= Solution
\b[Source correction:] equation 11 on the original page has $\Gamma(\alpha-1/2)$. The dimensionally regulated integral instead contains $\Gamma(\alpha-d/2)$. This is not an interchangeable convention: for $\alpha=2$ and $d=4-\epsilon$, the correct factor $\Gamma(\epsilon/2)$ has a logarithmic <ultraviolet pole>, whereas the printed $\Gamma(3/2)$ is finite.
To obtain the correct expression, shift the <loop momentum> and perform a <Wick rotation>. For positive Euclidean $\mathcal M^2$, introduce a <Schwinger parameterization>:
$$
\begin{aligned}
\int\frac{d^dq_E}{(2\pi)^d}\frac1{(q_E^2+\mathcal M^2)^\alpha}
&=\frac1{\Gamma(\alpha)}\int_0^\infty ds\,s^{\alpha-1}e^{-s\mathcal M^2}
\int\frac{d^dq_E}{(2\pi)^d}e^{-sq_E^2}\\
&=\frac{\Gamma(\alpha-d/2)}{(4\pi)^{d/2}\Gamma(\alpha)}(\mathcal M^2)^{d/2-\alpha}.
\end{aligned}
$$
The derivation first holds in the convergent range and defines the result elsewhere by <analytic continuation>. For integer $\alpha$, rotating back gives
$$
\boxed{\int\frac{d^dq}{(2\pi)^d}\frac1{(q^2+2q\cdot p-m^2+i0)^\alpha}
=\frac{i(-1)^\alpha}{(4\pi)^{d/2}}\frac{\Gamma(\alpha-d/2)}{\Gamma(\alpha)}
(p^2+m^2)^{d/2-\alpha},}
$$
with the branch inherited from $i0$. This is the Lorentzian continuation of the <Euclidean massive loop integral>.
Set $p=0$, $\alpha=1$, $d=4-\epsilon$. Since $\Gamma(-1+\epsilon/2)=-2/\epsilon+O(1)$,
$$
\int_k\frac1{k^2-m^2+i0}=\frac{im^2}{8\pi^2\epsilon}+O(1).
$$
Consequently the directed tadpole in part d gives
$$
\left.i\Gamma^{(2)}_{\mathrm{loop}}\right|_{\mathrm{div}}
=(-ig\mu^\epsilon)\left.i\int_k\frac1{k^2-m^2+i0}\right|_{\mathrm{div}}
=\frac{igm^2}{8\pi^2\epsilon}.
$$
The <counterterm> is $-i\delta_{m^2}$, so cancelling only the pole in the <minimal subtraction scheme> requires
$$
\boxed{\delta_{m^2}=\frac{gm^2}{8\pi^2\epsilon}.}
$$
This is the <quartic complex-scalar mass renormalization> result. The scale factors and logarithms contribute finite terms at this order and do not change the displayed pole coefficient. A real-field check gives a tadpole coefficient $\tfrac12(3g/2+g/2)=g$, in agreement with the oriented complex-field computation.
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