hyperbolic distance between two reals The hyperbolic distance between two points $z_1, z_2 \in D_1(0)$ is defined as 
\begin{equation}
d(z_1, z_2)= \inf_\gamma \int_0^1 \frac{|\gamma'(t)|} {1- |\gamma(t)|^2} dt \, ,
\end{equation}where the infimum is taken over all smooth curves $\gamma:[0,1] \to D_1(0)$ joining $z_1$ and $z_2$. Now I want to prove that 
\begin{equation}
d(0,s) = \frac{1}{2} \log \left( \frac{1+s}{1-s}\right)
\end{equation} for $s \in [0,1)$.
I think I know where I have to go, since
\begin{equation}
\int_0^s \frac{1}{1-x^2} dx = \frac{1}{2} \log \left( \frac{1+s}{1-s}\right) \,,
\end{equation} but I don't know how to get there. Any ideas?
 A: Have you considered letting $\gamma(t) = t$, for $0 \le t \le s$? 
Of course, doing this only proves that the distance is no more than the specified value. But that's a start at least. 
Then again, if you knew what all the geodesics looked like, you'd know that it was a shortest path...do you already know that? 
A: If you consider a smooth path $\gamma(t)=x(t)$ with $x(t)$ a real function satisfying $x(0)=0$ and $x(1)=s$, you can easily integrate and use the fundamental theorem of calculus to show that
$$\int_\gamma \frac{|\gamma '(t)|}{1-|\gamma(t)|^2}dt = \frac{1}{2}\log \left(\frac{1+s}{1-s} \right).$$
Notice that this works for any such smooth path. Now let $\gamma(t) = x(t) + iy(t)$ be a smooth path from $0$ to $s$ with $x(t)$ and $y(t)$ real functions. Then $x'(t) \leq \sqrt{x'(t)^2} \leq \sqrt{x'(t)^2 + y'(t)^2} = |\gamma'(t)|$. Furthermore, $1-x(t)^2 \geq 1-(x(t)^2 + y(t)^2) = 1-|\gamma(t)|^2$ since $x(t)^2 + y(t)^2 \leq 1$. Hence, 
$$\frac{1}{2}\log \left(\frac{1+s}{1-s} \right) = \int_0^1 \frac{x'(t)}{1-x(t)^2}dt \leq \int_0^1 \frac{|\gamma'(t)|}{1-|\gamma(t)|^2} dt$$
as required.
A: { Let $x,y \in (-1,1)$ and $\gamma$ a smooth curve joining $x,y$ $\gamma(0)=x,\gamma(1)=y$ in $\mathbb{D}$ $$\int_{\gamma}\lambda (j)|dj|= \int_{0}^{1}\frac{|\gamma'(t)|}{1-|\gamma(t)|^2}dt\geq \int_{0}^{1}\frac{Re(\gamma'(t))}{1-Re(\gamma(t))^2}dt=\frac{1}{2}\log(\frac{1+|\frac{x-y}{1-xy}|}{1-|\frac{x-y}{1-xy}|})=d_\mathbb{D}(x,y).$$
Let now $z_0,w_0 \in \mathbb{D}$  and $$T_{w_0}(z)=\frac{w_0-z}{1-\bar{w_0}z}$$ then by Schwarz-Pick lemma$$d_\mathbb{D}(z_0,w_0)=d_\mathbb{D}(T_{w_0}(z_0),T_{w_0}(w_0))=d_\mathbb{D}(0,T_{w_0}(z_0)=d_\mathbb{D}(0,|T_{w_0}(z_0)|=$$
$$\frac{1}{2}\log(\frac{1+|T_{w_0}(z_0)|}{1-|T_{w_0}(z_0)|})=\frac{1}{2}\log(\frac{1+|\frac{w-z}{1-\bar{w}z}|}{1-|\frac{w-z}{1-\bar{w}z}|})=arc\tanh(|\frac{w-z}{1-\bar{w}z}|),$$ since $z_0,w_0 \in \mathbb{D}$ were  arbitraries.}
