Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2020 / 3 / b / iii / Solution

For fixed $A$, the delta functional sets $\sigma (x)=F[A](x)$. Therefore its weighted functional integral isSince the action enters as $e^{iS}$, the gauge fixing contribution isThe Faddeev-Popov determinant does not depend on $\sigma$ and remains in the path integral. The limit $\xi \to 0$ enforces the original condition sharply; for nonzero $\xi$ this is a weighted average of the gauge slices.