\frac{ 2\pi}{\sqrt{n^2(n^2+2\xi r^2)} + n^2+\xi r^2} \simeq \frac{\pi}{2} \log \left(\frac{2n_0^2}{\xi r^2} \right)  - \frac{\pi}{4}$$ and $$2\pi \int_{n_0}^\infty  \mbox{d} n  \frac{J_0(n)}{2n} \simeq  -\pi \left( \log(n_0) - \log(2) + \gamma \right),$$ where $\gamma$ denotes the Euler’s constant. As a result, we obtain $$\begin{aligned}
 \nonumber
&& \int \mbox{d}^2 q_\perp  \frac{e^{i{\bf q} {\bf r}} }{\sqrt{q_\perp^2(q_\perp^2+2\xi)} + q_\perp^2+\xi} 
\\ \label{w1}
&& = \frac{3\pi}{2} \log(2) - \pi\gamma  - \frac{\pi}{2} \log(\xi r^2)  - \frac{\pi}{4}.\end{aligned}$$

Let us now analyze the second integral in Eq. (\[row2\]): $$\begin{aligned}
&& \sum_{q_z \neq 0} \int \mbox{d}^2 q_\perp \, \frac{ e^{i{\bf q} {{\bf r}}}}{2q^2} 
 = \pi \sum_{q_z \neq 0} \int_0^\infty q_\perp \mbox{d}q_\perp \, \frac{J_0(q_\perp r)}{q_\perp^2+q_z^2}
\\
&&
 = \pi \int_0^\infty q_\perp \mbox{d}q_\perp \, J_0(q_\perp r) \frac{\pi q_\perp \coth(q_\perp \pi) -1 }{q_\perp^2}.\end{aligned}$$ Now, we change the variables into $n = q_\perp r$, and take $\epsilon$ as the lower limit of the integral. The right-hand side is: $$\begin{aligned}
 \frac{\pi}{r}\int_{\epsilon }^\infty \mbox{d} n \,   J_0(n)\coth \left( \frac{n \pi}{r} \right)  
- \int_{\epsilon }^\infty \mbox{d} n \, \frac{J_0(n)}{n}.\end{aligned}$$ Now, the second term in this expression is: $$\begin{aligned}
 - \int_{\epsilon }^\infty \mbox{d} n \, \frac{J_0(n)}{n} \simeq \log(\epsilon/2) + \gamma,\end{aligned}$$ whereas the first gives: $$\begin{aligned}
&&\frac{\pi}{r}\int_{\epsilon }^\infty \mbox{d} n \,  J_0(n) \coth \left( \frac{n \pi}{r} \right)  
\simeq \frac{\pi}{r}\int_{\epsilon }^{n_0} \mbox{d} 