Veneziano amplitude pole residue
= Veneziano amplitude pole residue
{c}
{title2=$\operatorname*{Res}_{s=n-1}A=-\frac1{n!}\prod_{j=2}^{n+1}(t+j)$}
In units $\alpha'=1$, generic fixed $t$ gives <simple poles> at $s=n-1$, $n\geq0$, with <residue> $-\prod_{j=2}^{n+1}(t+j)/n!$. The empty product is one. A <residue> zero removes that pole, and fixed $t$ must be away from its own channel singularities.