On the limit of $f(n)$, specifically having to do with integration of an iterated $\arctan$

Question

Assume we are given that $A_n(x)$ denotes $n$ iterations of $\arctan(x)$, for example $A_2(x)=\arctan (\arctan(x))$

If $$f(n)=\int_{0}^n A_n(x)\space \text{d}x$$

I am looking for a rigorous proof that $\lim_{n\to\infty}f(n)$ diverges

I have worked on this problem for a while actually, and all I have gotten was a proof that $\forall x\in\Bbb{R},\lim_{x\to\infty}A_n(x)=0$

Beyond that information though, I am stuck. Thanks for any help.

Answer 1

Step 1. Applying the mean value theorem to the function $x \mapsto x - \arctan x$, for some $\xi \in [0, A_{k-1}]$ we have

$$ A_{k-1} - A_k = (A_{k-1} - \arctan A_{k-1}) = A_{k-1} \cdot \frac{\xi^2}{1 + \xi^2} \leq \frac{A_{k-1}^3}{1 + A_{k-1}^2}.$$

This shows that

$$ A_{k-1} - A_k \leq A_{k-1}^3 \qquad \text{and} \qquad \frac{A_{k-1}}{1+A_{k-1}^2} \leq A_k \leq A_{k-1}. \tag{*} $$

Step 2. From $\text{(*)}$, for $n \geq 2$ we have

\begin{align*} \frac{1}{A_n^2} &= \frac{1}{A_1^2} + \sum_{k=2}^{n} \left( \frac{1}{A_k^2} - \frac{1}{A_{k-1}^2} \right) \\ &= \frac{1}{A_1^2} + \sum_{k=2}^{n} \frac{(A_{k-1} + A_k)(A_{k-1} - A_k)}{A_k^2 A_{k-1}^2} \\ &\leq \frac{1}{A_1^2} + \sum_{k=2}^{n} \frac{(1+A_{k-1}^2)^2}{A_{k-1}^2} \frac{(2A_{k-1})(A_{k-1}^3)}{A_{k-1}^2} \\ &= \frac{1}{A_1^2} + \sum_{k=2}^{n} 2(1+A_{k-1}^2)^2 \\ &\leq \frac{1}{A_1^2} + Cn \end{align*}

for some absolute constant $C > 0$, where the last inequality follows from the fact that $A_i \leq \pi/2$ for $i \geq 1$. (Here, we may choose $C = 2(1+\frac{1}{4}\pi^2)^2$ though this particular value is not important.) This shows that

$$ \frac{\arctan x}{\sqrt{1 + Cn\arctan^2 x}} \leq A_n (x) $$

and in particular,

$$ A_n(x) \geq \frac{c}{\sqrt{1+C'n}} \quad \text{for } x \geq 1 $$

for some absolute constant $c, C > 0$. (Here, we may choose $c = \arctan 1$ and $C' = \frac{1}{2}\pi C$.) This gives

$$ \int_{0}^{n} A_n(x) \, dx \geq \int_{1}^{n} \frac{c}{\sqrt{1+C'n}} \, dx = \frac{c(n-1)}{\sqrt{1+C'n}}. $$

Therefore, as $n \to \infty$, the integral diverges to infinity.

Answer 2

I put $g(x)={\rm Arctan}(x)$. Note that $g$ and its iterates $g^{[n]}(x)$ are increasing. Hence for $x\geq 1$, we have $g^{[n]}(x)\geq g^{[n]}(1)=u_n$. The sequence $u_n$ is positive, and satisfy $u_{n+1}=g(u_n)$. As $g(x)\leq x$ for all $x$, $u_n$ is decreasing and $u_n\to L$, it is easy to show $L=0$. Now we can find a simple sequence $w_n$ such that $u_n\sim w_n$; this is a consequence of Cesaro's theorem : we study $\displaystyle \frac{1}{u_{n+1}^2}-\frac{1}{u_n^2}$, show that this has a limit $c$ (I think that this limit is $2/3$), and hence we get $\displaystyle u_n\sim \frac{d}{\sqrt{n}}$, with $\displaystyle d=1/\sqrt{c}$. Now for $0<m<d$ there exists an $N$ such that if $n\geq N$ we have $\displaystyle u_n\geq \frac{m}{\sqrt{n}}$, and hence $\displaystyle g^{[n]}(x)\geq \frac{m}{\sqrt{n}}$ for $x\geq 1$. It is easy now to finish.