 Solution

After integration by parts, the quadratic Proca action with an external source isSince $M>0$, this kernel is invertible with the Feynman i-epsilon prescription; the massless gauge redundancy is absent in this unitary gauge quadratic action. Complete the square by $A\mapsto A+K_F^{-1}J$ in the Gaussian functional integral. Its source-independent determinant cancels in the normalized generating functional:In Fourier space, the kernel isFor $k^2\ne 0$, use the transverse projector of a vector field $P_T^{\mu \nu }=g^{\mu \nu }-k^{\mu }{k}^{\nu }/k^2$ and the longitudinal projector of a vector field $P_L^{\mu \nu }=k^{\mu }{k}^{\nu }/k^2$. Their mixed-index products obey $P_T{P}_L=0$, soThe last expression also extends to $k^2=0$; that point only obstructs the intermediate projector decomposition. Two source functional derivatives yield the gauge-boson propagator $D_F=i{K}_F^{-1}$:The pole residue is the physical massive-vector massive vector polarization sum. The longitudinal term is essential for inverting the quadratic kernel, and makes the naive $M\to 0$ limit of this unitary gauge Proca propagator singular.