= Solution
{c}
After a <Wick rotation>, the term in question has radial behavior
$$
I_m=m^2\int\frac{d^4k}{(2\pi)^4}\frac1{(k^2+i0)(k^2-m^2+i0)^2}=-i\int\frac{d^4k_E}{(2\pi)^4}\frac{m^2}{k_E^2(k_E^2+m^2)^2}.
$$
The radial integrand is proportional to $k_E$ at the origin and to $k_E^{-3}$ at infinity. Thus it has neither an <infrared divergence> nor an <ultraviolet divergence>. In fact the area of the unit three-sphere gives
$$
I_m=-\frac{i m^2}{8\pi^2}\int_0^\infty\frac{r\,dr}{(r^2+m^2)^2}=\boxed{-\frac{i}{16\pi^2}}.
$$
This also follows from the <Feynman parameter> identity
$$
\frac1{AB^2}=2\int_0^1dx\,\frac{1-x}{[xA+(1-x)B]^3}
$$
with $A=k^2+i0$, $B=k^2-m^2+i0$. Although the parameterized denominator becomes massless at $x=1$, its numerator $1-x$ cancels the apparent endpoint singularity. Since $\beta\to-2$, multiplication by $\beta$ preserves finiteness. No artificial <photon> <mass> is needed.
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