# 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.

• If $L = \limsup\limits_{\xi\rightarrow +\infty}{\frac{\xi}{g(\xi)}} > 0$, it does not necessarily follow that there exists $a>R$ such that $\inf\limits_{t\ge a}{\frac{t}{g(t)}} > L/2$. However, since the integrands in both integrals are positive, I think your line of reasoning would work if you let $L = \inf\limits_{\xi\ge R}{\frac{\xi}{g(\xi)}}$, and look at whether $L=0$ or $L>0$. – Joey Zou Nov 24 '16 at 22:28

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.