The divergence is local in the induced boundary metric and is cancelled by the leading <holographic renormalization> counterterm
$$
\boxed{I_{\rm ct}=-\frac1{8\pi GR}\int_{\partial M}d^2y\sqrt{-h}.}
$$
On the regulated cylinder this is
$$
I_{\rm ct}=-\frac{R\Delta t}{4G}
\frac{\cos\epsilon}{\sin^2\epsilon}
=-\frac{R\Delta t}{4G}\left(\epsilon^{-2}-\frac16+O(\epsilon^2)\right),
$$
which cancels the $\epsilon^{-2}$ divergence of $I_{\rm GR}$. An overall sign changes if one uses the oppositely signed Euclidean generating functional; the local counterterm changes with it.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 354 / 2 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-354/2/b/ii/solution