Proving a function is Lebesgue integrable

Question

I need to prove that $$\frac{|x|^\alpha}{1+x^2}$$ is Lebesgue integrable for $\alpha \in [0,1)$ but I'm not sure how to do this. I first tried expanding this using the Taylor expansion to show it is bounded on $[-1,1]$ but I'm unsure about finding a dominating function for $|x|>1$.

Answer 1

You It is easily seen to be bounded on $[-1,1]$ because it is continuous there. For $|x|>1$ we have that

$${|x|^\alpha\over 1+x^2}\le {|x|^\alpha\over x^2}=|x|^{\alpha-2}.$$

The inequality comes from the fact that $1+x^2>x^2>0$ so that $0<{1\over 1+x^2}<{1\over x^2}$, and of course $x^2=|x|^2$ since we have real numbers.

Then this is easily seen to have integral on $(-\infty,-1)\cup(1,\infty)$

$$2\int_1^\infty x^{\alpha-2}\,dx={1\over 1-\alpha}<\infty$$

The factor of $2$ comes from the fact that $|x|^{\alpha-2}$ is an even function, so that the integral on $(-\infty, -1)$ is equal to the one on $(1,\infty)$. Of course the integral on $[-1,1]$ is of a bounded, continuous function, so it is some finite value as well.

If you want an explicit dominating function for all of $\Bbb R$, set

$$M=\sup_{-1\le x\le 1}{|x|^\alpha\over 1+x^2}$$ and set

$$g(x) = \begin{cases} M & |x|\le 1 \\ |x|^{\alpha-2} & |x|>1\end{cases}$$