Let $s=\log\zeta$. For an infinitesimal momentum shell,
$$
\frac{dI_d}{ds}=\frac{\Omega_{d-1}}{(2\pi)^d}\Lambda^{d-2}.
$$
Differentiating the inverse-coupling relation and using $d(1/g)/ds=-g^{-2}dg/ds$ gives the one-loop <beta function (physics)>
$$
\frac{dg}{ds}=-(d-2)g+\frac{\Omega_{d-1}}{(2\pi)^d}\Lambda^{d-2}(N-2)g^2.
$$
For $d=2+\epsilon$, $\Omega_{d-1}/(2\pi)^d=1/(2\pi)+O(\epsilon)$, so
$$
\boxed{\frac{dg}{ds}\simeq-\epsilon g+(N-2)\Lambda^\epsilon\frac{g^2}{2\pi}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 303 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-303/3/d/solution