Fix $z_1=0$, $z_2=1$, and $z_4=\infty$, and write $z_3=z$. Its scattering equation is
$$
\frac{p_3\cdot p_1}{z}+\frac{p_3\cdot p_2}{z-1}=0,
$$
so in the hard limit
$$
z=\frac{p_3\cdot p_1}{p_3\cdot p_1+p_3\cdot p_2}
=\frac{t}{t+u}=-\frac ts+O(m_{\rm tachyon}^2/s).
$$
Evaluating the <Koba-Nielsen factor> at this saddle and using $s+t+u=O(m_{\rm tachyon}^2)$ yields, with the logarithms defined by analytic continuation,
$$
\boxed{A^{(4,0)}\sim g_s^2\delta^{(26)}\!\left(\sum_i p_i\right)
\exp\left[-\frac{\alpha'}2(s\log s+t\log t+u\log u+\cdots)\right]}.
$$
The omitted terms grow more slowly than the displayed fixed-angle $s\log s$ terms.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 306 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-306/3/c/solution