Regard $\lambda$ as a <running coupling> $\lambda(M)$. At fixed $s,t,u$ and to order $\lambda^2$, differentiating the one-loop answer with respect to $\log M$ gives
$$
0=\frac{d\Gamma_4}{d\log M}
=\frac{d\lambda}{d\log M}
-\frac{\lambda^2}{32\pi^2}\int_0^1 6\,dx+O(\lambda^3),
$$
because $m\ll M$ implies $F(M^2)\simeq-x(1-x)M^2$. Thus the <beta function (physics)> is
$$
\beta(\lambda)=\frac{d\lambda}{d\log M}
=\frac{3\lambda^2}{16\pi^2}+O(\lambda^3).
$$
Solving this <ordinary differential equation> with $\lambda(M_0)=\lambda_0$ gives
$$
\frac1{\lambda(M)}=\frac1{\lambda_0}
-\frac3{16\pi^2}\log\frac{M}{M_0},
\qquad
\lambda(M)=\frac{\lambda_0}{1-\dfrac{3\lambda_0}{16\pi^2}\log(M/M_0)}
$$
to leading-logarithmic order.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-304/1/c/solution