 Solution

Continue the loop momentum to $D=4-\epsilon$ dimensions, keeping $\mathrm{tr}1=4$ and the external physical polarization vectors four dimensional. Introduce Feynman parameters $x,y\ge 0$, $x+y\le 1$, denoting this integration triangle by $\mathcal{T}$. Completing the square givesOdd terms in $k$ integrate to zero. Lorentz invariance of dimensional regularization gives $k_{\mu }{k}_{\nu }\mapsto g_{\mu \nu }{k}^2/D$. The general projected numerator therefore becomes, under the Feynman integral,Only the $k^2$ term can give an ultraviolet divergence. Define, with the conventional scale factor implicit,The gamma function formula givesThus the pole of the numerator integral, including $2{\int }_{\mathcal{T}}dxdy=1$, isThe coefficients $(4,1,1)$ satisfy this condition. A nonzero local divergence would have to be removed by a counterterm. The Ward identity forbids a local $h{A}_{\mu }{A}^{\mu }$ term, and the first electromagnetic gauge-invariant coupling with one Higgs boson and two photons has the form $h{F}_{\mu \nu }{F}^{\mu \nu }$, of engineering dimension five. It is not an independent counterterm of the Standard Model, a renormalizable quantum field theory. The top quark contribution is separately electromagnetic gauge-invariant; no additional W boson loop Feynman diagram is needed to cancel this pole.

The finite remainder must still be retained. For $A=4C$,Putting $D=4$ before evaluating the divergent Feynman integral would incorrectly set this finite contribution to zero.