= Six-dimensional cubic scalar counterterms
{title2=$\delta_Z=-Cg^2\log(\Lambda^2/m^2)/6$}
For a <real scalar field> in six spacetime dimensions with interaction $-g\phi^3/6$, the <mass dimension> of $g$ is zero. With Euclidean sharp <cutoff regularization>, $C=1/(128\pi^3)$ and $L=\log(\Lambda^2/m^2)$, choose the Minkowski <counterterm Lagrangian> $\delta_Z(\partial\phi)^2/2-\delta_m\phi^2/2-\delta_g\phi^3/6-\delta_h\phi$. Its one-loop divergent coefficients are
$$
\delta_Z=-\frac{Cg^2}{6}L,\qquad
\delta_m=\frac{Cg^2}{2}(\Lambda^2-2m^2L),\qquad
\delta_g=-Cg^3L,\qquad
\delta_h=-\frac{Cg}{2}(\Lambda^4/2-m^2\Lambda^2+m^4L).
$$
The proof uses the <Euclidean effective action> $\tfrac12\operatorname{Tr}\log(-\partial_E^2+m^2+g\phi)$. Its first three background derivatives have divergent parts $gI_1/2$, $-g^2B(P)/2$, $g^3I_3$. Radial integration gives $I_1=C(\Lambda^4/2-m^2\Lambda^2+m^4L)$, $I_3=CL$, and $B(P)=C(\Lambda^2-2m^2L-P^2L/3)$ modulo finite terms. Subtraction proves the coefficients. The <tadpole counterterm> is essential despite the absence of a linear term in the original <Lagrangian density>. A separate constant handles the <vacuum energy density>.
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