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Math Help - definite integral

  1. #1
    Super Member Random Variable's Avatar
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    definite integral

    Math Challenge Problem:

     \int^{2}_{1} \frac{\ln(x-1)}{x} \ dx
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Random Variable View Post
    Math Challenge Problem:

     \int^{2}_{1} \frac{\ln(x-1)}{x} \ dx
    Let's denote:

     I=\displaystyle\int^{2}_{1} \dfrac{\ln(x-1)}{x} \ dx

    with the substitution:

    t=\ln (x-1) we obtain:

     I=\displaystyle\int^{0}_{-\infty} \dfrac{te^t\;dt}{e^t+1}

    Now, apply twice integration by parts.

    Fernando Revilla
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  3. #3
    Super Member Random Variable's Avatar
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     \int \frac{t e^{t}}{e^{t}+1} \ dt does not have an antiderivative that can be expressed as a finite number of elementary functions. So it's not quite that straightforward.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Random Variable View Post
    Math Challenge Problem:

     \int^{2}_{1} \frac{\ln(x-1)}{x} \ dx
    Setting x-1= u the integral becomes...

    \displaystyle \int_{0}^{1} \frac{\ln u}{1+u}\ du (1)

    Now if You consider that is...

    \displaystyle \frac{1}{1+u} = \sum_{n=0}^{\infty} (-1)^{n}\ u^{n} (2)

    ... and...

    \displaystyle \int_{0}^{1} u^{n}\ \ln u = - \frac{1}{(n+1)^{2}} (3)

    ... You obtain...

    \displaystyle \int_{0}^{1} \frac{\ln u}{1+u}\ du = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} = - \frac{\pi ^{2}}{12} (4)



    Merry Christmas from Italy

    \chi \sigma
    Last edited by chisigma; December 22nd 2010 at 02:26 PM. Reason: mistake signalled by Ackbeet corrected!...
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Random Variable View Post
     \int \frac{t e^{t}}{e^{t}+1} \ dt does not have an antiderivative that can be expressed as a finite number of elementary functions. So it's not quite that straightforward.
    So what?. You can find I (definite integral) by means of an equation.

    Fernando Revilla
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  6. #6
    A Plied Mathematician
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    Quote Originally Posted by chisigma View Post
    Setting x-1= u the integral becomes...
    \displaystyle \int_{0}^{1} \frac{\ln u}{1+u}\ du = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2}} = \frac{\pi ^{2}}{12} (4)
    Mathematica and WolframAlpha both give the negative of your answer, so I'm guessing you have a sign error in there somewhere. Not sure where, though.
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  7. #7
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Mathematica and WolframAlpha both give the negative of your answer, so I'm guessing you have a sign error in there somewhere. Not sure where, though.
    I think he just left out the negative by mistake.
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  8. #8
    MHF Contributor chisigma's Avatar
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    Some years ago I 'found' the following formula...

    \displaystyle \int_{0}^{1} u^{n}\ \ln^{m} u\ du = (-1)^{m}\ \frac{m!}{(n+1)^{m+1}} (1)

    In that case is m=1 so that...

    \displaystyle \int_{0}^{1} u^{n}\ \ln u\ du = - \frac{1}{(n+1)^{2}} (2)

    Thank a lot to Ackbeet!...



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  9. #9
    Super Member Random Variable's Avatar
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    My solution:

    Let  x = 1+e^{-t}

     = -\int^{0}_{\infty} \frac{\ln(e^{-t})}{1+e^{-t}} \ e^{-t} \ dt

     =  - \int^{\infty}_{0} \frac{te^{-t}}{1+e^{-t}} \ dt

     = \int^{\infty}_{0} t \sum^{\infty}_{k=1} (-1)^{k} e^{-kt} \ dt (since e^{-t} < 1 for  0<t<\infty )

    now integrate by parts

      t \sum_{k=1}^{\infty} (-1)^{k+1} \frac{e^{-kt}}{k} \Big|^{\infty}_{0} - \int^{\infty}_{0} \sum^{\infty}_{k=1} (-1)^{k+1} \ \frac{e^{-kt}}{k} \ dt

     0 + \int^{\infty}_{0} \sum^{\infty}_{k=1} (-1)^{k} \ \frac{e^{-kt}}{k}

     = \sum^{\infty}_{k=1} \int^{\infty}_{0} (-1)^{k} \ \frac{e^{-kt}}{k}<br />
(assuming f_{k}(t) = (-1)^{k} \ \frac{e^{-kt}}{k} converges uniformly on  0 \le t<\infty )

     = -\sum^{\infty}_{k=1} (-1)^{k+1} \ \frac{1}{k^{2}}

     = \sum^{\infty}_{k=1} (-1)^{k} \ \frac{1}{k^{2}} = -\frac{\pi^{2}}{12}
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  10. #10
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by FernandoRevilla View Post
    So what?. You can find I (definite integral) by means of an equation.

    Fernando Revilla
    How do you expect to get \pi^2 in there? I'm not saying you're wrong, but I really don't see how you can find I by integration by parts or by "means of an equation" - perhaps you can be more precise and provide your solution?
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  11. #11
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Bruno J. View Post
    I'm not saying you're wrong,
    You can say it.


    but I really don't see how you can find I by integration by parts or by "means of an equation" - perhaps you can be more precise and provide your solution?
    I can't. I had an stupid computational mistake. Sorry.

    Fernando Revilla
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  12. #12
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    I'm able to show that:

    \begin{aligned} \int_{0}^{1}\frac{\ln\left(x+1\right)}{x}\;{dt} &=\int_{0}^{1}\bigg(\sum_{k = 0}^{\infty}\frac{(-x)^k}{k+1}\bigg) \;{dx} = \displaystyle  \sum_{k=0}^{\infty}\bigg(\int_{0}^{1}\frac{(-x)^k}{k+1} \;{dx}\bigg)\\ &= \sum_{k=0}^{\infty}\bigg[\frac{(-1)^{k}x^{k+1}}{(k+1)^2}\;{dx}\bigg]_{x = 0}^{x = 1} = \sum_{k=0}^{\infty}\frac{(-1)^{k}}{(k+1)^2} = \frac{\pi^2}{12}.\end{aligned}

    But I'm lacking the change of variable to show this:

    \displaystyle \int^{2}_{1} \frac{\ln(x-1)}{x} \ dx = -\int_{0}^{1}\frac{\ln\left(x+1\right)}{x}

    What 'substitution' could do the trick? Mental block!
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