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How to integrate $$\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$$

I tried both $t=\sqrt{x^2-1}$ and $t=\sin x$ but didn't reach the right result.

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    $\begingroup$ use that $$\cosh(t)^2-\sinh(t)^2=1$$ $\endgroup$ – Dr. Sonnhard Graubner Nov 26 '14 at 20:13
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    $\begingroup$ Or even simpler $x \mapsto 1/y$ $\endgroup$ – N3buchadnezzar Nov 26 '14 at 20:15
  • $\begingroup$ @N3buchadnezzar, How is that going to help? $\endgroup$ – AlonAlon Nov 26 '14 at 20:18
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\overbrace{\color{#66f}{\large% \int_{1}^{\infty}{\dd x \over x^{2}\root{x^{2} - 1}}}} ^{\ds{\dsc{x} = \dsc{1 \over t}\ \imp\ \dsc{t} = \dsc{1 \over x}}}\ =\ \int_{1}^{0}{-\,\dd t/t^{2} \over t^{-2}\root{1/t^{2} - 1}} =\int_{0}^{1}{t\,\dd t \over \root{1 - t^{2}}} =\left. -\root{1 - t^{2}}\right\vert_{0}^{1}=\color{#66f}{\LARGE 1} \end{align}

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Hint: Let $x \mapsto 1/y$, then we have $\mathrm{d}x = -\mathrm{d}y/y^2$ which implies that $-\mathrm{d}y = \mathrm{d}x/x^2$. Hence $$ \int_1^\infty \frac1{\sqrt{x^2-1\,}\,}\frac{\mathrm{d}x}{x^2} = \int_1^0 \frac{-\mathrm{d}y}{\sqrt{(1/y)^2-1\,}\,} = \int_0^1 \frac{y\,\mathrm{d}y}{\sqrt{1-y^2\,}\,} $$ From here a simple and (obvious) substitution will complete the work.

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  • $\begingroup$ I set $z=y^2$ and got $\sqrt{1-z}\Big|_0^1 = \sqrt{1-y^2}\Big|_0^1 = \sqrt{1-1/x^2}\Big|_1^\infty = 0$ $\endgroup$ – AlonAlon Nov 26 '14 at 20:48
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    $\begingroup$ Not quite, you forgot a minus sign. also I do not see why you switch back to $x$? There is no need for such rash decicions. We have $[ \sqrt{1-y^2} ]' = -y/\sqrt{1-y^2}$. Hence $$ \int_0^1 \frac{y \mathrm{d}y}{\sqrt{1-y^2}} = -\Bigl[ \sqrt{1-y^2} \Bigr]_0^1 = \Bigl[ \sqrt{1-y^2}\Bigr]_1^0 $$ $\endgroup$ – N3buchadnezzar Nov 26 '14 at 20:53
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Here's the easiest way:

$$\int \frac{dx}{x^2\sqrt{x^2-1}} = \int \frac{dx}{x^3 \sqrt{1 - \frac{1}{x^2}}}$$

Now substitute $u = 1 - \frac{1}{x^2}$

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  • $\begingroup$ This is very similar to N3buchadnezzar's answer, but it is very simple. (+1) $\endgroup$ – robjohn Dec 24 '14 at 15:33
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$$x=\sec u, \;\mathrm{d}x=\tan u\sec u\,\mathrm{d}u$$

If you decide to $x\mapsto \frac1x$ then substitute $y=1-x^2,\;\mathrm{d}y=-2x\,\mathrm{d}x$

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  • $\begingroup$ and then deal with a secant integral? No thanks. $\endgroup$ – Ali Caglayan Dec 24 '14 at 10:37
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$$ \begin{align} \int_1^\infty\frac{\mathrm{d}x}{x^2\sqrt{x^2-1}} &=\frac12\int_1^\infty\frac{\mathrm{d}x}{x^{3/2}\sqrt{x-1}}\\ &=\frac12\int_0^\infty\frac{x^{-1/2}\mathrm{d}x}{(1+x)^{3/2}}\\ &=\frac12\mathrm{B}\left(\frac12,1\right)\\ &=\frac12\frac{\Gamma\left(\frac12\right)\Gamma(1)}{\Gamma\left(\frac32\right)}\\ &=\frac12\frac{\Gamma\left(\frac12\right)\Gamma(1)}{\frac12\Gamma\left(\frac12\right)}\\[9pt] &=1 \end{align} $$

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we set $x=\cosh(t)$ then we have $$dx=sinh(t)dt$$ thus we get $$\int\frac{\sinh(t)}{\cosh(t)^2\sqrt{\cosh(t)^2-1}}\,dt$$

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I recommend regular trig-sub, you should get a pretty simple integral at the end. Set up a right triangle with some angle $\theta$, hypotenuse length $x$ and the side opposite $\theta$ with length $1$. You should get $$\frac{1}{x^2} = \sin^2(\theta), \quad \tan(\theta) = \frac{1}{\sqrt{x^2-1}}, \quad dx = -\csc(\theta)\cot(\theta)d\theta$$ Hence $$\int \frac{dx}{x^2\sqrt{x^2-1}} = \int \ -sin^2(\theta)\tan(\theta)\csc(\theta)\cot(\theta)d\theta \\ = \int -\sin(\theta)d\theta$$

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Apply the substitution $x=sec(u)$, then $dx=\frac{\tan \left(u\right)}{\cos \left(u\right)}du$.

$$\int \frac{1}{\sec ^2\left(u\right)\sqrt{\sec ^2\left(u\right)-1}}\frac{\tan \left(u\right)}{\cos \left(u\right)}du$$

$$\int \frac{1}{\sec ^2\left(u\right)\sqrt{\sec ^2\left(u\right)-1}}\frac{1}{\cos^2 \left(u\right)}du$$

Remember that $\frac {1}{\cos(u)}=\sec(u)$ :

$$\int \frac{\sin \left(u\right)}{\sqrt{\sec ^2\left(u\right)-1}}du$$

And $\sec ^2\left(x\right)=1+\tan ^2\left(x\right)$:

$$\int \frac{\sin \left(u\right)}{\sqrt{-1+1+\tan ^2\left(u\right)}}du$$

Then:

$$\int \frac{\sin \left(u\right)}{\sqrt{\tan ^2\left(u\right)}}du$$

$$if \int \frac{g\left(x\right)}{f\left(x\right)}=F\left(x\right)\mathrm{,\:then\:}\:\int \frac{g\left(x\right)}{\sqrt{f^2\left(x\right)}}=\frac{f\left(x\right)}{\sqrt{f^2\left(x\right)}}F\left(x\right)$$

$$\frac{\tan \left(u\right)}{\sqrt{\left(\tan \left(u\right)\right)^2}}\int \frac{\sin \left(u\right)}{\tan \left(u\right)}du$$

$$\frac{\tan \left(u\right)}{\sqrt{\left(\tan \left(u\right)\right)^2}}\int \cos \left(u\right)du$$

$$\frac{\tan \left(u\right)}{\sqrt{\left(\tan \left(u\right)\right)^2}}\sin \left(u\right)$$

Substitute back, $u=arcsec(u)$:

$$\frac{\tan \left(arcsec \left(x\right)\right)}{\sqrt{\left(\tan \left(arcsec \left(x\right)\right)\right)^2}}\sin \left(arcsec \left(x\right)\right)$$

Simplifies to:

$$\sqrt{1-\frac{1}{x^2}}+C$$

Now compute the boundaries and you will find $1-0=1$.

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