= Solution
Use the mostly-minus <Minkowski metric>, <quantum field theory propagator> $i/(p^2-m^2+i0)$ and dimensionally continued interaction $-\mu^{2\epsilon}\lambda\phi^4/4!$, with $d=4-2\epsilon$. Take all four external momenta incoming, so $\sum_jp_j=0$, and set $s=(p_1+p_2)^2$, $t=(p_1+p_3)^2$, $u=(p_1+p_4)^2$.
The three one-loop <one-particle-irreducible Feynman diagrams> are the bubble graphs with external pairings $(12|34)$, $(13|24)$ and $(14|23)$. Their external <quantum field theory propagators> are omitted by <amputation>. Each <Feynman diagram> has <symmetry factor> $1/2$ from exchanging the two internal lines; the fixed external labels are already accounted for by the three distinct channels.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-47-bubble-channels.png]
{title=The three labeled one-loop scalar four-point bubble channels with all external momenta incoming}
For a channel carrying momentum $q$, its amplitude is
$$
B(q^2)=\frac{(-i\mu^{2\epsilon}\lambda)^2}{2}
\int\frac{d^d\ell}{(2\pi)^d}
\frac{i}{\ell^2-m^2+i0}\frac{i}{(\ell+q)^2-m^2+i0}
=\frac{\mu^{4\epsilon}\lambda^2}{2}
\int\frac{d^d\ell}{(2\pi)^d}\frac1{AB}.
$$
The signs in the last equality follow from the two vertices and two <quantum field theory propagators>. Apply the <Feynman parameter> identity
$$
\frac1{AB}=\int_0^1\frac{dx}{[xA+(1-x)B]^2}.
$$
After shifting the <loop momentum>, the denominator is $(k^2-\Delta_q(x)+i0)^2$, with $\Delta_q(x)=m^2-x(1-x)q^2$. For positive Euclidean $\Delta$, <Wick rotation> and the Schwinger representation give
$$
\begin{aligned}
\int\frac{d^dk}{(2\pi)^d}\frac1{(k^2-\Delta+i0)^2}
&=i\int\frac{d^dk_E}{(2\pi)^d}\int_0^\infty d\alpha\,\alpha e^{-\alpha(k_E^2+\Delta)}\\
&=\frac{i}{(4\pi)^{d/2}}\int_0^\infty d\alpha\,\alpha^{1-d/2}e^{-\alpha\Delta}\\
&=\frac{i}{(4\pi)^{d/2}}\Gamma(2-d/2)\Delta^{d/2-2}.
\end{aligned}
$$
Analytic continuation to physical momenta keeps the prescription $\Delta_q-i0$. Thus <dimensional regularization> evaluates the channel as
$$
B(q^2)=\frac{i\mu^{4\epsilon}\lambda^2}{2(4\pi)^{2-\epsilon}}
\Gamma(\epsilon)\int_0^1dx\,[\Delta_q(x)-i0]^{-\epsilon}.
$$
Use $\Gamma(\epsilon)=\epsilon^{-1}-\gamma_E+O(\epsilon)$ and expand the other factors. The <minimal-subtraction scalar four-point bubble channels> result through the finite part is
$$
B(q^2)=\frac{i\mu^{2\epsilon}\lambda^2}{32\pi^2}
\left[\frac1\epsilon-\gamma_E+\log(4\pi)
-\int_0^1dx\,\log\frac{m^2-x(1-x)q^2-i0}{\mu^2}\right]+O(\epsilon).
$$
The parameter <integral> is left unevaluated, as requested. The three channel contributions are $B(s)+B(t)+B(u)$. Their pole is momentum independent and is cancelled by the four-point <counterterm> $-i\mu^{2\epsilon}\delta\lambda$, with
$$
\boxed{\delta\lambda=\frac{3\lambda^2}{32\pi^2\epsilon}+O(\lambda^3).}
$$
After pure <minimal subtraction scheme> subtraction, the complete vertex through one loop is the tree vertex $-i\lambda$ plus
$$
\frac{i\lambda^2}{32\pi^2}\sum_{r\in\{s,t,u\}}
\left[\log(4\pi)-\gamma_E-\int_0^1dx\,
\log\frac{m^2-x(1-x)r-i0}{\mu^2}\right].
$$
Here the $\epsilon\to0$ limit has been taken. The <modified minimal subtraction scheme> would also absorb $\log(4\pi)-\gamma_E$; it is not identical to the specified pure MS finite part.
The bare coupling is scale independent and satisfies
$$
\lambda_B=\mu^{2\epsilon}\left(\lambda+\frac{a\lambda^2}{\epsilon}+O(\lambda^3)\right),
\qquad a=\frac3{32\pi^2}.
$$
Writing $\beta_d=-2\epsilon\lambda+b\lambda^2+O(\lambda^3)$ and differentiating at fixed $\lambda_B$ gives
$$
0=2\epsilon\left(\lambda+\frac{a\lambda^2}{\epsilon}\right)
+\beta_d\left(1+\frac{2a\lambda}{\epsilon}\right).
$$
At order $\lambda^2$, the finite coefficient is $b-2a$, so the <one-loop quartic scalar beta function> is
$$
\boxed{\beta(\lambda)=\mu\frac{d\lambda}{d\mu}\bigg|_{\lambda_B}
=\frac{3\lambda^2}{16\pi^2}+O(\lambda^3).}
$$
The factor two relative to the pole residue follows from using $d=4-2\epsilon$. With $d=4-\epsilon$ the residue doubles, but the four-dimensional beta function is unchanged.
For pure nonabelian gauge theory, the corresponding <pure Yang-Mills one-loop beta function> has the opposite sign:
$$
\beta(g)=-\frac{11C_A}{3(16\pi^2)}g^3+O(g^5),
\qquad f^{acd}f^{bcd}=C_A\delta^{ab}.
$$
Gauge-boson self-interactions and the gauge-consistency ghost contributions give antiscreening. For a compact nonabelian simple factor $C_A>0$, the coupling decreases at high energy: <asymptotic freedom>. In contrast, positive scalar $\lambda$ increases toward the ultraviolet. Its one-loop extrapolation has a <Landau pole>, indicating breakdown of weak-coupling perturbation theory, not a proof by itself of nonperturbative triviality. Ultraviolet poles should be extracted at infrared-safe mass or nonexceptional kinematics; a scaleless massless zero-momentum <integral> cannot be used to read off this ultraviolet <counterterm> without separating its infrared behavior.
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