Put $I_r=\int_{\Lambda<|q|<\Lambda_0}d^6q\,(2\pi)^{-6}(q^2+m^2)^{-r}$, the <scalar shell integral in six dimensions>. Each of the three boxes is $g^4I_4$ as a connected insertion. To fix the effective-action sign and avoid a factorial ambiguity, use the <one-loop scalar effective action>:
$$
\frac12\operatorname{Tr}\log(D+g\phi)=\frac12\operatorname{Tr}\log D+\frac12\sum_{r\ge1}\frac{(-1)^{r+1}g^r}{r}\operatorname{Tr}(D^{-1}\phi)^r,\qquad D=-\partial^2+m^2.
$$
For a constant background the fourth-order density is $-g^4I_4\phi^4/8$. Thus the <cubic scalar box contribution to a quartic coupling> is $\delta g_4=-3g^4I_4$, whose vertex insertion is $-\delta g_4=+3g^4I_4$.

Using the supplied area of the unit five-sphere,
$$
I_4=\frac1{64\pi^3}\int_\Lambda^{\Lambda_0}\frac{q^5\,dq}{(q^2+m^2)^4}=\frac1{128\pi^3}\left[G(\Lambda^2+m^2)-G(\Lambda_0^2+m^2)\right],
$$
where $G(u)=u^{-1}-m^2u^{-2}+m^4/(3u^3)$. This follows by substituting $u=q^2+m^2$ and integrating $(u-m^2)^2/(2u^4)$. Hence
$$
\boxed{\delta g_4=-\frac{3g^4}{128\pi^3}\left[G(\Lambda^2+m^2)-G(\Lambda_0^2+m^2)\right].}
$$
As a check, the massless limit at positive $\Lambda$ is $-3g^4(\Lambda^{-2}-\Lambda_0^{-2})/(128\pi^3)$. Both versions have <mass dimension> minus two, as a six-dimensional local quartic coupling must.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 304 / 3 / iii / Solution

At fixed $m>0$ and $\Lambda>0$, the radial integrands in the <scalar shell integral in six dimensions> behave as $q^{5-2r}$ in the ultraviolet. Thus
$$
\begin{aligned}
I_4&\longrightarrow\text{a finite limit},\\
I_3&=\frac1{64\pi^3}\log(\Lambda_0/\Lambda)+O(1),\\
I_2&=\frac{\Lambda_0^2}{128\pi^3}+O(\log\Lambda_0),\\
I_1&=\frac{\Lambda_0^4}{256\pi^3}+O(\Lambda_0^2).
\end{aligned}
$$
Consequently the proper box quartic vertex is ultraviolet finite. The reducible triangle insertion contains the logarithmic cubic-vertex divergence. The internal bubble contains the mass divergence; its nonzero-momentum expansion additionally has a logarithmic kinetic-term divergence. An unsubtracted tadpole attachment has the quartic one-point divergence and is canceled by <tadpole subtraction>, or incorporated consistently in the chosen stationary background.

These divergent lower-point subgraphs require <counterterms>, not a new independent divergent $\phi^4$ coupling. More generally, cubic graph identities give the <superficial degree of divergence>
$$
D=6L-2I=6-2E,\qquad 3V=2I+E,\quad L=I-V+1.
$$
Vacuum, one-, two-, and three-point functions need vacuum-energy, linear, mass, <wave-function renormalization> and cubic-coupling <counterterms>. Proper higher-point functions have negative superficial degree and are finite after subtraction of divergent subgraphs. Hence the <perturbative renormalizability of cubic scalar theory in six dimensions> permits a continuum perturbative expansion with finitely many renormalized parameters; induced irrelevant higher-field and derivative interactions are finite predictions at finite scale, not an infinite list of ultraviolet parameters.

This perturbative statement does not establish a nonperturbative real Euclidean measure. For any nonzero real $g$, $m^2\phi^2/2+g\phi^3/6$ tends to $-\infty$ in one field direction, so even its constant-field integral is divergent. Thus $\boxed{\text{perturbatively renormalizable, with no convergent positive real Euclidean cubic measure}.}$ Stability or a specified <analytic continuation> would be additional input beyond these loop calculations.

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 304 / 3 / v / Solution

{title2=$I_r(\Lambda,\Lambda_0;m)$}

For $m>0$ and $0<\Lambda<\Lambda_0$, define
$$
I_r=\int_{\Lambda<|q|<\Lambda_0}\frac{d^6q}{(2\pi)^6}(q^2+m^2)^{-r}=\frac1{64\pi^3}\int_\Lambda^{\Lambda_0}\frac{q^5\,dq}{(q^2+m^2)^r}.
$$
The radial formula uses the area $\pi^3$ of the unit five-sphere. Its <mass dimension> is $6-2r$. As $\Lambda_0\to\infty$, $I_1$ diverges quartically, $I_2$ quadratically, $I_3$ logarithmically, and $I_r$ is ultraviolet finite for $r\ge4$ at fixed $\Lambda,m$. This follows directly by comparison with $q^{5-2r}$.

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 Incoming links: Scalar shell integral in six dimensions