Construct such function, that can't be integrated on an interval, but its square is integrable on the interval

Question

Okay, so there's the text of my problem on Russian:
Построить не интегрируемую функцию на отрезке, квадрат которой интегрируем на этом отрезке.
And its translation on English:
Construct such function, that can't be integrated on an interval, but its square is integrable on the interval.
So I decided to start with the $f(x) = \frac{1}{x^p}$, that doesn't converge with $p > 1$ and does vice versa on a segment $[0,1]$ but I found nothing out, because my $p$ has to be $> 1$, and thus $p^2$ is also $> 1$.
Then I tried to do something with the square root on some negative interval(e.g. $f(x) = \frac{1}{\sqrt{5-x^2}}$ on an $[-8,-2]$ interval. By this way, it makes some sense kinda. But I want to know, is there some other functions that can't be integrated, but its square is integrable.
Well, I forgot to say, the interval is closed one.

Answer 1

If you're allowed to use an unbounded interval, then $\frac1x$ on $[1,\infty)$ works. In any bounded interval, a good choice is a function that equals either $1$ or $-1$, depending whether $x$ is rational or irrational, or something like that. This was suggested in the comments by Clement C., who wrote it out as the formula $f(x)=2\mathbf{1}_{\Bbb Q}-1$.

Concerning your example, which is equivalent to integrating $\frac1{\sqrt x}$ on any interval of negative numbers, that feels like it works for the wrong reason somehow. We'd like to see $f(x)$ defined, but not integrable, as opposed to not being integrable because it's not even defined.

What I'm not finding right now is an example of a continuous function on a bounded interval that is not integrable, whose square is integrable on that bounded interval. If such an example cannot be produced, there must be a reason.....

Answer 2

$$\int^\infty_1 \frac1{x^{2/3}}dx$$ diverges by p-test.

$$\int^\infty_1 \frac1{x^{4/3}}dx$$ converges by p-test.