= Solution
For the Euclidean <quartic scalar field theory>, expansion of $e^{-S}$ gives the momentum-space rules
* an internal scalar line of momentum $p$ contributes $(p^2+m^2)^{-1}$;
* a four-scalar vertex contributes $-\lambda$ and a momentum-conserving delta function;
* each independent <loop momentum> is integrated with $\int d^4k/(2\pi)^4$;
* each graph is divided by its <Feynman-diagram symmetry factor>.
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 is
$$
\frac{\lambda_a^2}{2}I(P;m,\Lambda),
\qquad
I(P;m,\Lambda)=\int_{|k|<\Lambda}\frac{d^4k}{(2\pi)^4}
\frac1{(k^2+m^2)((k+P)^2+m^2)}.
$$
Using 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,
$$
I(P;m,\Lambda)=\frac1{16\pi^2}\int_0^1dx
\left[
\log\frac{\Lambda^2+\Delta_x}{\Delta_x}
+\frac{\Delta_x}{\Lambda^2+\Delta_x}-1
\right].
$$
Thus every channel has the logarithmic ultraviolet divergence
$$
I(P;m,\Lambda)=\frac1{16\pi^2}\log\Lambda^2+O(1).
$$
The complete one-loop vertex is
$$
V_a^{(4)}=-\lambda_a+\frac{\lambda_a^2}{2}
\bigl[I(p_1+p_2;m,\Lambda)+I(p_1+p_3;m,\Lambda)+I(p_1+p_4;m,\Lambda)\bigr]
+O(\lambda_a^3).
$$
Let $I_{m,\mathrm{os}}$ 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 is
$$
\boxed{\lambda_a=\lambda_{\mathrm{phys}}
+\frac{\lambda_{\mathrm{phys}}^2}{2}I_{m,\mathrm{os}}
+O(\lambda_{\mathrm{phys}}^3)}.
$$
The cutoff dependence of $\lambda_a$ is the coupling <counterterm> needed to hold the measured coupling fixed.
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