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March 25th, 2019, 08:21 AM   #1
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Improper integral

I'm trying to calculate the following improper integral:

(i) $\displaystyle \ \int_1^{+\infty} \!\frac{3x^2}{\sqrt{x^8 + 1}}\,dx$

Any help will be appreciated.

Last edited by skipjack; March 26th, 2019 at 12:59 AM.
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March 26th, 2019, 01:51 AM   #2
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Where did you get this problem from?
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March 26th, 2019, 04:49 AM   #3
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Let $\displaystyle u=\frac{3x^2 }{\sqrt{1+x^8 }}$ and $\displaystyle dv=dx$.
$\displaystyle \int_{1}^{\infty } udx =ux\mid_{1}^{\infty} -\int_{1}^{\infty} xu’dx $ .
Now just continue .
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March 26th, 2019, 02:52 PM   #4
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$u'$ looks a lot worse than $u$.
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March 26th, 2019, 03:30 PM   #5
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Yes $\displaystyle u’$ just makes the function inside integral more complicated.
How can this integral be solved ? Maybe a better substitution for partial integration.
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March 26th, 2019, 04:06 PM   #6
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This integral cannot be expressed in terms of elementary functions in order to take on a limiting value.
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March 26th, 2019, 08:52 PM   #7
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The root at the bottom may point toward a trig substitution like $x = (\tan{\theta})^{1/4}$ which would leave a secant in the bottom and $dx = (1/4)(\tan{\theta})^{-3/4} \cdot \sec^{2}{\theta}d\theta$
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March 26th, 2019, 09:27 PM   #8
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I'm not planning to work it out myself (too lazy!) but there are some possibilities for using contour integration. We would have a closed curve going from x = 1 out to infinity on the +x axis, a semi-circle with R (taken to the limit R to infinity) and then back from - infinity to -1 (the integrand is even so this adds nothing new.) The problem would be the integral from $\displaystyle -1 \leq x \leq 1$. But we might get away with a polynomial series near x = 0 for this.

Possible perhaps, but defininetly messy. But it does restrict the numerical approximation to the interval from -1 to 1. I don't know if that will help much.

-Dan
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March 27th, 2019, 08:43 AM   #9
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Using a trig substitution:

$\displaystyle x = (\tan{\theta})^{1/4}$

$\displaystyle \theta = \arctan{x^4}$

$\displaystyle dx = \frac{1}{4}(\tan{\theta})^{-3/4} \cdot \sec^{2}(\theta) d\theta$

Putting this into the integral leaves:

$\displaystyle 3\int_{\pi/4}^{\pi/2} \frac{\sqrt{\tan{\theta}}}{\sec{\theta}} \cdot \frac{1}{4}(\tan{\theta})^{-3/4} \cdot \sec^{2}(\theta)d\theta$

$\displaystyle = \frac{3}{4}\int_{\pi/4}^{\pi/2} \frac{\sec{\theta}}{(\tan{\theta})^{1/4}}d\theta$

This makes the bounds finite, but I'm not sure whether this goes anywhere. There might be some substitution here I'm not seeing.
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Last edited by skipjack; March 27th, 2019 at 04:23 PM.
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