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For incoming momenta, the Euclidean momentum-space Feynman rules are

The one-loop one-particle-irreducible diagrams with at most three external legs can be classified by their external species. There is a one-point $\chi$ tadpole diagram made from a $\varphi$ loop. There are two two-point bubble diagrams: the $\varphi$ self-energy has one internal $\varphi$ and one internal $\chi$, while the $\chi$ self-energy has two internal $\varphi$ lines and symmetry factor $1/2$. For three external legs, three vertices make a triangle: one triangle corrects the $\varphi \varphi \chi$ vertex and contains two internal $\varphi$ lines and one internal $\chi$ line; another has three external $\chi$ legs and a $\varphi$ loop, generating a ${\chi }^3$ interaction. The symmetry $\varphi \mapsto -\varphi$ forbids amplitudes with an odd number of external $\varphi$ legs. The classical $\varphi \varphi \chi$ vertex is the corresponding tree-level three-point diagram.

Adopt the self-energy conventionThis follows by summing the geometric series of exact propagators separated by amputated one-particle-irreducible two-point insertions. At one loop, after writing ${\lambda }_0={\mu }^{ϵ/2}\lambda +O({\lambda }^3)$,Introduce a Feynman parameter and shift the loop momentum. Withdimensional regularization givesUsing $\Gamma (-1+ϵ/2)=-2/ϵ+\gamma -1+O(ϵ)$ produceswhereand one convenient integral form of the finite part isChanging the definition of the dimensional-regularization scale only moves a finite constant between $C$ and the counterterm.

To make the two-point function finite, write ${\varphi }_0=Z_{\varphi }^{1/2}\varphi$, express $m_0^2$ in terms of a renormalized mass and a mass counterterm, and choose the pole parts of $\delta Z_{\varphi }$ and $\delta m^2$ to cancel $B{p}^2/ϵ$ and $A/ϵ$. In the minimal subtraction scheme no additional finite pieces are removed. The physical mass is the pole mass, so with the self-energy convention above it obeysThe explicit $\mu$ dependence of the finite self-energy cancels the running of $m(\mu )$, leaving $m_{\mathrm{phys}}$ independent of the renormalization scale.

The superficial degree of divergence counts the ultraviolet power before subdivergences and symmetry cancellations are considered. At $d=6$, a connected graph made from cubic vertices haswhere $E$ is its number of external legs. Hence one-, two-, and three-point functions can have quartic, quadratic, and logarithmic superficial divergences, whereas graphs with more external legs are superficially convergent.

Full renormalization also requires the $\chi$ mass and wave-function counterterms from its two-point function, a linear $\chi$ counterterm cancelling the tadpole, a coupling counterterm from the divergent $\varphi \varphi \chi$ triangle, and a ${\chi }^3$ counterterm from the three-$\chi$ triangle. A vacuum-energy counterterm removes divergent vacuum diagrams. These are precisely the local operators allowed by power counting in quantum field theory and the exact $\varphi \mapsto -\varphi$ symmetry.

For the Euclidean quartic scalar field theory, expansion of $e^{-S}$ gives the momentum-space rules

At one loop, the four-point one-particle-irreducible correlation function receives the three bubble diagrams in the $s$, $t$, and $u$ channels. If $P$ is the momentum through one channel, its contribution isUsing a Feynman parameter, shifting the loop momentum, and writing ${\Delta }_x=m^2+x(1-x)P^2$ gives, up to terms caused by shifting the boundary of a hard cutoff,Thus every channel has the logarithmic ultraviolet divergenceThe complete one-loop vertex isLet $I_{m,\mathrm{os}}$ denote the bracket evaluated at the chosen on-shell kinematic point. The on-shell renormalization scheme requires $V_a^{(4)}{|}_{\mathrm{os}}=-{\lambda }_{\mathrm{phys}}$, hence the perturbative solution isThe cutoff dependence of ${\lambda }_a$ is the coupling counterterm needed to hold the measured coupling fixed.

For one channel with momentum $p$, introduce a Feynman parameter and shift the loop momentum:In $d=4-ϵ$, with dimensional-regularization scale $\bar{\mu }$ before the usual modified-minimal-subtraction redefinition, the bubble at $p^2=-M^2$ isSince ${\int }_0^1\log \{x(1-x)\}dx=-2$, summing the three equal channels gives the form displayed in the question, beginning with $6/ϵ-3\gamma$.

ConsequentlyThe momentum-subtraction scheme condition ${\lambda }_{\mathrm{eff}}(M)=\lambda$ is enforced byChoosing $\bar{\mu }=M$ removes the logarithm. A minimal-subtraction scheme keeps only the pole and therefore defines a different finite renormalized coupling.

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

Introducing a Feynman parameter and subtracting at zero external momentum giveThe bare coupling ${\lambda }_0=\lambda +{\delta }_{\lambda }$ is independent of the renormalization scale. Differentiating its logarithmic counterterm at fixed ${\lambda }_0$ gives the leading beta function

For a bubble carrying momentum $p$, a Feynman parameter and a shift of the loop momentum give, after Wick rotation and cutoff regularization,where $F(p^2)=m^2-x(1-x)p^2-i0$. The quantum effective action therefore containsThe renormalization condition ${\Gamma }_4(M^2,M^2,M^2)=\lambda$ fixesSubstitution cancels both the cutoff and the scheme-dependent constant and leavesas required.

Use dimensional regularization with $d=4-ϵ$ and choose Feynman gauge. Combining the electron and photon denominators with a Feynman parameter, shifting the loop momentum, and discarding the odd term leaves the numeratorThe pole of the rotationally symmetric integral consequently givesWith the inverse-propagator conventionthe minimal subtraction scheme chooses the counterterm pole to equal ${\Sigma }_{\mathrm{pole}}$. Hence

In mostly-minus signature, take $D_j=(\ell +r_j{)}^2-m^2+i0$. Feynman parameters and a momentum shift giveFor massive internal lines this is ultraviolet finite in four dimensions. Physical external kinematics can still create threshold cuts, encoded by the Feynman i-epsilon prescription.

For $\mathrm{Re}A>0$ and $\mathrm{Re}v>0$, rescaling the defining Gamma function integral proves this representation, with compatible complex-power branches. It turns denominator powers into exponential factors, making Gaussian integrals possible. Combining two parameters by their sum and ratio gives the usual Feynman parameter identity. Outside its convergence domain a further continuation or pole prescription is needed.