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Math Help - Find the value of p, improper integral

  1. #1
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    Find the value of p, improper integral

    Find the value of p for which the integral converges and evaluate the integral for those values of p

    ∫ 1/(x(ln x)^p)

    upper limit : infinity
    lower limit : e
    Last edited by racewithferrari; February 25th 2010 at 06:16 PM.
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  2. #2
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    Our integral is

    \int_0^{\infty}\frac{1}{x(\ln x)^p}\,dx=\int_0^{\infty}\frac{1}{x}\cdot\frac{1}{  (\ln x)^p}\,dx.

    Hint: we may use a substitution of variables.
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  3. #3
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    I didn't really get it. can you explain it further. BTW there was a typo, the lower limit is e not 0.
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    well that really makes a difference, so make the substitution what do you get?

    what can we say about \int_1^\infty\frac{dt}{t^p} ?
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  5. #5
    MHF Contributor chisigma's Avatar
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    The integral proposed by Scott H , though different from the original one, is quite interesting. First we suppose that is p\ne 1. In this case the integral can be devided in two parts...

    \int_{0}^{\infty} \frac{dx}{x\cdot \ln^{p} x}= \int_{0}^{1} \frac{dx}{x\cdot \ln^{p} x} + \int_{1}^{\infty} \frac{dx}{x\cdot \ln^{p} x} (1)

    Now we can set in second integral t=\frac{1}{x} and obtain...

    \int_{1}^{\infty} \frac{dx}{x\cdot \ln^{p} x} = (-1)^{p} \int_{0}^{1} \frac{dt}{t\cdot \ln^{p} t} (2)

    ... so that is...

    \int_{0}^{\infty} \frac{dx}{x\cdot \ln^{p} x}= \{1 + (-1)^{p}\}\cdot \int_{0}^{1} \frac{dx}{x\cdot \ln^{p} x} = \lim_{\xi \rightarrow 1} \frac{\{1 + (-1)^{p}\}}{p-1} \cdot \ln ^{1-p} \xi (3)

    From (3) we deduct that for p odd with p\ne 1 is...

    \int_{0}^{\infty} \frac{dx}{x\cdot \ln^{p} x}= 0 (4)

    ... and for p even with p \ge 2 the integral diverges...

    Kind regards

    \chi \sigma
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