= Solution
Continue with the cubic <scalar field theory> of <1/c>, and use precisely its <counterterm> signs. Take $m^2>0$ to separate <ultraviolet divergences> from <infrared divergences>. Make a <Wick rotation> and impose the rotationally invariant <cutoff regularization> $|q_E|<\Lambda$. Put
$$
C=\frac{\operatorname{area}(S^5)}{2(2\pi)^6}
=\frac1{128\pi^3},\qquad L_\Lambda=\log\frac{\Lambda^2}{m^2},\qquad
I_r=\int_{|q_E|<\Lambda}\frac{d^6q_E}{(2\pi)^6}(q_E^2+m^2)^{-r}.
$$
For example, <spherical coordinates> give $I_1=C\int_0^{\Lambda^2} t^2/(t+m^2)\,dt$. Polynomial division and then differentiation with respect to $m^2$ give the divergent parts
$$
\begin{aligned}
I_1\big|_{\mathrm{div}}&=C\left(\frac{\Lambda^4}{2}-m^2\Lambda^2+m^4L_\Lambda\right),\\
I_2\big|_{\mathrm{div}}&=C(\Lambda^2-2m^2L_\Lambda),\qquad
I_3\big|_{\mathrm{div}}=CL_\Lambda.
\end{aligned}
$$
The derivative relations are $I_2=-\partial_{m^2}I_1$ and $I_3=\tfrac12\partial_{m^2}^2I_1$; differentiation also generates finite terms, which are irrelevant here. The first supplied integral identity is consistent with these divergent parts after <Wick rotation>.
\b[The second supplied expansion is incorrect.] Write $A=p^2-m^2$. Direct multiplication of the two denominators gives $(A+k^2)^2-4(p\cdot k)^2$, so the correct <Taylor expansion> is
$$
\frac1{[(p-k)^2-m^2][(p+k)^2-m^2]}
=\frac1{A^2}\left(1-\frac{2k^2}{A}+\frac{4(p\cdot k)^2}{A^2}+O(k^4)\right).
$$
A concrete counterexample is $p=0$, $m=1$, $k=(\epsilon,\mathbf0)$: the exact expression is $(1-\epsilon^2)^{-2}=1+2\epsilon^2+O(\epsilon^4)$, whereas the printed expansion gives $1+\epsilon^2/2+O(\epsilon^4)$. We therefore compute the <self-energy> without that identity.
For Euclidean external <momentum> $P$, define the <scalar bubble integral>
$$
B(P)=\int\frac{d^6q_E}{(2\pi)^6}
\frac1{(q_E^2+m^2)((q_E+P)^2+m^2)}.
$$
A <Feynman parameter> $x$ combines its denominators into one squared denominator with $K=m^2+x(1-x)P^2$. Applying the radial result above and integrating $\int_0^1x(1-x)\,dx=1/6$ yields
$$
B(P)\big|_{\mathrm{div}}
=C\left(\Lambda^2-2m^2L_\Lambda-\frac{P^2}{3}L_\Lambda\right).
$$
One can instead expand with the <ultraviolet cutoff> fixed on the unshifted <loop momentum>: the $P^2$ coefficient of the integrand is $-(q_E^2+m^2)^{-3}+\tfrac23q_E^2(q_E^2+m^2)^{-4}$, using rotational averaging in six dimensions. Its leading radial term is $-1/(3q_E^6)$, giving the same logarithm. Shifting a sharp integration boundary changes finite terms in this <scalar bubble integral>, and does not change the displayed divergent coefficients.
To fix all signs and <symmetry factors> together, expand the one-loop <Euclidean effective action> around a background <scalar field>:
$$
\Gamma_{1,E}[\phi]=\frac12\operatorname{Tr}\log(D+g\phi),\qquad D=-\partial_E^2+m^2,
$$
$$
\Gamma_{1,E}[\phi]-\Gamma_{1,E}[0]
=\frac g2\operatorname{Tr}(D^{-1}\phi)
-\frac{g^2}{4}\operatorname{Tr}(D^{-1}\phi D^{-1}\phi)
+\frac{g^3}{6}\operatorname{Tr}(D^{-1}\phi)^3+\cdots.
$$
These terms are the <tadpole diagram>, the two-point <bubble diagram>, and the <triangle Feynman diagram>. The <tadpole diagram> has factor $1/2$; the two-point <bubble diagram> also has factor $1/2$ after two differentiations with respect to the background <scalar field>. The local three-point <interaction vertex> is $g^3I_3$. Thus their divergent contributions to the one-, two-, and three-point Euclidean <one-particle irreducible> vertices are respectively
$$
\frac g2 I_1,\qquad -\frac{g^2}{2}B(P),\qquad g^3I_3.
$$
The Euclidean <counterterm Lagrangian> is $\delta_Z(\partial_E\phi)^2/2+\delta_m\phi^2/2+\delta_g\phi^3/6+\delta_h\phi+\delta_\mathcal E$. Cancelling each divergent <one-particle irreducible> vertex therefore gives
$$
\boxed{\begin{aligned}
\delta_h\big|_{\mathrm{div}}&=-\frac{Cg}{2}\left(\frac{\Lambda^4}{2}-m^2\Lambda^2+m^4L_\Lambda\right),\\
\delta_m\big|_{\mathrm{div}}&=\frac{Cg^2}{2}(\Lambda^2-2m^2L_\Lambda),\\
\delta_Z\big|_{\mathrm{div}}&=-\frac{Cg^2}{6}L_\Lambda,\\
\delta_g\big|_{\mathrm{div}}&=-Cg^3L_\Lambda.
\end{aligned}}
$$
For example, the two-point sum is $\delta_ZP^2+\delta_m-g^2B(P)/2$ and its divergent coefficients cancel separately. Finite <counterterms> depend on the chosen <renormalization conditions>. The zero-<mass> limit can use a positive subtraction scale in the logarithms after <infrared regularization>.
If unnormalized <vacuum functionals> are retained, specify the determinant measure by
$$
V_{1,E}(0)=\frac12\int_{|q_E|<\Lambda}\frac{d^6q_E}{(2\pi)^6}
\log\frac{q_E^2+m^2}{\mu^2}.
$$
Expanding its radial integrand at large $q_E$ fixes the additional <vacuum energy density> <counterterm>:
$$
\boxed{\delta_\mathcal E\big|_{\mathrm{div}}=-\frac C2\left[
\frac{\Lambda^6}{3}\log\frac{\Lambda^2}{\mu^2}-\frac{\Lambda^6}{9}
+\frac{m^2\Lambda^4}{2}-\frac{m^4\Lambda^2}{2}
+\frac{m^6}{3}L_\Lambda\right].}
$$
The purely <mass>-independent terms depend on the determinant normalization. Normalized <Green functions> eliminate this entire constant.
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