BTZ partition function

Question

I am unable to obtain the internal energy of the BTZ black hole. Recall its metric, which is given by \begin{align} ds^2=-N^2(r)dt^2+\frac{dr^2}{N^2(r)}+r^2\left(d\phi+N^\phi(r)dt\right)^2\,, \end{align} where the Lapse and shift function are \begin{align} N(r)=\left(-8MG+\frac{r^2}{\ell^2}+\frac{16G^2J^2}{r^2}\right)^{1/2},\quad N^\phi=-\frac{4GJ}{r^2}, \end{align} with $0<\phi\leq 2\pi$. Here $M$ and $J$ are integration constants that correspond to the mass and angular momentum, respectively. When $N(r_\pm)=0$, this black hole has an event horizon at $r=r_+$ and an inner horizon at $r=r_-$ \begin{align} r_\pm^2=4MG\ell^2\left(1\pm\sqrt{1-\left(\frac{J}{M\ell}\right)^2}\right). \end{align} We can write the mass $M$ and the angular momentum $J$ in terms of the event and inner horizon radius as \begin{align} M=\frac{r_+^2+r_-^2}{8G\ell^2},\quad J=\frac{r_+ r_-}{4G\ell}. \end{align} In this regard, the Euclidean Einstein-Hilbert action reads \begin{align} I_\mathrm{EEH}\approx -4\kappa\Lambda\int_0^\beta id\tau\int_{r_+}^{r_{\infty}}\int_0^{2\pi}d\phi r=-4i\pi\kappa\beta\Lambda\left(r_\infty^2-r_+^2\right),\quad \beta=\frac{4\pi r_+^3\ell^2}{2r_+^4-32G^2J^2\ell^2}\,, \end{align} where $\approx$ is an equality that holds on-shell, and we used the fact that in $(2+1)$-dimensions $R\approx 6\Lambda$. Here, $\kappa=(16\pi G)^{-1}$ and $r_\infty\to\infty$ is a radial cut-off. As we see, when $r_\infty\to\infty$ the action diverges quadratically. Thus, we have to sum the Gibbons-Hawking term \begin{align} I_\mathrm{GH}=\kappa \int d^2x\sqrt{|h|}\left(K-K_0\right), \end{align} where $K$ and $K_0$ are the trace of the boundary extrinsic curvature and $AdS_3$, respectively; and $h$ is the determinant of the induced metric \begin{align} h_{\mu\nu}=g_{\mu\nu}-n_\mu n_\nu. \end{align} With the radial-oriented unit normal vector $n=N_E(r)\partial_r$, we find \begin{align} {dh^2_E}={N^2_E}(r)d\tau^2+r^2\left(d\phi+{N^\phi_E}(r)d\tau\right)^2, \end{align} where the Euclidean lapse and shift functions are \begin{align} N_E(r)=\left(-8MG+\frac{r^2}{\ell^2}-\frac{16G^2J^2_E}{r^2}\right)^{1/2},\quad {N^\phi_E}(r)=-\frac{4GJ_E}{r^2},\quad J\to iJ_E, \end{align} respectively. With these ingredients, the Euclidean Gibbons-Hawking surface integral yields \begin{align} I_\mathrm{EGH}=\kappa \int_0^\beta (-i)d\tau\int_0^{2\pi}d\phi\left(-8MG+\frac{2r^2}{\ell^2}\right)\bigg\rvert_{r=r_\infty}=-2i\pi\kappa \beta\left(-8MG+\frac{2r_\infty^2}{\ell^2}\right). \end{align} Therefore, the quadratic divergence $r_\infty^2$ of the Euclidean Einstein-Hilbert action cancels with the Gibbons-Hawking one, so the renormalized action is \begin{align} I_\mathrm{ren}=I_\mathrm{EEH}+I_\mathrm{EGH}=4i\kappa\pi\left(4MG-\frac{r_+^2}{\ell^2}\right)\beta. \end{align} As I said at the beginning, I am unable to obtain the internal energy, since I cannot write the renormalized action in the form $\propto \left(M-\Omega_h J\right)$, where $\Omega_h$ is the velocity of the horizon $\Omega_h={N^\phi_E}(r_+)=-\frac{4GJ_E}{r_+}$. What I am doing wrong?