Inserting these expressions into Eq. (\[row1\]), we find $$\begin{aligned}
e_\mathrm{LHY}^\mathrm{2d}(\xi) = \frac{\pi}{4} \xi^2 \left( \log \xi + \frac{1}{2} + \log(2\pi^2)  + \frac{\pi^2}{3} \xi \right),\end{aligned}$$ which is Eq. (7) of the main paper.

Now, we show in details how to derive Eq. (\[row2\]). In order to simplify the calculation, we take $z=0$ in the vector ${{\bf r}}=(x,y,z)$, which gives $r = \sqrt{x^2+y^2}$.

We rewrite the first term from the left-hand side of Eq.(\[row2\]) in cylindrical coordinates: $$\begin{aligned}
&& \int \mbox{d}^2 q_\perp  \frac{e^{i{\bf q} {\bf r}} }{\sqrt{q_\perp^2(q_\perp^2+2\xi)} + q_\perp^2+\xi}  
\\
&&  = 2\pi \int_0^\infty q_\perp \mbox{d} q_\perp \,  \frac{J_0(q_\perp r)}{\sqrt{q_\perp^2(q_\perp^2+2\xi)} +q_\perp^2+\xi}.\end{aligned}$$ The change of variable into $n = q_\perp r$ yields: $$\begin{aligned}
&&
= 2\pi \int_0^\infty n \mbox{d} n  J_0(n) \frac{1}{\sqrt{n^2(n^2+2\xi r^2)} + n^2+\xi r^2}
\\
&& \simeq 2\pi \int_0^{n_0} n \mbox{d} n   \frac{1}{\sqrt{n^2(n^2+2\xi r^2)} + n^2+\xi r^2}
\\
&&\quad + 2\pi \int_{n_0}^\infty  \mbox{d} n  \, \frac{J_0(n)}{2n}. \end{aligned}$$ Here, we approximated $J_0(n)\approx 1$ in the first integral on right-hand side, and neglected $\xi r^2$ terms in the denominator in the second integral since the main contribution comes from large $n$. These approximations are valid as long as $n_0 \ll 1$. In these limit, the integrals can be evaluated: $$\int_0^{n_0} n \mbox{d} n   