# Stringy scaling of $n$-point Regge string scattering amplitudes

Sheng-Hong Lai<EMAIL_ADDRESS>Department of Electrophysics, National
Yang Ming Chiao Tung University, Hsinchu, Taiwan, R.O.C.  Jen-Chi Lee
<EMAIL_ADDRESS>Department of Electrophysics, National Yang Ming Chiao
Tung University, Hsinchu, Taiwan, R.O.C.  Institute of Physics, University of
Tokyo, Komaba, Meguro-ku, Tokyo 153-8902, Japan  Yi Yang
<EMAIL_ADDRESS>Department of Electrophysics, National Yang Ming Chiao
Tung University, Hsinchu, Taiwan, R.O.C.

###### Abstract

We discover a stringy scaling behavior for a class of $n$-point Regge string
scattering amplitudes (RSSA). The number of independent kinematics variables
is found to be reduced by dim$\mathcal{M}$.

##  I Introduction

Recent development of string scattering amplitudes (SSA) has shown that a
class of $4$-point SSA form representations of the $SL(K+3,C)$ group Group ;
LSSA . These are SSA with three tachyons and one arbitrary string states

$\left|r_{n}^{T},r_{m}^{P},r_{l}^{L}\right\rangle=\prod_{n>0}\left(\alpha_{-n}^{T}\right)^{r_{n}^{T}}\prod_{m>0}\left(\alpha_{-m}^{P}\right)^{r_{m}^{P}}\prod_{l>0}\left(\alpha_{-l}^{L}\right)^{r_{l}^{L}}|0,k\rangle$
(1.1)

where $e^{P}=\frac{1}{M_{2}}(E_{2},\mathrm{k}_{2},0)=\frac{k_{2}}{M_{2}}$ is
the momentum polarization, $e^{L}=\frac{1}{M_{2}}(\mathrm{k}_{2},E_{2},0)$ is
the longitudinal polarization and $e^{T}=(0,0,1)$ is the transverse
polarization on the $\left(2+1\right)$-dimensional scattering plane. Note that
SSA of three tachyons and one arbitrary string states with polarizations
orthogonal to the scattering plane vanish. In addition to the mass level
$M_{2}^{2}=2(N-1)$ with

$N=\sum_{\begin{subarray}{c}n,m,l>0\\\ \\{\text{ }r_{j}^{X}\neq
0\\}\end{subarray}}\left(nr_{n}^{T}+mr_{m}^{P}+lr_{l}^{L}\right),$ (1.2)

another important index $K$ was identified for the state in Eq.(1.1) LLY2

$K=\sum_{\begin{subarray}{c}n,m,l>0\\\ \\{\text{ }r_{j}^{X}\neq
0\\}\end{subarray}}\left(n+m+l\right)$ (1.3)

where $X=\left(T,P,L\right)$ and one has put $r_{n}^{T}=r_{m}^{P}=r_{l}^{L}=1$
in Eq.(1.2) in the definition of $K$ in Eq.(1.3). Intuitively, $K$ counts the
number of variety of the $\alpha_{-j}^{X}$ oscillators in Eq.(1.1).

The representation bases of the above subclass of $4$-point SSA was soon
extended to all $4$-point SSA with arbitrary four string states, and
eventually to all $n$-point SSA with arbitrary $n$ string states LLY3 ; exact
. It is thus important to study whether other known interesting
characteristics of the $4$-point SSA can be similarly extended to the
$n$-point SSA. One such characteristics of the $4$-point SSA is the existence
of infinite linear relations and their associated constant ratios, independent
of the scattering angle $\phi$, among hard SSA (HSSA) at each fixed mass level
of the open bosonic string spectrum. These infinite linear relations and their
associated constant ratios were first conjectured by Gross Gross ; Gross1  and
later explicitly calculated by the method of decoupling of zero-norm states in
ChanLee ; ChanLee2 ; CHL ; CHLTY2 ; CHLTY1 .

Indeed, in one of the authors’ previous publications NHard , we discovered a
general stringy scaling behavior for all $n$-point HSSA to all string loop
orders. For the simplest case of $n=4$, the stringy scaling behavior reduces
to the infinite linear relations and the constant ratios of HSSA at each mass
level mentioned above. For this case, the ratios are independent of $1$
scattering angle $\phi$ and thus the number of independent kinematics variable
reduced from $1$ to $0$ with dim$\mathcal{M=}$ $1-0=1$. In general higher
$n$-point HSSA, the stringy scaling behavior implies that the number of
independent kinematics variables of the ratios reduced by dim$\mathcal{M}$
NHard . See the definition of dim$\mathcal{M}$ in Eq.(2.35) and Eq.(2.40). As
a result, the linear relations and their associated constant ratios of
$4$-point HSSA persist only in the parameter spaces $\mathcal{M}$ for the
cases of higher $n$-point HSSA NHard . See the example of constant ratios
calculated among $6$-point HSSA in Eq.(2.39).

In this paper, we will extend our calculation of stringy scaling behavior of
HSSA to the case of Regge SSA (RSSA). We will demonstrate a stringy scaling
behavior for a class of $n$-point RSSA, and the number of independent
kinematics variables is again found to be reduced by dim$\mathcal{M}$.

This paper is organized as following. In section II, we review and give a
detailed calculation of the stringy scaling behavior of HSSA NHard . In
section III, we give a saddle point calculation of the hard stringy scaling to
justify the zero norm state (ZNS) calculation in section II. Section IV and V
are the main parts of this paper and we extend the calculation of HSSA to the
stringy scaling of RSSA. We will derive a stringy scaling behavior for a class
of $n$-point RSSA with arbitrary $n$ in section V. A brief conclusion was
given in section VI.

##  II The hard stringy scaling

A brief report on stringy scaling of $n$-point hard string scattering
amplitudes (HSSA) was recently given in NHard . In this section, we will first
give a detailed calculation of hard string scaling behavior. This can also be
served as a preparartion for the calculation of stringy scaling of Regge
string scattering amplitudes (RSSA) to be discussed in section IV and V.

###  II.1 Stringy scaling of $4$-point HSSA

The first stringy scaling was conjectured by Gross in 1988 Gross  which
claimed that all $4$-point HSSA ($E\rightarrow\infty$, fixed $\phi$) at each
fixed mass level share the same functional form. That is, all HSSA at each
fixed mass level are proportional to each other with constant ratios
independent of the scattering angle $\phi$.

$k_{1}$$k_{2}$$-k_{3}$$-k_{4}$$e^{T}(1)$$e^{T}(2)$$e^{T}(3)$$e^{T}(4)$$\phi$
Figure 1: Kinematic variables in the center of mass frame

To show this remarkable behavior, the starting point is to apply the $4$-point
$l$-loop stringy on-shell Ward identities ChanLee ; ChanLee2

$\left\langle V_{1}\chi V_{3}V_{4}\right\rangle_{l-loop}=0$ (2.1)

in the hard scattering limit. In Eq.(2.1) $V_{j}$ above can be any string
vertex and the second vertex $\chi$ is the vertex of a zero-norm state (ZNS).
In the hard scattering limit, components of polarization orthogonal to the
scattering plane are subleading order in energy. On the other hand, it can be
shown that at each fixed mass level $M^{2}=2(N-1)$ only states of the
following form CHLTY2 ; CHLTY1  (in the hard scattering limit $e^{P}\simeq$
$e^{L}$)

$\left|N,2m,q\right\rangle=\left(\alpha_{-1}^{T}\right)^{N-2m-2q}\left(\alpha_{-1}^{L}\right)^{2m}\left(\alpha_{-2}^{L}\right)^{q}\left|0;k\right\rangle$
(2.2)

are leading order in energy.

There are two types of physical ZNS in the old covariant first quantized open
bosonic string spectrum: GSW

$\text{Type I}:L_{-1}\left|y\right\rangle,\text{ where
}L_{1}\left|y\right\rangle=L_{2}\left|y\right\rangle=0,\text{
}L_{0}\left|y\right\rangle=0;$ (2.3) $\text{Type
II}:(L_{-2}+\frac{3}{2}L_{-1}^{2})\left|\widetilde{y}\right\rangle,\text{
where
}L_{1}\left|\widetilde{y}\right\rangle=L_{2}\left|\widetilde{y}\right\rangle=0,\text{
}(L_{0}+1)\left|\widetilde{y}\right\rangle=0.\text{(}D=26\text{ {only}).}$
(2.4)

(1) We first consider $\chi$ to be the type I hard ZNS (HZNS) calculated from
Type I ZNS

$\displaystyle L_{-1}|N-1,2m-1,q\rangle$
$\displaystyle=(M\alpha_{-1}^{L}+\alpha_{-2}^{L}\alpha_{1}^{L}+\underset{irrelevant}{\underbrace{\alpha_{-2}^{T}\alpha_{1}^{T}+\alpha_{-3}\cdot\alpha_{2}+\cdots}})|N-1,2m-1,q\rangle$
$\displaystyle\simeq M|N,2m,q\rangle+(2m-1)|N,2m-2,q+1\rangle$ (2.5)

where many terms are omitted because they are not of the form of Eq.(2.2).
This implies the following relation among $4$-point amplitudes

$\mathcal{T}^{(N,2m,q)}=-\frac{2m-1}{M}\mathcal{T}^{(N,2m-2,q+1)}.$ (2.6)

Using this relation repeatedly, we get

$\mathcal{T}^{(N,2m,q)}=\frac{(2m-1)!!}{(-M)^{m}}\mathcal{T}^{(N,0,m+q)}.$
(2.7)

(2) Next, we consider another class of HZNS calculated from type II ZNS

$\displaystyle L_{-2}|N-2,0,q\rangle$
$\displaystyle=(\frac{1}{2}\alpha_{-1}^{T}\alpha_{-1}^{T}+M\alpha_{-2}^{L}+\underset{irrelevant}{\underbrace{\alpha_{-3}\cdot\alpha_{1}+\cdots}})|N-2,0,q\rangle$
$\displaystyle\simeq\frac{1}{2}|N,0,q\rangle+M|N,0,q+1\rangle.$ (2.8)

Again, irrelevant terms are omitted here. From this we deduce that

$\mathcal{T}^{(N,0,q+1)}=-\frac{1}{2M}\mathcal{T}^{(N,0,q)},$ (2.9)

which leads to

$\mathcal{T}^{(N,0,q)}=\frac{1}{(-2M)^{q}}\mathcal{T}^{(N,0,0)}.$ (2.10)

In conclusion, the decoupling of ZNS in Eq.(2.7) and Eq.(2.10) leads to
constant ratios among $4$-point HSSA ChanLee ; ChanLee2 ; CHLTY2 ; CHLTY1

$\frac{\mathcal{T}^{\left(N,2m,q\right)}}{\mathcal{T}^{\left(N,0,0\right)}}=\frac{\left(2m\right)!}{m!}\left(\frac{-1}{2M}\right)^{2m+q}.\text{({independent
of }}\phi\text{ !)}$ (2.11)

In Eq.(2.11) $\mathcal{T}^{\left(N,2m,q\right)}$ is the $4$-point HSSA of any
string vertex $V_{j}$ with $j=1,3,4$ and $V_{2}$ is the high energy state in
Eq.(2.2); while $\mathcal{T}^{\left(N,0,0\right)}$ is the $4$-point HSSA of
any string vertex $V_{j}$ with $j=1,3,4$, and $V_{2}$ is the leading Regge
trajectory string state at mass level $N$. Note that we have omitted the
tensor indice of $V_{j}$ with $j=1,3,4$ and keep only those of $V_{2}$ in
$\mathcal{T}^{\left(N,2m,q\right)}$.

###  II.2 Examples of $4$-point stringy scaling

####  II.2.1 Bosonic open string

Since the ratios of the amplitudes in Eq.(2.11) are independent of the choices
of $V_{1}$, $V_{3}$ and $V_{4}$, we choose them to be tachyons and $V_{2}$ to
be Eq.(2.2). On the other hand, since the ratios are independent of the loop
order, we choose to calculate HSSA of $l=0$ loop. An explicit amplitude
calculation for $M^{2}=4$, $6$ and $8$ gives ChanLee ; ChanLee2 ; CHLTY2 ;
CHLTY1

$\mathcal{T}_{TTT}:\mathcal{T}_{(LLT)}:\mathcal{T}_{(LT)}:\mathcal{T}_{[LT]}=8:1:-1:-1,$
(2.12)

$\displaystyle\mathcal{T}_{(TTTT)}:\mathcal{T}_{(TTLL)}:\mathcal{T}_{(LLLL)}:\mathcal{T}_{TT,L}:\mathcal{T}_{(TTL)}:\mathcal{T}_{(LLL)}:\mathcal{T}_{(LL)}$
$\displaystyle=16:\frac{4}{3}:\frac{1}{3}:-\frac{2\sqrt{6}}{3}:-\frac{4\sqrt{6}}{9}:-\frac{\sqrt{6}}{9}:\frac{2}{3}$
(2.13)

and

$\displaystyle\mathcal{T}_{(TTTTT)}:\mathcal{T}_{(TTTL)}:\mathcal{T}_{(TTTLL)}:\mathcal{T}_{(TLLL)}:\mathcal{T}_{(TLLLL)}:\mathcal{T}_{(TLL)}:\mathcal{T}_{T,LL}:\mathcal{T}_{TLL,L}:\mathcal{T}_{TTT,L}$
$\displaystyle=32:\sqrt{2}:2:\frac{3\sqrt{2}}{16}:\frac{3}{8}:\frac{1}{3}:\frac{2}{3}:\frac{\sqrt{2}}{16}:3\sqrt{2},$
(2.14)

respectively. These are all remarkably consistent with Eq.(2.11) of ZNS
calculation review ; over .

It is important to note that for subleading order amplitudes, they are in
general not proportional to each other. For $M^{2}=4$, for example, one gets
$6$ subleading order amplitudes and $4$ linear relations (on-shell Ward
identities) in the ZNS calculation. An explicit subleading order amplitude
calculation gives ChanLee ; ChanLee2

$\displaystyle\mathcal{T}_{LLL}^{2}$
$\displaystyle\sim-4E^{8}\sin\phi\cos\phi,$
$\displaystyle\mathcal{T}_{LTT}^{2}$
$\displaystyle\sim-8E^{8}\sin^{2}\phi\cos\phi,$ (2.15)

which show that the proportional coefficients do depend on the scattering
angle $\phi$.

####  II.2.2 Bosonic closed string and D-particle

For closed string scatterings Closed ; bosonic2 , one can use the KLT formula
KLT , which expresses the relation between tree amplitudes of closed and two
channels of open string
$(\alpha_{\text{closed}}^{\prime}=4\alpha_{\text{open}}^{\prime}=2),$ to
obtain the closed string ratios which are the tensor product of two open
string ratios in Eq.(2.11). On the other hand, it is interesting to find that
the ratios of hard closed string D-particle scatterings are again given by the
tensor product of two open string ratios LMY

$\frac{T_{SD}^{\left(N;2m,2m^{{}^{\prime}};q,q^{{}^{\prime}}\right)}}{T_{SD}^{\left(N;0,0;0,0\right)}}=\left(-\frac{1}{M_{2}}\right)^{2(m+m^{{}^{\prime}})+q+q^{{}^{\prime}}}\left(\frac{1}{2}\right)^{m+m^{{}^{\prime}}+q+q^{{}^{\prime}}}(2m-1)!!(2m^{\prime}-1)!!,$
(2.16)

which came as a surprise since there is no physical picture for open string
D-particle tree scattering amplitudes and thus no factorization for closed
string D-particle scatterings into two channels of open string D-particle
scatterings, and hence no KLT-like formula there. However, these ratios are
consistent with the decoupling of high energy ZNS calculation.

####  II.2.3 Stringy scaling of Superstring

It turned out to be nontrivial to extend the linear relations and their
associated constant ratios of the HSSA of bosonic string to the case of $10D$
open superstring. First of all, in addition to the NS-sector, there are
massive fermionic states in the R-sector whose vertex operators are still
unknown except the leading Regge trajectory states in the spectrum Osch . So
the only known complete vertex operators so far are those for the mass level
$M^{2}=2$ RRR  which contains no off leading massive Regge trajectory
fermionic string states.

Secondly, in the NS-sector of $M^{2}=2$ it was surprised to note that New
there exists no ”inter-particle gauge transformation” induced by bosonic ZNS
for the two positive-norm physical propagating states, the symmetric spin
three and the anti-symmetric spin two states. However, the $4$-point HSSA
among these two positive-norm states are still related and are indeed again
proportional to each others. Presumably, this is due to the massive spacetime
SUSY and the existence of spacetime massive fermion string scattering
amplitudes of the R-sector of the theory spin .

Thirdly, it was noted that for the HSSA of the NS sector of superstring, there
existed leading order HSSA with polarizations orthogonal to the scattering
plane New . This was due to the ”worldsheet fermion exchange” susy  in the
correlation functions and was argued to be related to the HSSA of massive
spacetime fermion of R-sector of the theory spin .

The first calculation of the $4$-point superstringy scaling was performed for
the NS-sector of $10D$ open superstring theory. There are four classes of HSSA
of superstring which are all proportional to each other susy

$\displaystyle\left|N,2m,q\right\rangle\otimes\left|b_{-\frac{3}{2}}^{P}\right\rangle$
$\displaystyle=\left(-\frac{1}{2M_{2}}\right)^{q+m}\frac{\left(2m-1\right)!!}{\left(-M_{2}\right)^{m}}\left|N,0,0\right\rangle\otimes\left|b_{-\frac{3}{2}}^{P}\right\rangle,$
(2.17)
$\displaystyle\left|N+1,2m+1,q\right\rangle\otimes\left|b_{-\frac{1}{2}}^{P}\right\rangle$
$\displaystyle=\left(-\frac{1}{2M_{2}}\right)^{q+m}\frac{\left(2m+1\right)!!}{\left(-M_{2}\right)^{m+1}}\left|N,0,0\right\rangle\otimes\left|b_{-\frac{3}{2}}^{P}\right\rangle,$
(2.18)
$\displaystyle\left|N+1,2m,q\right\rangle\otimes\left|b_{-\frac{1}{2}}^{T}\right\rangle$
$\displaystyle=\left(-\frac{1}{2M_{2}}\right)^{q+m}\frac{\left(2m-1\right)!!}{\left(-M_{2}\right)^{m-1}}\left|N,0,0\right\rangle\otimes\left|b_{-\frac{3}{2}}^{P}\right\rangle,$
(2.19)
$\displaystyle\left|N-1,2m,q-1\right\rangle\otimes\left|b_{-\frac{1}{2}}^{T}b_{-\frac{1}{2}}^{P}b_{-\frac{3}{2}}^{P}\right\rangle$
$\displaystyle=\left(-\frac{1}{2M_{2}}\right)^{q+m}\frac{\left(2m-1\right)!!}{\left(-M_{2}\right)^{m}}\left|N,0,0\right\rangle\otimes\left|b_{-\frac{3}{2}}^{P}\right\rangle.$
(2.20)

Note that, in order to simplify the notation, we have only shown the second
state of the four point functions to represent the scattering amplitudes on
both sides of each equation above. Eqs.(2.17) to (2.20) are thus the SUSY
generalization of Eq.(2.11) for the bosonic string.

Moreover, a recent calculation showed that spin  among $2^{4}\times 2^{4}=256$
$4$-point polarized fermion SSA (PFSSA) in the R-sector of $M^{2}=2$ states,
only $16$ of them are of leading order in energy and all of them share the
same functional form in the hard scattering limit. On the other hand, the
ratios of the complete $4$-point HSSA in the NS sector of mass level $M^{2}=2$
which include HSSA with polarizations orthogonal to the sacttering plane are
New

$\displaystyle\left\langle
b_{\frac{-1}{2}}^{T},\alpha_{-1}^{T}b_{\frac{-1}{2}}^{T}\right\rangle$
$\displaystyle:\left\langle
b_{\frac{-1}{2}}^{T},\left(2b_{\frac{-1}{2}}^{L}\alpha_{-1}^{L}-Mb_{\frac{-3}{2}}^{L}\right)\right\rangle:\left\langle
b_{\frac{-1}{2}}^{T_{i}},\alpha_{-1}^{T}b_{\frac{-1}{2}}^{T_{j}}\right\rangle:\left\langle
b_{\frac{-1}{2}}^{T_{k}},b_{\frac{-1}{2}}^{L}b_{\frac{-1}{2}}^{T}b_{\frac{-1}{2}}^{T_{l}}\right\rangle$
(2.21)
$\displaystyle=-2k_{3}^{T}E^{2}:-2(\frac{2}{M^{2}}+1)k_{3}^{T}E^{2}:\delta_{ij}2k_{3}^{T}E^{2}:\delta_{lk}\frac{-2k_{3}^{T}E^{2}}{M}$
$\displaystyle=1:2:-\delta_{ij}:\frac{\delta_{lk}}{\sqrt{2}}.\text{ \
}(\text{\ }i,j,k,l=3,4,5,...,9)$ (2.22)

where we have, for simplicity, omitted the last two tachyon vertices in the
notation of each HSSA in Eq.(2.21). In sum, in the NS sector one gets
$1+1+7+7=16$ HSSA in Eq.(2.21). This result agrees with those of $16$ hard
massive PFSSA in the R-sector calculated recently spin .

####  II.2.4 Field theory

On the other hand, in field theory, as an example, the leading order process
of the elastic scattering of a spin-$\frac{1}{2}$ particle by a spin-$0$
particle such as $e^{-}\pi^{+}\longrightarrow e^{-}\pi^{+}$, the non-vanishing
amplitudes were shown to be JW

$\displaystyle\mathcal{T}\text{ }(e_{R}^{-}\pi^{+}$
$\displaystyle\longrightarrow e_{R}^{-}\pi^{+})=\mathcal{T}\text{
}(e_{L}^{-}\pi^{+}\longrightarrow e_{L}^{-}\pi^{+})\sim\text{
}\cos\frac{\phi}{2},$ (2.23) $\displaystyle\mathcal{T}\text{
}(e_{R}^{-}\pi^{+}$ $\displaystyle\longrightarrow
e_{L}^{-}\pi^{+})=\mathcal{T}\text{ }(e_{L}^{-}\pi^{+}\longrightarrow
e_{R}^{-}\pi^{+})\sim\text{ }\sin\frac{\phi}{2},$ (2.24)

which are not proportional to each other. In QED, as another example, for the
leading order process of $e^{-}e^{+}\longrightarrow\mu^{-}\mu^{+}$, there are
$4$ non-vanishing among $16$ hard polarized amplitudes Peskin

$\displaystyle\mathcal{T}\text{ }(e_{R}^{-}e_{L}^{+}$
$\displaystyle\longrightarrow\mu_{R}^{-}\mu_{L}^{+})=\mathcal{T}\text{
}(e_{L}^{-}e_{R}^{+}\longrightarrow\mu_{L}^{-}\mu_{R}^{+})\sim\text{
}(1+\cos\theta)=2\text{ }\cos^{2}\frac{\phi}{2},$ (2.25)
$\displaystyle\mathcal{T}\text{ }(e_{R}^{-}e_{L}^{+}$
$\displaystyle\longrightarrow\mu_{L}^{-}\mu_{R}^{+})=\mathcal{T}\text{
}(e_{L}^{-}e_{R}^{+}\longrightarrow\mu_{R}^{-}\mu_{L}^{+})\sim\text{
}(1-\cos\theta)=2\text{ }\sin^{2}\frac{\phi}{2},$ (2.26)

and they are not all proportional to each other.

###  II.3 Stringy scaling of higher point ($n\geq 5$) HSSA

It is tempted to extend the stringy scaling behavior of $4$-point SSA derived
in the previous subsection to the higher point SSA. The $n$-point stringy on-
shell Ward identities can be written as

$\left\langle V_{1}\chi V_{3}\cdots V_{n}\right\rangle_{l-loop}=0$ (2.27)

where $\chi$ again is the vertex of a ZNS. We begin the discussion with a
simple kinematics regime on the scattering plane.

####  II.3.1 On the scattering plane

In the hard scattering limit on the scattering plane, the space part of
momenta $k_{j}$ ( $j=3,4,\cdots,n$) form a closed $1$-chain with $(n-2)$ sides
due to momentum conservation. It turned out that all the $4$-point calculation
in the previous subsection persist and one ends up with Eq.(2.11) again NHard
. However, while for $n=4$ the ratios are independent of $1$ scattering angle
$\phi$, for $n=5$, the ratios are independent of $3$ kinematics variables ($2$
angles and $1$ fixed ratio of two infinite energies) or, for simplicity, $3$
scattering ”angles”. For $n=6$, there are $5$ scattering ”angles” etc..

####  II.3.2 Out of the scattering plane

The general high energy states at each fixed mass level $M^{2}=2(N-1)$ can be
written as NHard

$\left|\left\\{p_{i}\right\\},2m,q\right\rangle=\left(\alpha_{-1}^{T_{1}}\right)^{N+p_{1}}\left(\alpha_{-1}^{T_{2}}\right)^{p_{2}}\cdots\left(\alpha_{-1}^{T_{r}}\right)^{p_{r}}\left(\alpha_{-1}^{L}\right)^{2m}\left(\alpha_{-2}^{L}\right)^{q}\left|0;k\right\rangle$
(2.28)

where $\sum_{i=1}^{r}p_{i}=-2(m+q)$ with $r\leq 24$. In Eq.(2.28), $T_{j}$ is
the $j$th transverse direction orthogonal to $k_{2}$. For higher dimensional
scattering space, one generalizes the transverse polarization $e^{T}=(0,0,1)$
to $e^{\hat{T}}=(0,0,\vec{\omega})$ where

$\omega_{i}=\cos\theta_{i}\prod\limits_{\sigma=1}^{i-1}\sin\theta_{\sigma}\text{with
}i=1,\cdots,r,\text{ }\theta_{r}=0\text{ }$ (2.29)

are the solid angles in the transverse space spanned by $24$ transverse
directions $e^{T_{i}}$. Note that $\alpha_{-1}^{\hat{T}}=\alpha_{-1}\cdot
e^{\hat{T}}$ etc. With
$\left(\alpha_{-1}^{T_{i}}\right)=\left(\alpha_{-1}^{\hat{T}}\right)\omega_{i}$,
we easily obtain

$\displaystyle\left(\alpha_{-1}^{T_{1}}\right)^{N+p_{1}}\left(\alpha_{-1}^{T_{2}}\right)^{p_{2}}\cdots\left(\alpha_{-1}^{T_{r}}\right)^{p_{r}}\left(\alpha_{-1}^{L}\right)^{2m}\left(\alpha_{-2}^{L}\right)^{q}\left|0;k\right\rangle$
$\displaystyle=\left(\omega_{1}^{N}\prod_{i=1}^{r}\omega_{i}^{p_{i}}\right)\left(\alpha_{-1}^{\hat{T}}\right)^{N-2m-2q}\left(\alpha_{-1}^{L}\right)^{2m}\left(\alpha_{-2}^{L}\right)^{q}\left|0;k\right\rangle,$
(2.30)

which leads to the ratios of $n$-point HSSA NHard

$\frac{\mathcal{T}^{\left(\left\\{p_{i}\right\\},2m,q\right)}}{\mathcal{T}^{\left(\left\\{0_{i}\right\\},0,0\right)}}=\frac{\left(2m\right)!}{m!}\left(\frac{-1}{2M}\right)^{2m+q}\prod_{i=1}^{r}\omega_{i}^{p_{i}}$
(2.31)

where $\mathcal{T}^{\left(\left\\{0_{i}\right\\},0,0\right)}$ is the HSSA of
leading Regge trajectory state at mass level $M^{2}=2(N-1)$. It is important
to note that the number of kinematics variables dependence in the ratios of
Eq.(2.31) reduced. This stringy scaling behavior of $n$-point ($n\geq 5$) HSSA
is the generalization of that of $4$-point HSSA in Eq.(2.11). Since the result
of ZNS calculation in Eq.(2.31) is based on the stringy Ward identity in
Eq.(2.27), The ratios calculated in Eq.(2.31) are valid to all string loop
orders.

###  II.4 Degree of Stringy Scaling

We see in the previous section that for the simple case with $n=4$ and $r=1$,
one has two variables, $s$ and $t$ (or $E$, $\phi$). The ratios of all HSSA
are independent of the scattering angle $\phi$ and we will call the degree of
the scaling dim$\mathcal{M}=1$. The dependence of the number of kinematics
variable reduced from $1$ to $0$ and we have $1-0=$ dim$\mathcal{M}=1$. (see
the definition of $\mathcal{M}$ below)

For the general $n$-point HSSA with $r\leq 24$, $d=r+2$, we have $k_{j}$
vector with $j=1,\cdots,n$ and $k_{j}$ $\in R^{d-1,1}$. The number of
kinematics variables is $n\left(d-1\right)-\frac{d\left(d+1\right)}{2}$.
Indeed, as $p=E\rightarrow\infty$ , that implies $q_{j}\rightarrow\infty$ in
the hard limit, we define the $26$-dimensional momenta in the CM frame to be

$\displaystyle k_{1}$ $\displaystyle=\left(E,-E,0^{r}\right),$ $\displaystyle
k_{2}$ $\displaystyle=\left(E,+E,0^{r}\right),$ $\displaystyle\vdots$
$\displaystyle k_{j}$
$\displaystyle=\left(-q_{j},-q_{j}\Omega_{1}^{j},-q_{j}\Omega_{2}^{j},\cdots,-q_{j}\Omega_{r}^{j},-q_{j}\Omega_{r+1}^{j}\right)$
(2.32)

where $j=3,4,\cdots,n$, and

$\Omega_{i}^{j}=\cos\phi_{i}^{j}\prod\limits_{\sigma=1}^{i-1}\sin\phi_{\sigma}^{j}\text{
with }\phi_{j-1}^{j}=0,\text{ }\phi_{i>r}^{j}=0\text{ and
}r\leq\min\left\\{n-3,24\right\\}$ (2.33)

are the solid angles in the $\left(j-2\right)$-dimensional spherical space
with $\sum_{i=1}^{j-2}\left(\Omega_{i}^{j}\right)^{2}=1$. In Eq.(2.32),
$0^{r}$ denotes the $r$-dimensional null vector. The condition
$\phi_{j-1}^{j}=0$ in Eq.(2.33) was chosen to fix the frame by using the
rotational symmetry.

The independent kinematics variables can be chosen to be some
$\varphi_{i}^{j}$ and some fixed ratios of infinite $q_{j}$. For the
kinematics parameter space $\mathcal{M}$ defined by NHard

$\omega_{j}\left(\text{kinematics parameters with
}E\rightarrow\infty\right)=\text{fixed constant \ }(j=2,\cdots,r)\text{,}$
(2.34)

we can count the dimension of $\mathcal{M}$ to be NHard

$\text{dim}\mathcal{M}\text{ }\mathcal{=}\text{
}n\left(d-1\right)-\frac{d\left(d+1\right)}{2}-1-\left(r-1\right)=\frac{\left(r+1\right)\left(2n-r-6\right)}{2}$
(2.35)

where $r=d-2$ is the number of transverse directions $e^{T_{i}}$. In sum, the
ratios among $n$-point $HSSA$ with $r\leq 24$ are constants and independent of
the scattering ”angles” in the kinematic regime $\mathcal{M}$.

####  II.4.1 Examples

(1). For $n=5$ and $r=2$, $d=r+2=4$ and one has
$n\left(d-1\right)-\frac{d\left(d+1\right)}{2}=5$ parameters ($r_{1}$ is the
ratio of two infinite energies)

$E,\phi_{2}^{3},\phi_{2}^{4},\phi_{3}^{4},r_{1}.$ (2.36)

In the hard scattering limit $E\rightarrow\infty$, for $\theta_{1}=fixed$ we
get dim$\mathcal{M=}$ $3$.

(2). For $n=6$ and $r=3$, the ratios of $6$-point HSSA depends only on $2$
variables $\theta_{1}$ and $\theta_{2}$ instead of $8$ ”angles” and
dim$\mathcal{M=}$ $6$. For this case, $\mathcal{M}$ is defined by

$\theta_{j}\left(8\text{ kinematics parameters}\right)=\text{fixed constant, \
}j=1,2,$ (2.37)

and the ratios NHard

$\frac{\mathcal{T}^{\left(\left\\{p_{1},p_{2},p_{3}\right\\},2m,q\right)}}{\mathcal{T}^{\left(\left\\{0,0,0\right\\},0,0\right)}}=\frac{\left(2m\right)!}{m!}\left(\frac{-1}{2M}\right)^{2m+q}\left(\cos\theta_{1}\right)^{p_{1}}\left(\sin\theta_{1}\cos\theta_{2}\right)^{p_{2}}\left(\sin\theta_{1}\sin\theta_{2}\right)^{p_{3}}$
(2.38)

are independent of kinematics parameters in the space $\mathcal{M}$. For
example, for say $\theta_{1}=\frac{\pi}{4}$ and $\theta_{2}=\frac{\pi}{6}$, we
get the ratios among $6$-point HSSA

$\frac{\mathcal{T}^{\left(\left\\{p_{1},p_{2},p_{3}\right\\},2m,q\right)}}{\mathcal{T}^{\left(\left\\{0,0,0\right\\},0,0\right)}}=\left(-\frac{1}{M}\right)^{2m+q}(2m-1)!!\left(\frac{1}{2}\right)^{p_{2}+p_{3}}\left(\sqrt{3}\right)^{p_{3}}\text{.}$
(2.39)

These ratios for higher point HSSA are one example of generalization of
previous ratios calculated in Eq.(2.11) for the case $4$-point HSSA.

####  II.4.2 General cases

In general, in the hard scattering limit, the number of scattering ”angles”
dependence on ratios of $n$-point $HSSA$ with $r\leq 24$ reduces by
dim$\mathcal{M}$. For a given ($n,r$), we can calculate some examples of
dim$\mathcal{M}$ NHard

$\begin{tabular}[c]{lllll}$\text{dim}\mathcal{M}$&$r=1$&$r=2$&$r=3$&$r=4$\\\
$n=4$&$1$&&&\\\ $n=5$&$3$&$3$&&\\\ $n=6$&$5$&$6$&$6$&\\\
$n=7$&$7$&$9$&$10$&$10$\end{tabular}\ \ \ \ \ \ \ \ \ \ \ \ \ \ .$ (2.40)

Note that for the $n=4$ and $r=1$ case, one obtains the previous $4$-point
case in Eq.(2.11).

##  III Saddle point calculation

To justify the ZNS calculation in Eq.(2.11) and Eq.(2.31), we use the saddle
point calculation to explicitly calculate the HSSA. Since the ratios are
independent of the choices of $V_{j}$ ($J=1,3,4\cdots,n$), we choose them to
be tachyons and $V_{2}$ to be the high energy state in Eq.(2.2). On the other
hand, since the ratios are independent of the loop order, we choose to
calculate $l=0$ loop. We begin with the $4$-point case CHLTY2 ; CHLTY1 .

###  III.1 The four point calculation

The $t-u$ channel contribution to the stringy amplitude at tree level is
(after $SL(2,R)$ fixing)

$\displaystyle\mathcal{T}^{(N,2m,q)}$
$\displaystyle=\int_{1}^{\infty}dxx^{(1,2)}(1-x)^{(2,3)}\left[\frac{e^{T}\cdot
k_{1}}{x}-\frac{e^{T}\cdot k_{3}}{1-x}\right]^{N-2m-2q}$
$\displaystyle\cdot\left[\frac{e^{P}\cdot k_{1}}{x}-\frac{e^{P}\cdot
k_{3}}{1-x}\right]^{2m}\left[-\frac{e^{P}\cdot k_{1}}{x^{2}}-\frac{e^{P}\cdot
k_{3}}{(1-x)^{2}}\right]^{q}$ (3.1)

where $(1,2)=k_{1}\cdot k_{2}$ etc.

In order to apply the saddle-point method, we rewrite the amplitude above into
the following form

$\mathcal{T}^{(N,2m,q)}(K)=\int_{1}^{\infty}dx\mbox{ }u(x)e^{-Kf(x)},$ (3.2)

where

$\displaystyle K$ $\displaystyle\equiv-(1,2)\rightarrow\frac{s}{2}\rightarrow
2E^{2},$ (3.3) $\displaystyle\tau$
$\displaystyle\equiv-\frac{(2,3)}{(1,2)}\rightarrow-\frac{t}{s}\rightarrow\sin^{2}\frac{\phi}{2},$
(3.4) $\displaystyle f(x)$ $\displaystyle\equiv\ln x-\tau\ln(1-x),$ (3.5)
$\displaystyle u(x)$
$\displaystyle\equiv\left[\frac{(1,2)}{M}\right]^{2m+q}(1-x)^{-N+2m+2q}\underset{}{\underbrace{(f^{\prime})^{2m}}}(f^{\prime\prime})^{q}(-e^{T}\cdot
k_{3})^{N-2m-2q}.$ (3.6)

The saddle-point for the integration of moduli, $x=x_{0}$, is defined by

$f^{\prime}(x_{0})=0,$ (3.7)

and we have

$x_{0}=\frac{1}{1-\tau}=\sec^{2}\frac{\phi}{2},\hskip
28.45274pt1-x_{0}=-\frac{\tau}{1-\tau},\hskip
28.45274ptf^{\prime\prime}(x_{0})=(1-\tau)^{3}\tau^{-1}.$ (3.8)

Due to the factor $(f^{\prime})^{2m}$ in Eq.(3.6), it is easy to see that
CHLTY2 ; CHLTY1

$u(x_{0})=u^{\prime}(x_{0})=....=u^{(2m-1)}(x_{0})=0,$ (3.9)

and

$u^{(2m)}(x_{0})=\left[\frac{(1,2)}{M}\right]^{2m+q}(1-x_{0})^{-N+2m+2q}(2m)!(f_{0}^{\prime\prime})^{2m+q}(-e^{T}\cdot
k_{3})^{N-2m-2q}.$ (3.10)

With these inputs, one can easily evaluate the Gaussian integral associated
with the four-point amplitudes CHLTY2 ; CHLTY1

$\displaystyle\int_{1}^{\infty}dx\mbox{ }u(x)e^{-Kf(x)}$
$\displaystyle=\sqrt{\frac{2\pi}{Kf_{0}^{\prime\prime}}}e^{-Kf_{0}}\left[\frac{u_{0}^{(2m)}}{2^{m}\
m!\ (f_{0}^{\prime\prime})^{m}\ K^{m}}+O(\frac{1}{K^{m+1}})\right]$
$\displaystyle=\sqrt{\frac{2\pi}{Kf_{0}^{\prime\prime}}}e^{-Kf_{0}}\left[(-1)^{N-q}\frac{2^{N-2m-q}(2m)!}{m!\
{M}^{2m+q}}\
\tau^{-\frac{N}{2}}(1-\tau)^{\frac{3N}{2}}E^{N}+O(E^{N-2})\right].$ (3.11)

This result shows explicitly that with one tensor and three tachyons, the
energy and angle dependence for the four-point HSS amplitudes only depend on
the level $N$ CHLTY2 ; CHLTY1

$\displaystyle\lim_{E\rightarrow\infty}\frac{\mathcal{T}^{(N,2m,q)}}{\mathcal{T}^{(N,0,0)}}$
$\displaystyle=\frac{(-1)^{q}(2m)!}{m!(2M)^{2m+q}}$
$\displaystyle=(-\frac{2m-1}{M})....(-\frac{3}{M})(-\frac{1}{M})(-\frac{1}{2M})^{m+q},$
(3.12)

which is remarkably consistent with calculation of decoupling of high energy
ZNS obtained in Eq.(2.11).

###  III.2 The $n$-point HSSA with $r=1$

To illustrate the $n$-point HSSA calculation, we begin with $n$-point HSSA
with $r=1$. We want to calculate $n$-point HSSA with $(n-1)$ tachyons and $1$
high energy state in Eq.(2.2). With the change of variables
$z_{i}=\frac{x_{i}}{x_{i+1}}$ or $x_{i}=z_{i}\cdots z_{n-2}$, the HSSA can be
written as

$\displaystyle\mathcal{T}^{\left(\left\\{p_{i}\right\\},m,q\right)}$
$\displaystyle=\int_{0}^{1}dx_{n-2}\cdots\text{
}\int_{0}^{x_{4}}dx_{3}\int_{0}^{x_{3}}dx_{2}ue^{-Kf}$
$\displaystyle=\int_{0}^{1}dz_{n-2}\cdots\text{
}\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}\begin{vmatrix}z_{3}\cdots
z_{n-2}&z_{2}z_{4}\cdots z_{n-2}&\cdots&z_{2}\cdots z_{n-3}\\\ 0&z_{4}\cdots
z_{n-2}&\cdots&\\\ &&\ddots&\\\ 0&0&\cdots&1\end{vmatrix}ue^{-Kf}$
$\displaystyle=\left(\prod_{i=3}^{n-2}\int_{0}^{1}dz_{i}\text{
}z_{i}^{i-2-N}\right)\int_{0}^{1}dz_{2}ue^{-Kf}$ (3.13)

where

$\displaystyle f\left(x_{i}\right)$
$\displaystyle=-\underset{i<j}{\sum}\frac{k_{i}\cdot
k_{j}}{K}\ln\left(x_{j}-x_{i}\right)=-\underset{i<j}{\sum}\frac{k_{i}\cdot
k_{j}}{K}\ln\left(z_{j}\cdots z_{n-2}-z_{i}\cdots z_{n-2}\right)$
$\displaystyle=-\underset{i<j}{\sum}\frac{k_{i}\cdot
k_{j}}{K}\left[\ln(z_{j}\cdots z_{n-2})+\ln\left(1-z_{i}\cdots
z_{j-1}\right)\right],\text{ }K=-k_{1}\cdot k_{2},$ (3.14) $\displaystyle
u\left(x_{i}\right)$
$\displaystyle=\left(k^{T}\right)^{N-2m-q}\underbrace{\left(k^{L}\right)^{2m}}\left(k^{\prime
L}\right)^{q}.(k^{\prime L}=\frac{\partial k^{L}}{\partial x_{2}})$ (3.15)

In Eq.(3.15), we have defined

$k=\sum_{i\neq 2,n}\frac{k_{i}}{x_{i}-x_{2}}=\sum_{i\neq
2,n}\frac{k_{i}}{z_{i}\cdots z_{n-2}-z_{2}\cdots z_{n-2}},$ (3.16)

and
$k_{\perp}=\left|k_{\perp}\right|\sum_{i=1}^{r}e^{T_{i}}\omega_{i}=\left|k_{\perp}\right|e^{\hat{T}}$.

The saddle points $\left(\tilde{z}_{2},\cdots,\tilde{z}_{n-2}\right)$ are the
solution of

$\frac{\partial f}{\partial z_{2}}=0\text{, }\cdots\text{, }\frac{\partial
f}{\partial z_{n-2}}=0.$ (3.17)

Note that Eq.(3.17) implies

$\tilde{k}^{L}=\frac{\tilde{k}\cdot
k_{2}}{M}=\frac{k_{12}}{M}\left.\frac{\partial f}{\partial
x_{2}}\right|_{z_{i}=\tilde{z}_{i}}=\frac{k_{12}}{M}\left.\frac{\partial
z_{j}}{\partial x_{2}}\frac{\partial f}{\partial
z_{j}}\right|_{z_{i}=\tilde{z}_{i}}=0\text{ ,
}\left|\tilde{k}\right|=\left|\tilde{k}_{\perp}\right|\text{. }$ (3.18)

We also define

$f_{2}\equiv\frac{\partial f}{\partial z_{2}}\text{,
}f_{22}\equiv\frac{\partial^{2}f}{\partial z_{2}^{2}}\text{,
}\tilde{f}=f\left(\tilde{z}_{2},\cdots,\tilde{z}_{n-2}\right)\text{,
}\tilde{f}_{22}=\left.\frac{\partial^{2}f}{\partial
z_{2}^{2}}\right|_{\left(\tilde{z}_{2},\cdots,\tilde{z}_{n-2}\right)}.$ (3.19)

In view of the factor $\left(k^{L}\right)^{2m}$ in Eq.(3.15) and Eq.(3.18),
all up to $(2m)$-order differentiations of $u$ function in Eq.(3.15) at the
saddle point vanish except NHard

$\displaystyle\left.\frac{\partial^{2m}u}{\partial
z_{2}^{2m}}\right|_{\left(\tilde{z}_{2},\cdots,\tilde{z}_{n-2}\right)}$
$\displaystyle=\left(\frac{k_{12}}{M}\right)^{2m+q}\left(-\sum_{i\neq
2,n}\frac{k_{i}^{T}}{\tilde{x}_{i}-\tilde{x}_{2}}\right)^{N-2m-2q}\left(2m\right)!\left(\tilde{f}_{22}\right)^{q+2m}$
$\displaystyle=\left(\frac{k_{12}}{M}\right)^{2m+q}\left(\tilde{k}^{T}\right)^{N-2m-2q}\left(2m\right)!\left(\tilde{f}_{22}\right)^{q+2m}.$
(3.20)

Finally, with the saddle point, we can calculate the HSSA to be NHard

$\displaystyle\mathcal{T}^{\left(N,2m,2q\right)}$
$\displaystyle=\left(\prod_{i=3}^{n-2}\int_{0}^{1}dz_{i}\text{
}z_{i}^{i-2-N}\right)\int_{0}^{1}dz_{2}\left(\frac{\partial^{2m}\tilde{u}}{\partial
z_{2}^{2m}}\frac{\left(z_{2}-\tilde{z}_{2}\right)^{2m}}{\left(2m\right)!}\right)e^{-Kf}$
(3.21)
$\displaystyle\simeq\frac{1}{\left(2m\right)!}\frac{\partial^{2m}\tilde{u}}{\partial
z_{2}^{2m}}\left(\prod_{i=3}^{n-2}\tilde{z}_{i}^{i-2-N}\right)\int_{0}^{1}dz_{2}\left(z_{2}-\tilde{z}_{2}\right)^{2m}e^{-Kf\left(z_{2}\right)}$
(3.22)
$\displaystyle\simeq\frac{1}{\left(2m\right)!}\frac{\partial^{2m}\tilde{u}}{\partial
z_{2}^{2m}}\left(\prod_{i=3}^{n-2}\tilde{z}_{i}^{i-2-N}\right)\int_{0}^{\infty}dz_{2}\left(z_{2}-\tilde{z}_{2}\right)^{2m}e^{-Kf\left(z_{2}\right)}$
(3.23)
$\displaystyle=\frac{2\sqrt{\pi}}{m!}\left(\prod_{i=3}^{n-2}\tilde{z}_{i}^{i-2-N}\right)\frac{e^{-K\tilde{f}}}{\left|\tilde{k}\right|^{2m+1}}\left.\frac{\partial^{2m}u}{\partial
z_{2}^{2m}}\right|_{z_{i}=\tilde{z}_{i}}$ (3.24)
$\displaystyle=2\sqrt{\pi}e^{-K\tilde{f}}\left|\tilde{k}\right|^{N-1}\left(\prod_{i=3}^{n-2}\tilde{z}_{i}^{i-2-N}\right)\frac{\left(2m\right)!}{m!}\left(\frac{-1}{2M}\right)^{2m+q}\left(\frac{2K\tilde{f}_{22}}{\left(\sum_{i\neq
2,n}\frac{k_{i}^{T}}{\tilde{x}_{i}-\tilde{x}_{2}}\right)^{2}}\right)^{m+q}$
(3.25)

where
$f\left(z_{2}\right)=f\left(z_{2},\tilde{z}_{3},\cdots,\tilde{z}_{n-2}\right)$.
The ratios of $n$-point $HSSA$ with $r=1$ is

$\displaystyle\frac{\mathcal{T}^{\left(N,m,q\right)}}{\mathcal{T}^{\left(N,0,0\right)}}$
$\displaystyle=\frac{\left(2m\right)!}{m!}\left(\frac{-1}{2M}\right)^{2m+q}\left(\frac{2K\tilde{f}_{22}}{\left(\sum_{i\neq
2,n}\frac{k_{i}^{T}}{\tilde{x}_{i}-\tilde{x}_{2}}\right)^{2}}\right)^{m+q}$
(3.26)
$\displaystyle=\frac{\left(2m\right)!}{m!}\left(\frac{-1}{2M}\right)^{2m+q}$
(3.27)

where the second equality followed from the calculation of decoupling of ZNS
in Eq.(2.11).

This suggests the identity

$\frac{2K\tilde{f}_{22}}{\left(\sum_{i\neq
2,n}\frac{k_{i}^{T}}{\tilde{x}_{i}-\tilde{x}_{2}}\right)^{2}}=1.$ (3.28)

For the case of $n=4$, one can easily solve the saddle point
$\tilde{z}_{2}=\sec^{2}\frac{\phi}{2}$ to verify the identity. We have also
proved the identity for $n=5$ by using maple numerically. Similar proof can be
done by maple for the case of $n=6$.

###  III.3 The $n$-point HSSA with $r=2$

Now we calculate the case of $n$-point HSSA with $r=2$. We want to calculate
$n$-point HSSA with $(n-1)$ tachyons and $1$ high energy state

$\left(\alpha_{-1}^{T_{1}}\right)^{N+p_{1}}\left(\alpha_{-1}^{T_{2}}\right)^{p_{2}}\left(\alpha_{-1}^{L}\right)^{2m}\left(\alpha_{-2}^{L}\right)^{q}\left|0;k\right\rangle\text{,
\ }p_{1}+p_{2}=-2(m+q).$ (3.29)

The ratios of $n$-point HSSA with $r=2$ can be similarly calculated to be

$\displaystyle\frac{\mathcal{T}^{\left(p_{1},p_{2},m,q\right)}}{\mathcal{T}^{\left(N,0,0,0\right)}}$
$\displaystyle=\frac{\left(2m\right)!}{m!}\left(\frac{-1}{2M}\right)^{2m+q}\frac{\left(2K\tilde{f}_{22}\right)^{m+q}}{\left(\sum_{i\neq
2,n}\frac{k_{i}^{T_{1}}}{\tilde{x}_{i}-\tilde{x}_{2}}\right)^{2m+2q+p_{2}}\left(\sum_{i\neq
2,n}\frac{k_{i}^{T_{2}}}{\tilde{x}_{i}-\tilde{x}_{2}}\right)^{-p_{2}}}$
$\displaystyle=\frac{\left(2m\right)!}{m!}\left(\frac{-1}{2M}\right)^{2m+q}\frac{\left(\frac{\sum_{i\neq
2,n}\frac{k_{i}^{T_{2}}}{\tilde{x}_{i}-\tilde{x}_{2}}}{\sum_{i\neq
2,n}\frac{k_{i}^{T_{1}}}{\tilde{x}_{i}-\tilde{x}_{2}}}\right)^{p_{2}}}{\left(\frac{\sum_{i\neq
2,n}\frac{k_{i}^{T_{1}}}{\tilde{x}_{i}-\tilde{x}_{2}}}{\sqrt{2K\tilde{f}_{22}}}\right)^{2m+2q}}.$
(3.30)

On the other hand, the decoupling of ZNS calculated in Eq.(2.31) gives

$\frac{\mathcal{T}^{\left(p_{1},p_{2},m,q\right)}}{\mathcal{T}^{\left(N,0,0,0\right)}}=\frac{\left(2m\right)!}{m!}\left(\frac{-1}{2M}\right)^{2m+q}\omega_{1}^{p_{1}}\omega_{2}^{p_{2}}=\frac{\left(2m\right)!}{m!}\left(\frac{-1}{2M}\right)^{2m+q}\frac{(\tan\theta_{1})^{p_{2}}}{(\cos\theta_{1})^{2m+2q}}.$
(3.31)

Eq.(3.30) and Eq.(3.31) can be identified for any $p_{2}$, $m$ and $q$ if

$\left(\sum_{i\neq
2,n}\frac{k_{i}^{T_{1}}}{\tilde{x}_{i}-\tilde{x}_{2}}\right)=\sqrt{2K\tilde{f}_{22}}\cos\theta_{1}\text{,
}\left(\sum_{i\neq
2,n}\frac{k_{i}^{T_{2}}}{\tilde{x}_{i}-\tilde{x}_{2}}\right)=\sqrt{2K\tilde{f}_{22}}\sin\theta_{1},$
(3.32)

which implies the identity

$\left(\sum_{i\neq
2,n}\frac{k_{i}^{T_{1}}}{\tilde{x}_{i}-\tilde{x}_{2}}\right)^{2}+\left(\sum_{i\neq
2,n}\frac{k_{i}^{T_{2}}}{\tilde{x}_{i}-\tilde{x}_{2}}\right)^{2}=2K\tilde{f}_{22}.$
(3.33)

It is not surprising that Eq.(3.33) is a generalization of Eq.(3.28) to two
transverse directions $T_{1}$ and $T_{2}$.

###  III.4 The $n$-point HSSA with $r\leq 24$

It is now easy to generalize Eq.(3.33) to any $r$ (number of $T_{i}$) with
$r\leq 24$

$\left(\sum_{i\neq
2,n}\frac{k_{i}^{T_{1}}}{\tilde{x}_{i}-\tilde{x}_{2}}\right)^{2}+\left(\sum_{i\neq
2,n}\frac{k_{i}^{T_{2}}}{\tilde{x}_{i}-\tilde{x}_{2}}\right)^{2}+\cdots+\left(\sum_{i\neq
2,n}\frac{k_{i}^{T_{r}}}{\tilde{x}_{i}-\tilde{x}_{2}}\right)^{2}=2K\tilde{f}_{22}.$
(3.34)

By using Eq.(3.16) and Eq.(3.18), we see that the key identity Eq.(3.34) can
be written as NHard

$\tilde{k}^{2}+2M\tilde{k}^{\prime L}=0.$ (3.35)

The ratios in Eq.(2.31) are thus proved by the saddle point method.

##  IV Stringy scaling of Regge string scattering amplitudes

Another important high-energy regime of $4$-point SSA is the fixed momentum
transfer regime which contains complementary information of the theory. That
is in the kinematic regime

$s\rightarrow\infty,\qquad\sqrt{-t}=\text{f\/ixed,}\quad(\text{but}\
\sqrt{-t}\neq\infty).$ (4.1)

In this regime, the number of high-energy SSA is much more numerous than that
of the fixed angle regime. One of the reason is that in contrast to the
identification $e^{P}\simeq$ $e^{L}$ in the hard scattering limit, $e^{P}$
does not approach to $e^{L}$ in the Regge scattering limit. For example, at
mass level $M^{2}=4$ of open bosonic string, there are only $4$ HSSA while
there are $22$ RSSA bosonic ; KLY1 . On the other hand, in the Regge regime
both the saddle-point method and the method of decoupling of zero-norm states
adopted in the calculation of fixed angle regime do not apply.

The complete leading order high-energy open string states in the Regge regime
at each fixed mass level $N=\sum_{n,m,l>0}np_{n}+mq_{m}+lr_{l}$ are

$\left|v_{n},q_{m},r_{l}\right\rangle=\prod_{n>0}(\alpha_{-n}^{T})^{v_{n}}\prod_{m>0}(\alpha_{-m}^{P})^{q_{m}}\prod_{l>0}(\alpha_{-l}^{L})^{r_{l}}|0,k\rangle.$
(4.2)

It turned out that the $4$-pont RSSA of three tachyons and states in Eq.(4.2)
are NOT proportional to each other, and the ratios are $t$-dependent
functions. However, it was shown that for the RSSA $A^{(N,2m,q)}$ with
$v_{1}=N-m-q$, $r_{1}=2m$ and $r_{2}=q$ and all others $0$ in Eq.(4.2), one
can extract the ratios of hard string scatterings in Eq.(2.11) from
$A^{(N,2m,q)}$ RRsusy ; LYAM ; HLY . It is thus reasonable to expect that for
the $n$-point ($n\geq 5$) RSSA with $n-1$ tachyons and some subset of the
high-energy states in Eq.(2.28), the RSSA show similar stringy scaling
behavior as in Eq.(2.31) of HSSA.

In this paper, we will consider a class of $n$-point ($n\geq 5$) RSSA with
$n-1$ tachyons and one high-energy state at mass level $N$

$\left|\left\\{p_{i}\right\\},0,0\right\rangle=\left(\alpha_{-1}^{T_{1}}\right)^{N+p_{1}}\left(\alpha_{-1}^{T_{2}}\right)^{p_{2}}\cdots\left(\alpha_{-1}^{T_{r}}\right)^{p_{r}}\left|0;k\right\rangle,$
(4.3)

which is obtained by setting $m=q=0$ in Eq.(2.28). We will show that these
RSSA show stringy scaling behavior for arbitrary $n$ similar to that we
obtained for the HSSA in Eq.(2.31).

There are many different Regge regimes for the $n$-point ($n\geq 5$) RSSA. To
specify the Regge regime, we first discuss the system of kinematics variables
we will use. The standard kinematics variables commonly adopted for the
$n$-point scatterings can be defined as following. One first defines the
$(n-3)$ $s$ variables

$s_{12}=-\left(k_{1}+k_{2}\right)^{2},s_{123}=-\left(k_{1}+k_{2}+k_{3}\right)^{2},\cdots,s_{1,\cdots,n-2}=-\left(k_{1}+\cdots+k_{n-2}\right)^{2},$
(4.4)

and then defines the $\frac{\left(n-2\right)\left(n-3\right)}{2}$ $t$
variables

$\displaystyle t_{23}$
$\displaystyle=-\left(k_{2}+k_{3}\right)^{2},t_{24}=-\left(k_{2}+k_{4}\right)^{2},\cdots,t_{2,n-1}=-\left(k_{2}+k_{n-1}\right)^{2},$
$\displaystyle t_{34}$
$\displaystyle=-\left(k_{3}+k_{4}\right)^{2},\cdots,t_{3,n-1}=-\left(k_{3}+k_{n-1}\right)^{2},$
$\displaystyle\vdots$ $\displaystyle t_{n-2,n-1}$
$\displaystyle=-\left(k_{n-2}+k_{n-1}\right)^{2},$ (4.5)

which amount to $\frac{n\left(n-3\right)}{2}$ independent kinematics
variables.

For our purpose in the calculation of this paper, we will adopt another system
of independent kinematics variables. We use the notation $k_{ij}\equiv
k_{i}\cdot k_{j}$ to define the following $\frac{n\left(n-3\right)}{2}$
independent kinematics variables

$\displaystyle k_{12},k_{13},k_{14},\cdots,k_{1,n-2},$ $\displaystyle
k_{23},k_{24},k_{25},\cdots,k_{2,n-1},$ $\displaystyle
k_{34},k_{35},\cdots,k_{3,n-1},$ $\displaystyle\vdots$ $\displaystyle
k_{n-3,n-2},k_{n-3,n-1},$ $\displaystyle k_{n-2,n-1}.$ (4.6)

For later use, we also define

$k_{1,\cdots,i-1,i}=k_{1,\cdots,i-1}+\sum_{j=1}^{i-1}k_{ji},$ (4.7)

which means, for example,

$k_{123}=k_{12}+k_{13}+k_{23},k_{1234}=k_{123}+k_{14}+k_{24}+k_{34},k_{12345}=k_{1234}+k_{15}+k_{25}+k_{35}+k_{45}.$
(4.8)

###  IV.1 The $5$-point and $6$-point Regge stringy scaling

Let’s begin with the calculation of $5$-point RSSA with $r=2$ in Eq.(4.3). The
kinematics are

$\displaystyle k_{1}$
$\displaystyle=\left(\sqrt{p^{2}+M_{1}^{2}},-p,0,0\right),$ $\displaystyle
k_{2}$ $\displaystyle=\left(\sqrt{p^{2}+M_{2}^{2}},p,0,0\right),$
$\displaystyle k_{3}$
$\displaystyle=\left(-\sqrt{q_{3}^{2}+M_{3}^{2}},-q_{3}\cos\phi_{1}^{3},-q_{3}\sin\phi_{1}^{3},0\right),$
$\displaystyle k_{4}$
$\displaystyle=\left(-\sqrt{q_{4}^{2}+M_{4}^{2}},-q_{4}\cos\phi_{1}^{4},-q_{4}\sin\phi_{1}^{4}\cos\phi_{2}^{4},-q_{4}\sin\phi_{1}^{4}\sin\phi_{2}^{4}\right),$
$\displaystyle k_{5}$
$\displaystyle=\left(-\sqrt{q_{5}^{2}+M_{5}^{2}},-q_{5}\cos\phi_{1}^{5},-q_{5}\sin\phi_{1}^{5}\cos\phi_{2}^{5},-q_{5}\sin\phi_{1}^{5}\sin\phi_{2}^{5}\right).$
(4.9)

During the calculation, we will keep record of the notations used for each
step so that eventually we can generalize the calculation to the case of
$n$-point RSSA. The amplitude of state

$\left(\alpha_{-1}^{T_{1}}\right)^{N+p_{1}}\left(\alpha_{-1}^{T_{2}}\right)^{p_{2}}\left|0,k\right\rangle,p_{1}+p_{2}=0$
(4.10)

and $4$ tachyon states can be written as

$\displaystyle A^{\left\\{p_{1},p_{2}\right\\},0,0}$
$\displaystyle=\int_{0}^{1}dx_{3}\int_{0}^{x_{3}}dx_{2}\times
x_{2}^{k_{12}}x_{3}^{k_{13}}\left(x_{3}-x_{2}\right)^{k_{23}}\left(1-x_{2}\right)^{k_{24}}\left(1-x_{3}\right)^{k_{34}}$
$\displaystyle\times\left[\frac{k_{3}^{T_{1}}}{x_{3}-x_{2}}+\frac{k_{4}^{T_{1}}}{1-x_{2}}\right]^{N+p_{1}}\left[\underset{0}{\underbrace{\frac{k_{3}^{T_{2}}}{x_{3}-x_{2}}}}+\frac{k_{4}^{T_{2}}}{1-x_{2}}\right]^{p_{2}}.$
(4.11)

One can easily find that $k_{3}^{T_{2}}=0$. After doing the change of
variables

$x_{2}=z_{2}z_{3},x_{3}=z_{3},$ (4.12)

we can rewrite the above $5$-point amplitude as following

$\displaystyle A^{\left\\{p_{1},p_{2}\right\\},0,0}$
$\displaystyle=\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}z_{2}^{k_{12}}z_{3}^{k_{123}+1}\left(1-z_{2}\right)^{k_{23}}\left(1-z_{2}z_{3}\right)^{k_{24}}\left(1-z_{3}\right)^{k_{34}}$
$\displaystyle\times\left[\frac{k_{3}^{T_{1}}}{z_{3}-z_{2}z_{3}}+\frac{k_{4}^{T_{1}}}{1-z_{2}z_{3}}\right]^{N+p_{1}}\left[\frac{k_{4}^{T_{2}}}{1-z_{2}z_{3}}\right]^{p_{2}}$
(4.13)

where we have defined $k_{123}=k_{12}+k_{23}+k_{13}$. Next, let’s perform the
binomial expansion on the bracket to obtain

$\displaystyle
A^{\left\\{p_{1},p_{2}\right\\},0,0}=\sum_{J_{1}^{1}+J_{2}^{1}=N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{4}^{T_{2}}\right)^{p_{2}}$
$\displaystyle\times\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}z_{2}^{k_{12}}z_{3}^{k_{123}+1-J_{1}^{1}}\left(1-z_{2}\right)^{k_{23}-J_{1}^{1}}\left(1-z_{2}z_{3}\right)^{k_{24}-J_{2}^{1}-p_{2}}\left(1-z_{3}\right)^{k_{34}}.$
(4.14)

For the next step, we expand the crossing term
$\left(1-z_{2}z_{3}\right)^{k_{24}-J_{2}^{1}-p_{2}}$ to obtain

$\displaystyle
A^{\left\\{p_{1},p_{2}\right\\},0,0}=\sum_{J_{1}^{1}+J_{2}^{1}=N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{4}^{T_{2}}\right)^{p_{2}}$
$\displaystyle\times\sum_{m_{23}}\frac{\left(-k_{24}+p_{2}+J_{2}^{1}\right)_{m_{23}}}{m_{23}!}\int_{0}^{1}dz_{2}z_{2}^{k_{12}+m_{23}}\left(1-z_{2}\right)^{k_{23}-J_{1}^{1}}\int_{0}^{1}dz_{3}z_{3}^{k_{123}+1-J_{1}^{1}+m_{23}}\left(1-z_{3}\right)^{k_{34}}$
(4.15)

where the subscripts of $m_{23}$ keep record of the subscripts $z_{2}z_{3}$ in
$\left(1-z_{2}z_{3}\right)^{k_{24}-J_{2}^{1}-p_{2}}$. After the integration,
the amplitude can be written as

$\displaystyle A^{\left\\{p_{1},p_{2}\right\\},0,0}$
$\displaystyle=\sum_{J_{1}^{1}+J_{2}^{1}=N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{4}^{T_{2}}\right)^{p_{2}}$
$\displaystyle\times\sum_{m_{23}}\frac{\left(-k_{24}+p_{2}+J_{2}^{1}\right)_{m_{23}}}{m_{23}!}\frac{\Gamma\left(k_{12}+1+m_{23}\right)\Gamma\left(k_{23}+1-J_{1}^{1}\right)}{\Gamma\left(k_{12}+k_{23}+2+m_{23}-J_{1}^{1}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{123}+2+m_{23}-J_{1}^{1}\right)\Gamma\left(k_{34}+1\right)}{\Gamma\left(k_{123}+k_{34}+3+m_{23}-J_{1}^{1}\right)}.$
(4.16)

Now we choose to work on the Regge regime defined by

$k_{123}\sim s,k_{34}\sim s,k_{123}+k_{34}\sim t$ (4.17)

where $s\rightarrow\infty$ and $t=$ fixed. (we will use these notations to
define a Regge regime for the rest of the paper) In this Regge regime, the
amplitude can be approximated as

$\displaystyle A^{\left\\{p_{1},p_{2}\right\\},0,0}$
$\displaystyle\sim\sum_{J_{1}^{1}+J_{2}^{1}=N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{4}^{T_{2}}\right)^{p_{2}}$
$\displaystyle\times\sum_{m_{23}}\frac{\left(-k_{24}+p_{2}+J_{2}^{1}\right)_{m_{23}}}{m_{23}!}\frac{\Gamma\left(k_{12}+1+m_{23}\right)\Gamma\left(k_{23}+1-J_{1}^{1}\right)}{\Gamma\left(k_{12}+k_{23}+2+m_{23}-J_{1}^{1}\right)}$
$\displaystyle\times\frac{\left(k_{123}\right)^{m_{23}-J_{1}^{1}}\Gamma\left(k_{123}+2\right)\Gamma\left(k_{34}+1\right)}{\left(k_{123}+k_{34}+3\right)_{m_{23}-J_{1}^{1}}\Gamma\left(k_{123}+k_{34}+3\right)}.$
(4.18)

The leading power of $k_{123}$ occurs when $J_{1}^{1}=0$ which means
$J_{2}^{1}=N+p_{1}$. Since $p_{1}+p_{2}=0$, the leading term of the RSSA is

$\displaystyle A^{\left\\{p_{1},p_{2}\right\\},0,0}$
$\displaystyle\sim\left(k_{4}^{T_{1}}\right)^{N+p_{1}}\left(k_{4}^{T_{2}}\right)^{p_{2}}\sum_{m_{23}}\frac{\left(-k_{24}+N\right)_{m_{23}}}{m_{23}!}\frac{\Gamma\left(k_{12}+1+m_{23}\right)\Gamma\left(k_{23}+1\right)}{\Gamma\left(k_{12}+k_{23}+2+m_{23}\right)}$
$\displaystyle\times\frac{\left(k_{123}\right)^{m_{23}}\Gamma\left(k_{123}+2\right)\Gamma\left(k_{34}+1\right)}{\left(k_{123}+k_{34}+3\right)_{m_{23}}\Gamma\left(k_{123}+k_{34}+3\right)}.$
(4.19)

The ratio of $A^{\left\\{p_{1},p_{2}\right\\},0,0}$ and
$A^{\left\\{0,0\right\\},0,0}$ can be easily calculated to be

$\displaystyle\frac{A^{\left\\{p_{1},p_{2}\right\\},0,0}}{A^{\left\\{0,0\right\\},0,0}}$
$\displaystyle=\frac{\left(k_{4}^{T_{1}}\right)^{N+p_{1}}\left(k_{4}^{T_{2}}\right)^{p_{2}}}{\left(k_{4}^{T_{1}}\right)^{N}}=\left(k_{4}^{T_{1}}\right)^{p_{1}}\left(k_{4}^{T_{2}}\right)^{p_{2}}=\left(-q_{4}\sin\phi_{1}^{4}\cos\phi_{2}^{4}\right)^{p_{1}}\left(-q_{4}\sin\phi_{1}^{4}\sin\phi_{2}^{4}\right)^{p_{2}}$
$\displaystyle=\left(\cos\theta_{1}\right)^{p_{1}}\left(\sin\theta_{1}\right)^{p_{2}}=\left(\omega_{1}\right)^{p_{1}}\left(\omega_{2}\right)^{p_{2}},$
(4.20)

which is the same as Eq.(2.31) with $m=q=0$ and $r=2$.

Let’s now calculate the $6$-point RSSA with $r=3$ in Eq.(4.3). The kinematics
are

$\displaystyle k_{1}$
$\displaystyle=\left(\sqrt{p^{2}+M_{1}^{2}},-p,0,0,0\right),$ $\displaystyle
k_{2}$ $\displaystyle=\left(\sqrt{p^{2}+M_{2}^{2}},p,0,0,0\right),$
$\displaystyle k_{3}$
$\displaystyle=\left(-\sqrt{q_{3}^{2}+M_{3}^{2}},-q_{3}\cos\phi_{1}^{3},-q_{3}\sin\phi_{1}^{3},0,0\right),$
$\displaystyle k_{4}$
$\displaystyle=\left(-\sqrt{q_{4}^{2}+M_{4}^{2}},-q_{4}\cos\phi_{1}^{4},-q_{4}\sin\phi_{1}^{4}\cos\phi_{2}^{4},-q_{4}\sin\phi_{1}^{4}\sin\phi_{2}^{4},0\right),$
$\displaystyle k_{5}$
$\displaystyle=\left(-\sqrt{q_{5}^{2}+M_{5}^{2}},-q_{5}\cos\phi_{1}^{5},-q_{5}\sin\phi_{1}^{5}\cos\phi_{2}^{5},-q_{5}\sin\phi_{1}^{5}\sin\phi_{2}^{5}\cos\phi_{3}^{5},-q_{5}\sin\phi_{1}^{5}\sin\phi_{2}^{5}\sin\phi_{3}^{5}\right),$
$\displaystyle k_{6}$
$\displaystyle=\left(-\sqrt{q_{5}^{2}+M_{5}^{2}},-q_{6}\cos\phi_{1}^{6},-q_{6}\sin\phi_{1}^{6}\cos\phi_{2}^{6},-q_{6}\sin\phi_{1}^{6}\sin\phi_{2}^{6}\cos\phi_{3}^{6},-q_{6}\sin\phi_{1}^{6}\sin\phi_{2}^{6}\sin\phi_{3}^{6}\right).$
(4.21)

The amplitude of state

$\left(\alpha_{-1}^{T_{1}}\right)^{N+p_{1}}\left(\alpha_{-1}^{T_{2}}\right)^{p_{2}}\left(\alpha_{-1}^{T_{3}}\right)^{p_{3}}\left|0,k\right\rangle,p_{1}+p_{2}+p_{3}=0$
(4.22)

and $5$ tachyon states is

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3}\right\\},0,0}$
$\displaystyle=\int_{0}^{1}dx_{4}\int_{0}^{x_{4}}dx_{3}\int_{0}^{x_{3}}dx_{2}\text{
}x_{2}^{k_{12}}x_{3}^{k_{13}}x_{4}^{k_{14}}\left(x_{3}-x_{2}\right)^{k_{23}}\left(x_{4}-x_{2}\right)^{k_{24}}\left(1-x_{2}\right)^{k_{25}}\left(x_{4}-x_{3}\right)^{k_{34}}\left(1-x_{3}\right)^{k_{35}}\left(1-x_{4}\right)^{k_{45}}$
$\displaystyle\times\left[\frac{k_{3}^{T_{1}}}{x_{3}-x_{2}}+\frac{k_{4}^{T_{1}}}{x_{4}-x_{2}}+\frac{k_{5}^{T_{1}}}{\underset{x_{5}}{\underbrace{1}}-x_{2}}\right]^{N+p_{1}}\left[\underset{=0}{\underbrace{\frac{k_{3}^{T_{2}}}{x_{3}-x_{2}}}}+\frac{k_{4}^{T_{2}}}{x_{4}-x_{2}}+\frac{k_{5}^{T_{2}}}{1-x_{2}}\right]^{p_{2}}\left[\underset{=0\text{
}}{\underbrace{\frac{k_{3}^{T_{3}}}{x_{3}-x_{2}}}}+\underset{=0\text{
}}{\underbrace{\frac{k_{4}^{T_{3}}}{x_{4}-x_{2}}}}+\frac{k_{5}^{T_{3}}}{1-x_{2}}\right]^{p_{3}}.$
(4.23)

Since $k_{3}^{T_{2}}=k_{3}^{T_{3}}=k_{4}^{T_{3}}=0,$ we can rewrite the
amplitude as

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3}\right\\},0,0}$
$\displaystyle=\int_{0}^{1}dx_{4}\int_{0}^{x_{4}}dx_{3}\int_{0}^{x_{3}}dx_{2}\text{
}x_{2}^{k_{12}}x_{3}^{k_{13}}x_{4}^{k_{14}}\left(x_{3}-x_{2}\right)^{k_{23}}\left(x_{4}-x_{2}\right)^{k_{24}}\left(1-x_{2}\right)^{k_{25}}\left(x_{4}-x_{3}\right)^{k_{34}}\left(1-x_{3}\right)^{k_{35}}\left(1-x_{4}\right)^{k_{45}}$
$\displaystyle\times\left[\frac{k_{3}^{T_{1}}}{x_{3}-x_{2}}+\frac{k_{4}^{T_{1}}}{x_{4}-x_{2}}+\frac{k_{5}^{T_{1}}}{1-x_{2}}\right]^{N+p_{1}}\left[\frac{k_{4}^{T_{2}}}{x_{4}-x_{2}}+\frac{k_{5}^{T_{2}}}{1-x_{2}}\right]^{p_{2}}\left[\frac{k_{5}^{T_{3}}}{1-x_{2}}\right]^{p_{3}}.$
(4.24)

We can do the following change of variables

$x_{i}=z_{i}\cdots z_{n-2},$ (4.25)

or

$x_{2}=z_{2}z_{3}z_{4},x_{3}=z_{3}z_{4},x_{4}=z_{4}$ (4.26)

to obtain

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3}\right\\},0,0}$
$\displaystyle=\int_{0}^{1}dz_{4}\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}\times
z_{2}^{k_{12}}z_{3}^{k_{123}+1}z_{4}^{k_{1234}+2}\left(1-z_{2}\right)^{k_{23}}\left(1-z_{3}\right)^{k_{34}}\left(1-z_{4}\right)^{k_{45}}$
$\displaystyle\times\left(1-z_{2}z_{3}\right)^{k_{24}}\left(1-z_{2}z_{3}z_{4}\right)^{k_{25}}\left(1-z_{3}z_{4}\right)^{k_{35}}$
$\displaystyle\times\left[\frac{k_{3}^{T_{1}}}{z_{3}z_{4}-z_{2}z_{3}z_{4}}+\frac{k_{4}^{T_{1}}}{z_{4}-z_{2}z_{3}z_{4}}+\frac{k_{5}^{T_{1}}}{1-z_{2}z_{3}z_{4}}\right]^{N+p_{1}}\left[\frac{k_{4}^{T_{2}}}{z_{4}-z_{2}z_{3}z_{4}}+\frac{k_{5}^{T_{2}}}{1-z_{2}z_{3}z_{4}}\right]^{p_{2}}\left[\frac{k_{5}^{T_{3}}}{1-z_{2}z_{3}z_{4}}\right]^{p_{3}}$
(4.27)

where we have defined

$k_{123}=k_{12}+k_{13}+k_{23},k_{1234}=k_{12}+k_{13}+k_{14}+k_{23}+k_{24}+k_{34}.$
(4.28)

Next, let’s perform the binomial expansion on the brackets to obtain

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3}\right\\},0,0}$
$\displaystyle=\int_{0}^{1}dz_{4}\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}\times
z_{2}^{k_{12}}z_{3}^{k_{123}+1}z_{4}^{k_{1234}+2}\left(1-z_{2}\right)^{k_{23}}\left(1-z_{3}\right)^{k_{34}}\left(1-z_{4}\right)^{k_{45}}$
$\displaystyle\times\left(1-z_{2}z_{3}\right)^{k_{24}}\left(1-z_{2}z_{3}z_{4}\right)^{k_{25}}\left(1-z_{3}z_{4}\right)^{k_{35}}$
$\displaystyle\times\sum_{J_{1}^{1}+J_{2}^{1}+J_{3}^{1}=N+p_{1}}^{N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!J_{3}^{1}!}\left(\frac{k_{3}^{T_{1}}}{z_{3}z_{4}-z_{2}z_{3}z_{4}}\right)^{J_{1}^{1}}\left(\frac{k_{4}^{T_{1}}}{z_{4}-z_{2}z_{3}z_{4}}\right)^{J_{2}^{1}}\left(\frac{k_{5}^{T_{1}}}{1-z_{2}z_{3}z_{4}}\right)^{J_{3}^{1}}$
$\displaystyle\times\sum_{J_{1}^{2}+J_{2}^{2}=p_{2}}^{p_{2}}\frac{p_{2}!}{J_{1}^{2}!J_{2}^{2}!}\left(\frac{k_{4}^{T_{2}}}{z_{4}-z_{2}z_{3}z_{4}}\right)^{J_{1}^{2}}\left(\frac{k_{5}^{T_{2}}}{1-z_{2}z_{3}z_{4}}\right)^{J_{2}^{2}}\left[\frac{k_{5}^{T_{3}}}{1-z_{2}z_{3}z_{4}}\right]^{p_{3}}$
(4.29)

where $J_{1}^{1}$ ,$J_{2}^{1}$ ,$J_{3}^{1}$,$J_{1}^{2}$,$J_{2}^{2}$ are non-
negative integers with $J_{1}^{1}+J_{2}^{1}=N+p_{1}$ and
$J_{1}^{2}+J_{2}^{2}=p_{2}$. We then rearrange the above equation

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3}\right\\},0,0}$
$\displaystyle=\sum_{J_{1}^{1}+J_{2}^{1}+J_{3}^{1}=N+p_{1}}^{N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!J_{3}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{5}^{T_{1}}\right)^{J_{3}^{1}}\sum_{J_{1}^{2}+J_{2}^{2}=p_{2}}^{p_{2}}\frac{p_{2}!}{J_{1}^{2}!J_{2}^{2}!}\left(k_{4}^{T_{2}}\right)^{J_{1}^{2}}\left(k_{5}^{T_{2}}\right)^{J_{2}^{2}}\left(k_{5}^{T_{3}}\right)^{p_{3}}$
$\displaystyle\times\int_{0}^{1}dz_{4}\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}\times
z_{2}^{k_{12}}z_{3}^{k_{123}+1-J_{1}^{1}}z_{4}^{k_{1234}+2-J_{1}^{1}-\left(J_{2}^{1}+J_{1}^{2}\right)}\left(1-z_{2}\right)^{k_{23}-J_{1}^{1}}\left(1-z_{3}\right)^{k_{34}}\left(1-z_{4}\right)^{k_{45}}$
$\displaystyle\times\left(1-z_{2}z_{3}\right)^{k_{24}-\left(J_{2}^{1}+J_{1}^{2}\right)}\left(1-z_{2}z_{3}z_{4}\right)^{k_{25}-\left(J_{3}^{1}+J_{2}^{2}+p_{3}\right)}\left(1-z_{3}z_{4}\right)^{k_{35}},$
(4.30)

and expand the crossing terms to obain

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3}\right\\},0,0}$
$\displaystyle=\sum_{J_{1}^{1}+J_{2}^{1}+J_{3}^{1}=N+p_{1}}^{N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!J_{3}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{5}^{T_{1}}\right)^{J_{3}^{1}}\sum_{J_{1}^{2}+J_{2}^{2}=p_{2}}^{p_{2}}\frac{p_{2}!}{J_{1}^{2}!J_{2}^{2}!}\left(k_{4}^{T_{2}}\right)^{J_{1}^{2}}\left(k_{5}^{T_{2}}\right)^{J_{2}^{2}}\left(k_{5}^{T_{3}}\right)^{p_{3}}$
$\displaystyle\times\int_{0}^{1}dz_{4}\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}\times
z_{2}^{k_{12}}z_{3}^{k_{123}+1-J_{1}^{1}}z_{4}^{k_{1234}+2-J_{1}^{1}-\left(J_{2}^{1}+J_{1}^{2}\right)}\left(1-z_{2}\right)^{k_{23}-J_{1}^{1}}\left(1-z_{3}\right)^{k_{34}}\left(1-z_{4}\right)^{k_{45}}$
$\displaystyle\times\sum_{m_{23}=0}\frac{\left[-k_{24}+\left(J_{2}^{1}+J_{1}^{2}\right)\right]_{m_{23}}}{m_{23}!}\left(z_{2}z_{3}\right)^{m_{23}}\sum_{m_{24}=0}\frac{\left[-k_{25}+\left(J_{3}^{1}+J_{2}^{2}+p_{3}\right)\right]_{m_{24}}}{m_{24}!}\left(z_{2}z_{3}z_{4}\right)^{m_{24}}$
$\displaystyle\times\sum_{m_{34}=0}\frac{\left[-k_{35}\right]_{m_{34}}}{m_{34}!}\left(z_{3}z_{4}\right)^{m_{34}}$
(4.31)

where, for example, the subscripts of $m_{24}$ keep record of the first and
the last subscripts of $\left(z_{2}z_{3}z_{4}\right)$ etc.. We rearrange the
above equation again

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3}\right\\},0,0}$
$\displaystyle=\sum_{J_{1}^{1}+J_{2}^{1}+J_{3}^{1}=N+p_{1}}^{N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!J_{3}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{5}^{T_{1}}\right)^{J_{3}^{1}}\sum_{J_{1}^{2}+J_{2}^{2}=p_{2}}^{N+p_{1}}\left(k_{4}^{T_{2}}\right)^{J_{1}^{2}}\left(k_{5}^{T_{2}}\right)^{J_{2}^{2}}\left(k_{5}^{T_{3}}\right)^{p_{3}}$
$\displaystyle\times\sum_{m_{23}=0}\frac{\left[-k_{24}+\left(J_{2}^{1}+J_{1}^{2}\right)\right]_{m_{23}}}{m_{23}!}\sum_{m_{24}=0}\frac{\left[-k_{25}+\left(J_{3}^{1}+J_{2}^{2}+p_{3}\right)\right]_{m_{24}}}{m_{24}!}\sum_{m_{34}=0}\frac{\left[-k_{35}\right]_{m_{34}}}{m_{34}!}$
$\displaystyle\times\int_{0}^{1}dz_{2}z_{2}^{k_{12}+m_{23}+m_{24}}\left(1-z_{2}\right)^{k_{23}-J_{1}^{1}}$
$\displaystyle\times\int_{0}^{1}dz_{3}z_{3}^{k_{123}+1-J_{1}^{1}+m_{23}+m_{24}+m_{34}}\left(1-z_{3}\right)^{k_{34}}$
$\displaystyle\times\int_{0}^{1}dz_{4}z_{4}^{k_{1234}+2-J_{1}^{1}-\left(J_{2}^{1}+J_{1}^{2}\right)+m_{24}+m_{34}}\left(1-z_{4}\right)^{k_{45}},$
(4.32)

and perform the integration to obtain

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3}\right\\},0,0}$
$\displaystyle=\sum_{J_{1}^{1}+J_{2}^{1}+J_{3}^{1}=N+p_{1}}^{N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!J_{3}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{5}^{T_{1}}\right)^{J_{3}^{1}}\sum_{J_{1}^{2}+J_{2}^{2}=p_{2}}^{p_{2}}\frac{p_{2}!}{J_{1}^{2}!J_{2}^{2}!}\left(k_{4}^{T_{2}}\right)^{J_{1}^{2}}\left(k_{5}^{T_{2}}\right)^{J_{2}^{2}}\left(k_{5}^{T_{3}}\right)^{p_{3}}$
$\displaystyle\times\sum_{m_{23}=0}\frac{\left[-k_{24}+\left(J_{2}^{1}+J_{1}^{2}\right)\right]_{m_{23}}}{m_{23}!}\sum_{m_{24}=0}\frac{\left[-k_{25}+\left(J_{3}^{1}+J_{2}^{2}+p_{3}\right)\right]_{m_{24}}}{m_{24}!}\sum_{m_{34}=0}\frac{\left[-k_{35}\right]_{m_{34}}}{m_{34}!}$
$\displaystyle\times\frac{\Gamma\left(k_{12}+1+m_{23}+m_{24}\right)\Gamma\left(k_{23}+1-J_{1}^{1}\right)}{\Gamma\left(k_{12}+k_{23}+2-J_{1}^{1}+m_{23}+m_{24}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{123}+2-J_{1}^{1}+m_{23}+m_{24}+m_{34}\right)\Gamma\left(k_{34}+1\right)}{\Gamma\left(k_{123}+k_{34}+3-J_{1}^{1}+m_{23}+m_{24}+m_{34}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{1234}+3-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}\right)+m_{24}+m_{34}\right)\Gamma\left(k_{45}+1\right)}{\Gamma\left(k_{1234}+k_{23}+4-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}\right)+m_{24}+m_{34}\right)}.$
(4.33)

Now we choose to work on the Regge regime defined by

$k_{1234}\sim s,k_{1234}+k_{23}\sim t.$ (4.34)

In this Regge regime, the amplitude can be approximated as

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3}\right\\},0,0}$
$\displaystyle\sim\sum_{J_{1}^{1}+J_{2}^{1}+J_{3}^{1}=N+p_{1}}^{N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!J_{3}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{5}^{T_{1}}\right)^{J_{3}^{1}}\sum_{J_{1}^{2}+J_{2}^{2}=p_{2}}^{p_{2}}\frac{p_{2}!}{J_{1}^{2}!J_{2}^{2}!}\left(k_{4}^{T_{2}}\right)^{J_{1}^{2}}\left(k_{5}^{T_{2}}\right)^{J_{2}^{2}}\left(k_{5}^{T_{3}}\right)^{p_{3}}$
$\displaystyle\times\sum_{m_{23}=0}\frac{\left[-k_{24}+\left(J_{2}^{1}+J_{1}^{2}\right)\right]_{m_{23}}}{m_{23}!}\sum_{m_{24}=0}\frac{\left[-k_{25}+\left(J_{3}^{1}+J_{2}^{2}+p_{3}\right)\right]_{m_{24}}}{m_{24}!}\sum_{m_{34}=0}\frac{\left[-k_{35}\right]_{m_{34}}}{m_{34}!}$
$\displaystyle\times\frac{\Gamma\left(k_{12}+1+m_{23}+m_{24}\right)\Gamma\left(k_{23}+1-J_{1}^{1}\right)}{\Gamma\left(k_{12}+k_{23}+2-J_{1}^{1}+m_{23}+m_{24}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{123}+2-J_{1}^{1}+m_{23}+m_{24}+m_{34}\right)\Gamma\left(k_{34}+1\right)}{\Gamma\left(k_{123}+k_{34}+3-J_{1}^{1}+m_{23}+m_{24}+m_{34}\right)}$
$\displaystyle\times\frac{\left(k_{1234}\right)^{-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}\right)+m_{24}+m_{34}}}{\left(k_{1234}+k_{23}+4\right)_{-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}\right)+m_{24}+m_{34}}}\frac{\Gamma\left(k_{1234}+3\right)\Gamma\left(k_{45}+1\right)}{\Gamma\left(k_{1234}+k_{23}+4\right)}.$
(4.35)

We can now take $J_{1}^{1}=J_{2}^{1}=J_{1}^{2}=0$ to extract the leading order
term in $k_{1234}$. This inplies $J_{3}^{1}=N+p_{1}$ and $J_{2}^{2}=p_{2}$
which give

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3}\right\\},0,0}$
$\displaystyle\sim\left(k_{5}^{T_{1}}\right)^{N+p_{1}}\left(k_{5}^{T_{2}}\right)^{p_{2}}\left(k_{5}^{T_{3}}\right)^{p_{3}}$
$\displaystyle\times\sum_{m_{23}=0}\frac{\left[-k_{24}\right]_{m_{23}}}{m_{23}!}\sum_{m_{24}=0}\frac{\left[-k_{25}+N\right]_{m_{24}}}{m_{24}!}\sum_{m_{34}=0}\frac{\left[-k_{35}\right]_{m_{34}}}{m_{34}!}$
$\displaystyle\times\frac{\Gamma\left(k_{12}+1+m_{23}+m_{24}\right)\Gamma\left(k_{23}+1\right)}{\Gamma\left(k_{12}+k_{23}+2+m_{23}+m_{24}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{123}+2+m_{23}+m_{24}+m_{34}\right)\Gamma\left(k_{34}+1\right)}{\Gamma\left(k_{123}+k_{34}+3+m_{23}+m_{24}+m_{34}\right)}$
$\displaystyle\times\frac{\left(k_{1234}\right)^{m_{24}+m_{34}}}{\left(k_{1234}+k_{23}+4\right)_{m_{24}+m_{34}}}\frac{\Gamma\left(k_{1234}+3\right)\Gamma\left(k_{45}+1\right)}{\Gamma\left(k_{1234}+k_{23}+4\right)}.$
(4.36)

Finally, the ratios of the $7$-point RSSA can be easily calculated to be

$\displaystyle\frac{A^{\left\\{p_{1},p_{2},p_{3}\right\\},0,0}}{A^{\left\\{0,0,0\right\\},0,0}}$
$\displaystyle=\left(k_{5}^{T_{1}}\right)^{p_{1}}\left(k_{5}^{T_{2}}\right)^{p_{2}}\left(k_{5}^{T_{3}}\right)^{p_{3}}$
$\displaystyle=\left(\cos\phi_{2}^{5}\right)^{p_{1}}\left(\sin\phi_{2}^{5}\cos\phi_{3}^{5}\right)^{p_{2}}\left(\sin\phi_{2}^{5}\sin\phi_{3}^{5}\right)^{p_{3}}$
$\displaystyle=\left(\cos\theta_{1}\right)^{p_{1}}\left(\sin\theta_{1}\cos\theta_{2}\right)^{p_{2}}\left(\sin\theta_{1}\sin\theta_{2}\right)^{p_{3}}$
$\displaystyle=\left(\omega_{1}\right)^{p_{1}}\left(\omega_{2}\right)^{p_{2}}\left(\omega_{3}\right)^{p_{3}},$
(4.37)

which is the same as Eq.(2.31) with $m=q=0$ and $r=3$.

###  IV.2 The $7$-point Regge stringy scaling

In this section we calculate the $7$-point RSSA with $r=4$ in Eq.(4.3). The
kinematics are

$\displaystyle k_{1}$
$\displaystyle=\left(\sqrt{p^{2}+M_{1}^{2}},-p,0,0,0,0\right),$ $\displaystyle
k_{2}$ $\displaystyle=\left(\sqrt{p^{2}+M_{2}^{2}},p,0,0,0,0\right),$
$\displaystyle k_{3}$
$\displaystyle=\left(-\sqrt{q_{3}^{2}+M_{3}^{2}},-q_{3}\cos\phi_{1}^{3},-q_{3}\sin\phi_{1}^{3},0,0,0\right),$
$\displaystyle k_{4}$
$\displaystyle=\left(-\sqrt{q_{4}^{2}+M_{4}^{2}},-q_{4}\cos\phi_{1}^{4},-q_{4}\sin\phi_{1}^{4}\cos\phi_{2}^{4},-q_{4}\sin\phi_{1}^{4}\sin\phi_{2}^{4},0,0\right),$
$\displaystyle k_{5}$
$\displaystyle=\left(-\sqrt{q_{5}^{2}+M_{5}^{2}},-q_{5}\cos\phi_{1}^{5},-q_{5}\sin\phi_{1}^{5}\cos\phi_{2}^{5},-q_{5}\sin\phi_{1}^{5}\sin\phi_{2}^{5}\cos\phi_{3}^{5},-q_{5}\sin\phi_{1}^{5}\sin\phi_{2}^{5}\sin\phi_{3}^{5},0\right),$
$\displaystyle k_{6}$
$\displaystyle=\left(\begin{array}[c]{c}-\sqrt{q_{5}^{2}+M_{5}^{2}},-q_{6}\cos\phi_{1}^{6},-q_{6}\sin\phi_{1}^{6}\cos\phi_{2}^{6},-q_{6}\sin\phi_{1}^{6}\sin\phi_{2}^{6}\cos\phi_{3}^{6},\\\
-q_{6}\sin\phi_{1}^{6}\sin\phi_{2}^{6}\sin\phi_{3}^{6}\cos\phi_{4}^{6},-q_{6}\sin\phi_{1}^{6}\sin\phi_{2}^{6}\sin\phi_{3}^{6}\sin\phi_{4}^{6}\end{array}\right),$
(4.40) $\displaystyle k_{7}$
$\displaystyle=\left(\begin{array}[c]{c}-\sqrt{q_{5}^{2}+M_{5}^{2}},-q_{7}\cos\phi_{1}^{7},-q_{7}\sin\phi_{1}^{7}\cos\phi_{2}^{7},-q_{7}\sin\phi_{1}^{7}\sin\phi_{2}^{7}\cos\phi_{3}^{7},\\\
-q_{7}\sin\phi_{1}^{7}\sin\phi_{2}^{7}\sin\phi_{3}^{7}\cos\phi_{4}^{7},-q_{7}\sin\phi_{1}^{7}\sin\phi_{2}^{7}\sin\phi_{3}^{7}\sin\phi_{4}^{7}\end{array}\right).$
(4.43)

The tensor state we are going to consider is

$\left(\alpha_{-1}^{T_{1}}\right)^{N+p_{1}}\left(\alpha_{-1}^{T_{2}}\right)^{p_{2}}\left(\alpha_{-1}^{T_{3}}\right)^{p_{3}}\left(\alpha_{-1}^{T_{4}}\right)^{p_{4}}\left|0,k\right\rangle,p_{1}+p_{2}+p_{3}+p_{4}=0.$
(4.44)

We will use the notation defined in Eq.(4.6), so we have the following
$\frac{7\left(7-3\right)}{2}=14$ independent kinematics variables

$k_{12},k_{13},k_{14},k_{15},k_{23},k_{24},k_{25},k_{26},k_{34},k_{35},k_{36},k_{45},k_{46},k_{56}.$
(4.45)

The RSSA of one tensor state and $6$ tachyon states is

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3},p_{4}\right\\},0,0}$
$\displaystyle=\int_{0}^{1}dx_{5}\int_{0}^{x_{5}}dx_{4}\int_{0}^{x_{4}}dx_{3}\int_{0}^{x_{3}}dx_{2}\cdot
x_{2}^{k_{12}}x_{3}^{k_{13}}x_{4}^{k_{14}}x_{5}^{k_{15}}\left(x_{3}-x_{2}\right)^{k_{23}}\left(x_{4}-x_{2}\right)^{k_{24}}\left(x_{5}-x_{2}\right)^{k_{25}}\left(1-x_{2}\right)^{k_{26}}$
$\displaystyle\times\left(x_{4}-x_{3}\right)^{k_{34}}\left(x_{5}-x_{3}\right)^{k_{35}}\left(1-x_{3}\right)^{k_{36}}\left(x_{5}-x_{4}\right)^{k_{45}}\left(1-x_{4}\right)^{k_{46}}\left(1-x_{5}\right)^{k_{56}}$
$\displaystyle\times\left[\frac{k_{3}^{T_{1}}}{x_{3}-x_{2}}+\frac{k_{4}^{T_{1}}}{x_{4}-x_{2}}+\frac{k_{5}^{T_{1}}}{x_{5}-x_{2}}+\frac{k_{6}^{T_{1}}}{1-x_{2}}\right]^{N+p_{1}}$
$\displaystyle\times\left[\frac{k_{4}^{T_{2}}}{x_{4}-x_{2}}+\frac{k_{5}^{T_{2}}}{x_{5}-x_{2}}+\frac{k_{6}^{T_{2}}}{1-x_{2}}\right]^{p_{2}}$
$\displaystyle\times\left[\frac{k_{5}^{T_{3}}}{x_{5}-x_{2}}+\frac{k_{6}^{T_{3}}}{1-x_{2}}\right]^{p_{3}}\left[\frac{k_{6}^{T_{4}}}{1-x_{2}}\right]^{p_{4}}.$
(4.46)

Note that
$k_{3}^{T_{2}},k_{3}^{T_{3}},k_{3}^{T_{4}},k_{4}^{T_{3}},k_{4}^{T_{4}},k_{5}^{T_{4}}$
are all zeros. Let us make the following change of variables

$x_{2}=z_{2}z_{3}z_{4}z_{5},x_{3}=z_{3}z_{4}z_{5},x_{4}=z_{4}z_{5},x_{5}=z_{5},x_{6}=z_{6}=1$
(4.47)

to obtain

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3},p_{4}\right\\},0,0}$
$\displaystyle=\int_{0}^{1}dz_{5}\int_{0}^{1}dz_{4}\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}\cdot\left(z_{3}z_{4}^{2}z_{5}^{3}\right)\left(z_{2}z_{3}z_{4}z_{5}\right)^{k_{12}}\left(z_{3}z_{4}z_{5}\right)^{k_{13}}\left(z_{4}z_{5}\right)^{k_{14}}z_{5}^{k_{15}}$
$\displaystyle\times\left(z_{3}z_{4}z_{5}-z_{2}z_{3}z_{4}z_{5}\right)^{k_{23}}\left(z_{4}z_{5}-z_{2}z_{3}z_{4}z_{5}\right)^{k_{24}}\left(z_{5}-z_{2}z_{3}z_{4}z_{5}\right)^{k_{25}}\left(1-z_{2}z_{3}z_{4}z_{5}\right)^{k_{26}}$
$\displaystyle\times\left(z_{4}z_{5}-z_{3}z_{4}z_{5}\right)^{k_{34}}\left(z_{5}-z_{3}z_{4}z_{5}\right)^{k_{35}}\left(1-z_{3}z_{4}z_{5}\right)^{k_{36}}\left(z_{5}-z_{4}z_{5}\right)^{k_{45}}\left(1-z_{4}z_{5}\right)^{k_{46}}\left(1-z_{5}\right)^{k_{56}}$
$\displaystyle\times\left[\frac{k_{3}^{T_{1}}}{z_{3}z_{4}z_{5}-z_{2}z_{3}z_{4}z_{5}}+\frac{k_{4}^{T_{1}}}{z_{4}z_{5}-z_{2}z_{3}z_{4}z_{5}}+\frac{k_{5}^{T_{1}}}{z_{5}-z_{2}z_{3}z_{4}z_{5}}+\frac{k_{6}^{T_{1}}}{1-z_{2}z_{3}z_{4}z_{5}}\right]^{N+p_{1}}$
$\displaystyle\times\left[\frac{k_{4}^{T_{2}}}{z_{4}z_{5}-z_{2}z_{3}z_{4}z_{5}}+\frac{k_{5}^{T_{2}}}{z_{5}-z_{2}z_{3}z_{4}z_{5}}+\frac{k_{6}^{T_{2}}}{1-z_{2}z_{3}z_{4}z_{5}}\right]^{p_{2}}$
$\displaystyle\times\left[\frac{k_{5}^{T_{3}}}{z_{5}-z_{2}z_{3}z_{4}z_{5}}+\frac{k_{6}^{T_{3}}}{1-z_{2}z_{3}z_{4}z_{5}}\right]^{p_{3}}\left[\frac{k_{6}^{T_{4}}}{1-z_{2}z_{3}z_{4}z_{5}}\right]^{p_{4}}.$
(4.48)

We use the definition in Eq.(4.7) to obtain

$k_{123}=k_{12}+k_{13}+k_{23},k_{1234}=k_{123}+k_{14}+k_{24}+k_{34},k_{12345}=k_{1234}+k_{15}+k_{25}+k_{35}+k_{45}.$
(4.49)

After some calculation, we get

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3},p_{4}\right\\},0,0}$
$\displaystyle=\int_{0}^{1}dz_{5}\int_{0}^{1}dz_{4}\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}\cdot
z_{2}^{k_{12}}z_{3}^{k_{123}+1}z_{4}^{k_{1234}+2}z_{5}^{k_{12345}+3}$
$\displaystyle\times\left(1-z_{2}\right)^{k_{23}}\left(1-z_{2}z_{3}\right)^{k_{24}}\left(1-z_{2}z_{3}z_{4}\right)^{k_{25}}\left(1-z_{2}z_{3}z_{4}z_{5}\right)^{k_{26}}$
$\displaystyle\times\left(1-z_{3}\right)^{k_{34}}\left(1-z_{3}z_{4}\right)^{k_{35}}\left(1-z_{3}z_{4}z_{5}\right)^{k_{36}}$
$\displaystyle\times\left(1-z_{4}\right)^{k_{45}}\left(1-z_{4}z_{5}\right)^{k_{46}}$
$\displaystyle\times\left(1-z_{5}\right)^{k_{56}}$
$\displaystyle\times\left[\frac{k_{3}^{T_{1}}}{z_{3}z_{4}z_{5}-z_{2}z_{3}z_{4}z_{5}}+\frac{k_{4}^{T_{1}}}{z_{4}z_{5}-z_{2}z_{3}z_{4}z_{5}}+\frac{k_{5}^{T_{1}}}{z_{5}-z_{2}z_{3}z_{4}z_{5}}+\frac{k_{6}^{T_{1}}}{1-z_{2}z_{3}z_{4}z_{5}}\right]^{N+p_{1}}$
$\displaystyle\times\left[\frac{k_{4}^{T_{2}}}{z_{4}z_{5}-z_{2}z_{3}z_{4}z_{5}}+\frac{k_{5}^{T_{2}}}{z_{5}-z_{2}z_{3}z_{4}z_{5}}+\frac{k_{6}^{T_{2}}}{1-z_{2}z_{3}z_{4}z_{5}}\right]^{p_{2}}$
$\displaystyle\times\left[\frac{k_{5}^{T_{3}}}{z_{5}-z_{2}z_{3}z_{4}z_{5}}+\frac{k_{6}^{T_{3}}}{1-z_{2}z_{3}z_{4}z_{5}}\right]^{p_{3}}\left[\frac{k_{6}^{T_{4}}}{1-z_{2}z_{3}z_{4}z_{5}}\right]^{p_{4}}.$
(4.50)

The next step is to expand the brackets to get

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3},p_{4}\right\\},0,0}$
$\displaystyle=\int_{0}^{1}dz_{5}\int_{0}^{1}dz_{4}\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}\cdot
z_{2}^{k_{12}}z_{3}^{k_{123}+1}z_{4}^{k_{1234}+2}z_{5}^{k_{12345}+3}$
$\displaystyle\times\left(1-z_{2}\right)^{k_{23}}\left(1-z_{2}z_{3}\right)^{k_{24}}\left(1-z_{2}z_{3}z_{4}\right)^{k_{25}}\left(1-z_{2}z_{3}z_{4}z_{5}\right)^{k_{26}}$
$\displaystyle\times\left(1-z_{3}\right)^{k_{34}}\left(1-z_{3}z_{4}\right)^{k_{35}}\left(1-z_{3}z_{4}z_{5}\right)^{k_{36}}$
$\displaystyle\times\left(1-z_{4}\right)^{k_{45}}\left(1-z_{4}z_{5}\right)^{k_{46}}$
$\displaystyle\times\left(1-z_{5}\right)^{k_{56}}$
$\displaystyle\times\sum_{J_{1}^{1}+J_{2}^{1}+J_{3}^{1}+J_{4}^{1}=N+p_{1}}^{N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!J_{3}^{1}!J_{4}^{1}!}\left(\frac{k_{3}^{T_{1}}}{z_{3}z_{4}z_{5}-z_{2}z_{3}z_{4}z_{5}}\right)^{J_{1}^{1}}\left(\frac{k_{4}^{T_{1}}}{z_{4}z_{5}-z_{2}z_{3}z_{4}z_{5}}\right)^{J_{2}^{1}}\left(\frac{k_{5}^{T_{1}}}{z_{5}-z_{2}z_{3}z_{4}z_{5}}\right)^{J_{3}^{1}}\left(\frac{k_{6}^{T_{1}}}{1-z_{2}z_{3}z_{4}z_{5}}\right)^{J_{4}^{1}}$
$\displaystyle\times\sum_{J_{1}^{2}+J_{2}^{2}+J_{3}^{2}=p_{2}}^{p_{2}}\frac{P_{2}!}{J_{1}^{2}!J_{2}^{2}!J_{3}^{2}!}\left(\frac{k_{4}^{T_{2}}}{z_{4}z_{5}-z_{2}z_{3}z_{4}z_{5}}\right)^{J_{1}^{2}}\left(\frac{k_{5}^{T_{2}}}{z_{5}-z_{2}z_{3}z_{4}z_{5}}\right)^{J_{2}^{2}}\left(\frac{k_{6}^{T_{2}}}{1-z_{2}z_{3}z_{4}z_{5}}\right)^{J_{3}^{2}}$
$\displaystyle\times\sum_{J_{1}^{3}+J_{2}^{3}=p_{3}}^{p_{2}}\frac{P_{3}!}{J_{1}^{3}!J_{2}^{3}!}\left(\frac{k_{5}^{T_{3}}}{z_{5}-z_{2}z_{3}z_{4}z_{5}}\right)^{J_{1}^{3}}\left(\frac{k_{6}^{T_{3}}}{1-z_{2}z_{3}z_{4}z_{5}}\right)^{J_{2}^{3}}$
$\displaystyle\times\left(\frac{k_{6}^{T_{4}}}{1-z_{2}z_{3}z_{4}z_{5}}\right)^{p_{4}}$
(4.51)

where
$J_{1}^{1},J_{2}^{1},J_{3}^{1},J_{4}^{1},J_{1}^{2},J_{2}^{2},J_{3}^{2},J_{1}^{3},J_{2}^{3}$
are non-negative integers with
$J_{1}^{1}+J_{2}^{1}+J_{3}^{1}+J_{4}^{1}=N+p_{1},$
$J_{1}^{2}+J_{2}^{2}+J_{3}^{2}=p_{2}$ and $J_{1}^{3}+J_{2}^{3}=p_{3}$. Let us
rearrange the above equation as

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3},p_{4}\right\\},0,0}$
$\displaystyle=\sum_{J_{1}^{1}+J_{2}^{1}+J_{3}^{1}+J_{4}^{1}=N+p_{1}}^{N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!J_{3}^{1}!J_{4}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{5}^{T_{1}}\right)^{J_{3}^{1}}\left(k_{6}^{T_{1}}\right)^{J_{4}^{1}}$
$\displaystyle\times\sum_{J_{1}^{2}+J_{2}^{2}+J_{3}^{2}=p_{2}}^{p_{2}}\frac{P_{2}!}{J_{1}^{2}!J_{2}^{2}!J_{3}^{2}!}\left(k_{4}^{T_{2}}\right)^{J_{1}^{2}}\left(k_{5}^{T_{2}}\right)^{J_{2}^{2}}\left(k_{6}^{T_{2}}\right)^{J_{3}^{2}}$
$\displaystyle\times\sum_{J_{1}^{3}+J_{2}^{3}=p_{3}}^{p_{2}}\frac{P_{3}!}{J_{1}^{3}!J_{2}^{3}!}\left(k_{5}^{T_{3}}\right)^{J_{1}^{3}}\left(k_{6}^{T_{3}}\right)^{J_{2}^{3}}\left(k_{6}^{T_{4}}\right)^{p_{4}}.$
$\displaystyle\times\int_{0}^{1}dz_{5}\int_{0}^{1}dz_{4}\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}\cdot
z_{2}^{k_{12}}z_{3}^{k_{123}+1}z_{4}^{k_{1234}+2}z_{5}^{k_{12345}+3}$
$\displaystyle\times\left(1-z_{2}\right)^{k_{23}}\left(1-z_{2}z_{3}\right)^{k_{24}}\left(1-z_{2}z_{3}z_{4}\right)^{k_{25}}\left(1-z_{2}z_{3}z_{4}z_{5}\right)^{k_{26}}$
$\displaystyle\times\left(1-z_{3}\right)^{k_{34}}\left(1-z_{3}z_{4}\right)^{k_{35}}\left(1-z_{3}z_{4}z_{5}\right)^{k_{36}}$
$\displaystyle\times\left(1-z_{4}\right)^{k_{45}}\left(1-z_{4}z_{5}\right)^{k_{46}}$
$\displaystyle\times\left(1-z_{5}\right)^{k_{56}}$
$\displaystyle\times\left(z_{3}z_{4}z_{5}-z_{2}z_{3}z_{4}z_{5}\right)^{-\left(J_{1}^{1}\right)}\left(z_{4}z_{5}-z_{2}z_{3}z_{4}z_{5}\right)^{-\left(J_{2}^{1}+J_{1}^{2}\right)}$
$\displaystyle\times\left(z_{5}-z_{2}z_{3}z_{4}z_{5}\right)^{-\left(J_{3}^{1}+J_{2}^{2}+J_{1}^{3}\right)}\left(1-z_{2}z_{3}z_{4}z_{5}\right)^{-\left(J_{4}^{1}+J_{3}^{2}+J_{2}^{3}+p_{4}\right)},$
(4.52)

which means

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3},p_{4}\right\\},0,0}$
$\displaystyle=\sum_{J_{1}^{1}+J_{2}^{1}+J_{3}^{1}+J_{4}^{1}=N+p_{1}}^{N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!J_{3}^{1}!J_{4}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{5}^{T_{1}}\right)^{J_{3}^{1}}\left(k_{6}^{T_{1}}\right)^{J_{4}^{1}}$
$\displaystyle\times\sum_{J_{1}^{2}+J_{2}^{2}+J_{3}^{2}=p_{2}}^{p_{2}}\frac{P_{2}!}{J_{1}^{2}!J_{2}^{2}!J_{3}^{2}!}\left(k_{4}^{T_{2}}\right)^{J_{1}^{2}}\left(k_{5}^{T_{2}}\right)^{J_{2}^{2}}\left(k_{6}^{T_{2}}\right)^{J_{3}^{2}}$
$\displaystyle\times\sum_{J_{1}^{3}+J_{2}^{3}=p_{3}}^{p_{2}}\frac{P_{3}!}{J_{1}^{3}!J_{2}^{3}!}\left(k_{5}^{T_{3}}\right)^{J_{1}^{3}}\left(k_{6}^{T_{3}}\right)^{J_{2}^{3}}\left(k_{6}^{T_{4}}\right)^{p_{4}}.$
$\displaystyle\times\int_{0}^{1}dz_{5}\int_{0}^{1}dz_{4}\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}\cdot
z_{2}^{k_{12}}z_{3}^{k_{123}+1-\left(J_{1}^{1}\right)}z_{4}^{k_{1234}+2-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}\right)}z_{5}^{k_{12345}+3-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}+J_{3}^{1}+J_{2}^{2}+J_{1}^{3}\right)}$
$\displaystyle\times\left(1-z_{2}\right)^{k_{23}-J_{1}^{1}}\left(1-z_{2}z_{3}\right)^{k_{24}-\left(J_{2}^{1}+J_{1}^{2}\right)}\left(1-z_{2}z_{3}z_{4}\right)^{k_{25}-\left(J_{3}^{1}+J_{2}^{2}+J_{1}^{3}\right)}\left(1-z_{2}z_{3}z_{4}z_{5}\right)^{k_{26}-\left(J_{4}^{1}+J_{3}^{2}+J_{2}^{3}+p_{4}\right)}$
$\displaystyle\times\left(1-z_{3}\right)^{k_{34}}\left(1-z_{3}z_{4}\right)^{k_{35}}\left(1-z_{3}z_{4}z_{5}\right)^{k_{36}}$
$\displaystyle\times\left(1-z_{4}\right)^{k_{45}}\left(1-z_{4}z_{5}\right)^{k_{46}}$
$\displaystyle\times\left(1-z_{5}\right)^{k_{56}}.$ (4.53)

Then we expand the crossing terms to get

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3},p_{4}\right\\},0,0}$
$\displaystyle=\sum_{J_{1}^{1}+J_{2}^{1}+J_{3}^{1}+J_{4}^{1}=N+p_{1}}^{N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!J_{3}^{1}!J_{4}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{5}^{T_{1}}\right)^{J_{3}^{1}}\left(k_{6}^{T_{1}}\right)^{J_{4}^{1}}$
$\displaystyle\times\sum_{J_{1}^{2}+J_{2}^{2}+J_{3}^{2}=p_{2}}^{p_{2}}\frac{P_{2}!}{J_{1}^{2}!J_{2}^{2}!J_{3}^{2}!}\left(k_{4}^{T_{2}}\right)^{J_{1}^{2}}\left(k_{5}^{T_{2}}\right)^{J_{2}^{2}}\left(k_{6}^{T_{2}}\right)^{J_{3}^{2}}$
$\displaystyle\times\sum_{J_{1}^{3}+J_{2}^{3}=p_{3}}^{p_{2}}\frac{P_{3}!}{J_{1}^{3}!J_{2}^{3}!}\left(k_{5}^{T_{3}}\right)^{J_{1}^{3}}\left(k_{6}^{T_{3}}\right)^{J_{2}^{3}}\left(k_{6}^{T_{4}}\right)^{p_{4}}.$
$\displaystyle\times\int_{0}^{1}dz_{5}\int_{0}^{1}dz_{4}\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}\cdot
z_{2}^{k_{12}}z_{3}^{k_{123}+1-\left(J_{1}^{1}\right)}z_{4}^{k_{1234}+2-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}\right)}z_{5}^{k_{12345}+3-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}+J_{3}^{1}+J_{2}^{2}+J_{1}^{3}\right)}$
$\displaystyle\times\left(1-z_{2}\right)^{k_{23}-J_{1}^{1}}\left(1-z_{3}\right)^{k_{34}}\left(1-z_{4}\right)^{k_{45}}\left(1-z_{5}\right)^{k_{56}}$
$\displaystyle\times\sum_{m_{23}}\frac{\left[-k_{24}+\left(J_{2}^{1}+J_{1}^{2}\right)\right]_{m_{23}}}{m_{23}!}\left(z_{2}z_{3}\right)^{m_{23}}\sum_{m_{24}}\frac{\left[-k_{25}+\left(J_{3}^{1}+J_{2}^{2}+J_{1}^{3}\right)\right]_{m_{24}}}{m_{24}!}\left(z_{2}z_{3}z_{4}\right)^{m_{24}}$
$\displaystyle\times\sum_{m_{25}}\frac{\left[-k_{26}+\left(J_{4}^{1}+J_{3}^{2}+J_{2}^{3}+p_{4}\right)\right]_{m_{25}}}{m_{25}!}\left(z_{2}z_{3}z_{4}z_{5}\right)^{m_{25}}$
$\displaystyle\times\sum_{m_{34}}\frac{\left[-k_{35}\right]_{m_{34}}}{m_{34}!}\left(z_{3}z_{4}\right)^{m_{34}}\sum_{m_{35}}\frac{\left[-k_{36}\right]_{m_{34}}}{m_{35}!}\left(z_{3}z_{4}z_{5}\right)^{m_{35}}\sum_{m_{45}}\frac{\left[-k_{46}\right]_{m_{34}}}{m_{45}!}\left(z_{4}z_{5}\right)^{m_{45}},$
(4.54)

which gives

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3},p_{4}\right\\},0,0}$
$\displaystyle=\sum_{J_{1}^{1}+J_{2}^{1}+J_{3}^{1}+J_{4}^{1}=N+p_{1}}^{N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!J_{3}^{1}!J_{4}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{5}^{T_{1}}\right)^{J_{3}^{1}}\left(k_{6}^{T_{1}}\right)^{J_{4}^{1}}$
$\displaystyle\times\sum_{J_{1}^{2}+J_{2}^{2}+J_{3}^{2}=p_{2}}^{p_{2}}\frac{P_{2}!}{J_{1}^{2}!J_{2}^{2}!J_{3}^{2}!}\left(k_{4}^{T_{2}}\right)^{J_{1}^{2}}\left(k_{5}^{T_{2}}\right)^{J_{2}^{2}}\left(k_{6}^{T_{2}}\right)^{J_{3}^{2}}$
$\displaystyle\times\sum_{J_{1}^{3}+J_{2}^{3}=p_{3}}^{p_{2}}\frac{P_{3}!}{J_{1}^{3}!J_{2}^{3}!}\left(k_{5}^{T_{3}}\right)^{J_{1}^{3}}\left(k_{6}^{T_{3}}\right)^{J_{2}^{3}}\left(k_{6}^{T_{4}}\right)^{p_{4}}$
$\displaystyle\times\sum_{m_{23}}\frac{\left[-k_{24}+\left(J_{2}^{1}+J_{1}^{2}\right)\right]_{m_{23}}}{m_{23}!}\sum_{m_{24}}\frac{\left[-k_{25}+\left(J_{3}^{1}+J_{2}^{2}+J_{1}^{3}\right)\right]_{m_{24}}}{m_{24}!}\sum_{m_{25}}\frac{\left[-k_{26}+\left(J_{4}^{1}+J_{3}^{2}+J_{2}^{3}+p_{4}\right)\right]_{m_{25}}}{m_{25}!}$
$\displaystyle\times\sum_{m_{34}}\frac{\left[-k_{35}\right]_{m_{34}}}{m_{34}!}\sum_{m_{35}}\frac{\left[-k_{36}\right]_{m_{35}}}{m_{35}!}\sum_{m_{45}}\frac{\left[-k_{46}\right]_{m_{45}}}{m_{45}!}$
$\displaystyle\times\int_{0}^{1}dz_{2}\cdot
z_{2}^{k_{12}+m_{23}+m_{24}+m_{25}}\left(1-z_{2}\right)^{k_{23}-J_{1}^{1}}$
$\displaystyle\times\int_{0}^{1}dz_{3}\cdot
z_{3}^{k_{123}+1-\left(J_{1}^{1}\right)+m_{23}+m_{24}+m_{25}+m_{34}+m_{35}}\left(1-z_{3}\right)^{k_{34}}$
$\displaystyle\times\int_{0}^{1}dz_{4}\cdot
z_{4}^{k_{1234}+2-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}\right)+m_{24}+m_{25}+m_{34}+m_{35}+m_{45}}\left(1-z_{4}\right)^{k_{45}}$
$\displaystyle\times\int_{0}^{1}dz_{5}z_{5}^{k_{12345}+3-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}+J_{3}^{1}+J_{2}^{2}+J_{1}^{3}\right)+m_{25}+m_{35}+m_{45}}\left(1-z_{5}\right)^{k_{56}}.$
(4.55)

After integration, we obtain

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3},p_{4}\right\\},0,0}$
$\displaystyle=\sum_{J_{1}^{1}+J_{2}^{1}+J_{3}^{1}+J_{4}^{1}=N+p_{1}}^{N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!J_{3}^{1}!J_{4}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{5}^{T_{1}}\right)^{J_{3}^{1}}\left(k_{6}^{T_{1}}\right)^{J_{4}^{1}}$
$\displaystyle\times\sum_{J_{1}^{2}+J_{2}^{2}+J_{3}^{2}=p_{2}}^{p_{2}}\frac{P_{2}!}{J_{1}^{2}!J_{2}^{2}!J_{3}^{2}!}\left(k_{4}^{T_{2}}\right)^{J_{1}^{2}}\left(k_{5}^{T_{2}}\right)^{J_{2}^{2}}\left(k_{6}^{T_{2}}\right)^{J_{3}^{2}}$
$\displaystyle\times\sum_{J_{1}^{3}+J_{2}^{3}=p_{3}}^{p_{2}}\frac{P_{3}!}{J_{1}^{3}!J_{2}^{3}!}\left(k_{5}^{T_{3}}\right)^{J_{1}^{3}}\left(k_{6}^{T_{3}}\right)^{J_{2}^{3}}\left(k_{6}^{T_{4}}\right)^{p_{4}}$
$\displaystyle\times\sum_{m_{23}}\frac{\left[-k_{24}+\left(J_{2}^{1}+J_{1}^{2}\right)\right]_{m_{23}}}{m_{23}!}\sum_{m_{24}}\frac{\left[-k_{25}+\left(J_{3}^{1}+J_{2}^{2}+J_{1}^{3}\right)\right]_{m_{24}}}{m_{24}!}\sum_{m_{25}}\frac{\left[-k_{26}+\left(J_{4}^{1}+J_{3}^{2}+J_{2}^{3}+p_{4}\right)\right]_{m_{25}}}{m_{25}!}$
$\displaystyle\times\sum_{m_{34}}\frac{\left[-k_{35}\right]_{m_{34}}}{m_{34}!}\sum_{m_{35}}\frac{\left[-k_{36}\right]_{m_{35}}}{m_{35}!}\sum_{m_{45}}\frac{\left[-k_{46}\right]_{m_{45}}}{m_{45}!}$
$\displaystyle\times\frac{\Gamma\left(k_{12}+1+m_{23}+m_{24}+m_{25}\right)\Gamma\left(k_{23}+1-J_{1}^{1}\right)}{\Gamma\left(k_{12}+k_{23}+2+m_{23}+m_{24}+m_{25}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{123}+2-J_{1}^{1}+m_{23}+m_{24}+m_{25}+m_{34}+m_{35}\right)\Gamma\left(k_{34}+1\right)}{\Gamma\left(k_{123}+k_{34}+3-J_{1}^{1}+m_{23}+m_{24}+m_{25}+m_{34}+m_{35}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{1234}+3-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}\right)+m_{24}+m_{25}+m_{34}+m_{35}+m_{45}\right)\Gamma\left(k_{45}+1\right)}{\Gamma\left(k_{1234}+k_{45}+4-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}\right)+m_{24}+m_{25}+m_{34}+m_{35}+m_{45}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{12345}+4-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}+J_{3}^{1}+J_{2}^{2}+J_{1}^{3}\right)+m_{25}+m_{35}+m_{45}\right)\Gamma\left(k_{56}+1\right)}{\Gamma\left(k_{12345}+k_{56}+5-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}+J_{3}^{1}+J_{2}^{2}+J_{1}^{3}\right)+m_{25}+m_{35}+m_{45}\right)}.$
(4.56)

Now we choose to work on the Regge regime defined by

$k_{12345}\sim s,k_{12345}+k_{56}\sim t.$ (4.57)

In this regime, the RSSA can be approximated as

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3},p_{4}\right\\},0,0}$
$\displaystyle\sim\sum_{J_{1}^{1}+J_{2}^{1}+J_{3}^{1}+J_{4}^{1}=N+p_{1}}^{N+p_{1}}\frac{\left(N+p_{1}\right)!}{J_{1}^{1}!J_{2}^{1}!J_{3}^{1}!J_{4}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\left(k_{5}^{T_{1}}\right)^{J_{3}^{1}}\left(k_{6}^{T_{1}}\right)^{J_{4}^{1}}$
$\displaystyle\times\sum_{J_{1}^{2}+J_{2}^{2}+J_{3}^{2}=p_{2}}^{p_{2}}\frac{P_{2}!}{J_{1}^{2}!J_{2}^{2}!J_{3}^{2}!}\left(k_{4}^{T_{2}}\right)^{J_{1}^{2}}\left(k_{5}^{T_{2}}\right)^{J_{2}^{2}}\left(k_{6}^{T_{2}}\right)^{J_{3}^{2}}$
$\displaystyle\times\sum_{J_{1}^{3}+J_{2}^{3}=p_{3}}^{p_{2}}\frac{P_{3}!}{J_{1}^{3}!J_{2}^{3}!}\left(k_{5}^{T_{3}}\right)^{J_{1}^{3}}\left(k_{6}^{T_{3}}\right)^{J_{2}^{3}}\left(k_{6}^{T_{4}}\right)^{p_{4}}$
$\displaystyle\times\sum_{m_{23}}\frac{\left[-k_{24}+\left(J_{2}^{1}+J_{1}^{2}\right)\right]_{m_{23}}}{m_{23}!}\sum_{m_{24}}\frac{\left[-k_{25}+\left(J_{3}^{1}+J_{2}^{2}+J_{1}^{3}\right)\right]_{m_{24}}}{m_{24}!}\sum_{m_{25}}\frac{\left[-k_{26}+\left(J_{4}^{1}+J_{3}^{2}+J_{2}^{3}+p_{4}\right)\right]_{m_{25}}}{m_{25}!}$
$\displaystyle\times\sum_{m_{34}}\frac{\left[-k_{35}\right]_{m_{34}}}{m_{34}!}\sum_{m_{35}}\frac{\left[-k_{36}\right]_{m_{35}}}{m_{35}!}\sum_{m_{45}}\frac{\left[-k_{46}\right]_{m_{45}}}{m_{45}!}$
$\displaystyle\times\frac{\Gamma\left(k_{12}+1+m_{23}+m_{24}+m_{25}\right)\Gamma\left(k_{23}+1-J_{1}^{1}\right)}{\Gamma\left(k_{12}+k_{23}+2-J_{1}^{1}+m_{23}+m_{24}+m_{25}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{123}+2-J_{1}^{1}+m_{23}+m_{24}+m_{25}+m_{34}+m_{35}\right)\Gamma\left(k_{34}+1\right)}{\Gamma\left(k_{123}+k_{34}+3-J_{1}^{1}+m_{23}+m_{24}+m_{25}+m_{34}+m_{35}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{1234}+3-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}\right)+m_{24}+m_{25}+m_{34}+m_{35}+m_{45}\right)\Gamma\left(k_{45}+1\right)}{\Gamma\left(k_{1234}+k_{45}+4-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}\right)+m_{24}+m_{25}+m_{34}+m_{35}+m_{45}\right)}$
$\displaystyle\times\frac{\left(k_{12345}\right)^{-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}+J_{3}^{1}+J_{2}^{2}+J_{1}^{3}\right)+m_{25}+m_{35}+m_{45}}}{\left(k_{12345}+k_{56}+5\right)_{-\left(J_{1}^{1}+J_{2}^{1}+J_{1}^{2}+J_{3}^{1}+J_{2}^{2}+J_{1}^{3}\right)+m_{25}+m_{35}+m_{45}}}\frac{\Gamma\left(k_{12345}+4\right)\Gamma\left(k_{56}+1\right)}{\Gamma\left(k_{12345}+k_{56}+5\right)}.$
(4.58)

To get the leading order in $k_{12345}$, we take

$J_{1}^{1}=J_{2}^{1}=J_{3}^{1}=J_{1}^{2}=J_{2}^{2}=J_{1}^{3}=0,$

which implies

$J_{4}^{1}=N+p_{1},J_{3}^{2}=p_{2},J_{2}^{3}=p_{3}.$ (4.59)

With $p_{1}+p_{2}+p_{3}+p_{4}=0$, the leading term is

$\displaystyle A^{\left\\{p_{1},p_{2},p_{3},p_{4}\right\\},0,0}$
$\displaystyle\sim\left(k_{6}^{T_{1}}\right)^{N+p_{1}}\left(k_{6}^{T_{2}}\right)^{p_{2}}\left(k_{6}^{T_{3}}\right)^{p_{3}}\left(k_{6}^{T_{4}}\right)^{p_{4}}$
$\displaystyle\times\sum_{m_{23}}\frac{\left[-k_{24}\right]_{m_{23}}}{m_{23}!}\sum_{m_{24}}\frac{\left[-k_{25}\right]_{m_{24}}}{m_{24}!}\sum_{m_{25}}\frac{\left[-k_{26}+N\right]_{m_{25}}}{m_{25}!}\sum_{m_{34}}\frac{\left[-k_{35}\right]_{m_{34}}}{m_{34}!}\sum_{m_{35}}\frac{\left[-k_{36}\right]_{m_{35}}}{m_{35}!}\sum_{m_{45}}\frac{\left[-k_{46}\right]_{m_{45}}}{m_{45}!}$
$\displaystyle\times\frac{\Gamma\left(k_{12}+1+m_{23}+m_{24}+m_{25}\right)\Gamma\left(k_{23}+1\right)}{\Gamma\left(k_{12}+k_{23}+2+m_{23}+m_{24}+m_{25}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{123}+2+m_{23}+m_{24}+m_{25}+m_{34}+m_{35}\right)\Gamma\left(k_{34}+1\right)}{\Gamma\left(k_{123}+k_{34}+3+m_{23}+m_{24}+m_{25}+m_{34}+m_{35}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{1234}+3+m_{24}+m_{25}+m_{34}+m_{35}+m_{45}\right)\Gamma\left(k_{45}+1\right)}{\Gamma\left(k_{1234}+k_{45}+4+m_{24}+m_{25}+m_{34}+m_{35}+m_{45}\right)}$
$\displaystyle\times\frac{\left(k_{12345}\right)^{+m_{25}+m_{35}+m_{45}}}{\left(k_{12345}+k_{56}+5\right)_{m_{25}+m_{35}+m_{45}}}\frac{\Gamma\left(k_{12345}+4\right)\Gamma\left(k_{56}+1\right)}{\Gamma\left(k_{12345}+k_{56}+5\right)}.$
(4.60)

So the ratios of the $7$-point RSSA is

$\displaystyle\frac{A^{\left\\{p_{1},p_{2},p_{3},p_{4}\right\\},0,0}}{A^{\left\\{0,0,0,0\right\\},0,0}}$
$\displaystyle=\left(k_{6}^{T_{1}}\right)^{p_{1}}\left(k_{6}^{T_{2}}\right)^{p_{2}}\left(k_{6}^{T_{3}}\right)^{p_{3}}\left(k_{6}^{T_{4}}\right)^{p_{4}}$
$\displaystyle=\left(\cos\phi_{2}^{6}\right)^{p_{1}}\left(\sin\phi_{2}^{6}\cos\phi_{3}^{6}\right)^{p_{2}}\left(\sin\phi_{2}^{6}\sin\phi_{3}^{6}\cos\phi_{4}^{6}\right)^{p_{3}}\left(\sin\phi_{2}^{6}\sin\phi_{3}^{6}\sin\phi_{4}^{6}\right)^{p_{4}}$
$\displaystyle=\left(\cos\theta_{1}\right)^{p_{1}}\left(\sin\theta_{1}\cos\theta_{2}\right)^{p_{2}}\left(\sin\theta_{1}\sin\theta_{2}\cos\theta_{3}\right)^{p_{3}}\left(\sin\theta_{1}\sin\theta_{2}\sin\theta_{3}\right)^{p_{4}}$
$\displaystyle=\left(\omega_{1}\right)^{p_{1}}\left(\omega_{2}\right)^{p_{2}}\left(\omega_{3}\right)^{p_{3}}\left(\omega_{4}\right)^{p_{4}},$
(4.61)

which is the same as Eq.(2.31) with $m=q=0$ and $r=4$.

##  V The $n$-point Regge stringy scaling

In this section, we generalize the previous calculations to the case of
$n$-point RSSA. We first define the $26$-dimensional momenta in the CM frame
to be

$\displaystyle k_{1}$
$\displaystyle=\left(\sqrt{p^{2}+M_{1}^{2}},-p,0^{r}\right),$ $\displaystyle
k_{2}$ $\displaystyle=\left(\sqrt{p^{2}+M_{2}^{2}},p,0^{r}\right),$
$\displaystyle\vdots$ $\displaystyle k_{j}$
$\displaystyle=\left(-\sqrt{q_{j}^{2}+M_{j}^{2}},-q_{j}\Omega_{1}^{j},-q_{j}\Omega_{2}^{j},\cdots,-q_{j}\Omega_{r}^{j},-q_{j}\Omega_{r+1}^{j}\right)$
(5.1)

where $j=3,4,\cdots,n$, and

$\Omega_{i}^{j}=\cos\phi_{i}^{j}\prod\limits_{\sigma=1}^{i-1}\sin\phi_{\sigma}^{j}\text{
with }\phi_{j-1}^{j}=0,\text{ }\phi_{i>r}^{j}=0\text{ and
}r\leq\min\left\\{n-3,24\right\\}$ (5.2)

are the solid angles in the $\left(j-2\right)$-dimensional spherical space
with $\sum_{i=1}^{j-2}\left(\Omega_{i}^{j}\right)^{2}=1$. In Eq.(5.1), $0^{r}$
denotes the $r$-dimensional null vector. The amplitude of one tensor state

$\left(\alpha_{-1}^{T_{1}}\right)^{N+p_{1}}\left(\alpha_{-1}^{T_{2}}\right)^{p_{2}}\cdots\left(\alpha_{-1}^{T_{r}}\right)^{p_{r}}\left|0,k\right\rangle,p_{1}+p_{2}+\cdots+p_{r}=0$
(5.3)

and $n-1$ tachyon states is

$\displaystyle A^{\left\\{p_{1},p_{2},\cdots,p_{r}\right\\},0,0}$
$\displaystyle=\int_{0}^{1}dx_{n-2}\int_{0}^{x_{n-2}}dx_{n-3}\cdots\int_{0}^{x_{4}}dx_{3}\int_{0}^{x_{3}}dx_{2}\times\prod_{0\leq
i<j\leq
n-1}\left(x_{j}-x_{i}\right)^{k_{ij}}\prod_{\sigma=1}^{r}\left[\sum_{j=\sigma+2}^{n-1}\left(\frac{k_{j}^{T_{\sigma}}}{x_{j}-x_{2}}\right)\right]^{\mathcal{P}_{\sigma}}$
(5.4)

where we have defined

$\mathcal{P}_{1}=N+p_{1},\mathcal{P}_{\sigma\neq 1}=p_{\sigma}.$ (5.5)

Now, let’s explicitly write down the second product part of Eq.(5.4) as

$\displaystyle A^{\left\\{p_{1},p_{2},\cdots,p_{r}\right\\},0,0}$
$\displaystyle=\int_{0}^{1}dx_{n-2}\int_{0}^{x_{n-2}}dx_{n-3}\cdots\int_{0}^{x_{4}}dx_{3}\int_{0}^{x_{3}}dx_{2}\times\prod_{0\leq
i<j\leq n-1}\left(x_{j}-x_{i}\right)^{k_{ij}}$
$\displaystyle\times\left[\frac{k_{3}^{T_{1}}}{x_{3}-x_{2}}+\frac{k_{4}^{T_{1}}}{x_{4}-x_{2}}+\frac{k_{5}^{T_{1}}}{x_{5}-x_{2}}\cdots+\frac{k_{n-1}^{T_{1}}}{1-x_{2}}\right]^{\mathcal{P}_{1}}$
$\displaystyle\times\left[\frac{k_{4}^{T_{2}}}{x_{4}-x_{2}}+\frac{k_{5}^{T_{2}}}{x_{5}-x_{2}}+\cdots+\frac{k_{n-1}^{T_{2}}}{1-x_{2}}\right]^{\mathcal{P}_{2}}$
$\displaystyle\vdots$
$\displaystyle\times\left[\frac{k_{r+2}^{T_{r}}}{x_{r+2}-x_{2}}+\cdots+\frac{k_{n-1}^{T_{r}}}{1-x_{2}}\right]^{\mathcal{P}_{r}}.$
(5.6)

For convience, from now on we add trivial terms with $\mathcal{P}_{\sigma}=0$
$(r+1\leq\sigma\leq n-3)$ to the amplitude and obtain

$\displaystyle A^{\left\\{p_{1},p_{2},\cdots,p_{r}\right\\},0,0}$
$\displaystyle=\int_{0}^{1}dx_{n-2}\int_{0}^{x_{n-2}}dx_{n-3}\cdots\int_{0}^{x_{4}}dx_{3}\int_{0}^{x_{3}}dx_{2}\times\prod_{0\leq
i<j\leq n-1}\left(x_{j}-x_{i}\right)^{k_{ij}}$
$\displaystyle\times\left[\frac{k_{3}^{T_{1}}}{x_{3}-x_{2}}+\frac{k_{4}^{T_{1}}}{x_{4}-x_{2}}+\frac{k_{5}^{T_{1}}}{x_{5}-x_{2}}+\frac{k_{5}^{T_{1}}}{x_{6}-x_{2}}+\cdots+\frac{k_{n-1}^{T_{1}}}{1-x_{2}}\right]^{\mathcal{P}_{1}}$
$\displaystyle\times\left[\frac{k_{4}^{T_{2}}}{x_{4}-x_{2}}+\frac{k_{5}^{T_{2}}}{x_{5}-x_{2}}+\frac{k_{6}^{T_{2}}}{x_{6}-x_{2}}+\cdots+\frac{k_{n-1}^{T_{2}}}{1-x_{2}}\right]^{\mathcal{P}_{2}}$
$\displaystyle\vdots$
$\displaystyle\times\left[\frac{k_{r+2}^{T_{r}}}{x_{r+2}-x_{2}}+\frac{k_{r+3}^{T_{r}}}{x_{r+3}-x_{2}}+\cdots+\frac{k_{n-1}^{T_{r}}}{1-x_{2}}\right]^{\mathcal{P}_{r}}$
$\displaystyle\times\left[\frac{k_{r+3}^{T_{r+1}}}{x_{r+3}-x_{2}}+\cdots+\frac{k_{n-1}^{T_{r+1}}}{1-x_{2}}\right]^{\mathcal{P}_{r+1}}$
$\displaystyle\vdots$
$\displaystyle\times\left[\frac{k_{n-1}^{T_{n-3}}}{1-x_{2}}\right]^{\mathcal{P}_{n-3}}.$
(5.7)

Now we can expand the brackets

$\displaystyle A^{\left\\{p_{1},p_{2},\cdots,p_{r}\right\\},0,0}$
$\displaystyle=\int_{0}^{1}dx_{n-2}\int_{0}^{x_{n-2}}dx_{n-3}\cdots\int_{0}^{x_{4}}dx_{3}\int_{0}^{x_{3}}dx_{2}\times\prod_{0\leq
i<j\leq n-1}\left(x_{j}-x_{i}\right)^{k_{ij}}$
$\displaystyle\times\sum_{J_{1}^{1}+J_{2}^{1}+\cdots+J_{n-4}^{1}+J_{n-3}^{1}\equiv\mathcal{P}_{1}}\frac{\mathcal{P}_{1}!}{J_{1}^{1}!J_{2}^{1}!\cdots
J_{n-4}^{1}!J_{n-3}^{1}!}\left(\frac{k_{3}^{T_{1}}}{x_{3}-x_{2}}\right)^{J_{1}^{1}}\left(\frac{k_{4}^{T_{1}}}{x_{4}-x_{2}}\right)^{J_{2}^{1}}\cdots\left(\frac{k_{n-1}^{T_{1}}}{1-x_{2}}\right)^{J_{n-3}^{1}}$
$\displaystyle\times\sum_{J_{1}^{2}+J_{2}^{2}+\cdots+J_{n-5}^{2}+J_{n-4}^{2}=\mathcal{P}_{2}}\frac{\mathcal{P}_{2}!}{J_{1}^{2}!J_{2}^{2}!\cdots
J_{n-5}^{2}!J_{n-4}^{2}!}\left(\frac{k_{4}^{T_{2}}}{x_{4}-x_{2}}\right)^{J_{1}^{2}}\cdots\left(\frac{k_{n-1}^{T_{2}}}{1-x_{2}}\right)^{J_{n-4}^{2}}$
$\displaystyle\vdots$ $\displaystyle\times$
$\displaystyle\sum_{J_{1}^{n-3}=\mathcal{P}_{n-3}=0}\frac{\mathcal{P}_{n-3}!}{J_{1}^{n-3}!}\left(\frac{k_{n-1}^{T_{n-3}}}{1-x_{2}}\right)^{J_{1}^{n-3}}$
(5.8)

where all $J$ are non-negative integers. (Note that all $J_{j}^{\sigma\geq
r+1}=0$ due to $\mathcal{P}_{\sigma\geq r+1}=0.$)

After some rearrangements, we can derive

$\displaystyle A^{\left\\{p_{1},p_{2},\cdots,p_{n-3}\right\\},0,0}$
$\displaystyle=\sum_{J_{1}^{1}+J_{2}^{1}+\cdots+J_{n-4}^{1}+J_{n-3}^{1}\equiv\mathcal{P}_{1}}\frac{\mathcal{P}_{1}!}{J_{1}^{1}!J_{2}^{1}!\cdots
J_{n-4}^{1}!J_{n-3}^{1}!}\left(k_{3}^{T_{1}}\right)^{J_{1}^{1}}\left(k_{4}^{T_{1}}\right)^{J_{2}^{1}}\cdots\left(k_{n-1}^{T_{1}}\right)^{J_{n-3}^{1}}$
$\displaystyle\times\sum_{J_{1}^{2}+J_{2}^{2}+\cdots+J_{n-5}^{2}+J_{n-4}^{2}=\mathcal{P}_{2}}\frac{\mathcal{P}_{2}!}{J_{1}^{2}!J_{2}^{2}!\cdots
J_{n-5}^{2}!J_{n-4}^{2}!}\left(k_{4}^{T_{2}}\right)^{J_{1}^{2}}\left(k_{5}^{T_{2}}\right)^{J_{2}^{2}}\cdots\left(k_{n-1}^{T_{2}}\right)^{J_{n-4}^{2}}$
$\displaystyle\vdots$
$\displaystyle\times\sum_{J_{1}^{n-3}=\mathcal{P}_{n-3}=0}\frac{\mathcal{P}_{n-3}!}{J_{1}^{n-3}!}\left(k_{n-1}^{T_{n-3}}\right)^{J_{1}^{n-3}}$
$\displaystyle\times\int_{0}^{1}dx_{n-2}\int_{0}^{x_{n-2}}dx_{n-3}\cdots\int_{0}^{x_{4}}dx_{3}\int_{0}^{x_{3}}dx_{2}\times\prod_{0\leq
i<j\leq n-1}\left(x_{j}-x_{i}\right)^{k_{ij}}$
$\displaystyle\times\left(\frac{1}{x_{3}-x_{2}}\right)^{J_{1}^{1}}\left(\frac{1}{x_{4}-x_{2}}\right)^{J_{2}^{1}}\cdots\left(\frac{1}{1-x_{2}}\right)^{J_{n-3}^{1}}$
$\displaystyle\times\left(\frac{1}{x_{4}-x_{2}}\right)^{J_{1}^{2}}\cdots\left(\frac{1}{1-x_{2}}\right)^{J_{n-4}^{2}}$
$\displaystyle\vdots$
$\displaystyle\times\left(\frac{1}{1-x_{2}}\right)^{J_{1}^{n-3}}.$ (5.9)

We can collect terms with the same sum of subscripts and superscripts to get

$\displaystyle A^{\left\\{p_{1},p_{2},\cdots,p_{n-3}\right\\},0,0}$
$\displaystyle=\prod_{\sigma=1}^{n-3}\left[\sum_{\sum_{j=1}^{n-2-\sigma}J_{j}^{\sigma}=\mathcal{P}_{\sigma}}\left(\mathcal{P}_{\sigma}!\prod_{j=1}^{n-2-\sigma}\frac{\left(k_{j+\sigma+1}^{T_{\sigma}}\right)^{J_{j}^{\sigma}}}{J_{j}^{\sigma}!}\right)\right]$
(5.10)
$\displaystyle\times\int_{0}^{1}dx_{n-2}\int_{0}^{x_{n-2}}dx_{n-3}\cdots\int_{0}^{x_{4}}dx_{3}\int_{0}^{x_{3}}dx_{2}\times\prod_{0\leq
i<j\leq
n-1}\left(x_{j}-x_{i}\right)^{k_{ij}-\delta_{i2}\left(J_{j-2}^{1}+\cdots+J_{1}^{j-2}\right)}.$
(5.11)

To perform the integral of the last line in Eq.(5.11), let’s do the following
change of variables

$x_{i}=\prod_{k=i}^{n-2}z_{k},\text{ or }x_{2}=z_{2}z_{3}\cdots
z_{n-2},x_{3}=z_{3}z_{4}\cdots
z_{n-2},\cdots,x_{n-2}=z_{n-2},x_{n-1}=z_{n-1}=1$ (5.12)

to make all the integral intervals from $0$ to $1$. The integral becomes

$\displaystyle\int_{0}^{1}dx_{n-2}\int_{0}^{x_{n-2}}dx_{n-3}\cdots\int_{0}^{x_{4}}dx_{3}\int_{0}^{x_{3}}dx_{2}\times\prod_{0\leq
i<j\leq
n-1}\left(x_{j}-x_{i}\right)^{k_{ij}-\delta_{i2}\left(J_{j-2}^{1}+\cdots+J_{1}^{j-2}\right)}$
$\displaystyle=\int_{0}^{1}dz_{n-2}\cdots\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}\prod_{i=1}^{n-4}\left(z_{i+2}\right)^{i}\prod_{0\leq
i<j\leq
n-1}\left(\prod_{k=j}^{n-2}z_{k}-\prod_{k=i}^{n-2}z_{k}\right)^{k_{ij}-\delta_{i2}\left(J_{j-2}^{1}+\cdots+J_{1}^{j-2}\right)}$
$\displaystyle=\int_{0}^{1}dz_{n-2}\cdots\int_{0}^{1}dz_{2}\times\prod_{i=1}^{n-4}\left(z_{i+2}\right)^{i}\prod_{0\leq
i<j\leq
n-1}\left[\prod_{k=j}^{n-2}z_{k}\left(1-\prod_{k=i}^{j-1}z_{k}\right)\right]^{k_{ij}-\delta_{i2}\left(J_{j-2}^{1}+\cdots+J_{1}^{j-2}\right)}.$
(5.13)

Now, the amplitude can be explicitly written as

$\displaystyle A^{\left\\{p_{1},p_{2},\cdots,p_{n-3}\right\\},0,0}$
$\displaystyle=\prod_{\sigma=1}^{n-3}\left[\sum_{\sum_{j=1}^{n-2-\sigma}J_{j}^{\sigma}=\mathcal{P}_{\sigma}}\left(\mathcal{P}_{\sigma}!\prod_{j=1}^{n-2-\sigma}\frac{\left(k_{j+\sigma+1}^{T_{\sigma}}\right)^{J_{j}^{\sigma}}}{J_{j}^{\sigma}!}\right)\right]$
$\displaystyle\times\int_{0}^{1}dz_{n-2}\int_{0}^{1}dz_{n-3}\cdots\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}$
$\displaystyle\times
z_{2}^{k_{12}}z_{3}^{k_{123}+1-J_{1}^{1}}z_{4}^{k_{1234}+2-J_{1}^{1}-\left(J_{1}^{2}+J_{2}^{1}\right)}\cdots
z_{n-2}^{k_{1,\cdots,n-2}+\left(n-4\right)-J_{1}^{1}-\left(J_{1}^{2}+J_{2}^{1}\right)-\cdots-\left(J_{1}^{n-4}+\cdots+J_{n-4}^{1}\right)}$
$\displaystyle\times\left(1-z_{2}\right)^{k_{23}-J_{1}^{1}}\left(1-z_{2}z_{3}\right)^{k_{24}-\left(J_{1}^{2}+J_{2}^{1}\right)}\cdots\left(1-z_{2}z_{3}z_{4}\cdots
z_{n-2}\right)^{k_{2,n-1}-\left(J_{1}^{n-3}+\cdots+J_{n-3}^{1}\right)}$
$\displaystyle\times\left(1-z_{3}\right)^{k_{34}}\left(1-z_{3}z_{4}\right)^{k_{35}}\cdots\left(1-z_{3}z_{4}\cdots
z_{n-2}\right)^{k_{3,n-1}}$ $\displaystyle\vdots$
$\displaystyle\times\left(1-z_{n-3}\right)^{k_{n-3,n-2}}\left(1-z_{n-3}z_{n-2}\right)^{k_{n-3,n-1}}$
$\displaystyle\times\left(1-z_{n-2}\right)^{k_{n-2,n-1}}.$ (5.14)

Let’s rearrange the above equation to get a more symmetric form in the
following

$\displaystyle A^{\left\\{p_{1},p_{2},\cdots,p_{n-3}\right\\},0,0}$
$\displaystyle=\prod_{\sigma=1}^{n-3}\left[\sum_{\sum_{j=1}^{n-2-\sigma}J_{j}^{\sigma}=\mathcal{P}_{\sigma}}\left(\mathcal{P}_{\sigma}!\prod_{j=1}^{n-2-\sigma}\frac{\left(k_{j+\sigma+1}^{T_{\sigma}}\right)^{J_{j}^{\sigma}}}{J_{j}^{\sigma}!}\right)\right]$
$\displaystyle\times\int_{0}^{1}dz_{n-2}\int_{0}^{1}dz_{n-3}\cdots\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}\cdot$
$\displaystyle\times
z_{2}^{k_{12}}z_{3}^{k_{123}+1-J_{1}^{1}}z_{4}^{k_{1234}+2-J_{1}^{1}-\left(J_{1}^{2}+J_{2}^{1}\right)}\cdots
z_{n-2}^{k_{1,\cdots,n-2}+\left(n-4\right)-J_{1}^{1}-\left(J_{1}^{2}+J_{2}^{1}\right)-\cdots-\left(J_{1}^{n-4}+\cdots+J_{n-4}^{1}\right)}$
$\displaystyle\times\left(1-z_{2}\right)^{k_{23}-J_{1}^{1}}\left(1-z_{3}\right)^{k_{34}}\left(1-z_{4}\right)^{k_{45}}\cdots\left(1-z_{n-2}\right)^{k_{n-2,n-1}}$
$\displaystyle\times\left(1-z_{2}z_{3}\right)^{k_{24}-\left(J_{1}^{2}+J_{1}^{2}\right)}\left(1-z_{2}z_{3}z_{4}\right)^{k_{25}-\left(J_{1}^{3}+J_{2}^{2}+J_{3}^{1}\right)}\cdots\left(1-z_{2}z_{3}z_{4}\cdots
z_{n-2}\right)^{k_{2,n-1}-\left(J_{1}^{n-3}+\cdots+J_{n-3}^{1}\right)}$
$\displaystyle\times\left(1-z_{3}z_{4}\right)^{k_{35}}\cdots\left(1-z_{3}z_{4}\cdots
z_{n-2}\right)^{k_{3,n-1}}$ $\displaystyle\vdots$
$\displaystyle\times\left(1-z_{n-3}z_{n-2}\right)^{k_{n-3,n-1}}.$ (5.15)

Then we expand the crossing terms to get

$\displaystyle A^{\left\\{p_{1},p_{2},\cdots,p_{n-3}\right\\},0,0}$
$\displaystyle=\prod_{\sigma=1}^{n-3}\left[\sum_{\sum_{j=1}^{n-2-\sigma}J_{j}^{\sigma}=\mathcal{P}_{\sigma}}\left(\mathcal{P}_{\sigma}!\prod_{j=1}^{n-2-\sigma}\frac{\left(k_{j+\sigma+1}^{T_{\sigma}}\right)^{J_{j}^{\sigma}}}{J_{j}^{\sigma}!}\right)\right]$
$\displaystyle\times\int_{0}^{1}dz_{n-2}\int_{0}^{1}dz_{n-3}\cdots\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}$
$\displaystyle\times
z_{2}^{k_{12}}z_{3}^{k_{123}+1-J_{1}^{1}}z_{4}^{k_{1234}+2-J_{1}^{1}-\left(J_{1}^{2}+J_{2}^{1}\right)}\cdots
z_{n-2}^{k_{1,\cdots,n-2}+\left(n-4\right)-J_{1}^{1}-\left(J_{1}^{2}+J_{2}^{1}\right)-\cdots-\left(J_{1}^{n-4}+\cdots+J_{n-4}^{1}\right)}$
$\displaystyle\times\left(1-z_{2}\right)^{k_{23}-J_{1}^{1}}\left(1-z_{3}\right)^{k_{34}}\left(1-z_{4}\right)^{k_{45}}\cdots\left(1-z_{n-2}\right)^{k_{n-2,n-1}}$
$\displaystyle\times\sum_{m_{23}}\frac{\left[-k_{24}+\left(J_{1}^{2}+J_{2}^{1}\right)\right]_{m_{23}}}{m_{23}!}\left(z_{2}z_{3}\right)^{m_{23}}\cdots\sum_{m_{24}}\frac{\left[-k_{2,n-1}+\left(J_{1}^{n-3}+\cdots+J_{n-3}^{1}\right)\right]_{m_{2,n-2}}}{m_{2,n-2}!}\left(z_{2}z_{3}z_{4}\cdots
z_{n-2}\right)^{m_{2,n-2}}$
$\displaystyle\times\sum_{m_{34}}\frac{\left(-k_{35}\right)_{m_{34}}}{m_{34}!}\left(z_{3}z_{4}\right)^{m_{34}}\cdots\sum_{m_{3,n-2}}\frac{\left(-k_{3,n-1}\right)_{m_{3,n-2}}}{m_{3,n-2}!}\left(z_{3}z_{4}\cdots
z_{n-2}\right)^{m_{3,n-2}}$ $\displaystyle\vdots$
$\displaystyle\times\sum_{m_{n-3,n-2}}\frac{\left(-k_{n-3,n-1}\right)_{m_{3,n-2}}}{m_{n-3,n-2}!}\left(z_{n-3}z_{n-2}\right)^{m_{n-3,n-2}}$
(5.16)

where the subscripts of $m_{ij}$ keep record of the first and the last
subscripts of $\left(z_{i}z_{i+1}\cdots z_{j-1}z_{j}\right)$ etc.. The
amplitude becomes

$\displaystyle A^{\left\\{p_{1},p_{2},\cdots,p_{n-3}\right\\},0,0}$
$\displaystyle=\prod_{\sigma=1}^{n-3}\left[\sum_{\sum_{j=1}^{n-2-\sigma}J_{j}^{\sigma}=\mathcal{P}_{\sigma}}\left(\mathcal{P}_{\sigma}!\prod_{j=1}^{n-2-\sigma}\frac{\left(k_{j+\sigma+1}^{T_{\sigma}}\right)^{J_{j}^{\sigma}}}{J_{j}^{\sigma}!}\right)\right]$
$\displaystyle\times\sum_{m_{23}}\frac{\left[-k_{24}+\left(J_{1}^{2}+J_{2}^{1}\right)\right]_{m_{23}}}{m_{23}!}\cdots\sum_{m_{24}}\frac{\left[-k_{2,n-1}+\left(J_{1}^{n-3}+\cdots+J_{n-3}^{1}\right)\right]_{m_{2,n-2}}}{m_{2,n-2}!}$
$\displaystyle\times\sum_{m_{34}}\frac{\left(-k_{35}\right)_{m_{34}}}{m_{34}!}\cdots\sum_{m_{3,n-2}}\frac{\left(-k_{3,n-1}\right)_{m_{3,n-2}}}{m_{3,n-2}!}$
$\displaystyle\vdots$
$\displaystyle\times\sum_{m_{n-3,n-2}}\frac{\left(-k_{n-3,n-1}\right)_{m_{3,n-2}}}{m_{n-3,n-2}!}$
$\displaystyle\times\int_{0}^{1}dz_{n-2}\int_{0}^{1}dz_{n-3}\cdots\int_{0}^{1}dz_{3}\int_{0}^{1}dz_{2}\cdot$
$\displaystyle\times z_{2}^{k_{12}+\sum_{i\leq 2\leq
j}m_{ij}}z_{3}^{k_{123}+1-\sum_{i+j\leq 2}J_{j}^{i}+\sum_{i\leq 3\leq
j}m_{ij}}\cdots z_{n-2}^{k_{1,\cdots,n-2}+\left(n-4\right)-\sum_{i+j\leq
n-3}J_{j}^{i}+\sum_{i\leq n-2\leq j}m_{ij}}$
$\displaystyle\times\left(1-z_{2}\right)^{k_{23}-J_{1}^{1}}\left(1-z_{3}\right)^{k_{34}}\cdots\left(1-z_{n-2}\right)^{k_{n-2,n-1}}.$
(5.17)

After integration, we can write it as

$\displaystyle A^{\left\\{p_{1},p_{2},\cdots,p_{n-3}\right\\},0,0}$
$\displaystyle=\prod_{\sigma=1}^{n-3}\left[\sum_{\sum_{j=1}^{n-2-\sigma}J_{j}^{\sigma}=\mathcal{P}_{\sigma}}\left(\mathcal{P}_{\sigma}!\prod_{j=1}^{n-2-\sigma}\frac{\left(k_{j+\sigma+1}^{T_{\sigma}}\right)^{J_{j}^{\sigma}}}{J_{j}^{\sigma}!}\right)\right]$
$\displaystyle\times\sum_{m_{23}}\frac{\left[-k_{24}+\left(J_{1}^{2}+J_{2}^{1}\right)\right]_{m_{23}}}{m_{23}!}\sum_{m_{24}}\frac{\left[-k_{25}+\left(J_{1}^{3}+J_{2}^{2}+J_{3}^{1}\right)\right]_{m_{24}}}{m_{24}!}\cdots\sum_{m_{24}}\frac{\left[-k_{2,n-1}+\left(J_{1}^{n-3}+\cdots+J_{n-3}^{1}\right)\right]_{m_{2,n-2}}}{m_{2,n-2}!}$
$\displaystyle\times\sum_{m_{34}}\frac{\left(-k_{35}\right)_{m_{34}}}{m_{34}!}\cdots\sum_{m_{3,n-2}}\frac{\left(-k_{3,n-1}\right)_{m_{3,n-2}}}{m_{3,n-2}!}$
$\displaystyle\vdots$
$\displaystyle\times\sum_{m_{n-3,n-2}}\frac{\left(-k_{n-3,n-1}\right)_{m_{3,n-2}}}{m_{n-3,n-2}!}$
$\displaystyle\times\frac{\Gamma\left(k_{12}+1+\sum_{i\leq 2\leq
j}m_{ij}\right)\Gamma\left(k_{23}+1-J_{1}^{1}\right)}{\Gamma\left(k_{12}+k_{23}+2-J_{1}^{1}+\sum_{i\leq
2\leq j}m_{ij}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{123}+2-\sum_{i+j\leq
2}J_{j}^{i}+\sum_{i\leq 3\leq
j}m_{ij}\right)\Gamma\left(k_{34}+1\right)}{\Gamma\left(k_{123}+k_{34}+3-\sum_{i+j\leq
2}J_{j}^{i}+\sum_{i\leq 3\leq j}m_{ij}\right)}$ $\displaystyle\vdots$
$\displaystyle\times\frac{\Gamma\left(k_{1,\cdots,n-2}+\left(n-3\right)-\sum_{i+j\leq
n-3}J_{j}^{i}+\sum_{i\leq n-2\leq
j}m_{ij}\right)\Gamma\left(k_{n-2,n-1}+1\right)}{\Gamma\left(k_{1,\cdots,n-2}+k_{n-2,n-1}+\left(n-2\right)-\sum_{i+j\leq
n-3}J_{j}^{i}+\sum_{i\leq n-2\leq j}m_{ij}\right)}.$ (5.18)

Now we choose to work on the Regge regime defined by

$k_{1,\cdots,n-2}\sim s,k_{1,\cdots,n-2}+k_{n-2,n-1}\sim t.$ (5.19)

In this regime, the RSSA can be approximated as

$\displaystyle A^{\left\\{p_{1},p_{2},\cdots,p_{n-3}\right\\},0,0}$
$\displaystyle\sim\prod_{\sigma=1}^{n-3}\left[\sum_{\sum_{j=1}^{n-2-\sigma}J_{j}^{\sigma}=\mathcal{P}_{\sigma}}\left(\mathcal{P}_{\sigma}!\prod_{j=1}^{n-2-\sigma}\frac{\left(k_{j+\sigma+1}^{T_{\sigma}}\right)^{J_{j}^{\sigma}}}{J_{j}^{\sigma}!}\right)\right]$
$\displaystyle\times\sum_{m_{23}}\frac{\left[-k_{24}+\left(J_{1}^{2}+J_{2}^{1}\right)\right]_{m_{23}}}{m_{23}!}\sum_{m_{24}}\frac{\left[-k_{25}+\left(J_{1}^{3}+J_{2}^{2}+J_{3}^{1}\right)\right]_{m_{24}}}{m_{24}!}\cdots\sum_{m_{2,n-2}}\frac{\left[-k_{2,n-1}+\left(J_{1}^{n-3}+\cdots+J_{n-3}^{1}\right)\right]_{m_{2,n-2}}}{m_{2,n-2}!}$
$\displaystyle\times\sum_{m_{34}}\frac{\left(-k_{35}\right)_{m_{34}}}{m_{34}!}\cdots\sum_{m_{3,n-2}}\frac{\left(-k_{3,n-1}\right)_{m_{3,n-2}}}{m_{3,n-2}!}$
$\displaystyle\vdots$
$\displaystyle\times\sum_{m_{n-3,n-2}}\frac{\left(-k_{n-3,n-1}\right)_{m_{3,n-2}}}{m_{n-3,n-2}!}$
$\displaystyle\times\frac{\Gamma\left(k_{12}+1+\sum_{i\leq 2\leq
j}m_{ij}\right)\Gamma\left(k_{23}+1-J_{1}^{1}\right)}{\Gamma\left(k_{12}+k_{23}+2-J_{1}^{1}+\sum_{i\leq
2\leq j}m_{ij}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{123}+2-\sum_{i+j\leq
2}J_{j}^{i}+\sum_{i\leq 3\leq
j}m_{ij}\right)\Gamma\left(k_{34}+1\right)}{\Gamma\left(k_{123}+k_{34}+3-\sum_{i+j\leq
2}J_{j}^{i}+\sum_{i\leq 3\leq j}m_{ij}\right)}$ $\displaystyle\vdots$
$\displaystyle\times\frac{\left(k_{1,\cdots,n-2}\right)^{-\sum_{i+j\leq
n-3}J_{j}^{i}+\sum_{i\leq n-2\leq
j}m_{ij}}\Gamma\left(k_{1,\cdots,n-2}+\left(n-3\right)\right)\Gamma\left(k_{n-2,n-1}+1\right)}{\left(k_{1,\cdots,n-2}+k_{n-2,n-1}+\left(n-2\right)\right)_{-\sum_{i+j\leq
n-3}J_{j}^{i}+\sum_{i\leq n-2\leq
j}m_{ij}}\Gamma\left(k_{1,\cdots,n-2}+k_{n-2,n-1}+\left(n-2\right)\right)}.$
(5.20)

To get the leading order in $k_{1,\cdots,n-2}\char 126\relax$ $s$, we take

$J_{j}^{i}=0,\text{ (for all }i+j\leq n-3\text{)}$ (5.21)

or

$\displaystyle J_{1}^{1}$ $\displaystyle=J_{2}^{1}=\cdots=J_{n-4}^{1}=0,$
$\displaystyle J_{1}^{2}$ $\displaystyle=\cdots=J_{n-5}^{2}=0,$ $\displaystyle
J_{1}^{r}$ $\displaystyle=\cdots=J_{n-r-3}^{r}=0$ (5.22)

which imply

$J_{n-3}^{1}=N+p_{1},J_{n-4}^{2}=p_{2},\cdots,J_{n-r-2}^{r}=p_{r}.$ (5.23)

Finally, the leading order term of the amplitude is

$\displaystyle A^{\left\\{p_{1},p_{2},\cdots,p_{r}\right\\},0,0}$
$\displaystyle\sim\prod_{\sigma=1}^{r}\left[\left(k_{n-1}^{T_{\sigma}}\right)^{\mathcal{P}_{\sigma}}\right]$
$\displaystyle\times\sum_{m_{23}}\frac{\left[-k_{24}\right]_{m_{23}}}{m_{23}!}\sum_{m_{24}}\frac{\left[-k_{25}\right]_{m_{24}}}{m_{24}!}\cdots\sum_{m_{24}}\frac{\left[-k_{2,n-1}\right]_{m_{2,n-2}}}{m_{2,n-2}!}$
$\displaystyle\times\sum_{m_{34}}\frac{\left(-k_{35}\right)_{m_{34}}}{m_{34}!}\cdots\sum_{m_{3,n-2}}\frac{\left(-k_{3,n-1}\right)_{m_{3,n-2}}}{m_{3,n-2}!}$
$\displaystyle\times\sum_{m_{n-3,n-2}}\frac{\left(-k_{n-3,n-1}\right)_{m_{3,n-2}}}{m_{n-3,n-2}!}$
$\displaystyle\times\frac{\Gamma\left(k_{12}+1+\sum_{i\leq 2\leq
j}m_{ij}\right)\Gamma\left(k_{23}+1\right)}{\Gamma\left(k_{12}+k_{23}+2+\sum_{i\leq
2\leq j}m_{ij}\right)}$
$\displaystyle\times\frac{\Gamma\left(k_{123}+2+\sum_{i\leq 3\leq
j}m_{ij}\right)\Gamma\left(k_{34}+1\right)}{\Gamma\left(k_{123}+k_{34}+3-\sum_{i+j\leq
2}J_{j}^{i}+\sum_{i\leq 3\leq j}m_{ij}\right)}$ $\displaystyle\vdots$
$\displaystyle\times\frac{\left(k_{1,\cdots,n-2}\right)^{\sum_{i\leq n-2\leq
j}m_{ij}}\Gamma\left(k_{1,\cdots,n-2}+\left(n-3\right)\right)\Gamma\left(k_{n-2,n-1}+1\right)}{\left(k_{1,\cdots,n-2}+k_{n-2,n-1}+\left(n-2\right)\right)_{\sum_{i\leq
n-2\leq
j}m_{ij}}\Gamma\left(k_{1,\cdots,n-2}+k_{n-2,n-1}+\left(n-2\right)\right)}$
$\displaystyle=\prod_{\sigma=1}^{r}\left[\left(k_{n-1}^{T_{\sigma}}\right)^{\mathcal{P}_{\sigma}}\right]\times(\text{factors
independent of }J_{q}^{r}\text{'}s\text{ }).$ (5.24)

The ratios of the amplitudes are

$\displaystyle\frac{A^{\left\\{p_{1},p_{2},\cdots,p_{r}\right\\},0,0}}{A^{\left\\{0,0,\cdots,0\right\\},0,0}}$
$\displaystyle=\left(k_{n-1}^{T_{1}}\right)^{p_{1}}\left(k_{n-1}^{T_{2}}\right)^{p_{2}}\cdots\left(k_{n-1}^{T_{r}}\right)^{p_{r}},$
$\displaystyle=\left(\Omega_{2}^{n-1}\right)^{p_{1}}\left(\Omega_{3}^{n-1}\right)^{p_{2}}\cdots\left(\Omega_{r+1}^{n-1}\right)^{p_{r}},$
$\displaystyle=\left(\omega_{1}\right)^{p_{1}}\left(\omega_{2}\right)^{p_{2}}\cdots\left(\omega_{r}\right)^{p_{r}},$
(5.25)

which is the same as Eq.(2.31) with $m=q=0$.

##  VI Conclusion

In this paper, we first give a review with detailed calculations of ratios
among HSSA at each fixed mass level to demonstrate the stringy scaling
behavior in the hard scattering limit. We then extend the calculations and
discover a similar stringy scaling behavior for a class of $n$-point RSSA. The
number of independent kinematics variables of these RSSA is found to be
reduced by dim$\mathcal{M}$, similar to those of the HSSA.

These stringy scaling behaviors are reminiscent of deep inelastic scattering
of electron and proton where the two structure functions $W_{1}(Q^{2},\nu)$
and $W_{2}(Q^{2},\nu)$ scale, and become not functions of $2$ kinematics
variables $Q^{2}$ and $\nu$ independently but only of their ratio $Q^{2}/\nu$.
Thus the number of independent kinematics variables reduces from $2$ to $1$.
Indeed, it is now well-known that the structure functions scale as bs

$MW_{1}(Q^{2},\nu)\rightarrow F_{1}(x),\text{ \ \ }\nu
W_{2}(Q^{2},\nu)\rightarrow F_{2}(x)$ (6.1)

where $x$ is the Bjorken variable and $M$ is the proton mass. Moreover, due to
the spin-$\frac{1}{2}$ assumption of quark, Callan and Gross derived the
following relation cg

$2xF_{1}(x)=F_{2}(x).$ (6.2)

Both of these scaling behaviors, the reduction of the number of kinematics
variables in Eq.(6.1) and the number of structure functions in Eq.(6.2) in the
hard scattering limit of quark-parton model in QCD seem to persist in some way
in the HSSA and some RSSA of string theory. We believe that, comparing to hard
QCD scaling, high energy stringy scaling in general has not been well studied
yet in the literature oleg . More new phenomena of stringy scaling remain to
be uncovered.

###### Acknowledgements.

This work is supported in part by the National Science and Technology Council
(NSTC) and S.T. Yau center of National Yang Ming Chiao Tung University (NYCU),
Taiwan. We thank H. Kawai and Y. Okawa for givng many valuable comments on
stringy scaling before the publication.

## References

  * (1) S. H. Lai, J. C. Lee and Yi Yang, ”The $SL(K+3,\mathbb{C})$ Symmetry of the Bosonic String Scattering Amplitudes”, Nucl. Phys. B 941 (2019) 53-71.
  * (2) S. H. Lai, J. C. Lee and Yi Yang, ”Recent developments of the Lauricella string scattering amplitudes and their exact $SL(K+3,\mathbb{C})$ Symmetry”, Symmetry 13 (2021) 454, arXiv:2012.14726 [hep-th].
  * (3) Sheng-Hong Lai, Jen-Chi Lee, and Yi Yang.  The Lauricella functions and exact string scattering amplitudes.  Journal of High Energy Physics, 2016(11):62, 2016.
  * (4) Sheng-Hong Lai, Jen-Chi Lee, and Yi Yang, ”Residues of bosonic string scattering amplitudes and the Lauricella functions”, arXiv:2109.08601 [hep-th].
  * (5) Sheng-Hong Lai, Jen-Chi Lee, and Yi Yang, ”The exact $SL(K+3,\mathbb{C})$ symmetry of string theory”, Phys. Lett. B 832 (2022) 137257. arXiv:2108.06326 [hep-th].
  * (6) David J. Gross.  High-Energy Symmetries of String Theory.  Phys. Rev. Lett., 60:1229, 1988.
  * (7) D. J. Gross.  Strings at superPlanckian energies: In search of the string symmetry.  In _In *London 1988, Proceedings, Physics and mathematics of strings* 83-95. (Philos. Trans. R. Soc. London A329 (1989) 401-413)._ , 1988.
  * (8) Chuan-Tsung Chan and Jen-Chi Lee.  Stringy symmetries and their high-energy limits.  Phys. Lett. B, 611(1):193–198, 2005.
  * (9) Chuan-Tsung Chan and Jen-Chi Lee.  Zero-norm states and high-energy symmetries of string theory.  Nucl. Phys. B, 690(1):3–20, 2004.
  * (10) Chuan-Tsung Chan, Pei-Ming Ho, and Jen-Chi Lee.  Ward identities and high energy scattering amplitudes in string theory.  Nucl. Phys. B, 708(1):99–114, 2005.
  * (11) Chuan-Tsung Chan, Pei-Ming Ho, Jen-Chi Lee, Shunsuke Teraguchi, and Yi Yang.  High-energy zero-norm states and symmetries of string theory.  Phys. Rev. Lett., 96(17):171601, 2006.
  * (12) Chuan-Tsung Chan, Pei-Ming Ho, Jen-Chi Lee, Shunsuke Teraguchi, and Yi Yang.  Solving all 4-point correlation functions for bosonic open string theory in the high-energy limit.  Nucl. Phys. B, 725(1):352–382, 2005.
  * (13) Sheng-Hong Lai, Jen-Chi Lee, and Yi Yang, ”Stringy scaling of $n$-point Hard string scattering amplitudes”, e-Print: 2207.09236 [hep-th].
  * (14) MB Green, JH Schwarz, and E Witten.  Superstring theory, v. 1.  Cambridge University, Cambridge, 1987.
  * (15) Jen-Chi Lee and Yi Yang.  Review on high energy string scattering amplitudes and symmetries of string theory.  arXiv preprint arXiv:1510.03297, 2015.
  * (16) Jen-Chi Lee and Yi Yang.  Overview of high energy string scattering amplitudes and symmetries of string theory.  Symmetry, 11(8):1045, 2019.
  * (17) Chuan-Tsung Chan, Jen-Chi Lee, and Yi Yang.  Notes on high-energy limit of bosonic closed string scattering amplitudes.  Nucl. Phys. B, 749(1):280–290, 2006.
  * (18) Jen-Chi Lee and Yi Yang.  Regge closed string scattering and its implication on fixed angle closed string scattering.  Phys. Lett. B, 687(1):84–88, 2010.
  * (19) H. Kawai, D.C. Lewellen, and S-HH Tye.  A relation between tree amplitudes of closed and open strings.  Nucl. Phys. B, 269(1):1–23, 1986.
  * (20) Jen-Chi Lee, Yoshihiro Mitsuka, and Yi Yang.  Higher spin string states scattered from d-particle in the Regge regime and factorized ratios of fixed angle scatterings.  Progress of Theoretical Physics, 126(3):397–417, 2011.
  * (21) O. Schlotterer, ”Scattering amplitudes in open superstring theory”, Fortschr. Phys.60,No.5, 373-691 (2012).
  * (22) I.G. Koh, W. Troost and A. Van Proeyen, Nucl. Phys. B292, 201 (1987).
  * (23) S. H. Lai, J. C. Lee and Yi Yang, ”Spin polarization independence of hard polarized fermion string scattering amplitudes”, Phys. Lett. B 797 (2019) 134812.
  * (24) S. H. Lai, J. C. Lee and Yi Yang, ”New bosonic hard string scattering amplitudes and extended Gross conjecture in superstring theory”, Phys. Lett. B 816 (2021) 136264.
  * (25) Chuan-Tsung Chan, Jen-Chi Lee, and Yi Yang.  High energy scattering amplitudes of superstring theory.  Nucl. Phys. B, 738(1):93–123, 2006.
  * (26) M. Jacob and G. C. Wick, Annals of Physics 281, 774-799 (2000).
  * (27) M. Peskin, ”An Introduction to Quantum Field Theory”, Addison-Wesley Publishing Company (1995).
  * (28) Sheng-Lan Ko, Jen-Chi Lee, and Yi Yang.  Patterns of high energy massive string scatterings in the Regge regime.  JHEP, 2009(06):028, 2009.
  * (29) Sheng-Lan Ko, Jen-Chi Lee, and Yi Yang.  Kummer function and high energy string scatterings.  arXiv preprint arXiv:0811.4502, 2008.
  * (30) Song He, Jen-Chi Lee, Keijiro Takahashi, and Yi Yang.  Massive superstring scatterings in the Regge regime.  Phys. Rev. D, 83(6):066016, 2011.
  * (31) Jen-Chi Lee, Catherine H Yan, and Yi Yang.  High-energy string scattering amplitudes and signless Stirling number identity.  SIGMA. Symmetry, Integrability and Geometry: Methods and Applications, 8:045, 2012.
  * (32) Song He, Jen-Chi Lee, and Yi Yang.  Exponential fall-off behavior of Regge scatterings in compactified open string theory.  Prog. of th. phys., 128(5):887–901, 2012.
  * (33) J. D. Bjorken, Phys. Rev. 179 (1969) 1547.
  * (34) C. G. Callan and D. Gross, Phys. Rev. Lett. 22 (1969) 156.
  * (35) Oleg Andreev, Scaling laws in hadronic processes and string theory”, hep-th/0209256 (2002).