I know that the Faddeev-Popov path integral is gauge invariant. But how does one show that \begin{equation} I = \int \mathcal{D}\mathcal{A}_\mu \bigg|\frac{\delta\mathcal{G}}{\delta{\omega}}\bigg|\delta(\mathcal{G}) \exp(- S_\text{YM}[A_\mu]) \end{equation} is independent of the gauge choice $\mathcal{G}$? Here $\omega$ is the infinitesimal gauge parameter and $S_\text{YM}$ is the Euclidean Yang-Mills action. I came across this while reading Ramond's Field Theory: A Modern Primer but haven't found the explicit proof and I am unable to do. This is the first time I am seeing this material.
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