For one channel with momentum $p$, introduce a <Feynman parameter> and shift the loop momentum:
$$
\frac1{\ell^2(\ell+p)^2}
=\int_0^1dx\,
\frac1{\{(\ell+xp)^2+x(1-x)(-p^2)\}^2}.
$$
In $d=4-\epsilon$, with dimensional-regularization scale $\bar\mu$ before the usual modified-minimal-subtraction redefinition, the bubble at $p^2=-M^2$ is
$$
B(M)=\frac{i}{16\pi^2}
\left[
\frac2\epsilon-\gamma+\log4\pi
-\log\frac{M^2}{\bar\mu^2}
-\int_0^1dx\,\log\{x(1-x)\}
+O(\epsilon)
\right].
$$
Since $\int_0^1\log\{x(1-x)\}\,dx=-2$, summing the three equal channels gives the form displayed in the question, beginning with $6/\epsilon-3\gamma$.

Consequently
$$
-i\lambda_{\mathrm{eff}}
=-i\lambda
+\frac{i\,3\lambda^2}{32\pi^2}
\left[
\frac2\epsilon-\gamma+\log4\pi
-\log\frac{M^2}{\bar\mu^2}+2
\right]
-i\delta\lambda+O(\lambda^3).
$$
The <momentum-subtraction scheme> condition $\lambda_{\mathrm{eff}}(M)=\lambda$ is enforced by
$$
\delta\lambda
=\frac{3\lambda^2}{32\pi^2}
\left[
\frac2\epsilon-\gamma+\log4\pi
-\log\frac{M^2}{\bar\mu^2}+2
\right].
$$
Choosing $\bar\mu=M$ removes the logarithm. A minimal-subtraction scheme keeps only the pole and therefore defines a different finite renormalized coupling.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 304 / 2 / b / Solution

{wiki=Renormalization_scheme#Momentum_subtraction_schemes}

A momentum-subtraction scheme defines renormalized masses, fields, and couplings by fixing correlation functions at specified external momenta.


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 Incoming links: Momentum-subtraction scheme