Introducing a <Feynman parameter> and subtracting at zero external momentum give
$$
\Gamma_4(s,t,u)=\lambda+\frac{\lambda^2}{32\pi^2}
\sum_{q^2=s,t,u}\int_0^1dx\,
\log\frac{m^2}{m^2-x(1-x)q^2-i\epsilon}+O(\lambda^3).
$$
The bare coupling $\lambda_0=\lambda+\delta_\lambda$ is independent of the <renormalization scale>. Differentiating its logarithmic counterterm at fixed $\lambda_0$ gives the leading <beta function (physics)>
$$
\boxed{\beta_\lambda=\mu\frac{d\lambda}{d\mu}=\frac{3\lambda^2}{16\pi^2}+O(\lambda^3)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 304 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-304/1/c/solution