Fix $z_1=0$, $z_2=1$, and $z_3=\infty$, and write $z_4=z$. For the closed-string tachyon, $k_i^2=4/\alpha'$. Hence
$$
\alpha'k_1\mathbin\cdot k_4=-\frac{\alpha'u}{2}-4,
\qquad
\alpha'k_2\mathbin\cdot k_4=-\frac{\alpha't}{2}-4,
$$
where momentum conservation was used in the second identity. Taking the limit at infinity together with the ghost factor gives the <Virasoro–Shapiro amplitude>
$$
\boxed{A_4=\delta^{26}\!\left(\sum_{i=1}^4k_i\right)g_c^2
\int_{\mathbb C}d^2z\,
|z|^{-\alpha'u/2-4}|1-z|^{-\alpha't/2-4}}.
$$
Choosing a different set of three punctures to fix merely applies a Möbius transformation and relabels identical external tachyons. Exchanging the roles of the relevant punctures interchanges $s$ and $t$ without changing the integral, proving crossing symmetry under $s\leftrightarrow t$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 306 / 3 / 3 / Solution

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The Virasoro–Shapiro amplitude is the crossing-symmetric tree-level four-point amplitude of closed-string tachyons. Its worldsheet integral is a complex beta integral, and its poles display the closed-string spectrum.


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