 Solution

Let ${\mathcal{V}}_4=i{\Gamma }^{(4)}$ denote the amputated Feynman vertex, with tree term $v_4=-i\lambda$. A one-loop one-particle-irreducible Feynman diagram has $V_3+2V_4=4$, allowing three families: two quartic vertices, two cubic vertices and one quartic vertex, or four cubic vertices. DefineUsing $I_2,I_3$ from part c and writing $P_{ij}=p_i+p_j$, the complete one-loop result isIn each term of the middle sum, $k,l$ are the two complementary indices. Their order does not change the integral after a loop-momentum redefinition.

The first line contains three bubble channels, one for each partition of the four labelled external legs into two pairs. Each has $S=2$ from interchanging its two identical internal lines. In the second line, the quartic vertex carries a chosen pair of external legs, giving $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ mixed triangles. Each has $S=1$. In the last line, cyclic orderings of four labelled external legs are identified under rotation and reversal, leaving $(4-1)!/2=3$ boxes, again with $S=1$ each. The figure in part b shows one representative of each family and its multiplicity. External-leg corrections and tadpoles connected by bridges are excluded because ${\Gamma }^{(4)}$ is a one-particle-irreducible vertex. All integrals are left unevaluated, as appropriate here.