close to the horizon, we take $$\label{eps def}
z_{\textrm{max}}  \equiv z_0- \varepsilon \zeta_{\textrm{max}} 
\hspace{1.5cm}
\varepsilon \,\rightarrow\,0
\hspace{1.5cm}
\textrm{finite $\zeta_{\textrm{max}}$}$$ and change the integration variable in (\[calArea\]) according to this expansion, i.e. we set $w = z_0- \varepsilon \zeta$, where $0 < \zeta_{\textrm{max}} < \zeta$. Then, the finite term can be written as follows $$\label{A0 expanded}
\mathcal{A}_d(z_{\textrm{max}},0)\,=\,
\sum_{k \in B }
\varepsilon^{k}
 \int_{\frac{z_0}{\varepsilon}}^{\zeta_{\textrm{max}}} 
\mathcal{I}_{k}(\zeta,\zeta_{\textrm{max}}) d\zeta
-\frac{2}{(d-1)z_{\textrm{max}}^{d-1}}$$ where $B \subset [k_{\textrm{min}}, \infty) \subset \mathbb{Q}$ is some discrete set of increasing rational numbers, which are not necessarily positive ($k_{\textrm{min}} <0$). For instance, in the case of the charged black hole with $d=2$ we have $k \in \{ -1/2,1/2,1,3/2,\dots\}$. In order to write $\mathcal{A}_d(z_{\textrm{max}},0)$ as an expansion in terms of powers of $z_0- z_{\textrm{max}}$, we have to compute the definite integrals occurring at each $k$ and then expand each of them for small $\varepsilon$. Then this expansion can be written in powers of $z_0- z_{\textrm{max}}$ by using the definition $\varepsilon=(z_0- z_{\textrm{max}})/\zeta_{\textrm{max}}$ from (\[eps def\]).\
In all examples we have considered we find that this method provides only the divergent term as $z_{\textrm{max}} \rightarrow z_0$. This is due to the fact that all the integrals occurring in (\[A0 expanded\]) give a contribution to the finite 