Show an improper integral converges

I would like to show that the following improper integral converges, but it's been a while since I've done this sort of calculus and I'm drawing a blank: $$\int_0^1 \frac{dx}{\sqrt{x^3-x}}$$ My first thought was to try and simplify the expression under the radical. I tried to factor out $\sqrt{x}$ and use partial fractions, but that didn't really lead anywhere. A similar thing happened when I tried to factor out $x$. Next I thought about ways I could say that the integral converged without actually evaluating it, but again, I didn't really get anywhere with any of the standard integration techniques. Any thoughts?

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Try comparing it to some known convergent integral. – Gerry Myerson Oct 15 '12 at 22:58
Over the reals, the function is not defined. Do you mean $\sqrt{x-x^3}$ at the bottom? – André Nicolas Oct 15 '12 at 22:58
@GerryMyerson That was one of my other thoughts, but I wasn't able to find anything that worked. – chris Oct 15 '12 at 23:00
@AndréNicolas Unfortunately no. This is from a former real analysis qualifying exam, and the fact that it isn't defined over the reals is part of the reason why I'm stumped. – chris Oct 15 '12 at 23:01
@AndréNicolas Are you suggesting that I compare the given integral with that integral? Like factor out $\frac{1}{i}$ as a constant and claim that since $\int_0^1\frac{dx}{\sqrt{x-x^3}}$ converges, so must the other as a constant multiple of it? – chris Oct 15 '12 at 23:07

When $x \in (0, 1/\sqrt{2})$, we have $x^3 - x/2 < 0$. It follows that

$$\frac{1}{\sqrt{x-x^3}} < \frac{1}{\sqrt{x - x/2}} = \frac{\sqrt{2}}{\sqrt{x}}$$

on the same interval. Since $\int_0^1 dx/\sqrt{x}$ converges, your integral also converges on $(0, 1/\sqrt{2})$.

For the other side of the improper integral, use the following similar comparison:

$$\frac{1}{\sqrt{x-x^3}} < \frac{\sqrt{2}}{\sqrt{1-x}}$$

on $x \in \left(\frac{-1+\sqrt{5}}{2}, 1\right)$.

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Thanks. This looks like the explicit computation of the hint given by Andre below. Again, this is very helpful. – chris Oct 15 '12 at 23:52
@chris Happy to help! – Ayman Hourieh Oct 15 '12 at 23:52

Let $I = \displaystyle \int_0^1 \dfrac{dx}{\sqrt{x-x^3}}$.

$x = t^2 \implies I = \displaystyle \int_0^1 \dfrac{dx}{\sqrt{x-x^3}} = \displaystyle \int_0^1 \dfrac{2t dt}{\sqrt{t^2-t^6}}= 2 \int_0^1 \dfrac{dt}{\sqrt{1-t^4}}$

$$\dfrac1{\sqrt{1-t^4}} = \dfrac1{\sqrt{1-t} \sqrt{1+t} \sqrt{1+t^2}} \leq \dfrac1{\sqrt{1-t}}$$ for all $t \in [0,1]$.

Hence, $$I \leq 2 \int_0^1 \dfrac{dt}{\sqrt{1-t}} = - \left. 2 \dfrac{\left(1-t\right)^{1/2}}{1/2} \right\vert_0^1 = 4$$

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$$\int_0^1\frac{dx}{\sqrt{x^3-x}} = \int_0^1 \frac{dx}{i\sqrt{x-x^3}} = \frac{1}{i}\int_0^1 \frac{dx}{\sqrt{x-x^3}}$$ Since $x-x^3\geq0\;\forall\;x\in[0,1]$, the integral clearly converges.

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I disagree with my statement that the last integral clearly converges. Is there some sort of justification I can offer that makes this proof a little better? – chris Oct 15 '12 at 23:21
I think this would also "prove" that $\int_0^1x^{-1}\,dx$ converges (which it doesn't). Really: do you know any convergent improper integral to which you could compare this one? – Gerry Myerson Oct 15 '12 at 23:25
Again, none of the functions I've thought of compare correctly. It's been a long time since I've worked on these kinds of problems and I would appreciate a little more direction. – chris Oct 15 '12 at 23:30
@Chris: Split into two parts at say $1/2$. In left interval compare with $\frac{1}{\sqrt{x/2}}$. On right half, let $u=1-x$. – André Nicolas Oct 15 '12 at 23:50
Thank you. That gives me something to work with. I appreciate your help. – chris Oct 15 '12 at 23:51