May 18th, 2018, 06:43 PM  #1 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6  Improper integrals
Hello everyone, Can anyone help me with the antiderivative of this integtral? I graphically proved it is convergent to zero but algebraicly I could't find the antiderivative of the integral. I tried usub and then arctan. but couldn't get out of it. 
May 18th, 2018, 07:24 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,344 Thanks: 2466 Math Focus: Mainly analysis and algebra 
Are you trying to evaluate the integral of just prove convergence? In the latter case you can use $$\frac1{\sqrt{x^5+2}} < \frac1{\sqrt{x+2}}$$ on the interval. 
May 19th, 2018, 06:01 AM  #3  
Math Team Joined: Jul 2011 From: Texas Posts: 2,761 Thanks: 1416  Quote:
$$\frac1{\sqrt{x^5+2}} < \frac1{\sqrt{x^4+2}}$$ since $\displaystyle \int_1^\infty \frac{1}{\sqrt{x+2}} \, dx$ diverges  
May 19th, 2018, 10:29 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,042 Thanks: 1064 
before you all rush off to try and show this converges to zero... WolframAlpha: Computational Intelligence 
May 19th, 2018, 11:24 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,344 Thanks: 2466 Math Focus: Mainly analysis and algebra 
The integral clearly doesn't, the integrand does.

May 19th, 2018, 01:19 PM  #6 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6 
smaller function diverges mean larger function (above that line) also diverges or larger function converges mean smaller function (line below that) also converges. How can I prove that original function is diverging by proving a larger function is diverging? Shouldn't I pick smaller function then that and prove it diverges instead of picking larger one which diverges?! Last edited by skipjack; May 19th, 2018 at 10:02 PM. 
May 19th, 2018, 04:14 PM  #7  
Math Team Joined: Jul 2011 From: Texas Posts: 2,761 Thanks: 1416  Quote:
Last edited by skipjack; May 19th, 2018 at 10:13 PM.  

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