= Solution
For $M>0$, integrate the vector action by parts, dropping the boundary term, and write it as a quadratic functional:
$$
S_0[A]=\frac12\int d^4x\,A_\mu K^{\mu\nu}A_\nu,\qquad
K^{\mu\nu}=(\Box-M^2)g^{\mu\nu}-\partial^\mu\partial^\nu.
$$
The explicit <mass> makes this Proca <kernel> invertible: the displayed action alone has no gauge redundancy left to fix. It is interpreted as <unitary gauge> when obtained from a Higgs or Stückelberg completion with its would-be Goldstone field removed.
Introduce a source $J_\mu$ and complete the square, $A=A'-K^{-1}J$. The normalized Gaussian <path integral> is
$$
\boxed{\frac{Z[J]}{Z[0]}=\exp\left[-\frac i2\int d^4x\,d^4y\,
J_\mu(x)(K^{-1})^{\mu\nu}(x,y)J_\nu(y)\right],\qquad
\widetilde D_{\mu\nu}=i(\widetilde K^{-1})_{\mu\nu}.}
$$
The source-independent <functional determinant> cancels in $Z[J]/Z[0]$. For Fourier <momentum> $k$, the mixed-index <kernel> is $\widetilde K^\mu{}_{\nu}=-(k^2+M^2)\delta^\mu{}_{\nu}+k^\mu k_\nu$. For $k^2\ne0$, use the <projectors>
$$
P_L{}^\mu{}_{\nu}=\frac{k^\mu k_\nu}{k^2},\qquad P_T=I-P_L.
$$
Their orthogonality gives
$$
\widetilde K=-(k^2+M^2)P_T-M^2P_L,\qquad
\widetilde K^{-1}=-\frac{P_T}{k^2+M^2}-\frac{P_L}{M^2}.
$$
Combining the <projectors> removes their apparent massless denominator. Adding the Feynman prescription gives the unitary-gauge propagator
$$
\boxed{\widetilde D_{\mu\nu}(k)=\frac{-i}{k^2+M^2-i0}
\left(g_{\mu\nu}+\frac{k_\mu k_\nu}{M^2}\right).}
$$
Multiplication by $\widetilde K$ verifies that it inverts the quadratic operator. On the physical pole $k^2=-M^2$, the numerator is transverse to $k$, as required for the three physical massive-vector polarizations.
The drawback is its off-shell ultraviolet behaviour: at large spacelike <momentum> the transverse term falls as $1/k^2$, but the longitudinal tensor term remains of order $1/M^2$. Naive loop <power counting> therefore looks worse and <renormalizability> is not manifest. In a finite $R_\xi$ gauge the corresponding expression is
$$
D_\xi=-i\left[\frac{P_T}{k^2+M^2-i0}+\frac{\xi P_L}{k^2+\xi M^2-i0}\right],
$$
which falls as $1/k^2$ at fixed finite $\xi$. Its $\xi\to\infty$ limit gives the unitary-gauge longitudinal term, so taking that limit before the ultraviolet limit changes <power counting>.
This is not a claim that a renormalizable Higgs theory ceases to be physically renormalizable in this gauge. Nor does it mean that the free Proca <Gaussian integral> is inconsistent. For conserved external sources $k\cdot J=0$, the longitudinal numerator drops out of $J^\mu D_{\mu\nu}J^\nu$ and the tree exchange again falls as $1/k^2$. The difficulty concerns off-shell <Green functions> and manifest control of loop divergences, not the number of physical polarizations.
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