For what real $\alpha >0$, does $\int_{1}^{\infty}{\frac{1}{\sqrt{x^\alpha-1}}\, dx}$ converge?

Question

I want to check my compute.

I split the exercise in two integrals: $\int_{1}^{A}{\frac{1}{\sqrt{x^\alpha-1}}\, dx}$ and $\int_{A}^{\infty}{\frac{1}{\sqrt{x^\alpha-1}}\, dx}$ for some positive $A>1,A\in \mathbb{R}$.

From $\int_{A}^{\infty}{\frac{1}{\sqrt{x^\alpha-1}}\, dx}=\int_{A}^{\infty}{\frac{1}{x^{\frac{\alpha}{2}}\sqrt{1-\frac{1}{x^\alpha}}}\, dx} \sim \int_{A}^{\infty}{\frac{1}{x^{\frac{\alpha}{2}}}\, dx}$ for $x\to \infty$; I find that it converges for $\alpha >2$.

With the substitution $x^\alpha-1=y^\alpha$, the second part becames: $\int_{0}^{A}{\frac{(y^\alpha+1)^{\frac{1-\alpha}{\alpha}}y^{\alpha-1}}{\sqrt{y^\alpha}}\, dy}= \int_{0}^{A}{\frac{(y^\alpha+1)^{\frac{1-\alpha}{\alpha}}}{y^{1-\frac{\alpha}{2}}}\, dy}\sim \int_{0}^{A}{\frac{1}{y^{1-\frac{\alpha}{2}}}\,dx}$ for $y \to 0$; I find thah it converges for $\alpha>0$.

At the end, it converges for $\alpha >2$.

Answer 1

hint

near $1$, use the equivalence

$$x^\alpha-1=e^{\alpha \ln (x)}-1$$ $$\sim \alpha \ln (x) \sim \alpha (x-1)$$ but $\int_1\frac {dx}{\sqrt {x-1}} $ converges.

near $+\infty $,

$$\sqrt {x^\alpha-1}\sim x^\frac {\alpha}{2} $$ and $$\int^{+\infty }\frac {dx}{x^\frac \alpha 2}$$ converges if $$\alpha>2.$$

as a conclusion, your integral converges

if $\alpha>2$.