= Solution
Use the <metric signature> $(-,+,\ldots,+)$, so $p^2=-p_0^2+\mathbf p^2$, and the weight $e^{iS}$. The momentum-space <Feynman rules> for the massless <Phi-fourth theory> are: an internal scalar line contributes $-i/(k^2-i\epsilon)$; a quartic <interaction vertex> contributes $-i\lambda$ with its momentum-conservation delta function; every independent loop is integrated with $d^dk/(2\pi)^d$; and each graph is multiplied by its <Feynman-diagram symmetry factor>. All external <momenta> can be taken incoming. These rules follow by expanding the interaction exponential and contracting free fields with the <Wick theorem>.
For a connected graph with $I$ internal lines, $V$ <interaction vertices> and $L$ loops,
$$
4V=2I+n,\qquad L=I-V+1,\qquad
D=dL-2I=d+(d-4)V-\frac{d-2}{2}n.
$$
Here $D$ is the <superficial degree of divergence>. In four dimensions,
$$
\boxed{D=4-n.}
$$
Odd $n$ vanish by <Z2 symmetry>. Thus only the nonvacuum two-point and four-point <one-particle-irreducible Feynman diagrams> can have overall ultraviolet divergences: degree two for $n=2$, degree zero for $n=4$. Vacuum graphs with $n=0$ may also diverge but cancel from normalized correlators. This is an overall power-counting statement; subgraphs must be subtracted before using it to conclude finiteness for $n\ge6$.
For the full <one-particle-irreducible vertices>, the tree-level quadratic kernel and quartic interaction give
$$
\boxed{\widehat\tau_2^{(0)}(p,-p)=-p^2,\qquad \widehat\tau_4^{(0)}=-\lambda.}
$$
If $\widehat\tau_2$ is instead reserved for interaction-generated <self-energy> insertions, its tree value is zero and $-p^2$ is displayed separately as the free inverse kernel. The distinction is only that convention; the loop insertions below are unchanged.
The requested one-loop diagrams are the two-point <tadpole diagram> and the three four-point <bubble diagrams>. Each has symmetry factor $1/2$. The three bubble channels correspond to $P_s=p_1+p_2$, $P_t=p_1+p_3$, $P_u=p_1+p_4$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-50-one-loop-graphs.png]
{title=One-loop tadpole and the three four-point bubble channels in massless phi-fourth theory}
The tadpole insertion obeys
$$
i\widehat\tau_2^{(1)}=\frac{-i\lambda}{2}\int\frac{d^dk}{(2\pi)^d}\frac{-i}{k^2-i\epsilon}.
$$
In massless <dimensional regularization> this scaleless integral is zero. A mass or cutoff regulator exposes the <ultraviolet divergence> allowed by power counting; its vanishing in the scaleless convention does not remove the need to include this diagram.
For a bubble, the vertex and <quantum field theory propagator> factors give $(-i\lambda)^2(-i)^2/2=\lambda^2/2$. The PDF defines $f$ with the factor $1/i$ outside the Minkowski integral, so the integral itself is $if(P^2)$. Therefore
$$
\boxed{\widehat\tau_4=-\lambda+\frac{\lambda^2}{2}
\bigl[f(P_s^2)+f(P_t^2)+f(P_u^2)\bigr]+O(\lambda^3).}
$$
The converted TeX misplaces this $i$ as though it were part of the exponent of $(2\pi)$; keeping the printed $1/i$ is essential to this real pole coefficient.
For nonexceptional Euclidean external <momentum>, <Wick rotation> cancels that factor $1/i$. Use a <Feynman parameter> and translate the integration variable:
$$
f_d(P^2)=\int_0^1dx\int\frac{d^d\ell}{(2\pi)^d}
\frac1{\bigl(\ell^2+x(1-x)P^2\bigr)^2}
=\frac{\Gamma(2-d/2)}{(4\pi)^{d/2}}\int_0^1dx\,[x(1-x)P^2]^{d/2-2}.
$$
For example, represent the inverse square as $\int_0^\infty d\alpha\,\alpha e^{-\alpha(\ell^2+\Delta)}$, do the Gaussian <momentum> integral, and then use the <Gamma function>. This also derives the formula rather than assuming a loop-integration table. At $d=4-\varepsilon$, $\Gamma(\varepsilon/2)=2/\varepsilon+O(1)$ and the parameter integral tends to one, giving
$$
\boxed{f_d(P^2)=\frac1{16\pi^2}\frac2{4-d}+O(1).}
$$
Take $P^2>0$ before continuation back to Minkowski <momenta>. At $P=0$, a massless scaleless integral mixes ultraviolet and infrared issues, so it cannot be used to read off this pole.
Introduce a <renormalization scale> $\mu$ and a quartic <counterterm>. With $\lambda$ denoting a dimensionless renormalized <coupling constant (physics)>,
$$
\lambda_B=\mu^{\varepsilon}(\lambda+\delta\lambda),\qquad
\delta\lambda=\frac{3\lambda^2}{16\pi^2\varepsilon}+O(\lambda^3).
$$
After factoring out $\mu^\varepsilon$ from the vertex, each loop contains $\mu^\varepsilon f_d$. The <counterterm> contributes $-\delta\lambda$ to $\widehat\tau_4$ and cancels the pole from all three channels. More explicitly,
$$
\mu^\varepsilon f_d(P^2)=\frac1{16\pi^2}
\left[\frac2\varepsilon-\gamma_E+\log4\pi+2-\log\frac{P^2}{\mu^2}\right]+O(\varepsilon).
$$
In <modified minimal subtraction>, replace the pole-only counterterm by
$$
\delta\lambda_{\overline{\mathrm{MS}}}=\frac{3\lambda^2}{32\pi^2}
\left(\frac2\varepsilon-\gamma_E+\log4\pi\right)+O(\lambda^3),
$$
which subtracts the first three terms inside the brackets. The finite <renormalized quartic scalar vertex> is then
$$
\boxed{\widehat\tau_{4,R}=-\lambda+\frac{\lambda^2}{32\pi^2}
\sum_{r=s,t,u}\left[2-\log\frac{P_r^2}{\mu^2}\right]+O(\lambda^3).}
$$
Another <renormalization condition> can change the finite constant, but not the pole cancellation or the momentum-dependent logarithms. The corresponding timelike expressions follow by the same Feynman continuation.
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