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Improper integral help

  1. Mar 4, 2007 #1
    Hi, I'm having a bit of trouble showing that the integral from 1 to infinity of x/((1+x^6))^1/2 converges or diverges by the comparative property.

    I'm not sure if I'm setting it up right, but so far I have that 1/rad(1+x^6) is less than or equal to x/rad(1+x^6) which is less than 1/rad(x^6).

    I don't know if this is right, or where to go from here if it is right.

    Thanks for your help!
     
  2. jcsd
  3. Mar 4, 2007 #2

    Dick

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    Do you think it converges or diverges? What is your gut feeling. If x is large what does it look like (ie ignoring the 1)?
     
  4. Mar 5, 2007 #3

    Gib Z

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    [tex]\sqrt{x^6+1}>\sqrt{x^6}[/tex]. (Trivial)


    [tex]\frac{x}{\sqrt{x^6+1}}<\frac{x}{\sqrt{x^6}}[/tex]
    For every value of x within our bounds of integration, that is true.
    [tex]\int^{\infty}_1 \frac{x}{\sqrt{x^6+1}} dx< \int^{\infty}_1 \frac{1}{x^2} dx[/tex]. Since we know, and can show, the 2nd part converges, the 1st part does as well.
     
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