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Improper integral

  1. Sep 24, 2006 #1
    basically I'm stuck with this problem........

    Integral the upper limit is positive infinite and the lower limit is 1
    X/(sqrt(1+x^6) dx......

    can someone give me an idea on how to start this?...I really don't know. Thanks.
     
  2. jcsd
  3. Sep 25, 2006 #2

    dextercioby

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    Are you sure it's this...?

    [tex] \int_{1}^{\infty} \frac{x}{\sqrt{1+x^{6}}} \ dx \ = ...? [/tex]

    Daniel.
     
  4. Sep 25, 2006 #3

    dextercioby

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    If so, then

    [tex] \int_{1}^{\infty} \frac{x}{\sqrt{1+x^{6}}} \ dx = {}_{2}F_{1}\left(\frac{1}{6},\frac{1}{2}; \frac{7}{6}; -1\right) [/tex]

    , with F Gauss' hypergeometric function.

    Daniel.
     
    Last edited: Sep 25, 2006
  5. Sep 25, 2006 #4
    I don't know how to do that. How do you use the comparison theorem to do this? Thanks.
     
  6. Sep 25, 2006 #5

    dextercioby

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    What comparison theorem are you referring to ...?

    A simple substitution [itex] x^{2}=t [/itex] will bring your integral to a simpler form, however that wouldn't help too much.

    Daniel.
     
  7. Sep 25, 2006 #6
  8. Sep 25, 2006 #7

    shmoe

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    It's always a good idea to state your question in full. I suspect yours is

    Prove the integral [tex] \int_{1}^{\infty} \frac{x}{\sqrt{1+x^{6}}} \ dx[/tex] is convergent.

    Specifically you don't care about finding a value for this integral. When x is large, what simple polynomial is the denominator kinda like? This should suggest a function for your comparison test.
     
  9. Sep 25, 2006 #8
    When X is large, x/(sqrt(1+x^6)) is getting close to the x-axis?
    x/(sqrt(1+x^2)) is greater than x/(sqrt(1+x^6))..is my logic right?
     
  10. Sep 25, 2006 #9
    I'm sorry that I didn't state my question clearly at the beginning.
     
  11. Sep 25, 2006 #10
    ..........................
     
  12. Sep 25, 2006 #11
    [tex] \frac{x}{\sqrt{1+x^{6}}} [/tex] is about [tex] \frac{x}{x^{3}} = \frac{1}{x^{2}} [/tex]. Since [tex] \frac{1}{x^{2}} [/tex] is convergent ([tex] p > 1 [/tex]), then so is the original integral by the comparison test (the original function is less than the new function)
     
    Last edited: Sep 25, 2006
  13. Sep 25, 2006 #12
    but why did you choose x/(x^3)?
     
  14. Sep 25, 2006 #13
    [tex]\frac{1}{\sqrt{1+ x^{6}}}= \frac{1}{ \sqrt{x^{6}}} = \frac{1}{x^{3}} [/tex]. The 1 doesnt matter for large enough x.
     
    Last edited: Sep 25, 2006
  15. Sep 25, 2006 #14

    shmoe

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    I don't consider pointing this out to be picky- you never have [tex]\frac{1}{\sqrt{1+ x^{6}}}= \frac{1}{ \sqrt{x^{6}}}[/tex].

    Much better to write [tex]\frac{1}{\sqrt{1+ x^{6}}}\sim \frac{1}{ \sqrt{x^{6}}}[/tex] with a precise asymptotic meaning behind [tex]\sim[/tex].
     
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