Veneziano amplitude pole residue (${\mathrm{Res}}_{s=n-1}A=-\frac{1}{n!}{\prod }_{j=2}^{n+1}(t+j)$)

In units ${\alpha }^{\prime }=1$, generic fixed $t$ gives simple poles at $s=n-1$, $n\ge 0$, 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.

For identical open-string tachyons, $s+t+u=-4/{\alpha }^{\prime }$. At ${\alpha }^{\prime }s=n-1$, the ordered Veneziano amplitude residues in the $s,t$ and $s,u$ terms differ by $(-1{)}^n$, so their symmetric sum cancels odd-level $s$-channel poles at generic $t$. In particular the massless vector does not couple to this identical-scalar singlet amplitude. It remains a state of the open-string spectrum and is visible in suitable ordered or color-dependent amplitudes. Pole absence due to an external-state selection rule must not be interpreted as absence of the exchanged state from the theory.

The pole at $s=0$ has residue $-(t+2)=-(t-u)/2$ for unit-slope tachyons. It is proportional to conserved scalar-pair vector currents and linear in the scattering-angle cosine, exhibiting a massless spin-one intermediate string state.