= Solution
For a bosonic quadratic <action> $S_2=\tfrac12\int\phi K\phi$, <completing the square> in the <generating functional> gives $Z_0[J]/Z_0[0]=\exp(-iJK^{-1}J/2)$. Two <functional derivatives> therefore give the <Feynman propagator> $\Delta=iK^{-1}$, with boundary conditions and a pole prescription included in the inverse. Zero modes must be removed or fixed before an inverse exists.
Take Minkowski metric $\eta=\operatorname{diag}(1,-1,-1,-1)$, Fourier convention $e^{-ikx}$, and the <Yang-Mills field strength> $F_{\mu\nu}^a=\partial_\mu A_\nu^a-\partial_\nu A_\mu^a+gf^{abc}A_\mu^bA_\nu^c$. The interaction terms do not contribute to the quadratic <action>. <Integration by parts> in $-F^a_{\mu\nu}F^{a\mu\nu}/4-(\partial\cdot A^a)^2/(2\alpha)$ yields
$$
S_2=\frac12\int d^dx\,A_\mu^a\left[\eta^{\mu\nu}\Box-(1-\alpha^{-1})\partial^\mu\partial^\nu\right]A_\nu^a.
$$
The color kernel is diagonal. For $k^2\ne0$, use the <projectors> $P_L^{\mu\nu}=k^\mu k^\nu/k^2$ and $P_T^{\mu\nu}=\eta^{\mu\nu}-P_L^{\mu\nu}$. They obey $P_L^2=P_L$, $P_T^2=P_T$, $P_LP_T=0$, so
$$
K^{ab\mu\nu}(k)=-\delta^{ab}k^2(P_T^{\mu\nu}+\alpha^{-1}P_L^{\mu\nu}),\qquad (K^{-1})_{\mu\nu}^{ab}=-\frac{\delta^{ab}}{k^2}(P_{T\mu\nu}+\alpha P_{L\mu\nu}).
$$
Multiplication explicitly gives the identity on both transverse and longitudinal components. Continuing this inverse with the <Feynman propagator> prescription gives the <gluon propagator>
$$
\boxed{\Delta_{\mu\nu}^{ab}(x)=\delta^{ab}\int\frac{d^dk}{(2\pi)^d}e^{-ikx}\left[\frac{-i\eta_{\mu\nu}}{k^2+i0}+\frac{i(1-\alpha)k_\mu k_\nu}{(k^2+i0)^2}\right].}
$$
The second term has the causal double-pole prescription. Away from the pole this is the usual $-i\delta^{ab}[\eta_{\mu\nu}-(1-\alpha)k_\mu k_\nu/k^2]/(k^2+i0)$. For $\alpha=1$ it becomes the <Feynman-gauge adjoint propagator>; $\alpha\to0$ gives a transverse <covariant Landau gauge> <propagator>.
\b[The limit $\alpha\to\infty$ removes <gauge fixing> and has no finite inverse on the full field space.] The longitudinal eigenvalue $-k^2/\alpha$ tends to zero and its inverse diverges. A contraction with conserved external currents kills the longitudinal term, but that does not define a gauge-unfixed <Gaussian integral>. This massless limit is not the massive-vector unitary-gauge limit: here it exposes the unfixed <gauge orbit> degeneracy.
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