Convergence of improper integral $\int_{0}^{+\infty}\frac{1}{\arctan ^\alpha x} dx$

Question

I have to analyse the convergence of $\displaystyle \int_{0}^{+\infty}\frac{1}{\arctan ^\alpha x} dx$, $\alpha \in R$

I have written: $\displaystyle \int_{0}^{c}\frac{1}{\arctan ^\alpha x} dx+\int_{c}^{\infty} \frac{1}{\arctan ^\alpha x} dx $

The first integral converges for $0<\alpha<1$

But I don't know how I can conclude something about the second integral.. the only thing that I know is that for $x \to \infty, \arctan x \to \pi/2$...

Answer 1

Since as $x \to \infty$, $\arctan{x} \to \pi/2$, for a given $\varepsilon>0$ there is $n$ such that for every $x>n$, $\pi/2>\arctan{x}>\pi/2-\varepsilon$. Hence $$ \int_n^{\infty} (\arctan{x})^{-\alpha} \, dx > \int_n^{\infty} (\pi/2-\varepsilon)^{-\alpha} \, dx = \infty $$ if $\alpha>0$, and $$\int_n^{\infty} (\arctan{x})^{-\alpha} \, dx>\int_n^{\infty} (\pi/2)^{-\alpha} \, dx = \infty$$ if $\alpha<0$, so either way the integral diverges.

Answer 2

Carrying on from your ending statement, that means that $\displaystyle\frac{1}{\arctan^\alpha x}\ge\frac{1}{(\pi/2)^\alpha},$ so I am assuming your integral diverges for $\alpha>1$ by comparison.