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Math Help - Improper Integral/ Converge-Diverge.

  1. #1
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    Improper Integral/ Converge-Diverge.

    Hello All,

    I am trying to figure out how to solve the Improper Integral and then decide if it converges or diverges.

    Upper limit 2, lower limit 0 (x-1)^-1/3

    Any help you can give me would be greatly appreciated.

    Thanks!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by CHAMPAN View Post
    Hello All,

    I am trying to figure out how to solve the Improper Integral and then decide if it converges or diverges.

    Upper limit 2, lower limit 0 (x-1)^-1/3

    Any help you can give me would be greatly appreciated.

    Thanks!
    Put:

    I_{a,b}= \int_0^a (x-1)^{-1/3}\; dx+ \int_b^2 (x-1)^{-1/3}\; dx,\ a<1,\ b>1

    For the integral to converge requires that:

    I=\lim_{a\to 1_-, b \to 1_+}I_{a,b}

    exists, which is eqivalent to requiring that:

    \lim_{a \to 1_-}\int_0^a (x-1)^{-1/3}\; dx

    and

    \lim_{b \to 1_+}\int_b^2 (x-1)^{-1/3}\; dx

    both exist.

    CB
    Last edited by CaptainBlack; April 7th 2010 at 04:16 AM. Reason: to make the limits approach 1 from the right directions
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  3. #3
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    Thanks for the explanation, do you happen to know how to tell where it converges?

    Thank you again for your help! It is greatly appreciated.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by CHAMPAN View Post
    Thanks for the explanation, do you happen to know how to tell where it converges?

    Thank you again for your help! It is greatly appreciated.
    Where it converges? I'm not sure I know what you mean. Both the constituent limits exist so the improper integral converges (the only point that there is a problem is around x=1 where the integrand misbehaves, which is why that is the point about which we break up the integral)

    CB
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