$q_x=q_y=0$ term. In the limit $r\to 0$, the first term is: $$\begin{aligned}
&& \int \mbox{d} q_z  \frac{ e^{i{\bf q} {\bf r}}}{\sqrt{q_z^2(q_z^2+2\xi)} + q_z^2+\xi}  
\\
&& \simeq \int \mbox{d} q_z  \frac{1}{\sqrt{q_z^2(q_z^2+2\xi)} + q_z^2+\xi}
=  \frac{4\sqrt{2}}{3 \sqrt{\xi}},\end{aligned}$$ the second term is: $$\begin{aligned}
&& \sum_{q_x,q_y}{}\!' \int \mbox{d} q_z \, \frac{ e^{i{\bf q} {{\bf r}}} }{2q^4} 
\simeq \sum_{q_x,q_y}{}\!' \int \mbox{d} q_z \, \frac{ 1 }{2q^4}
\\
&& =  \frac{\pi}{4} \sum_{q_x,q_y}{}\!' \frac{1}{(q_x^2+q_y^2)^{3/2}} = 2c_3,\end{aligned}$$ and the third is: $$\label{c2}
 \sum_{q_x,q_y}{}\!' \int \mbox{d} q_z \, \frac{ e^{i{\bf q} {{\bf r}}}}{2q^2}
\simeq \frac{\pi^2}{r} - 2c_2,$$ where $c_2$ is a constant given by $$\begin{aligned}
  \nonumber
  -\frac{4}{\pi}c_2 &=& \lim_{q_c \rightarrow \infty} \left( \sum_{|q_x|,|q_y| \leq q_c}{}\!\!\!\!\!\!\!'\quad \frac{1}{\sqrt{q_x^2+q_y^2}} \right.
  \\ 
  & & 
  \left.
  - \int_{-q_c-1/2}^{q_c+1/2} 
  \mbox{d}q_x   \int_{-q_c-1/2}^{q_c+1/2}  \mbox{d}q_y \, \frac{1}{\sqrt{q_x^2+q_y^2}} \right)\label{c22}.\end{aligned}$$ In these expression, the sum $\sum'$ means that we exclude the term with $q_x=q_y=0$. Collecting the terms altogether, we recover Eq. (15) of the main paper.

Below, we show in details how to derive Eqs. (\[c2\]) and (\[c22\]). To start, we assume ${{\bf r}}= (r,0,0)$ for simplicity, and, then, we obtain $$\begin{aligned}
\sum_{q_x,q_y}{}\!' \int \mbox{d} q_z \, \frac{ e^{i q_x r}}{2q^2}
= \sum_{q_x,q_y}{}\!' \frac{\pi}{2\sqrt{q_x^2+q_y^2}} e^{ -r\sqrt{q_x^2+q_y^2} }.\end{aligned}$$ In the limit $r \to 0$, the exponent is important for ensuring the 