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Math Help - Find values for which this integral diverges

  1. #1
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    Find values for which this integral diverges

    I have the following question, I have to find the values of a for which the indefinite integral of x-a from 1 to infinity diverge.

    Substitution h for infinity I get: limit h->0 (1/(a-1)-h1-y/(a-1))

    So it diverges only for a=1 because there the limit is not defined! Is this correct?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Find values for which this integral diverges

    If a=1 then, \int_0^b\dfrac{dx}{x}=\log b. If b\to +\infty, then \log b\to +\infty and the integral is divergent. If a\neq 1 then, \int_0^b\dfrac{dx}{x^2}=\ldots=\dfrac{b^{-a+1}}{-a+1}-\dfrac{1}{-a+1}. Easily proved:

    \int_0^{+\infty}\dfrac{dx}{x^a}=\begin{Bmatrix} +\infty & \mbox{ if }& a<1\\\dfrac{1}{a-1}& \mbox{if}& a>1\end{matrix}
    Thanks from infernalmich
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    Re: Find values for which this integral diverges

    Thank you! I didn't thought about writing x-a as 1/xa
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Find values for which this integral diverges

    Quote Originally Posted by FernandoRevilla View Post
    If a\neq 1 then, \int_0^b\dfrac{dx}{x^2}=\ldots=\dfrac{b^{-a+1}}{-a+1}-\dfrac{1}{-a+1}.
    Of course I meant \int_0^b\dfrac{dx}{x^a} instead of \int_0^b\dfrac{dx}{x^2} .
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    Re: Find values for which this integral diverges

    If I do the same exercise with the Integral from 0 to infinity, it diverges for every "a" - is this true?
    Last edited by infernalmich; June 17th 2012 at 08:04 AM.
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    Re: Find values for which this integral diverges

    What a satisfaction, I found the solution... I was wrong.

    It converges for a<1 and diverges for a >1
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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    Re: Find values for which this integral diverges

    Quote Originally Posted by FernandoRevilla View Post
    If a=1 then, \int_0^b\dfrac{dx}{x}=\log b. If b\to +\infty, then \log b\to +\infty and the integral is divergent. If a\neq 1 then, \int_0^b\dfrac{dx}{x^2}=\ldots=\dfrac{b^{-a+1}}{-a+1}-\dfrac{1}{-a+1}. Easily proved:

    \int_0^{+\infty}\dfrac{dx}{x^a}=\begin{Bmatrix} +\infty & \mbox{ if }& a<1\\\dfrac{1}{a-1}& \mbox{if}& a>1\end{matrix}
    Of course the lower limit is 1 instead of 0.
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Re: Find values for which this integral diverges

    Quote Originally Posted by FernandoRevilla View Post
    Of course I meant \int_0^b\dfrac{dx}{x^a} instead of \int_0^b\dfrac{dx}{x^2} .
    Of course (again) the lower limit is 1 instead of 0.
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