= Solution
Fix the uniform angular mode, for example by fixing the mean phase in a large box, and introduce a microscopic <ultraviolet cutoff> $a$. The quadratic <Goldstone-mode effective free energy> becomes, under a <Fourier transform>,
$$
\beta H=\frac{\bar K}{2}\int\frac{d^dq}{(2\pi)^d}\,q^2\theta(q)\theta(-q).
$$
To evaluate the <Gaussian functional integral>, add a source $j$. Completing the square in every nonzero <Fourier transform> mode gives
$$
\frac{Z[j]}{Z[0]}=\exp\left[\frac1{2\bar K}\int\frac{d^dq}{(2\pi)^d}\frac{j(q)j(-q)}{q^2}\right].
$$
Taking two source derivatives yields the <correlation function>
$$
\langle\theta(q)\theta(q')\rangle=\frac{(2\pi)^d\delta^{(d)}(q+q')}{\bar Kq^2},\qquad
G(x)=\langle\theta(x)\theta(0)\rangle=\frac1{\bar K}\int\frac{d^dq}{(2\pi)^d}\frac{e^{iq\cdot x}}{q^2}.
$$
The cutoff and the treatment of the uniform mode regulate this expression. At separations much larger than $a$, its nonconstant part is the <Green function> satisfying $-\bar K\nabla^2G=\delta^{(d)}(x)$. Rotation invariance and the radial <Laplacian> give $(r^{d-1}G')'=0$ away from the origin. Integrating the <Green function> equation over a small sphere, whose unit-sphere area is $S_d=2\pi^{d/2}/\Gamma(d/2)$, fixes the flux:
$$
-\bar K S_dr^{d-1}G'(r)=1.
$$
Thus for $d\ne2$ the required long-distance <correlation function> is
$$
\boxed{G(r)=-\frac{r^{2-d}}{(2-d)S_d\bar K}+C.}
$$
At $d=2$, the finite separation-dependent limit is
$$
\boxed{G(r)=-\frac1{2\pi\bar K}\log(r/a)+C.}
$$
The constant absorbs the divergent constant term in the $d\to2$ limit. In low dimension the individual $G(r)$ has no regulator-independent infinite-volume value; its differences do. This is why the <phase-difference variance>, rather than an absolute phase variance, is the useful observable.
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