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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. Tags improper, integral Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post DavidSzalai Calculus 1 March 19th, 2017 11:30 AM jones1234 Calculus 1 December 23rd, 2012 08:51 AM citbquinn Calculus 2 March 15th, 2011 11:42 AM utj98 Calculus 5 May 18th, 2010 07:00 PM izseekzu Calculus 1 April 13th, 2010 04:37 PM

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