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Finding convergence/divergence of improper integral

  1. Mar 18, 2012 #1
    Determine the convergence or divergence of this following improper integral:

    [tex]\int_2^∞ \frac {1}{(x^3+7)^{\frac{1}{3}}}[/tex]

    So I'm trying to find something easy to compare this to, any help?
     
  2. jcsd
  3. Mar 18, 2012 #2
    Well, as I'm sure you can tell, it is a lot like 1/x. So, I would try to find something that is close to 1/x to which you can compare this.
     
  4. Mar 18, 2012 #3
    Why can't I just compare it to 1/x?
     
  5. Mar 18, 2012 #4
    I don't know if you have gone over this, but you can compare it to

    [tex]\int\limits_1^\infty {\frac{1}{{{x^n}}}} [/tex]

    If n<2, it is divergent. If n≥2, then it is convergent.
     
    Last edited: Mar 18, 2012
  6. Mar 18, 2012 #5
    Well,
    [tex]\frac{1}{(x^3+7)^{\frac{1}{3}}} < \frac{1}{x} [/tex]

    isn't it? Since [
    [tex]\int_2^∞ \frac {1}{x}[/tex]
    diverges, this isn't any help.

    So, I'd try showing that
    [tex](x^3+7)^{1/3} < x + b[/tex]

    where b is a number. (In particular, b is a number that appears in your problem).

    Then realize that

    [tex]\int_2^∞ \frac {1}{x+b}[/tex]

    diverges, and you're done.
     
  7. Mar 19, 2012 #6
    How can I show that [tex]\int_2^∞ \frac {1}{x+7}[/tex] is divergent?
     
  8. Mar 19, 2012 #7
    First of all, I was kind of wrong when I said that "b" was in your problem. Use [tex]7^{1/3}[/tex] in stead of [tex]7[/tex]. As for showing that

    [tex]\int_2^∞ \frac {1}{x+7^{1/3}}[/tex]

    is divergent, is there some theorem in your book that would help? I don't know if there is or not. But, you could say that

    [tex]lim \frac{1}{x+7^{1/3}} = \frac{1}{x}[/tex]

    and argue from there. Or, you could just integrate


    [tex]\int_2^∞ \frac {1}{x+7^{1/3}}[/tex]

    You have done so to show that


    [tex]\int_2^∞ \frac {1}{x}[/tex]

    is divergent, right? Just use a change of variables.
     
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