= Solution
{c}
Keep $m>0$ and set $p=p'=0$, which is an <off shell> choice. The numerator is
$$
N^\mu=\gamma^\nu(-\not k+m)\gamma^\mu(-\not k+m)\gamma_\nu.
$$
Terms odd in $k$ integrate to zero. <Lorentz covariance> and <dimensional regularization> give, inside this integral,
$$
k_\rho k_\sigma\longmapsto\frac{k^2}{D}g_{\rho\sigma},\qquad
\not k\gamma^\mu\not k\longmapsto\frac{2-D}{D}k^2\gamma^\mu.
$$
Applying the <gamma matrices> identity from part (ii) once more yields
$$
N^\mu\longmapsto\left[\frac{(D-2)^2}{D}k^2+(2-D)m^2\right]\gamma^\mu.
$$
Consequently
$$
\boxed{iV^\mu_{\mathrm{loop}}(0,0)=-e_D^3\gamma^\mu\int\frac{d^Dk}{(2\pi)^D}\frac{\alpha k^2+\beta m^2}{(k^2+i0)(k^2-m^2+i0)^2},\quad\alpha=\frac{(2-\epsilon)^2}{4-\epsilon},\quad\beta=\epsilon-2.}
$$
This <massive zero-momentum QED vertex integral> uses an integrated tensor identity for the replacement of $N^\mu$, not an equality for each fixed $k$. The physical <photon> remains massless. The retained <fermion> <mass> supplies a scale and makes this zero-external-<four-momentum> calculation suitable for extracting the <ultraviolet pole> without a <scaleless integral>.
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