# Residues of bosonic string scattering amplitudes and the Lauricella
functions

Sheng-Hong Lai<EMAIL_ADDRESS>Department of Electrophysics, National
Yang Ming Chiao Tung University, Hsinchu, Taiwan, R.O.C.  Department of
Electrophysics, National 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.  Department of
Electrophysics, National Chiao-Tung University, Hsinchu, Taiwan, R.O.C.  Yi
Yang<EMAIL_ADDRESS>Department of Electrophysics, National Yang Ming
Chiao Tung University, Hsinchu, Taiwan, R.O.C.  Department of Electrophysics,
National Chiao-Tung University, Hsinchu, Taiwan, R.O.C.

###### Abstract

We calculate explicitly residues of all $n$-point Koba-Nielsen (KN) amplitudes
by using on-shell recursion relation of string scattering amplitudes (SSA). In
addition, we show that the residues of all SSA including the KN amplitudes can
be expressed in terms of the Lauricella functions. This result demonstrates
the exact $SL(K+3,\mathbb{C})$ symmetry of the tree-level open bosonic string
theory. Moreover, we derive an iteration relation among the residues of a
given SSA. This iteration relation is related to the $SL(K+3,\mathbb{C})$
symmetry and can presumably be used to soften the well-known hard SSA.

###### Contents

  1. I Introduction
  2. II The Lauricella string scattering amplitudes
  3. III Residues of the Koba-Nielsen (KN) amplitudes
    1. III.1 The Veneziano amplitude
    2. III.2 The five-point KN amplitude
    3. III.3 The six-point KN amplitude
    4. III.4 The seven-point KN amplitude
    5. III.5 The n-point KN amplitude
      1. III.5.1 Method of subamplitudes
      2. III.5.2 Direct calculation
  4. IV Iteration relations among residues
  5. V Expressing n-point SSA in terms of the LSSA
    1. V.1 Four-point SSA with tensor legs and the Shifting principle
    2. V.2 Five-point SSA with tensor legs
    3. V.3 Six-point SSA with tensor legs
    4. V.4 Seven-point SSA with tensor legs
    5. V.5 N-point SSA with tensor legs
  6. VI Conclusion
  7. A Generating function

##  I Introduction

It is known that a string scattering amplitude (SSA) can be expressed in terms
of an infinite sum of simple pole terms with residues which are well-organized
that its hard high energy behavior is very soft and decays as an exponential
fall-off. A well-known simple example is the four tachyon Veneziano amplitude
GSW . Presumably, there is a mechanism similar to the symmetry principle in
quantum field theory (QFT) which triggers this soft behavior for hard SSA. In
contrast to QFT, string theory as a quantum theory of an extended string
consists of an infinite number of particles with arbitrary high spins and
masses in the spectrum. It is thus reasonable to believe that there exists a
huge symmetry group Moore ; Moore1 ; CKT  which relates these infinite number
of couplings and soften the SSA in the hard scattering limit.

Recently, the present authors calculated a subset of exact $4$-point SSA,
namely, amplitudes of three tachyons and one arbitrary string states, and
expressed them in terms of the $D$-type Lauricella functions LLY2 . In
addition, it was shown that these Lauricella SSA (LSSA) can be expressed in
terms of the basis functions in the infinite dimensional representation of the
$SL(K+3,C)$ group Group . For any fixed positive integer $K$, one has infinite
number of LSSA in the $SL(K+3,C)$ representations slkc . Moreover, it was
further shown that there existed $K+2$ recurrence relations among the $D$-type
Lauricella functions. These recurrence relations can be used to reproduce the
Cartan subalgebra and simple root system of the $SL(K+3,\mathbb{C})$ group
with rank $K+2$. As a result, the $SL(K+3,\mathbb{C})$ group with its
corresponding stringy Ward identities (recurrence relations) can be used to
solve solve  all the LSSA and express them in terms of one amplitude. See the
recent review LSSA .

The LSSA can be used to rederive the solvability of SSA in various scattering
limits LLY2 ,0705 ; RR1 ; RR2 ; RR3 ; RR4 ; RR5 ; RR6 ; RR7 . One important
application of this solvability was in the hard string scattering limit. In
particular, it was shown that in the hard scattering limit the LSSA can be
used to reproduce LLY2  infinite linear relations with constant coefficients
among all hard SSA and solve the ratios among them. These behaviors of high
energy string scatterings GM ; GM1 ; GrossManes  were first conjectured by
Gross Gross ; Gross1  and later corrected ChanLee ; ChanLee2 ; CHL  and proved
ChanLee ; ChanLee2 ; CHLTY2 ; CHLTY1  by the method of decoupling of zero norm
states (ZNS) Lee ; lee-Ov ; LeePRL . For details, see the recent review papers
review ; over . More importantly, since the decoupling of ZNS and thus the
infinite linear relations in the hard scattering limit persist to all string
loop orders, it was believed that the proposed $SL(K+3,\mathbb{C})$ symmetry
at string-tree level is also valid for string loop amplitudes. On the other
hand, one believes that by keeping $M$ fixed as a finite constant in the ZNS
calculation, one can obtain more information about the high energy behavior of
string theory in contrast to the tensionless string
($\alpha^{\prime}\rightarrow\infty$) approach less1 ; less2 ; less3 ; less5 ;
less6 ; less7 ; less8 ; less9 ; less10 ; less11 ; less12  in which all string
states are massless.

In this paper, we will apply the string theory extension stringbcfw , bcfw3 ;
bcfw4  of field theory BCFW on-shell recursion relations bcfw1 ; bcfw2  to
calculate explicitly residues of all $n$-point Koba-Nielsen (KN) amplitudes.
In addition, we will show that the residues of all SSA including the KN
amplitudes can be expressed in terms of the Lauricella functions. This result
demonstrates the exact $SL(K+3,\mathbb{C})$ symmetry of the tree-level open
bosonic string theory. On the other hand, since the LSSA of three tachyons and
one arbitrary string states can be rederived LLYT  from the deformed cubic
string field theory (SFT) Taejin , we conjecture that the proposed
$SL(K+3,\mathbb{C})$ symmetry can be obtained from Witten SFT.

This paper is organized as following. In section II, we give a brief review of
the calculation of the LSSA of three tachyons and one arbitrary string states.
In addition, we review the $SL(K+3,\mathbb{C})$ symmetry group and the
corresponding recurrence relations of the LSSA. We will then show that SSA of
four arbitrary string states can also be written in terms of the Lauricella
functions. (details will be given in section V) In section III, we first
explicitly calculate the residues of $5$, $6$ and $7$-point KN amplitudes. We
then extend the calculation of residues to all $n$-point KN amplitudes. In
section IV, we derive an iteration relation among the residues of a given KN
amplitude. This iteration relation is related to the $SL(K+3,\mathbb{C})$
symmetry and can presumably be used to soften the well-known hard SSA. In
section V, we will first use the shifting method to demonstrate that all
$5$-point SSA with arbitrary five tensor legs can be expressed in terms of the
LSSA. In general, we can use mathematical induction, together with the on-
shell recursion and the shifting principle, to show that all $n$-point SSA can
be expressed in terms of the LSSA. We will see that this multi-tensor
calculation is closely related to the calculation of applying multi-step
string on-shell recursion calculation to KN amplitudes. In section VI, a
conclusion and some open questions are given. Finally, in the end an appendix
is given to prove some mathematical identities used in the text.

##  II The Lauricella string scattering amplitudes

In this section we give a brief review of the LSSA of three tachyons and one
arbitrary string states in the $26D$ open bosonic string theory and its
associated $SL(K+3,C)$ group. The general states at mass level
$M_{2}^{2}=2(N_{2}-1)$,
$N_{2}=\sum_{n,m,l>0}\left(nr_{n}^{T}+mr_{m}^{P}+lr_{l}^{L}\right)$ with
polarizations on the scattering plane can be written as LSSA

$\left|r_{n}^{T},r_{m}^{P},r_{l}^{L}\right\rangle=\prod_{n>0}\left(\alpha_{-n}^{T}\right)^{r_{n}^{T}}\prod_{m>0}\left(\alpha_{-m}^{P}\right)^{r_{m}^{P}}\prod_{l>0}\left(\alpha_{-l}^{L}\right)^{r_{l}^{L}}|0,k\rangle$
(2.1)

where $e^{P}=\frac{1}{M_{2}}(E_{2},\mathrm{k}_{2},0)=\frac{k_{2}}{M_{2}}$ the
momentum polarization, $e^{L}=\frac{1}{M_{2}}(\mathrm{k}_{2},E_{2},0)$ the
longitudinal polarization on and $e^{T}=(0,0,1)$ the transverse polarization.
In the CM frame, we define the kinematics as

$\displaystyle k_{1}$
$\displaystyle=\left(\sqrt{M_{1}^{2}+|\vec{k_{1}}|^{2}},-|\vec{k_{1}}|,0\right),$
(2.2) $\displaystyle k_{2}$
$\displaystyle=\left(\sqrt{M_{2}+|\vec{k_{1}}|^{2}},+|\vec{k_{1}}|,0\right),$
(2.3) $\displaystyle k_{3}$
$\displaystyle=\left(-\sqrt{M_{3}^{2}+|\vec{k_{3}}|^{2}},-|\vec{k_{3}}|\cos\phi,-|\vec{k_{3}}|\sin\phi\right),$
(2.4) $\displaystyle k_{4}$
$\displaystyle=\left(-\sqrt{M_{4}^{2}+|\vec{k_{3}}|^{2}},+|\vec{k_{3}}|\cos\phi,+|\vec{k_{3}}|\sin\phi\right)$
(2.5)

with $M_{1}^{2}=M_{3}^{2}=M_{4}^{2}=-2$ and $\phi$ is the scattering angle.
The Mandelstam variables are defined to be $s=-\left(k_{1}+k_{2}\right)^{2}$,
$t=-\left(k_{2}+k_{3}\right)^{2}$ and $u=-\left(k_{1}+k_{3}\right)^{2}$. We
will also use the notation $k_{i}^{X}\equiv e^{X}\cdot k_{i}$ for
$X=\left(T,P,L\right).$ It is important to note that SSA of three tachyons and
one arbitrary string states with polarizations orthogonal to the scattering
plane vanish. It is interesting to note that the $\left(s,t\right)$ channel of
the LSSA can be calculated and expressed in terms of the Lauricella functions
LSSA

$\displaystyle A_{st}^{(r_{n}^{T},r_{m}^{P},r_{l}^{L})}$
$\displaystyle=\prod_{n=1}\left[-(n-1)!k_{3}^{T}\right]^{r_{n}^{T}}\cdot\prod_{m=1}\left[-(m-1)!k_{3}^{P}\right]^{r_{m}^{P}}\prod_{l=1}\left[-(l-1)!k_{3}^{L}\right]^{r_{l}^{L}}$
$\displaystyle\cdot
B\left(-\frac{t}{2}-1,-\frac{s}{2}-1\right)F_{D}^{(K)}\left(-\frac{t}{2}-1;R_{n}^{T},R_{m}^{P},R_{l}^{L};\frac{u}{2}+2-N;\tilde{Z}_{n}^{T},\tilde{Z}_{m}^{P},\tilde{Z}_{l}^{L}\right)$
(2.6)

where we have defined

$R_{k}^{X}\equiv\left\\{-r_{1}^{X}\right\\}^{1},\cdots,\left\\{-r_{k}^{X}\right\\}^{k}\text{
\ with \ }\left\\{a\right\\}^{n}=\underset{n}{\underbrace{a,a,\cdots,a}}.$
(2.7)

and

$Z_{k}^{X}\equiv\left[z_{1}^{X}\right],\cdots,\left[z_{k}^{X}\right]\text{ \ \
with \ \ }\left[z_{k}^{X}\right]=z_{k0}^{X},\cdots,z_{k\left(k-1\right)}^{X}.$
(2.8)

In Eq.(2.8), we have defined

$z_{k}^{X}=\left|\left(-\frac{k_{1}^{X}}{k_{3}^{X}}\right)^{\frac{1}{k}}\right|,\
z_{kk^{\prime}}^{X}=z_{k}^{X}e^{\frac{2\pi ik^{\prime}}{k}},\
\tilde{z}_{kk^{\prime}}^{X}\equiv 1-z_{kk^{\prime}}^{X}\text{ \ \ for \ \
}k^{\prime}=0,\cdots,k-1.$ (2.9)

In addition to the mass level $M_{2}$ and $N_{2}$, the integer $K$ in Eq.(2.6)
is defined to be

$\text{ }K=\underset{\\{\text{for all }r_{j}^{T}\neq 0\\}}{\sum
j}+\underset{\\{\text{for all }r_{j}^{P}\neq 0\\}}{\sum
j}+\underset{\\{\text{for all }r_{j}^{L}\neq 0\\}}{\sum j}.$ (2.10)

The $D$-type Lauricella function $F_{D}^{(K)}$ in Eq.(2.6) is one of the four
extensions of the Gauss hypergeometric function to $K$ variables and is
defined to be

$\displaystyle
F_{D}^{(K)}\left(\alpha;\beta_{1},...,\beta_{K};\gamma;x_{1},...,x_{K}\right)$
$\displaystyle=\sum_{n_{1},\cdots,n_{K}=0}^{\infty}\frac{\left(\alpha\right)_{n_{1}+\cdots+n_{K}}}{\left(\gamma\right)_{n_{1}+\cdots+n_{K}}}\frac{\left(\beta_{1}\right)_{n_{1}}\cdots\left(\beta_{K}\right)_{n_{K}}}{n_{1}!\cdots
n_{K}!}x_{1}^{n_{1}}\cdots x_{K}^{n_{K}}$ (2.11)

where
$(\alpha)_{n}=\alpha\cdot\left(\alpha+1\right)\cdots\left(\alpha+n-1\right)$
is the Pochhammer symbol. There was an important integral representation of
the Lauricella function Appell

$\displaystyle
F_{D}^{(K)}\left(\alpha;\beta_{1},...,\beta_{K};\gamma;x_{1},...,x_{K}\right)$
$\displaystyle=\frac{\Gamma(\gamma)}{\Gamma(\alpha)\Gamma(\gamma-\alpha)}\int_{0}^{1}dt\,t^{\alpha-1}(1-t)^{\gamma-\alpha-1}\cdot(1-x_{1}t)^{-\beta_{1}}(1-x_{2}t)^{-\beta_{2}}...(1-x_{K}t)^{-\beta_{K}},$
(2.12)

which was used to calculate Eq.(2.6). One can use the LSSA to calculate
various scattering limits of the SSA LSSA .

In the following, we discuss the hard scattering limit of the LSSA. For more
details, see the recent review LSSA . It was first observed that for each
fixed mass level $N$ with $M^{2}=2(N-1)$, the following states are of leading
order in energy at the hard scattering limit CHLTY2 ; CHLTY1

$\left|N,2m,q\right\rangle\equiv(\alpha_{-1}^{T})^{N-2m-2q}(\alpha_{-1}^{L})^{2m}(\alpha_{-2}^{L})^{q}|0,k\rangle.$
(2.13)

Note that in Eq.(2.13), only even powers $2m$ in $\alpha_{-1}^{L}$ ChanLee ;
ChanLee2  survive, and the naive energy order of the amplitudes will drop by
an even number of energy powers in general. One important application of the
LSSA presented in Eq.(2.6) in the hard string scattering limit is to reproduce
LLY2  the infinite linear relations with constant coefficients among all hard
SSA and solve the ratios among them LSSA

$\frac{T^{(N,2m,q)}}{T^{(N,0,0)}}=\left(-\frac{1}{M}\right)^{2m+q}\left(\frac{1}{2}\right)^{m+q}(2m-1)!!.$
(2.14)

The infinite linear relations with constant coefficients and the solvability
of SSA in the hard scattering limit were first conjectured by Gross GM ; GM1
and later corrected ChanLee ; ChanLee2  and proved CHLTY2 ; CHLTY1  by the
method of decoupling of zero norm states (ZNS) Lee ; lee-Ov ; LeePRL .

To reproduce the infinite linear relations in Eq.(2.14) from LSSA, we first
note that, in the hard scattering limit, the relevant kinematics reduce to

$\displaystyle k_{1}^{T}$ $\displaystyle=0\text{, \ \
}k_{3}^{T}\simeq-E\sin\phi,$ (2.15) $\displaystyle k_{1}^{L}$
$\displaystyle\simeq-\frac{2p^{2}}{M_{2}}\simeq-\frac{2E^{2}}{M_{2}},$ (2.16)
$\displaystyle k_{3}^{L}$
$\displaystyle\simeq\frac{2E^{2}}{M_{2}}\sin^{2}\frac{\phi}{2}$ (2.17)

where $E$ and $\phi$ are the CM frame energy and scattering angle,
respectively. One can calculate

$\tilde{z}_{kk^{\prime}}^{T}=1,\
\tilde{z}_{kk^{\prime}}^{L}=1-\left(-\frac{s}{t}\right)^{1/k}e^{\frac{i2\pi
k^{\prime}}{k}}\sim O\left(1\right),$ (2.18)

and the LSSA in Eq.(2.6) reduces to LLY2

$\displaystyle
A_{st}^{(r_{n}^{T},r_{l}^{L})}=B\left(-\frac{t}{2}-1,-\frac{s}{2}-1\right)$
$\displaystyle\cdot\prod_{n=1}\left[(n-1)!E\sin\phi\right]^{r_{n}^{T}}\prod_{l=1}\left[-(l-1)!\frac{2E^{2}}{M_{2}}\sin^{2}\frac{\phi}{2}\right]^{r_{l}^{L}}$
$\displaystyle\cdot
F_{D}^{(K)}\left(-\frac{t}{2}-1;R_{n}^{T},R_{l}^{L};\frac{u}{2}+2-N;\left(1\right)_{n},\tilde{Z}_{l}^{L}\right).$
(2.19)

Since in the hard scattering limit, there was a difference between the naive
energy order and the real energy order corresponding to the
$\left(\alpha_{-1}^{L}\right)^{r_{1}^{L}}$ operator in Eq.(2.1), it is
important to pay attention to the corresponding summation and write

$\displaystyle
A_{st}^{(r_{n}^{T},r_{l}^{L})}=B\left(-\frac{t}{2}-1,-\frac{s}{2}-1\right)$
$\displaystyle\cdot\prod_{n=1}\left[(n-1)!E\sin\phi\right]^{r_{n}^{T}}\prod_{l=1}\left[-(l-1)!\frac{2E^{2}}{M_{2}}\sin^{2}\frac{\phi}{2}\right]^{r_{l}^{L}}$
$\displaystyle\cdot\sum_{k_{r}}\frac{\left(-\frac{t}{2}-1\right)_{k_{r}}}{\left(\frac{u}{2}+2-N\right)_{k_{r}}}\frac{\left(-r_{1}^{L}\right)_{k_{r}}}{k_{r}!}\left(1+\frac{s}{t}\right)^{k_{r}}\cdot\left(\cdots\right)$
(2.20)

where $\left(a\right)_{n+m}=\left(a\right)_{n}\left(a+n\right)_{m}$ and
$\left(\cdots\right)$ are terms which are not relevant to the following
discussion. To proceed, the following formula was proposed LLY2

$\displaystyle\sum_{k_{r}=0}^{r_{1}^{L}}\frac{\left(-\frac{t}{2}-1\right)_{k_{r}}}{\left(\frac{u}{2}+2-N\right)_{k_{r}}}\frac{\left(-r_{1}^{L}\right)_{k_{r}}}{k_{r}!}\left(1+\frac{s}{t}\right)^{k_{r}}$
$\displaystyle=$ $\displaystyle
0\cdot\left(\frac{tu}{s}\right)^{0}\\!+0\cdot\left(\frac{tu}{s}\right)^{-1}\\!+\dots+0\cdot\left(\frac{tu}{s}\right)^{-\left[\frac{r_{1}^{L}+1}{2}\right]-1}$
$\displaystyle+C_{r_{1}^{L}}\left(\frac{tu}{s}\right)^{-\left[\frac{r_{1}^{L}+1}{2}\right]}+\mathit{O}\left\\{\left(\frac{tu}{s}\right)^{-\left[\frac{r_{1}^{L}+1}{2}\right]+1}\right\\}.$
(2.21)

where $[$ $]$ stands for the Gauss symbol, $C_{r_{1}^{L}}$ is independent of
energy $E$ and depends on $r_{1}^{L}$ and possibly the scattering angle
$\phi$. When $r_{1}^{L}=2m$ is an even number, one further proposes that
$C_{r_{1}^{L}}=\frac{\left(2m\right)!}{m!}$ and is $\phi$ independent. It was
verified that Eq.(2.21) is valid for $r_{1}^{L}=0,1,2,\cdots,10$ LLY2 .

It is interesting to Note that Eq.(2.21) reduces to the Stirling number
identity by taking the Regge limit ($s\rightarrow\infty$ with $t$ fixed) and
setting $r_{1}^{L}=2m$,

$\displaystyle\sum_{k_{r}=0}^{2m}\frac{\left(-\frac{t}{2}-1\right)_{k_{r}}}{\left(-\frac{s}{2}\right)_{k_{r}}}\frac{\left(-2m\right)_{k_{r}}}{k_{r}!}\left(\frac{s}{t}\right)^{k_{r}}\simeq\sum_{k_{r}=0}^{2m}\left(-2m\right)_{k_{r}}\left(-\frac{t}{2}-1\right)_{k_{r}}\frac{\left(-2/t\right)^{k_{r}}}{k_{r}!}$
$\displaystyle=0\cdot\left(-t\right)^{0}\\!+0\cdot\left(-t\right)^{-1}\\!+\dots+0\cdot\left(-t\right)^{-m+1}+\frac{(2m)!}{m!}\left(-t\right)^{-m}+\mathit{O}\left\\{\left(\frac{1}{t}\right)^{m+1}\right\\},$
(2.22)

which was proposed in KLY  and proved in LYAM . The zero terms in Eq.(2.21)
correspond to the naive leading energy orders in the hard SSA calculation. In
the hard scattering limit, the true leading order SSA can then be identified

$\displaystyle A_{st}^{(r_{n}^{T},r_{l}^{L})}\simeq
B\left(-\frac{t}{2}-1,-\frac{s}{2}-1\right)$
$\displaystyle\cdot\prod_{n=1}\left[(n-1)!E\sin\phi\right]^{r_{n}^{T}}\prod_{l=1}\left[-(l-1)!\frac{2E^{2}}{M_{2}}\sin^{2}\frac{\phi}{2}\right]^{r_{l}^{L}}$
$\displaystyle\cdot
C_{r_{1}^{L}}\left(E\sin\phi\right)^{-2\left[\frac{r_{1}^{L}+1}{2}\right]}\cdot\left(\cdots\right)$
$\displaystyle\sim E^{N-\sum_{n\geq
2}nr_{n}^{T}-\left(2\left[\frac{r_{1}^{L}+1}{2}\right]-r_{1}^{L}\right)-\sum_{l\geq
3}lr_{l}^{L}},$ (2.23)

which means that SSA reaches its highest energy when $r_{n\geq 2}^{T}=r_{l\geq
3}^{L}=0$ and $r_{1}^{L}=2m$, an even number. This result is consistent with
the previous result presented in Eq.(2.13).

Finally, the leading order SSA in the hard scattering limit, i.e.,
$r_{1}^{T}=N-2m-2q$, $r_{1}^{L}=2m$ and $r_{2}^{L}=q$, can be calculated to be
LLY2

$\displaystyle A_{st}^{(N-2m-2q,2m,q)}$ $\displaystyle\simeq
B\left(-\frac{t}{2}-1,-\frac{s}{2}-1\right)\left(E\sin\phi\right)^{N}\frac{\left(2m\right)!}{m!}\left(-\frac{1}{2M_{2}}\right)^{2m+q}$
$\displaystyle=(2m-1)!!\left(-\frac{1}{M_{2}}\right)^{2m+q}\left(\frac{1}{2}\right)^{m+q}A_{st}^{(N,0,0)}$
(2.24)

which reproduces the ratios in Eq.(2.14).

To obtain the recurrence relations of the Lauricella functions, we consider
the type-$1$ Appell functions. For the case of $K=2$, the $D$-type Lauricella
functions
$F_{D}^{(K)}\left(\alpha;\beta_{1},...,\beta_{K};\gamma;x_{1},...,x_{K}\right)$
reduce to the type-$1$ Appell functions
$F_{1}\left(\alpha;\beta_{1},\beta_{2};\gamma,x,y\right)$, and one has $4$
known recurrence relations. It was shown that one can generalize the $4=2+2$
fundamental recurrence relations of the Appell functions $F_{1}$ and prove the
following $K+2$ recurrence relations for the $D$-type Lauricella functions
$(m=1,2,...,K)$ Group

$\displaystyle\left(\alpha-\underset{i}{\sum}\beta_{i}\right)F_{D}^{(K)}-\alpha
F_{D}^{(K)}\left(\alpha+1\right)+\beta_{1}F_{D}^{(K)}\left(\beta_{1}+1\right)+...+\beta_{K}F_{D}^{(K)}\left(\beta_{K}+1\right)$
$\displaystyle=0,$ (2.25) $\displaystyle\gamma
F_{D}^{(K)}-\left(\gamma-\alpha\right)F_{D}^{(K)}\left(\gamma+1\right)-\alpha
F_{D}^{(K)}\left(\alpha+1;\gamma+1\right)$ $\displaystyle=0,$ (2.26)
$\displaystyle\gamma
F_{D}^{(K)}+\gamma(x_{m}-1)F_{D}^{(K)}\left(\beta_{m}+1\right)+(\alpha-\gamma)x_{m}F_{D}^{(K)}\left(\beta_{m}+1,;\gamma+1\right)$
$\displaystyle=0$ (2.27)

where for simplicity we have omitted those arguments of $F_{D}^{(K)}$ which
remain the same in the relations. Moreover, these recurrence relations can be
used to reproduce the Cartan subalgebra and simple root system of the
$SL(K+3,\mathbb{C})$ group with rank $K+2$Group . As a result, the
$SL(K+3,\mathbb{C})$ group with its corresponding stringy Ward identities
(recurrence relations) can be used to solve solve  all the LSSA and express
them in terms of one amplitude.

In particular, it was shown that one can use the above recurrence relations to
decrease the value of $K$ step by step and reduce all the Lauricella functions
$F_{D}^{(K)}$ in the LSSA to the Gauss hypergeometry functions $F_{D}^{(1)}=$
${}_{2}F_{1}(\alpha,\beta,\gamma,x)$. One can further reduce the Gauss
hypergeometry functions by deriving a multiplication theorem for them, and
then solve all the LSSA in terms of one single $4$-tachyon amplitude. One of
the reason of this solvability is that all $\beta_{J}$ in the Lauricella
functions of the LSSA take very special values, namely, nonpositive integers.
See the recent review LSSA .

On the other hand, to obtain the $SL(K+3,\mathbb{C})$ symmetry group of the
LSSA, we introduce the variables $a,b_{1},\cdots,b_{K},c$ and define the basis
functions slkc

$\displaystyle f_{ac}^{b_{1}\cdots
b_{K}}\left(\alpha;\beta_{1},\cdots,\beta_{K};\gamma;x_{1},\cdots,x_{K}\right)$
$\displaystyle=B\left(\gamma-\alpha,\alpha\right)F_{D}^{\left(K\right)}\left(\alpha;\beta_{1},\cdots,\beta_{K};\gamma;x_{1},\cdots,x_{K}\right)a^{\alpha}b_{1}^{\beta_{1}}\cdots
b_{K}^{\beta_{K}}c^{\gamma},$ (2.28)

so that the LSSA in Eq.(2.6) can be written as Group

$A_{st}^{(r_{n}^{T},r_{m}^{P},r_{l}^{L})}=f_{11}^{-(n-1)!k_{3}^{T},-(m-1)!k_{3}^{P},-(l-1)!k_{3}^{L}}\left(-\frac{t}{2}-1;R_{n}^{T},R_{m}^{P},R_{l}^{L};\frac{u}{2}+2-N;\tilde{Z}_{n}^{T},\tilde{Z}_{m}^{P},\tilde{Z}_{l}^{L}\right).$
(2.29)

We then define and introduce the $(K+3)^{2}-1$ generators of $SL(K+3,C)$ group
slkc ; Group . These are $1$ $E^{\alpha}$, $K$ $E^{\beta_{k}}$, $1$
$E^{\gamma}$,$1$ $E^{\alpha\gamma}$,$K$ $E^{\beta_{k}\gamma}$ and $K$
$E^{\alpha\beta_{k}\gamma}$ which sum up to $3K+3$ raising generators. There
are also $3K+3$ lowering operators. In addition, there are $K\left(K-1\right)$
$E_{\beta_{p}}^{\beta_{k}}$ and $K+2$ $\ J$ ,
$\left\\{J_{\alpha},J_{\beta_{k}},J_{\gamma}\right\\}$, the Cartan subalgebra.
In sum, the total number of generators are $2(3K+3)+K(K-1)+K+2=(K+3)^{2}-1$.
The explicit forms of these generators are

$\displaystyle E^{\alpha}$
$\displaystyle=a\left(\underset{j}{{\displaystyle\sum}}x_{j}\partial_{j}+a\partial_{a}\right),$
$\displaystyle E^{\beta_{k}}$
$\displaystyle=b_{k}\left(x_{k}\partial_{k}+b_{k}\partial_{b_{k}}\right),$
$\displaystyle E^{\gamma}$
$\displaystyle=c\left(\underset{j}{{\displaystyle\sum}}\left(1-x_{j}\right)\partial_{x_{j}}+c\partial_{c}-a\partial_{a}-\underset{j}{{\displaystyle\sum}}b_{j}\partial_{b_{j}}\right),$
$\displaystyle E^{\alpha\gamma}$
$\displaystyle=ac\left(\underset{j}{{\displaystyle\sum}}\left(1-x_{j}\right)\partial_{x_{j}}-a\partial_{a}\right),$
$\displaystyle E^{\beta_{k}\gamma}$
$\displaystyle=b_{k}c\left[\left(x_{k}-1\right)\partial_{x_{k}}+b_{k}\partial_{b_{k}}\right],$
$\displaystyle E^{\alpha\beta_{k}\gamma}$
$\displaystyle=ab_{k}c\partial_{x_{k}},$ $\displaystyle E_{\alpha}$
$\displaystyle=\frac{1}{a}\left[\underset{j}{{\displaystyle\sum}}x_{j}\left(1-x_{j}\right)\partial_{x_{j}}+c\partial_{c}-a\partial_{a}-\underset{j}{{\displaystyle\sum}}x_{j}b_{j}\partial_{b_{j}}\right],$
$\displaystyle E_{\beta_{k}}$
$\displaystyle=\frac{1}{b_{k}}\left[x_{k}\left(1-x_{k}\right)\partial_{x_{k}}+x_{k}\underset{j\neq
k}{{\displaystyle\sum}}\left(1-x_{j}\right)x_{j}\partial_{x_{j}}+c\partial_{c}-x_{k}a\partial_{a}-\underset{j}{{\displaystyle\sum}}b_{j}\partial_{u_{j}}\right],$
$\displaystyle E_{\gamma}$
$\displaystyle=-\frac{1}{c}\left(\underset{j}{{\displaystyle\sum}}x_{j}\partial_{x_{j}}+c\partial_{c}-1\right),$
$\displaystyle E_{\alpha\gamma}$
$\displaystyle=\frac{1}{ac}\left[\underset{j}{{\displaystyle\sum}}x_{j}\left(1-x_{j}\right)\partial_{x_{j}}-\underset{j}{{\displaystyle\sum}}x_{j}b_{j}\partial_{b_{j}}+c\partial_{c}-1\right],$
$\displaystyle E_{\beta_{k}\gamma}$
$\displaystyle=\frac{1}{b_{k}c}\left[x_{k}\left(x_{k}-1\right)\partial_{x_{k}}+\underset{j\neq
k}{{\displaystyle\sum}}\left(x_{j}-1\right)x_{j}\partial_{x_{j}}+x_{k}a\partial_{a}-c\partial_{c}+1\right],$
$\displaystyle E_{\alpha\beta_{k}\gamma}$
$\displaystyle=\frac{1}{ab_{k}c}\left[\underset{j}{{\displaystyle\sum}}x_{j}\left(x_{j}-1\right)\partial_{x_{j}}-c\partial_{c}+x_{k}a\partial_{a}+\underset{j}{{\displaystyle\sum}}x_{j}b_{j}\partial_{b_{j}}-x_{k}+1\right],$
$\displaystyle E_{\beta_{p}}^{\beta_{k}}$
$\displaystyle=\frac{b_{k}}{b_{p}}\left[\left(x_{k}-x_{p}\right)\partial_{z_{k}}+b_{k}\partial_{b_{k}}\right],(k\neq
p),$ $\displaystyle J_{\alpha}$ $\displaystyle=a\partial_{a},$ $\displaystyle
J_{\beta_{k}}$ $\displaystyle=b_{k}\partial_{b_{k}},$ $\displaystyle
J_{\gamma}$ $\displaystyle=c\partial_{c}.$ (2.30)

It is straightforward to calculate the operation of these generators on the
basis functions ($k=1,2,...K$) $f_{ac}^{b_{1}\cdots
b_{K}}\left(\alpha;\beta_{1},\cdots,\beta_{K};\gamma;x_{1},\cdots,x_{K}\right)$

$\displaystyle E^{\alpha}f_{ac}^{b_{1}\cdots b_{K}}\left(\alpha\right)$
$\displaystyle=\left(\gamma-\alpha-1\right)f_{ac}^{b_{1}\cdots
b_{K}}\left(\alpha+1\right),$ $\displaystyle E^{\beta_{k}}f_{ac}^{b_{1}\cdots
b_{K}}\left(\beta_{k}\right)$ $\displaystyle=\beta_{k}f_{ac}^{b_{1}\cdots
b_{K}}\left(\beta_{k}+1\right),$ $\displaystyle E^{\gamma}f_{ac}^{b_{1}\cdots
b_{K}}\left(\gamma\right)$
$\displaystyle=\left(\gamma-\underset{j}{{\displaystyle\sum}}\beta_{j}\right)f_{ac}^{b_{1}\cdots
b_{K}}\left(\gamma+1\right),$ $\displaystyle
E^{\alpha\gamma}f_{ac}^{b_{1}\cdots b_{K}}\left(\alpha;\gamma\right)$
$\displaystyle=\left(\underset{j}{{\displaystyle\sum}}\beta_{j}-\gamma\right)f_{ac}^{b_{1}\cdots
b_{K}}\left(\alpha+1;\gamma+1\right),$ $\displaystyle
E^{\beta_{k}\gamma}f_{ac}^{b_{1}\cdots b_{K}}\left(\beta_{k};\gamma\right)$
$\displaystyle=\beta_{k}f_{ac}^{b_{1}\cdots
b_{K}}\left(\beta_{k}+1;\gamma+1\right),$ $\displaystyle
E^{\alpha\beta_{k}\gamma}f_{ac}^{b_{1}\cdots
b_{K}}\left(\alpha;\beta_{k};\gamma\right)$
$\displaystyle=\beta_{k}f_{ac}^{b_{1}\cdots
b_{K}}\left(\alpha+1;\beta_{k}+1;\gamma+1\right),$ $\displaystyle
E_{\alpha}f_{ac}^{b_{1}\cdots b_{K}}\left(\alpha\right)$
$\displaystyle=\left(\alpha-1\right)f_{ac}^{b_{1}\cdots
b_{K}}\left(\alpha-1\right),$ $\displaystyle E_{\beta_{k}}f_{ac}^{b_{1}\cdots
b_{K}}\left(\beta_{k}\right)$
$\displaystyle=\left(\gamma-\underset{j}{{\displaystyle\sum}}\beta_{j}\right)f_{ac}^{b_{1}\cdots
b_{K}}\left(\beta_{k}-1\right),$ $\displaystyle E_{\gamma}f_{ac}^{b_{1}\cdots
b_{K}}\left(\gamma\right)$
$\displaystyle=\left(\alpha-\gamma+1\right)f_{ac}^{b_{1}\cdots
b_{K}}\left(\gamma-1\right),$ $\displaystyle
E_{\alpha\gamma}f_{ac}^{b_{1}\cdots b_{K}}\left(\alpha;\gamma\right)$
$\displaystyle=\left(\alpha-1\right)f_{ac}^{b_{1}\cdots
b_{K}}\left(\alpha-1;\gamma-1\right),$ $\displaystyle
E_{\beta_{k}\gamma}f_{ac}^{b_{1}\cdots b_{K}}\left(\beta_{k};\gamma\right)$
$\displaystyle=\left(\alpha-\gamma+1\right)f_{ac}^{b_{1}\cdots
b_{K}}\left(\beta_{k}-1;\gamma-1\right),$ $\displaystyle
E_{\alpha\beta_{k}\gamma}f_{ac}^{b_{1}\cdots
b_{K}}\left(\alpha;\beta_{k};\gamma\right)$
$\displaystyle=\left(1-\alpha\right)f_{ac}^{b_{1}\cdots
b_{K}}\left(\alpha-1;\beta_{k}-1;\gamma-1\right),$ $\displaystyle
E_{\beta_{p}}^{\beta_{k}}f_{ac}^{b_{1}\cdots
b_{K}}\left(\beta_{k};\beta_{p}\right)$
$\displaystyle=\beta_{k}f_{ac}^{b_{1}\cdots
b_{K}}\left(\beta_{k}+1;\beta_{p}-1\right),$ $\displaystyle
J_{\alpha}f_{ac}^{b_{1}\cdots b_{K}}\left(\alpha;\beta_{k};\gamma\right)$
$\displaystyle=\alpha f_{ac}^{b_{1}\cdots
b_{K}}\left(\alpha;\beta_{k};\gamma\right),$ $\displaystyle
J_{\beta_{k}}f_{ac}^{b_{1}\cdots b_{K}}\left(\alpha;\beta_{k};\gamma\right)$
$\displaystyle=\beta_{k}f_{ac}^{b_{1}\cdots
b_{K}}\left(\alpha;\beta_{k};\gamma\right),$ $\displaystyle
J_{\gamma}f_{ac}^{b_{1}\cdots b_{K}}\left(\alpha;\beta_{k};\gamma\right)$
$\displaystyle=\gamma f_{ac}^{b_{1}\cdots
b_{K}}\left(\alpha;\beta_{k};\gamma\right)$ (2.31)

and show explicitly the $SL(K+3,\mathbb{C})$ symmetry of the LSSA LSSA . In
Eq.(2.31), we have omitted those arguments in $f_{ac}^{b_{1}\cdots b_{K}}$
that remain the same after the operation. Indeed, with the following
identifications slkc

$\displaystyle E^{\alpha}$
$\displaystyle=\mathcal{E}_{12},E_{\alpha}=\mathcal{E}_{21},E^{\beta_{k}}=\mathcal{E}_{k+3,3},E_{\beta}=\mathcal{E}_{3,k+3},$
$\displaystyle E^{\gamma}$
$\displaystyle=\mathcal{E}_{31},E_{\gamma}=\mathcal{E}_{13},E^{\alpha\gamma}=\mathcal{E}_{32},E_{\alpha\gamma}=\mathcal{E}_{23},$
$\displaystyle E^{\beta_{k}\gamma}$
$\displaystyle=-\mathcal{E}_{k+3,1},E_{\beta_{k}\gamma}=-\mathcal{E}_{1,k+3},E_{\alpha\beta_{k}\gamma}=-\mathcal{E}_{k+3,2},$
$\displaystyle E_{\alpha\beta_{k}\gamma}$
$\displaystyle=-\mathcal{E}_{2,k+3},J_{\alpha}^{\prime}=\frac{1}{2}\left(\mathcal{E}_{11}-\mathcal{E}_{22}\right),J_{\beta_{k}}^{\prime}=\frac{1}{2}\left(\mathcal{E}_{k+3,k+3}-\mathcal{E}_{33}\right),J_{\gamma}^{\prime}=\frac{1}{2}\left(\mathcal{E}_{33}-\mathcal{E}_{11}\right),$
(2.32)

one can calculate the Lie algebra commutation relations of the
$SL(K+3,\mathbb{C})$

$\left[\mathcal{E}_{ij},\mathcal{E}_{kl}\right]=\delta_{jk}\mathcal{E}_{il}-\delta_{li}\mathcal{E}_{kj};\text{
\ }1\leqslant i,j\leqslant K+3.$ (2.33)

It is important to note that instead of finite dimensional representation,
here we encounter infinite dimensional representation of the noncompact Lie
group $SL(K+3,C)$. This means that for any fixed positive integer $K$, one has
infinite number of LSSA in the $SL(K+3,C)$ representations.

Finally, it is straightforward to generalize the LSSA of three tachyons and
one arbitrary string states and show that all SSA of four arbitrary string
states can still be expressed in terms of the $D$-type Lauricella functions
$F_{D}^{(K)}.$ See section V for more details. Indeed, for the general multi-
tensor $4$-point functions, there are new terms with finite number of
contractions among $\partial^{n}X...$ and $\partial^{m}X...$, and one obtains
more $D$-type Lauricella functions with different values of $K$. In general,
the LSSA of four arbitrary string states can be expressed as sum over finite
terms of the $D$-type Lauricella functions and can be written as

$LSSA=\underset{J}{\sum}c_{J}(k_{i})\int_{0}^{1}dt\,t^{\alpha-1}(1-t)^{\gamma-\alpha-1}\cdot
P_{J}(t)$ (2.34)

where $P_{J}(t)$ is a polynomial with coefficients depending on $k_{i}$ and
polarizations $\zeta$, and $c_{J}(k_{i})$ is a $k_{i}$ and $\zeta$ dependent
coefficient. One simple example KLT  is the two vectors, two tachyons bosonic
open SSA

$\displaystyle A\left(\zeta_{1},k_{1};k_{2};\zeta_{3},k_{3};k_{4}\right)$
$\displaystyle=\left(\zeta_{1}\cdot k_{2}\right)\left(\zeta_{3}\cdot
k_{4}\right)\frac{\Gamma\left(\frac{-s}{2}-1\right)\Gamma\left(\frac{-t}{2}-1\right)}{\Gamma\left(-\frac{s}{2}-\frac{t}{2}-2\right)}$
$\displaystyle\times
F_{D}^{\left(2\right)}\left(\frac{-s}{2}-1;-1,-1;-\frac{s}{2}-\frac{t}{2}-2;-\frac{\zeta_{1}\cdot
k_{3}}{\zeta_{1}\cdot k_{2}},-\frac{\zeta_{3}\cdot k_{1}}{\zeta_{3}\cdot
k_{4}}\right)$
$\displaystyle+\left(\zeta_{1}\cdot\zeta_{3}\right)\frac{\Gamma\left(\frac{-s}{2}\right)\Gamma\left(\frac{-t}{2}\right)}{\Gamma\left(-\frac{s}{2}-\frac{t}{2}\right)}F_{D}^{\left(0\right)}\left(\frac{-s}{2};;-\frac{s}{2}-\frac{t}{2};\right),$
(2.35)

which contains two Lauricella functions with $K=0,2$. Note that in contrast to
Eq.(2.6), all physical polarizations $\zeta_{i}^{\mu}$ except ZNS contribute
for the case of multi-tensor SSA as in Eq.(2.35). More examples will be given
in section V.

##  III Residues of the Koba-Nielsen (KN) amplitudes

In this section, we will use the string theory extension bcfw3 ; bcfw4  of
field theory BCFW on-shell recursion relations bcfw1 ; bcfw2  to calculate in
details the residues of all $n$-point Koba-Nielsen (KN) amplitudes. In section
V, we will show that the residues of all $n$-point KN amplitudes calculated in
this section can be expressed in terms of the Lauricella functions. Moreover,
we will use the shifting principle to show that all SSA including the KN
amplitudes can be expressed in terms of the Lauricella functions.

###  III.1 The Veneziano amplitude

We begin with the discussion on four tachyon amplitude which can be written as

$A_{4}=\int_{0}^{1}dzz^{k_{1}\cdot k_{2}}\left(1-z\right)^{k_{2}\cdot
k_{3}}=\int_{0}^{1}dzz^{k_{1}\cdot k_{2}}e^{k_{2}\cdot
k_{3}\ln\left(1-z\right)}=\int_{0}^{1}dzz^{k_{1}\cdot k_{2}}e^{-k_{2}\cdot
k_{3}\left(z+\frac{z^{2}}{2}+\frac{z^{3}}{3}+\cdots\right)},$ (3.1)

or

$\displaystyle A_{4}=\int_{0}^{1}dzz^{k_{1}\cdot
k_{2}}\exp\left[\begin{array}[c]{c}1+T\left(z+\frac{z^{2}}{2}+\frac{z^{3}}{3}+\cdots\right)+\frac{T^{2}}{2}\left(z+\frac{z^{2}}{2}+\frac{z^{3}}{3}+\cdots\right)^{2}\\\
+\frac{T^{3}}{3!}\left(z+\frac{z^{2}}{2}+\frac{z^{3}}{3}+\cdots\right)^{3}+\cdots\end{array}\right]$
(3.4) $\displaystyle=\underset{1}{\underbrace{\int_{0}^{1}dzz^{k_{1}\cdot
k_{2}}}}+\underset{\alpha_{-1}}{\underbrace{T}}\int_{0}^{1}dzz^{k_{1}\cdot
k_{2}+1}+\left(\underset{\alpha_{-2}}{\underbrace{\frac{T}{2}}}+\underset{\alpha_{-1}^{2}}{\underbrace{\frac{T^{2}}{2!}\frac{1}{1^{2}}}}\right)\int_{0}^{1}dzz^{k_{1}\cdot
k_{2}+2}$
$\displaystyle+\left(\underset{\alpha_{-3}}{\underbrace{\frac{T}{3}}}+\underset{\alpha_{-1}\alpha_{-2}}{\underbrace{\frac{T^{2}}{1!1!}\frac{1}{1^{1}}\frac{1}{2^{1}}}}+\underset{\alpha_{-1}^{3}}{\underbrace{\frac{T^{3}}{3!}}}\right)\int_{0}^{1}dzz^{k_{1}\cdot
k_{2}+3}$
$\displaystyle+\left(\underset{\alpha_{-4}}{\underbrace{\frac{T}{4}}}+\underset{\alpha_{-3}\alpha_{-1}}{\underbrace{\frac{T^{2}}{1!1!}\frac{1}{1^{1}}\frac{1}{3^{1}}}}+\underset{\alpha_{-2}^{2}}{\underbrace{\frac{T^{2}}{2!}\frac{1}{2^{2}}}}+\underset{\alpha_{-1}^{2}\alpha_{-2}}{\underbrace{\frac{T^{3}}{2!1!}\frac{1}{1^{2}2^{1}}}}+\underset{\alpha_{-1}^{4}}{\underbrace{\frac{T^{4}}{4!}\frac{1}{1^{4}}}}\right)\int_{0}^{1}dzz^{k_{1}\cdot
k_{2}+4}+\cdots$ (3.5)

where $T=-k_{2}\cdot k_{3}$. With the definition of the Mandelstam variables
$s=-\left(k_{1}+k_{2}\right)^{2}$, $t=-\left(k_{2}+k_{3}\right)^{2}$, we can
express the amplitude in terms of a series of simple pole terms with residues

$\displaystyle
A_{4}=\underset{1}{\underbrace{\frac{2}{-s-2}}}+\underset{\alpha_{-1}}{\underbrace{T\frac{2}{-s+0}}}+(\underset{\alpha_{-2}}{\underbrace{\frac{T}{2}}}+\underset{\alpha_{-1}^{2}}{\underbrace{\frac{T^{2}}{2!}\frac{1}{1^{2}})}}\frac{2}{-s+2}$
$\displaystyle+\underset{\alpha_{-3}}{\underbrace{(\frac{T}{3}}}+\underset{\alpha_{-1}\alpha_{-2}}{\underbrace{\frac{T^{2}}{1!1!}\frac{1}{1^{1}}\frac{1}{2^{1}}}}+\underset{\alpha_{-1}^{3}}{\underbrace{\frac{T^{3}}{3!})}}\frac{2}{-s+4}$
$\displaystyle\underset{\alpha_{-4}}{\underbrace{(\frac{T}{4}}}+\underset{\alpha_{-3}\alpha_{-1}}{\underbrace{\frac{T^{2}}{1!1!}\frac{1}{1^{1}}\frac{1}{3^{1}}}}+\underset{\alpha_{-2}^{2}}{\underbrace{\frac{T^{2}}{2!}\frac{1}{2^{2}}}}+\underset{\alpha_{-1}^{2}\alpha_{-2}}{\underbrace{\frac{T^{3}}{2!1!}\frac{1}{1^{2}2^{1}}}}+\underset{\alpha_{-1}^{4}}{\underbrace{\frac{T^{4}}{4!}\frac{1}{1^{4}})}}\frac{2}{-s+6}+\cdots$
(3.6)
$\displaystyle=\overset{\infty}{\underset{n=0}{\sum}}-\frac{(\alpha(t)+1)(\alpha(t)+2)\cdots(\alpha(t)+n)}{n!}\frac{1}{\alpha(s)-n}$
(3.7)

where $\alpha(t)=\alpha^{\prime}t+\alpha(0)$ with $\alpha^{\prime}=1/2$ and
$\alpha(0)=1$ GSW . In Eq.(3.6) or Eq.(3.5), we can see various residues at
each pole of the Veneziano amplitude. We will find generalization of these
residues for higher point KN amplitudes in section IV.

###  III.2 The five-point KN amplitude

For the $5$-point KN amplitude, a simple generalization of Eq.(3.6) is to use
the string theory extension stringbcfw  of field theory BCFW on-shell
recursion relations bcfw1 ; bcfw2  to calculate the infinite number of
residues. Moreover, we will show that all residues can be expressed in terms
of the Lauricella functions. The $5$-point KN amplitude can be written as
stringbcfw

$\displaystyle A_{5}^{KN}$
$\displaystyle=\int_{0}^{1}dz_{3}\int_{0}^{z_{3}}dz_{2}z_{2}^{k_{2}\cdot
k_{1}}z_{3}^{k_{3}\cdot k_{1}}\left(1-z_{3}\right)^{k_{4}\cdot
k_{3}}\left(1-z_{2}\right)^{k_{4}\cdot
k_{2}}\left(z_{3}-z_{2}\right)^{k_{3}\cdot k_{2}}$ (3.8)
$\displaystyle=\underset{a,b,c=0}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{3}\cdot
k_{4}\right)_{c}}{c!}\frac{2}{-s_{12}+2\left(a+b-1\right)}\frac{2}{-s_{123}+2\left(b+c-1\right)}$
(3.9)

where
$s_{12}=-\left(k_{1}+k_{2}\right)^{2},s_{123}=-\left(k_{1}+k_{2}+k_{3}\right)^{2}$.
We now consider the BCFW deformation with pair $(k_{1},k_{5})=(1,5)$, and set

$\hat{k}_{1}\left(z\right)=k_{1}+zq,\hat{k}_{5}\left(z\right)=k_{5}-zq,q^{2}=k_{1}\cdot
q=k_{5}\cdot q=0.$ (3.10)

For $s_{12}$ and $s_{123}$, the poles are located at

$\displaystyle z_{2,M}$
$\displaystyle=\frac{\left(k_{1}+k_{2}\right)^{2}+2\left(M-1\right)}{-2q\cdot\left(k_{1}+k_{2}\right)},M=a+b=0,1,2,...;$
(3.11) $\displaystyle z_{3,M}$
$\displaystyle=\frac{\left(k_{1}+k_{2}+k_{3}\right)^{2}+2\left(M-1\right)}{-2q\cdot\left(k_{1}+k_{2}+k_{3}\right)},M=b+c=0,1,2,...$
(3.12)

respectively. By using string on-shell recursion relation, we get stringbcfw

$A_{5}^{KN}=\underset{M}{\sum}\frac{2}{-s_{12}+2\left(M-1\right)}R_{3,4}^{M}+\underset{{}_{M}}{\sum}\frac{2}{-s_{123}+2\left(M-1\right)}R_{4,3}^{M}$
(3.13)

where the residues $R_{3,4}^{M}$ and $R_{4,3}^{M}$ can be calculated to be

$\displaystyle R_{3,4}^{M}$
$\displaystyle=\underset{a,b,c=0,M=a+b}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{3}\cdot
k_{4}\right)_{c}}{c!}\frac{2}{-\hat{s}_{123}\left(z_{N}\right)+2\left(b+c-1\right)},$
(3.14) $\displaystyle R_{4,3}^{M}$
$\displaystyle=\underset{a,b,c=0,M=b+c}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{3}\cdot
k_{4}\right)_{c}}{c!}\frac{2}{-\hat{s}_{12}\left(w_{M}\right)+2\left(a+b-1\right)}.$
(3.15)

The residues $R_{3,4}^{M}$ and $R_{4,3}^{M}$ above can be further calculated
to be

$\displaystyle R_{3,4}^{M}$
$\displaystyle=\underset{a,b=0,M=a+b}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\int_{0}^{1}dz_{2}z_{2}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}-a}\left(1-z_{2}\right)^{k_{3}\cdot k_{4}},$ (3.16) $\displaystyle
R_{4,3}^{M}$
$\displaystyle=\underset{b,c=0,M=b+c}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{3}\cdot
k_{4}\right)_{c}}{c!}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}+b}\left(1-z_{2}\right)^{k_{2}\cdot k_{3}}.$ (3.17)

We can now use the identity derived in the appendix

$\underset{b+c=M,b,c=0}{\sum}\frac{\left(T_{1}\right)_{b}}{b!}\frac{\left(T_{2}\right)_{c}}{c!}x_{1}^{b}x_{2}^{c}=\underset{\left\\{N_{r}\right\\}}{\sum}\prod_{r=1}\left(\frac{1}{N_{r}!r^{N_{r}}}\left[T_{1}x_{1}^{r}+T_{2}x_{2}^{r}\right]^{N_{r}}\right)$
(3.18)

to obtain

$\displaystyle R_{3,4}^{M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{2}\cdot
k_{3}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}-M}\left(1-z_{2}\right)^{k_{3}\cdot
k_{4}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(-\frac{k_{2}\cdot
k_{4}}{k_{2}\cdot k_{3}}\right)z_{2}^{r}\right]^{N_{r}},$ (3.19)
$\displaystyle R_{4,3}^{M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{3}\cdot
k_{4}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(-\frac{k_{2}\cdot
k_{4}}{k_{3}\cdot k_{4}}\right)z_{2}^{r}\right]^{N_{r}}.$ (3.20)

In Eq.(3.19) and Eq.(3.20) the sums are over partitions of $N$ into
$\left\\{N_{r}\right\\}$ with $M=\underset{r=1}{\sum}rN_{r}$ and
$N=\underset{r=1}{\sum}N_{r}$. Thus the string on-shell recursion relation
reduces the $5$-point KN amplitude into two products of subamplitudes
$3\otimes 4$ and $4\otimes 3$ with $3$-point and $4$-point functions. One can
now easily express the residues in Eq.(3.19) and Eq.(3.20) in terms of sums of
the Lauricella functions

$\displaystyle R_{3,4}^{M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{2}\cdot
k_{3}\right)^{N}B\left(\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}-M+1,k_{3}\cdot
k_{4}+1\right)\left(\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\right)$
$\displaystyle\cdot
F_{D}^{(K)}\left(\begin{array}[c]{c}\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}-M+1;\cdots,\left\\{-N_{r}\right\\}^{r},\cdots;\\\
\left(\hat{k}_{1}+k_{2}\right)\cdot k_{3}+k_{3}\cdot
k_{4}-M+2;\cdots,\left[\left(-\frac{k_{2}\cdot k_{4}}{k_{2}\cdot
k_{3}}\right)_{r}\right],\cdots\end{array}\right),$ (3.23) $\displaystyle
R_{4,3}^{M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{3}\cdot
k_{4}\right)^{N}B\left(\hat{k}_{1}\cdot k_{2}+1,k_{2}\cdot
k_{3}+1\right)\left(\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\right)$
$\displaystyle\cdot F_{D}^{(K)}\left(\begin{array}[c]{c}\hat{k}_{1}\cdot
k_{2}+1;\cdots,\left\\{-N_{r}\right\\}^{r},\cdots;\\\ \hat{k}_{1}\cdot
k_{2}+k_{2}\cdot k_{3}+2;\cdots,\left[\left(-\frac{k_{2}\cdot
k_{4}}{k_{3}\cdot k_{4}}\right)_{r}\right],\cdots\end{array}\right).$ (3.26)

where $K=\underset{\\{\text{for all }N_{r}\neq 0\\}}{\sum r}$.

###  III.3 The six-point KN amplitude

For the higher $n$-point KN amplitudes, we define the residues
$R_{m+1,n-m+1}^{M}$ ($2\leq m\leq n-2$) of $n$-point KN amplitude and express
the KN amplitude as

$A_{n}^{KN}=\underset{m=2}{\overset{n-2}{\sum}}\underset{M=0}{\overset{\infty}{\sum}}\frac{2R_{m+1,n-m+1}^{M}}{\left(k_{1}+k_{2}+...+k_{m}\right)^{2}+2(M-1)}$
(3.27)

where the pole locations $z_{m,M}$ are given by solutions of

$\left(\hat{k}_{1}+k_{2}+....+k_{m}\right)^{2}+2(M-1)=0,M=0,1,2...\text{.}$
(3.28)

Figure 1: Residues of KN amplitudes

For the case of $6$-point KN amplitude. There are three types of residues
$R_{3,5}^{M}$, $R_{4,4}^{M}$ and $R_{5,3}^{M}$. For the residue $R_{3,5}^{M}$,
we propose the left and right subamplitudes as following

$A\left(\hat{1},2,-P\right)=R_{3,5}^{P,L}=\prod_{r=1}\frac{\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\epsilon_{r}^{(\sigma)}\cdot
k_{2}}{\sqrt{N_{r}!r^{N_{r}}}},$ (3.29) $\displaystyle
A\left(P,3,4,5,\hat{6}\right)=R_{3,5}^{P,R}$
$\displaystyle=\int_{0}^{1}dw_{4}\int_{0}^{w_{4}}dw_{3}w_{3}^{k_{P}\cdot
k_{3}}w_{4}^{k_{P}\cdot k_{4}}\left(1-w_{3}\right)^{k_{3}\cdot
k_{5}}\left(1-w_{4}\right)^{k_{4}\cdot
k_{5}}\left(w_{4}-w_{3}\right)^{k_{3}\cdot k_{4}}$
$\displaystyle\times\prod_{r=1}\frac{1}{\sqrt{N_{r}!r^{N_{r}}}}\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\left(-\epsilon_{r}^{(\sigma)}\right)\cdot\left[k_{3}\left(\frac{1}{w_{3}}\right)^{r}+k_{4}\left(\frac{1}{w_{4}}\right)^{r}+k_{5}\left(\frac{1}{1}\right)^{r}\right].$
(3.30)

where $k_{P}=\hat{k}_{1}+k_{2}$ and $P$ is a normalized arbitrary string fock
state

$\left|P\right\rangle=\prod_{r=1}\frac{\left(\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\epsilon_{r}^{(\sigma)}\cdot\alpha_{-r}\right)}{\sqrt{N_{r}!r^{Nr}}}\left|0,k_{P}\right\rangle.$
(3.31)

In Eq.(3.31), it is understood, for example, that the state
$\epsilon_{1}^{(1)}\cdot\alpha_{-1}\epsilon_{1}^{(2)}\cdot\alpha_{-1}\epsilon_{1}^{(3)}\cdot\alpha_{-1}$
means
$\epsilon_{\mu\nu\lambda}\alpha_{-1}^{\mu}\alpha_{-1}^{\nu}\alpha_{-1}^{\lambda}$.
In Eq.(3.29), we have chosen $z_{1}=0$, $z_{2}=1$ and $z_{P}=\infty$ to fix
the $SL(2,R)$ gauge. Similarly, in Eq.(3.30), we set $w_{P}=0$, $w_{5}=1$ and
$w_{6}=\infty$. After summing over all $P$, we obtain the residue

$\displaystyle
R_{3,5}^{M}=\underset{\left\\{N_{r}\right\\}}{\sum}\int_{0}^{1}dw_{4}\int_{0}^{w_{4}}dw_{3}w_{3}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}}w_{4}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{4}}\left(1-w_{3}\right)^{k_{3}\cdot k_{5}}\left(1-w_{4}\right)^{k_{4}\cdot
k_{5}}\left(w_{4}-w_{3}\right)^{k_{3}\cdot k_{4}}$
$\displaystyle\times\prod_{r=1}\frac{\left[\left(-k_{2}\cdot
k_{3}\right)\left(\frac{1}{w_{3}}\right)^{r}+\left(-k_{2}\cdot
k_{4}\right)\left(\frac{1}{w_{4}}\right)^{r}+\left(-k_{2}\cdot
k_{5}\right)\left(\frac{1}{1}\right)^{r}\right]}{N_{r}!r^{N_{r}}}^{N_{r}}.$
(3.32)

For the residue $R_{4,4}^{M}$, we propose the left and right subamplitudes as
following

$\displaystyle A\left(\hat{1},2,3,-P\right)$
$\displaystyle=R_{4,4}^{P,L}=\prod_{r=1}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}}\frac{\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\epsilon_{r}^{(\sigma)}\cdot\left[k_{2}\left(z_{2}\right)^{r}+k_{3}\left(1\right)^{r}\right]}{\sqrt{N_{r}!r^{N_{r}}}},$
(3.33) $\displaystyle A\left(P,4,5,\hat{6}\right)$
$\displaystyle=R_{4,4}^{P,R}=\prod_{r=1}\int_{0}^{1}dw_{4}w_{4}^{k_{P}\cdot
k_{4}}\left(1-w_{4}\right)^{k_{4}\cdot
k_{5}}\frac{\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\left(-\epsilon_{r}^{(\sigma)}\right)\cdot\left[k_{4}\left(\frac{1}{w_{4}}\right)^{r}+k_{5}\left(\frac{1}{1}\right)^{r}\right]}{\sqrt{N_{r}!r^{N_{r}}}}$
(3.34)

where $k_{P}=\hat{k}_{1}+k_{2}+k_{3}$. In Eq.(3.33), we have chosen $z_{1}=0$,
$z_{3}=1$ and $z_{P}=\infty$ to fix the $SL(2,R)$ gauge. Similarly, in
Eq.(3.34), we set $w_{P}=0$, $w_{5}=1$ and $w_{6}=\infty$. After summing over
all $P$, we obtain the residue

$\displaystyle R_{4,4}^{M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\int_{0}^{1}dw_{4}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}}w_{4}^{\left(\hat{k}_{1}+k_{2}+k_{3}\right)\cdot
k_{4}}\left(1-w_{4}\right)^{k_{4}\cdot k_{5}}$
$\displaystyle\times\prod_{r=1}\frac{\left[\left(-k_{2}\cdot
k_{4}\right)\left(\frac{z_{2}}{w_{4}}\right)^{r}+\left(-k_{2}\cdot
k_{5}\right)\left(z_{2}\right)^{r}+\left(-k_{3}\cdot
k_{4}\right)\left(\frac{1}{w_{4}}\right)^{r}+\left(-k_{3}\cdot
k_{5}\right)1^{r}\right]^{N_{r}}}{N_{r}!r^{N_{r}}},$ (3.35)

which is a sum of products of the Lauricella functions after term by term
integrations for a given $M$. On the other hand, to prove that the residue
$R_{3,5}^{M}$ in Eq.(3.32) (and the residue $R_{5,3}^{M}$ to be discussed
below) can be expressed in terms of LSSA, one needs to do a second on-shell
recursion process. This will be done in section V.

For the residue $R_{5,3}^{M}$, we propose the left and right subamplitudes as
following

$\displaystyle
A\left(-P,5,\hat{6}\right)=R_{5,3}^{P,R}=\prod_{r=1}\frac{\left(-\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\epsilon_{r}^{(\sigma)}\cdot
k_{5}\right)}{\sqrt{N_{r}!r^{N_{r}}}}$ (3.36) $\displaystyle
A\left(\hat{1},2,3,4,P\right)=R_{5,3}^{P,L}$
$\displaystyle=\int_{0}^{1}dz_{3}\int_{0}^{z_{3}}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}z_{3}^{\hat{k}_{1}\cdot k_{3}}\left(z_{3}-z_{2}\right)^{k_{2}\cdot
k_{3}}\left(1-z_{2}\right)^{k_{2}\cdot k_{4}}\left(1-z_{3}\right)^{k_{3}\cdot
k_{4}}$
$\displaystyle\times\prod_{r=1}\frac{\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\epsilon_{r}^{(\sigma)}\cdot\left[k_{2}\left(z_{2}\right)^{r}+k_{3}\left(z_{3}\right)^{r}+k_{4}\left(1\right)^{r}\right]}{\sqrt{N_{r}!r^{N_{r}}}}$
(3.37)

where we have chosen $z_{1}=0$, $z_{4}=1$ and $z_{P}=\infty$ to fix the
$SL(2,R)$ gauge. Similarly, in Eq.(3.36), we set $w_{P}=0$, $w_{5}=1$ and
$w_{6}=\infty$. After summing over all $P$, we can get

$\displaystyle R_{5,3}^{M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\int_{0}^{1}dz_{3}\int_{0}^{z_{3}}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}z_{3}^{\hat{k}_{1}\cdot k_{3}}\left(z_{3}-z_{2}\right)^{k_{2}\cdot
k_{3}}\left(1-z_{2}\right)^{k_{2}\cdot k_{4}}\left(1-z_{3}\right)^{k_{3}\cdot
k_{4}}$ $\displaystyle\times\prod_{r=1}\frac{\left[\left(-k_{2}\cdot
k_{5}\right)\left(z_{2}\right)^{r}+\left(-k_{3}\cdot
k_{5}\right)\left(z_{3}\right)^{r}+\left(-k_{4}\cdot
k_{5}\right)\left(1\right)^{r}\right]^{N_{r}}}{N_{r}!r^{N_{r}}}.$ (3.38)

To justify the results in Eq.(3.32), Eq.(3.35) and Eq.(3.38), we follow the
standard string on-shell recursion calculation. The $6$-point KN amplitude can
be written as

$\displaystyle
A_{6}^{KN}=\int_{0}^{1}dz_{4}\int_{0}^{z_{4}}dz_{3}\int_{0}^{z_{3}}dz_{2}z_{2}^{k_{1}\cdot
k_{2}}z_{3}^{k_{1}\cdot k_{3}}z_{4}^{k_{1}\cdot k_{4}}$
$\displaystyle\cdot\left(z_{3}-z_{2}\right)^{k_{2}\cdot
k_{3}}\left(z_{4}-z_{2}\right)^{k_{2}\cdot
k_{4}}\left(1-z_{2}\right)^{k_{2}\cdot
k_{5}}\left(z_{4}-z_{3}\right)^{k_{3}\cdot
k_{4}}\left(1-z_{3}\right)^{k_{3}\cdot k_{5}}$
$\displaystyle=\underset{a,b,c,d,e,f=0}{\sum}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}\frac{\left(-k_{3}\cdot
k_{4}\right)_{d}}{d!}\frac{\left(-k_{3}\cdot
k_{5}\right)_{e}}{e!}\frac{\left(-k_{4}\cdot k_{5}\right)_{f}}{f!}$
$\displaystyle\times\frac{2}{\left(-s_{12}\right)+2\left(a+b+c-1\right)}\frac{2}{\left(-s_{123}\right)+2\left(b+c+d+e-1\right)}\frac{2}{\left(-s_{1234}\right)+2\left(c+e+f-1\right)}$
(3.39)

where

$s_{12}=-\left(k_{1}+k_{2}\right)^{2},s_{123}=-\left(k_{1}+k_{2}+k_{3}\right)^{2},s_{1234}=-\left(k_{1}+k_{2}+k_{3}+k_{4}\right)^{2}.$
(3.40)

The next step is to do the on-shell recursion deformation

$\hat{k}_{1}\left(z\right)=k_{1}+zq,q^{2}=k_{1}\cdot q=k_{6}\cdot
q=0,\hat{k}_{6}\left(z\right)=k_{6}-zq$ (3.41)

to obtain the poles

$\displaystyle z_{2,M}$
$\displaystyle=\frac{\left(k_{1}+k_{2}\right)^{2}+2\left(M-1\right)}{-2q\cdot\left(k_{1}+k_{2}\right)},M=a+b+c=0,1,2,...;$
(3.42) $\displaystyle z_{3,M}$
$\displaystyle=\frac{\left(k_{1}+k_{2}+k_{3}\right)^{2}+2\left(M-1\right)}{-2q\cdot\left(k_{1}+k_{2}+k_{3}\right)},M=b+c+d+e=0,1,2,...;$
(3.43) $\displaystyle z_{4,M}$
$\displaystyle=\frac{\left(k_{1}+k_{2}+k_{3}+k_{4}\right)^{2}+2\left(M-1\right)}{-2q\cdot\left(k_{1}+k_{2}+k_{3}+k_{4}\right)},M=c+e+f=0,1,2,....$
(3.44)

The amplitude can then be written as

$A_{6}^{KN}=\underset{M}{\sum}\frac{2}{\left(-s_{12}\right)+2\left(M-1\right)}R_{3,5}^{M}+\underset{M}{\sum}\frac{2}{\left(-s_{123}\right)+2\left(M-1\right)}R_{4,4}^{M}+\underset{M}{\sum}\frac{2}{\left(-s_{1234}\right)+2\left(M-1\right)}R_{5,3}^{M}$
(3.45)

where the residues can be calculated to be

$\displaystyle R_{3,5}^{M}=\underset{\begin{array}[c]{c}a,b,c,d,e,f=0\\\
M=a+b+c\end{array}}{\sum}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}\frac{\left(-k_{3}\cdot
k_{4}\right)_{d}}{d!}\frac{\left(-k_{3}\cdot
k_{5}\right)_{e}}{e!}\frac{\left(-k_{4}\cdot k_{5}\right)_{f}}{f!}$ (3.48)
$\displaystyle\times\frac{2}{\left(-\hat{s}_{123}\left(z_{2,M}\right)\right)+2\left(b+c+d+e-1\right)}\frac{2}{\left(-\hat{s}_{1234}\left(z_{2,M}\right)\right)+2\left(c+e+f-1\right)},$
(3.49) $\displaystyle
R_{4,4}^{M}=\underset{\begin{array}[c]{c}a,b,c,d,e,f=0\\\
M=b+c+d+e\end{array}}{\sum}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}\frac{\left(-k_{3}\cdot
k_{4}\right)_{d}}{d!}\frac{\left(-k_{3}\cdot
k_{5}\right)_{e}}{e!}\frac{\left(-k_{4}\cdot k_{5}\right)_{f}}{f!}$ (3.52)
$\displaystyle\times\frac{2}{\left(-\hat{s}_{12}\left(z_{3,M}\right)\right)+2\left(a+b+c-1\right)}\frac{2}{\left(-\hat{s}_{1234}\left(z_{3,M}\right)\right)+2\left(c+e+f-1\right)},$
(3.53) $\displaystyle
R_{5,3}^{M}=\underset{\begin{array}[c]{c}a,b,c,d,e,f=0\\\
M=c+e+f\end{array}}{\sum}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}\frac{\left(-k_{3}\cdot
k_{4}\right)_{d}}{d!}\frac{\left(-k_{3}\cdot
k_{5}\right)_{e}}{e!}\frac{\left(-k_{4}\cdot k_{5}\right)_{f}}{f!}$ (3.56)
$\displaystyle\times\frac{2}{\left(-\hat{s}_{12}\left(z_{4,M}\right)\right)+2\left(a+b+c-1\right)}\frac{2}{\left(-\hat{s}_{123}\left(z_{4,M}\right)\right)+2\left(b+c+d+e-1\right)}.$
(3.57)

Finally, we need to identify Eq.(3.32) with Eq.(3.49), Eq.(3.35) with
Eq.(3.53) and Eq.(3.38) with Eq.(3.57) respectively. To do this on the residue
$R_{3,5}^{M}$, we apply Eq.(A.16) for the case of $n=3$ and rewrite Eq.(3.32)
as

$\displaystyle
R_{3,5}^{M}=\int_{0}^{1}dw_{4}\int_{0}^{w_{4}}dw_{3}w_{3}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}}w_{4}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{4}}\left(1-w_{3}\right)^{k_{3}\cdot k_{5}}\left(1-w_{4}\right)^{k_{4}\cdot
k_{5}}\left(w_{4}-w_{3}\right)^{k_{3}\cdot k_{4}}$
$\displaystyle\times\underset{\begin{array}[c]{c}M=a+b+c\\\
a,b,c=0\end{array}}{\sum}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}w_{3}^{-a}w_{4}^{-b}1^{-c},$ (3.60)

which after binormal expansions and integrations can be further written as

$\displaystyle R_{3,5}^{M}$
$\displaystyle=\underset{\begin{array}[c]{c}M=a+b+c\\\
a,b,c,d,e,f=0\end{array}}{\sum}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}\frac{\left(-k_{3}\cdot
k_{4}\right)_{d}}{d!}\frac{\left(-k_{3}\cdot
k_{5}\right)_{e}}{e!}\frac{\left(-k_{4}\cdot k_{5}\right)_{f}}{f!}$ (3.63)
$\displaystyle\times\frac{1}{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}+(d+e-a+1)}\frac{1}{\left(\hat{k}_{1}+k_{2}\right)\cdot\left(k_{3}+k_{4}\right)+k_{3}\cdot
k_{4}+\left(e+f-a+2\right)}.$ (3.64)

We see that Eq.(3.64) above is the same as Eq.(3.49) since

$2\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}+2\left(d+e-a+1\right)=\left(-\hat{s}_{123}\left(z_{2,M}\right)\right)+2\left(b+c+d+e-1\right)$
(3.65)

and

$2\left(\hat{k}_{1}+k_{2}\right)\cdot\left(k_{3}+k_{4}\right)+2k_{3}\cdot
k_{4}+2\left(e+f-a+2\right)=\left(-\hat{s}_{1234}\left(z_{2,M}\right)\right)+2\left(c+e+f-1\right).$
(3.66)

This completes the proof that the proposals of the subamplitudes in Eq.(3.29),
Eq.(3.30) and thus the result of the residue calculated in Eq.(3.32) are
indeed the correct ones.

For the residue $R_{4,4}^{M}$, we apply Eq.(A.16) for the case of $n=4$ and
rewrite Eq.(3.35) as

$\displaystyle R_{4,4}^{M}$
$\displaystyle=\int_{0}^{1}dw_{4}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}}w_{4}^{\left(\hat{k}_{1}+k_{2}+k_{3}\right)\cdot
k_{4}}\left(1-w_{4}\right)^{k_{4}\cdot k_{5}}$
$\displaystyle\underset{\begin{array}[c]{c}b,c,d,e=0\\\
M=b+c+d+e\end{array}}{\sum}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}\frac{\left(-k_{3}\cdot
k_{4}\right)_{d}}{d!}\frac{\left(-k_{3}\cdot
k_{5}\right)_{e}}{e!}\left(\frac{z_{2}}{w_{4}}\right)^{b}\left(z_{2}\right)^{c}\left(\frac{1}{w_{4}}\right)^{d}$
(3.69)

which after binomial expansions and integrations can be further written as

$\displaystyle R_{4,4}^{M}$
$\displaystyle=\underset{\begin{array}[c]{c}M=b+c+d+e\\\
a,b,c,d,e,f=0\end{array}}{\sum}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}\frac{\left(-k_{3}\cdot
k_{4}\right)_{d}}{d!}\frac{\left(-k_{3}\cdot
k_{5}\right)_{e}}{e!}\frac{\left(-k_{4}\cdot k_{5}\right)_{f}}{f!}$ (3.72)
$\displaystyle\times\frac{1}{\hat{k}_{1}\cdot
k_{2}+\left(a+b+c+1\right)}\frac{1}{\left(\hat{k}_{1}+k_{2}+k_{3}\right)\cdot
k_{4}+\left(f-b-d+1\right)}.$ (3.73)

We see that Eq.(3.73) above is the same as Eq.(3.53) since

$-\hat{s}_{12}\left(z_{3,M}\right)+2\left(a+b+c-1\right)=2\hat{k}_{1}\cdot
k_{2}+2\left(a+b+c+1\right)$ (3.74)

and

$-\hat{s}_{1234}(z_{3,M})+2\left(c+e+f-1\right)=2\left(\hat{k}_{1}+k_{2}+k_{3}\right)\cdot
k_{4}+2\left(f-b-d+1\right).$ (3.75)

This completes the proof that the proposals of the subamplitudes in Eq.(3.33),
Eq.(3.34) and thus the result of the residue calculated in Eq.(3.35) are
indeed the correct ones.

For the residue $R_{5,3}^{M}$, we apply Eq.(A.16) for the case of $n=3$ and
rewrite Eq.(3.38) as

$\displaystyle R_{5,3}^{M}$
$\displaystyle=\int_{0}^{1}dz_{3}\int_{0}^{z_{3}}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}z_{3}^{\hat{k}_{1}\cdot k_{3}}\left(z_{3}-z_{2}\right)^{k_{2}\cdot
k_{3}}\left(1-z_{2}\right)^{k_{2}\cdot k_{4}}\left(1-z_{3}\right)^{k_{3}\cdot
k_{4}}$ $\displaystyle\underset{\begin{array}[c]{c}M=c+e+f\\\
c,e,f=0\end{array}}{\sum}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}\frac{\left(-k_{3}\cdot
k_{5}\right)_{e}}{e!}\frac{\left(-k_{4}\cdot
k_{5}\right)_{f}}{f!}\left(z_{2}\right)^{c}\left(z_{3}\right)^{e}\left(1\right)^{f},$
(3.78)

which after binomial expansions and integrations can be further written as

$\displaystyle R_{5,3}^{M}$
$\displaystyle=\underset{\begin{array}[c]{c}M=c+e+f\\\
a,b,c,d,e,f=0\end{array}}{\sum}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}\frac{\left(-k_{3}\cdot
k_{4}\right)_{c}}{d!}\frac{\left(-k_{3}\cdot
k_{5}\right)_{e}}{e!}\frac{\left(-k_{4}\cdot k_{5}\right)_{f}}{f!}$ (3.81)
$\displaystyle\times\frac{1}{\hat{k}_{1}\cdot
k_{2}+a+b+c+1}\frac{1}{\hat{k}_{1}\cdot k_{2}+\hat{k}_{1}\cdot
k_{3}+k_{2}\cdot k_{3}+b+c+d+e+2}$ (3.82)

We see that Eq.(3.82) above is the same as Eq.(3.57) since

$-\hat{s}_{12}\left(z_{4,M}\right)+2\left(a+b+c-1\right)=2\hat{k}_{1}\cdot
k_{2}+2\left(a+b+c+1\right)$ (3.83)

and

$-\hat{s}_{123}\left(z_{4,M}\right)+2\left(b+c+d+e-1\right)=2\hat{k}_{1}\cdot
k_{2}+2\hat{k}_{1}\cdot k_{3}+2k_{2}\cdot k_{3}+2(b+c+d+e+2).$ (3.84)

This completes the proof that the proposals of the subamplitudes in Eq.(3.37),
Eq.(3.36) and thus the result of the residue calculated in Eq.(3.38) are
indeed the correct ones.

###  III.4 The seven-point KN amplitude

For the case of $7$-point KN amplitude. There are four types of residues
$R_{3,6}^{M}$, $R_{4,5}^{M}$, $R_{5,4}^{M}$ and $R_{6,3}^{M}$. We will work
out residues $R_{3,6}^{M}$and $R_{4,5}^{M}$. Similar calculation can be done
for $R_{5,4}^{M}$ and $R_{6,3}^{M}$. For the residue $R_{3,6}^{M}$, we propose
the left and right subamplitudes as following

$A\left(\hat{1},2,-P\right)=R_{3,6}^{P,L}=\prod_{r=1}\frac{\left(\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\epsilon_{r}^{(\sigma)}\cdot
k_{2}\right)}{\sqrt{N_{r}!r^{N_{r}}}},$ (3.85) $\displaystyle
A\left(P,3,4,5,6,\hat{7}\right)=R_{3,6}^{P,R}$
$\displaystyle=\int_{0}^{1}dw_{5}\int_{0}^{w_{5}}dw_{4}\int_{0}^{w_{4}}dw_{3}w_{3}^{k_{P}\cdot
k_{3}}w_{4}^{k_{P}\cdot k_{4}}w_{5}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{5}}\left(1-w_{3}\right)^{k_{3}\cdot k_{6}}\left(1-w_{4}\right)^{k_{4}\cdot
k_{6}}$ $\displaystyle\cdot\left(1-w_{5}\right)^{k_{5}\cdot
k_{6}}\left(w_{4}-w_{3}\right)^{k_{3}\cdot
k_{4}}\left(w_{5}-w_{3}\right)^{k_{3}\cdot
k_{5}}\left(w_{5}-w_{4}\right)^{k_{4}\cdot k_{5}}$
$\displaystyle\times\prod_{r=1}\frac{1}{\sqrt{N_{r}!r^{N_{r}}}}\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\left(-\epsilon_{r}^{(\sigma)}\right)\cdot\left[k_{3}\left(\frac{1}{w_{3}}\right)^{r}+k_{4}\left(\frac{1}{w_{4}}\right)^{r}+k_{5}\left(\frac{1}{w_{5}}\right)^{r}+k_{6}\left(\frac{1}{1}\right)^{r}\right]$
(3.86)

where $P$ is a normalized arbitrary string fock state defined in Eq.(3.31) and
$k_{P}=\hat{k}_{1}+k_{2}$. In Eq.(3.85), we have chosen $z_{1}=0$, $z_{2}=1$
and $z_{P}=\infty$ to fix the $SL(2,R)$ gauge. Similarly, in Eq.(3.86), we set
$w_{P}=0$, $w_{6}=1$ and $w_{7}=\infty$. After summing over all $P$, we obtain
the residue

$\displaystyle
R_{3,6}^{M}=\underset{\left\\{N_{r}\right\\}}{\sum}\int_{0}^{1}dw_{5}\int_{0}^{w_{5}}dw_{4}\int_{0}^{w_{4}}dw_{3}w_{3}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}}w_{4}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{4}}w_{5}^{\left(\hat{k}_{1}+k_{2}\right)\cdot k_{5}}$
$\displaystyle\cdot\left(1-w_{3}\right)^{k_{3}\cdot
k_{5}}\left(1-w_{4}\right)^{k_{4}\cdot
k_{5}}\left(w_{4}-w_{3}\right)^{k_{3}\cdot k_{4}}$
$\displaystyle\times\prod_{r=1}\frac{\left[\left(-k_{2}\cdot
k_{3}\right)\left(\frac{1}{w_{3}}\right)^{r}+\left(-k_{2}\cdot
k_{4}\right)\left(\frac{1}{w_{4}}\right)^{r}+\left(-k_{2}\cdot
k_{5}\right)\left(\frac{1}{w_{5}}\right)^{r}+\left(-k_{2}\cdot
k_{6}\right)\left(\frac{1}{1}\right)^{r}\right]}{N_{r}!r^{N_{r}}}^{N_{r}}.$
(3.87)

To justify the result in Eq.(3.87), we follow the standard string on-shell
recursion calculation. First of all, we use the identity Eq.(A.16) to
transform the last line of Eq.(3.87) to a summation, and then do binomial
expansions before performing integrations. Finally we end up with

$\displaystyle R_{3,6}^{M}$
$\displaystyle=\underset{\begin{array}[c]{c}M=a+b+c+d\\\
a,b,c,d,e,f,g,h,i,j=0\end{array}}{\sum}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}\frac{\left(-k_{2}\cdot k_{6}\right)_{d}}{d!}$ (3.90)
$\displaystyle\cdot\frac{\left(-k_{3}\cdot
k_{4}\right)_{e}}{e!}\frac{\left(-k_{3}\cdot
k_{5}\right)_{f}}{f!}\frac{\left(-k_{3}\cdot
k_{6}\right)_{g}}{g!}\frac{\left(-k_{4}\cdot
k_{5}\right)_{h}}{h!}\frac{\left(-k_{4}\cdot
k_{6}\right)_{i}}{i!}\frac{\left(-k_{5}\cdot k_{6}\right)_{j}}{j!}$
$\displaystyle\times\frac{1}{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}+(e+f+g-a+1)}\frac{1}{\left(\hat{k}_{1}+k_{2}\right)\cdot\left(k_{3}+k_{4}\right)+k_{3}\cdot
k_{4}+\left(f+g+h+i-a-b+2\right)}$
$\displaystyle\times\frac{1}{\left(\hat{k}_{1}+k_{2}\right)\cdot\left(k_{3}+k_{4}+k_{5}\right)+k_{3}\cdot
k_{4}+k_{3}\cdot k_{5}+k_{4}\cdot k_{5}+\left(g+i+j-a-b-c+3\right)}.$ (3.91)

On the other hand, the residue $R_{3,6}^{M}$ can be extracted from the pole of
the $7$-point KN amplitude

$z_{2,M}=\frac{\left(k_{1}+k_{2}\right)^{2}+2\left(M-1\right)}{-2q\cdot\left(k_{1}+k_{2}\right)},M=a+b+c+d=0,1,2,...\text{,}$
(3.92)

and calculated to be

$\displaystyle R_{3,6}^{M}$
$\displaystyle=\underset{\begin{array}[c]{c}M=a+b+c+d\\\
a,b,c,d,e,f,g,h,i,j=0\end{array}}{\sum}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}\frac{\left(-k_{2}\cdot k_{6}\right)_{d}}{d!}$ (3.95)
$\displaystyle\cdot\frac{\left(-k_{3}\cdot
k_{4}\right)_{e}}{e!}\frac{\left(-k_{3}\cdot
k_{5}\right)_{f}}{f!}\frac{\left(-k_{3}\cdot
k_{6}\right)_{g}}{g!}\frac{\left(-k_{4}\cdot
k_{5}\right)_{h}}{h!}\frac{\left(-k_{4}\cdot
k_{6}\right)_{i}}{i!}\frac{\left(-k_{5}\cdot k_{6}\right)_{j}}{j!}$
$\displaystyle\times\frac{2}{-\hat{s}_{123}\left(z_{2,M}\right)+2\left(b+c+d+e+f+g-1\right)}\frac{2}{-\hat{s}_{1234}\left(z_{2,M}\right)+2\left(c+d+f+g+h+i-1\right)}$
$\displaystyle\times\frac{2}{-\hat{s}_{12345}\left(z_{2,M}\right)+2\left(d+g+i+j-1\right)}.$
(3.96)

It is straightforwad to verify

$\displaystyle-\hat{s}_{123}\left(z_{2,M}\right)+2\left(b+c+d+e+f+g-1\right)$
$\displaystyle=2\left(\hat{k}_{1}+k_{2}\right)\cdot k_{3}+2(e+f+g-a+1),$
(3.97)
$\displaystyle-\hat{s}_{1234}\left(z_{2,M}\right)+2\left(c+d+f+g+h+i-1\right)$
$\displaystyle=2\left(\hat{k}_{1}+k_{2}\right)\cdot\left(k_{3}+k_{4}\right)+2k_{3}\cdot
k_{4}$ $\displaystyle+2\left(f+g+h+i-a-b+2\right),$ (3.98)
$\displaystyle-\hat{s}_{12345}\left(z_{2,M}\right)+2\left(d+g+i+j-1\right)$
$\displaystyle=2\left(\hat{k}_{1}+k_{2}\right)\cdot\left(k_{3}+k_{4}+k_{5}\right)+2k_{3}\cdot
k_{4}+2k_{3}\cdot k_{5}$ $\displaystyle+2k_{4}\cdot k_{5}+2\left(g+i+j-a-
b-c+3\right).$ (3.99)

This shows that Eq.(3.91) is the same as Eq.(3.96), and the proposal in
Eq.(3.85) and Eq.(3.86) are consistent ones.

For the residue $R_{4,5}^{M}$, we propose the left and right subamplitudes as
following

$\displaystyle A\left(\hat{1},2,3,-P\right)$
$\displaystyle=R_{4,5}^{P,L}=\prod_{r=1}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}}\frac{\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\epsilon_{r}^{(\sigma)}\cdot\left[k_{2}\left(z_{2}\right)^{r}+k_{3}\left(1\right)^{r}\right]}{\sqrt{N_{r}!r^{N_{r}}}},$
(3.100) $\displaystyle A\left(P,4,5,6,\hat{7}\right)$
$\displaystyle=R_{4,5}^{P,R}=\prod_{r=1}\int_{0}^{1}dw_{4}\int_{0}^{w_{4}}dw_{5}w_{4}^{k_{P}\cdot
k_{4}}w_{5}^{k_{P}\cdot k_{5}}\left(1-w_{4}\right)^{k_{4}\cdot
k_{6}}\left(1-w_{5}\right)^{k_{5}\cdot
k_{6}}\left(w_{5}-w_{4}\right)^{k_{4}\cdot k_{5}}$
$\displaystyle\times\frac{\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\left(-\epsilon_{r}^{(\sigma)}\right)\cdot\left[k_{4}\left(\frac{1}{w_{4}}\right)^{r}+k_{5}\left(\frac{1}{w_{5}}\right)^{r}+k_{6}\left(\frac{1}{1}\right)^{r}\right]}{\sqrt{N_{r}!r^{N_{r}}}}$
(3.101)

where $k_{P}=\hat{k}_{1}+k_{2}+k_{3}$. After summing over all $P$, we obtain
the residue

$\displaystyle R_{4,5}^{M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\int_{0}^{1}dw_{4}\int_{0}^{w_{4}}dw_{5}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot k_{3}}w_{4}^{k_{P}\cdot
k_{4}}w_{5}^{k_{P}\cdot k_{5}}\left(1-w_{4}\right)^{k_{4}\cdot
k_{6}}\left(1-w_{5}\right)^{k_{5}\cdot
k_{6}}\left(w_{5}-w_{4}\right)^{k_{4}\cdot k_{5}}$
$\displaystyle\times\prod_{r=1}\frac{\left[\begin{array}[c]{c}\left(-k_{2}\cdot
k_{4}\right)\left(\frac{z_{2}}{w_{4}}\right)^{r}+\left(-k_{2}\cdot
k_{5}\right)\left(\frac{z_{2}}{w_{5}}\right)^{r}+\left(-k_{2}\cdot
k_{6}\right)\left(z_{2}\right)^{r}\\\ +\left(-k_{3}\cdot
k_{4}\right)\left(\frac{1}{w_{4}}\right)^{r}+\left(-k_{3}\cdot
k_{5}\right)\left(\frac{1}{w_{5}}\right)^{r}+\left(k_{3}\cdot
k_{6}\right)\left(\frac{1}{1}\right)^{r}\end{array}\right]^{N_{r}}}{N_{r}!r^{N_{r}}}.$
(3.104)

The next step is to use the identity in Eq.(A.16) and then do binomial
expansions before performing integrations on Eq.(3.104), we finally end up
with

$\displaystyle R_{4,5}^{M}$
$\displaystyle=\underset{\begin{array}[c]{c}M=b+c+d+e+f+g\\\
a,b,c,d,e,f,g,h,i,j=0\end{array}}{\sum}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}\frac{\left(-k_{2}\cdot k_{6}\right)_{d}}{d!}$ (3.107)
$\displaystyle\cdot\frac{\left(-k_{3}\cdot
k_{4}\right)_{e}}{e!}\frac{\left(-k_{3}\cdot
k_{5}\right)_{f}}{f!}\frac{\left(-k_{3}\cdot
k_{6}\right)_{g}}{g!}\frac{\left(-k_{4}\cdot
k_{5}\right)_{h}}{h!}\frac{\left(-k_{4}\cdot
k_{6}\right)_{i}}{i!}\frac{\left(-k_{5}\cdot k_{6}\right)_{j}}{j!}$
$\displaystyle\times\frac{1}{\left(\hat{k}_{1}+k_{2}+k_{3}\right)\cdot
k_{4}+(h+i-b-e+1)}$
$\displaystyle\times\frac{1}{\left(\hat{k}_{1}+k_{2}+k_{3}\right)\cdot\left(k_{4}+k_{5}\right)+k_{4}\cdot
k_{5}+\left(i+j-b-c-e-f+2\right)}$
$\displaystyle\times\frac{1}{\hat{k}_{1}\cdot k_{2}+\left(a+b+c+d+1\right)}.$
(3.108)

On the other hand, the residue $R_{4,5}^{M}$ can be extracted from the pole of
the $7$-point KN amplitude

$z_{3,M}=\frac{\left(k_{1}+k_{2}+k_{3}\right)^{2}+2\left(M-1\right)}{-2q\cdot\left(k_{1}+k_{2}+k_{3}\right)},M=b+c+d+e+f+g=0,1,2,...\text{,}$
(3.109)

and calculated to be

$\displaystyle R_{4,5}^{M}$
$\displaystyle=\underset{\begin{array}[c]{c}M=b+c+d+e+f+g\\\
a,b,c,d,e,f,g,h,i,j=0\end{array}}{\sum}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}\frac{\left(-k_{2}\cdot k_{6}\right)_{d}}{d!}$ (3.112)
$\displaystyle\cdot\frac{\left(-k_{3}\cdot
k_{4}\right)_{e}}{e!}\frac{\left(-k_{3}\cdot
k_{5}\right)_{f}}{f!}\frac{\left(-k_{3}\cdot
k_{6}\right)_{g}}{g!}\frac{\left(-k_{4}\cdot
k_{5}\right)_{h}}{h!}\frac{\left(-k_{4}\cdot
k_{6}\right)_{i}}{i!}\frac{\left(-k_{5}\cdot k_{6}\right)_{j}}{j!}$
$\displaystyle\times\frac{1}{-\hat{s}_{1234}\left(z_{3,M}\right)+2\left(c+d+f+g+h+i-1\right)}$
$\displaystyle\times\frac{1}{-\hat{s}_{12345}\left(z_{3,M}\right)+2\left(d+g+i+j-1\right)}\times\frac{1}{-\hat{s}_{12}\left(z_{3,M}\right)+2\left(a+b+c+d-1\right)}.$
(3.113)

It is straightforward to verify

$\displaystyle-\hat{s}_{1234}\left(z_{3,M}\right)+2\left(c+d+f+g+h+i-1\right)$
$\displaystyle=2\left(\hat{k}_{1}+k_{2}+k_{3}\right)\cdot k_{4}+2(h+i-b-e+1),$
(3.114)
$\displaystyle-\hat{s}_{12345}\left(z_{3,M}\right)+2\left(d+g+i+j-1\right)$
$\displaystyle=2\left(\hat{k}_{1}+k_{2}+k_{3}\right)\cdot\left(k_{4}+k_{5}\right)+2k_{4}\cdot
k_{5}$ $\displaystyle+2\left(i+j-b-c-e-f+2\right),$ (3.115)
$\displaystyle-\hat{s}_{12}\left(z_{3,M}\right)+2\left(a+b+c+d-1\right)$
$\displaystyle=2\hat{k}_{1}\cdot k_{2}+2\left(a+b+c+d+1\right).$ (3.116)

This shows that Eq.(3.108) is the same as Eq.(3.113), and the proposal in
Eq.(3.100) and Eq.(3.101) are consistent ones.

###  III.5 The n-point KN amplitude

In this section, we will adopt two methods to calculate the residues of
general $n$-point KN amplitudes. We will first use the method of subamplitudes
similar to the previous subsections. We then use a direct calculation method.
The result that residues of all $n$-point SSA can be expressed in terms of
Lauricella functions will be discussed in section V.

####  III.5.1 Method of subamplitudes

To calculate the residues $R_{m+1,n-m+1}^{M}$ of $A_{n}$ in Eq.(3.27), we can
generalize the previous calculations and propose the following two
subamplitudes $R_{m+1,n-m+1}^{P,L}\left(\hat{1},2,...,m,-P\right)$ and
$R_{m+1,n-m+1}^{P,R}\left(P,m+1,\cdots,n-1,\hat{n}\right)$

$\displaystyle
R_{m+1,n-m+1}^{P,L}\left(\underset{z_{1}=0}{\underbrace{\hat{1}}},\underset{z_{2}}{\underbrace{2}},\cdots,\underset{z_{m-1}}{\underbrace{m-1}},\underset{z_{m}=1}{\underbrace{m}},\underset{\infty}{\underbrace{-P}}\right)$
$\displaystyle=\int_{0}^{1}dz_{m-1}\cdots\int_{0}^{z_{3}}dz_{2}\underset{1\leq
j<i\leq m}{\prod}\left(z_{i}-z_{j}\right)^{k_{i}\cdot
k_{j}}\prod_{r=1}\frac{\left[\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\epsilon_{r}^{(\sigma)}\cdot\underset{2\leq
i\leq m}{\sum}k_{i}\left(z_{i}\right)^{r}\right]}{\sqrt{N_{r}!r^{N_{r}}}},$
(3.117) $\displaystyle
R_{m+1,n-m+1}^{P,R}\left(\underset{w_{m}=0}{\underbrace{P}},\underset{{}_{w_{m+1}}}{\underbrace{m+1}},\cdots,\underset{w_{n-2}}{\underbrace{n-2}},\underset{w_{n-1}=1}{\underbrace{n-1}},\underset{w_{n}=\infty}{\underbrace{\hat{n}}}\right)$
$\displaystyle=\int_{0}^{1}dw_{n-2}\cdots\int_{0}^{w_{m+2}}dw_{m+1}\underset{P=m\leq
j<i\leq n-1}{\prod}\left(w_{i}-w_{j}\right)^{k_{i}\cdot
k_{j}}\prod_{r=1}\frac{\left[\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\left(-\epsilon_{r}^{(\sigma)}\right)\cdot\underset{m+1\leq
i\leq
n-1}{\sum}k_{i}\left(\frac{1}{w_{i}}\right)^{r}\right]}{\sqrt{N_{r}!r^{N_{r}}}}$
(3.118)

where $k_{P}=\hat{k}_{1}+\underset{i=2}{\overset{m}{\sum}}k_{i}$. It is
understood that $k_{1}$ in Eq.(3.117) should be replaced by $\hat{k}_{1}$.

Figure 2: The left and right subamplitudes

We then sum over all $P$ to obtain the residue

$\displaystyle
R_{m+1,n-m+1}^{M}\left(\hat{1},2,\cdots,m,m+1,\cdots,n-1,\hat{n}\right)$
$\displaystyle=\int_{0}^{1}dz_{m-1}\cdots\int_{0}^{z_{3}}dz_{2}\int_{0}^{1}dw_{n-2}\cdots\int_{0}^{w_{m+2}}dw_{m+1}\underset{1\leq
j<i\leq m}{\prod}\left(z_{i}-z_{j}\right)^{k_{i}\cdot k_{j}}$
$\displaystyle\times\underset{P=m\leq j<i\leq
n-1}{\prod}\left(w_{i}-w_{j}\right)^{k_{i}\cdot
k_{j}}\underset{\left\\{N_{r}\right\\}}{\sum}\prod_{r=1}\frac{\left[\underset{2\leq
j\leq m,m+1\leq i\leq n-1}{\sum}\left(-k_{j}\cdot
k_{i}\right)\left(\frac{z_{j}}{w_{i}}\right)^{r}\right]}{N_{r}!r^{N_{r}}}^{N_{r}}.$
(3.119)

We will show in section V that all residues $R_{m+1,n-m+1}^{M}$ in Eq.(3.119)
can be expressed in terms of sum over LSSA.

####  III.5.2 Direct calculation

In the previous method of subamplitudes calculation, we did not do a
consistency check similar to Eq.(3.114), Eq.(3.115) and Eq.(3.116). In this
section, we adopt a direct calculation starting from the general $n$-point KN
amplitude. The $n$-point KN amplitude is

$A_{n}^{KN}=\int_{0}^{1}dz_{n-2}\int_{0}^{z_{n-2}}dz_{n-3}\cdots\int_{0}^{z_{3}}dz_{2}\text{
}\prod_{i>j=1}^{n-2}\left(z_{i}-z_{j}\right)^{k_{i}\cdot k_{j}}\text{,
}z_{i}\geq z_{j}$ (3.120)

which has $\left(n-3\right)$ series of poles for each integral on
$z_{2},\cdots,z_{n-2}$.

To investigate the poles for the integral on $z_{m}$, which splits the
$n$-point Koba-Nielsen amplitude to a $\left(m+1\right)$-point and a
$\left(n-m+1\right)$-point amplitudes connected by a string propagator, we
express $A_{n}^{KN}$ in the following form

$\displaystyle A_{n}^{KN}$
$\displaystyle=\int_{0}^{1}dz_{n-2}\cdots\int_{0}^{z_{m+2}}dz_{m+1}\int_{0}^{z_{m+1}}dz_{m}\int_{0}^{z_{m}}dz_{m-1}\cdots\int_{0}^{z_{3}}dz_{2}$
$\displaystyle\cdot\prod_{m\geq i>j}\left(z_{i}-z_{j}\right)^{k_{i}\cdot
k_{j}}\prod_{i>j>m}\left(z_{i}-z_{j}\right)^{k_{i}\cdot k_{j}}\prod_{i>m\geq
j}\left(z_{i}-z_{j}\right)^{k_{i}\cdot k_{j}}$ (3.121)

where we have taken $\left(z_{0},z_{n-1},z_{n}\right)=\left(0,1,\infty\right)$
to fix the $SL\left(2,R\right)$ invariance.

By defining the new coordinates,

$\displaystyle\text{for }i$ $\displaystyle>m:y_{i}=z_{i}\text{ with
}\left(y_{m}\equiv y_{P},y_{n-1},y_{n}\right)=\left(0,1,\infty\right),$
(3.122) $\displaystyle\text{for }i$
$\displaystyle=m:z_{P}=\frac{z_{m}}{z_{m+1}},$ (3.123) $\displaystyle\text{for
}i$ $\displaystyle<m:x_{i}=\frac{z_{i}}{z_{m}}\text{ with
}\left(x_{1},x_{m},x_{m+1}\right)=\left(0,1,\infty\right),$ (3.124)

Eq.(3.121) becomes

$\displaystyle A_{n}^{KN}$
$\displaystyle=\int_{0}^{1}dy_{n-2}\cdots\int_{0}^{y_{m+2}}dy_{m+1}\text{
}\prod_{i=m+1}^{n-2}y_{i}^{k_{i}\cdot
k_{P}}\prod_{i>j=m+1}^{n-2}\left(y_{i}-y_{j}\right)^{k_{i}\cdot k_{j}}$
$\displaystyle\times\int_{0}^{1}dx_{m-1}\cdots\int_{0}^{x_{3}}dx_{2}\text{
}\prod_{i>j=1}^{m-1}\left(x_{i}-x_{j}\right)^{k_{i}\cdot k_{j}}$
$\displaystyle\times\int_{0}^{1}dz_{P}\text{
}\left(z_{P}\right)^{\sum\limits_{m\geq i>j}k_{i}\cdot
k_{j}+m-2}\left(y_{m+1}\right)^{\sum\limits_{m\geq i>j}k_{i}\cdot
k_{j}+m-1}\prod_{2\leq j\leq m,m+1\leq i\leq
n-1}\left(1-\frac{y_{m+1}z_{P}x_{j}}{y_{i}}\right)^{k_{i}\cdot k_{j}}.$
(3.125)

We then expand the last product in Eq.(3.125) by using Binomial theorem

$\left(1-\frac{y_{m+1}z_{P}x_{j}}{y_{i}}\right)^{k_{i}\cdot
k_{j}}=\sum_{a_{ij}}\frac{\left(-k_{i}\cdot
k_{j}\right)_{a_{ij}}}{a_{ij}!}\left(\frac{y_{m+1}z_{P}x_{j}}{y_{i}}\right)^{a_{ij}}$
(3.126)

where $a_{ij}$’s are integers, to get

$\displaystyle A_{n}^{KN}$
$\displaystyle=\int_{0}^{1}dy_{n-2}\cdots\int_{0}^{y_{m+2}}dy_{m+1}\text{
}\underset{P=m\leq j<i\leq n-1}{\prod}\left(y_{i}-y_{j}\right)^{k_{i}\cdot
k_{j}}$
$\displaystyle\times\int_{0}^{1}dx_{m-1}\cdots\int_{0}^{x_{3}}dx_{2}\text{
}\underset{1\leq j<i\leq m}{\prod}\left(x_{i}-x_{j}\right)^{k_{i}\cdot
k_{j}}\sum_{a_{ij}}\left(y_{m+1}\right)^{\sum\limits_{m\geq i>j}k_{i}\cdot
k_{j}+\sum\limits_{2\leq j\leq m,m+1\leq i\leq n-1}a_{ij}+m-1}$
$\displaystyle\times\prod_{2\leq j\leq m,m+1\leq i\leq
n-1}\frac{\left(-k_{i}\cdot
k_{j}\right)_{a_{ij}}}{a_{ij}!}\left(\frac{x_{j}}{y_{i}}\right)^{a_{ij}}\int_{0}^{1}dz_{P}\text{
}\left(z_{P}\right)^{\sum\limits_{m\geq i>j}k_{i}\cdot k_{j}\sum\limits_{2\leq
j\leq m,m+1\leq i\leq n-1}a_{ij}+m-2}.$ (3.127)

In the above calculation, we have defined the momentum of the propagator as
$k_{P}=\sum\limits_{j=1}^{m}k_{j}$.

By performing the integral on $z_{P}$, we finally express $A_{n}^{KN}$ in the
following form

$\displaystyle A_{n}^{KN}$
$\displaystyle=\int_{0}^{1}dy_{n-2}\cdots\int_{0}^{y_{m+2}}dy_{m+1}\text{
}\underset{P=m\leq j<i\leq n-1}{\prod}\left(y_{i}-y_{j}\right)^{k_{i}\cdot
k_{j}}$
$\displaystyle\times\int_{0}^{1}dx_{m-1}\cdots\int_{0}^{x_{3}}dx_{2}\text{
}\underset{1\leq j<i\leq m}{\prod}\left(x_{i}-x_{j}\right)^{k_{i}\cdot
k_{j}}\sum_{a_{ij}}\left(y_{m+1}\right)^{\sum\limits_{m\geq i>j}k_{i}\cdot
k_{j}+\sum\limits_{2\leq j\leq m,m+1\leq i\leq n-1}a_{ij}+m-1}$
$\displaystyle\times\prod_{2\leq j\leq m,m+1\leq i\leq
n-1}\frac{\left(-k_{i}\cdot
k_{j}\right)_{a_{ij}}}{a_{ij}!}\left(\frac{x_{j}}{y_{i}}\right)^{a_{ij}}\frac{2}{-s_{1\cdots
m}+2\left(\sum\limits_{2\leq j\leq m,m+1\leq i\leq n-1}a_{ij}-1\right)}$
(3.128)

where we have defined

$s_{1\cdots m}=-\left(k_{1}+\cdots+k_{m}\right)^{2}.$ (3.129)

Next, we make the momentum shift as in BCFW,

$k_{1}\left(z\right)=k_{1}+zq,k_{n}\left(z\right)=k_{n}-zq\text{ with
}q^{2}=k_{1}\cdot q=k_{n}\cdot q=0.$ (3.130)

The KN amplitudes can be written as in Eq.(3.27)

$\displaystyle A_{n}^{KN}$
$\displaystyle=\sum_{m=2}^{n-2}\underset{M=0}{\overset{\infty}{\sum}}\left[\frac{A_{n}^{KN}(z)}{z}(z-z_{m,M})\right]_{z=z_{m,M}}$
$\displaystyle=\sum_{m=2}^{n-2}\underset{M=0}{\overset{\infty}{\sum}}\frac{2}{-s_{1\cdots
m}+2\left(M-1\right)}R_{m+1,n-m+1}^{M}$ (3.131)

where we have defined the residue $R_{m+1,n-m+1}^{M}$. The poles $z_{m,M}$ can
be calculated from the equation

$-s_{1\cdots m}\left(z_{m,M}\right)+2\left(\sum_{2\leq j\leq m,m+1\leq i\leq
n-1}a_{ij}-1\right)=0.$ (3.132)

For a fixed mass $M=\sum\limits_{2\leq j\leq m,m+1\leq i\leq n-1}a_{ij}$, the
poles are

$z_{m,M}=\frac{-s_{1\cdots m}+2\left(M-1\right)}{-2\sum\limits_{i=1}^{m}q\cdot
k_{i}}.$ (3.133)

One of the term in Eq.(3.131) for the split $m$ can be calculated as

$\displaystyle\underset{M=0}{\overset{\infty}{\sum}}\frac{2}{-s_{1\cdots
m}+2\left(M-1\right)}R_{m+1,n-m+1}^{M}$
$\displaystyle=\underset{M=0}{\overset{\infty}{\sum}}\int_{0}^{1}dy_{n-2}\cdots\int_{0}^{y_{m+2}}dy_{m+1}\text{
}\underset{P=m\leq j<i\leq n-1}{\prod}\left(y_{i}-y_{j}\right)^{k_{i}\cdot
k_{j}}\int_{0}^{1}dx_{m-1}\cdots\int_{0}^{x_{3}}dx_{2}\text{ }\underset{1\leq
j<i\leq m}{\prod}\left(x_{i}-x_{j}\right)^{k_{i}\cdot k_{j}}$
$\displaystyle\times\frac{2}{-s_{12\cdots n}+2\left(M-1\right)}\prod_{2\leq
j\leq m,m+1\leq i\leq n-1}\sum_{\Sigma a_{ij}=M}\frac{\left(-k_{i}\cdot
k_{j}\right)_{a_{ij}}}{a_{ij}!}\left(\frac{x_{j}}{y_{i}}\right)^{a_{ij}}.$
(3.134)

In the above calculation, we have used the identity

$\left(\sum\limits_{m\geq i>j}k_{i}\cdot
k_{j}\right)\left(z_{m,M}\right)+M+m-1=0.$ (3.135)

A special case of Eq.(3.135) corresponds to the consistency check of
Eq.(3.114), Eq.(3.115) and Eq.(3.116) in the calculation of the residue
$R_{4,5}^{M}$ of $7$-point KN amplitude. Using the identity

$\prod_{i=1}^{n}\sum_{\Sigma
b_{i}=M}\frac{\left(T_{i}\right)_{b_{i}}}{b_{i}!}X_{i}^{b_{i}}=\sum_{\left\\{N_{r}\right\\}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left(\sum_{i=1}^{n}T_{i}X_{i}^{r}\right)^{N_{r}},$
(3.136)

which was proved in the appendix, and identifying the parameter $a_{ij}$ as
$b_{i}$, the residue becomes

$\displaystyle R_{m+1,n-m+1}^{M}$
$\displaystyle=\int_{0}^{1}dx_{m-1}\cdots\int_{0}^{x_{3}}dx_{2}\int_{0}^{1}dy_{n-2}\cdots\int_{0}^{y_{m+2}}dy_{m+1}\text{
}\underset{1\leq j<i\leq m}{\prod}\left(x_{i}-x_{j}\right)^{k_{i}\cdot k_{j}}$
$\displaystyle\times\text{ }\underset{P=m\leq j<i\leq
n-1}{\prod}\left(y_{i}-y_{j}\right)^{k_{i}\cdot
k_{j}}\sum_{\left\\{N_{r}\right\\}}\prod_{r=1}\frac{\left[\sum\limits_{2\leq
j\leq m,m+1\leq i\leq n-1}\left(-k_{i}\cdot
k_{j}\right)\left(\frac{x_{j}}{y_{i}}\right)^{r}\right]^{N_{r}}}{N_{r}!r^{N_{r}}}.$
(3.137)

The residue in Eq.(3.137) can be written as

$R_{m+1,n-m+1}^{M}=\sum_{P}A_{m+1}\left(1,\cdots,m,-P\right)A_{n-m+1}\left(P,m+1,\cdots,n\right)$
(3.138)

where

$\displaystyle A_{m+1}\left(1,\cdots,m,-P\right)$
$\displaystyle=\int_{0}^{1}dx_{m-1}\cdots\int_{0}^{x_{3}}dx_{2}$
$\displaystyle\times\text{ }\underset{1\leq j<i\leq
m}{\prod}\left(x_{i}-x_{j}\right)^{k_{i}\cdot
k_{j}}\prod_{r=1}\frac{\left[\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\epsilon_{r}^{(\sigma)}\cdot\sum\limits_{2\leq
j\leq m}k_{j}\left(x_{j}\right)^{r}\right]}{\sqrt{N_{r}!r^{N_{r}}}},$ (3.139)
$\displaystyle A_{n-m+1}\left(P,m+1,\cdots,n\right)$
$\displaystyle=\int_{0}^{1}dy_{n-2}\cdots\int_{0}^{y_{m+2}}dy_{m+1}$
$\displaystyle\times\text{ }\underset{P=m\leq j<i\leq
n-1}{\prod}\left(y_{i}-y_{j}\right)^{k_{i}\cdot
k_{j}}\prod_{r=1}\frac{\left[\underset{\sigma=1}{\overset{N_{r}}{{\displaystyle\prod}}}\left(-\epsilon_{r}^{(\sigma)}\right)\cdot\sum\limits_{m+1\leq
i\leq
n-1}k_{i}\left(\frac{1}{y_{i}}\right)^{r}\right]}{\sqrt{N_{r}!r^{N_{r}}}}$
(3.140)

are the $m+1$ and $\left(n-m+1\right)$-point subamplitudes of $m$ and
$\left(n-m\right)$ tachyons and a string fock state $P$. Note that the sum
over $P$ in Eq.(3.138) reduces to $\sum_{\left\\{N_{r}\right\\}}$ in
Eq.(3.137). The results in Eq.(3.137), Eq.(3.139) and Eq.(3.140) are
consistent with Eq.(3.119), Eq.(3.117) and Eq.(3.118) respectively proposed
and calculated in the last subsection.

##  IV Iteration relations among residues

To get some feeling of the residue calculation, we give the explicit form of
$R_{4,3}^{4}$ in Eq.(3.20) as one example of residue calculation

$\displaystyle R_{4,3}^{4}$
$\displaystyle=\frac{1}{4}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot k_{3}}\left[-k_{3}\cdot
k_{4}-(-k_{2}\cdot k_{4})z_{2}^{4}\right]$
$\displaystyle+\frac{1}{1!1!}\left(\frac{1}{3}\right)\left(\frac{1}{1}\right)\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot k_{3}}\left[-k_{3}\cdot
k_{4}-(-k_{2}\cdot k_{4})z_{2}^{3}\right]\left[-k_{3}\cdot k_{4}-(-k_{2}\cdot
k_{4})z_{2}\right]$
$\displaystyle+\frac{1}{2!}\left(\frac{1}{2}\right)^{2}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot k_{3}}\left[-k_{3}\cdot
k_{4}-(-k_{2}\cdot k_{4})z_{2}^{2}\right]^{2}$
$\displaystyle+\frac{1}{2!1!}\left(\frac{1}{1}\right)^{2}\left(\frac{1}{2}\right)\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot k_{3}}\left[-k_{3}\cdot
k_{4}-(-k_{2}\cdot k_{4})z_{2}\right]^{2}\left[-k_{3}\cdot k_{4}-(-k_{2}\cdot
k_{4})z_{2}^{2}\right]$
$\displaystyle+\frac{1}{4!}\left(\frac{1}{1}\right)^{4}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot k_{3}}\left[-k_{3}\cdot
k_{4}-(-k_{2}\cdot k_{4})z_{2}\right]^{4}.$ (4.1)

There are $5$ terms in $R_{4,3}^{4}$ which are all in the form of Eq.(2.34)
and thus can be expressed in terms of the Lauricella functions. It is
interesting to note that Eq.(4.1) is a generalization of the last line of
Eq.(3.5) of Veneziano amplitude to $5$-point KN amplitude. In particular, the
coefficients of each of the $5$ terms in both expressions are the same.
Similarly, $R_{4,3}^{1}$, $R_{4,3}^{2}$ and $R_{4,3}^{3}$ can all be expressed
in terms of the Lauricella functions. Moreover, because of the solvability or
the $SL(K+3,C)$ symmetry of the LSSA, we expect relations among various
residues $R_{4,3}^{J}$. Indeed, if we define

$\displaystyle F_{M}\left(z_{2},z_{1}\right)$
$\displaystyle=\underset{b+c=M,b,c=0}{\sum}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{3}\cdot
k_{4}\right)_{c}}{c!}\left(z_{2}\right)^{b}\left(z_{1}\right)^{c}$ (4.2)
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{3}\cdot
k_{4}\right)^{N}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[z_{1}^{r}-\left(-\frac{k_{2}\cdot
k_{4}}{k_{3}\cdot k_{4}}\right)z_{2}^{r}\right]^{N_{r}},$ (4.3)

then by Eq.(3.20)

$R_{4,3}^{M}=\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot k_{3}}F_{M}\left(z_{2},1\right).$ (4.4)

Moreover, one can show the following iteration relation

$F_{M}\left(z_{2},z_{1}\right)=\frac{1}{M}\underset{N=0}{\overset{M-1}{\sum}}F_{N}\left(z_{2},z_{1}\right)\left[\left(-k_{2}\cdot
k_{4}\right)\left(z_{2}\right)^{M-N}+\left(-k_{3}\cdot
k_{4}\right)\left(z_{1}\right)^{M-N}\right],$ (4.5)

which expresses $F_{M}$ in terms of $F_{M-1}$, $F_{M-2}$, $....$, $F_{1}$,
$F_{0}\equiv 1$. Similar iteration relation holds for the residue
$R_{3,4}^{M}$. So one obtains an iteration relation among all residues of
$5$-point KN amplitude. Presumably, these kind of relations among various
residues $R_{4,3}^{J}$ $($or $R_{3,4}^{M})$ resulting from the $SL(K+3,C)$
symmetry soften the SSA in the hard scattering limit.

We believe that this iteration relation and its implication on the softness of
some kind of hard $5$-point SSA are generalization of the well-known soft
$4$-point Vaneziano amplitude in the hard scattering limit. To further study
this issue, one first needs to define and identify the specific hard
scattering limit of $5$-point SSA as there are $5$ kinematic variables for the
$5$-point function instead of $2$ for the well-known $4$-point function.
Indeed this iteration relation persists for all higher point SSA. We give one
more example here. The explicit form of $R_{5,3}^{3}$ of the $6$-point KN
amplitude can be calculated from Eq.(3.38)

$\displaystyle R_{5,3}^{3}$
$\displaystyle=\frac{1}{1!3}\int_{0}^{1}dz_{3}\int_{0}^{z_{3}}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}z_{3}^{\hat{k}_{1}\cdot k_{3}}\left(z_{3}-z_{2}\right)^{k_{2}\cdot
k_{3}}\left(1-z_{2}\right)^{k_{2}\cdot k_{4}}\left(1-z_{3}\right)^{k_{3}\cdot
k_{4}}$ $\displaystyle\times\left[\left(-k_{2}\cdot
k_{5}\right)\left(z_{2}\right)^{3}+\left(-k_{3}\cdot
k_{5}\right)\left(z_{3}\right)^{3}+\left(-k_{4}\cdot
k_{5}\right)\left(1\right)^{3}\right]$
$\displaystyle+\frac{1}{1!1}\frac{1}{1!2}\int_{0}^{1}dz_{3}\int_{0}^{z_{3}}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}z_{3}^{\hat{k}_{1}\cdot k_{3}}\left(z_{3}-z_{2}\right)^{k_{2}\cdot
k_{3}}\left(1-z_{2}\right)^{k_{2}\cdot k_{4}}\left(1-z_{3}\right)^{k_{3}\cdot
k_{4}}$ $\displaystyle\times\left[\left(-k_{2}\cdot
k_{5}\right)\left(z_{2}\right)+\left(-k_{3}\cdot
k_{5}\right)\left(z_{3}\right)+\left(-k_{4}\cdot
k_{5}\right)\left(1\right)\right]^{2}$
$\displaystyle\times\left[\left(-k_{2}\cdot
k_{5}\right)\left(z_{2}\right)^{2}+\left(-k_{3}\cdot
k_{5}\right)\left(z_{3}\right)^{2}+\left(-k_{4}\cdot
k_{5}\right)\left(1\right)^{2}\right]$
$\displaystyle+\frac{1}{3!1}\int_{0}^{1}dz_{3}\int_{0}^{z_{3}}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}z_{3}^{\hat{k}_{1}\cdot k_{3}}\left(z_{3}-z_{2}\right)^{k_{2}\cdot
k_{3}}\left(1-z_{2}\right)^{k_{2}\cdot k_{4}}\left(1-z_{3}\right)^{k_{3}\cdot
k_{4}}$ $\displaystyle\times\left[\left(-k_{2}\cdot
k_{5}\right)\left(z_{2}\right)+\left(-k_{3}\cdot
k_{5}\right)\left(z_{3}\right)+\left(-k_{4}\cdot
k_{5}\right)\left(1\right)\right]^{3}.$ (4.6)

There are $3$ terms in $R_{5,3}^{3}$ which can all be expressed in terms of
the Lauricella functions as will be shown in section V. It is interesting to
note that Eq.(4.6) is a generalization of the third line of Eq.(3.5) of
Veneziano amplitude to $6$-point KN amplitude. In particular, the coefficients
of each of the $3$ terms in both expressions are the same. Similarly,
$R_{5,3}^{1}$ and $R_{5,3}^{2}$ can all be expressed in terms of the
Lauricella functions. If we define

$F_{M}\left(z_{2,}z_{3},z_{1}\right)=\underset{\begin{array}[c]{c}M=c+e+f\\\
c,e,f=0\end{array}}{\sum}\frac{\left(-k_{2}\cdot
k_{5}\right)_{c}}{c!}\frac{\left(-k_{3}\cdot
k_{5}\right)_{e}}{e!}\frac{\left(-k_{4}\cdot
k_{5}\right)_{f}}{f!}\left(z_{2}\right)^{c}\left(z_{3}\right)^{e}\left(z_{1}\right)^{f},$
(4.7)

then

$R_{5,3}^{M}=\int_{0}^{1}dz_{3}\int_{0}^{z_{3}}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}z_{3}^{\hat{k}_{1}\cdot k_{3}}\left(z_{3}-z_{2}\right)^{k_{2}\cdot
k_{3}}\left(1-z_{2}\right)^{k_{2}\cdot k_{4}}\left(1-z_{3}\right)^{k_{3}\cdot
k_{4}}F_{M}\left(z_{2,}z_{3},1\right).$ (4.8)

One can show the following iteration relation

$F_{M}\left(z_{2},z_{3},z_{1}\right)=\frac{1}{M}\underset{N=0}{\overset{M-1}{\sum}}F_{N}\left(z_{2},z_{3},z_{1}\right)\left[\left(-k_{2}\cdot
k_{5}\right)\left(z_{2}\right)^{M-N}+\left(-k_{3}\cdot
k_{5}\right)\left(z_{3}\right)^{M-N}+\left(-k_{4}\cdot
k_{5}\right)\left(z_{1}\right)^{M-N}\right],$ (4.9)

which again expresses $F_{M}$ in terms of $F_{M-1}$, $F_{M-2}$, $....$,
$F_{1}$, $F_{0}\equiv 1$.

One can easily generalize this iteration property to any residue
$R_{m+1,n-m+1}^{M}$ of $n$-point KN amplitudes. Indeed, by using Eq.(3.119)
and Eq.(A.16), we can express the residue $R_{m+1,n-m+1}^{M}$ in terms of
$F_{M}$

$\displaystyle
R_{m+1,n-m+1}^{M}\left(\hat{1},2,\cdots,m,m+1,\cdots,n-1,\hat{n}\right)$
$\displaystyle=\int_{0}^{1}dz_{m-1}\cdots\int_{0}^{z_{3}}dz_{2}\int_{0}^{1}dw_{n-2}\cdots\int_{0}^{w_{m+2}}dw_{m+1}\underset{1\leq
j<i\leq m}{\prod}\left(z_{i}-z_{j}\right)^{k_{i}\cdot k_{j}}$
$\displaystyle\times\underset{P=m\leq j<i\leq
n-1}{\prod}\left(w_{i}-w_{j}\right)^{k_{i}\cdot
k_{j}}\underset{\left\\{N_{r}\right\\}}{\sum}\prod_{r=1}\frac{\left[\underset{2\leq
j\leq m,m+1\leq i\leq n-1}{\sum}\left(-k_{j}\cdot
k_{i}\right)\left(\frac{z_{j}}{w_{i}}\right)^{r}\right]}{N_{r}!r^{N_{r}}}^{N_{r}}$
$\displaystyle=\int_{0}^{1}dz_{m-1}\cdots\int_{0}^{z_{3}}dz_{2}\int_{0}^{1}dw_{n-2}\cdots\int_{0}^{w_{m+2}}dw_{m+1}\underset{1\leq
j<i\leq m}{\prod}\left(z_{i}-z_{j}\right)^{k_{i}\cdot k_{j}}$
$\displaystyle\times\underset{P=m\leq j<i\leq
n-1}{\prod}\left(w_{i}-w_{j}\right)^{k_{i}\cdot k_{j}}F_{M}\text{
}(\frac{z_{j}}{w_{i}})$ (4.10)

where $j=2,\cdots,m;i=m+1,\cdots,n-1$ and $z_{m}=w_{n-1}=1$. Similarly, one
can show the following iteration relation

$F_{M}=\frac{1}{M}\underset{N=0}{\overset{M-1}{\sum}}F_{N}\left[\underset{2\leq
j\leq m,m+1\leq i\leq n-1}{\sum}\left(-k_{j}\cdot
k_{i}\right)\left(\frac{z_{j}}{w_{i}}\right)^{M-N}\right],$ (4.11)

which again expresses $F_{M}$ in terms of $F_{M-1}$, $F_{M-2}$, $....$,
$F_{1}$, $F_{0}\equiv 1$. More investigation needs to be done on this
interesting issue.

##  V Expressing n-point SSA in terms of the LSSA

We note from Eq.(3.13) that the $5$-point KN amplitude can be expressed in
terms of residues $R_{3,4}^{M}$ and $R_{4,3}^{M}$ after doing $1$-step
recursion. These residues can then be written as $4$-point SSA with one tensor
leg (excited string state) or the LSSA. In this section, we will first use the
shifting method to demonstrate that all $5$-point SSA with arbitrary five
tensor legs can be expressed in terms of the LSSA. In general, we can use
mathematical induction, together with the on-shell recursion and the shifting
principle, to show that all $n$-point SSA including the KN amplitudes can be
expressed in terms of the LSSA. We begins with the discussion of the $4$-point
case and introduce the shifting method.

###  V.1 Four-point SSA with tensor legs and the Shifting principle

We first consider the two tachyons and two vectors amplitude KLT

$\displaystyle A\left(\zeta_{1},k_{1};k_{2};\zeta_{3},k_{3};k_{4}\right)$
$\displaystyle=\int dz_{2}\left|z_{13}z_{14}z_{34}\right|\left\langle
e^{ik_{1}X_{1}+i\zeta_{1}\partial_{1}X_{1}}e^{ik_{2}X_{2}}e^{ik_{3}X_{3}+i\zeta_{3}\partial_{3}X_{3}}e^{ik_{4}X_{4}}\right\rangle|_{m.l.}$
$\displaystyle=\int
dz_{2}\left|z_{13}z_{14}z_{34}\right|\left|z_{12}\right|^{k_{1}\cdot
k_{2}}\left|z_{13}\right|^{k_{1}\cdot k_{3}}\left|z_{14}\right|^{k_{1}\cdot
k_{4}}\left|z_{23}\right|^{k_{2}\cdot k_{3}}\left|z_{24}\right|^{k_{2}\cdot
k_{4}}\left|z_{34}\right|^{k_{3}\cdot k_{4}}$
$\displaystyle\times\exp\left[\zeta_{1}\cdot\left(\frac{k_{2}}{z_{21}}+\frac{k_{3}}{z_{31}}+\frac{k_{4}}{z_{41}}\right)+\zeta_{3}\cdot\left(\frac{k_{1}}{z_{13}}+\frac{k_{2}}{z_{23}}+\frac{k_{4}}{z_{43}}\right)+\frac{\zeta_{1}\cdot\zeta_{3}}{z_{13}^{2}}\right].$
(5.1)

After the standard $SL(2,R)$ gauge fixing, we obtain

$\displaystyle A\left(\zeta_{1},k_{1};k_{2};\zeta_{3},k_{3};k_{4}\right)$
$\displaystyle=\int_{0}^{1}dz_{2}z_{2}^{k_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot k_{3}}\left[\left(\frac{\zeta_{1}\cdot
k_{2}}{z_{2}}+\frac{\zeta_{1}\cdot k_{3}}{1}\right)\left(\frac{-\zeta_{3}\cdot
k_{1}}{1}-\frac{\zeta_{3}\cdot
k_{2}}{1-z_{2}}\right)+\frac{\zeta_{1}\cdot\zeta_{3}}{1}\right].$ (5.2)

Plug in the kinematic $k_{1}\cdot k_{2}=\frac{-s}{2}-1$, $k_{2}\cdot
k_{3}=\frac{-t}{2}-1$ and after some algebra, we get

$\displaystyle A\left(\zeta_{1},k_{1};k_{2};\zeta_{3},k_{3};k_{4}\right)$
$\displaystyle=$
$\displaystyle\int_{0}^{1}dz_{2}z_{2}^{\frac{-s}{2}-1}\left(1-z_{2}\right)^{\frac{-t}{2}-1}\left[-\left(\frac{\zeta_{1}\cdot
k_{2}}{z_{2}}+\frac{\zeta_{1}\cdot k_{3}}{1}\right)\left(\frac{\zeta_{3}\cdot
k_{1}}{1}+\frac{\zeta_{3}\cdot
k_{2}}{1-z_{2}}\right)+\frac{\zeta_{1}\cdot\zeta_{3}}{1}\right]$
$\displaystyle=\left(\zeta_{1}\cdot k_{2}\right)\left(\zeta_{3}\cdot
k_{4}\right)\int_{0}^{1}dz_{2}z_{2}^{\frac{-s}{2}-2}\left(1-z_{2}\right)^{\frac{-t}{2}-2}\left(1-\left(-\frac{\zeta_{1}\cdot
k_{3}}{\zeta_{1}\cdot
k_{2}}z_{2}\right)\right)\left(1-\left(-\frac{\zeta_{3}\cdot
k_{1}}{\zeta_{3}\cdot k_{4}}z_{2}\right)\right)$ (5.3)
$\displaystyle+\zeta_{1}\cdot\zeta_{3}\int_{0}^{1}dz_{2}z_{2}^{\frac{-s}{2}-1}\left(1-z_{2}\right)^{\frac{-t}{2}-1},$
(5.4)

which can be written in terms of the Lauricella functions with $K=0,2$
presented in Eq.(2.35) of section II. It is important to note that in order to
get the $F_{D}^{\left(2\right)}$ term in the above amplitude calculation, we
avoid doing expansion for the product of binomial in the integrand in
Eq.(5.3). This is a great simplification for amplitude calculation with higher
$K$.

The calculation of the above $4$-point SSA with tensor legs are similar to
that of the $4$-tachyon Veneziano amplitude except shifting some appropriate
kinematic variables. This is not surprising as the similar shiftings already
show up in the more familiar $4$-point SSA with three tachyons and one vector
SSA which can be calculated to be
($\left(z_{1},z_{3},z_{4}\right)=\left(0,1,\infty\right)$)

$\displaystyle A_{4}\left(k_{1};\zeta_{2},k_{2};k_{3};k_{4}\right)$
$\displaystyle=\int_{0}^{1}dz_{2}z_{2}^{k_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}}\times(-\zeta_{2})\cdot\left[\frac{k_{1}}{z_{2}}-\frac{k_{3}}{1-z_{2}}\right]$
$\displaystyle=(-\zeta_{2}\cdot k_{1})A_{4}^{(1)}-(-\zeta_{2}\cdot
k_{3})A_{4}^{(2)}$ (5.5)

where ($T=-k_{2}\cdot k_{3}$, $s=-2k_{1}\cdot k_{2}-2$)

$\displaystyle
A_{4}^{(1)}=\underset{1}{\underbrace{\frac{2}{-s-2}}}+\underset{\alpha_{-1}}{\underbrace{T\frac{2}{-s+0}}}+(\underset{\alpha_{-2}}{\underbrace{\frac{T}{2}}}+\underset{\alpha_{-1}^{2}}{\underbrace{\frac{T^{2}}{2!}\frac{1}{1^{2}})}}\frac{2}{-s+2}$
$\displaystyle+\underset{\alpha_{-3}}{\underbrace{(\frac{T}{3}}}+\underset{\alpha_{-1}\alpha_{-2}}{\underbrace{\frac{T^{2}}{1!1!}\frac{1}{1^{1}}\frac{1}{2^{1}}}}+\underset{\alpha_{-1}^{3}}{\underbrace{\frac{T^{3}}{3!})}}\frac{2}{-s+4}+.....$
(5.6)

and

$\displaystyle
A_{4}^{(2)}=\underset{\alpha_{-1}}{\underbrace{\frac{2}{-s+2}}}+\underset{\alpha_{-2},\alpha_{-1}^{2}}{\underbrace{(T+1)\frac{2}{-s+4}}}+\underset{\alpha_{-3},\alpha_{-1}\alpha_{-2},\alpha_{-1}^{3}}{\underbrace{\left[\frac{(T+1)}{2}+\frac{(T+1)^{2}}{2!}\frac{1}{1^{2}}\right]\frac{2}{-s+6}}}$
$\displaystyle+\underset{\alpha_{-4},\alpha_{-1}\alpha_{-3},\alpha_{-2}^{2},\alpha_{-1}^{2}\alpha_{-2},\alpha_{-1}^{4}}{\underbrace{\left[\frac{(T+1)}{3}+\frac{(T+1)^{2}}{1!1!}\frac{1}{1^{1}}\frac{1}{2^{1}}+\frac{(T+1)^{3}}{3!}\right]\frac{2}{-s+8}}}+.....\text{.}$
(5.7)

These results correspond to shift the kinematic variables $T$ and $s$ of
results of the original $4$-point Veneziano amplitude presented in Eq.(3.6) to
$T+1$ and $s+4$, and they agree with results obtained by operator method
calculation in stringbcfw .

The SSA of three vectors and one tachyon can be similarly calculated to be

$\displaystyle
A\left(\zeta_{1},k_{1};\zeta_{2},k_{2};\zeta_{3},k_{3};k_{4}\right)$
$\displaystyle=\left(\zeta_{1}\cdot k_{2}\right)\left(\zeta_{2}\cdot
k_{2}\right)\left(\zeta_{3}\cdot
k_{4}\right)\frac{\Gamma\left(\frac{-s}{2}-2\right)\Gamma\left(\frac{-t}{2}-2\right)}{\Gamma\left(-\frac{s}{2}-\frac{t}{2}-4\right)}$
$\displaystyle\times
F_{D}^{\left(3\right)}\left(\frac{-s}{2}-2;-1,-1,-1;-\frac{s}{2}-\frac{t}{2}-4;\frac{\zeta_{1}\cdot
k_{3}}{\zeta_{1}\cdot k_{2}},-\frac{\zeta_{2}\cdot k_{4}}{\zeta_{2}\cdot
k_{1}},-\frac{\zeta_{3}\cdot k_{1}}{\zeta_{3}\cdot k_{4}}\right)$
$\displaystyle+\left(\zeta_{1}\cdot\zeta_{2}\right)\left(-\zeta_{3}\cdot
k_{4}\right)\frac{\Gamma\left(\frac{-s}{2}-2\right)\Gamma\left(\frac{-t}{2}-1\right)}{\Gamma\left(-\frac{s}{2}-\frac{t}{2}-3\right)}F_{D}^{\left(1\right)}\left(\frac{-s}{2}-2;-1;-\frac{s}{2}-\frac{t}{2}-3;-\frac{\zeta_{3}\cdot
k_{1}}{\zeta_{3}\cdot k_{4}}\right)$
$\displaystyle+\left(\zeta_{2}\cdot\zeta_{3}\right)\left(-\zeta_{1}\cdot
k_{2}\right)\frac{\Gamma\left(\frac{-s}{2}-1\right)\Gamma\left(\frac{-t}{2}-2\right)}{\Gamma\left(-\frac{s}{2}-\frac{t}{2}-3\right)}F_{D}^{\left(1\right)}\left(\frac{-s}{2}-1;-1;-\frac{s}{2}-\frac{t}{2}-3;\frac{\zeta_{1}\cdot
k_{3}}{\zeta_{1}\cdot k_{2}}\right)$
$\displaystyle+\left(\zeta_{1}\cdot\zeta_{3}\right)\left(-\zeta_{2}\cdot
k_{1}\right)\frac{\Gamma\left(\frac{-s}{2}-1\right)\Gamma\left(\frac{-t}{2}-1\right)}{\Gamma\left(-\frac{s}{2}-\frac{t}{2}-2\right)}F_{D}^{\left(1\right)}\left(\frac{-s}{2}-1;-1;-\frac{s}{2}-\frac{t}{2}-2;-\frac{\zeta_{2}\cdot
k_{4}}{\zeta_{2}\cdot k_{1}}\right)$ (5.8)

which can be written in terms of the Lauricella functions with $K=1,3$. The
expression with $4$ terms in Eq.(5.8) is to be compared with expression with
$12$ terms calculated in KLT .

As another example, the SSA of one rank-$2$ tensor, one vector and one tachyon
can be calculated to be ($\zeta_{\mu\nu}=\zeta_{\mu}^{(1)}\zeta_{\nu}^{(2)}$)

$\displaystyle A\left(\zeta_{\mu\nu},k_{1};k_{2};\zeta,k_{3};k_{4}\right)$
$\displaystyle=-\zeta_{\mu\nu}k_{2}^{\mu}k_{2}^{\nu}\left(\zeta\cdot
k_{4}\right)\frac{\Gamma\left(\frac{-s}{2}-2\right)\Gamma\left(\frac{-t}{2}-2\right)}{\Gamma\left(-\frac{s}{2}-\frac{t}{2}-4\right)}$
$\displaystyle\times
F_{D}^{\left(3\right)}\left(\frac{-s}{2}-2;-1,-1,-1;-\frac{s}{2}-\frac{t}{2}-4;-\frac{\zeta^{(1)}\cdot
k_{3}}{\zeta^{(1)}\cdot k_{2}},-\frac{\zeta^{(2)}\cdot k_{3}}{\zeta^{(2)}\cdot
k_{2}},-\frac{\zeta\cdot k_{1}}{\zeta\cdot k_{4}}\right)$
$\displaystyle+\zeta_{\mu\nu}\zeta^{\mu}k_{2}^{\nu}\left(\zeta\cdot
k_{4}\right)\frac{\Gamma\left(\frac{-s}{2}-1\right)\Gamma\left(\frac{-t}{2}-1\right)}{\Gamma\left(-\frac{s}{2}-\frac{t}{2}-2\right)}F_{D}^{\left(1\right)}\left(\frac{-s}{2}-1;-1;-\frac{s}{2}-\frac{t}{2}-2;-\frac{\zeta^{(1)}\cdot
k_{3}}{\zeta^{(1)}\cdot k_{2}}\right)$
$\displaystyle+\zeta_{\mu\nu}\zeta^{\mu}k_{2}^{\nu}\left(\zeta\cdot
k_{4}\right)\frac{\Gamma\left(\frac{-s}{2}-1\right)\Gamma\left(\frac{-t}{2}-1\right)}{\Gamma\left(-\frac{s}{2}-\frac{t}{2}-2\right)}F_{D}^{\left(1\right)}\left(\frac{-s}{2}-1;-1;-\frac{s}{2}-\frac{t}{2}-2;-\frac{\zeta^{(2)}\cdot
k_{3}}{\zeta^{(2)}\cdot k_{2}}\right),$ (5.9)

which again can be written in terms of the Lauricella functions with $K=1,3$.
In conclusion, we see that the calculation of SSA expressed in terms of the
Lauricella functions is much simpler than the traditional calculation.

Finally for the 26D open bosonic string, a general state at mass level $N$

$M^{2}=2\left(N-1\right),\text{ and }N=\sum_{r>0}rn_{r},$ (5.10)

is of the form

$\left|P\right\rangle=\prod_{r>0}\prod_{\sigma=1}^{n_{r}}\frac{\varepsilon_{r}^{\left(\sigma\right)}\cdot\alpha_{-r}}{\sqrt{n_{r}!r^{n_{r}}}}|0,k\rangle$
(5.11)

where $\varepsilon_{r}^{\left(\sigma\right)}$ are polarizations with
$\sigma=1,\cdots n_{r}$ for each operator $\alpha_{-r}$. The corresponding
string vertex is

$V\left(k,z\right)=\prod_{r>0}\prod_{\sigma=1}^{n_{r}}\varepsilon_{r}^{\left(\sigma\right)}\cdot\partial_{z}^{r}X\left(z\right)e^{ik\cdot
X\left(z\right)}.$ (5.12)

For 4-point amplitude $i=1,2,3,4$, let

$X_{i}=X\left(z_{i}\right)\text{ and
}k_{i}=\left(E_{i},\vec{p}_{i}\right)\text{ with }\sum
k_{i}=\sum\vec{p}_{i}=0,$ (5.13)

and we define the Mandelstam variables as $s=-\left(k_{1}+k_{2}\right)^{2}$,
$t=-\left(k_{2}+k_{3}\right)^{2}$. The $4$-point SSA with four general string
states can be calculated as

$\displaystyle A$ $\displaystyle=\int
dz_{2}\left|z_{13}z_{14}z_{34}\right|\left\langle\prod_{i=1}^{4}V_{i}\left(k_{i},z_{i}\right)\right\rangle$
$\displaystyle=\int
dz_{2}\left|z_{13}z_{14}z_{34}\right|\left\langle\prod_{i=1}^{4}\exp\left(ik_{i}\cdot
X_{i}+i\sum_{r_{i}>0}\sum_{\sigma_{i}=1}^{n_{r_{i}}}\varepsilon_{r_{i}}^{\left(\sigma_{i}\right)}\cdot\partial_{i}^{r_{i}}X_{i}\right)\right\rangle_{m.l.}$
$\displaystyle=\int
dz_{2}\left|z_{13}z_{14}z_{34}\right|\left|z_{12}\right|^{k_{1}\cdot
k_{2}}\left|z_{13}\right|^{k_{1}\cdot k_{3}}\left|z_{14}\right|^{k_{1}\cdot
k_{4}}\left|z_{23}\right|^{k_{2}\cdot k_{3}}\left|z_{24}\right|^{k_{2}\cdot
k_{4}}\left|z_{34}\right|^{k_{3}\cdot k_{4}}$
$\displaystyle\cdot\exp\left(\sum_{r_{i}>0}\sum_{\sigma_{i}=1}^{n_{r_{i}}}\sum_{i}^{4}\sum_{j\neq
i}\frac{-\varepsilon_{r_{i}}^{\left(\sigma_{i}\right)}\cdot
k_{j}}{z_{ji}^{r_{i}}}+\sum_{r_{i},r_{j}>0}\sum_{\sigma_{i}=1}^{n_{r_{i}}}\sum_{\sigma_{j}=1}^{n_{r_{j}}}\sum_{i<j=2}^{4}\frac{-\varepsilon_{r_{i}}^{\left(\sigma_{i}\right)}\varepsilon_{r_{j}}^{\left(\sigma_{j}\right)}}{z_{ji}^{r_{i}}z_{ij}^{r_{j}}}\right)_{m.l.}$
(5.14)

where the lower label $m.l.$ means that we only keep multi-linear terms with
each polarization $\varepsilon_{r_{i}}^{\left(\sigma_{i}\right)}$. The
amplitude can be expressed as

$\displaystyle A$ $\displaystyle=\int_{0}^{1}dz_{2}z_{2}^{k_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot k_{3}}$
$\displaystyle\lim_{z_{4}\rightarrow\infty}\cdot\sum_{\left\\{\varepsilon_{r_{i}}^{\left(\sigma_{i}\right)}\right\\}}\left[\prod\limits_{i=1}^{4}\prod\limits_{\left\\{r_{i},\sigma_{i}\right\\}}\left(\sum_{j\neq
i}\frac{\varepsilon_{r_{i}}^{\left(\sigma_{i}\right)}\cdot
k_{j}}{z_{ji}^{r_{i}}}\right)\cdot\prod\limits_{i<j=2}^{4}\prod\limits_{\left\\{r_{i},\sigma_{i};r_{j},\sigma_{j}\right\\}}\frac{\varepsilon_{r_{i}}^{\left(\sigma_{i}\right)}\varepsilon_{r_{j}}^{\left(\sigma_{j}\right)}}{z_{ji}^{r_{i}}z_{ij}^{r_{j}}}\right]_{z_{1}=0,z_{3}=1}$
(5.15)

where the configurations
$\left\\{\varepsilon_{r_{i}}^{\left(\sigma_{i}\right)}\right\\}$ satisfy

$\prod\limits_{i=1}^{4}\prod\limits_{\left\\{r_{i},\sigma_{i}\right\\}}\varepsilon_{r_{i}}^{\left(\sigma_{i}\right)}\cdot\prod\limits_{i<j=2}^{4}\prod\limits_{\left\\{r_{i},\sigma_{i};r_{j},\sigma_{j}\right\\}}\left(\varepsilon_{r_{i}}^{\left(\sigma_{i}\right)}\varepsilon_{r_{j}}^{\left(\sigma_{j}\right)}\right)=\prod_{i=1}^{4}\prod_{r_{i}>0}\prod_{\sigma_{i}=1}^{n_{r_{i}}}\varepsilon_{r_{i}}^{\left(\sigma_{i}\right)},$
(5.16)

which ensures the multi-linear condition. For each configuration
$\left\\{\varepsilon_{r_{i}}^{\left(\sigma_{i}\right)}\right\\}$, it is
straightforward to transform Eq.(5.15) to the standard integral form of the
Lauricella function. In the first line of Eq.(5.15), we have fixed
$SL\left(2,R\right)$ by choosing
$\left(z_{1},z_{3},z_{4}\right)=\left(0,1,\infty\right)$ and drop a factor
which is independent of momentum $k_{i}$. It is much more tricky to take the
$z_{4}\rightarrow\infty$ in the second line of Eq.(5.15). For
$V_{4}\left(k_{i},z_{i}\right)$ to be a higher spin state, one needs to apply
stringy massive Ward identities or decoupling of massive ZNS to obtain a
consistent result LLY3 .

We conclude this subsection by noting that the calculation of $4$-point SSA
with tensor legs are similar to that of the $4$-tachyon Veneziano amplitude
except shifting some appropriate kinematic variables. It is easy to generalize
this shifting method to arbitrary $n$-point SSA and obtain the following

Shifting principle : If the $n$-point KN amplitude can be expressed in terms
of the LSSA, then one can use the shifting method to calculate all $n$-point
SSA with tensor legs (excited string states) and express them in terms of the
LSSA.

We will present more examples to demonstrate the shifting method in the
following subsections. We will see that while the string on-shell recursion
relation can be used to reduce higher point KN amplitudes to the lower SSA and
express them in terms of the LSSA, the shifting method can be used to express
SSA with tensor legs in terms of the LSSA.

###  V.2 Five-point SSA with tensor legs

In section III, we have shown that the two residues $R_{3,4}^{M}$ and
$R_{4,3}^{M}$ of $5$-point KN amplitude in Eq.(3.13) can be expressed in terms
of the Lauricella functions. In this subsection, we will use the shifting
method to calculate more examples of $5$-point SSA with tensor legs and
express them in terms of the LSSA. In general, according to the shifting
principle, all $5$-point SSA can be expressed in terms of LSSA. Unfortunately,
to explicitly calculate the higher point ($n\geq 5$) SSA with multi-tensor
legs is quite lengthy. The case of $5$-point SSA with $1$-tensor legs, for
example, will show up after doing $1$-step recursion on the $6$-point KN
amplitude. To begin with, let’s consider the four tachyons, one vector SSA
($\left(z_{1},z_{4},z_{5}\right)=\left(0,1,\infty\right)$)

$\displaystyle A_{5}\left(\zeta_{1},k_{1};k_{2};k_{3};k_{4};k_{5}\right)$
$\displaystyle=\int_{0}^{1}dz_{3}\int_{0}^{z_{3}}dz_{2}z_{2}^{k_{2}\cdot
k_{1}}z_{3}^{k_{3}\cdot k_{1}}\left(1-z_{3}\right)^{k_{4}\cdot
k_{3}}\left(1-z_{2}\right)^{k_{4}\cdot
k_{2}}\left(z_{3}-z_{2}\right)^{k_{3}\cdot k_{2}}$
$\displaystyle\times(-\zeta_{1})\cdot\left[\frac{k_{2}}{z_{2}}+\frac{k_{3}}{z_{3}}+\frac{k_{4}}{1}\right]$
$\displaystyle=A_{5}^{(1)}+A_{5}^{(2)}+A_{5}^{(3)}$ (5.17)

where we have used the result of Eq.(3.30) by taking $N=1=M$. Mathematically,
all three terms of Eq.(5.17) are similar to the $5$-point KN amplitude and we
can apply the shifting principle to do the calculation. We will explicitly
calculate the three terms separately. We first note that the original
$5$-point KN amplitude in Eq.(3.9) can be rewritten as

$\displaystyle A_{5}^{KN}$
$\displaystyle=\underset{a,b,c=0}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{3}\cdot k_{4}\right)_{c}}{c!}$
$\displaystyle\times\frac{2}{2k_{1}\cdot
k_{2}+2\left(a+b+1\right)}\frac{2}{2(k_{1}\cdot k_{2}+k_{2}\cdot
k_{3}+k_{1}\cdot k_{3})+2\left(b+c+2\right)}.$ (5.18)

Similarly, the three terms in Eq.(5.17) can be calculated to be

$\displaystyle A_{5}^{(1)}$ $\displaystyle=(-\zeta_{1}\cdot
k_{2})\int_{0}^{1}dz_{3}\int_{0}^{z_{3}}dz_{2}z_{2}^{k_{2}\cdot
k_{1}-1}z_{3}^{k_{3}\cdot k_{1}}\left(1-z_{3}\right)^{k_{4}\cdot
k_{3}}\left(1-z_{2}\right)^{k_{4}\cdot
k_{2}}\left(z_{3}-z_{2}\right)^{k_{3}\cdot k_{2}}$
$\displaystyle=(-\zeta_{1}\cdot
k_{2})\underset{a,b,c=0}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{3}\cdot k_{4}\right)_{c}}{c!}$
$\displaystyle\times\frac{2}{2k_{1}\cdot
k_{2}+2\left(a+b\right)}\frac{2}{2(k_{1}\cdot k_{2}+k_{2}\cdot
k_{3}+k_{1}\cdot k_{3})+2\left(b+c+1\right)},$ (5.19) $\displaystyle
A_{5}^{(2)}$ $\displaystyle=(-\zeta_{1}\cdot
k_{3})\int_{0}^{1}dz_{3}\int_{0}^{z_{3}}dz_{2}z_{2}^{k_{2}\cdot
k_{1}}z_{3}^{k_{3}\cdot k_{1}-1}\left(1-z_{3}\right)^{k_{4}\cdot
k_{3}}\left(1-z_{2}\right)^{k_{4}\cdot
k_{2}}\left(z_{3}-z_{2}\right)^{k_{3}\cdot k_{2}}$
$\displaystyle=(-\zeta_{1}\cdot
k_{3})\underset{a,b,c=0}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{3}\cdot k_{4}\right)_{c}}{c!}$
$\displaystyle\times\frac{2}{2k_{1}\cdot
k_{2}+2\left(a+b+1\right)}\frac{2}{2(k_{1}\cdot k_{2}+k_{2}\cdot
k_{3}+k_{1}\cdot k_{3})+2\left(b+c-1\right)}$ (5.20)

and

$\displaystyle A_{5}^{(3)}$ $\displaystyle=(-\zeta_{1}\cdot
k_{4})\int_{0}^{1}dz_{3}\int_{0}^{z_{3}}dz_{2}z_{2}^{k_{2}\cdot
k_{1}}z_{3}^{k_{3}\cdot k_{1}}\left(1-z_{3}\right)^{k_{4}\cdot
k_{3}}\left(1-z_{2}\right)^{k_{4}\cdot
k_{2}}\left(z_{3}-z_{2}\right)^{k_{3}\cdot k_{2}}$
$\displaystyle(-\zeta_{1}\cdot
k_{4})\underset{a,b,c=0}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{3}\cdot k_{4}\right)_{c}}{c!}$ (5.21)
$\displaystyle\times\frac{2}{2k_{1}\cdot
k_{2}+2\left(a+b+1\right)}\frac{2}{2(k_{1}\cdot k_{2}+k_{2}\cdot
k_{3}+k_{1}\cdot k_{3})+2\left(b+c+2\right)}.$ (5.22)

We can now apply string on-shell recursion relation to obtain

$\displaystyle A_{5}^{(1)}$ $\displaystyle=(-\zeta_{1}\cdot
k_{2})\underset{{}_{M}}{\sum}\frac{2}{2k_{1}\cdot k_{2}+2M}R_{3,4}^{(1),M}$
$\displaystyle+(-\zeta_{1}\cdot
k_{2})\underset{{}_{M}}{\sum}\frac{2}{2(k_{1}\cdot k_{2}+k_{2}\cdot
k_{3}+k_{1}\cdot k_{3})+2\left(M+1\right)}R_{4,3}^{(1),M}$ (5.23)

where the two residues

$R_{3,4}^{(1),M}=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{2}\cdot
k_{3}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}-M}\left(1-z_{2}\right)^{k_{3}\cdot
k_{4}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(-\frac{k_{2}\cdot
k_{4}}{k_{2}\cdot k_{3}}\right)z_{2}^{r}\right]^{N_{r}}$ (5.24)

and

$R_{4,3}^{(1),M}=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{3}\cdot
k_{4}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}-1}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(-\frac{k_{2}\cdot
k_{4}}{k_{3}\cdot k_{4}}\right)z_{2}^{r}\right]^{N_{r}}$ (5.25)

can all be expressed in terms of the Lauricella functions. Similarly, we can
calculate

$\displaystyle A_{5}^{(2)}$ $\displaystyle=(-\zeta_{1}\cdot
k_{3})\underset{{}_{M}}{\sum}\frac{2}{2k_{1}\cdot
k_{2}+2\left(M+1\right)}R_{3,4}^{(2),M}$ $\displaystyle+(-\zeta_{1}\cdot
k_{2})\underset{{}_{M}}{\sum}\frac{2}{2(k_{1}\cdot k_{2}+k_{2}\cdot
k_{3}+k_{1}\cdot k_{3})+2\left(M+1\right)}R_{4,3}^{(2),M}$ (5.26)

where

$\displaystyle R_{3,4}^{(2),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{2}\cdot
k_{3}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}-M-1}\left(1-z_{2}\right)^{k_{3}\cdot
k_{4}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(-\frac{k_{2}\cdot
k_{4}}{k_{2}\cdot k_{3}}\right)z_{2}^{r}\right]^{N_{r}},$ (5.27)
$\displaystyle R_{4,3}^{(2),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{3}\cdot
k_{4}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(-\frac{k_{2}\cdot
k_{4}}{k_{3}\cdot k_{4}}\right)z_{2}^{r}\right]^{N_{r}};$ (5.28)

and

$\displaystyle A_{5}^{(3)}$ $\displaystyle=(-\zeta_{1}\cdot
k_{4})\underset{{}_{M}}{\sum}\frac{2}{2k_{1}\cdot
k_{2}+2\left(M+1\right)}R_{3,4}^{(3),M}$ $\displaystyle+(-\zeta_{1}\cdot
k_{4})\underset{{}_{M}}{\sum}\frac{2}{2(k_{1}\cdot k_{2}+k_{2}\cdot
k_{3}+k_{1}\cdot k_{3})+2\left(M+2\right)}R_{4,3}^{(3),M}$ (5.29)

where

$\displaystyle R_{3,4}^{(3),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{2}\cdot
k_{3}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}-M}\left(1-z_{2}\right)^{k_{3}\cdot
k_{4}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(-\frac{k_{2}\cdot
k_{4}}{k_{2}\cdot k_{3}}\right)z_{2}^{r}\right]^{N_{r}},$ (5.30)
$\displaystyle R_{4,3}^{(3),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{3}\cdot
k_{4}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}-1}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(-\frac{k_{2}\cdot
k_{4}}{k_{3}\cdot k_{4}}\right)z_{2}^{r}\right]^{N_{r}}.$ (5.31)

We note that the calculations above are straightforward and the results
correspond to shift the appropriate kinematic variables of results of the
original $5$-point KN amplitude. According to the shifting principle, one can
generalize the above calculation to the four tachyons and one higher tensor
SSA and express it in terms of the LSSA.

It is lengthy to explicitly calculate $5$-point SSA with three tachyons and
two general tensor legs SSA. However, it is straightforward though still
lengthy to explicitly calculate the three tachyons, two vectors SSA

$\displaystyle
A_{5}\left(\zeta_{1},k_{1};k_{2};k_{3};\zeta_{4},k_{4};k_{5}\right)$
$\displaystyle=\int_{0}^{1}dz_{3}\int_{0}^{z_{3}}dz_{2}z_{2}^{k_{2}\cdot
k_{1}}z_{3}^{k_{3}\cdot k_{1}}\left(1-z_{3}\right)^{k_{4}\cdot
k_{3}}\left(1-z_{2}\right)^{k_{4}\cdot
k_{2}}\left(z_{3}-z_{2}\right)^{k_{3}\cdot k_{2}}$
$\displaystyle\times\left[\zeta_{1}\cdot\left(\frac{k_{2}}{z_{2}}+\frac{k_{3}}{z_{3}}+\frac{k_{4}}{1}\right)(-\zeta_{4})\cdot\left(\frac{k_{1}}{1}+\frac{k_{2}}{1-z_{2}}+\frac{k_{3}}{1-z_{3}}\right)+\frac{\zeta_{1}\cdot\zeta_{4}}{1}\right]$
$\displaystyle=A_{5}^{(i,j)}+A_{5}^{(0,0)},\text{ \ }i,j=1,2,3,$ (5.32)

and express it in terms of the sum of LSSA. To do this, one needs to calculate
$3\times 3+1=10$ $5$-point SSA. Mathematically, all $10$ amplitude
calculations are similar to the calculation of $5$-point KN amplitude in
Eq.(5.18). The first $3$ amplitudes can be identified to be

$\displaystyle A_{5}^{(1,1)}$ $\displaystyle=(\zeta_{4}\cdot
k_{1})A_{5}^{(1)},$ (5.33) $\displaystyle A_{5}^{(2,1)}$
$\displaystyle=(\zeta_{4}\cdot k_{1})A_{5}^{(2)},$ (5.34) $\displaystyle
A_{5}^{(3,1)}$ $\displaystyle=(\zeta_{4}\cdot k_{1})A_{5}^{(3)}$ (5.35)

where $k_{4}$ of a tachyon with $k_{4}^{2}=2$ in $A_{5}^{(j)}$ is replaced by
$k_{4}$ of a vector with $k_{4}^{2}=0$. The other $7$ amplitudes correspond to
shift some appropriate kinematic variables of Eq.(5.18) and can be calculated
to be

$\displaystyle A_{5}^{(1,2)}$ $\displaystyle=(-\zeta_{1}\cdot
k_{2})(\zeta_{4}\cdot
k_{2})\underset{a,b,c=0}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}+1\right)_{b}}{b!}\frac{\left(-k_{3}\cdot k_{4}\right)_{c}}{c!}$
$\displaystyle\times\frac{2}{2k_{1}\cdot
k_{2}+2\left(a+b\right)}\frac{2}{2(k_{1}\cdot k_{2}+k_{2}\cdot
k_{3}+k_{1}\cdot k_{3})+2\left(b+c+1\right)},$ (5.36) $\displaystyle
A_{5}^{(2,2)}$ $\displaystyle=(-\zeta_{1}\cdot k_{3})(\zeta_{4}\cdot
k_{2})\underset{a,b,c=0}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}+1\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}+1\right)_{b}}{b!}\frac{\left(-k_{3}\cdot k_{4}\right)_{c}}{c!}$
$\displaystyle\times\frac{2}{2k_{1}\cdot
k_{2}+2\left(a+b+1\right)}\frac{2}{2(k_{1}\cdot k_{2}+k_{2}\cdot
k_{3}+k_{1}\cdot k_{3})+2\left(b+c+1\right)},$ (5.37) $\displaystyle
A_{5}^{(3,2)}$ $\displaystyle=(-\zeta_{1}\cdot k_{4})(\zeta_{4}\cdot
k_{2})\underset{a,b,c=0}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}+1\right)_{b}}{b!}\frac{\left(-k_{3}\cdot k_{4}\right)_{c}}{c!}$
$\displaystyle\times\frac{2}{2k_{1}\cdot
k_{2}+2\left(a+b+1\right)}\frac{2}{2(k_{1}\cdot k_{2}+k_{2}\cdot
k_{3}+k_{1}\cdot k_{3})+2\left(b+c+2\right)},$ (5.38) $\displaystyle
A_{5}^{(1,3)}$ $\displaystyle=(-\zeta_{1}\cdot k_{2})(\zeta_{4}\cdot
k_{3})\underset{a,b,c=0}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{3}\cdot k_{4}+1\right)_{c}}{c!}$
$\displaystyle\times\frac{2}{2k_{1}\cdot
k_{2}+2\left(a+b\right)}\frac{2}{2(k_{1}\cdot k_{2}+k_{2}\cdot
k_{3}+k_{1}\cdot k_{3})+2\left(b+c+1\right)},$ (5.39) $\displaystyle
A_{5}^{(2,3)}$ $\displaystyle=(-\zeta_{1}\cdot k_{3})(\zeta_{4}\cdot
k_{3})\underset{a,b,c=0}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}+1\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{3}\cdot k_{4}+1\right)_{c}}{c!}$
$\displaystyle\times\frac{2}{2k_{1}\cdot
k_{2}+2\left(a+b+1\right)}\frac{2}{2(k_{1}\cdot k_{2}+k_{2}\cdot
k_{3}+k_{1}\cdot k_{3})+2\left(b+c+1\right)},$ (5.40) $\displaystyle
A_{5}^{(3,3)}$ $\displaystyle=(-\zeta_{1}\cdot k_{4})(\zeta_{4}\cdot
k_{3})\underset{a,b,c=0}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{3}\cdot k_{4}+1\right)_{c}}{c!}$
$\displaystyle\times\frac{2}{2k_{1}\cdot
k_{2}+2\left(a+b+1\right)}\frac{2}{2(k_{1}\cdot k_{2}+k_{2}\cdot
k_{3}+k_{1}\cdot k_{3})+2\left(b+c+2\right)}$ (5.41)

and

$\displaystyle A_{5}^{(0,0)}$
$\displaystyle=(\zeta_{1}\cdot\zeta_{4})\underset{a,b,c=0}{\overset{\infty}{{\displaystyle\sum}}}\frac{\left(-k_{2}\cdot
k_{3}\right)_{a}}{a!}\frac{\left(-k_{2}\cdot
k_{4}\right)_{b}}{b!}\frac{\left(-k_{3}\cdot k_{4}\right)_{c}}{c!}$
$\displaystyle\times\frac{2}{2k_{1}\cdot
k_{2}+2\left(a+b+1\right)}\frac{2}{2(k_{1}\cdot k_{2}+k_{2}\cdot
k_{3}+k_{1}\cdot k_{3})+2\left(b+c+2\right)}.$ (5.42)

We can now apply string on-shell recursion relation to obtain

$\displaystyle A_{5}^{(1,2)}$
$\displaystyle=\underset{{}_{M}}{\sum}\frac{2(-\zeta_{1}\cdot
k_{2})(\zeta_{4}\cdot k_{2})}{2k_{1}\cdot k_{2}+2M}R_{3,4}^{(1,2),M}$
$\displaystyle+\underset{{}_{M}}{\sum}\frac{2(-\zeta_{1}\cdot
k_{2})(\zeta_{4}\cdot k_{2})}{2(k_{1}\cdot k_{2}+k_{2}\cdot k_{3}+k_{1}\cdot
k_{3})+2\left(M+1\right)}R_{4,3}^{(1,2),M}$ (5.43)

where the two residues

$R_{3,4}^{(1,2),M}=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{2}\cdot
k_{3}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}-M}\left(1-z_{2}\right)^{k_{3}\cdot
k_{4}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(\frac{-k_{2}\cdot
k_{4}+1}{k_{2}\cdot k_{3}}\right)z_{2}^{r}\right]^{N_{r}}$ (5.44)

and

$R_{4,3}^{(1,2),M}=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{3}\cdot
k_{4}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}-1}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(\frac{-k_{2}\cdot
k_{4}+1}{k_{3}\cdot k_{4}}\right)z_{2}^{r}\right]^{N_{r}}$ (5.45)

can all be expressed in terms of the Lauricella functions. Similarly, we can
calculate

$\displaystyle A_{5}^{(2,2)}$
$\displaystyle=\underset{{}_{M}}{\sum}\frac{2(-\zeta_{1}\cdot
k_{2})(\zeta_{4}\cdot k_{2})}{2k_{1}\cdot k_{2}+2(M+1)}R_{3,4}^{(2,2),M}$
$\displaystyle+\underset{{}_{M}}{\sum}\frac{2(-\zeta_{1}\cdot
k_{2})(\zeta_{4}\cdot k_{2})}{2(k_{1}\cdot k_{2}+k_{2}\cdot k_{3}+k_{1}\cdot
k_{3})+2\left(M+2\right)}R_{4,3}^{(2,2),M}$ (5.46)

where

$\displaystyle R_{3,4}^{(2,2),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{2}\cdot
k_{3}+1\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}-M-1}\left(1-z_{2}\right)^{k_{3}\cdot
k_{4}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(\frac{-k_{2}\cdot
k_{4}+1}{k_{2}\cdot k_{3}-1}\right)z_{2}^{r}\right]^{N_{r}},$ (5.47)
$\displaystyle R_{4,3}^{(2,2),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{3}\cdot
k_{4}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}-1}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}-1}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(\frac{-k_{2}\cdot
k_{4}+1}{k_{3}\cdot k_{4}}\right)z_{2}^{r}\right]^{N_{r}};$ (5.48)
$\displaystyle A_{5}^{(3,2)}$
$\displaystyle=\underset{{}_{M}}{\sum}\frac{2(-\zeta_{1}\cdot
k_{4})(\zeta_{4}\cdot k_{2})}{2k_{1}\cdot k_{2}+2(M+1)}R_{3,4}^{(3,2),M}$
$\displaystyle+\underset{{}_{M}}{\sum}\frac{2(-\zeta_{1}\cdot
k_{4})(\zeta_{4}\cdot k_{2})}{2(k_{1}\cdot k_{2}+k_{2}\cdot k_{3}+k_{1}\cdot
k_{3})+2\left(M+2\right)}R_{4,3}^{(3,2),M}$ (5.49)

where

$\displaystyle R_{3,4}^{(3,2),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{2}\cdot
k_{3}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}-M}\left(1-z_{2}\right)^{k_{3}\cdot
k_{4}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(\frac{-k_{2}\cdot
k_{4}+1}{k_{2}\cdot k_{3}}\right)z_{2}^{r}\right]^{N_{r}},$ (5.50)
$\displaystyle R_{4,3}^{(3,2),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{3}\cdot
k_{4}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(\frac{-k_{2}\cdot
k_{4}+1}{k_{3}\cdot k_{4}}\right)z_{2}^{r}\right]^{N_{r}}.$ (5.51)

The next $3$ amplitudes can be calculated to be

$\displaystyle A_{5}^{(1,3)}$
$\displaystyle=\underset{{}_{M}}{\sum}\frac{2(-\zeta_{1}\cdot
k_{2})(\zeta_{4}\cdot k_{3})}{2k_{1}\cdot k_{2}+2(M+1)}R_{3,4}^{(1,3),M}$
$\displaystyle+\underset{{}_{M}}{\sum}\frac{2(-\zeta_{1}\cdot
k_{2})(\zeta_{4}\cdot k_{3})}{2(k_{1}\cdot k_{2}+k_{2}\cdot k_{3}+k_{1}\cdot
k_{3})+2\left(M+2\right)}R_{4,3}^{(1,3),M}$ (5.52)

where

$\displaystyle R_{3,4}^{(1,3),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{2}\cdot
k_{3}+1\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}-M}\left(1-z_{2}\right)^{k_{3}\cdot
k_{4}-1}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(\frac{-k_{2}\cdot
k_{4}+1}{k_{2}\cdot k_{3}-1}\right)z_{2}^{r}\right]^{N_{r}},$ (5.53)
$\displaystyle R_{4,3}^{(1,3),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{3}\cdot
k_{4}+1\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}-1}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(\frac{-k_{2}\cdot
k_{4}+1}{k_{3}\cdot k_{4}-1}\right)z_{2}^{r}\right]^{N_{r}};$ (5.54)
$\displaystyle A_{5}^{(2,3)}$
$\displaystyle=\underset{{}_{M}}{\sum}\frac{2(-\zeta_{1}\cdot
k_{3})(\zeta_{4}\cdot k_{3})}{2k_{1}\cdot k_{2}+2(M+1)}R_{3,4}^{(2,3),M}$
$\displaystyle+\underset{{}_{M}}{\sum}\frac{2(-\zeta_{1}\cdot
k_{3})(\zeta_{4}\cdot k_{3})}{2(k_{1}\cdot k_{2}+k_{2}\cdot k_{3}+k_{1}\cdot
k_{3})+2\left(M+2\right)}R_{4,3}^{(2,3),M}$ (5.55)

where

$\displaystyle R_{3,4}^{(2,3),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{2}\cdot
k_{3}+1\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}-M-1}\left(1-z_{2}\right)^{k_{3}\cdot
k_{4}-1}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(\frac{-k_{2}\cdot
k_{4}}{k_{2}\cdot k_{3}-1}\right)z_{2}^{r}\right]^{N_{r}},$ (5.56)
$\displaystyle R_{4,3}^{(2,3),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{3}\cdot
k_{4}+1\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}-1}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(\frac{-k_{2}\cdot
k_{4}}{k_{3}\cdot k_{4}-1}\right)z_{2}^{r}\right]^{N_{r}};$ (5.57)

and

$\displaystyle A_{5}^{(3,3)}$
$\displaystyle=\underset{{}_{M}}{\sum}\frac{2(-\zeta_{1}\cdot
k_{4})(\zeta_{4}\cdot k_{3})}{2k_{1}\cdot k_{2}+2(M+1)}R_{3,4}^{(3,3),M}$
$\displaystyle+\underset{{}_{M}}{\sum}\frac{2(-\zeta_{1}\cdot
k_{4})(\zeta_{4}\cdot k_{3})}{2(k_{1}\cdot k_{2}+k_{2}\cdot k_{3}+k_{1}\cdot
k_{3})+2\left(M+2\right)}R_{4,3}^{(3,3),M}$ (5.58)

where

$\displaystyle R_{3,4}^{(3,3),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{2}\cdot
k_{3}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}-M}\left(1-z_{2}\right)^{k_{3}\cdot
k_{4}-1}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(\frac{-k_{2}\cdot
k_{4}}{k_{2}\cdot k_{3}-1}\right)z_{2}^{r}\right]^{N_{r}},$ (5.59)
$\displaystyle R_{4,3}^{(3,3),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{3}\cdot
k_{4}+1\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(\frac{-k_{2}\cdot
k_{4}}{k_{3}\cdot k_{4}-1}\right)z_{2}^{r}\right]^{N_{r}}.$ (5.60)

Finally, the last one is

$\displaystyle A_{5}^{(0,0)}$
$\displaystyle=\underset{{}_{M}}{\sum}\frac{2\zeta_{1}\cdot\zeta_{4}}{2k_{1}\cdot
k_{2}+2(M+1)}R_{3,4}^{(0,0),M}$
$\displaystyle+\underset{{}_{M}}{\sum}\frac{2\zeta_{1}\cdot\zeta_{4}}{2(k_{1}\cdot
k_{2}+k_{2}\cdot k_{3}+k_{1}\cdot k_{3})+2\left(M+2\right)}R_{4,3}^{(0,0),M}$
(5.61)

where

$\displaystyle R_{3,4}^{(0,0),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{2}\cdot
k_{3}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}-M}\left(1-z_{2}\right)^{k_{3}\cdot
k_{4}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(\frac{-k_{2}\cdot
k_{4}}{k_{2}\cdot k_{3}}\right)z_{2}^{r}\right]^{N_{r}},$ (5.62)
$\displaystyle R_{4,3}^{(0,0),M}$
$\displaystyle=\underset{\left\\{N_{r}\right\\}}{\sum}\left(-k_{3}\cdot
k_{4}\right)^{N}\int_{0}^{1}dz_{2}z_{2}^{\hat{k}_{1}\cdot
k_{2}}\left(1-z_{2}\right)^{k_{2}\cdot
k_{3}}\prod_{r=1}\frac{1}{N_{r}!r^{N_{r}}}\left[1-\left(\frac{-k_{2}\cdot
k_{4}}{k_{3}\cdot k_{4}}\right)z_{2}^{r}\right]^{N_{r}}.$ (5.63)

This completes the long calculation of Eq.(5.32), and we have successfully
expressed the $5$-point SSA with three tachyons and two vectors in terms of
the $4$-point LSSA. According to the shifting principle, one can generalize
the above calculation to the three tachyons and two tensor SSA and express it
in terms of the LSSA. In general, it is easy to see that the above
calculations can be generalized to $5$-point SSA with arbitrary five tensor
legs and express them in terms of the LSSA.

###  V.3 Six-point SSA with tensor legs

One important application of the above calculations is to prove that the
$6$-point KN amplitude can be expressed in terms of the Lauricella functions.
Indeed, we note that Eq.(5.17) can be obtained from Eq.(3.30) by taking
$N=1=M$. So what we have just shown above is that $R_{3,5}^{1}$, the $M=1$
case of the residue $R_{3,5}^{M}$ in Eq.(3.30) can be expressed in terms of
the Lauricella functions. In fact, by using Eq.(3.32) and Eq.(5.17), we have
the identification

$R_{3,5}^{1}(\hat{k}_{1}+k_{2},k_{3},k_{4},k_{5},k_{6})=A_{5}(\zeta_{1}=k_{2},\hat{k}_{1}+k_{2};k_{3};k_{4};k_{5};k_{6})$
(5.64)

and the following expression

$\displaystyle R_{3,5}^{1}(k=\hat{k}_{1}+k_{2},k_{3},k_{4},k_{5},k_{6})$
$\displaystyle=\underset{{}_{M}}{\sum}\frac{-2k_{2}\cdot k_{3}}{2k\cdot
k_{3}+2M}R_{3,4}^{(1),M}+\underset{{}_{M}}{\sum}\frac{-2k_{2}\cdot
k_{3}}{2(k\cdot k_{3}+k_{3}\cdot k_{4}+k\cdot k_{4})+2(M+1)}R_{4,3}^{(1),M}$
$\displaystyle+\underset{{}_{M}}{\sum}\frac{-2k_{2}\cdot k_{4}}{2k\cdot
k_{3}+2(M+1)}R_{3,4}^{(2),M}+\underset{{}_{M}}{\sum}\frac{-2k_{2}\cdot
k_{4}}{2(k\cdot k_{3}+k_{3}\cdot k_{4}+k\cdot k_{4})+2(M+1)}R_{4,3}^{(2),M}$
$\displaystyle+\underset{{}_{M}}{\sum}\frac{-2k_{2}\cdot k_{5}}{2k\cdot
k_{3}+2(M+1)}R_{3,4}^{(3),M}+\underset{{}_{M}}{\sum}\frac{-2k_{2}\cdot
k_{5}}{2(k\cdot k_{3}+k_{3}\cdot k_{4}+k\cdot k_{4})+2(M+1)}R_{4,3}^{(3),M},$
(5.65)

which expresses one of the residue of the $6$-point KN amplitude $R_{3,5}^{1}$
in terms of the LSSA. Similar consideration applies to $M=2,3,4\cdots$ cases.
For example, for $M=2$ case, by Eq.(3.32), it is straightforward to calculate

$\displaystyle
R_{3,5}^{2}=\int_{0}^{1}dw_{4}\int_{0}^{w_{4}}dw_{3}w_{3}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{3}}w_{4}^{\left(\hat{k}_{1}+k_{2}\right)\cdot
k_{4}}\left(1-w_{3}\right)^{k_{3}\cdot k_{5}}\left(1-w_{4}\right)^{k_{4}\cdot
k_{5}}\left(w_{4}-w_{3}\right)^{k_{3}\cdot k_{4}}$
$\displaystyle\times\left\\{\begin{array}[c]{c}\frac{1}{2!1^{2}}\left[\left(-k_{2}\cdot
k_{3}\right)\left(\frac{1}{w_{3}}\right)+\left(-k_{2}\cdot
k_{4}\right)\left(\frac{1}{w_{4}}\right)+\left(-k_{2}\cdot
k_{5}\right)\left(\frac{1}{1}\right)\right]^{2}\\\
+\frac{1}{1!2^{1}}\left[\left(-k_{2}\cdot
k_{3}\right)\left(\frac{1}{w_{3}^{2}}\right)+\left(-k_{2}\cdot
k_{4}\right)\left(\frac{1}{w_{4}^{2}}\right)+\left(-k_{2}\cdot
k_{5}\right)\left(\frac{1}{1^{2}}\right)\right]\end{array}\right\\},$ (5.68)

and express it in terms of the LSSA. The calculation of each term in Eq.(5.68)
is similar to that of the $5$-point KN amplitude except shifting some
kinematic variables. Thus $R_{3,5}^{M}$ are LSSA and so are $R_{5,3}^{M}$.

Figure 3: Expressing the 6-point KN amplitude in terms of the LSSA by the
first recursion Figure 4: Expressing the 6-point KN amplitude in terms of the
LSSA by the second recursion

On the other hand, the result of Eq.(3.35) tells us that the residue
$R_{4,4}^{M}$ is a sum of products of the Lauricella functions after term by
term integrations for a given $M$. We conclude that the $6$-point KN amplitude
is a LSSA. It is interesting to note that to obtain the LSSA form of the
residue $R_{4,4}^{M}$, one needs only do one-step recursion from the $6$-point
KN amplitude. But to obtain the LSSA form of the residue $R_{3,5}^{M}$
($R_{5,3}^{M}$), one needs to do $2$-step recursion or $(recursion)^{2}$. See
Fig.3 and Fig.4.

In fact, to show that the $6$-point KN amplitude is a LSSA, one needs only do
$1$-step recursion to express it in terms of the lower $5$-point and $4$-point
amplitudes as was shown in Fig.3. Since we have shown that all $5$-point and
$4$-point SSA are LSSA, the $6$-point KN amplitude is a LSSA. However, to
explicitly calculate the LSSA form of the $6$-point KN amplitude, one needs to
do the second recursion as was shown in Fig.4, and the calculation will be
very lengthy.

The next step is to use the shifting principle to argue that all $6$-point SSA
with arbitrary six tensor legs can be expressed in terms of the LSSA. The
argument is obvious although the detailed calculation will again be very
lengthy.

###  V.4 Seven-point SSA with tensor legs

For the case of $7$-point SSA, we first do $1$-step recursion on $7$-point KN
amplitude. The subamplitude $R_{3,6}^{P,R}$ of the residue $R_{3,6}^{M}$ we