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What integration technique to use?

  1. Dec 24, 2007 #1
    1. The problem statement, all variables and given/known data
    What technique should I use to evaluate

    [tex] \int_0^1 dx \frac{x^{m-1}}{\ln x} [/tex]

    where m is a positive integer


    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 24, 2007 #2
    [tex]x^{m-1}=x^{m}\times x^{-1}[/tex]
  4. Dec 24, 2007 #3

    Gib Z

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    Don't go around trying to find the anti-derivative of that in elementary functions, the indefinite integral evaluates to [tex]Ei (m \ln x) + C, \frac{d}{dx}C = 0[/tex] where Ei(x) is the exponential integral. This result follows because after the substitution u= ln x, the original integral becomes [tex]\int^0_{-\infty} \frac{ (e^u)^m}{u} du[/tex].

    Best you can do is to get an infinite series solution.
    Last edited: Dec 24, 2007
  5. Dec 25, 2007 #4

    Gib Z

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    Ahh I think I've found out how to solve it, because those bounds looked quite nice =] Try Differentiation under the integral.
  6. Dec 25, 2007 #5
    Yes. That works. My original question was a mistake, however. I wanted to use parameter differentiation to find

    [tex]H(m) = \int_0^1 dx \frac{x^{m}}{\ln x} [/tex]

    So, I differentiated erroneously and got

    [tex]H(m) = \int_0^1 dx (m-1)\frac{x^{m-1}}{\ln x} [/tex]

    instead of

    [tex]H(m) = \int_0^1 dx x^{m} [/tex]

    By, the way, [tex]H(m) = \ln{m+1}[/tex].
  7. Dec 25, 2007 #6

    Gib Z

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    Ah no it's not as easy as that...You have forgotten to deal with the Constant associated with the integrals..
  8. Dec 25, 2007 #7

    Gib Z

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    Well I might as well give another clue, try looking at H(-1) to evaluate the constant. If my mind was a bit clear yesterday, I could have told you that; looking at the natural logs series between 0 and 1, that integral is......
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