How to prove this integration is finite

Question

I got stuck in verifying the condition that the following integration is finite:

$$\int_{-\infty }^{\infty }\frac{\left | x \right |^\alpha }{\pi (1+x^2)}dx \tag{1}$$

The answer is that the integration is finite when $-1<\alpha<1$

What I have tried is that when $\alpha=1$, this integration diverge by using integration by part.
And I'm guessing, the integration will diverge for $\alpha>1$ under the similar manners.

However, when I tried to look into $\alpha<1$, I started doing integration by part on (1), like: $$\int_{-\infty }^{\infty }\frac{\left | x \right |^\alpha }{\pi (1+x^2)}dx=\frac{1}{\pi }(\left | x \right |^\alpha \tan^{-1}x|_{-\infty }^{\infty })-\alpha \int_{-\infty }^{\infty }\left | x \right |^{\alpha-1}\tan^{-1}xdx$$ and when I apply the integration by part again on the last term, $$-\alpha \int_{-\infty }^{\infty }\left | x \right |^{\alpha-1}\tan^{-1}xdx=-\alpha\left(\tan^{-1}x\cdot \frac{1}{\alpha }\left | x \right |^{\alpha}-\frac{1}{\alpha }\int_{-\infty }^{\infty }\frac{\left | x \right |^{\alpha}}{1+x^2}dx\right)$$

then it turns out everything cancelled out.....
I think I did something stupid, but don't know where when wrong.....

Could anyone help me this out? Thank you!!

Answer 1

Don't try explicit integration; use (Limit) Comparison of improper integrals.

There are two potential issues of convergence here: near $x=0$ and at infinity. You may want to split the integral into three: $(-\infty,-1]$, $[-1,1]$, and $[1,\infty)$.

• Near $0$, the integrand is comparable to $|x|^\alpha$. Thus, the integral over $[-1,1]$ converges if and only if $\alpha>-1$.
• For large $x$, the integrand is comparable to $|x|^{\alpha-2}$. Thus, the integrals over the two unbounded intervals converge if and only if $\alpha-2<-1$; that is, if $\alpha<1$.