March 25th, 2019, 07:21 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 25th, 2019 at 11:59 PM. 
March 26th, 2019, 12:51 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,939 Thanks: 2210 
Where did you get this problem from?

March 26th, 2019, 03:49 AM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 605 Thanks: 88 
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 . 
March 26th, 2019, 01:52 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,807 Thanks: 717 
$u'$ looks a lot worse than $u$.

March 26th, 2019, 02:30 PM  #5 
Senior Member Joined: Dec 2015 From: somewhere Posts: 605 Thanks: 88 
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, 03:06 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, 07:52 PM  #7 
Member Joined: Oct 2018 From: USA Posts: 88 Thanks: 61 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, 08:27 PM  #8 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,258 Thanks: 929 Math Focus: Wibbly wobbly timeywimey 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 semicircle 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, 07:43 AM  #9 
Member Joined: Oct 2018 From: USA Posts: 88 Thanks: 61 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. Last edited by skipjack; March 27th, 2019 at 03:23 PM. 

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