 Solution

The Ward identity implies $q^{\mu }{\Pi }_{\mu \nu }=0$. Lorentz invariance then gives ${\Pi }_{\mu \nu }=(q^2{g}_{\mu \nu }-q_{\mu }{q}_{\nu })\Pi$, so $g^{\mu \nu }{\Pi }_{\mu \nu }=(d-1)q^2\Pi$. The gamma matrices satisfy ${\gamma }_{\mu }/k{\gamma }^{\mu }=(2-d)/k$. Odd traces vanish, and thereforeSubstituting gives the requested scalar coefficient:To extract the ultraviolet divergence without introducing a Feynman parameter, set $A=(k+q{)}^2-m^2$, $B=k^2-m^2$ and useTranslation invariance of dimensional regularization gives ${\int }_k{A}^{-1}={\int }_k{B}^{-1}={\mathcal{I}}_1$. ConsequentlyIn the pole coefficient one may set $d=4$: the bracket becomesThus the ultraviolet pole of QED vacuum polarization isThe cancellation of the apparent $m^2/q^2$ divergence is essential: gauge invariance forbids a photon mass counterterm. The scalar result at $q^2=0$ is understood by continuation after this cancellation.