 Solution

The amputated connected four-point function through one loop contains the tree quartic vertex, the quartic-counterterm vertex, and three bubble diagrams in the $s$, $t$, and $u$ channels. The unamputated connected function also has one-loop self-energy and two-point-counterterm insertions on each external propagator, which do not determine $\delta g$.

For channel momentum $P$, the bubble has symmetry factor $1/2$ and the Feynman parameter identity gives an integral proportional toIn dimensional regularization, $\Gamma (ϵ/2)=2/ϵ+O(1)$, so the pole from one channel is $g^2/(16{\pi }^2ϵ)$. Summing the three crossing channels gives $3g^2/(16{\pi }^2ϵ)$. The minimal subtraction scheme cancels only this pole, hence