Question about Faddeev-Popov gauge-fixing in Schwartz textbook I am trying to understand equation (25.91) from Schwartz's Quantum Field Theory textbook. The goal is to gauge-fix the path integral for Quantum chromodynamics using the Faddeev-Popov trick. Briefly, the argument boils down to multiplying the integral by:
$$1=C\sqrt{\det(\partial_{\mu}D^{\mu})^2}\int {\cal D}\pi~ e^{-i\int d^{4}x \frac{1}{2\zeta}(\partial_{\mu}D^{\mu}\pi-\partial_{\mu}A^{\mu})^2}$$ where $C$ is some numerical coefficient.
Now, in the second line of (25.91) the author redefines $$A\rightarrow A+ D\pi ,$$ where $D$ is the gauge covariant derivative in the adjoint representation. He claims that this shift results in the dependence of the integrand on $\pi$ dropping out leading to an extra factor $\int {\cal D}\pi$ which is not significant.
I do not understand how the shift $A\rightarrow A+D\pi$ leads to the expression in (25.91). Shouldnt we also shift the $D$ in the factor $\partial D \pi$ living in the argument of the exponential?
 A: OP has a point. Ref. 1 transforms the integration variables$^1$
$$ A^b_{\nu}\quad\longrightarrow\quad A^{\prime a}_{\mu}~=~A^a_{\mu} - \partial_{\mu}\pi^a - gf^{abc} A^b_{\mu}\pi^c~=~A^a_{\mu} - D_{\mu}^{ab}(A)\pi^b$$
upstairs in the exponential function of eq. (25.91) but forgets to also transform the factor $\frac{1}{f[A]}$ downstairs. The result (25.93) is certainly the well-known correct result for the Faddeev-Popov path integral in $R_{\xi}$-gauge, but the derivation$^2$ leading to eq. (25.93) is flawed. The Faddeev-Popov trick usually starts by considering an identity of the form "delta-function times determinant" rather than the identity $\frac{f[A]}{f[A]}=1$. A correct derivation is given in e.g. Ref. 2.
References:

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*M.D. Schwartz, QFT & the standard model, 2014; eq. (25.91).


*M. Srednicki, QFT, 2007; chapter 71. A prepublication draft PDF file is available here.
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$^1$ Notice the minus sign. It should probably also be mentioned that the change of integration variables induces a Jacobian determinant
$$ \det\frac{\delta A^{\prime a}_{\mu}}{\delta A^b_{\nu}}~=~\text{function of } \pi \text{ but not a function of }A,$$
which in principle has to be included in the path integral $\int\!{\cal D}\pi$.
$^2$ The corresponding abelian derivation in section 14.5 of Ref. 1 is fine because $f$ then doesn't depend on $A$.
