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Thread: Prove the values of the improper integrals.

  1. #1
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    Prove the values of the improper integrals.

    Prove that

    $\displaystyle
    \int_{0}^{\infty}x\frac{ ln(1+x^3)}{1+x^3}dx
    = ln\ 3 \ \frac{\pi}{\sqrt{3}} + \left(\frac{\pi}{3}\right)^2$

    $\displaystyle
    \int_{0}^{\infty} \frac{ln(1+x^3)}{1+x^3} dx
    = ln\ 3 \ \frac{\pi}{\sqrt{3}} - \left(\frac{\pi}{3}\right)^2$

    EDIT: Yep, I know the first one is a duplicate of the improper integral I posted last time, but what about the second one?
    Last edited by mathwizard; Jun 18th 2008 at 09:48 PM.
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  2. #2
    Super Member PaulRS's Avatar
    Joined
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    Quote Originally Posted by mathwizard View Post
    Prove that

    $\displaystyle
    \int_{0}^{\infty}x\frac{ ln(1+x^3)}{1+x^3}dx
    = ln\ 3 \ \frac{\pi}{\sqrt{3}} + \left(\frac{\pi}{3}\right)^2$

    $\displaystyle
    \int_{0}^{\infty} \frac{ln(1+x^3)}{1+x^3} dx
    = ln\ 3 \ \frac{\pi}{\sqrt{3}} - \left(\frac{\pi}{3}\right)^2$

    EDIT: Yep, I know the first one is a duplicate of the improper integral I posted last time, but what about the second one?
    Let $\displaystyle
    J\left( \theta \right) = \int_0^\infty {\tfrac{{\ln \left( {1 + \theta ^3 \cdot x^3 } \right)}}
    {{1 + x^3 }}dx}
    $ differentiating by Leibniz rule: $\displaystyle
    J'\left( \theta \right) = 3 \cdot \theta ^{2} \cdot \int_0^\infty {\tfrac{{x^3 }}
    {{\left( {1 + \theta ^3\cdot x^3 } \right) \cdot \left( {1 + x^3 } \right)}}dx}
    $

    Remember that: $\displaystyle
    \int_0^\infty {\tfrac{{dx}}
    {{1 + x^3 }}} = \tfrac{{2 \cdot \pi }}
    {{3 \cdot \sqrt 3 }}
    $ , calculate the derivative and integrate (considering that $\displaystyle
    J\left( 0 \right) = 0
    $) to get $\displaystyle J(1)$
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