$$\begin{aligned}
 \label{row1}
&& - \frac{2}{\xi^2} e_\mathrm{LHY}^\mathrm{2d}(\xi) 
\\ \nonumber
&& = \frac{\partial }{\partial r} \left (r 
\sum_{q_z} \int \mbox{d}^2 q_\perp  \frac{1}{\sqrt{q^2(q^2+2\xi)} + q^2+\xi}  e^{i{\bf q} {{\bf r}}} \right)\!\Bigg|_{r=0}. \end{aligned}$$ Now, we expand the right hand side in the power series in $\xi$, and we find $$\begin{aligned}
&& \sum_{q_z} \int \mbox{d}^2 q_\perp  \frac{1}{\sqrt{q^2(q^2+2\xi)} + q^2+\xi}  e^{i{\bf q} {{\bf r}}} 
\\
&& \simeq \int \mbox{d}^2 q_\perp  \frac{1}{\sqrt{q_\perp^2(q_\perp^2+2\xi)} + q_\perp^2+\xi}  e^{i{\bf q} {\bf r}}
\\
&&
+ \sum_{q_z \neq 0} \int \mbox{d}^2 q_\perp \, \frac{1}{2q^2} e^{i{\bf q} {{\bf r}}}
- \xi \sum_{q_z \neq 0} \int \mbox{d}^2 q_\perp \, \frac{1}{2q^4} e^{i{\bf q} {{\bf r}}}.\end{aligned}$$

As we show in details below, in the limit $r\to 0$, we find that the first two terms are of the following form: $$\begin{aligned}
 \nonumber
&& \int \mbox{d}^2 q_\perp  \frac{e^{i{\bf q} {\bf r}} }{\sqrt{q_\perp^2(q_\perp^2+2\xi)} + q_\perp^2+\xi}  
+ \sum_{q_z \neq 0} \int \mbox{d}^2 q_\perp \, \frac{ e^{i{\bf q} {{\bf r}}}}{2q^2}  
\\ \nonumber
\\ \label{row2}
&& =
\frac{\pi^2}{r}  - \frac{\pi}{2} \left( \log \xi + \frac{1}{2} + \log(2\pi^2) \right),\end{aligned}$$ whereas the third term is given by $$\begin{aligned}
\sum_{q_z \neq 0} \int \mbox{d}^2 q_\perp \, \frac{ e^{i{\bf q} {{\bf r}}}}{2q^4} 
\simeq \sum_{q_z \neq 0} \int \mbox{d}^2 q_\perp \, \frac{1}{2q^4}
 = \sum_{n_z > 0} \frac{\pi}{n_z^2}= \frac{\pi^3}{6}.\end{aligned}$$ 