 Solution

Expanding $e^{i{S}_{\mathrm{int}}}$ gives the Feynman vertexThere is one scalar leg and a continuous oriented fermion line; no extra factor $i$ is absorbed into ${\gamma }^5$. The Dirac propagator is $S_F(k)=i(/k+M)/(k^2-M^2+i0)$, where $/k={\gamma }^{\mu }{k}_{\mu }$. Each closed fermion loop supplies a minus sign.

The tadpole diagram is proportional tobecause both numerator traces vanish by a. This is zero before the loop momentum integration.

For either orientation of the triangle Feynman diagram, let $a,b,c$ be its three internal four-momenta. Its numerator trace isAfter using cyclic property of the trace, each expanded term is a trace of ${\gamma }^5$ times at most three ordinary gamma matrices. Such traces vanish in four dimensions. For an odd number of ordinary gamma matrices, conjugation by ${\gamma }^5$ changes the sign. For two ordinary gamma matrices, the contracted Clifford algebra identity ${\gamma }_{\rho }{\gamma }^{\mu }{\gamma }^{\nu }{\gamma }^{\rho }=4{\eta }^{\mu \nu }I$ givesThe zero-matrix case is a. Hence both the one-point and three-point one-loop bosonic vertices vanish, independently of their external momenta. No cancellation between different loop orientations or delicate shift of a divergent integral is needed.