Show convergence of $\int_1^\infty{\frac{dx}{\sqrt{x^3-x}}}$ How can I show that $I=\int_1^\infty{\frac{dx}{\sqrt{x^3-x}}}$ converges. I've had no success with finding any bounding function.
I also tried tried trig-substitution so $I=\int_0^{\pi/2}{\frac{dx}{\sqrt{\cos{x}}}}=\int_0^{\pi/2}{\frac{dx}{\sqrt{\sin{x}}}}$. But this did not help me.
 A: In cases like these, it may be useful to look at the asymptotic behaviour. You say you couldn't find a bounding function, but if it were $\frac{1}{\sqrt{x^3+x}}$ I'm quite certain you could.
The fact is that you almost immediately want to compare with some kind of $\frac{1}{\sqrt{x^3}}$. So what you can do is show that this is asymptotically the same as $\frac{1}{\sqrt{x^3+x}}$, in the sense that $\frac{1/\sqrt{x^3+x}}{1/\sqrt{x^3}} \to 1$ as $x \to +\infty$. With this fact at hand, you now have that for $x$ greater than a certain $K$, it is true that $\frac{1/\sqrt{x^3+x}}{1/\sqrt{x^3}} < 2$, say. Therefore, $\frac{1}{\sqrt{x^3+x}}<\frac{2}{\sqrt{x^3}}$ for those $x>K$, which gives you enough to conclude the exercise. This strategy also is easily generalizable for other cases. (Note that it isn't necessary for the limit to actually be $1$. It suffices for it to exist as a real number.)
A: Note that for $x>1$, we have that $2<x(x+1)$ and $(x-1)^2<x(x+1)$ therefore
$$\frac{1}{\sqrt{x^3-x}}=\frac{1}{((x-1)x(x+1))^{1/2}}<\frac{1}{(2(x-1))^{1/2}}$$
and
$$\frac{1}{\sqrt{x^3-x}}=\frac{1}{((x-1)x(x+1))^{1/2}}<\frac{1}{((x-1)^3)^{1/2}}.$$
Can you take if from here?
A: You don't need a bounding function if you've found an  equivalent function  near $1$ and  near $\infty$ (in general they're not the same) for which the integral functions $\int_1^a$ and $\int_a^\infty$ converge.


*

*Near $\infty$, $x^3-x\sim x^3$, so $\;\dfrac1{\sqrt{x^3-x}}\sim_\infty \dfrac1{x^{3/2}}$, and the integral to $\infty$ of this function converges.

*Near $1$, set $x=t-1$. The integral becomes
$$\int_1^a\frac{\mathrm d x}{\sqrt{x^3-x}}=\int_0^{a+1}\!\!\!\frac{\mathrm dt}{\sqrt{2t-3t^2+t^3}}$$
and near $t=0$, we have $\;\dfrac{\mathrm 1}{\sqrt{2t-3t^2+t^3}}\sim\dfrac{\mathrm 1}{\sqrt{2t}}$, which has a convergent integral.

A: $I_1=\int_1^\infty{\frac{dx}{\sqrt{x^3-x^2}}}$ converges. The integrand lies above the integrad of $I_2=\int_1^\infty{\frac{dx}{\sqrt{x^3-x}}}$ for all $x>1$.
So the area under the first on is greater than the area under the second one. So your integral must be less than the first one. Since area under the curve is equal to the integral.
$I_1$ can easily be computed and is $2arctan{(x-1)}^{0.5}$ which is equal to $pi$ which means that your integral is less than $pi$ but greater $0$. 
A: I believe the simplest way is just to enforce the substitution $x\mapsto x+1$ and then break the integral into simpler pieces:
$$\int_{1}^{+\infty}\frac{dx}{\sqrt{x^3-x}} = \int_{0}^{+\infty}\frac{dx}{\sqrt{x^3+3x^2+2x}} \leq \int_{0}^{1}\frac{dx}{\sqrt{2x}}+\int_{1}^{+\infty}\frac{dx}{\sqrt{x^3}} = \sqrt{2}+2.$$
You may also notice that the second term is an elliptic integral:
$$ \int_{0}^{+\infty}\frac{dx}{\sqrt{x(x^2+3x+2)}}\stackrel{x\mapsto x^2}{=}2\int_{0}^{+\infty}\frac{dx}{\sqrt{x^4+3x^2+2}}=2\int_{0}^{+\infty}\frac{dx}{\sqrt{(x^2+a^2)(x^2+b^2)}} $$
and the last term equals $\frac{\pi}{\text{AGM}(a,b)} $ where $a=1$ and $b=\sqrt{2}$, so
$$\boxed{2.60258\ldots= \frac{2\pi}{1+\sqrt{2}}\leq\int_{1}^{+\infty}\frac{dx}{\sqrt{x^3-x}}\leq \frac{\pi}{2^{1/4}}=2.64175\ldots.}$$
By the relations between $K$ (the complete elliptic integral of the first kind) and the $\text{AGM}$ the exact value of the integral is $\frac{\Gamma\left(\frac{1}{4}\right)^2}{2\sqrt{2\pi}}=2.62205755429211981\ldots.$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
{1 \over \root{x^{3} - x}} & \sim
\left\{\begin{array}{ll}
\ds{{1 \over \root{2}}\,\pars{x - 1}^{-1/2}} &
\mbox{as}\ds{\ x \to 1^{+}}
\\[2mm]
\ds{x^{-3/2}} & \mbox{as}\ds{\ x \to \infty}
\end{array}\right.
\\[5mm]
&\mbox{That's}\ \underline{enough}\ \mbox{to ensure the}\
\underline{convergence}\ \mbox{!!!.}
\end{align}
