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 March 25th, 2019, 08:21 AM #1 Newbie   Joined: Jan 2019 From: Athens Posts: 5 Thanks: 0 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.
 March 26th, 2019, 01:51 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,105 Thanks: 2324 Where did you get this problem from?
 March 26th, 2019, 04:49 AM #3 Senior Member   Joined: Dec 2015 From: Earth Posts: 820 Thanks: 113 Math Focus: Elementary Math 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 . Thanks from topsquark
 March 26th, 2019, 02:52 PM #4 Global Moderator   Joined: May 2007 Posts: 6,852 Thanks: 743 $u'$ looks a lot worse than $u$.
 March 26th, 2019, 03:30 PM #5 Senior Member   Joined: Dec 2015 From: Earth Posts: 820 Thanks: 113 Math Focus: Elementary Math Yes $\displaystyle u’$ just makes the function inside integral more complicated. How can this integral be solved ? Maybe a better substitution for partial integration.
 March 26th, 2019, 04:06 PM #6 Member     Joined: Feb 2019 From: United Kingdom Posts: 44 Thanks: 3 This integral cannot be expressed in terms of elementary functions in order to take on a limiting value.
 March 26th, 2019, 08:52 PM #7 Senior Member     Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 Math Focus: Algebraic Geometry 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 $dx = (1/4)(\tan{\theta})^{-3/4} \cdot \sec^{2}{\theta}d\theta$
 March 26th, 2019, 09:27 PM #8 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,340 Thanks: 983 Math Focus: Wibbly wobbly timey-wimey stuff. 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
 March 27th, 2019, 08:43 AM #9 Senior Member     Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 Math Focus: Algebraic Geometry 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. Thanks from topsquark Last edited by skipjack; March 27th, 2019 at 04:23 PM.

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