Let $f(x)=\frac{\arctan(x)}{x}$. Let $F$ be the anti-derivative of $f$ in $\mathbb{R}$. Prove $F$ is not bounded above in $\mathbb{R}$.

Question

Let $f(x)=\frac{\arctan(x)}{x}$. Let $F$ be the anti-derivative of $f$ in $\mathbb{R}$. Prove $F$ is not bounded above in $\mathbb{R}$.

First, I had already proven that for all $x\gt 0\Rightarrow \arctan(x)\lt x$ using Lagrange's theorem, if that helps. Then I proved that $x=0$ is a removable discontinuity point and defined the function as $f(x)= \frac{arctan(x)}{x}$ if $ x\neq 0, 1$ if $x=0$, so $F$ is well defined. I found that at $x=0$, $F$ has an inflection point. now I need to prove that $F$ has no upper bound in $\mathbb{R}$. I don't really know how to do it. Do I need to show that $\int_{1}^{\infty}\frac{\arctan(x)}{x}$ diverges? I cannot find the antiderivative obviously. How do I prove this?

Answer 1

Hint:

For large values of $x$ (not even that large, i.e. for $x>\tan 1$) you have the inequality $$\frac{\arctan(x)}{x} > \frac{1}{x}$$ and the antiderivative of $\frac1x$ is not bounded above in $\mathbb R$.

Answer 2

Note that for $ x\ge 1$

$$f(x)=\frac{arctan(x)}{x} \ge \frac { (\frac {\pi }{4})}{x}$$

and

$$ \int _1^{\infty}\frac { (\frac {\pi }{4})}{x} dx $$ is not bounded above.

By comparison the $$ \int _1^{\infty}f(x) dx $$ is not bounded above.

Answer 3

Overkill: by the Shafer-Fink inequality, for any $x>0$ $$ \frac{\arctan x}{x}\geq \frac{3}{1+2\sqrt{1+x^2}}$$ hence $\int_{0}^{x}\frac{\arctan u}{u}\,du \geq \frac{3}{2}\text{arcsinh}(x)-O(1)$ (actually $\int_{0}^{x}\frac{\arctan u}{u}\,du \geq \frac{3}{2}\text{arcsinh}(x)-\frac{\pi}{2\sqrt{3}}$)
implies that $F(x)$ is unbounded. The LHS is $\text{Im}\,\text{Li}_2(ix)$.