n \,  \coth \left( \frac{n \pi}{r} \right)
\\
&&    
 +  \frac{\pi}{r} \int_{n_0}^\infty \mbox{d} n \,   J_0(n)
 \simeq \log \left( \frac{r}{2\pi \epsilon} \right)   + \frac{\pi n_0}{r} + \frac{\pi}{r}(1-n_0),
 \end{aligned}$$ where $ \epsilon \ll n_0 \ll 1$ and $n_0 \gg r$. Finally, from these results altogether, we obtain $$\label{w2}
\sum_{q_z \neq 0} \int \mbox{d}^2 q_\perp \, \frac{ e^{i{\bf q} {{\bf r}}}}{2q^2} = \frac{\pi^2}{r} + \pi \log \left( \frac{r}{4\pi} \right) + \pi \gamma.$$ Therefore, the sum of the expressions from Eqs. (\[w1\]) and (\[w2\]) recovers Eq. (\[row2\]).

Crossover from 3D to quasi-1D {#crossover-from-3d-to-quasi-1d .unnumbered}
-----------------------------

From Eq. (14) of the main paper, we find $$\begin{aligned}
 \label{row21}
&& - \frac{2}{\xi^2} e_\mathrm{LHY}^\mathrm{1d}(\xi) 
\\ \nonumber
&& = \frac{\partial }{\partial r} \left (r 
\sum_{q_x,q_y} \int \mbox{d} q_z  \frac{1}{\sqrt{q^2(q^2+2\xi)} + q^2+\xi}  e^{i{\bf q} {{\bf r}}} \right). \end{aligned}$$ As before, we this expression in powers of $\xi$, and we get: $$\begin{aligned}
&& \sum_{q_x,q_y} \int \mbox{d} q_z  \frac{e^{i{\bf q} {{\bf r}}} }{\sqrt{q^2(q^2+2\xi)} + q^2+\xi}   
\\
&& \simeq \int \mbox{d} q_z  \frac{ e^{i{\bf q} {\bf r}} }{\sqrt{q_z^2(q_z^2+2\xi)} + q_z^2+\xi}  
\\
&&
+ \sum_{q_x,q_y}{}\!' \int \mbox{d} q_z \, \frac{ e^{i{\bf q} {{\bf r}}}}{2q^2} 
- \xi \sum_{q_x,q_y}{}\!' \int \mbox{d} q_z \, \frac{ e^{i{\bf q} {{\bf r}}}}{2q^4},\end{aligned}$$ where $\sum_{q_x,q_y}'$ denotes the sum without the 