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Math Help - integration with factorials

  1. #1
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    integration with factorials

    im having a lot of trouble showing that the

    definite integral from 0 to 1 (ln x)^n dx = n!(-1)^n
    if someone could help get me started cause i have no idea how to get the factorial in the answer ... all i get is this infinite loop of integration by parts which keeps going on and on
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  2. #2
    Math Engineering Student
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    Put u=-\ln x and your integral becomes {{(-1)}^{n}}\int_{0}^{\infty }{{{u}^{n}}{{e}^{-u}}\,du}={{(-1)}^{n}}n!.\quad\blacksquare

    See some of

    Gamma function - Wikipedia, the free encyclopedia
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  3. #3
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    i am not familiar with gamma functions ... is there an easier way of solving this problem ... maybe this will help

    there was a previous part to this question which i was able to solve

    I(subscript)n = integral (ln x)^n dx implies that
    I(subscript)n = x (ln x)^n - n*(I(subscript)n-1)

    maybe i have to somehow use this to solve the second question ... but i cannot see anyway to use it ...
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  4. #4
    Eater of Worlds
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    The way Kriz showed is about the easiest. If you don't know the Gamma, then get to know it. It's a great tool to have in the arsenal.

    Here's another subtle difference in the sub Kriz mentioned.

    \int_{0}^{1}(ln(x))^{n}dx

    Let x=e^{-t}, \;\ dx=-e^{-t}dt

    Make the subs and change the limits of integration.

    When x=1, then t=0. When x=0, then t\to {\infty}

    -\int_{\infty}^{0}(-t)^{n}e^{-t}dt

    Factor out the (-1)^{n} and reverse the limits of integration to shed the negative sign.

    \int_{0}^{\infty}(-1)^{n}t^{n}e^{-t}dt

    Now, since {\Gamma}(n)=\int_{0}^{\infty}t^{n-1}e^{-t}dt

    Then we have for ours, {\Gamma}(n+1) which equals n!

    So, we get (-1)^{n}n!
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  5. #5
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    Now, if you make an iterative integration by parts for your integral, you'll get the following:

    \begin{aligned}<br />
   \int_{0}^{1}{{{\ln }^{n}}(x)\,dx}&=\int_{0}^{1}{(x)'{{\ln }^{n}}(x)\,dx} \\ <br />
 & =-n\int_{0}^{1}{{{\ln }^{n-1}}(x)\,dx} \\ <br />
 & =n(n-1)\int_{0}^{1}{{{\ln }^{n-2}}(x)\,dx} \\ <br />
  &~\,\vdots  \\ <br />
 & ={{(-1)}^{n}}n!. <br />
\end{aligned}
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  6. #6
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    and can u show me how to get that (-1) in your answer ... as when i replace x with 1 and 0 they end up cancelling out and giving me 0
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Induction, I assume since you put that you have n! and not \Gamma(n+1) that n\in\mathbb{N}.

    Proposition: \int_0^1 \ln^n(x)~dx=n!(-1)^n

    Base case: \int_0^1 \ln(x)~dx=\left[x\ln(x)-x\right]\bigg|_{x=0}^{x=1}=-1=1!\cdot (-1)^1

    Inductive hypothesis: \int_0^1 \ln^n(x)~dx=n!(-1)^n

    Inductive step:

    \begin{aligned}\int_0^1 \ln^{n+1}(x)~dx&=\left[x\ln^{n+1}(x)\right]\bigg|_{x=0}^{x=n}-(n+1)\int_0^1 \ln^n(x)~dx{\color{red}\star}\\<br />
&=0-(n+1)\overbrace{n!(-1)^n}^{IH}\\<br />
&=(n+1)!(-1)^{n+1}\quad\blacksquare\end{aligned}

    \color{red}\star: This is gotten by parts letting u=\ln^{n+1}(x)\implies du=(n+1)\frac{\ln^n(x)}{x}~dx and dv=dx\implies v=x


    Hope this makes sense
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  8. #8
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    thanks a lot everyone
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