= Solution
The <superficial degree of divergence> is the power of a common large-momentum scale obtained by scaling all independent internal momenta together, before cancellations or subtraction of divergent subdiagrams. In a <scalar field theory> with a standard quadratic kinetic term and non-derivative interactions, each <loop momentum> integral supplies $d$ powers of momentum and each internal <propagator> supplies minus two. Thus
$$
D=dL-2I.
$$
For a connected <Feynman diagram>, counting vertex half-edges and applying the stated identity for the <loop order> gives
$$
2I+E=\sum_n nV_n,\qquad I=L+V-1,\qquad V=\sum_n V_n.
$$
It follows that $D=(d-2)L-2V+2$. Meanwhile, $\sum_n(n-4)V_n=2I+E-4V=2L-2V+E-2$. Substitution proves
$$
\boxed{D=(d-4)L+\sum_n(n-4)V_n-E+4.}
$$
This <power counting in quantum field theory> concerns the overall scaling: even a <Feynman diagram> with $D<0$ can contain divergent subdiagrams, and cancellations may improve convergence when $D\geq0$.
The canonical <mass dimensions> are $[\phi]=(d-2)/2$ and $[g_n]=d-n(d-2)/2$. A finite polynomial <scalar field theory> has power-counting <renormalizability> when its interactions have nonnegative <mass dimensions> and the allowed local <counterterms> form a finite set closed under subtraction of divergent subdiagrams. For $d>2$, non-derivative interactions therefore satisfy $n\leq2d/(d-2)$. Equivalently,
$$
D=d-\frac{d-2}{2}E-\sum_n[g_n]V_n,
$$
so adding vertices of such interactions cannot force an indefinitely increasing number of external legs or derivatives in divergent <counterterms>. In dimensions $d\leq2$ a fixed finite polynomial still has this favorable <power counting in quantum field theory>; the displayed upper bound on $n$ is intended only for $d>2$.
For <six-dimensional cubic scalar field theory>, $3V=2I+E$ and $L=I-V+1$ yield
$$
\boxed{D=6-2E.}
$$
The coupling $g$ has zero <mass dimension>. The possible overall divergences have $E\leq3$: vacuum energy, a linear <tadpole diagram> term, a mass term, a kinetic term, and a cubic interaction. These are finitely many local <counterterms>. Divergent subdiagrams of higher-point <Feynman diagrams> are removed by the same <counterterms>, establishing <perturbative renormalizability of cubic scalar theory in six dimensions>. A real cubic potential is unbounded below, so this claim is one of <perturbative quantum field theory>, not a claim that the polynomial defines a stable nonperturbative vacuum.
After <Wick rotation>, use the <Euclidean action> with positive kinetic and mass terms and interaction $g\phi^3/3!$. The Euclidean momentum-space <Feynman rules> are: an internal <scalar propagator> $1/(k^2+m^2)$, a vertex $-g$, momentum conservation at each vertex, integration $d^6k/(2\pi)^6$ for each independent loop, and division by the <Feynman-diagram symmetry factor>. Strip the overall momentum-conserving delta function and external <propagators> when computing an <amputated Green's function>.
The <one-particle-irreducible two-point vertex> at one loop is the bubble:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-48-self-energy.png]
{title=One-loop cubic scalar two-point bubble with symmetry factor two}
Interchanging its two internal lines is an automorphism, giving <Feynman-diagram symmetry factor> two. In the normalization of the printed Minkowski <amputated Green's function>, the two vertices contribute $(-ig)^2$, the two internal <propagators> contribute $(-i)^2$, and <Wick rotation> supplies the overall $i$ already factored out there. Thus the Euclidean bubble contribution is
$$
\boxed{\widehat F_2^{(1)}(p)=\frac{g^2}{2}\int\frac{d^6k}{(2\pi)^6}\frac1{(k^2+m^2)((k+p)^2+m^2)}.}
$$
For this <self-energy> calculation we impose a vanishing <one-point function>, so <tadpole diagram> insertions are canceled by the linear <counterterm>. Without that convention a connected but <one-particle-reducible Feynman diagram> with a tadpole insertion is also possible: both external legs attach to one vertex, a zero-momentum line joins it to a second vertex, and that second vertex has a self-loop. Its amputated contribution is $g^2/(2m^2)\int d^6k/[(2\pi)^6(k^2+m^2)]$, independent of $p$. It is not the bubble drawn above; keeping it without a tadpole subtraction would add this constant to the full amputated answer.
To find the bubble's local divergence, put $K=k^2+m^2$ and expand at large $|k|$:
$$
\frac1{K(K+2k\cdot p+p^2)}=\frac1{K^2}-\frac{2k\cdot p+p^2}{K^3}+\frac{(2k\cdot p+p^2)^2}{K^4}+\cdots.
$$
Odd terms vanish under the angular integral. Rotational symmetry gives $\langle(k\cdot p)^2\rangle=k^2p^2/6$. The terms through order $p^2$ are therefore
$$
\frac1{K^2}+p^2\left(-\frac1{K^3}+\frac{2k^2}{3K^4}\right),
$$
and the $p^2$ coefficient approaches $-1/(3k^6)$. Terms of order $p^4$ and higher are integrable at infinity in six dimensions; subtracting the two displayed terms leaves an ultraviolet-convergent integral. Hence the divergent part is \b[a local polynomial $A p^2+B$], with momentum-independent coefficients. For example, when $m^2>0$ its logarithmically divergent coefficient is
$$
A_{\rm div}=-\frac{g^2}{768\pi^3}\log\frac{\Lambda^2}{m^2}.
$$
Changing the reference scale in this logarithm only changes a finite local term.
At zero external momentum the spherical <ultraviolet cutoff> and $\operatorname{Vol}(S^5)=\pi^3$ give
$$
B_\Lambda=\widehat F_2^{(1)}(0)=\frac{g^2}{128\pi^3}\int_0^\Lambda\frac{k^5\,dk}{(k^2+m^2)^2}.
$$
With $u=k^2$, integrate $u^2/(u+m^2)^2=1-2m^2/(u+m^2)+m^4/(u+m^2)^2$. This gives the <momentum-cutoff two-point function in six-dimensional cubic scalar theory>
$$
\boxed{B_\Lambda=\frac{g^2}{256\pi^3}\left[\Lambda^2-2m^2\log\left(1+\frac{\Lambda^2}{m^2}\right)+\frac{m^2\Lambda^2}{\Lambda^2+m^2}\right].}
$$
In particular, the divergent part of this constant is
$$
\boxed{B_{\rm div}=\frac{g^2}{256\pi^3}\left[\Lambda^2-2m^2\log\frac{\Lambda^2}{m^2}\right],}
$$
while the last term in $B_\Lambda$ tends to a finite constant. For $m=0$ the zero-momentum integral gives $B_\Lambda=g^2\Lambda^2/(256\pi^3)$ directly. The finite parts of a local subtraction can depend on the <regularization> prescription; the conclusion about mass and kinetic <counterterms> does not.
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