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Determine whether $$\int^{\infty}_{2} \frac{1}{x\sqrt{x^2-4}} dx$$ converges.

Doing some rough work, I realize that this function near $\infty$, behaves like $$\frac{1}{x\sqrt{x^2}} = \frac{1}{x^2}$$ I know this function converges, but I am having a hard time finding a $larger$ function that converges too.

I know that:

$$\frac{1}{x\sqrt{x^2-4}} \leq \frac{1}{\sqrt{x^2-4}}$$

but the bigger function diverges, so it doesn't work.

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    $\begingroup$ the integral converges the Limit is $$\frac{\pi}{4}$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 4 '17 at 18:22
  • $\begingroup$ I know it converges, but did you do that using the convergence test, or evaluating the integral directly? $\endgroup$ – K Split X Mar 4 '17 at 18:23
  • $\begingroup$ @KSplitX He evaluated it directly. $\endgroup$ – Simply Beautiful Art Mar 4 '17 at 18:32
  • $\begingroup$ You know, $\infty$ is not the only problematic part here. Even though the integral $\int_3^\infty$ coverges, the integral $\int_2^3$ may not be convergent $\endgroup$ – ThePortakal Mar 4 '17 at 18:52
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Hint:

For $x>2$, $$x^2-4>x^2-4(x-2)-4=(x-2)^2$$

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Let $x=2\sec t$. \begin{align} \int^{\infty}_{2} \frac{1}{x\sqrt{x^2-4}} dx = \frac12\int_0^{\pi/2} \frac{\sec t\tan t\, dt}{\sec t\sqrt{\sec^2 t-1}}=\frac12\int_0^{\pi/2}dt = \frac\pi4 \end{align}

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Easier and more accurate to evaluate directly. I took a look at that integral and it reminded me of something to do with secant or was that inverse secant? Tried an integral table quick search but failed to find the one needed. Paper and pen a couple of minutes gives:

By trig substitution

Let $x/2 = \sec u;\ \ x = 2 \sec u$

$dx = 2 \sec u \tan u \ du; \ \ \sqrt{x^2 - 4} = \sqrt{4\sec^2 u -4} = 2 \tan u $

$$ \int\frac{dx}{x\sqrt{x^2 - 4}} = \int \frac{2 \sec u \tan u \ du}{ 2 \sec u \ 2 \tan u} = \frac{1}{2}\int du = u = \frac{1}{2}\sec^{-1}(\frac{x}{2}) $$

$$ \int_2^{\infty}\frac{dx}{x\sqrt{x^2 - 4}}=\lim_{b \rightarrow \infty} \frac{1}{2}\sec^{-1}(\frac{x}{2})]_{x=2}^b $$

$$ = \frac{1}{2}(\lim_{b \rightarrow \infty} \sec^{-1}(\frac{b}{2}) - \sec^{-1}(\frac{2}{2})) $$

$$ = \frac{1}{2} (\pi/2 - 0) = \pi/4$$

So yes it converges and we have the limit.

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As $x\to2^+$ :

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

and we know that $\int_2^3\frac{dx}{\sqrt{x-2}}$ converges.

As $x\to+\infty$ :

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

and we know that $\int_3^{+\infty}\frac{dx}{x^2}$ converges.

Hence the requested integral converges.

Remark Finding equivalent of (nonnegative) functions usually helps to find whether an integral is convergent or not (I mean : it's sometimes easier than finding a sufficient inequality).

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Let $x-2=u$. Then $dx=du$. The integral becomes \begin{align*} \int_0^\infty \frac{du}{(u+2)\sqrt{(u+2)^2-4}}&=\int_0^\infty \frac{du}{(u+2)\sqrt{u^2+4u}}=\int_0^\infty \frac{du}{\sqrt{u^3+6u^2+8u}} \\ \\ &= \underbrace{\int_0^1 \frac{du}{\sqrt{u^3+6u^2+8u}}}_{=I_1} +\underbrace{\int_1^\infty \frac{du}{\sqrt{u^3+6u^2+8u}}}_{=I_2} \end{align*}

$\bullet$ Now, since $\displaystyle \frac{1}{\sqrt{u^3+6u^2+8u}} \geq \frac{1}{\sqrt{u^3}} = \frac{1}{u^{3/2}}$, $\ I_2$ converges by $p-$test ($3/2>1)$.

$\bullet$ Since, $\displaystyle \frac{1}{\sqrt{u^3+6u^2+8u}} \geq \frac{1}{\sqrt{8u}} = \frac{1}{\sqrt 8 u^{1/2}}$, $ \ \displaystyle \frac{I_1}{\sqrt 8}$ converges and hence $I_1$ converges by

$p-$test ($1/2<1)$.

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  • $\begingroup$ Shouldn't the second equality be less and the two $\ge$ be $\le$? $\endgroup$ – farruhota Jun 19 '17 at 5:02

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