<s>[INST] 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.
 [/INST] 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 $\frac{\Gamma\left(\frac{1}{4}\right)^2}{2\sqrt{2\pi}}=2.62205755429211981\ldots.$
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