F is integrable but f^2 is not integrable

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Homework Statement



Give an example of a sequence of integrable functions f:R -> R such that f is integrable but f^2 is not integrable. Prove your results.


The Attempt at a Solution




would 1/sqrt(x) for x ∈ [0,1] work? any other suggestions?
how would you prove that its does work? woudl you just plug it in and show it doesn't work or is there something more to it?

Thanks! I really appreciate your help!
 
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Sure f(x)=1/sqrt(x) on [0,1] works. Just integrate it. It's improper, but that shouldn't be a problem.
 
super. I thought proofs had to be more complicated, somehow.
Is there any theorem that explains integrability?
 
pulin816 said:
super. I thought proofs had to be more complicated, somehow.
Is there any theorem that explains integrability?

There's a lot of theorems about integrability. There's no one that explains everything. If you can do it as an elementary integral, like this one, you don't need anything complicated. Just a little imagination.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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