Holographic entanglement entropy (Thermal case) I'm trying to calculate the entanglement entropy in CFT2/AdS3 in the thermal case for a finite interval (-a,a). I'm reading the paper of Takayanagi and Rangamani (2016): https://arxiv.org/abs/1609.01287 (page 68) but they don't have the explicit calculation. The result for this case has to be:
\begin{equation}S_{A}= \frac{c}{3}log\left(\frac{\beta}{\pi\epsilon}\sinh\left(\frac{2\pi a}{\beta}\right)\right).
\end{equation}
where $\epsilon$ is a cutoff.
My problem is that I have those 2 equations:
The metric of BTZ BH:
\begin{equation}
ds^{2}=-\frac{(r^{2}-r_{+}^{2})}{l^{2}_{AdS}}dt^{2}+ \frac{dr^{2}}{(r^{2}-r_{+}^{2})}+ \frac{r^{2}}{l^{2}_{AdS}}dx^{2}
\end{equation}
and from geodesics equations:
\begin{equation}
\frac{dr}{dx}= \frac{r}{l^{2}_{AdS}}\sqrt{(r^{2}-r_{+}^{2})\left(\frac{r^{2}}{r^{2}_{\ast}}-1\right)}.
\end{equation}
I've tried the same procedure that I used with the normal case (without temperature):
\begin{align}
\textrm{Length}& =\int \textrm{d}x\sqrt{r^{2}+r^{2}\left(r^{2}-r_{+}^{2}\right)\left(\frac{r^{2}}{r_{\ast}^{2}}-1\right) \frac{1}{r^{2}-r_{+}^{2}}}= \int\textrm{d}x\sqrt{r^{2}+\frac{r^{2}r^{2}}{r_{\ast}^{2}}-r^{2}}\\
&= \int \textrm{d}x\frac{r^{2}}{r_{\ast}}=\int_{r_{\ast}}^{\infty} \frac{\textrm{d}r}{r}\frac{1}{\sqrt{(r^{2}-r_{+}^{2})\left(\frac{r^{2}}{r_{\ast}^{2}}-1\right)}}\frac{r^{2}}{r^{\ast}}= 2\ln\left( \sqrt{r^{2}-r_{+}^{2}}+\sqrt{r^{2}-r_{\ast}^{2}}\right)\Bigg|_{r_{\ast}}^{\infty}
\end{align}
Now I don't know how to get the result that they obtained... because I don't know how to continue. The only thing they especified in their article is that
\begin{equation}
r_{\ast}=r_{+}\textrm{coth}\left(ar_{+}\right).
\end{equation}
 A: First, introduce a UV cutoff $r_b\equiv (1/\epsilon)\to\infty$. Then perform the length integral on $[r_*,r_b]$ which you'll get a linear combination of natural logarithms in terms of $r_+, r_*$ and $r_b$ (let's denote the result by $\mathcal{L}\,$). Now by integrating the following equation $$\frac{dr}{dx}= r\,\sqrt{(r^{2}-r_{+}^{2})\left(\frac{r^{2}}{r^{2}_{\ast}}-1\right)}$$ on the half interval for $x$, you'll get an expression for $r_*$ in terms of $a$, $r_+$ and $r_b\,$ and you can approximate it in the limit $r_b\to \infty$ which will give you $$r_*=r_+\coth\,(a\, r_+)$$ Substituting this result in $\mathcal{L}$ and again approximating the obtained result in $r_b\to \infty$ limit yields $$2\ln\,\left(\frac{2\,r_b}{r_+}\,\sinh(a\,r_+) \right)$$ Finally, by substituting $r_b=1/\epsilon\,$, $r_+=2\pi/\beta\,$ and $c=\frac{3}{2G_N}$ together with $S=\frac{\mathcal{L}}{4G_N}\,$, you'll will get the desired result. Please note that for convenience, I set the $\small AdS$ radius to unity. I hope this helps you.
