How to prove $\int_1^{+\infty} x^{1-a}\mathrm{d}x/\sqrt{x^2-1}$ converges when $a>1$

Question

I want to prove:

$$ \int_1^{+\infty} \frac{x^{1-a}}{\sqrt{x^2-1}} \mathrm{d}x< +\infty,\quad a>1 $$

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Update:

I use $\sqrt{x^2-1}<\sqrt{x^2}=x$ and:

$$ \int_1^{+\infty} \frac{x^{1-a}}{\sqrt{x^2-1}} \mathrm{d}x > \int_1^{+\infty} x^{-a} \mathrm{d}x $$

If $a=1$, we have:

$$ \int_1^{+\infty} \frac{1}{\sqrt{x^2-1}} \mathrm{d}x > \int_1^{+\infty} x^{-1} \mathrm{d}x = \ln(x)|_1^{+\infty} = +\infty $$

If $a<1$:

$$ \int_1^{+\infty} \frac{x^{1-a}}{\sqrt{x^2-1}} \mathrm{d}x > \int_1^{+\infty} x^{-a} \mathrm{d}x = \frac{x^{1-a}}{1-a}|_1^{+\infty} = +\infty $$

But is $a>1$ really converge? It maybe converge at $a>1.xxx$

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I tried to change this to: ($x=1/\cos t$)

$$ \int_0^{\pi/2} (\cos t)^{a-2}\mathrm{d}t $$

and when $a > 2$, I can use $(\cos t)^{a-2} < (2-t)^{a-2}$

but when $1 < a < 2$, I don't know what to do.

Thanks for any help!

Answer 1

The integral is doubly improper, so to determine convergence we consider $2$ integrals: $$\int_1^\infty \frac{x^{1 - a} \,dx}{\sqrt{x^2 - 1}} = \int_1^b \frac{x^{1 - a} \,dx}{\sqrt{x^2 - 1}} + \int_b^\infty \frac{x^{1 - a} \,dx}{\sqrt{x^2 - 1}} ,$$ where $b \in (1, \infty)$ is arbitrary.

Expanding in a series about $x = 1$ gives that the integrand is $$\frac{1}{\sqrt{2 (x - 1)}} + R(x), \qquad R(x) \in O(\sqrt{x - 1}),$$

which is integrable over $[1, b]$ (for all $a$, which does not appear in the expansion).

Expanding in a series about $x = \infty$ gives that the integrand is $$\frac{1}{x^a} + S(x), \qquad S(x) \in O\left(\frac{1}{x^{a + 2}}\right),$$

which is integrable over $[b, \infty)$ iff $a > 1$.

So, the integral converges iff

$a > 1$.

Remark Using the equivalent form $\int_0^{\pi / 2} \cos^{a - 2} t \,dt$ derived in the question statement, we can express the exact value of the integral in terms of the Beta or Gamma functions: $$\frac{1}{2} \mathrm{B}\left(\frac{1}{2}, \frac{a - 1}{2}\right) = \frac{\sqrt\pi \,\Gamma\left(\frac{a - 1}{2}\right)}{2 \Gamma\left(\frac{a}{2}\right)}$$

Answer 2

The only possible singularity is $t=\frac{\pi}{2}$, where $\cos t=0$.

When $t$ is close to $\frac{\pi}{2}$, we have$$ \cos t=\sin(\frac{\pi}{2}-t)=\frac{\pi}{2}-t+o(\frac{\pi}{2}-t)$$ You may now see why $\alpha-2>-1$ ($\alpha>1$) could guarantee the convergence.

For convenience, let $m=\frac{\pi}{2}-t$ and the integral becomes $$ \int_0^{\frac{\pi}{2}} (\sin m)^{\alpha-2}\,\mathrm{d} m$$ The singularity is $m=0$ and $$(\sin m)^{\alpha-2}=\left(m(1+o(1))\right)^{\alpha-2}=m^{\alpha-2}(1+o(1))$$ Hence $\alpha-2>-1$ if and only if the integral converges.

Answer 3

You have to study the behaviour of the function near $1$ and at infinity: near $1$ it converges for all $a\in \Bbb R$, whereas at infinity that's not true. Considering the asymptotic behaviour, you get: $$ \int_1^{+\infty} \frac{x^{1-a}}{\sqrt{x^2-1}} \mathrm{d}x\sim \int_1^{+\infty} \frac{x^{1-a}}{\sqrt{x^2}} \mathrm{d}x=\int_1^{+\infty} \frac{x^{1-a}}{x} \mathrm{d}x=\int_1^{+\infty} \frac{1}{x^a} \mathrm{d}x $$ The last integral converges iff $a>1$.