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Apply a Feynman parameter and shift the loop momentum to $\ell =p+xk$. The common denominator becomes $[{\ell }^2+M_x^2{]}^2$, whereFor positive $M_x^2$, the Gamma-integral representation $a^{-2}={\int }_0^{\infty }dss{e}^{-sa}$ and a Gaussian integral giveThis formula initially converges for $\mathrm{Re}d<4$ and defines the Euclidean massive loop integral at other dimensions by analytic continuation. Consequently,With $d=6-ϵ$, the Gamma function factor is $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$. Only its pole matters: the other factors can be evaluated at $ϵ=0$ when extracting that pole. Since ${\int }_0^{1}x(1-x)dx=1/6$,This is the one-loop two-point divergence in six-dimensional cubic scalar theory. Its polynomial momentum dependence is precisely what permits subtraction by local counterterms.

Use the Euclidean path integral with weight $e^{-S}$ and Fourier transform convention $\varphi (x)=\int d^{d}p(2\pi {)}^{-d}{e}^{ip\cdot x}\varphi (p)$. The momentum-space Feynman rules are: an internal scalar propagator $1/(p^2+m^2)$; a quartic vertex $-\lambda$ together with $(2\pi {)}^d{\delta }^{(d)}(\sum p_i)$ for incoming momenta; and an integral $\int d^{d}k/(2\pi {)}^d$ over each independent loop momentum. Multiply by the Feynman-diagram symmetry factor; external propagators are retained for a full correlation function and removed for an amputated connected correlation function.

To enumerate the requested one-particle-irreducible Feynman diagrams, let $V$ be the number of quartic vertices and $I$ the number of internal edges. Four external legs imply $4V=2I+4$, while one loop order implies $I-V+1=1$. Thus $V=I=2$. Both internal edges must connect the two vertices: an alternative with a tadpole and one connecting edge would disconnect on cutting that edge. The only graphs are therefore the three pairings of labelled external legs, each with Feynman-diagram symmetry factor $1/2$.

The corresponding channel momenta are $q_s=p_1+p_2$, $q_t=p_1+p_3$, and $q_u=p_1+p_4$, with all external momenta incoming. This exhausts the connected one-loop four-point one-particle-irreducible Feynman diagrams.

For the Euclidean massive loop integral, first take $m^2>0$ and an integer $n>d/2$, where the integral converges. Schwinger parameterization givesThe supplied Gaussian integral then yieldsIntegrating over $s$, and using the Gamma function recurrence to obtain $\Gamma (n)=(n-1)!$, provesOutside its initial convergence range, the right-hand side defines the meromorphic continuation used in dimensional regularization; the divergent ordinary integral is not being assigned a convergent value.

For the scalar bubble integral, the Feynman parameter identity $1/(AB)={\int }_0^{1}dx[xA+(1-x)B{]}^{-2}$ followed by a translation of loop momentum givesWith $d=4-ϵ$, the Gamma function satisfies $\Gamma (ϵ/2)=2/ϵ+O(1)$, and the parameter integral tends to one. Hence the ultraviolet pole is independent of external momentum:One can also see why the same pole occurs without introducing a Feynman parameter: at large $k$, the difference between this integrand and $(k^2+m^2{)}^{-2}$ is ultraviolet integrable near four dimensions, so both have the same dimensional regularization pole. The positive mass avoids an infrared ambiguity in this argument.

In the quantum effective action convention, the tree-level four-point vertex is $+\lambda$, the negative of the amputated connected correlation function tree vertex. The three bubble corrections giveThus the pole counterterm is $\delta \lambda =3{\lambda }^2/(16{\pi }^2ϵ)$. The modified minimal subtraction scheme also subtracts the conventional finite $\log 4\pi -{\gamma }_E$ combination, or equivalently absorbs it into the subtraction-scale convention; that does not change the one-loop renormalization-group beta function. Using that scale convention, writeHere $\lambda$ is dimensionless, $[{\lambda }_B]=ϵ$, and the conventional constant scale factor is implicit. The one-loop tadpole is momentum independent, so wave-function renormalization does not contribute at this order. Differentiating at fixed ${\lambda }_B$, with $a=3/(16{\pi }^2)$, gives $0=ϵ(\lambda +a{\lambda }^2/ϵ)+\beta (1+2a\lambda /ϵ)+O({\lambda }^3)$. ThereforeFor the stable quartic scalar field theory, with $\lambda >0$, the one-loop quartic scalar beta function is positive: the running coupling increases toward the ultraviolet and decreases toward the infrared. The theory is not asymptotically free; extrapolating the one-loop flow gives a Landau pole.