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Math Help - improper integral

  1. #1
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    improper integral

    I'm given an integral whose bounds are from e to infinity of dx/(x*(lnx)^a) and I have to find for which values of a it converges or diverges. How do you deal with the log in there? the only test I know is a comparison test but it didn't help when I tried it
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  2. #2
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    Do you know for what values of a \displaystyle \int_1^\infty  {\frac{{du}}{{u^a }}} converges?
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  3. #3
    Junior Member BariMutation's Avatar
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    http://orion.math.iastate.edu/vika/cal3_files/Lec22.pdf

    That should help - look at the last page. Essentially, if a>1 the infinite series is convergent. So for a<a1 the infinite series is divergent. The proof is within the link I provided.
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by BariMutation View Post
    http://orion.math.iastate.edu/vika/cal3_files/Lec22.pdf

    That should help - look at the last page. Essentially, if a>1 the infinite series is convergent. So for a<a1 the infinite series is divergent. The proof is within the link I provided.
    What is a primitive of \displaystyle \frac{1}{x\ \ln^{a} x} ?... consulting the book You find...

    \displaystyle \int \frac{dx}{x\ \ln^{a} x} = \left\{\begin{array}{ll}\frac {\ln ^{1-a} x}{1-a},\,\, a\ne 1\\{}\\ \ln |\ln x| ,\,\, a=1\end{array}\right. (1)

    Now if You valuate the limit x \rightarrow \infty of (1), You 'discover' that the definite integral converges for a>1 and diverges for a \le 1...



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  5. #5
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    thanks for the answers!
    I have another one now :S
    integral from 0 to infinity of the function (t^(a-1))*(e^(-t)), again find which values of a it converges/diverges
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  6. #6
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    and another still. I'm having a lot of trouble with these ]:
    integral from 0 to infinity of (x^(a-1))/(1+x)
    help would be greatly appreciated
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  7. #7
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by giygaskeptpraying View Post
    thanks for the answers!
    I have another one now :S
    integral from 0 to infinity of the function (t^(a-1))*(e^(-t)), again find which values of a it converges/diverges
    The question is a little controverisial... one way to 'attack' the problem is the following: let suppose to define the function \Gamma(a) as...

    \displaystyle \Gamma(a)= \int_{0}^{\infty} t^{a-1}\ e^{-t}\ dt (1)

    Now integrating by parts (1) You obtain...

    \displaystyle  \int_{0}^{\infty} t^{a-1}\ e^{-t}\ dt = \frac{1}{a}\ \int_{0}^{\infty} t^{a}\ e^{-t}\ dt (2)

    ... or equivalently...

    \displaystyle  \Gamma(a)= \frac{\Gamma(a+1)}{a} (3)

    Observing (3) You conclude that...

    \displaystyle \lim_{a \rightarrow 0+} \Gamma(a)= +\infty (4)

    ... so that may seem obvious that the integral converges for a>0 and diverges for a \le 0... therefore for someone [a little suprisingly!...] that's not obvious at all and that's a long story ...




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  8. #8
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by giygaskeptpraying View Post
    and another still. I'm having a lot of trouble with these ]:
    integral from 0 to infinity of (x^(a-1))/(1+x)
    help would be greatly appreciated
    Integrating by parts You obtain...

    \displaystyle \int_{0}^{\infty} \frac{x^{a-1}}{1+x}\ dx= \frac{1}{a}\ \{\lim_{x \rightarrow \infty} \frac{x^{a}}{1+x} + \int_{0}^{\infty} \frac{x^{a}}{(1+x)^{2}}\ dx \} (1)

    ... and observing the first tem of (1) You can conclude that the integral converges for 0<a<1...




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