= Solution
Return to the mostly-minus <Minkowski metric> of Question 3, and write $f=\partial_aA^a$. The field $\xi(x)$ is assumed real and nonzero wherever its inverse occurs. Under the usual Abelian <gauge transformation> $A_a\mapsto A_a+\partial_a\lambda$, with $\xi$ inert, the <electromagnetic field tensor> is invariant but $f\mapsto f+\Box\lambda$. The change in the added density is
$$
\Delta\mathcal L=\frac{f\Box\lambda}{\xi}+\frac{(\Box\lambda)^2}{2\xi}.
$$
This is not generally a total derivative. \b[The action is not invariant under arbitrary gauge transformations.] It retains <residual Lorenz gauge symmetry> for transformations satisfying $\Box\lambda=0$, with the same <boundary conditions> imposed before and after the transformation. The action itself is undefined at $\xi=0$; that value can only be considered as a limiting gauge.
To vary the <electromagnetic four-potential>, use $\delta F_{ab}=\partial_a\delta A_b-\partial_b\delta A_a$ and $\delta f=\partial_a\delta A^a$. The <integration by parts> of both terms gives
$$
\delta_A S=\int d^4x\left[\partial_aF^{ab}-\partial^b\left(\frac f\xi\right)\right]\delta A_b,
$$
with surface terms removed by the variational <boundary conditions>. Hence
$$
\boxed{\partial_aF^{ab}-\partial^b\left(\frac{\partial_cA^c}{\xi}\right)=0}.
$$
For variable $\xi$, the derivative must act on $1/\xi$ as well as on $f$:
$$
\Box A^b-\left(1+\frac1\xi\right)\partial^bf+\frac f{\xi^2}\partial^b\xi=0.
$$
Since $\xi$ enters algebraically, its <Euler-Lagrange equation> is
$$
\boxed{-\frac{f^2}{2\xi^2}=0}.
$$
For a real field this gives $f=0$. Thus a <dynamical gauge-fixing parameter> imposes a <constraint equation in field theory>; it is not an ordinary propagating scalar. On configurations satisfying both equations, $\partial_aF^{ab}=0$, $\partial_aA^a=0$, and therefore $\Box A^b=0$. There is no independent kinetic equation that determines $\xi$.
For the momentum-space equation at a prescribed constant $\xi\ne0$, use $A_a(x)=\int d^4p\,e^{-ip\cdot x}A_a(p)/(2\pi)^4$. Then $\partial_a\mapsto-ip_a$, and the linear $A$ equation is
$$
\boxed{K^{ab}(p)A_b(p)=0},\qquad
K^{ab}(p)=-p^2\eta^{ab}+\left(1+\frac1\xi\right)p^ap^b.
$$
Equivalently, $-p^2A^a+(1+1/\xi)p^a(p\cdot A)=0$. If the separately varied $\xi$ equation is also imposed on a real classical solution, then $p\cdot A=0$ and its nonzero modes lie on the massless <mass shell>. For constructing a full-field <Green function>, however, invert the fixed-background quadratic operator before imposing that on-shell constraint. Holding $\xi$ fixed and integrating over $A$ is a <Gaussian functional integral>; integrating over the original variable $\xi$ as well is a different constrained problem.
For $p^2\ne0$, the Lorentzian versions of the <transverse projector of a vector field> and <longitudinal projector of a vector field> are
$$
(P_L)^a{}_b=\frac{p^ap_b}{p^2},\qquad
(P_T)^a{}_b=\delta^a_b-(P_L)^a{}_b.
$$
They satisfy $P_T^2=P_T$, $P_L^2=P_L$, $P_TP_L=0$ and $P_T+P_L=I$. The mixed-index operator is
$$
K^a{}_b=-p^2(P_T)^a{}_b+\frac{p^2}{\xi}(P_L)^a{}_b.
$$
Its inverse follows by inverting these two scalar eigenvalues. Define $G_{ab}$ by $K^{ac}G_{cb}=\delta^a_b$. The <positive-sign covariant gauge-fixing inverse> is
$$
\boxed{G_{ab}(p)=-\frac{\eta_{ab}}{p^2}+\frac{1+\xi}{(p^2)^2}p_ap_b},\qquad
G^{ab}(p)=-\frac{\eta^{ab}}{p^2}+\frac{1+\xi}{(p^2)^2}p^ap^b.
$$
This is the algebraic inverse of the kinetic operator. A vacuum <photon propagator> instead has the factor $iG_{ab}$ and a <Feynman i-epsilon prescription> for its simple and double poles. If the name $G$ is used for the vacuum two-point function, include that factor $i$ consistently in its defining source equation. A <retarded Green function> would use different boundary conditions at the same poles. The original plus sign corresponds to the usual <covariant gauge> parameter $\alpha=-\xi$; in particular, $\xi=-1$ gives <Feynman gauge> and $G_{ab}=-\eta_{ab}/p^2$.
Contracting gives the <longitudinal gauge propagator contraction>
$$
\boxed{p^aG_{ab}(p)=\frac{\xi p_b}{p^2}}.
$$
Thus the contraction is not identically zero as a function of <four-momentum> for any $\xi\ne0$, and it is a nonzero vector at every nonzero non-null <four-momentum>. A particular component may vanish when $p_b=0$. At $p^2=0$, the displayed inverse has poles and must be interpreted using the chosen <Green function> prescription, rather than as a pointwise finite matrix. In the limiting <covariant Landau gauge>, $\xi\to0$, the longitudinal part vanishes. \b[A covariant photon Green function may have a longitudinal component even though physical photon polarizations are transverse.]
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