Evaluating $\int_{1}^\infty\frac{1}{t^a\sqrt{t^2-1}}dt$ for $a\geq 1$

Question

I know that the this integral converges, but I can't show it. And, how can I proceed to calculate its value?

$$\displaystyle \int_{1}^\infty \dfrac{1}{t^a\sqrt{t^2-1}} dt\qquad (a \geq 1)$$

Answer 1

Hint: I would do the substitution $$t=\frac1{\cos x}$$ When $t=1$, you have $x=0$, and for $t=\infty$ you get $x=\pi/2$. You also then have $\sin x\ge 0$ in this interval, so $$\sqrt{t^2-1}=\frac{\sin x}{\cos x}=\tan x$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\left.\int_{1}^{\infty} {\dd t \over t^{a}\root{t^{2} - 1}}\,\right\vert _{\ \Re\pars{a}\ >\ 0}} \,\,\,\stackrel{t\ \mapsto\ 1/t}{=}\,\,\, \int_{1}^{0} {-\dd t/t^{2} \over \pars{1/t}^{a}\root{1/t^{2} - 1}} \\[5mm] = &\ \int_{0}^{1}t^{a - 1}\pars{1 - t^{2}}^{-1/2}\,\dd t \,\,\,\stackrel{t^{2}\ \mapsto\ t}{=}\,\,\, {1 \over 2}\int_{0}^{1}t^{a/2 - 1}\pars{1 - t}^{-1/2}\,\dd t \\[5mm] = &\ {1 \over 2}\,{\Gamma\pars{a/2}\Gamma\pars{1/2} \over\Gamma\pars{a/2 + 1/2}} = \bbx{{\Gamma\pars{a/2} \over\Gamma\pars{a/2 + 1/2}}\, {\root{\pi} \over 2}} \end{align}