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Math Help - integral a la the Kriz way.

  1. #1
    Eater of Worlds
    galactus's Avatar
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    integral a la the Kriz way.

    Here is an integral from the site that isomorphism posted. I thought I would try my hand at this one using Kriz's double integration technique.

    \int_{0}^{1}\frac{ln(x)}{1+x}dx=\frac{-{\pi}^{2}}{12}

    We know that ln(x)=\int_{1}^{x}\frac{1}{u}du

    Then we get:

    \int_{0}^{1}\int_{1}^{x}\frac{1}{u(x+1)}dudx

    Now, here is where I am hung up on switching the limits of integration, coming up with the new limits, and thus coming up with a nice workable integral.

    It looks similar to \int_{0}^{1}\frac{ln(x+1)}{x^{2}+1}dx=\frac{\pi}{8  }ln(2)
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  2. #2
    Math Engineering Student
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    Well in that case, it won't work.

    To reverse integration order, we rewrite the double integral as -\int_{0}^{1}\!{\int_{x}^{1}{\frac{du\,dx}{u(x+1)}}  }.

    Now play with inequalities: 0\le x\le1,\,x\le u\le1. Hence 0\le x\le u,\,0\le u\le1. After reversing integration order the double integral equals -\int_{0}^{1}\!{\int_{0}^{u}{\frac{dx\,du}{u(x+1)}}  }, which is doable. (Only the inner one.)

    But we have a problem, that will yield another undoable integral which result is exactly -\frac{\pi^2}{12}, so the only way to tackle this, it's a power series application.

    ------

    If you want to try a double integral trick, see \int_{0}^{\pi /2}{\frac{\ln (\sec x)}{\tan x}\,dx}. You can watch my solution here.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Here is an integral from the site that isomorphism posted. I thought I would try my hand at this one using Kriz's double integration technique.

    \int_{0}^{1}\frac{ln(x)}{1+x}dx=\frac{-{\pi}^{2}}{12}

    We know that ln(x)=\int_{1}^{x}\frac{1}{u}du

    Then we get:

    \int_{0}^{1}\int_{1}^{x}\frac{1}{u(x+1)}dudx

    Now, here is where I am hung up on switching the limits of integration, coming up with the new limits, and thus coming up with a nice workable integral.

    It looks similar to \int_{0}^{1}\frac{ln(x+1)}{x^{2}+1}dx=\frac{\pi}{8  }ln(2)
    \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n

    So we have

    \sum_{n=0}^{\infty}(-1)^n\int_0^{1}\ln(x)x^ndx

    Now

    \int_0^{1}\ln(x)x^n=\frac{-1}{(n+1)^2}

    So we have

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

    Now we know that

    \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^x}=(1-2^{1-x})\zeta(x)

    So we have( Lets hope this works)

    -(1-2^{1-2})\zeta(2)=-\left(\frac{1}{2}\right)\frac{\pi^2}{6}=\frac{-\pi^2}{12}


    Hooray!
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  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
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    I dont know how to do it using double integrals. I always have a tough time switching the order of integration.

    I have a series way of doing it:

    For my ease, I started with x = e^t

    So I = \int_0^1 \frac{\ln x}{1+x} dx = \int_{-\infty}^0 \frac{t e^t}{1+e^t} dt =  -\int_{-\infty}^0 \left(\sum_{n=1}^{n = \infty} t (-e^t)^n\right) dt

    Now switching integration and summation,

    I =  -\sum_{n=1}^{n = \infty}(-1)^n \left(\int_{-\infty}^0 t e^{nt}\right) dt

    \int_{-\infty}^0 t e^{nt} \, dt = \frac{te^{nt}}{n}\bigg{|}_{-\infty}^0 - \frac{e^{nt}}{n^2}\bigg{|}_{-\infty}^0 = 0 - \frac1{n^2}

    Thus:

    I =  -\sum_{n=1}^{n = \infty}(-1)^n \left(\int_{-\infty}^0 t e^{nt}\right) dt  =  \sum_{n=1}^{n = \infty} \frac{(-1)^n}{n^2}

    = -\left(\sum_{n=1}^{n = \infty}\frac1{n^2} - \frac12 \sum_{n=1}^{n = \infty} \frac1{n^2}\right)

     =- \left(\frac{\pi^2}6 - \frac{\pi^2}{12}\right) = -\frac{\pi^2}{12}
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  5. #5
    Eater of Worlds
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    Thanks for all the replies, kids. I was thinking it may not work, but thought it was me.

    Actually, I worked it mathstud's way and got it OK because I recognized the \frac{1}{1+x}=\sum_{k=0}^{\infty}(-1)^{k}x^{k}. I was just wondering about that double integral thing.

    I like all your solutions.
    Last edited by galactus; June 22nd 2008 at 09:48 AM.
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