= Solution
{c}
Continue the <loop momentum> to $D=4-\varepsilon$ dimensions, keeping $\operatorname{tr}1=4$ and the external physical <polarization vectors> four dimensional. Introduce <Feynman parameters> $x,y\geq0$, $x+y\leq1$, denoting this integration triangle by $\mathcal T$. Completing the square gives
$$
\frac1{D_0D_1D_2}=2\int_{\mathcal T}dx\,dy\,\frac1{(k^2-\Delta+i0)^3},\qquad
k=l+xp_1-yp_2,\qquad \Delta=m^2-sxy.
$$
Odd 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>,
$$
g_{\mu\nu}\left[\left(\frac AD-C\right)k^2+C(m^2-s/2+sxy)\right]
+(B-Axy)p_{2,\mu}p_{1,\nu}.
$$
Only the $k^2$ term can give an <ultraviolet divergence>. Define, with the conventional scale factor implicit,
$$
J_n(\Delta)=\mu_R^{\varepsilon}\int\frac{d^Dk}{(2\pi)^D}(k^2-\Delta+i0)^{-n},\qquad
I_2=J_2+\Delta J_3.
$$
The <gamma function> formula gives
$$
J_2=\frac{i}{16\pi^2}\frac2\varepsilon+O(1),\qquad
J_3=-\frac{i}{32\pi^2\Delta}+O(\varepsilon).
$$
Thus the pole of the numerator integral, including $2\int_{\mathcal T}dx\,dy=1$, is
$$
g_{\mu\nu}\left(\frac A4-C\right)\frac{i}{8\pi^2\varepsilon},
\qquad \boxed{A=4C.}
$$
The 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 $hA_\mu A^\mu$ term, and the first electromagnetic <gauge-invariant> coupling with one <Higgs boson> and two <photons> has the form $hF_{\mu\nu}F^{\mu\nu}$, 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.
\b[The finite remainder must still be retained.] For $A=4C$,
$$
\frac AD-C=\frac{C\varepsilon}{4-\varepsilon},\qquad
\lim_{\varepsilon\to0}\left(\frac AD-C\right)I_2=\frac{iC}{32\pi^2}.
$$
Putting $D=4$ before evaluating the divergent <Feynman integral> would incorrectly set this finite contribution to zero.
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