{1})\ll
U_{0}^{2\mu}X_{0}^{2\mu+\beta+1}(U_{0}X_{0})^{3\varepsilon}(1+|t_{2}|)^{-\varepsilon}(1+|t_{3}|)^{-\varepsilon}.$

(Case 3).
$t_{2}\notin\left((1-2\delta_{1})U_{0}X_{0},(1+4\delta_{1})U_{0}X_{0}\right)$,
$t_{3}\in\left((1-2\delta_{1})U_{0}X_{0},(1+4\delta_{1})U_{0}X_{0}\right)$.

We can use a similar approach as (Case 2) to estimate $I(s,s_{1})$ and get the
same estimate as (7.23).

(Case 4).
$t_{2},t_{3}\in\left((1-2\delta_{1})U_{0}X_{0},(1+4\delta_{1})U_{0}X_{0}\right)$.

Note that $\beta+1+i(t_{3}-t_{2})\neq 0$. We first transform $I(s,s_{1})$ into
the expression (7.14). Then we use the same approach as $I_{2j}$ in (Case 2)
to treat $I_{2j}$ and estimate $I_{1j}$ respectively. Lastly by the second
derivative test (7.2) to each integral in $G(X_{j};t_{2},t_{3})$, we get

$\displaystyle
G(X_{j};t_{2},t_{3})\ll(U_{0}X_{0})^{\mu+\frac{1}{2}}(U_{0}X_{0})^{\mu+\frac{1}{2}}\ll(U_{0}X_{0})^{2\mu+1},\quad(j=0,1).$

Combining these estimates and (7.14) we obtain

$\displaystyle
I(s,s_{1})\ll\frac{U_{0}^{2\mu+1}X_{0}^{\beta+2\mu+2}}{1+|t_{2}-t_{3}|}(U_{0}X_{0})^{\varepsilon}$

Now let $h(Px)^{\frac{1}{2\alpha}}=T$. Note that $2\mu+\beta+1=2\alpha-2$.
Combining four estimates in the above cases and (7.11) we get

(7.24)
$\displaystyle\int_{x}^{(1+\delta)x}\left|I(\lambda,P,Q,y)\right|^{2}dy=\int_{(\sigma^{*})}\int_{(\sigma^{*})}\frac{\psi(s)\psi(\bar{s}_{1})}{s\bar{s}_{1}}I(s,s_{1})dsd\bar{s}_{1}$
$\displaystyle\ll
x^{1-\frac{1}{\alpha}}P^{\frac{\mu}{\alpha}}T^{3\varepsilon}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{|\psi(\sigma^{*}+it)\psi(\sigma^{*}-it_{1})|}{(1+|t|)(1+|t_{1}|)}\frac{dtdt_{1}}{(1+|t|)^{\varepsilon}(1+|t_{1}|)^{\varepsilon}}$
$\displaystyle\quad\quad+x^{1-\frac{1}{2\alpha}}P^{\frac{2\mu+1}{2\alpha}}T^{\varepsilon}\int_{(1-2\delta_{1})T}^{(1+4\delta_{1})T}\int_{(1-2\delta_{1})T}^{(1+4\delta_{1})T}\frac{|\psi(\sigma^{*}+it)\psi(\sigma^{*}-it_{1})|}{tt_{1}}\frac{dtdt_{1}}{1+|t-t_{1}|}$
$\displaystyle=:x^{1-\frac{1}{\alpha}}P^{\frac{\mu}{\alpha}}T^{3\varepsilon}W_{1}+x^{1-\frac{1}{2\alpha}}P^{\frac{2\mu+1}{2\alpha}}T^{\varepsilon}W_{2}.$

It follows from (7.6) that

(7.25) $\displaystyle
W_{1}=\int_{-\infty}^{\infty}\frac{|\psi(\sigma^{*}+it)|}{(1+|t|)^{1+\varepsilon}}dt\times\int_{-\infty}^{\infty}\frac{|\psi(\sigma^{*}-it_{1})|}{(1+|t_{1}|)^{1+\varepsilon}}dt_{1}\ll
1.$

Making the change of variables $t=u+v,t_{1}=u-v$, using Cauchy’s inequality
and noting (7.5), it is easy to check

(7.26) $\displaystyle W_{2}$ $\displaystyle\ll
T^{-2}\int_{0}^{2T}\int_{0}^{2T}\frac{|\psi(\sigma^{*}+it)\psi(\sigma^{*}-it_{1})|}{1+|t-t_{1}|}dtdt_{1}$
$\displaystyle\ll
T^{-2}\int_{-T}^{T}\frac{1}{1+|v|}\left(\int_{0}^{2T}|\psi(\sigma^{*}+i(u+v))\psi(\sigma^{*}-i(u-v))|du\right)dv$
$\displaystyle\ll
T^{-2}\int_{-T}^{T}\frac{1}{1+|v|}\left(\int_{0}^{2T}|\psi(\sigma^{*}+i(u+v))|^{2}du\int_{0}^{2T}|\psi(\sigma^{*}-i(u-v))|^{2}du\right)^{\frac{1}{2}}dv$
$\displaystyle\ll T^{-1+\varepsilon}\int_{-T}^{T}\frac{1}{1+|v|}dv\ll
T^{-1+2\varepsilon}.$

Now substituting (7.25) and (7.26) into (7.24) we get with
$\varepsilon_{1}=\frac{3\varepsilon}{2\alpha}$

(7.27)
$\displaystyle\int_{x}^{(1+\delta)x}\left|I(\lambda,P,Q,y)\right|^{2}dy\ll
x^{1-\frac{1}{\alpha}+\varepsilon_{1}}P^{\frac{\mu}{\alpha}+\varepsilon_{1}}.$

Applying a simple splitting argument to the interval $[M,N]$ with
$M_{0}=M,M_{r}=N,M_{j+1}\leq(1+\delta)M_{j}\ (j=0,\ldots,r-1;r\ll\log x)$ and
Cauchy’s inequality, it follows that from (7.27)

(7.28)
$\displaystyle\int_{x}^{(1+\delta)x}\left|I(\lambda,M,N,y)\right|^{2}dy\ll\int_{x}^{(1+\delta)x}\left(\sum_{j=0}^{r-1}\left|I(\lambda,M_{j},M_{j+1},y)\right|\right)^{2}dy$
$\displaystyle=\sum_{j=0}^{r-1}\sum_{l=0}^{r-1}\int_{x}^{(1+\delta)x}\left|I(\lambda,M_{j},M_{j+1},y)\right|\left|I(\lambda,M_{l},M_{l+1},y)\right|dy$
$\displaystyle\ll\sum_{j=0}^{r-1}\sum_{l=0}^{r-1}\sqrt{\int_{x}^{(1+\delta)x}\left|I(\lambda,M_{j},M_{j+1},y)\right|^{2}dy\times\int_{x}^{(1+\delta)x}\left|I(\lambda,M_{l},M_{l+1},y)\right|^{2}dy}$
$\displaystyle=\left(\sum_{j=0}^{r-1}\sqrt{\int_{x}^{(1+\delta)x}\left|I(\lambda,M_{j},M_{j+1},y)\right|^{2}dy}\right)^{2}\ll
x^{1-\frac{1}{\alpha}+\varepsilon_{1}}\left(\sum_{j=0}^{r-1}M_{j}^{\frac{\mu}{2\alpha}+\frac{\varepsilon_{1}}{2}}\right)^{2}.$

Recall $\mu=2\alpha(\lambda+\sigma^{*}+1)-1$. Now Lemma 9 follows from (7.28)
at once.∎

######  Lemma 10.

Let $2(\lambda+\sigma^{*})\neq-1$, $2(\lambda+\sigma^{*}+1)<\frac{1}{\alpha}$,
$M\geq 1$ , and $\delta>0$ with $(1+\delta)^{\frac{1}{\alpha}}-1<\frac{1}{4}$.
Then we have

(7.29)
$\displaystyle\int_{x}^{(1+\delta)x}\left|I(\lambda,M,\infty,y)\right|^{2}dy\ll
x^{1-\frac{1}{\alpha}+\varepsilon^{\prime}}M^{2(\lambda+\sigma^{*}+1)-\frac{1}{\alpha}}.$

###### Proof.

We take $M_{j}=M(1+\delta_{1})^{j},\ (j=0,1,\cdots,r-1)$. The lemma is
immediately obtained from (7.28) and let $r\rightarrow\infty$. ∎

######  Remark 11.

In Tong’s original proof (similar to Lemma 9) he used the second mean value
theorem for complex integral, but we don’t. Moreover, in both Lemma 9 and
Lemma 10 the condition $2(\lambda+\sigma^{*})\neq-1$ is not essential. In
fact, we can use the second mean value theorem to remove this restriction.

##  8 The proof of Theorem 1

In this section we shall prove Theorem 1.

Suppose that ${\cal L}(s)\in{\cal S}_{real}^{\theta}$ is a function of degree
$d\geq 2$ such that $0\leq\theta\leq 1/2-1/2d.$ So ${\cal L}(s)$ satisfies the
functional equation (1.1) in Section 1. We assume $\sum_{j=1}^{L}\beta_{j}$ is
real where $\beta_{j}$ are the constants in (1.2). From (2.3) and (3.3)–(3.6),
we have obviously
$d=2\alpha,\mu=\mu^{\prime},\nu=\nu^{\prime},\lambda=\lambda^{\prime},\theta_{\varrho}=1/2-1/2d+\varrho(1-1/d)$.
We also have $\varphi(s)={\cal L}(s)$ and $\psi(s)=\omega Q^{2s-1}{\cal
L}(s)$, hence $b(n)=\omega Q^{-1}a(n)$, $\lambda_{n}=n$ and $\mu_{n}=Q^{-2}n$.
Suppose that $1/2\leq\sigma^{*}<1$ satisfies (1.5) and (1.6).

Suppose $T$ is a large parameter and $\delta>0$ is a small positive constant.
We shall evaluate the integral $\int_{T}^{(1+\delta)T}E^{2}(y)dy.$

###  8.1 Tong’s formula of $E(y)$

We first show that

(8.1) $\int_{1}^{T}|E(y)|dy\ll T^{1+\frac{d^{2}-1}{2d^{2}}+\varepsilon}.$

Since ${\cal L}(s)$ is a function in ${\cal S}_{real}^{\theta}$ of degree $d$,
we have trivially that ${\cal L}(it)\ll(1+|t|)^{d/2+\varepsilon/2}$ and thus

(8.2) $\int_{0}^{T}|{\cal L}(it)|^{2}dt\ll T^{d+1+\varepsilon}.$

From (1.5), (8.2) and Lemma 8.3 of Ivić [19] we get for
$0\leq\sigma\leq\sigma^{*}$

(8.3) $\int_{0}^{T}|{\cal L}(\sigma+it)|^{2}dt\ll
T^{\frac{(d+1)(\sigma^{*}-\sigma)}{\sigma^{*}}+\frac{\sigma}{\sigma^{*}}+\varepsilon}.$

Define $\hat{E}(y)=A(y)-\operatorname*{Res}_{s=1}{\cal L}(s)y^{s}s^{-1}.$ Then

$E(y)=\hat{E}(y)-{\cal L}(0).$

By Perron’s formula we have for some $\sigma^{*}<\sigma<1$ that

$\hat{E}(y)=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}{\cal
L}(s)y^{s}s^{-1}ds.$

From (8.3) and integration by parts we see that

$\int_{-\infty}^{\infty}|{\cal L}(\sigma_{0}+it)|^{2}|\sigma_{0}+it|^{-2}dt\ll
1$

for $\sigma_{0}=\frac{(d-1)\sigma^{*}+2\sigma^{*}\varepsilon}{d}$. So we have

$\hat{E}(y)=\frac{1}{2\pi
i}\int_{\sigma_{0}-i\infty}^{\sigma_{0}+i\infty}{\cal L}(s)y^{s}s^{-1}ds.$

Replacing $y$ in the above formula by $1/y$, and then using Parseval’s
identity (see the formula (A.5) of Ivić [19]) we have

$\frac{1}{2\pi}\int_{-\infty}^{\infty}|{\cal
L}(\sigma_{0}+it)|^{2}|\sigma_{0}+it|^{-2}dt=\int_{0}^{\infty}|\hat{E}(1/y)|^{2}y^{2\sigma_{0}-1}dy=\int_{0}^{\infty}|\hat{E}(y)|^{2}y^{-2\sigma_{0}-1}dy.$

Thus we get for any $D>1$ that

$\int_{D}^{2D}|\hat{E}(y)|^{2}dy\ll D^{2\sigma_{0}+1},$

which with Cauchy’s inequality gives

$\int_{D}^{2D}|\hat{E}(y)|dy\ll D^{\sigma_{0}+1}.$

By the splitting argument we get

$\int_{0}^{T}|\hat{E}(y)|dy\ll T^{\sigma_{0}+1}.$

Recalling $E(y)=\hat{E}(y)-{\cal L}(0)$ and (1.5) we get

$\int_{0}^{T}|E(y)|dy\ll\int_{0}^{T}|\hat{E}(y)|dy+T\ll T^{\sigma_{0}+1}\ll
T^{1+\frac{d^{2}-1}{2d^{2}}+\varepsilon},$

namely (8.1) holds.

Now we use Theorem 7. Take $\varrho=0.$ From the first estimate of (1.4) we
take $\omega_{0}=\theta+\varepsilon$. From (8.1) we take
$\omega_{1}=(d^{2}-1)/2d^{2}<1/2$. Suppose $k>1$ is a large but fixed integer
which satisfies (6.2). Take $x=T,N=[T^{2d-1-\varepsilon}]$. $M$ is a real
number such that $T^{\varepsilon}\ll M\ll\sqrt{N}$ and $\hat{E}(M)=0$ if
$b(n)\geq 0\ \ (n\geq 1)$ and $\hat{E}(M)\ll M^{\theta+\varepsilon}$
otherwise.

The truncated Tong-type formula of $E(y)$ was given in Theorem 7. Since we are
not interested in the exact value of constants like $\kappa_{0}$, we may
assume $Q=1$ for simplicity of notation. Hence we may take $\mu_{n}=n$ in the
rest of this section. Then from (6.5)–(6.11) of Theorem 7 we have

(8.4) $E(y)=\sum_{j=1}^{7}R_{j}(y),$

where

$\displaystyle R_{1}(y)$
$\displaystyle=\kappa_{0}y^{\frac{d-1}{2d}}\sum_{n\leq
M}\frac{a(n)}{n^{\frac{d+1}{2d}}}\cos\left(h(yn)^{1/d}+c_{0}\pi\right),$
$\displaystyle R_{2}(y)$ $\displaystyle=y^{\frac{d+1}{2d}}{\rm
Re\,}\left(c_{00}I\left(-\frac{3}{2}+\frac{1}{2d},M,N,y\right)\right),$
$\displaystyle R_{3}(y)$ $\displaystyle=\sum_{\begin{subarray}{c}0\leq l,m\leq
J\\\ l+m>0\end{subarray}}{\rm
Re\,}\left(c_{lm}I\left(-\frac{3}{2}+\frac{1}{2d}+\frac{l-m}{d},M,N,y\right)\right)T^{-l}y^{-l+\frac{1}{2}+\frac{1}{2d}+\frac{l-m}{d}},$
$\displaystyle R_{4}(y)$ $\displaystyle=\sum_{0\leq l,m\leq k}{\rm
Re\,}\left(c_{lm}^{\prime}I\left(-\frac{3}{2}+\frac{1}{2d}-\frac{k+m}{d},N,\infty,y+\frac{l}{T}\right)\right)T^{k}\left(y+\frac{l}{T}\right)^{k+\frac{1}{2}+\frac{1}{2d}-\frac{k+m}{d}},$
$\displaystyle R_{5}(y)$ $\displaystyle\ll
T^{\frac{1}{2}-\frac{3}{2d}}M^{\max(\frac{1}{2}-\frac{3}{2d},0)}\log^{A}T+T^{-\frac{3}{2}+\frac{1}{2d}}M^{\frac{1}{2}+\frac{1}{2d}}$
$\displaystyle\quad+T^{-\frac{1}{2d}}M^{\omega_{1}-1-\frac{1}{2d}}+T^{(2d-1)(1+\omega_{1})-2k+1+4k/d-3d}$
$\displaystyle\ll
T^{\frac{1}{2}-\frac{3}{2d}}M^{\max(\frac{1}{2}-\frac{3}{2d},0)}\log^{A}T,$
$\displaystyle R_{6}(y)$ $\displaystyle\begin{cases}=0&\mbox{if $b(n)\geq
0$}\\\ \ll
T^{\frac{1}{2}-\frac{1}{2d}}M^{\theta-\frac{1}{2}-\frac{1}{2d}+\varepsilon}&\mbox{otherwise},\end{cases}$
$\displaystyle R_{7}(y)$
$\displaystyle=E(y)-\int_{\mathbf{E}_{k}}E(\tilde{y})d\mathrm{Y}_{k}.$

For simplicity, we write

$E(y)=K_{1}+K_{2},$

where

$\displaystyle K_{1}=R_{1}(y)+R_{2}(y),$ $\displaystyle
K_{2}=R_{3}(y)+R_{4}(y)+R_{5}(y)+R_{6}(y)+R_{7}(y).$

###  8.2 Evaluation of $\int_{T}^{(1+\delta)T}K_{1}^{2}dy$

In this subsection we shall evaluate $\int_{T}^{(1+\delta)T}K_{1}^{2}dy$.

We write

(8.5) $\displaystyle R_{1}^{2}(y)$
$\displaystyle=\kappa_{0}^{2}y^{\frac{d-1}{d}}\sum_{n,m\leq
M}\frac{a(n)a(m)}{(nm)^{\frac{d+1}{2d}}}\cos\left(h(yn)^{1/d}+c_{0}\pi\right)\cos\left(h(ym)^{1/d}+c_{0}\pi\right)$
$\displaystyle=\frac{\kappa_{0}^{2}}{2}y^{\frac{d-1}{d}}\sum_{n,m\leq
M}\frac{a(n)a(m)}{(nm)^{\frac{d+1}{2d}}}\left(\cos\left(hy^{1/d}(n^{1/d}-m^{1/d})\right)+\cos\left(hy^{1/d}(n^{1/d}+m^{1/d})+2c_{0}\pi\right)\right)$
$\displaystyle=W_{1}(y)+W_{2}(y)+W_{3}(y),$

where

$\displaystyle W_{1}(y)$
$\displaystyle=\frac{\kappa_{0}^{2}}{2}y^{\frac{d-1}{d}}\sum_{n\leq
M}\frac{a^{2}(n)}{n^{\frac{d+1}{d}}},$ $\displaystyle W_{2}(y)$
$\displaystyle=\frac{\kappa_{0}^{2}}{2}y^{\frac{d-1}{d}}\sum_{\begin{subarray}{c}n,m\leq
M\\\
n\not=m\end{subarray}}\frac{a(n)a(m)}{(nm)^{\frac{d+1}{2d}}}\cos\left(hy^{1/d}(n^{1/d}-m^{1/d})\right),$
$\displaystyle W_{3}(y)$
$\displaystyle=\frac{\kappa_{0}^{2}}{2}y^{\frac{d-1}{d}}\sum_{n,m\leq
M}\frac{a(n)a(m)}{(nm)^{\frac{d+1}{2d}}}\cos\left(hy^{1/d}(n^{1/d}+m^{1/d})+2c_{0}\pi\right).$

For $W_{1}(y),$ we have

(8.6) $\displaystyle\int_{T}^{(1+\delta)T}W_{1}(y)dy$
$\displaystyle=\frac{\kappa_{0}^{2}}{2}\sum_{n\leq
M}\frac{a^{2}(n)}{n^{\frac{d+1}{d}}}\int_{T}^{(1+\delta)T}y^{\frac{d-1}{d}}dy$
$\displaystyle=\frac{\kappa_{0}^{2}}{2}\sum_{n=1}^{\infty}\frac{a^{2}(n)}{n^{\frac{d+1}{d}}}\int_{T}^{(1+\delta)T}y^{\frac{d-1}{d}}dy+O(T^{2-1/d+\varepsilon}M^{-1/d}),$

where in the last step we used the second estimate of (1.4).

By the first derivative test we have

$\displaystyle\int_{T}^{(1+\delta)T}W_{2}(y)dy$ $\displaystyle\ll
T^{2-2/d}\sum_{\begin{subarray}{c}n,m\leq M\\\
n\not=m\end{subarray}}\frac{|a(n)a(m)|}{(nm)^{\frac{d+1}{2d}}|n^{1/d}-m^{1/d}|}$
$\displaystyle=T^{2-2/d}(\Sigma_{1}+\Sigma_{2}),$

say, where

$\displaystyle\Sigma_{1}$ $\displaystyle=\sum_{\begin{subarray}{c}n,m\leq M\\\
|n^{1/d}-m^{1/d}|\geq(mn)^{1/2d}/10d\end{subarray}}\frac{|a(n)a(m)|}{(nm)^{\frac{d+1}{2d}}|n^{1/d}-m^{1/d}|},$
$\displaystyle\Sigma_{2}$ $\displaystyle=\sum_{\begin{subarray}{c}n,m\leq M\\\
0<|n^{1/d}-m^{1/d}|<(mn)^{1/2d}/10d\end{subarray}}\frac{|a(n)a(m)|}{(nm)^{\frac{d+1}{2d}}|n^{1/d}-m^{1/d}|}.$

For $\Sigma_{1}$ we have

$\displaystyle\Sigma_{1}$ $\displaystyle\ll\sum_{\begin{subarray}{c}n,m\leq
M\\\
|n^{1/d}-m^{1/d}|\geq(mn)^{1/2d}/10d\end{subarray}}\frac{|a(n)a(m)|}{(nm)^{\frac{d+2}{2d}}}$
$\displaystyle\ll\left(\sum_{n\leq
M}\frac{|a(n)|}{n^{\frac{d+2}{2d}}}\right)^{2}\ll M^{1-2/d+\varepsilon},$

where we used the estimate $\sum_{n\leq y}|a(n)|\ll y^{1+\varepsilon/2},$
which follows from the second estimate of (1.4) and Cauchy’s inequality.

Next we consider the sum $\Sigma_{2}$. By Lagrange’s mean value theorem we
have $|n^{1/d}-m^{1/d}|\gg n_{0}^{1/d-1}|n-m|$ for some $n_{0}$ between $n$
and $m$. If $m,n$ are contained in the sum $\Sigma_{2},$ then $m\asymp n$ and
hence

$|n^{1/d}-m^{1/d}|\gg(mn)^{1/2d-1/2}|m-n|.$

Thus

$\displaystyle\Sigma_{2}$
$\displaystyle\ll\sum_{n\not=m}\frac{|a(n)a(m)|}{(nm)^{1/d}|n-m|}$
$\displaystyle\ll\sum_{n\neq
m}\frac{1}{|n-m|}\left\\{\left(\frac{a(n)}{n^{1/d}}\right)^{2}+\left(\frac{a(m)}{m^{1/d}}\right)^{2}\right\\}$
$\displaystyle\ll\sum_{n\neq m}\frac{1}{|n-m|}\frac{a^{2}(n)}{n^{2/d}}$
$\displaystyle\ll\sum_{n\leq M}\frac{a^{2}(n)}{n^{2/d}}\sum_{m\neq
n}\frac{1}{|n-m|}$ $\displaystyle\ll M^{1-2/d+\varepsilon}.$

From the above two estimates we get

(8.7) $\displaystyle\int_{T}^{(1+\delta)T}W_{2}(y)dy\ll
T^{2-2/d+\varepsilon}M^{1-2/d}.$

We have similarly that

(8.8) $\displaystyle\int_{T}^{(1+\delta)T}W_{3}(y)dy\ll
T^{2-2/d+\varepsilon}M^{1-2/d}.$

From (8.5), (8.6), (8.7) and (8.8) we have

(8.9) $\displaystyle\int_{T}^{(1+\delta)T}R_{1}^{2}(y)dy=$
$\displaystyle\frac{\kappa_{0}^{2}}{2}\sum_{n=1}^{\infty}\frac{a^{2}(n)}{n^{\frac{d+1}{d}}}\int_{T}^{(1+\delta)T}y^{\frac{d-1}{d}}dy$
$\displaystyle+O(T^{2-1/d+\varepsilon}M^{-1/d}+T^{2-2/d+\varepsilon}M^{1-2/d}).$

From (1.6) we have $2\sigma^{*}-1-1/d<0,$ so by Lemma 9 we have

(8.10) $\displaystyle\int_{T}^{(1+\delta)T}R_{2}^{2}(y)dy$ $\displaystyle\ll
T^{\frac{d+1}{d}}\int_{T}^{(1+\delta)T}\left|I\left(-\frac{3}{2}+\frac{1}{2d},M,N,y\right)\right|^{2}dy$
$\displaystyle\ll T^{2-1/d+\varepsilon}\max_{M\leq P\leq
N}P^{2\sigma^{*}-1-1/d}$ $\displaystyle\ll
T^{2-1/d+\varepsilon}M^{2\sigma^{*}-1-1/d}.$

By the definitions of $R_{1}(y)$ and $R_{2}(y)$ we have

$\displaystyle\int_{T}^{(1+\delta)T}R_{1}(y)R_{2}(y)dy$ $\displaystyle={\rm
Re\,}\kappa_{0}c_{00}\int_{T}^{(1+\delta)T}yI\left(-\frac{3}{2}+\frac{1}{2d},M,N,y\right)\sum_{n\leq
M}\frac{a(n)}{n^{\frac{d+1}{2d}}}\cos(h(yn)^{1/d}+c_{0}\pi)dy$
$\displaystyle={\rm Re\,}\kappa_{0}c_{00}(I_{1}+I_{2}),$

where

$\displaystyle I_{1}$
$\displaystyle=\int_{T}^{(1+\delta)T}yI\left(-\frac{3}{2}+\frac{1}{2d},M,N,y\right)\sum_{n\leq
M/2}\frac{a(n)}{n^{\frac{d+1}{2d}}}\cos(h(yn)^{1/d}+c_{0}\pi)dy,$
$\displaystyle I_{2}$
$\displaystyle=\int_{T}^{(1+\delta)T}yI\left(-\frac{3}{2}+\frac{1}{2d},M,N,y\right)\sum_{M/2<n\leq
M}\frac{a(n)}{n^{\frac{d+1}{2d}}}\cos(h(yn)^{1/d}+c_{0}\pi)dy.$

By Lemma 8 we get

$\displaystyle I_{1}$ $\displaystyle=\sum_{n\leq
M/2}\frac{a(n)}{n^{\frac{d+1}{2d}}}\int_{T}^{(1+\delta)T}yI\left(-\frac{3}{2}+\frac{1}{2d},M,N,y\right)\cos(h(yn)^{1/d}+c_{0}\pi)dy$
$\displaystyle\ll\sum_{n\leq
M/2}\frac{|a(n)|}{n^{\frac{d+1}{2d}}}\left|\int_{T}^{(1+\delta)T}yI\left(-\frac{3}{2}+\frac{1}{2d},M,N,y\right)\cos(h(yn)^{1/d}+c_{0}\pi)dy\right|$
$\displaystyle\ll\sum_{n\leq
M/2}\frac{|a(n)|}{n^{\frac{d+1}{2d}}}T^{2-3/2d+\varepsilon}\max_{M\leq P\leq
N}P^{\sigma^{*}-1/2-1/d}$ $\displaystyle\ll\sum_{n\leq
M/2}\frac{|a(n)|}{n^{\frac{d+1}{2d}}}T^{2-3/2d+\varepsilon}M^{\sigma^{*}-1/2-1/d}$
$\displaystyle\ll T^{2-3/2d+\varepsilon}M^{\sigma^{*}-3/2d}.$

By Cauchy’s inequality

$I_{2}\ll T(V_{1}V_{2})^{1/2},$

where

$\displaystyle
V_{1}=\int_{T}^{(1+\delta)T}\left|I\left(-\frac{3}{2}+\frac{1}{2d},M,N,y\right)\right|^{2}dy,$
$\displaystyle V_{2}=\int_{T}^{(1+\delta)T}\left(\sum_{M/2<n\leq
M}\frac{a(n)}{n^{\frac{d+1}{2d}}}\cos(h(yn)^{1/d}+c_{0}\pi)\right)^{2}dy.$

By Lemma 9 we get

$V_{1}\ll T^{1-2/d+\varepsilon}M^{2\sigma^{*}-1-1/d}.$

By using the same approach of $\int_{T}^{(1+\delta)T}R_{1}^{2}(y)dy,$ we get

$V_{2}\ll T^{1+\varepsilon}M^{-1/d}+T^{1-1/d+\varepsilon}M^{1-2/d}.$

Thus

$I_{2}\ll
T^{2-1/d+\varepsilon}M^{\sigma^{*}-1/2-1/d}+T^{2-3/2d+\varepsilon}M^{\sigma^{*}-3/2d}.$

From the estimates of $I_{1}$ and $I_{2}$ we get

(8.11) $\int_{T}^{(1+\delta)T}R_{1}(y)R_{2}(y)dy\ll
T^{2-1/d+\varepsilon}M^{\sigma^{*}-1/2-1/d}+T^{2-3/2d+\varepsilon}M^{\sigma^{*}-3/2d}.$

Combining (8.9), (8.10) and (8.11) we get

(8.12) $\displaystyle\int_{T}^{(1+\delta)T}K_{1}^{2}dy$
$\displaystyle=\frac{\kappa_{0}^{2}}{2}\sum_{n=1}^{\infty}\frac{a^{2}(n)}{n^{\frac{d+1}{d}}}\int_{T}^{(1+\delta)T}y^{\frac{d-1}{d}}dy$
$\displaystyle\quad+O(T^{2-1/d+\varepsilon}M^{2\sigma^{*}-1-1/d}+T^{2-2/d+\varepsilon}M^{1-2/d}),$

where we used the following estimates (since $\sigma^{*}\geq 1/2$)

$\displaystyle T^{2-1/d+\varepsilon}M^{-1/d}\ll
T^{2-1/d+\varepsilon}M^{2\sigma^{*}-1-1/d},$ $\displaystyle
T^{2-1/d+\varepsilon}M^{\sigma^{*}-1/2-1/d}\ll
T^{2-1/d+\varepsilon}M^{2\sigma^{*}-1-1/d},$ $\displaystyle
T^{2-3/2d+\varepsilon}M^{\sigma^{*}-3/2d}\ll\left(T^{2-1/d+\varepsilon}M^{2\sigma^{*}-1-1/d}\right)^{1/2}\left(T^{2-2/d+\varepsilon}M^{1-2/d}\right)^{1/2}.$

###  8.3 Upper bound of $\int_{T}^{(1+\delta)T}K_{2}^{2}dy$

By Cauchy’s inequality and Lemma 9, we have

$\displaystyle\int_{T}^{(1+\delta)T}R_{3}^{2}(y)dy$
$\displaystyle\ll\sum_{\begin{subarray}{c}0\leq l,m\leq J\\\\[2.0pt]
l+m>0\end{subarray}}T^{-2l}\int_{T}^{(1+\delta)T}y^{-2l+1+\frac{1}{d}+\frac{2l-2m}{d}}\left|I\left(-\frac{3}{2}+\frac{1}{2d}+\frac{l-m}{d},M,N,y\right)\right|^{2}dy$
$\displaystyle\ll\sum_{\begin{subarray}{c}0\leq l,m\leq J\\\\[2.0pt]
l+m>0\end{subarray}}T^{-4l+1+\frac{1}{d}+\frac{2l-2m}{d}}\int_{T}^{(1+\delta)T}\left|I\left(-\frac{3}{2}+\frac{1}{2d}+\frac{l-m}{d},M,N,y\right)\right|^{2}dy$
$\displaystyle\ll\sum_{\begin{subarray}{c}0\leq l,m\leq J\\\\[2.0pt]
l+m>0\end{subarray}}T^{-4l+1+\frac{1}{d}+\frac{2l-2m}{d}+1-\frac{2}{d}+\varepsilon}\max_{M\leq
P\leq N}P^{2\sigma^{*}-1-\frac{1}{d}+\frac{2l-2m}{d}}$
$\displaystyle=\Sigma_{3}+\Sigma_{4},$

say, where

$\displaystyle\Sigma_{3}$ $\displaystyle=\sum_{\begin{subarray}{c}0\leq l\leq
m\leq J\\\\[2.0pt]
l+m>0\end{subarray}}T^{-4l+1+\frac{1}{d}+\frac{2l-2m}{d}+1-\frac{2}{d}+\varepsilon}\max_{M\leq
P\leq N}P^{2\sigma^{*}-1-\frac{1}{d}+\frac{2l-2m}{d}},$
$\displaystyle\Sigma_{4}$ $\displaystyle=\sum_{0\leq m<l\leq
J}T^{-4l+1+\frac{1}{d}+\frac{2l-2m}{d}+1-\frac{2}{d}+\varepsilon}\max_{M\leq
P\leq N}P^{2\sigma^{*}-1-\frac{1}{d}+\frac{2l-2m}{d}}.$

For $\Sigma_{3}$ we have by (1.6) that

$\displaystyle\Sigma_{3}$ $\displaystyle\ll\sum_{\begin{subarray}{c}0\leq
l\leq m\leq J\\\\[2.0pt]
l+m>0\end{subarray}}T^{-4l+1+\frac{1}{d}+\frac{2l-2m}{d}+1-\frac{2}{d}+\varepsilon}M^{2\sigma^{*}-1-\frac{1}{d}+\frac{2l-2m}{d}}$
$\displaystyle\ll T^{2-1/d+\varepsilon}M^{2\sigma^{*}-1-1/d}\sum_{1\leq m\leq
J}T^{-2m/d}M^{-2m/d}$
$\displaystyle\quad+T^{2-1/d+\varepsilon}M^{2\sigma^{*}-1-1/d}\sum_{1\leq
l\leq J}T^{-4l+2l/d}M^{2l/d}\sum_{l\leq m\leq J}T^{-2m/d}M^{-2m/d}$
$\displaystyle\ll T^{2-3/d+\varepsilon}M^{2\sigma^{*}-1-3/d}.$

For $\Sigma_{4}$ we have (since $\sigma^{*}\geq 1/2$ and recall
$N=[T^{2d-1-\varepsilon}]$)

$\displaystyle\Sigma_{4}$ $\displaystyle\ll\sum_{0\leq m<l\leq
J}T^{-4l+1+\frac{1}{d}+\frac{2l-2m}{d}+1-\frac{2}{d}+\varepsilon}N^{2\sigma^{*}-1-\frac{1}{d}+\frac{2l-2m}{d}}$
$\displaystyle\ll T^{2-1/d+\varepsilon}N^{2\sigma^{*}-1-1/d}\sum_{0\leq m\leq
J}T^{-2m/d}N^{-2m/d}\sum_{m+1\leq l\leq
J}T^{-4l+\frac{2l}{d}}N^{\frac{2l}{d}}$ $\displaystyle\ll
T^{2-1/d+\varepsilon}N^{2\sigma^{*}-1-1/d}.$

From the above estimates we get

(8.13) $\displaystyle\int_{T}^{(1+\delta)T}R_{3}^{2}(y)dy\ll
T^{2-3/d+\varepsilon}M^{2\sigma^{*}-1-3/d}+T^{2-1/d+\varepsilon}N^{2\sigma^{*}-1-1/d}.$

By Cauchy’s inequality and Lemma 10, we get (recall that
$N=[T^{2d-1-\varepsilon}]$)

(8.14) $\displaystyle\int_{T}^{(1+\delta)T}R_{4}^{2}(y)dy$
$\displaystyle\ll\sum_{0\leq l,m\leq
k}T^{2k}\int_{T}^{(1+\delta)T}(y+\frac{l}{T})^{2k+1+\frac{1}{d}-\frac{2k+2m}{d}}\left|I\left(-\frac{3}{2}+\frac{1}{2d}-\frac{k+m}{d},N,\infty,y+\frac{l}{T}\right)\right|^{2}dy$
$\displaystyle\ll\sum_{0\leq l,m\leq
k}T^{4k+1+\frac{1}{d}-\frac{2k+2m}{d}+1-\frac{2}{d}+\varepsilon}N^{2\sigma^{*}-1-\frac{1}{d}-\frac{2k+2m}{d}}$
$\displaystyle\ll
T^{4k+2-\frac{1}{d}-\frac{2k}{d}+\varepsilon}N^{2\sigma^{*}-1-\frac{1}{d}-\frac{2k}{d}}\ll
T^{2-\frac{1}{d}+\varepsilon}N^{2\sigma^{*}-1-\frac{1}{d}}.$

For $R_{5}(y)$ we have trivially that

(8.17)
$\displaystyle\int_{T}^{(1+\delta)T}R_{5}^{2}(y)dy\ll\left\\{\begin{array}[]{ll}T^{1/2},&\mbox{if
$d=2,$}\\\\[4.30554pt] T^{2-3/d+\varepsilon}M^{1-3/d},&\mbox{if $d\geq
3.$}\end{array}\right.$

For $R_{6}(y)$ we have trivially that

(8.18) $\displaystyle\int_{T}^{(1+\delta)T}R_{6}^{2}(y)dy\ll
T^{2-1/d+\varepsilon}M^{2\theta-1-1/d+\varepsilon}.$

Next we estimate $\int_{T}^{(1+\delta)T}R_{7}^{2}(y)dy.$ Trivially we have

$\displaystyle\int_{T}^{(1+\delta)T}R_{7}^{2}(y)dy$
$\displaystyle\ll\sum_{T\leq n\leq(1+\delta)T+1}\int_{n-1}^{n}R_{7}^{2}(y)dy$
$\displaystyle\ll\sum_{T\leq
n\leq(1+\delta)T+1}\int_{n-1}^{n-k/T}R_{7}^{2}(y)dy+\sum_{T\leq
n\leq(1+\delta)T+1}\int_{n-k/T}^{n}R_{7}^{2}(y)dy.$

It is easy to see that

$\displaystyle
R_{7}(y)=E(y)-\int_{\mathbf{E}_{k}}E(\tilde{y})d\mathrm{Y}_{k}=\int_{\mathbf{E}_{k}}(E(y)-E(\tilde{y}))d\mathrm{Y}_{k}.$

Suppose $n-1<y\leq n.$ If $n-1<y\leq\tilde{y}<n,$ then by the definition of
$E(u)$ we get

$E(\tilde{y})-E(y)=Q(y)-Q(\tilde{y})\ll T^{-1}\max_{y\ll
u\ll\tilde{y}}Q^{\prime}(u)\ll T^{-1}\log^{m_{\cal L}}T;$

otherwise if $\tilde{y}\geq n,$ then

$E(\tilde{y})-E(y)=Q(y)-Q(\tilde{y})+a(n)/2\ll T^{-1}\log^{m_{\cal
L}}T+|a(n)|.$

Thus we have

$\displaystyle\sum_{T\leq
n\leq(1+\delta)T+1}\int_{n-1}^{n-k/T}R_{7}^{2}(y)dy\ll T^{-1}\log^{2m_{\cal
L}}T,$ $\displaystyle\sum_{T\leq
n\leq(1+\delta)T+1}\int_{n-k/T}^{n}R_{7}^{2}(y)dy\ll T^{-1}\log^{2m_{\cal
L}}T+T^{-1}\sum_{n\ll T}a^{2}(n)\ll T^{\varepsilon}.$

Hence we get

(8.19) $\int_{T}^{(1+\delta)T}R_{7}^{2}(y)dy\ll T^{\varepsilon}.$

The first term in the right hand side of (8.13) is absorbed in the right hand
side of (8.17). Furthermore from the assumption $M\ll N^{1/2}$ and
$1/2\leq\sigma^{*}<1/2+1/2d$, we have

$N^{2\sigma^{*}-1-1/d}\ll M^{2(2\sigma^{*}-1-1/d)}$

and also

$M^{2\theta-1-1/d}\ll M^{2(2\sigma^{*}-1-1/d)}$

by $\theta<1/2-1/2d$. Hence from (8.13)–(8.19) we have

(8.22) $\displaystyle\int_{T}^{(1+\delta)T}K_{2}^{2}dy$ $\displaystyle\ll
T^{2-1/d+\varepsilon}M^{2(2\sigma^{\ast}-1-1/d)}+\left\\{\begin{array}[]{ll}T^{1/2},&\mbox{if
$d=2,$}\\\\[4.30554pt] T^{2-3/d+\varepsilon}M^{1-3/d},&\mbox{if $d\geq
3.$}\end{array}\right.$

###  8.4 Proof of Theorem 1

Now we take $M$ such that

(8.23) $\displaystyle M\asymp T^{\frac{1}{2d(1-\sigma^{*})-1}},$ and (8.26)
$\displaystyle\hat{E}(M)\left\\{\begin{array}[]{ll}=0,&\mbox{if $b(n)\geq 0\
(n\geq 1)$,}\\\\[4.30554pt] \ll
T^{\theta+\varepsilon},&\mbox{otherwise.}\end{array}\right.$

By noting that $\sigma^{*}=1/2$ for $d=2$, we can see easily that $M\ll
N^{1/2}$ for any $d\geq 2$. Then the formula (8.12) becomes

(8.27)
$\int_{T}^{(1+\delta)T}K_{1}^{2}dy=\frac{\kappa_{0}^{2}}{2}\sum_{n=1}^{\infty}\frac{a^{2}(n)}{n^{\frac{d+1}{d}}}\int_{T}^{(1+\delta)T}y^{\frac{d-1}{d}}dy+O(T^{2-\frac{3-4\sigma^{*}}{2d(1-\sigma^{*})-1}+\varepsilon}).$

From (8.22), (8.27) and Cauchy’s inequality we get

(8.30) $\displaystyle\int_{T}^{(1+\delta)T}K_{1}K_{2}dy$ $\displaystyle\ll
T^{2-1/d+\varepsilon}M^{(2\sigma^{*}-1-1/d)}+\left\\{\begin{array}[]{ll}T^{\varepsilon},&\mbox{if
$d=2$}\\\\[4.30554pt] T^{2-2/d+\varepsilon}M^{(1-3/d)/2},&\mbox{if $d\geq
3$}\end{array}\right.$ $\displaystyle\ll
T^{2-\frac{3-4\sigma^{*}}{2d(1-\sigma^{*})-1}+\varepsilon},$

where the last inequality follows from the choice of $M$ (8.23). It is also
seen easily that

(8.31) $\displaystyle\int_{T}^{(1+\delta)T}K_{2}^{2}dy\ll
T^{2-\frac{3-4\sigma^{*}}{2d(1-\sigma^{*})-1}+\varepsilon}$

by (8.22) and the choice of $M$. Hence from (8.27), (8.30) and (8.31) we get

(8.32)
$\int_{T}^{(1+\delta)T}E^{2}(y)dy=\frac{\kappa_{0}^{2}}{2}\sum_{n=1}^{\infty}\frac{a^{2}(n)}{n^{\frac{d+1}{d}}}\int_{T}^{(1+\delta)T}y^{\frac{d-1}{d}}dy+O(T^{2-\frac{3-4\sigma^{*}}{2d(1-\sigma^{*})-1}+\varepsilon}).$

Finally from (8.32) we get

$\displaystyle\int_{0}^{T}E^{2}(y)dy$ $\displaystyle=\sum_{0\leq
j\leq\frac{\log
T}{\delta}}\int_{\frac{T}{(1+\delta)^{j+1}}}^{\frac{T}{(1+\delta)^{j}}}E^{2}(y)dy+O(T^{2\delta})$
$\displaystyle=\sum_{0\leq j\leq\frac{\log
T}{\delta}}\frac{\kappa_{0}^{2}}{2}\sum_{n=1}^{\infty}\frac{a^{2}(n)}{n^{\frac{d+1}{d}}}\int_{\frac{T}{(1+\delta)^{j+1}}}^{\frac{T}{(1+\delta)^{j}}}y^{\frac{d-1}{d}}dy$
$\displaystyle\quad+\sum_{0\leq j\leq\frac{\log
T}{\delta}}O\left(\left(\frac{T}{(1+\delta)^{j}}\right)^{2-\frac{3-4\sigma^{*}}{2d(1-\sigma^{*})-1}+\varepsilon}\right)+O(T^{2\delta})$
$\displaystyle=\frac{\kappa_{0}^{2}}{2}\sum_{n=1}^{\infty}\frac{a^{2}(n)}{n^{\frac{d+1}{d}}}\int_{0}^{T}y^{\frac{d-1}{d}}dy+O\left(T^{2-\frac{3-4\sigma^{*}}{2d(1-\sigma^{*})-1}+\varepsilon}\right).$

###  8.5 Proofs of Corollaries 1 and 2

We first prove Corollary 1. Suppose that $0\leq\theta\leq 1/4$ is a real
number and ${\cal L}(s)\in{\cal S}_{real}^{\theta}$ is a function of degree
$2$. In this case it is well-known that

(8.33) $\int_{0}^{T}|{\cal L}(1/2+it)|^{2}dt\ll T^{1+\varepsilon},$

namely, we have $\sigma^{*}=1/2<3/4$. So Corollary 1 follows from Theorem 1.

Next we prove Corollary 2. Suppose that $0\leq\theta\leq 1/3$ is a real number
and ${\cal L}(s)\in{\cal S}_{real}^{\theta}$ is a function of degree $3$ which
can be written as ${\cal L}(s)={\cal L}_{1}(s){\cal L}_{2}(s)$, where ${\cal
L}_{1}(s)\in{\cal S}_{real}^{\theta}$ is a function of degree $1$ and ${\cal
L}_{2}(s)\in{\cal S}_{real}^{\theta}$ is a function of degree $2.$ We can show
that

(8.34) $\int_{0}^{T}|{\cal L}(5/8+it)|^{2}dt\ll T^{1+\varepsilon}.$

Hence we can take $\sigma^{*}=5/8$ in (1.5) and the assertion (1.9) follows
from Theorem 1. Furthermore if we assume (1.10), we can show that

(8.35) $\int_{0}^{T}|{\cal L}(7/12+it)|^{2}dt\ll T^{1+\varepsilon}.$

Hence we can take $\sigma^{*}=7/12$ in this case and we get a better assertion
(1.11).

In order to prove (8.34) or (8.35), we follow a method of Ivić [22]. In [22],
Ivić treated the automorphic $L$-function attached to a cusp form over
$SL_{2}({\mathbb{Z}})$. Our proof for the general case is due to Ivić with
some modifications. So here we give a detailed proof.

######  Lemma 11.

Let $t_{1}<\cdots<t_{R}$ be real numbers such that $T\leq t_{r}\leq 2T$ for
$r=1,2,\ldots,R$ and $|t_{r}-t_{s}|\geq\log^{4}T$ for $1\leq s\not=r\leq R.$
Suppose $V$ and $1\ll M\ll T^{C}\ (C>0)$ are large parameters such that

$T^{\varepsilon}<V\leq\left|\sum_{M<n\leq 2M}a(n)n^{-\sigma-it_{r}}\right|$

with

(8.36) $\sum_{M\leq n\leq 2M}|a(n)|^{2}\ll M^{1+\varepsilon},$

then we have

$R\ll(M^{2-2\sigma}V^{-2}+TV^{-\frac{2}{3-4\sigma}})T^{\varepsilon}$

for $1/2<\sigma\leq 2/3$.

Remark to Lemma 11. This is the case $1/2<\sigma\leq 2/3$ of Lemma 8.2 of Ivić