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Choosing constants to make integral convergent

  1. Aug 1, 2012 #1
    I want to integrate γβγx/(β + x)γ+1 from 0 to ∞ (given β and γ are both > 0)

    So for large x the integrand is approximately proportional to x

    So for which values of γ is the integral defined? Surely for any γ > 0 the integrand tends to zero as x tends to infinity?

  2. jcsd
  3. Aug 1, 2012 #2


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    You need γ > 1 to integrate to ∞. The integral of 1/x is divergent for x -> ∞.
  4. Aug 1, 2012 #3
    But 1/x2 etc. is ok?
  5. Aug 1, 2012 #4


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    Yes, in fact ∫x-(1+ε) converges for any ε > 0. Even ∫1/(x ln1+ε(x)) converges, but not ∫1/(x ln(x)) , nor ∫1/(x ln(x)ln(ln(x))) etc.
  6. Aug 2, 2012 #5
    Thanks that's really helpful. Is it because the ones that don't exist aren't going to zero fast enough?

    And what about non-negative powers - are there any special cases where these can be integrated to infinity that I should be aware of?

    e.g. ∫x2
  7. Aug 2, 2012 #6


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    Hey sr3056.

    For this problem I'm assuming that you are integrating with respect to x and that the other parameters are independent from x.

    It might help if you make a change of variable u = β + x, change your limits and then integrate with respect to this.

    The reason is that you will get (u-β)/u^[y+1] which can be analytically integrated under which you can get a lower and upper limit and decide exactly what your parameter must be for it to converge.
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