Type II four-graviton tree amplitude and alpha-prime expansion

Apply [F4] to $s,t,u$ and add the three logarithms. The linear contribution is $2\gamma (s+t+u)=0$. The cubic and quintic contributions are $\frac{2\zeta (3)}{3}(s^3+t^3+u^3)$ and $\frac{2\zeta (5)}{5}(s^5+t^5+u^5)$. Using $s+t+u=0$ gives $s^3+t^3+u^3=3stu$ and $s^5+t^5+u^5=\frac{5}{2}stu(s^2+t^2+u^2)$, so the summed logarithm begins $2\zeta (3)stu+\zeta (5)stu(s^2+t^2+u^2)+O({\rho }^7)$, where $\rho =\max (\mathrm{\mid }s\mathrm{\mid },\mathrm{\mid }t\mathrm{\mid },\mathrm{\mid }u\mathrm{\mid })$. More generally, for every odd $m\ge 3$, $s^m+t^m+u^m$ vanishes on each of the lines $s=0$, $t=0$, $u=0$ in the plane $s+t+u=0$; the factor theorem makes each such symmetric power sum divisible by $stu$. The locally convergent logarithm series is consequently divisible by $stu$.

Set $F(s,t):=G(s,t,-s-t)-1$, which is holomorphic near $(0,0)$ by [F4]. The factor $R(w):=\mathrm{\Gamma }(1-w)\mathrm{/}\mathrm{\Gamma }(1+w)$ satisfies $R(w)R(-w)=1$ near $w=0$, so $F$ vanishes on each of the lines $s=0$, $t=0$, and $s+t=0$. To see the product divisibility holomorphically, write $F(s,t)=s{F}_1(s,t)$ with $F_1(s,t)={\int }_0^1{\mathrm{\partial }}_{s}F(rs,t)dr$. Since $F(s,0)=0$, one has $F_1(s,0)=0$ for $s\mathrm{\ne }0$, and then for all $s$ by continuity; the same one-variable division formula gives $F_1=t{F}_2$ with $F_2$ holomorphic. On $s+t=0$ with $s\mathrm{\ne }0$, this implies $F_2(s,-s)=0$, and continuity gives the same at $s=0$. In coordinates $(s,y=s+t)$, one-variable division by $y$ therefore gives $F_2(s,y)=y{F}_3(s,y)$ with $F_3$ holomorphic. Hence $F=st(s+t)F_3=-stu{F}_3$, and $H=F\mathrm{/}(stu)$ is analytic. From step 1.1 and $s+t+u=0$, the degree-three and degree-five parts of $L=\log G$ are $L_3=2\zeta (3)stu$ and $L_5=\zeta (5)stu(s^2+t^2+u^2)$; all remaining terms of $L$ have degree at least seven. Thus $G-1=L_3+L_5+\frac{1}{2}{L}_3^2$ plus terms of degree at least seven. Dividing by $stu$ gives the displayed constant, quadratic and cubic terms of $H$; because $H$ is analytic, the remaining Taylor terms have degree at least four. This establishes the contact expansion without a nonuniform big-O assertion at a channel pole.

The invariants $s,t,u$ are dimensionless and proportional to ${\alpha }^{\mathrm{\prime }}$ times quadratic momentum invariants. The massless pole is proportional to $(stu{)}^{-1}$, so a constant contact term relative to it carries the factor $stu=O(({\alpha }^{\mathrm{\prime }}{)}^3)$ at fixed momenta. The prefactor ${\kappa }_{10}^2{g}_s^{-2}{\mathcal{R}}^4$ is common to both terms by [A1]; therefore the first analytic term has the stated relative ${\alpha }^{\mathrm{\prime }3}$ order. The distinct sphere-normalized-vertex power in [F3] is not used to alter the source normalization.