For what values of $\alpha > 0$ and $\beta\in\mathbb{R}$ is the
$$ f:(1,+\infty) \to \mathbb{R}: f(x) = \frac{\arctan(x^\alpha)}{x^\beta}$$
function integrable ?
I know that the above, an example from my text, is integrable if and only if $ 1 < \beta < \alpha + 1$. But the text uses Landau notation to estimate the bounds and left me confused as to what is being done.
It states and I quote,
Note that $\lim_{x \to 0}\frac{\arctan(x)}{x} = 1$ and $\lim_{x\to+\infty}\arctan(x) = \pi/2$. Hence, $f(x) = \Theta(x^{\alpha - \beta})$ for $x \to 0$ and $f(x) = \Theta(\frac{1}{x^\beta})$ for $x\to+\infty$. Hence, $f$ is integrable if and only if $ \beta - \alpha < 1$ and $ \beta > 1$.
I know the above 2 limits exists and have the respective values. But, I have been unable to work out what happened to the $\alpha$ and $\beta$ in $f$ ? Also, how does this show $f(x) = \Theta(x^{\alpha - \beta})$ ?
A bit of handholding will be appreciated! Thanks in advance.
Hints: $$ \alpha>0\implies \lim_{x\to+\infty}x^\alpha=+\infty\implies \lim_{x\to+\infty}\arctan{x^\alpha}=\frac\pi2, $$ $$\alpha<0\implies \lim_{x\to+\infty}x^\alpha=0\implies\lim_{x\to+\infty}\frac{\arctan{x^\alpha}}{x^\alpha}=1.$$