 Solution

A gauge-invariant path integral counts all representatives of each gauge orbit. In a perturbative region with a regular gauge fixing slice, insert the Faddeev-Popov gauge-orbit identityThe Faddeev-Popov determinant is the functional Jacobian converting integration over the gauge transformation parameter into integration over the condition. The redundant gauge group volume then factors out of normalized correlation functions.

Choose $A_{\mu }^{\omega }=A_{\mu }+{\partial }_{\mu }\omega$. Its infinitesimal variation givesThere is no derivative of $A$ in this expression: only $A_{\mu }$ is varied in $A^2$. For $\lambda =0$, the determinant is field independent and can be absorbed into normalization. For $\lambda \ne 0$, it is a functional of the gauge field, so dropping it changes the path integral, even though the gauge group is Abelian. Constant residual gauge transformations are removed by boundary conditions or by dividing out their volume, as usual in perturbative gauge fixing.