###### Abstract

We evaluate the cutoff dependence of the Casimir energy that is usually
neglected. We consider the global Casimir energy in a five-dimensional (5D)
model on $S^{1}/Z_{2}$. If the cutoff scale of the 5D theory is not far from
the compactification scale, the UV physics may affect the low energy result.
We work in the cutoff regularization scheme by introducing two independent
cutoff scales for the spatial momentum in the non-compact space and for the
Kaluza-Klein masses. The effects of the UV physics are incorporated as a
damping effect of the contributions to the vacuum energy around the cutoff
scales. We numerically calculate the Casimir energy and evaluate the deviation
from the conventional result. We find that the result well agrees with the
conventional one for the Gaussian-type damping, while it can deviate for the
kink-type damping.

KEK-TH-2376

WU-HEP-21-04

UV sensitivity of Casimir energy

Yu Asai${}^{1}\\!$ 111E-mail address<EMAIL_ADDRESS>and Yutaka
Sakamura${}^{2,3}\\!$ 222E-mail address<EMAIL_ADDRESS>
1Department of Physics, Waseda University,
3-4-1 Ookubo, Shinjuku-ku, Tokyo 169-8555, Japan

2KEK Theory Center, Institute of Particle and Nuclear Studies, KEK,
1-1 Oho, Tsukuba, Ibaraki 305-0801, Japan

3Department of Particles and Nuclear Physics,
SOKENDAI (The Graduate University for Advanced Studies),
1-1 Oho, Tsukuba, Ibaraki 305-0801, Japan

##  1 Introduction

The Casimir effect is a well-known macroscopic quantum effect [1]. It has been
observed by various experiments, and the observed values well agree with the
theoretical predictions [2]-[5]. The Casimir effect can also play an important
role in the context of the extra dimensions. It generates the scalar potential
for the volume modulus of the compact extra space, and can stabilize the
modulus to a finite value [6]-[9].

It is believed that the Casimir effect is an IR phenomenon, and the Casimir
energy is insensitive to the UV physics. There are several ways to calculate
the Casimir energy, and the same result is obtained by those methods
[10]-[18]. On the other hands, the conventional calculations neglect the
contributions from the modes near the UV cutoff scale of the theory, as we
will explicitly show in the text. For the four-dimensional (4D)
electrodynamics, the fact that the experimental values are in agreement with
the theoretical predictions indicates that such UV contributions are
negligible. However, this does not mean that they can always be negligible in
any theories.

Since the extra-dimensional models are non-renormalizable, it should be
regarded as effective theories of more fundamental theories. In other words,
they have UV cutoff scales. In this paper, we discuss a five-dimensional (5D)
scalar theory compactified on $S^{1}/Z_{2}$. The theory behaves as a 5D theory
between the cutoff scale $\Lambda$ and the compactification scale $1/R$, where
$R$ is the radius of $S^{1}$. In particular, when the cutoff $\Lambda$ is not
far from $1/R$, effects of the UV physics might give non-negligible
contributions to the result. Such effects will appear as a cutoff-dependence
of the Casimir energy.

The purpose of this work is to investigate the UV contribution to the Casimir
energy, and to clarify the situation in which it is non-negligible.111 The
cutoff-dependence of the Casimir energy has been also discussed in Ref. [19]
with a different motivation. They discussed a possibility that the
cosmological constant originates from the Casimir energy in the 4D context,
and focused on a case that the spacing between the plates, which corresponds
to the size of the extra dimension $\pi R$ in our case, is shorter than the
cutoff length $1/\Lambda$. Another different point from their work is that we
take into account effects of the momentum cutoff, which is in principle
independent of the cutoff for the Kaluza-Klein masses.  In order to simplify
the discussion, we consider a 5D real scalar theory with one compact
dimension. For our purpose, we work in the cutoff regularization scheme, do
not take the limit $\Lambda\to\infty$, and numerically evaluate the deviation
from the result obtained by the conventional methods.

The paper is organized as follows. In the next section, we provide a brief
review of the conventional calculations for the Casimir energy. In Sec. 3, we
introduce the cutoff scales for the momentum and the Kaluza-Klein masses, and
derive the expression of the Casimir energy. In Sec. 4, we numerically
calculate the Casimir energy, and evaluate the deviation from the conventional
result. Sec. 5 is devoted to the summary and discussions. In the appendices,
we collect some formulae used in our calculations.

##  2 Conventional calculations for Casimir energy

In this section, we provide a brief review of the conventional way to
calculate the Casimir energy [2]-[9]. To simplify the discussion, we consider
a real scalar theory in the flat $D$-dimensional spacetime,222 We leave the
spacetime dimension $D$ unspecified in this and the next sections. In
numerical calculations performed in Sec. 4, we focus on the case of $D=5$.
and one of the spatial dimensions is compactified on $S^{1}/Z_{2}$.

$\displaystyle{\cal L}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{1}{2}\partial^{\mu}\Phi\partial_{\mu}\Phi-\frac{1}{2}M^{2}\Phi^{2},$
(2.1)

where $\mu=0,1,\cdots,D-1$, $M$ is the bulk mass parameter. The coordinate of
the compact dimension is denoted as $y$. The fundamental region of
$S^{1}/Z_{2}$ is chosen as $0\leq y\leq\pi R$, where $R$ is the radius of
$S^{1}$. The real scalar field $\Phi$ is assumed to be $Z_{2}$ odd. Thus, it
satisfies the Dirichlet boundary conditions at the boundaries of
$S^{1}/Z_{2}$, in addition to the periodic boundary condition. As a result of
these boundary conditions, the Kaluza-Klein (KK) masses are determined by

$\sin\left(\pi R\sqrt{m_{n}^{2}-M^{2}}\right)=0,$ (2.2)

whose solutions are

$m_{n}=\sqrt{M^{2}+\frac{n^{2}}{R^{2}}}.\;\;\;\;\;\left(n=1,2,\cdots\right)$
(2.3)

Then, the vacuum energy density in the $(D-1)$-dimensional effective theory is
expressed as

$E_{\rm
vac}=\sum_{n=1}^{\infty}\int\frac{d^{d}k}{2(2\pi)^{d}}\;\omega_{n}(k),$ (2.4)

where $d\equiv D-2$ is the dimension of the non-compact space, and

$\omega_{n}(k)\equiv\sqrt{k^{2}+m_{n}^{2}}\;\;\;\;\;\left(k^{2}\equiv|\vec{k}|^{2}\right)$
(2.5)

is the energy of a KK mode with the $d$-dimensional momentum $\vec{k}$ and the
mass $m_{n}$.

To perform the $\vec{k}$-integral, we work in the dimensional regularization,
and obtain

$\displaystyle E_{\rm vac}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{\Gamma(-\frac{d+1}{2})}{2(4\pi)^{(d+1)/2}}\sum_{n=1}^{\infty}m_{n}^{d+1},$
(2.6)

where $\Gamma(\alpha)$ is the Euler gamma function. The infinite sum over the
KK modes is evaluated by means of the zeta function regularization technique
[8, 13, 14]. Using the formula (B.19), the above expression becomes

$\displaystyle E_{\rm vac}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{\Gamma(-\frac{d+1}{2})}{2(4\pi)^{(d+1)/2}R^{d+1}}\left\\{-\frac{\bar{M}^{d+1}}{2}+\frac{\sqrt{\pi}\Gamma(-\frac{d+2}{2})}{2\Gamma(-\frac{d+1}{2})}\bar{M}^{d+2}\right.$
(2.7) $\displaystyle\hskip
99.58464pt\left.+\frac{2\bar{M}^{\frac{d+2}{2}}}{\pi^{\frac{d+1}{2}}\Gamma(-\frac{d+1}{2})}\sum_{n=1}^{\infty}n^{-\frac{d+2}{2}}K_{\frac{d+2}{2}}\left(2\pi
n\bar{M}\right)\right\\}$ $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{\Gamma(-\frac{d+1}{2})M^{d+1}}{4(4\pi)^{(d+1)/2}}-\frac{\Gamma(-\frac{d+2}{2})}{8(4\pi)^{d/2}}RM^{d+2}$
$\displaystyle-\frac{M^{\frac{d+2}{2}}}{(2\pi)^{d+1}R^{d/2}}\sum_{n=1}^{\infty}n^{-\frac{d+2}{2}}K_{\frac{d+2}{2}}\left(2\pi
nMR\right),$

where $\bar{M}\equiv MR$. The first term is irrelevant to the Casimir force
because it is independent of $R$. The second term gives a constant Casimir
force, which is also present even in the case of $R\to\infty$. Since the
physically relevant Casimir force is the difference between the above $E_{\rm
vac}$ and that of the non-compact case (i.e., $R\to\infty$), the contribution
from the second term in (2.7) is cancelled. Thus, the Casimir energy $E_{\rm
cas}$ is given by

$E_{\rm
cas}=-\frac{M^{\frac{d+2}{2}}}{(2\pi)^{d+1}R^{d/2}}\sum_{n=1}^{\infty}n^{-\frac{d+2}{2}}K_{\frac{d+2}{2}}\left(2\pi
nMR\right).$ (2.8)

In the massless case ($M=0$), (2.8) reduces to

$\displaystyle E_{\rm cas}^{(M=0)}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{1}{2^{d+1}\pi^{\frac{3}{2}d+2}R^{d+1}}\sum_{n=1}^{\infty}\frac{1}{n^{d+2}}\lim_{M\to
0}\left(\pi nMR\right)^{\frac{d+2}{2}}K_{\frac{d+2}{2}}\left(2\pi nMR\right)$
(2.9) $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{1}{2^{d+1}\pi^{\frac{3}{2}d+2}R^{d+1}}\sum_{n=1}^{\infty}\frac{1}{n^{d+2}}\frac{\Gamma(\frac{d+2}{2})}{2}$
$\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{\Gamma(\frac{d+2}{2})\zeta(d+2)}{2^{d+2}\pi^{\frac{3}{2}d+2}R^{d+1}},$

where $\zeta(s)$ is the Riemann zeta function. We have used the formula:

$\lim_{z\to
0}z^{\alpha}K_{\alpha}(2z)=\frac{\Gamma(\alpha)}{2}.\;\;\;\;\;\left(\alpha>0\right)$
(2.10)

Namely, we have

$E_{\rm
cas}^{(M=0)}=\begin{cases}\displaystyle-\frac{\zeta(3)}{16\pi^{3}R^{2}}&\left(d=1\right)\\\
\displaystyle\rule{0.0pt}{24.0pt}-\frac{1}{1440\pi R^{3}}&\left(d=2\right)\\\
\displaystyle\rule{0.0pt}{24.0pt}-\frac{3\zeta(5)}{128\pi^{6}R^{4}}&\left(d=3\right)\\\
\displaystyle\rule{0.0pt}{24.0pt}-\frac{1}{30240\pi^{2}R^{5}}&\left(d=4\right)\end{cases}.$
(2.11)

The Casimir energy (2.8) can be also expressed in the following integral form
by using the formula (B.12).

$\displaystyle E_{\rm cas}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{\Gamma(-\frac{d+1}{2})}{2(4\pi)^{(d+1)/2}R^{d+1}}\cdot\frac{d+1}{\pi}\sin\frac{(d+1)\pi}{2}\int_{0}^{\infty}dw\;w^{d}\ln\left(1-e^{-2\pi\sqrt{w^{2}+\bar{M}^{2}}}\right)$
(2.12) $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{(4\pi)^{(d+1)/2}\Gamma(\frac{d+1}{2})R^{d+1}}\int_{0}^{\infty}dw\;w^{d}\ln\left(1-e^{-2\pi\sqrt{w^{2}+\bar{M}^{2}}}\right).$

We have used the reflection formula for the gamma function in the second line.

The bulk mass $M$ exponentially suppresses the Casimir Energy as shown in Fig.
1.

Figure 1: $R^{4}E_{\rm cas}$ in the case of $D=5$ as a function of $MR$.

##  3 Cutoff regularization

Since the expression (2.4) diverges, we need some physical condition that
extracts a finite physically-sensible contribution, i.e., the Casimir energy
(density) $E_{\rm cas}$, from the divergent quantity $E_{\rm vac}$. To see
this extraction explicitly, the dimensional regularization and the zeta
function regularization are not suitable. Thus we work in the cutoff
regularization. We introduce the momentum cutoff $\Lambda$ and the cutoff for
the KK masses $\tilde{\Lambda}$, independently. In contrast to the
conventional calculations, we keep them finite.

As a condition to extract the finite energy density, we define the Casimir
energy $E_{\rm cas}$ in such a way that it vanishes when the size of the
compactification radius is taken to infinite. Namely, the energy density (in
the $D$-dimensional spacetime) is measured from that in the non-compact limit
[20]. In this work, we discuss the global Casimir energy, for simplicity.
Thus, this condition is adopted to the averaged energy density over the
compact space $E_{\rm vac}/\pi R$. Therefore, the Casimir energy $E_{\rm
cas}$, which is defined as the vacuum energy density in the
$(D-1)$-dimensional effective theory, is defined as

$\frac{E_{\rm cas}(R)}{\pi R}\equiv\frac{E_{\rm vac}(R)}{\pi
R}-\lim_{R\to\infty}\frac{E_{\rm vac}(R)}{\pi R}.$ (3.1)

###  3.1 Momentum cutoff

First, we perform the $\vec{k}$-integral.

$\displaystyle\int\frac{d^{d}k}{2(2\pi)^{d}}\;\left(k^{2}+m_{n}^{2}\right)^{1/2}\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{2\pi^{d/2}}{\Gamma(\frac{d}{2})}\int_{0}^{\Lambda}\frac{dk}{2(2\pi)^{d}}\;k^{d-1}\left(k^{2}+m_{n}^{2}\right)^{1/2}$
(3.2) $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{m_{n}^{d+1}}{2(4\pi)^{d/2}\Gamma(\frac{d}{2})}\int_{\epsilon_{n}}^{1}ds\;s^{-\frac{d+1}{2}-1}\left(1-s\right)^{\frac{d}{2}-1}$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{m_{n}^{d+1}}{2(4\pi)^{d/2}\Gamma(\frac{d}{2})}\beta_{\epsilon_{n}}\left(-\frac{d+1}{2},\frac{d}{2}\right),$

where $\Lambda$ is the $d$-dimensional momentum cutoff scale,
$2\pi^{d/2}/\Gamma(\frac{d}{2})$ is the area of the $(d-1)$-dimensional sphere
with a unit radius,

$s\equiv\frac{m_{n}^{2}}{k^{2}+m_{n}^{2}},\;\;\;\;\;\epsilon_{n}\equiv\frac{m_{n}^{2}}{\Lambda^{2}+m_{n}^{2}},$
(3.3)

and $\beta_{z}(\alpha,\beta)$ is the incomplete beta function defined in
(A.3). Thus, (2.4) is expressed as

$\displaystyle E_{\rm vac}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{2(4\pi)^{d/2}\Gamma(\frac{d}{2})}\lim_{N\to\infty}\sum_{n=1}^{N}m_{n}^{d+1}\beta_{\epsilon_{n}}\left(-\frac{d+1}{2},\frac{d}{2}\right).$
(3.4)

The explicit forms of $\beta_{\epsilon_{n}}$ in (3.4) for various values of
$d$ are collected in Appendix A.2.

If we take the limit $\Lambda\to\infty$ before taking the limit $N\to\infty$,
the above expression reduces to (2.6) in the previous section since 333 Notice
that $B\left(-\frac{d+1}{2},\frac{d}{2}\right)$ is well-defined only when
$\frac{d+1}{2}$ is not a positive integer. In contrast,
$\beta_{\epsilon_{n}}\left(-\frac{d+1}{2},\frac{d}{2}\right)$ can be defined
even in such a case as long as $\epsilon_{n}$ is kept non-zero.

$\displaystyle\lim_{\epsilon_{n}\to
0}\beta_{\epsilon_{n}}\left(-\frac{d+1}{2},\frac{d}{2}\right)=B\left(-\frac{d+1}{2},\frac{d}{2}\right)=\frac{\Gamma(-\frac{d+1}{2})\Gamma(\frac{d}{2})}{\Gamma(-\frac{1}{2})}.$
(3.5)

This indicates that, in the derivation of the Casimir energy shown in the
previous section, we have implicitly assumed that the contributions from the
KK modes near the cutoff $m_{n}\sim\Lambda$ are negligibly small. In the next
section, we will numerically check whether this is true or not.

###  3.2 Cutoff for KK masses

Here we introduce the cutoff scale $\tilde{\Lambda}$ for the KK masses.
Naively, the regularized vacuum energy is written as

$E_{\rm
vac}=\frac{1}{2(4\pi)^{d/2}\Gamma(\frac{d}{2})}\sum_{n=1}^{N}m_{n}^{d+1}\beta_{\epsilon_{n}}\left(-\frac{d+1}{2},\frac{d}{2}\right),$
(3.6)

where

$N\equiv{\rm floor}\,\left(R\sqrt{\tilde{\Lambda}^{2}-M^{2}}\right)$ (3.7)

is the cutoff for the number of the KK modes. However, it seems natural to
assume that the contributions of the KK modes near the cutoff are suppressed
by the UV physics.444 In fact, as we will see later, a sharp cutoff for the KK
summation like this leads to a divergent Casimir energy in the limit of
$\Lambda,\tilde{\Lambda}\to\infty$.  So we introduce the damping function that
has the property:

$g_{N}(x)\simeq\begin{cases}1&\mbox{for $x\ll N$}\\\ 0&\mbox{for $x\gg
N$}\end{cases},$ (3.8)

and consider the following quantity instead of (3.6).

$\displaystyle E_{\rm vac}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{2(4\pi)^{d/2}\Gamma(\frac{d}{2})}\sum_{n=1}^{\infty}m_{n}^{d+1}\beta_{\epsilon_{n}}\left(-\frac{d+1}{2},\frac{d}{2}\right)g_{N}(n).$
(3.9)

As examples of the damping function, we can take

$g_{N}(x)=\exp\left(-\frac{x^{2}}{2N^{2}}\right),$ (3.10)

or

$g_{N}(x)=\frac{1}{2}\left[1+\tanh\left\\{A\left(1-\frac{x}{N}\right)\right\\}\right],$
(3.11)

where $A$ is a positive constant that controls the steepness around the cutoff
(see Fig. 2).

Figure 2: The profiles of the damping function $g_{N}(x)$ for (3.10) (left
plot) and for (3.11) (right plot). The cutoff $N$ and the constant $A$ are
chosen as $10^{2}$ and $10$, respectively.

We cannot choose too small values of $A$. Otherwise the damping effect reaches
around $x=0$, and the property (3.8) is no longer satisfied. As we will see in
the next section, $A\gtrsim 10$ is required.

We rewrite (3.9) in terms of the dimensionless constants:

$a\equiv\frac{1}{\Lambda R},\;\;\;\;\;\hat{M}\equiv\frac{M}{\Lambda},$ (3.12)

as

$E_{\rm
vac}=\frac{\Lambda^{d+1}}{2(4\pi)^{d/2}\Gamma(\frac{d}{2})}\sum_{n=1}^{\infty}F(an),$
(3.13)

where

$\displaystyle F(x)\\!\\!\\!$ $\displaystyle\equiv\\!\\!\\!$ $\displaystyle
f(X(x))g_{N}(x),$ $\displaystyle X(x)\\!\\!\\!$ $\displaystyle\equiv\\!\\!\\!$
$\displaystyle\sqrt{\hat{M}^{2}+x^{2}},$ $\displaystyle f(X)\\!\\!\\!$
$\displaystyle\equiv\\!\\!\\!$
$\displaystyle\beta_{\frac{X^{2}}{1+X^{2}}}\left(-\frac{d+1}{2},\frac{d}{2}\right)X^{d+1}.$
(3.14)

From (A.8)-(A.11), the explicit forms of the function $f(X)$ are given by

$f(X)=\begin{cases}\displaystyle\sqrt{1+X^{2}}+X^{2}\ln\frac{1+\sqrt{1+X^{2}}}{X}&D=3\;\;(d=1)\\\
\displaystyle\rule{0.0pt}{23.0pt}\frac{2}{3}\left\\{\left(1+X^{2}\right)^{\frac{3}{2}}-X^{3}\right\\}&D=4\;\;(d=2)\\\
\displaystyle\rule{0.0pt}{27.0pt}\frac{1}{4}\left\\{\left(2+X^{2}\right)\sqrt{1+X^{2}}-X^{4}\ln\frac{1+\sqrt{1+X^{2}}}{X}\right\\}&D=5\;\;(d=3)\\\
\displaystyle\rule{0.0pt}{23.0pt}\frac{2}{15}\left\\{\left(3-2X^{2}\right)\left(1+X^{2}\right)^{\frac{3}{2}}+2X^{5}\right\\}&D=6\;\;(d=4)\end{cases}.$
(3.15)

Although the explicit functional forms of $f(X)$ for different values of $D$
are quite different, they have a similar behavior. In fact, we find that

$f(0)=\frac{2}{D-1},$ (3.16)

and

$f(X)\sim\frac{2}{D-2}X.\;\;\;\left(\mbox{for $X\gg 1$}\right)$ (3.17)

Fig. 3 shows the profile of $f(X)$ in each dimension.

Figure 3: The profiles of the function $f(X)$ in (3.15) for various
dimensions.

Due to the damping function $g_{N}(x)$, the function $F(x)$ satisfies the
boundary conditions,

$\lim_{x\to\infty}F(x)=0,\;\;\;\;\;\lim_{x\to\infty}F^{(p)}(x)=0.\;\;\;(p=1,2,3,\cdots)$
(3.18)

###  3.3 Non-compact limit

Next consider the non-compact limit ($R\to\infty$) in order to evaluate the
Casimir energy given in (3.1). When we take this limit, we have to treat the
contribution of the zero-mode carefully because it is projected out by the
orbifold boundary conditions. Let us first consider the case of the $S^{1}$
compactification instead of $S^{1}/Z_{2}$. Then, the sum in (3.13) is taken
over the whole integers. In this case, we can easily take the non-compact
limit.

$\displaystyle\lim_{R\to\infty}\frac{E^{S^{1}}_{\rm vac}(R)}{\pi R}\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{\Lambda^{d+2}}{2\pi(4\pi)^{d/2}\Gamma(\frac{d}{2})}\lim_{a\to
0}\sum_{n=-\infty}^{\infty}aF(an)$ (3.19) $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{\Lambda^{d+2}}{2\pi(4\pi)^{d/2}\Gamma(\frac{d}{2})}\int_{-\infty}^{\infty}dx\;F(x).$

Thus, we have

$\displaystyle\frac{E_{\rm cas}^{S^{1}}}{\pi R}\\!\\!\\!$
$\displaystyle=\\!\\!\\!$ $\displaystyle\frac{E^{S^{1}}_{\rm vac}(R)}{\pi
R}-\lim_{R\to\infty}\frac{E^{S^{1}}_{\rm vac}(R)}{\pi R}$ (3.20)
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{\Lambda^{d+2}}{2\pi(4\pi)^{d/2}\Gamma(\frac{d}{2})}\left\\{\sum_{n=-\infty}^{\infty}aF(an)-\int_{-\infty}^{\infty}dx\;F(x)\right\\}$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{\Lambda^{d+2}a}{2\pi(4\pi)^{d/2}\Gamma(\frac{d}{2})}\left\\{\sum_{n=-\infty}^{\infty}F(an)-\int_{-\infty}^{\infty}dx\;F(ax)\right\\}.$

We have used that

$\frac{1}{a}\int_{-\infty}^{\infty}dx\;F(x)=\int_{-\infty}^{\infty}dx\;F(ax).$
(3.21)

Since $F(x)$ is an even function, the brace part in (3.20) can be rewritten as

$\sum_{n=-\infty}^{\infty}F(an)-\int_{-\infty}^{\infty}dx\;F(ax)=2\sum_{n=1}^{\infty}F(an)+F(0)-2\int_{0}^{\infty}dx\;F(ax).$
(3.22)

Now let us come back to the case of $S^{1}/Z_{2}$. Noting that (3.20) is
finite in the limit of $\Lambda\to\infty$, we should identify the non-compact
limit of the energy density as 555 For a $Z_{2}$-even field, this becomes
$\lim_{R\to\infty}\frac{E_{\rm vac}(R)}{\pi
R}=\frac{\Lambda^{d+2}}{2\pi(4\pi)^{d/2}\Gamma(\frac{d}{2})}\left\\{\int_{0}^{\infty}dx\;F(x)+\frac{1}{2}F(0)\right\\}.$
(3.23)

$\lim_{R\to\infty}\frac{E_{\rm vac}(R)}{\pi
R}=\frac{\Lambda^{d+2}}{2\pi(4\pi)^{d/2}\Gamma(\frac{d}{2})}\left\\{\int_{0}^{\infty}dx\;F(x)-\frac{1}{2}F(0)\right\\}.$
(3.24)

Thus the Casimir energy in this case is expressed as

$\displaystyle E_{\rm cas}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{\Lambda^{d+1}}{2(4\pi)^{d/2}\Gamma(\frac{d}{2})}\Delta(a)=\frac{\Delta(a)/a^{d+1}}{2(4\pi)^{d/2}\Gamma(\frac{d}{2})R^{d+1}},$
(3.25)

where

$\Delta(a)\equiv\sum_{n=1}^{\infty}F(an)-\int_{0}^{\infty}dx\;F(ax)+\frac{1}{2}F(0).$
(3.26)

##  4 Calculation of Casimir energy

In this section, we numerically calculate the Casimir energy in the cutoff
regularization scheme, and evaluate the deviation from the conventional
result. As mentioned in Sec. 3.2, the behaviors of $f(X)$ in (3.15) are
similar for different $D$. Thus the qualitative features of the Casimir energy
do not depend on $D$ very much. Therefore we focus on the case of $D=5$
($d=3$) in this section.

###  4.1 Expression of Casimir energy

In (3.12), we have normalized the bulk mass $M$ by the cutoff scale $\Lambda$.
Since $\hat{M}$ is independent of $R$, it can be treated as a constant when we
take the non-compact limit ($R\to\infty$). However, once we obtain the
expression (3.25) with (3.26), it is more convenient to normalize $M$ by the
compactification scale $1/R$ rather than the cutoff scale $\Lambda$. Hence we
define

$\bar{M}\equiv RM,$ (4.1)

and rewrite $\Delta(a)$ defined in (3.26) as

$\Delta(a)=\sum_{n=1}^{\infty}\bar{F}(n)-\int_{0}^{\infty}dx\;\bar{F}(x)+\frac{1}{2}\bar{F}(0),$
(4.2)

where

$\displaystyle\bar{F}(x)\\!\\!\\!$ $\displaystyle\equiv\\!\\!\\!$
$\displaystyle f(a\bar{X}(x))g_{N}(x)$ $\displaystyle\bar{X}(x)\\!\\!\\!$
$\displaystyle\equiv\\!\\!\\!$ $\displaystyle\sqrt{\bar{M}^{2}+x^{2}}.$ (4.3)

In order to evaluate $\Delta(a)$ in (4.2), the Euler-Maclaurin formula is
useful (see Appendix C.1) [10, 19, 21]. Using the formula (C.1), $\Delta(a)$
is expressed as

$\displaystyle\Delta(a)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\lim_{n_{\rm max}\to\infty}\left\\{\sum_{n=1}^{n_{\rm
max}}\bar{F}(n)-\int_{0}^{n_{\rm
max}}dx\;\bar{F}(x)+\frac{1}{2}\bar{F}(0)\right\\}$ (4.4)
$\displaystyle=\\!\\!\\!$ $\displaystyle\lim_{n_{\rm
max}\to\infty}\left\\{\sum_{n=0}^{n_{\rm max}}\bar{F}(n)-\int_{0}^{n_{\rm
max}}dx\;\bar{F}(x)\right\\}-\frac{1}{2}\bar{F}(0)$ $\displaystyle=\\!\\!\\!$
$\displaystyle\lim_{n_{\rm
max}\to\infty}\left[\frac{1}{2}\left\\{\bar{F}(0)+\bar{F}(n_{\rm
max})\right\\}+\sum_{p=1}^{\lfloor
q/2\rfloor}\frac{B_{2p}}{(2p)!}\left\\{\bar{F}^{(2p-1)}(n_{\rm
max})-\bar{F}^{(2p-1)}(0)\right\\}+R_{q}\right]$
$\displaystyle-\frac{1}{2}\bar{F}(0)$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\sum_{p=1}^{\lfloor
q/2\rfloor}\frac{B_{2p}}{(2p)!}\bar{F}^{(2p-1)}(0)+R_{q},$

where $B_{2q}$ are the Bernoulli numbers (see (C.2)). We have used the
condition (3.18) at the last equality. The remainder term $R_{q}$ is defined
as

$R_{q}\equiv(-1)^{q-1}\int_{0}^{\infty}dx\;\frac{B_{q}(x-\lfloor
x\rfloor)}{q!}\bar{F}^{(q)}(x),$ (4.5)

where $B_{q}(x)$ is the Bernoulli polynomial (see (C.4)), and the symbol
$\lfloor\cdots\rfloor$ denotes the floor function. The integer $q$ can be
chosen to an arbitrary value greater than 1. Here we set it as $q=2$. Then,
(4.4) becomes

$\Delta(a)=-\frac{1}{12}\bar{F}^{(1)}(0)+R_{2},$ (4.6)

and the remainder term is expressed as

$\displaystyle R_{2}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\int_{0}^{\infty}dx\;\frac{B_{2}(x-\lfloor
x\rfloor)}{2}\bar{F}^{(2)}(x)$ (4.7) $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{1}{2}\sum_{k=0}^{\infty}\int_{k}^{k+1}dx\;B_{2}(x-k)\bar{F}^{(2)}(x).$

Due to the property (3.8), we have $g_{N}(0)\simeq 1$, and thus the first term
in (4.6) vanishes because

$\bar{F}^{(1)}(0)\simeq\left\\{\partial_{x}f(a\bar{X}(x))\right\\}_{x=0}=\left\\{\frac{ax}{\bar{X}(x)}f^{(1)}(a\bar{X}(x))\right\\}_{x=0}=0.$
(4.8)

###  4.2 Case of Gaussian damping

Here we choose the Gaussian-type damping function (3.10). In order to see the
deviation from the conventional result, we define the ratio:

$r_{E}\equiv\frac{E_{\rm cas}}{E_{\rm cas}^{\rm conv}},$ (4.9)

where $E_{\rm cas}^{\rm conv}$ is the Casimir energy calculated in the
conventional methods, such as (2.8) or (2.12). Fig. 4 shows the ratio $r_{E}$
as a function of the bulk mass $\bar{M}=MR$. In this plot, we have chosen the
cutoff scales as $\tilde{\Lambda}=\Lambda$.

Figure 4: The ratio (4.9) as a function of the bulk mass $\bar{M}=MR$. The
blue solid and orange dashed lines represent the case of $\Lambda
R=\tilde{\Lambda}R=10$ and $100$, respectively.

As this plot shows, the deviation is small and can be neglected if $\Lambda R$
is larger than $10^{2}$. Therefore, the cutoff-dependence of $E_{\rm cas}$ is
negligible, and the conventional result is reliable even in the case of a
finite cutoff scale $\Lambda$ unless $\Lambda R={\cal O}(10)$. This is true
also in the case of $\tilde{\Lambda}\neq\Lambda$ as long as they are in the
same order of magnitude.

###  4.3 Case of kink-like damping

####  4.3.1 General properties

Next we consider the case of the kink-type damping function (3.11). Fig. 5
shows the ratio $r_{E}$ as a function of the parameter $A$ in the damping
function. The cutoff scale are chosen as $\Lambda R=100$, and
$\tilde{\Lambda}=\Lambda$ (blue solid), $2\Lambda$ (orange dashed) and
$0.5\Lambda$ (green dotted). The bulk mass is chosen as $\bar{M}=MR=0.1$.

Figure 5: The ratio (4.9) as a function of $A$ in the damping function (3.11).
The cutoff scale is chosen as $a=10^{-2}$, and the bulk mass is chosen as
$\bar{M}=0.1$. The blue solid, the orange dashed and the green dotted lines
represent the case of $\tilde{\Lambda}=\Lambda$, $2\Lambda$ and $0.5\Lambda$,
respectively.

For $A\lesssim 10$, the ratio $r_{E}$ deviates from one because the damping
effect reaches near the origin and the property (3.8) is no longer satisfied.
Namely, $g_{N}(x)$ is not suitable for the damping function when $A\lesssim
10$. In the case of $\tilde{\Lambda}=\Lambda$, the result agrees with the
conventional one for $10\lesssim A\lesssim 35$. However, for $A\gtrsim 35$,
the ratio $r_{E}$ starts to deviate from one, flips the sign and becomes
negatively larger and larger. We can also see that the upper limit of the
region in which the result agrees with the conventional one is sensitive to
the ratio of the cutoffs $\tilde{\Lambda}/\Lambda$.

####  4.3.2 The limit of $A\to\infty$

In order to understand the behavior of $r_{E}$ for large values of $A$, we
consider the limit of $A\to\infty$. Since

$\displaystyle\bar{F}^{(2)}(x)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\partial_{x}^{2}\left\\{{\cal F}(x)g_{N}(x)\right\\}$ (4.10)
$\displaystyle=\\!\\!\\!$ $\displaystyle{\cal F}^{(2)}(x)g_{N}(x)+2{\cal
F}^{(1)}(x)g_{N}^{(1)}(x)+{\cal F}(x)g_{N}^{(2)}(x),$

where ${\cal F}(x)\equiv f(a\bar{X}(x))$, and $g_{N}^{(1)}(x)$ and
$g_{N}^{(2)}(x)$ sharply localizes around $x=N$ when $A$ is large, (4.7) is
rewritten as

$\displaystyle R_{2}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{1}{2}\sum_{k=0}^{\infty}\int_{k}^{k+1}dx\;B_{2}(x-k){\cal
F}^{(2)}(x)g_{N}(x)$ (4.11)
$\displaystyle-\frac{1}{2}\sum_{k=N-1}^{N}\int_{k}^{k+1}dx\;B_{2}(x-k)\left\\{2{\cal
F}^{(1)}(x)g_{N}^{(1)}(x)+{\cal F}(x)g_{N}^{(2)}(x)\right\\}.$

In the limit of $A\to\infty$, the damping function $g_{N}(x)$ and its
derivatives become

$\displaystyle g_{N}(x)\\!\\!\\!$ $\displaystyle\to\\!\\!\\!$
$\displaystyle\Theta(N-x),$ $\displaystyle g_{N}^{(1)}(x)\\!\\!\\!$
$\displaystyle\to\\!\\!\\!$ $\displaystyle-\delta(x-N),$ $\displaystyle
g_{N}^{(2)}(x)\\!\\!\\!$ $\displaystyle\to\\!\\!\\!$
$\displaystyle-\delta^{(1)}(x-N),$ (4.12)

where $\Theta(x)$ is the Heaviside step function. Therefore, we have

$\displaystyle\sum_{k=N-1}^{N}\int_{k}^{k+1}dx\;B_{2}(x-k){\cal
F}^{(1)}(x)g_{N}^{(1)}(x)$ (4.13) $\displaystyle=\\!\\!\\!$
$\displaystyle\int_{N-1}^{N}dx\;B_{2}(x-N+1){\cal
F}^{(1)}(x)g_{N}^{(1)}(x)+\int_{N}^{N+1}dx\;B_{2}(x-N){\cal
F}^{(1)}(x)g_{N}^{(1)}(x)$ $\displaystyle\to\\!\\!\\!$
$\displaystyle-\frac{1}{2}B_{2}(1){\cal F}^{(1)}(N)-\frac{1}{2}B_{2}(0){\cal
F}^{(1)}(N)=-\frac{1}{6}{\cal F}^{(1)}(N),$

and

$\displaystyle\sum_{k=N-1}^{N}\int_{k}^{k+1}dx\;B_{2}(x-k){\cal
F}(x)g_{N}^{(2)}(x)$ (4.14) $\displaystyle=\\!\\!\\!$
$\displaystyle\sum_{N-1}^{N}\left\\{\left[B_{2}(x-k){\cal
F}(x)g_{N}^{(1)}(x)\right]_{k}^{k+1}-\int_{k}^{k+1}dx\;\partial_{x}\left\\{B_{2}(x-k){\cal
F}(x)\right\\}g_{N}^{(1)}(x)\right\\}$ $\displaystyle=\\!\\!\\!$
$\displaystyle B_{2}\left\\{{\cal F}(N+1)g_{N}^{(1)}(N+1)-{\cal
F}(N-1)g_{N}^{(1)}(N-1)\right\\}$
$\displaystyle-\sum_{N-1}^{N}\int_{k}^{k+1}dx\;\left\\{2B_{1}(x-k){\cal
F}(x)+B_{2}(x-k){\cal F}^{(1)}(x)\right\\}g_{N}^{(1)}(x)$
$\displaystyle\to\\!\\!\\!$ $\displaystyle\frac{1}{2}\left\\{2B_{1}(1){\cal
F}(N)+B_{2}(1){\cal F}(N)\right\\}+\frac{1}{2}\left\\{2B_{1}(0){\cal
F}^{(1)}(N)+B_{2}(0){\cal F}^{(1)}(N)\right\\}$ $\displaystyle=\\!\\!\\!$
$\displaystyle B_{2}{\cal F}^{(1)}(N)=\frac{1}{6}{\cal F}^{(1)}(N).$

We have used that

$\displaystyle\lim_{A\to\infty}\int_{N-1}^{N}dx\;{\cal
G}(x)g_{N}^{(1)}(x)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{1}{2}{\cal G}(N),$
$\displaystyle\lim_{A\to\infty}\int_{N}^{N+1}dx\;{\cal
G}(x)g_{N}^{(1)}(x)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{1}{2}{\cal G}(N),$ (4.15)

for an arbitrary function ${\cal G}(x)$, and

$\displaystyle\partial_{x}B_{2}(x)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle 2B_{1}(x),\;\;\;\;\;\lim_{A\to\infty}g_{N}^{(1)}(N\pm 1)=0,$
$\displaystyle B_{1}(0)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$ $\displaystyle-
B_{1}(1)=-\frac{1}{2},$ $\displaystyle B_{2}(0)\\!\\!\\!$
$\displaystyle=\\!\\!\\!$ $\displaystyle B_{2}(1)=B_{2}=\frac{1}{6}.$ (4.16)

Therefore, since

$\bar{F}^{(1)}(0)={\cal
F}^{(1)}(0)=\left\\{a\bar{X}^{(1)}(x)f^{(1)}(a\bar{X}(x))\right\\}_{x=0}=\left\\{\frac{ax}{\bar{X}(x)}f^{(1)}(a\bar{X}(x))\right\\}_{x=0}=0,$
(4.17)

(4.6) is expressed as

$\displaystyle\Delta(a)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$ $\displaystyle
R_{2}=-\frac{1}{2}\sum_{k=0}^{N-1}\int_{k}^{k+1}dx\;B_{2}(x-k){\cal
F}^{(2)}(x)+\frac{1}{6}{\cal F}^{(1)}(N)-\frac{1}{2}\left\\{\frac{1}{6}{\cal
F}^{(1)}(N)\right\\}$ (4.18) $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{1}{2}\sum_{k=0}^{N-1}\int_{0}^{1}dx\;B_{2}(x){\cal
F}^{(2)}(x+k)+\frac{1}{12}{\cal F}^{(1)}(N).$

In the $D=5$ case, we have

$\displaystyle{\cal F}^{(1)}(N)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle
a^{2}N\left\\{\sqrt{1+a^{2}(\bar{M}^{2}+N^{2})}-a^{2}\left(\bar{M}^{2}+N^{2}\right)\ln\frac{1+\sqrt{1+a^{2}(\bar{M}^{2}+N^{2})}}{a\sqrt{\bar{M}^{2}+N^{2}}}\right\\},$
$\displaystyle{\cal F}^{(2)}(x)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{a^{2}}{\sqrt{1+a^{2}(\bar{M}^{2}+x^{2})}}\bigg{[}1$
$\displaystyle\hskip
28.45274pt+a^{2}\left(\bar{M}^{2}+3x^{2}\right)\left\\{1-\sqrt{1+a^{2}(\bar{M}^{2}+x^{2})}\ln\frac{1+\sqrt{1+a^{2}(\bar{M}^{2}+x^{2})}}{a\sqrt{\bar{M}^{2}+x^{2}}}\right\\}\bigg{]}.$

Fig. 6 shows the Casimir energy $E_{\rm cas}$ as a function of $\ln(\Lambda
R)$. From this plot, we can see that $E_{\rm cas}$ grows as ${\cal
O}(10^{-5}\Lambda R)$ for $\Lambda R>10^{3}$.

Figure 6: The Casimir energy $E_{\rm cas}$ and $E_{\rm cas}/(\Lambda R)$ in
the limit of $A\to\infty$ as functions of $\ln(\Lambda R)$.

Namely, the Casimir energy calculated by using the sharp kink damping function
diverges as $\Lambda\to\infty$. We should also notice that the Casimir energy
in this limit $\lim_{A\to\infty}E_{\rm cas}$ is positive, which has the
opposite sign to the conventional result.

####  4.3.3 Cutoff dependence

As we mentioned in Sec. 3.1, the deviation from the conventional result comes
from the contributions of the KK modes near the cutoff scale
$m_{n}\sim\Lambda$. The Casimir energy $E_{\rm cas}$ is expressed as

$E_{\rm cas}=\frac{1}{8\pi^{2}R^{4}}\frac{\Delta(a)}{a^{4}},$ (4.20)

where

$\displaystyle\Delta(a)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\sum_{n=1}^{\infty}\bar{F}(n)-\int_{0}^{\infty}dx\;\bar{F}(x)+\frac{1}{2}\bar{F}(0).$
(4.21)

Fig. 7 shows the profile of $\bar{F}(x)$ in the case of the Gaussian and the
kink-like damping functions (the kink-parameter $A$ is chosen as 10, 30 and
50).

Figure 7: The profile of $\bar{F}(x)$ in (4.3) in the case of the Gaussian
damping function (top left), $A=10$ (top right), $A=30$ (bottom left) and
$A=50$ (bottom right). The bulk mass and the cutoff scale are chosen as
$\bar{M}=MR=0.1$ and $\Lambda R=100$, respectively.

From these figures, we can see that the contributions from the modes near the
cutoff scale $m_{n}\sim\Lambda$ are well suppressed in the case of the
Gaussian damping (3.10). Thus the result almost agrees with the conventional
one (2.8) or (2.12). In contrast, the kink-like damping function (3.11) takes
account of those contributions with only small suppression. If the kink shape
is not very steep, the deviation from the conventional result still remains
tiny. However, if the damping function becomes close to the step function, the
deviation becomes large and $E_{\rm cas}$ changes the sign as discussed in
Sec. 4.3.2.

The range of $A$ in which the conventional result is reproduced depends on the
hierarchy between the cutoff scale and the compactification scale $1/R$. The
plots in Fig. 8 show the ratio $r_{E}$ in the cases of $a=3\times 10^{-2}$
(the left plot) and of $a=3\times 10^{-3}$ (the right plot). We can see that
the conventional result is reproduced in a wider range of $A$ for larger
cutoff scales. When the cutoff is not high enough, we cannot obtain the
conventional result for any values of $A$.

Figure 8: The ratio (4.9) as a function of $A$ in the damping function (3.11).
The cutoff scale is chosen as $a=(\Lambda R)^{-1}=3\times 10^{-2}$ (left plot)
and $a=3\times 10^{-3}$ (right plot). The bulk mass is chosen as
$\bar{M}=0.1$. The blue solid, the orange dashed and the green dotted lines
represent the case of $\tilde{\Lambda}=\Lambda$, $2\Lambda$ and $0.5\Lambda$,
respectively.

##  5 Summary and discussions

We have calculated the global Casimir energy in a simple model whose one
spatial dimension is compactified on $S^{1}/Z_{2}$ using the cutoff
regularization scheme. We keep the cutoff scales finite, and evaluate the
deviation from the conventional result. This deviation originates from the
contribution of the KK modes near the cutoff scale $m_{n}\sim\Lambda$. We can
easily show that the sharp cutoff for the KK masses leads to a large
deviation, which changes the sign of the Casimir energy. When $\Lambda
R>10^{3}$, this energy is of ${\cal O}(10^{-5}\Lambda R)$. Instead of adopting
such a sharp cutoff, we introduce the damping function and insert it into the
infinite KK summation. We consider two types of the damping function. One is
the Gaussian-type and the other is the kink-type. In the former case, the
contributions near the cutoff scale are well suppressed, and the result almost
agrees with the conventional one. In the latter case, the calculations take
into account the modes near the cutoff, but the deviation from the
conventional result still remains tiny unless the kink is extremely steep.
When the hierarchy between the compactification scale $1/R$ and the cutoff
scales $\Lambda$, $\tilde{\Lambda}$ is large, the conventional result is
reproduced for a wide range of the kink parameter $A$ defined in (3.11).
However, if the hierarchy is not large enough, such a region becomes narrow
and the contributions from the KK modes near the cutoff tend to give a non-
negligible contribution to the Casimir energy.

It is commonly believed that the Casimir energy is UV insensitive, and is
determined only by the low-lying KK modes. However, the above-mentioned
results show that it is not always true. Since the 5D extra-dimensional model
is non-renormalizable and should be regarded as an effective theory, it is
expected that the contribution of the KK modes to the vacuum energy becomes
suppressed as the KK mass approaches to the cutoff scale of the 5D theory due
to the effect of the UV physics. Our result indicates that the Casimir energy
can receive sizable contribution from the modes near the cutoff scale,
depending on how much the contribution to the vacuum energy is damped near the
cutoff. In particular, such a deviation from the conventional result tends to
appear when the cutoff scale is not far from the compactification scale.

The damping function $g_{N}(x)$ introduced in this work mimics the suppression
of the contributions from higher KK modes by the UV physics. For example, let
us consider a spontaneously broken supersymmetric (SUSY) model. The low-energy
effective theory is non-SUSY, and a nonvanishing vacuum energy is induced by
the non-SUSY field content. Above the SUSY-breaking scale $\Lambda_{\rm SB}$,
however, the superpartners start to contribute to the vacuum energy, and their
contributions have opposite signs. In the non-SUSY effective theory, these
negative contributions of the superpartners are represented by the damping
function $g_{N}(x)$ inserted into the KK summation for the vacuum energy.

It is interesting to investigate this damping effect in models that have UV-
completed theories. For example, we can calculate the Casimir energy in a 6D
theory with two extra dimensions compactified on a torus. If there is a
hierarchy between the two radii of the torus, an effective 5D theory appears
at intermediate scales. Then we can explicitly see how the contribution to the
vacuum energy from each KK mode behaves near the cuoff of the effective 5D
theory, and evaluate the deviation from the conventional Casimir energy.

We have worked in the flat spacetime. Although the KK mass spectrum depends on
the geometry of the extra space, the level spacing for the KK masses becomes
almost the same in the UV region, i.e., $\Delta m_{n}\equiv m_{n+1}-m_{n}\sim
1/R$. Hence the contribution from the KK modes around the cutoff does not
depend the geometry very much. Our result is common in various 5D models.
However, if the extra space has more dimensions, the distributions of the KK
mass eigenvalues around the cutoff scale are quite different from that of 5D
theories. Therefore, it is also interesting to extend our work to higher extra
dimensions.

In this work, we have adopted the sharp cutoff for the momentum $\vec{k}$, for
simplicity. To investigate the contributions of the modes near the cutoff in
more detail, we should also introduce the damping function for the momentum
integral. But we expect that the qualitative features obtained in this work do
not change very much.

We will discuss these issues in separate papers.

##  Appendix A Incomplete beta and gamma functions

###  A.1 Definitions and properties

The integral expressions of the complete beta and the gamma functions are
given by

$\displaystyle B(\alpha,\beta)\\!\\!\\!$ $\displaystyle\equiv\\!\\!\\!$
$\displaystyle\int_{0}^{\infty}dx\;x^{\alpha-1}(1-x)^{\beta-1}=B(\beta,\alpha),$
$\displaystyle\Gamma(\alpha)\\!\\!\\!$ $\displaystyle\equiv\\!\\!\\!$
$\displaystyle\int_{0}^{\infty}dt\;t^{\alpha-1}e^{-t}.$ (A.1)

These expressions are valid only for ${\rm Re}\,\alpha>0$ and ${\rm
Re}\,\beta>0$. They are related as

$B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}.$
(A.2)

This relation holds over the whole domain of the beta function.

The incomplete beta functions are defined as

$\displaystyle B_{z}(\alpha,\beta)\\!\\!\\!$ $\displaystyle\equiv\\!\\!\\!$
$\displaystyle\int_{0}^{z}dx\;x^{\alpha-1}(1-x)^{\beta-1},$
$\displaystyle\beta_{z}(\alpha,\beta)\\!\\!\\!$ $\displaystyle\equiv\\!\\!\\!$
$\displaystyle
B(\alpha,\beta)-B_{z}(\alpha,\beta)=\int_{z}^{1}dx\;x^{\alpha-1}\left(1-x\right)^{\beta-1},$
(A.3)

and the upper and the lower incomplete gamma functions are defined as

$\displaystyle\Gamma_{z}(\alpha)\\!\\!\\!$ $\displaystyle\equiv\\!\\!\\!$
$\displaystyle\int_{z}^{\infty}dt\;t^{\alpha-1}e^{-t},$
$\displaystyle\gamma_{z}(\alpha)\\!\\!\\!$ $\displaystyle\equiv\\!\\!\\!$
$\displaystyle\Gamma(\alpha)-\Gamma_{z}(\alpha)=\int_{0}^{z}dt\;t^{\alpha-1}e^{-t}.$
(A.4)

For $\alpha+\beta+1>0$ and $\beta>0$, it follows that

$\beta_{\epsilon}(\alpha,\beta)=\frac{1}{\beta\Gamma(\alpha+\beta)}\int_{0}^{\infty}d\tau\;\tau^{\beta}e^{-\tau}\Gamma_{\tau\epsilon/(1-\epsilon)}(\alpha)+\frac{1}{\beta}\epsilon^{\alpha}(1-\epsilon)^{\beta}.$
(A.5)

This reduces to (A.2) in the limit of $\epsilon\to 0$ if $\alpha>0$.

###  A.2 Explicit forms of Incomplete beta functions

The explicit forms of $\beta_{\epsilon_{n}}(-\frac{d+1}{2},\frac{d}{2})$,
where $\epsilon_{n}\equiv m_{n}^{2}/(\Lambda^{2}+m_{n}^{2})$, are obtained as
follows. Since

$\displaystyle\partial_{\Lambda}\beta_{\epsilon_{n}}\left(\alpha,\beta\right)\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle-\epsilon_{n}^{\alpha-1}\left(1-\epsilon_{n}\right)^{\beta-1}\partial_{\Lambda}\epsilon_{n}$
(A.6) $\displaystyle=\\!\\!\\!$
$\displaystyle\left(\frac{m_{n}^{2}}{\Lambda^{2}+m_{n}^{2}}\right)^{\alpha-1}\left(\frac{\Lambda^{2}}{\Lambda^{2}+m_{n}^{2}}\right)^{\beta-1}\frac{2\Lambda
m_{n}^{2}}{\left(\Lambda^{2}+m_{n}^{2}\right)^{2}}$ $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{2\Lambda^{2\beta-1}m_{n}^{2\alpha}}{\left(\Lambda^{2}+m_{n}^{2}\right)^{\alpha+\beta}},$

and $\beta_{\epsilon_{n}}(\alpha,\beta)|_{\Lambda=0}=0$, we have

$\beta_{\epsilon_{n}}\left(-\frac{d+1}{2},\frac{d}{2}\right)=\int_{0}^{\Lambda}d\Lambda^{\prime}\;\partial_{\Lambda^{\prime}}\beta_{\epsilon_{n}}\left(-\frac{d+1}{2},\frac{d}{2}\right)=\int_{0}^{\Lambda}d\Lambda^{\prime}\;\frac{2\Lambda^{\prime
d-1}m_{n}^{-d-1}}{\left(\Lambda^{\prime 2}+m_{n}^{2}\right)^{-\frac{1}{2}}}.$
(A.7)

Performing the integral, we obtain

$D=3\;(d=1)$

$\displaystyle\beta_{\epsilon_{n}}\left(-1,\frac{1}{2}\right)\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{\Lambda\sqrt{\Lambda^{2}+m_{n}^{2}}}{m_{n}^{2}}+\ln\frac{\Lambda+\sqrt{\Lambda^{2}+m_{n}^{2}}}{m_{n}},$
(A.8)

$D=4\;(d=2)$

$\displaystyle\beta_{\epsilon_{n}}\left(-\frac{3}{2},1\right)\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{2\left(\Lambda^{2}+m_{n}^{2}\right)^{\frac{3}{2}}}{3m_{n}^{3}}-\frac{2}{3}.$
(A.9)

$D=5\;(d=3)$

$\displaystyle\beta_{\epsilon_{n}}\left(-2,\frac{3}{2}\right)\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{\Lambda}{4m_{n}^{4}}\left(2\Lambda^{2}+m_{n}^{2}\right)\sqrt{\Lambda^{2}+m_{n}^{2}}-\frac{1}{4}\ln\frac{\Lambda+\sqrt{\Lambda^{2}+m_{n}^{2}}}{m_{n}}.$
(A.10)

$D=6\;(d=4)$

$\displaystyle\beta_{\epsilon_{n}}\left(-\frac{5}{2},2\right)\\!\\!\\!$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{2}{15m_{n}^{5}}\left(3\Lambda^{2}-2m_{n}^{2}\right)\left(\Lambda^{2}+m_{n}^{2}\right)^{\frac{3}{2}}+\frac{4}{15}.$
(A.11)

##  Appendix B Zeta function regularization

###  B.1 Integral form

Here we review the zeta function regularization technique [6, 7, 8, 9, 14].
Let us consider an infinite sum:

$S_{p}\equiv\sum_{n=1}^{\infty}x_{n}^{p},$ (B.1)

where $x_{n}$ is a solution of

${\cal F}(x_{n})=0.$ (B.2)

We assume that all the solutions are located on the positive real axis in the
complex plane, and

$x_{n}\sim cn,\;\;\;\;\;\left(\mbox{$c$: a positive constant}\right)$ (B.3)

when $n$ is large enough. Thus, the infinite sum $S_{p}$ converges only when
${\rm Re}\,p<-1$. In such a region, $S_{p}$ can be expressed by a contour
integral.

$\displaystyle S_{p}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{2\pi i}\oint_{C}dz\;z^{p}\frac{{\cal
F}^{\prime}(z)}{{\cal F}(z)}$ (B.4) $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{p}{2\pi i}\oint_{C}dz\;z^{p-1}\ln{\cal F}(z),$

where $C$ is a contour that encircles all the solutions of (B.2).

Suppose that the function ${\cal F}(x)$ has the following asymptotic behavior.

${\cal F}(z)\sim\begin{cases}{\cal F}_{+}^{\rm asp}(z)&\mbox{for ${\rm
Im}\,z>0$}\\\ {\cal F}_{-}^{\rm asp}(z)&\mbox{for ${\rm
Im}\,z<0$}\end{cases},$ (B.5)

where ${\cal F}_{\pm}^{\rm asp}(z)$ are analytic functions that do not have
poles inside the contour $C$. Then, the expression (B.4) can be rewritten as

$\displaystyle S_{p}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{p}{2\pi
i}\sum_{\sigma=\pm}\int_{C_{\sigma}}dz\;z^{p-1}\ln\frac{{\cal F}(z)}{{\cal
F}_{\sigma}^{\rm asp}(z)}-\frac{p}{2\pi
i}\sum_{\sigma=\pm}\int_{C_{\sigma}}dz\;z^{p-1}\ln{\cal F}^{\rm
asp}_{\sigma}(z),$ (B.6)

where $C_{+}$ and $C_{-}$ denote the upper and lower halves of $C$. Then, the
divergent part is pushed into the second term, which is irrelevant to the
physics since ${\cal F}^{\rm asp}_{\pm}(z)$ do not have poles inside $C$.666
In the case of the Randall-Sundrum background, we need to further extract
divergent contribution from the first term in (B.6), which will be
renormalized to the brane tensions [6]-[9].

In the case of

${\cal F}(z)=\sin\left(\pi\sqrt{z^{2}-\bar{M}^{2}}\right),$ (B.7)

where $\bar{M}$ is a positive constant, The contour $C$ consists of

$\displaystyle C_{1}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\left\\{z={\cal R}e^{i\theta}|\theta:0\to\frac{\pi}{2}\right\\}$
$\displaystyle C_{2}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\left\\{z=iw|w:{\cal R}\to\delta\right\\}$ $\displaystyle
C_{3}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\left\\{z=x+i\delta|x:0\to\bar{M}\right\\}$ $\displaystyle
C_{4}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\left\\{z=\bar{M}+\delta e^{i\theta}|\theta:\frac{\pi}{2}\to
0\right\\}$ $\displaystyle C_{5}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\left\\{z=\bar{M}+\delta
e^{i\theta}|\theta:0\to-\frac{\pi}{2}\right\\}$ $\displaystyle C_{6}\\!\\!\\!$
$\displaystyle=\\!\\!\\!$ $\displaystyle\left\\{z=x-i\delta|x:\bar{M}\to
0\right\\}$ $\displaystyle C_{7}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\left\\{z=iw|w:-\delta\to-{\cal R}\right\\}$ $\displaystyle
C_{8}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$ $\displaystyle\left\\{z={\cal
R}e^{i\theta}|\theta:-\frac{\pi}{2}\to 0\right\\}.$ (B.8)

The asymptotic function ${\cal F}^{\rm asp}(z)$ is chosen as

${\cal F}^{\rm asp}(z)=\begin{cases}\displaystyle{\cal F}_{+}^{\rm
asp}\equiv\frac{i}{2}\exp\left(-i\pi\sqrt{z^{2}-\bar{M}^{2}}\right)&\mbox{for
${\rm Im}\,z>0$}\\\ \displaystyle\rule{0.0pt}{25.0pt}{\cal F}_{-}^{\rm
asp}\equiv-\frac{i}{2}\exp\left(i\pi\sqrt{z^{2}-\bar{M}^{2}}\right)&\mbox{for
${\rm Im}\,z<0$}\end{cases}.$ (B.9)

We have chosen the branch of the square root is chosen as

$\sqrt{re^{i\theta}}=\sqrt{r}e^{i\theta/2},$ (B.10)

for $-\pi<\theta\leq\pi$. Then, since

$\frac{{\cal F}(z)}{{\cal F}_{\pm}^{\rm asp}(z)}=1-\exp\left(\pm 2\pi
i\sqrt{z^{2}-\bar{M}^{2}}\right),$ (B.11)

we can check that only the contributions from $C_{2}$ and $C_{7}$ survive in
the limit of ${\cal R}\to\infty$ and $\delta\to 0$. Thus, we have

$\displaystyle S_{p}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{p}{2\pi i}\left\\{\int_{C_{2}}dz\;z^{p-1}\ln\frac{{\cal
F}(z)}{{\cal F}_{+}^{\rm asp}(z)}+\int_{C_{7}}dz\;z^{p-1}\ln\frac{{\cal
F}(z)}{{\cal F}_{-}^{\rm asp}(z)}\right\\}+\cdots$ (B.12)
$\displaystyle=\\!\\!\\!$ $\displaystyle-\frac{p}{2\pi
i}\left\\{\int_{\infty}^{0}d(iw)\;(iw)^{p-1}\ln\left(1-e^{-2\pi\sqrt{w^{2}+\bar{M}^{2}}}\right)\right.$
$\displaystyle\hskip
36.98857pt\left.+\int_{0}^{-\infty}d(iw)\;(iw)^{p-1}\ln\left(1-e^{-2\pi\sqrt{w^{2}+\bar{M}^{2}}}\right)\right\\}+\cdots$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{p}{\pi}\sin\frac{p\pi}{2}\int_{0}^{\infty}dw\;w^{p-1}\ln\left(1-e^{-2\pi\sqrt{w^{2}+\bar{M}^{2}}}\right)+\cdots,$

where the ellipsis denotes divergent terms that are irrelevant to the physics.

###  B.2 Infinite-sum form

In the case of (B.7), there is another expression for $S_{p}$ [11]-[15]. In
this case, $S_{p}$ becomes

$S_{p}=\sum_{n=1}^{\infty}\left(n^{2}+\bar{M}^{2}\right)^{p/2}.$ (B.13)

This converges only when ${\rm Re}\,p<-1$.

Using the integral representation for the gamma function:

$\left(n^{2}+\bar{M}^{2}\right)^{p/2}\Gamma\left(-\frac{p}{2}\right)=\int_{0}^{\infty}dt\;t^{-\frac{p}{2}-1}e^{-(n^{2}+\bar{M}^{2})t},$
(B.14)

which is valid for ${\rm Re}\,p<0$, it is rewritten as

$\displaystyle S_{p}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{\Gamma(-\frac{p}{2})}\sum_{n=1}^{\infty}\int_{0}^{\infty}dt\;t^{-\frac{p}{2}-1}e^{-(n^{2}+\bar{M}^{2})t}$
(B.15) $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{\Gamma(-\frac{p}{2})}\int_{0}^{\infty}dt\;t^{-\frac{p}{2}-1}e^{-t\bar{M}^{2}}\vartheta(t),$

where $\vartheta(t)\equiv\sum_{n=1}^{\infty}e^{-n^{2}t}$ is the Jacobi theta
function, and has the property:

$\vartheta(t)=-\frac{1}{2}+\frac{1}{2}\sqrt{\frac{\pi}{t}}+\sqrt{\frac{\pi}{t}}\vartheta\left(\frac{\pi^{2}}{t}\right).$
(B.16)

Substituting this into (B.15), we obtain

$\displaystyle S_{p}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle-\frac{\bar{M}^{p}}{2}+\frac{\sqrt{\pi}\Gamma(-\frac{p+1}{2})}{2\Gamma(-\frac{p}{2})}\bar{M}^{p+1}$
(B.17)
$\displaystyle+\frac{\sqrt{\pi}}{\Gamma(-\frac{p}{2})}\sum_{n=1}^{\infty}\int_{0}^{\infty}dt\;t^{-\frac{p+3}{2}}\exp\left(-t\bar{M}^{2}-\frac{\pi^{2}n^{2}}{t}\right).$

Note that the $t$-integrals of the contributions from the first and the second
terms in (B.16) converge only when ${\rm Re}\,p<0$ and ${\rm Re}\,p<-1$,
respectively. However, once they are expressed with respect to the gamma
function, they can be analytically connected to the whole complex plane except
for the points $p=1,3,5,\cdots$.777 At the points $p=0,2,4,\cdots$, the second
term in (B.17) simply vanises.  The $t$-integral in the last term of (B.17)
converges for any values of $p$, thanks to the exponential factor
$\exp\left(-t\bar{M}^{2}-\frac{\pi^{2}n^{2}}{t}\right)$. In fact, it is
expressed with respect to the modified Bessel function by using the integral
expression:

$K_{\alpha}(z)=\frac{1}{2}\left(\frac{z}{2}\right)^{\alpha}\int_{0}^{\infty}dt\;t^{-\alpha-1}\exp\left(-t-\frac{z^{2}}{4t}\right),$
(B.18)

which is valid for ${\rm Re}\,\alpha>-1/2$ and $\left|\arg z\right|<\pi/4$.
Thus, we can analytically continue the above expression to the expression,

$S_{p}=-\frac{\bar{M}^{p}}{2}+\frac{\sqrt{\pi}\Gamma(-\frac{p+1}{2})}{2\Gamma(-\frac{p}{2})}\bar{M}^{p+1}+\frac{2\bar{M}^{\frac{p+1}{2}}}{\pi^{\frac{p}{2}}\Gamma(-\frac{p}{2})}\sum_{n=1}^{\infty}n^{-\frac{p+1}{2}}K_{\frac{p+1}{2}}\left(2\pi
n\bar{M}\right),$ (B.19)

which is valid for any values of $p$ except for $p=1,3,5,\cdots$. In this
expression, the infinite sum in the last term rapidly converges.

##  Appendix C Subtraction Scheme

###  C.1 Euler-Maclaurin formula

For arbitrary integers $n_{\rm min}$ and $n_{\rm max}$ ($n_{\rm min}<n_{\rm
max}$), a $q$ times continuously differentiable function $h(x)$ on the
interval $[n_{\rm min},n_{\rm max}]$ satisfies

$\displaystyle\sum_{n=n_{\rm min}}^{n_{\rm max}}h(n)-\int_{n_{\rm
min}}^{n_{\rm max}}dx\;h(x)$ (C.1) $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{2}\left\\{h(n_{\rm min})+h(n_{\rm
max})\right\\}-\sum_{k=2}^{q}(-1)^{k-1}\frac{B_{k}}{k!}\left\\{h^{(k-1)}(n_{\rm
max})-h^{(k-1)}(n_{\rm min})\right\\}+R_{q}$ $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{2}\left\\{h(n_{\rm min})+h(n_{\rm
max})\right\\}+\sum_{p=1}^{\lfloor
q/2\rfloor}\frac{B_{2p}}{(2p)!}\left\\{h^{(2p-1)}(n_{\rm
max})-h^{(2p-1)}(n_{\rm min})\right\\}+R_{q},$

where $B_{k}$ ($k=2,3,\cdots$) are the Bernoulli numbers,

$\displaystyle B_{2}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{6},\;\;\;\;\;B_{4}=-\frac{1}{30},\;\;\;\;\;B_{6}=\frac{1}{42},\;\;\;\;\;\cdots$
$\displaystyle B_{2p+1}\\!\\!\\!$ $\displaystyle=\\!\\!\\!$ $\displaystyle
0,\;\;\;\;\;\left(p=1,2,3,\cdots\right)$ (C.2)

and

$R_{q}\equiv(-1)^{q-1}\int_{n_{\rm min}}^{n_{\rm
max}}dx\;\frac{B_{q}(x-\lfloor x\rfloor)}{q!}h^{(q)}(x).$ (C.3)

The symbole $\lfloor\cdots\rfloor$ denotes the floor function. Here,
$B_{q}(x)$ is the Bernoulli polynomial whose explicit expression is given by

$\displaystyle B_{0}(x)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$ $\displaystyle 1,$
$\displaystyle B_{1}(x)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$ $\displaystyle
x-\frac{1}{2},$ $\displaystyle B_{2}(x)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle x^{2}-x+\frac{1}{6},$ $\displaystyle B_{3}(x)\\!\\!\\!$
$\displaystyle=\\!\\!\\!$ $\displaystyle x^{3}-\frac{3}{2}x^{2}+\frac{1}{2}x,$
$\displaystyle B_{4}(x)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$ $\displaystyle
x^{4}-2x^{3}+x^{2}-\frac{1}{30},$ $\displaystyle B_{5}(x)\\!\\!\\!$
$\displaystyle=\\!\\!\\!$ $\displaystyle
x^{5}-\frac{5}{2}x^{4}+\frac{5}{3}x^{3}-\frac{1}{6}x,$ $\displaystyle
B_{6}(x)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$ $\displaystyle
x^{6}-3x^{5}+\frac{5}{2}x^{4}-\frac{1}{2}x^{2}+\frac{1}{42},$ (C.4)
$\displaystyle\vdots$

These polynomials satisfy the relation

$B_{q}(1-x)=(-1)^{q}B_{q}(x).$ (C.5)

###  C.2 Simple examples

Let us define

$H[f](N)\equiv\sum_{n=0}^{N}f(n)-\int_{0}^{N}dx\;f(x),$ (C.6)

where

$f(x)=x,\;\;x^{2},\;\;x^{3}.$ (C.7)

Then we can explicitly calculate $H[f](N)$ as

$\displaystyle H[x](N)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\sum_{n=0}^{N}n-\int_{0}^{N}dx\;x=\frac{N(N+1)}{2}-\frac{N^{2}}{2}=\frac{N}{2},$
$\displaystyle H[x^{2}](N)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\sum_{n=0}^{N}n^{2}-\int_{0}^{N}dx\;x^{2}=\frac{N(N+1)(2N+1)}{6}-\frac{N^{3}}{3}=\frac{3N^{2}+N}{6},$
$\displaystyle H[x^{3}](N)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\sum_{n=0}^{N}n^{3}-\int_{0}^{N}dx\;x^{3}=\frac{N^{2}(N+1)^{2}}{4}-\frac{N^{4}}{4}=\frac{2N^{3}+N^{2}}{4}.$
(C.8)

These results can be reproduced by means of the Euler-Maclaurin formula. From
(C.1), we have

$H[f](N)=\frac{1}{2}\left\\{f(0)+f(N)\right\\}+\sum_{p=1}^{\lfloor(m+1)/2\rfloor}\frac{B_{2p}}{(2p)!}\left\\{f^{(2p-1)}(N)-f^{(2p-1)}(0)\right\\}+R_{m+1},$
(C.9)

where we have chosen the integer $q$ in (C.1) as $q=m+1$. Since
$f^{(m+1)}(x)=0$, the remainder term $R_{m+1}$ vanishes. Therefore, we have

$\displaystyle H[x](N)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{2}\left\\{f(0)+f(N)\right\\}+\frac{B_{2}}{2}\left\\{f^{(1)}(N)-f^{(1)}(0)\right\\}$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{2}\left(0+N\right)+\frac{1}{12}\left(1-1\right)=\frac{N}{2},$
$\displaystyle H[x^{2}](N)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{2}\left\\{f(0)+f(N)\right\\}+\frac{B_{2}}{2}\left\\{f^{(1)}(N)-f^{(1)}(0)\right\\}$
$\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{2}\left(0+N^{2}\right)+\frac{1}{12}\left(2N-0\right)=\frac{N^{2}}{2}+\frac{N}{6},$
$\displaystyle H[x^{3}](N)\\!\\!\\!$ $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{2}\left\\{f(0)+f(N)\right\\}+\frac{B_{2}}{2}\left\\{f^{(1)}(N)-f^{(1)}(0)\right\\}+\frac{B_{4}}{4!}\left\\{f^{(3)}(N)-f^{(3)}(0)\right\\}$
(C.10) $\displaystyle=\\!\\!\\!$
$\displaystyle\frac{1}{2}\left(0+N^{3}\right)+\frac{1}{12}\left(3N^{2}-0\right)-\frac{1}{720}\left(6-6\right)=\frac{N^{3}}{2}+\frac{N^{2}}{4}.$

These agrees with (C.8).

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##  Appendix A Some extra material

Here we present one quarter cuts of the Kepler light curves for all stars from
our set (Fig. 33). In addition there is shown an example of the full light
curve with outburst (Fig. 34). Finally, examples of TD periodograms for stars
with incoherent signal are plotted in Figs. 35-39.

Figure 33: One quarter ($\sim 90$ days) cuts of Kepler light curves for all
program stars. Figure 34: The full Kepler light curve of KIC 4939281. Figure
35: Time dependent periodogram of KIC 4939281. Figure 36: The same as in Fig.
35 but for KIC 5477601. Figure 37: The same as in Fig. 35 but for KIC 9278405.
Figure 38: The same as in Fig. 35 but for KIC 9715425. Figure 39: The same as
in Fig. 35 but for KIC 10790075.