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Math Help - Today's calculation of integral #9

  1. #1
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    Today's calculation of integral #9

    Find \int_0^1\frac{x^2}{(1-x)\ln^2(1-x)}\,dx.
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    Ok

    Quote Originally Posted by Krizalid View Post
    Find \int_0^1\frac{x^2}{(1-x)\ln^2(1-x)}\,dx.
    You can rewrite it as \int_0^1{\sum_{n=0}^{\infty}x^{n+2}}\cdot\bigg(-\sum_{n=0}^{\infty}\frac{x^{n}}{n}\bigg)^{-2}...which then is equal to \int_0^1{\sum_{n=0}^{\infty}x^{n+2}\cdot{2\sum_{n=  0}^{\infty}\sum_{n=0}^{\infty}\frac{n^2}{x^{2n}}}}...and then 2\int_0^1{\sum_{n=0}^{\infty}\sum_{n=0}^{\infty}\s  um_{n=0}^{\infty}\frac{x^{2-n}}{n}}...so wed get 2\sum_{n=0}^{\infty}\sum_{n=0}^{\infty}\sum_{n=0}^  {\infty}\frac{x^{3-n}}{n(3-n)} evaluated from 0 to 1...I know I know I did it completely wrong...but will someone point out where? I just did this for novelty
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    I just did this for novelty
    Then just don't do it if you don't know how to tackle what it comes.

    It does exist a quick solution.
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    I did not solve it yet. Here is the first step I was thinking. Let t = -\ln (1-x).
    This gives us, \int_0^{\infty} \left( \frac{1 - e^{-t}}{t} \right)^2 dt.
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  5. #5
    Moo
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    Hello,

    Let u=\ln(1-x)

    --------------
    du=-\frac{dx}{1-x}

    --------------
    e^u=1-x

    x=1-e^u

    x^2=(1-e^u)^2

    --------------
    If x=0 then u=0

    If x=1 then u=-\infty

    --------------
    Inverting the extremities of the integral with the - sign coming from du

    \int_0^1\frac{x^2}{(1-x)\ln^2(1-x)} dx = \int_{-\infty}^0 \frac{(1-e^u)^2}{u^2} du


    Where did i go wrong ? :'(
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    Quote Originally Posted by Moo View Post
    Where did i go wrong ? :'(
    You did not solve it, look at my post above. This only changes the integral into a (probably) simpler one. Now you need to work with this new integral.
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  7. #7
    Moo
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    Actually, i've been trying but gave up an hour ago... The first parts of integration by parts were divergent oO

    And ok, when i posted i saw the first version of your integral and got anxious because it was not the same...
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    So is it this?

    Quote Originally Posted by ThePerfectHacker View Post
    I did not solve it yet. Here is the first step I was thinking. Let t = -\ln (1-x).
    This gives us, \int_0^{\infty} \left( \frac{1 - e^{-t}}{t} \right)^2 dt.
    is that then equal to \int{\bigg(-2\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n-1}}{n!}+\sum_{n=0}^{\infty}\frac{(-2)^{n}x^{n-2}}{n!}+1\bigg)dx} =-2\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n}}{n\cdot{n!}}+\sum_{n=0}^{\infty}\frac{  (-2)^{n}x^{n-1}}{(n-1)n!}+x evaluated from 0 to ∞?
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  9. #9
    Moo
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    Yop,

    Where did \frac{1}{t^2} go ? (it's it, instead of +1)

    But your method doesn't simplify the problem... or i've missed a point ^^
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    oops

    Quote Originally Posted by Moo View Post
    Yop,

    Where did \frac{1}{t^2} go ? (it's it, instead of +1)

    But your method doesn't simplify the problem... or i've missed a point ^^
    what I said except change x to \frac{-1}{x}
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  11. #11
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    Quote Originally Posted by ThePerfectHacker View Post
    I did not solve it yet. Here is the first step I was thinking. Let t = -\ln (1-x).
    This is a good start.

    Now find the other one.
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  12. #12
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    Hint: construct a double integral. (Following on post #4.)
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  13. #13
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    What we need is \frac{1}{t^{2}}=\int_{0}^{\infty }{xe^{-tx}\,dx},\,t>0. From there you may observe that the conclusion follows.
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  14. #14
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    Quote Originally Posted by Krizalid View Post
    What we need is \frac{1}{t^{2}}=\int_{0}^{\infty }{xe^{-tx}\,dx},\,t>0. From there you may observe that the conclusion follows.
    Thank you for that. I had a similar idea: \frac1{t^2} = \int_0^{\infty} e^{-t^2x} dx but that gave a nasty integral.
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  15. #15
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    In general, \frac{(n-1)!}{t^n} = \int_0^{\infty} x^{n-1} e^{-tx} dx for t>0 and n\geq 1. This seems to be a useful fact to know in double integration.
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