w_{n,V_\varphi}(y_1,\dots, y_n) + \sum_i \log (1+|\varphi(y_i)|^2)  - \sum_i \log (1+|y_i|^2),\quad x_i = \varphi(y_i).$$

It follows from that an upper bound for $\min w_{n,V}$ can be computed by using a minimizer for $w_{n,V_\varphi}$ as a test function. But now we recall that $\mu_{V_\varphi}$ has compact support, hence the results of [@2DSandier] apply and we find, using the fact that for such a minimizer $\frac1n\sum_i\delta_{y_i}$ converges to $\mu_{V_\varphi}$, $$\label{mm}\min w_{n,V} \le  n^2 I_{V_\varphi}(\mu_{V_\varphi}) - \hal n \log n + \left(\alpha_{V_\varphi}  + \int \log \left(\frac{1+|\varphi(x)|^2}{1+|x|^2}\right)\,d\mu_{V_\varphi}(x)\right)n + o(n),$$ where $$\alpha_{V_\varphi} = \frac{\alpha_1}{\pi}-\hal \int_{\Sigma_{V_\varphi}} m_{V_\varphi}(x)\log  m_{V_\varphi}(x)\, dx, \quad  \alpha_1:=\min_{\mathcal A_1} W.$$ We remark that $I_{V_\varphi}(\mu_{V_\varphi})=I_V(\mu_V)$ because $\mu_{V_\varphi}=\varphi^{-1}\sharp \mu_V$. Moreover, it follows from that $$m_{V_\varphi}(x) = m_V(\varphi(x))\left(\frac{1+|\varphi(x)|^2}{1+|x|^2}\right)^2,$$ which plugged in the expression for $\alpha_{V_\varphi}$ and then in yields, $$\min w_{n,V} \le n^2 I_V(\mu_V) - \hal n \log n + \alpha_{V} n +o(n),$$ which matches the lower-bound we already obtained and thus proves Theorem \[thintro\] in the case where the support of $\mu_V$ is not the full plane.

Proof of the upper bound by compactification and conclusion
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Here we assume that $\Sigma_V={\mathbb R}^2$. Let $$\varphi(z):=-\frac{1}{z}=\varphi^{-1}(z).$$ Then, using the notations of the previous section, we deduce from that $$V_\varphi(z)=V(\varphi(z))+2\log|z|.$$ To simplify exposition and notation, we assume that 