Prove $\int_R^{\infty}\frac{dt}{f^{-1}(t)}=\int_r^{\infty}\frac{f(s)}{s^2}-\frac{R}{r}$ Preambule: The proof given below is erroneous, so you can skip down to the actual question.
Let $r,R>0$ and suppose we are given unbounded strictly increasing continuously differentiable function $f:[r,+\infty]\to[R,+\infty]$. I need to show that either the following equality 
$$
\int_R^{\infty}\frac{dt}{f^{-1}(t)}=\int_r^{+\infty}\frac{f(s)}{s^2}-\frac{R}{r}
$$
holds or both integrals in it are infinite. Here $f^{-1}$ stands for the inverse function.
Below is my proof. Denote $g(t)=f^{-1}(t)$, then $g(f(s))=s$ and $f(g(t))=t$. Since $f$ is strictly increasing, then
$$
\lim\limits_{s\to+\infty}f(s)=+\infty
\qquad\implies\qquad
\lim\limits_{t\to+\infty}g(t)=+\infty.
$$ 
For any $\xi>R$ we have
$$
\begin{align}
\int_{R}^{\xi}\frac{dt}{g(t)}
&=\{t=f(s)\}\\
&=\int_{f^{-1}(R)}^{f^{-1}(\xi)}\frac{d(f(s))}{g(f(s))}\\
&=\int_{r}^{g(\xi)}\frac{f'(s)ds}{s}\\
&=\int_{r}^{g(\xi)}\frac{f'(s)ds}{s}\\
&=\frac{f(s)}{s}\Bigl|_{r}^{g(\xi)}-\int_{r}^{g(\xi)}f(s)\frac{-ds}{s^2}\\
&=\frac{f(g(\xi))}{g(\xi)}-\frac{f(r)}{r}+\int_{r}^{g(\xi)}\frac{f(s)ds}{s^2}\\
&=\frac{\xi}{g(\xi)}-\frac{R}{r}+\int_{r}^{g(\xi)}\frac{f(s)ds}{s^2}\\
\end{align}
$$
Clearly, $g$ is positive, so $L:=\limsup\limits_{\xi\to+\infty}\frac{\xi}{g(\xi)}\geq 0$.
If $\limsup\limits_{\xi\to+\infty}\frac{\xi}{g(\xi)}=0$, then a fortiori $\lim\limits_{\xi\to+\infty}\frac{\xi}{g(\xi)}=0$ and
$$
\begin{align}
\int_{R}^{\infty}\frac{dt}{f^{-1}(t)}
&=\lim\limits_{\xi\to\infty}\int_{R}^{\xi}\frac{dt}{g(t)}\\
&=\lim\limits_{\xi\to\infty}\left(\frac{\xi}{g(\xi)}-\frac{R}{r}+\int_{r}^{g(\xi)}\frac{f(s)ds}{s^2}\right)\\
&=\int_{r}^{g(\xi)}\frac{f(s)ds}{s^2}-\frac{R}{r}
\end{align}
$$
So we got the desired equality. Now suppose $L>0$, then we shall show that both integrals are infinite. Since $L>0$, then there exist $a>R$ such  that $\inf\limits_{t\geq a}\frac{t}{g(t)}>\frac{L}{2}$, so
$$
\begin{aligned}
\int_{R}^{\infty}\frac{dt}{g(t)}
&\geq\int_{a}^{\infty}\frac{dt}{g(t)}\\
&=\lim\limits_{\xi\to+\infty}\int_{a}^{\xi}\frac{dt}{t}\frac{t}{g(t)}\\
&\geq\lim\limits_{\xi\to+\infty}\int_{a}^{\xi}\frac{dt}{t}\inf_{t\geq a}\frac{t}{g(t)}\\
&\geq\lim\limits_{\xi\to+\infty}\int_{a}^{\xi}\frac{dt}{t}\frac{L}{2}\\
&=\lim\limits_{\xi\to+\infty}\frac{L}{2}\ln\frac{\xi}{a}\\
&=+\infty
\end{aligned}
$$
Since $L=\limsup\limits_{\xi\to+\infty}\frac{\xi}{g(\xi)}=\limsup\limits_{\xi\to+\infty}\frac{f(s)}{s}$, similarly, we have $b>r$ such that $\inf\limits_{s\geq b}\frac{f(s)}{s}>\frac{L}{2}$. Therefore
$$
\begin{aligned}
\int_{R}^{\infty}\frac{f(s)ds}{s^2}
&\geq\int_{b}^{\infty}\frac{f(s)ds}{s^2}\\
&=\lim\limits_{\xi\to+\infty}\int_{b}^{\xi}\frac{ds}{s}\frac{f(s)}{s}\\
&\geq\lim\limits_{\xi\to+\infty}\int_{b}^{\xi}\frac{ds}{s}\inf_{s\geq a}\frac{f(s)}{s}\\
&\geq\lim\limits_{\xi\to+\infty}\int_{b}^{\xi}\frac{ds}{s}\frac{L}{2}\\
&=\lim\limits_{\xi\to+\infty}\frac{L}{2}\ln\frac{\xi}{b}\\
&=+\infty
\end{aligned}
$$
Thus we are done with the second option.
I was told this proof have a mistake. Could someone double check it? Even more the same person told me there is a simpler proof without usage of $\limsup$'s.
 A: Since $f$ is a bijection it is injective and, since it is continuous, it is either strictly increasing or strictly decreasing. Suppose for instance $f$ is strictly increasing (otherwise consider $-f$). The reciprocal map, $f^{-1}$, is also strictly increasing and therefore $\frac{1}{f^{-1}}$ is strictly decreasing. Suppose now that the integral $\int_R^{+\infty}\frac{1}{f^{-1}}$ is finite. Therefore $f^{-1}(t) $ must diverge to $\infty$ as $t \to \infty$ (it possesses a limit in $\infty$ since it is an increasing map). Then, if $\xi \geq 2R$ :  
\begin{equation*}
2\int_{\xi/2}^{\xi}\frac{dt}{f^{-1}(t)} \geq \frac{\xi}{f^{-1}(\xi)} \geq 0
\end{equation*}
The term on the left converges to $0$, therefore $\frac{\xi}{f^{-1}(\xi)} \underset{\xi \to \infty}{\longrightarrow} 0 $. 
Using your computation you may conclude now.
Now suppose that : 
\begin{equation*}
\int_R^{+\infty}\frac{dt}{f^{-1}(t)} = + \infty
\end{equation*}
If $\underset{\xi \to \infty}{\liminf} \frac{\xi}{f^{-1}(\xi)} = 0$ ; there exists a sequence $(\xi_n)$ of real numbers diverging to $\infty$ such that $\frac{\xi_n}{f^{-1}(\xi_n)} \underset{n \to \infty}{\longrightarrow} 0$. Then we have, using your computation : 
\begin{equation*}
\int_r^{f^{-1}(\xi_n)} \frac{f(s)}{s^2} ds = \frac{R}{r} - \frac{\xi_n}{f^{-1}(\xi_n)} +  \int_R^{\xi_n}\frac{dt}{f^{-1}(t)} \underset{n \to \infty}{\sim} \int_R^{\xi_n}\frac{dt}{f^{-1}(t)} \underset{n \to \infty}{\longrightarrow} + \infty
\end{equation*}
Therefore, $\int_r^{+ \infty} \frac{f(s)}{s^2} ds$ is not finite. 
Suppose now $L = \underset{\xi \to \infty}{\liminf} \frac{\xi}{f^{-1}(\xi)} = \underset{\xi \to \infty}{\liminf} \frac{f(\xi)}{\xi} >0$. Now your claim according to which there exists $b>r$ such that $\underset{s \geq b}{\inf}\frac{f(s)}{s} > \frac{L}{2}$ is correct. And by your last computation, the result follows.
I hope to have answered your question!
