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Math Help - Primitives of the natural logarithm

  1. #1
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    Primitives of the natural logarithm

    All right, the idea of what I'm trying to do is find an easy way to generate the primitives of the natural logarithm without having to perform iteration after iteration of integration:
    P_{k+1}[f(x)]=\int P_k dx
    P_0[ln{x}]=ln{x}
    P_n[ln{x}]=?
    After performing a few integrations, I seem to have found a nice solution:
    P_n[ln{x}]=\frac{x^n}{n!} (ln(x)-H(n))
    Where H(n) is the Harmonic Series:
    H(n)=\sum_{k=0}^{n}{\frac{1}{k}}
    I'd appreciate it if anyone could verify or disprove my result.
    If anyone would like to see my work, I'd gladly send them a .pdf of it.
    Last edited by m.walunga; December 28th 2009 at 02:42 PM. Reason: typos
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  2. #2
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    If you find a general term that you think it is correct , it is time for you to prove your hypothesis by Mathematical Induction .


    but at the same time , i have another method to do so


    Consider


     \ln(x) = \lim_{a\to 0} \frac{ x^a - 1}{a}

    If we integrate it  n times ,

    we can get

     \lim_{a\to 0} \frac{1}{a} \cdot \left( \frac{x^{a + n}}{ ( a+1)(a+2)...(a+n)} - \frac{x^n}{ n! } \right )

    then evaluate the limit by L'hospital rule

    here the numerator is  \left( \frac{x^{a + n}}{ ( a+1)(a+2)...(a+n)} - \frac{x^n}{ n! } \right )

    and the denominator , obviously , is  a


    so  P_n =  D[ \left( \frac{x^{a + n}}{ ( a+1)(a+2)...(a+n)} - \frac{x^n}{ n! } \right ) ]_{a\to 0}

     = D[  \frac{x^{a + n}}{ ( a+1)(a+2)...(a+n)}  ]

    (  D is the differential operator , wrt a )



    Then recall the logarithmic dervative

     P_n = \left [ \frac{x^{a + n}}{ ( a+1)(a+2)...(a+n)} \cdot [ \ln(x) - \frac{1}{a + 1} - \frac{1}{a+2} - ... - \frac{1}{a + n} ] \right ]_{a\to 0 }

     =\frac{x^n}{n!} (ln(x)-H(n))
    Last edited by simplependulum; December 28th 2009 at 11:37 PM.
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