 Solution

The superficial degree of divergence of this scalar electrodynamics three-point vertex is one. Its ultraviolet pole is consequently a local expression at most linear in the external four-momenta. The allowed counterterm Lagrangian, with the charge-conjugation and Hermiticity properties of scalar electrodynamics, contains a single one-derivative scalar electrodynamics three-point vertex:Thus the coefficient ${\delta }_1$ is independent of the chosen external four-momenta. Setting $k^{\prime }=0$ leaves $-ie{\delta }_1{k}^{\mu }$, so a nonzero generic $k$ still determines that coefficient. It does not determine the full finite scattering amplitude at other four-momenta.

Keep $m\ne 0$ and generic off shell $k$ while extracting the ultraviolet pole. These choices avoid spurious infrared divergences; setting all scales to zero would instead produce scaleless integrals that can conceal the ultraviolet pole.