Divergent Coulomb integrals in superfluid fluctuations In Chapter 3 of Kardar's statistical physics of fields, in the context of lower critical dimension, he works out an example about superfluids where starting from the probablity  of a particular configuration in the ordered phase, ie
$$\mathscr{P}[\theta(\mathbf{x})] \propto \exp\left\lbrace-\frac{K}{2} \int \mathrm{d}^d\mathbf{x} (\nabla\theta)^2\right\rbrace,$$
where $\theta$ is the phase of the wavefunction, he goes on to calculate the correlation functions, obtaining
$$ \left\langle \theta(\mathbf{x})\theta(\mathbf{x}')\right\rangle = - \frac{C_d(\mathbf{x}-\mathbf{x}')}{K},$$ 
where 
$$ C_d(\mathbf{x}) = - \int \frac{\mathrm{d}^d\mathbf{q}}{(2\pi)^d} \frac{e^{i\mathbf{q}\cdot\mathbf{x}}}{\mathbf{q}^2}$$
is presumebly the Coulomb potential in $d$ dimensions since
$$\nabla^2 C_d(\mathbf{x}) = \delta^d(\mathbf{x}).$$
He then uses the Gauss theorem, ie
$$ \int \mathrm{d}^d\mathbf{x} \nabla^2 C_d = \oint \mathrm{d}S\cdot\nabla C_d$$
together with $\nabla C_d = (\mathrm{d}C_d/\mathrm{d}r)\hat{r}$, which simply is due to the spherical symmetry, to build the solution
\begin{equation}
C_d(x) = \frac{x^{2-d}}{(2-d)S_d} + c_0,
\end{equation}
where $S_d$ is the area of unit sphere in $d$ dimensions and $c_0$ a constant to be determined. Then it is argued that this clearly diverges for $d \leq 2$ dimesnions as $x\to\infty$, while approaches $c_0$ for $d>2$. This is however related to the celebrated Mermin-Wagner theorem.
Now my question is that it seems to me $C_d(\mathbf{x})$ is a divergent integral as the integrand doesn't decay fast enough when $\mathbf{q}\to\infty$, except in $d=1$. So I would naively guess the integral is convergent in $d=1$, while in $d \geq 2$ it diverges, which is exactly opposite what Kardar claims. 
I guess there is some type of regularization scheme at work in Kardar's approach (maybe momentum cutoff or dimensional regularization), but I don't understand it exactly and in which of the steps outlined above it's been applied. 
I tried calculating the integral directly (in the $d$ dimensional sphereical coordinates) hoping that I can apply the momentum cutoff on the final result and compare it with Kardar's solution, but have no clue how to do the calculation since the angle between $\mathbf{q}$ and $\mathbf{x}$ introduced by $e^{i\mathbf{q}\cdot\mathbf{x}}$ makes it complicated. 
EDIT
I see the integral needs probably both lower and upper cutoffs (say $\Lambda_l$ and $\Lambda_u$) to converge, depending on the dimensionality of the space so
\begin{equation}
C_d(\mathbf{x}) = \lim_{\substack{\Lambda_u\to\infty \\ \Lambda_l \to 0}} \left\lbrace\mathcal{F}(|\mathbf{x}|) + \mathcal{G}(\Lambda_{u}, \Lambda_{l}) + c_0\right\rbrace.
\end{equation}
(The above form probably needs justification)
Then one argues the function $\mathcal{G}$ doesn't affect the correlations meaningfully (it is just a constant that approaches infinity) and we only study the behavior of function $\mathcal{F}$. My question is how to make sure the manipulation in Kardar's book is exactly doing this. 
 A: When you say the integrals diverge i guess you are refering to the following:
$$ |C_d| \leq \int \frac{d^d\mathbf{q}}{(2\pi)^d} \frac{1}{\mathbf{q}^2} = \text{const.} \times
\int_0^\infty q^{d-3} dq $$
which is UV-divergent unless if $d \geq 2$ and IR-divergent if $d \leq 2$.
However, this doesn't mean the integral is ill-defined, just that this way of bounding the integral doesn't produce an useful result.
Let's try to improve this bound:
$$ C_d = -\int \frac{d^d\mathbf{q}}{(2\pi)^d} \frac{e^{i\mathbf{q \cdot x}}}{\mathbf{q}^2} = - \frac{1}{(2\pi)^d} \int_0^\infty q^{d-3} \left[ \int_{S_{d-1}} e^{i q \ \mathbf{\hat{q}\cdot x}}  \ d \Omega_{d-1} \right] dq  \ ,$$
here $\mathbf{\hat{q}} = \frac{\mathbf{q}}{q}$ is the unit vector in $\mathbf{q}$ direction and $d \Omega_{d-1}$ is the surface element on the $(d-1)$ sphere.
For handling the integral over the $d-1$-sphere, pick spherical coordinates such that the $q_{n}$-axis coincides with the $\mathbf{x}$-axis, so that $\mathbf{q \cdot x} = q x \cos(\theta)$ and 
$$ d \Omega_{d-1} = \sin(\theta)^{d-2} d \theta d \Omega_{d-2} \ .$$
For the case $d=2$, one has to look at the integral
$$ \int_{S_{1}} e^{i q \ \mathbf{\hat{q}\cdot x}}  \ d \Omega_{1} =  \ \int_{-\pi}^\pi e^{i q x \cos(\theta)} d \theta $$
instead. One may look this up to be $2\pi J_0(qx)$, where $J_0$ is the Bessel function of the first kind. It has the asymptotics
$$J_0(qx) \sim \text{const.} \times q^{-1/2} \ , $$
so
$$ C_2 = - \frac{1}{2\pi} \int_{0}^\infty \frac{J_0(qx)}{q} dq $$ 
does not suffer from UV-divergencies.
Start with the case $d > 2$ even. Then
$$\int_{S_{d-1}} e^{i q \ \mathbf{\hat{q}\cdot x}}  \ d \Omega_{d-1} = \text{Vol}(S_{d-2}) \ \int_{0}^\pi e^{i q x \cos(\theta)} \sin(\theta)^{d-2}  d \theta = 
\\
=(2\pi)^{\frac{d}{2}}\frac{J_{\frac{d-2}{2}}(qx)}{(qx)^{\frac{d-2}{2}}} \ .  $$
Hence
$$ C_{d} = -\frac{1}{(2\pi)^{\frac{d}{2}}} \frac{1}{x^{d-2}} \int_0^\infty q^{d/2-2} J_{\frac{d}{2}-1}(q) dq \ .$$
for the case of $d=4$ it gives
$$C_4 = -\frac{1}{(2\pi)^2 x^2} \int_0^\infty J_1(q) dq = -\frac{1}{(2\pi)^2 x^2} \ .$$
for all other $d$ the integral is divergent and one has to introduce a regularization. 
The integral was convergent for $d=4$ because the oscillations of the Bessel function are leading to destructive interference. It is very similar to how $\sum_{n>0}n^{-1}$ is diverging while $\sum_{n>0} (-1)^n n^{-1}$ is not.
Now for $d = 3$ odd:
$$\int_{S_{2}} e^{i q \ \mathbf{\hat{q}\cdot x}}  \ d \Omega_{2} = 2\pi \ \int_{0}^\pi e^{i q x \cos(\theta)} \sin(\theta)  d \theta = 4\pi \frac{\sin(qx)}{qx} \ . $$
Thus
$$ C_3 = - \frac{1}{2\pi^2 x} \int_0^\infty \frac{\sin(q)}{q}  dq = - \frac{1}{4\pi x} \ . $$
$d = 1$:
$$C_1 = \frac{1}{\pi} \int_0^\infty \frac{e^{iqx}}{q^2} d q $$
This has no UV divergences, but clearly suffers from an IR divergence which will be discussed below.
The general $d$ case is a bit awkward, so i am not discussing it here.
So we see that there are no UV-divergencies really for $d \leq 4$. One should remark that in $d=2$ there is however an IR divergence. Naively, one could perform a change of variables $qx =p$, which would lead to $C_2$ being constant. However, the integral as written diverges since $J_0(0) \neq 0$ hence this manipulations are not allowed.
Now turn to the question of the IR-divergence in two dimension. This is real, so one needs to regularize the integral, for example by introducing a cutoff in the following way:
$$C_{2,\mu} := - \frac{1}{2\pi} \int_{0}^\infty \frac{J_0(qx)-f(qx)}{q} dq -  \frac{1}{2\pi} \int_{\mu}^\infty \frac{f(qx)}{q} dq  \ ,$$
where $f$ is some function such that $f(q) \sim 1 $ as $q\rightarrow 0$. Then the first integral is well-defined since $J_0(qx)-f(qx) \in \mathcal{O}(q^2)$. Furthermore, it does not depend on $x$ by a simple change of variables. Hence
$$C_{2,\mu} = -  \frac{1}{2\pi} \int_{\mu}^\infty \frac{f(qx)}{q} dq + C(f) \ , \quad C(f) := - \frac{1}{2\pi} \int_{0}^\infty \frac{J_0(q)-f(q)}{q} dq \ ,$$
We may now pick any function. A convenient choice is $f(x) = 1$ for $x < \Lambda$ and zero otherwise. Then:
$$C_{2,\mu} = -\frac{1}{2\pi} \int_\mu^{\Lambda/x} \frac{dq}{q} + C(f) = \frac{1}{2\pi} \log(x) + \frac{1}{2\pi} \log(\mu/\Lambda) + C(f) \ .$$ 
One can now check that the $x$-dependence of $C_2(x)$ is in fact universal in the limit $\mu \rightarrow 0$: it does not depend on the choice of $f$. Indeed, consider two different such functions $f_1,f_2$. Then their difference goes to zero at $q=0$. Hence in the integral
$$C_{2,f_1} - C_{2,f_2} = \frac{1}{2\pi} \int_0^\infty \frac{f_1(qx)-f_2(qx)}{q} dq + \text{const.} $$
the lower bound of integration may be extended to $\mu = 0$ and the resulting integral does not depend on position $x$.
Now in one dimensions: let f(q) be a function with $f(0) = 1$.
Then the following is well-defined:
$$ C_{1,f} = \frac{1}{\pi} \int_\mu^\infty \frac{f(qx)}{q^2} dq + \frac{1}{\pi} \int_0^\infty \frac{\cos(qx) - f(qx)}{q^2}dq \ . $$
The first term is in fact independent of $x$ while in the second you introduce a factor sign$(x)$ when changing variables $qx\rightarrow q$. Hence
$$ C_{1,f} = A_f + \text{sign}(x) B_f \ ,$$
with $A_f,B_f$ some constants depending on $f$.
A: Indeed, if one uses textbook measure theory, this is nonsense since the function is not integrable. The proper mathematical setting for understanding the correctness of this kind of computations with full rigor is the theory of Schwartz distributions. 
See for example 
https://math.stackexchange.com/questions/3120284/how-to-calculate-c-a-where-leftf-mapsto-int-mathbbr-fracft-f0t
where a rigorous computation of the Fourier transform of the OP's question is done.
