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Math Help - Integral-ln

  1. #1
    Super Member dhiab's Avatar
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    Integral-ln

    Calculate :
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  2. #2
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    Quote Originally Posted by dhiab View Post
    Calculate :

    Is it a definite integral actually ?

    If we replace x by  e^t
    then dx = e^t dt

    the integral  = \int e^t \sqrt{t} ~dt

    I am afraid that i dont have confidence to solve it !

    If the u. limit is 1 and l. limit is 0 , then this integral

     \int_0^1 \sqrt{-\ln{x}}~dx ~~ can be written as

     \int_0^{\infty} e^{-t} \sqrt{t} ~dt = \Gamma(3/2) = \frac{\sqrt{\pi}}{2}
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  3. #3
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    Quote Originally Posted by dhiab View Post
    Calculate :
    According to Mathematica, the answer is

    x\sqrt{\ln{x}} - \frac{1}{2}\sqrt{\pi}\,\textrm{erfi}(\sqrt{\ln{x}}  )


    Erfi -- from Wolfram MathWorld
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  4. #4
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    There is an identity which states:

    \int ln^{n}(x)dx=(-1)^{n}\cdot x\sum_{k=0}^{n}\frac{n!}{k!}(-ln(x))^{k}

    This would be interesting to prove.

    Also, since Ex(x,n)=\sum_{k=0}^{n}\frac{x^{k}}{k!},

    the above can be written as (-1)^{n}n!xEx(-ln(x),n)

    and Ex(x,n)=\frac{e^{x}\Gamma(n+1,x)}{\Gamma(n+1)}

    where {\Gamma}(x,n) is the incomplete gamma function.

    Just some food for thought.
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  5. #5
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    Quote Originally Posted by galactus View Post
    There is an identity which states:

    \int ln^{n}(x)dx=(-1)^{n}\cdot x\sum_{k=0}^{n}\frac{n!}{k!}(-ln(x))^{k}

    This would be interesting to prove.
    I might try letting x = e^t (as mention earlier by SP) giving

     <br />
\int t^n e^t\,dt<br />
    then try repeated integration by parts.
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