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I wish to check if the integral

$$ \int_2^{\infty}\frac{1}{x\sqrt{x^2-4}}\,dx $$

converges.

Here is what I did:

Using substitutions:

$$ \int_2^{\infty}\frac{1}{x\sqrt{x^2-4}}\,dx=\frac{1}{2}\int_4^{\infty}\frac{1}{t\sqrt{t-4}}\,dt= \int_0^{\infty} \frac{1}{(u+4)\sqrt u}\,du. $$

Separating:

$$\int_0^{\infty} \frac{1}{(u+4)\sqrt u}\,du=\int_0^{1} \frac{1}{(u+4)\sqrt u}\,du+\int_1^{\infty} \frac{1}{(u+4)\sqrt u}\,du.$$

For the first term: $\displaystyle \int_0^{1} \frac{1}{(u+4)\sqrt u}\,du:$

$$\int_0^{1} \frac{1}{(u+4)\sqrt u}\,du=\lim_{\epsilon \rightarrow 0}\int_{\epsilon}^{1} \frac{1}{(u+4)\sqrt u}\,du.$$

Using comperison test to $\dfrac{1}{\sqrt u},$ this term coverges.

For the second term: $\displaystyle\int_1^{\infty} \frac{1}{(u+4)\sqrt u}\,du,$

using the comparison test to $\dfrac{1}{ u^{3/2}},$ this term converges.

I would like to know if I made any mistakes. Also, let me know if you have any suggestions for other ways to solve this.

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    $\begingroup$ Everything is OK. $\endgroup$ – Julián Aguirre Jun 4 '18 at 14:52
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    $\begingroup$ A couple of typesetting tips (which I've edited in for you): 1. Use \displaystyle whenever possible, except in titles. This is especially helpful for integrals and fractions. 2. Always use a thinspace \, immediately before any differential inside an integral. $\endgroup$ – Adrian Keister Jun 4 '18 at 14:54
  • $\begingroup$ @JuliánAguirre thank you! $\endgroup$ – segevp Jun 4 '18 at 14:55
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    $\begingroup$ +1) Always good to see the work! $\endgroup$ – imranfat Jun 4 '18 at 15:27
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$$ \int_2^3 \frac {dx} {x\sqrt{x-2\,}\sqrt{x+2\,}} \le \int_2^3 \frac{dx}{\sqrt{x-2}} < +\infty $$ and $$ \int_3^\infty \frac{dx}{x\sqrt{x^2-4}} \le \int_3^\infty \frac{2\,dx}{x^2} < +\infty. $$ For this second inequality you need to show that $\sqrt{x^2-4\,} \ge \dfrac x 2$ when $x\ge3.$

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Yes it is correct.

As an alternative, without change of variables, we can observe that for $x$ large

$$\frac{1}{x\sqrt{x^2-4}}\sim\frac1{x^\frac32}$$

and refer to limit comparison test with $\int_3^{\infty} \frac1{x^\frac32}\, dx$ and for $x\to 2^+$

$$\frac{1}{x\sqrt{x^2-4}}=\frac{1}{x \sqrt{x+2}\sqrt{x-2} }\le\frac14 \frac{1}{\sqrt{x-2} }$$

and refer to comparison test with $\int_2^3 \frac{1}{\sqrt{x-2} }\, dx$.

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