 Solution

For a bosonic quadratic action $S_2=\frac{1}{2}\int \varphi K\varphi$, completing the square in the generating functional gives $Z_0[J]/Z_0[0]=\exp (-iJ{K}^{-1}J/2)$. Two functional derivatives therefore give the Feynman propagator $\Delta =i{K}^{-1}$, 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 $\eta =\mathrm{diag}(1,-1,-1,-1)$, Fourier convention $e^{-ikx}$, and the Yang-Mills field strength $F_{\mu \nu }^a={\partial }_{\mu }{A}_{\nu }^a-{\partial }_{\nu }{A}_{\mu }^a+g{f}^{abc}{A}_{\mu }^b{A}_{\nu }^c$. The interaction terms do not contribute to the quadratic action. Integration by parts in $-F_{\mu \nu }^a{F}^{a\mu \nu }/4-(\partial \cdot A^a{)}^2/(2\alpha )$ yieldsThe color kernel is diagonal. For $k^2\ne 0$, 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 $P_L^2=P_L$, $P_T^2=P_T$, $P_L{P}_T=0$, soMultiplication explicitly gives the identity on both transverse and longitudinal components. Continuing this inverse with the Feynman propagator prescription gives the gluon propagatorThe 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; $\alpha \to 0$ gives a transverse covariant Landau gauge propagator.

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.