How can one prove which of the following functions are $\in \mathcal{L}_2(-\infty,\infty)$?
- $f(x)=\frac{1}{1+x^2}$
- $f(x)=\frac{1}{x^2}$
- $f(x)=\frac{1}{1-x^2}$
I've run numerical simulations which suggest that the first function is square integrable and the second and third functions are not; but I haven't figured out how to prove this for any of them.
I thought I could prove that the second function is square integrable, because the integral of $\frac{1}{x^4}$ over $a<x<b$ is $\frac{1}{3a^3}-\frac{1}{3b^3}$, which appears to be finite as long as neither of the limits of integration are $0$; but this does not appear to be sensible for several reasons, not least because it would imply that the integral of a non-negative function over the real line is 0.
My intuition tells me that the above reasoning is flawed because of the singularity of the second function at 0, but that doesn't help me prove that it is not square integrable.
EDIT: Can someone offer any advice on when a function with a singularity is $\in \mathcal{L}_2(-\infty,\infty)$? Because $\sqrt{\delta(x)}\in \mathcal{L}_2(-\infty,\infty)$, but the reasoning below suggests that many functions with singularities are not square integrable.
For the second function, split the integral into two integrals for $-\infty<x<-\delta$ and $\delta<x<\infty$, which integrates to $\frac{2}{3\delta^3}$, which is unbounded as $\delta\rightarrow0$, so this function is not square-integrable over the real line.
For the first function, for a fixed value of $\delta$, the integral of both tails is bounded above by the integral over the same range of the second function, which is finite. The integral over $-\delta<x<\delta$ is bounded above by $2\delta$, because $|f(x)|^2\le1$, so the integral over the real line of the squared modulus of the second function is finite.
For the third function, use similar reasoning to the second function (can someone make this rigorous?)