= Solution
For $m^2>0$ and initially $\Re n>d/2$, <Schwinger parameterization> and the supplied <Gaussian integral> give
$$
I_{d,n}(m)=\frac1{\Gamma(n)}\int_0^\infty d\alpha\,\alpha^{n-1}e^{-\alpha m^2}\int\frac{d^dk}{(2\pi)^d}e^{-\alpha k^2}=\frac1{(4\pi)^{d/2}\Gamma(n)}\int_0^\infty d\alpha\,\alpha^{n-d/2-1}e^{-\alpha m^2}.
$$
Substitute $u=\alpha m^2$ to obtain the <Euclidean massive loop integral>
$$
\boxed{I_{d,n}(m)=\frac{\Gamma(n-d/2)}{(4\pi)^{d/2}\Gamma(n)}(m^2)^{d/2-n}.}
$$
For a positive integer $n$, $\Gamma(n)=(n-1)!$. Outside the convergence region this expression defines its <meromorphic continuation>, used in <dimensional regularization>.
Combine the two denominators with a <Feynman parameter> and shift $k$ by $(1-x)p$. This gives
$$
J_d(p)=\int_0^1dx\,I_{d,2}\bigl(\sqrt{m^2+x(1-x)p^2}\bigr)=\frac{\Gamma(\epsilon/2)}{(4\pi)^{2-\epsilon/2}}\int_0^1dx\,[m^2+x(1-x)p^2]^{-\epsilon/2}.
$$
The <gamma function> has $\Gamma(\epsilon/2)=2/\epsilon+O(1)$, and the remaining factors tend to $(4\pi)^{-2}$ and one. Hence the <massive scalar bubble pole in four dimensions> is
$$
\boxed{J_{4-\epsilon}(p)=\frac1{8\pi^2\epsilon}+O(1).}
$$
The pole is independent of the external <momentum>, as required for a local quartic <counterterm>.
Use the mostly-plus <Minkowski metric> and expand $e^{iS}$. The momentum-space <Feynman rules> are: an internal <scalar propagator> $i/(-p^2-m^2+i0)$; a four-scalar <interaction vertex> $-i\mu^\epsilon\lambda$; a <momentum conservation> delta function at each vertex; and $\int d^dk/(2\pi)^d$ per independent loop momentum. Divide by the <Feynman-diagram symmetry factor>. All momenta can be taken incoming. External propagators are omitted when computing an amputated <one-particle-irreducible vertex>.
The connected one-loop <one-particle-irreducible Feynman diagrams> are the two-point <tadpole diagram> and the three four-point <bubble diagrams>, in the $s,t,u$ channels. Each has <symmetry factor> $1/2$. A tadpole attached to an external leg through a single bridge would be reducible, so is excluded here.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-51-scalar-loops.png]
{title=One-loop two-point and four-point one-particle-irreducible graphs in quartic scalar theory}
The <Euclidean massive loop integral> at $n=1$ has $\Gamma(-1+\epsilon/2)=-2/\epsilon+O(1)$, so $I_{d,1}=-m^2/(8\pi^2\epsilon)+O(1)$. <Wick rotation> and the propagator/vertex factors give the two-point loop amplitude $-i\lambda I_{d,1}/2$, whose pole is $+i\lambda m^2/(16\pi^2\epsilon)$. A mass <counterterm> contributes $-i\delta m^2$ and cancels it when $\delta m^2=\lambda m^2/(16\pi^2\epsilon)$.
Each four-point <bubble diagram> has pole $+i\lambda^2J_d/2=i\lambda^2/(16\pi^2\epsilon)$, with the external coupling-dimension factor restored in the local vertex. Summing its three channels gives the quartic counterterm $\delta\lambda=3\lambda^2/(16\pi^2\epsilon)$. Therefore the <one-loop massive phi-fourth counterterms> are exactly
$$
\boxed{\mathcal L_{\rm ct}=-\frac12\frac{\lambda m^2}{16\pi^2\epsilon}\phi^2-\frac{\mu^\epsilon}{4!}\frac{3\lambda^2}{16\pi^2\epsilon}\phi^4.}
$$
There is no one-loop <wave-function renormalization>, since the tadpole is independent of external momentum. Vacuum graphs would additionally require a vacuum-energy subtraction.
In the <minimal subtraction scheme>, the bare coupling is
$$
\lambda_0=\mu^\epsilon\left(\lambda+\frac{3\lambda^2}{16\pi^2\epsilon}+O(\lambda^3)\right).
$$
Its dimension is $\epsilon$; the renormalized $\lambda$ is dimensionless. Differentiate at fixed $\lambda_0$, writing $b=3/(16\pi^2)$:
$$
0=\epsilon\left(\lambda+\frac{b\lambda^2}{\epsilon}\right)+\widehat\beta(\lambda)\left(1+\frac{2b\lambda}{\epsilon}\right)+O(\lambda^3).
$$
Solving through quadratic order gives the <engineering term in a dimensionally continued beta function> and the <one-loop quartic scalar beta function>
$$
\boxed{\widehat\beta(\lambda)=-\epsilon\lambda+\frac{3\lambda^2}{16\pi^2}+O(\lambda^3),\qquad\beta(\lambda)=\frac{3\lambda^2}{16\pi^2}+O(\lambda^3).}
$$
The first term comes from the coupling's <engineering dimension>; the second is the quantum contribution to the <renormalization-group beta function>.
For fixed $\epsilon>0$, the whole $\widehat\beta$ is not literally $O(\lambda^2)$ as $\lambda\to0$: its linear term is present. Interpret the final instruction as keeping terms through quadratic order. Set $\lambda_*=\epsilon/b=16\pi^2\epsilon/3$. Then $\widehat\beta=b\lambda(\lambda-\lambda_*)$ is an upward parabola, negative between its two zeros and positive above the interacting zero.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-51-quartic-beta.png]
{title=Quartic beta function below four dimensions, with arrows toward increasing renormalization scale}
For positive initial coupling $\lambda_r=\lambda(\mu_r)$, the <quartic running coupling below four dimensions> follows by solving the linear equation for $1/\lambda$:
$$
\boxed{\lambda(\mu)=\frac{\lambda_*}{1+(\lambda_*/\lambda_r-1)(\mu/\mu_r)^\epsilon}.}
$$
If $0<\lambda_r<\lambda_*$, then $\lambda(\mu)\to\lambda_*$ as $\mu\to0$ and $\lambda(\mu)\to0$ as $\mu\to\infty$. If $\lambda_r=\lambda_*$ it stays fixed; the zero solution also stays fixed. If $\lambda_r>\lambda_*$, the infrared limit is still $\lambda_*$, but the ultraviolet evolution reaches a <Landau pole> at the finite scale
$$
\boxed{\mu_L=\mu_r\left(\frac{\lambda_r}{\lambda_r-\lambda_*}\right)^{1/\epsilon}.}
$$
Thus this branch cannot be continued physically to $\mu=\infty$ using the one-loop approximation. These are statements about the quartic coupling direction. The <Wilson-Fisher fixed point> of the full scalar theory also has a <thermal relevant direction at the Wilson-Fisher fixed point>; approaching the critical theory requires mass tuning. For negative initial coupling, the quartic potential is unstable: the same formula tends to zero from below in the ultraviolet and reaches a pole at finite scale toward the infrared.
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