= Solution
For an <infrared-regulated phase covariance>, fix the uniform <order-parameter phase> zero mode and introduce ultraviolet and <infrared cutoffs> before taking limits. With the Fourier convention
$$
\theta(\mathbf x)=\int\frac{d^dq}{(2\pi)^d}\,\theta_{\mathbf q}e^{i\mathbf q\cdot\mathbf x},
$$
the dimensionless quadratic <Hamiltonian> diagonalizes as
$$
\beta H=\frac{\bar K}{2}\int\frac{d^dq}{(2\pi)^d}q^2\theta_{\mathbf q}\theta_{-\mathbf q}.
$$
Completing the square with a source gives $Z[J]/Z[0]=\exp[\tfrac12\int JGJ]$, with inverse kernel
$$
\langle\theta_{\mathbf q}\theta_{\mathbf q'}\rangle
=\frac{(2\pi)^d\delta^{(d)}(\mathbf q+\mathbf q')}{\bar Kq^2},\qquad
G(\mathbf x)=\frac1{\bar K}\int\frac{d^dq}{(2\pi)^d}\frac{e^{i\mathbf q\cdot\mathbf x}}{q^2}.
$$
There is no additional $k_BT$ factor: it is already included in $\bar K$ because the supplied <Hamiltonian> is $\beta H$.
It is simplest to evaluate the real-space dependence by its defining Green equation, $-\bar K\nabla^2G=\delta^{(d)}$. Rotational invariance gives $r^{d-1}G'(r)=$ constant for $r>0$. Integrating over a small <sphere> fixes $-\bar KS_dr^{d-1}G'(r)=1$, where $S_d=2\pi^{d/2}/\Gamma(d/2)$. Integrating once more,
$$
\boxed{G(r)=-\frac{r^{2-d}}{(2-d)S_d\bar K}+C,\qquad d\ne2.}
$$
In two dimensions, take the limit after absorbing the divergent $r$-independent piece into $C$:
$$
\boxed{G(r)=-\frac1{2\pi\bar K}\log(r/a)+C.}
$$
For $d\le2$, $G$ itself has an infrared-divergent constant in the infinite system; it is not a finite thermodynamic <covariance> without a regulator. The physical <phase-difference variance> is finite at fixed separation and cancels that constant. These are smooth <spin wave> <correlations>; compact-field <vortex> sectors are considered separately below.
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