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Math Help - Prove the values of the improper integrals.

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

    Prove that

    <br />
\int_{0}^{\infty}x\frac{ ln(1+x^3)}{1+x^3}dx<br />
= ln\ 3 \ \frac{\pi}{\sqrt{3}} + \left(\frac{\pi}{3}\right)^2

    <br />
\int_{0}^{\infty} \frac{ln(1+x^3)}{1+x^3} dx<br />
= 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; June 18th 2008 at 09:48 PM.
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  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
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    Quote Originally Posted by mathwizard View Post
    Prove that

    <br />
\int_{0}^{\infty}x\frac{ ln(1+x^3)}{1+x^3}dx<br />
= ln\ 3 \ \frac{\pi}{\sqrt{3}} + \left(\frac{\pi}{3}\right)^2

    <br />
\int_{0}^{\infty} \frac{ln(1+x^3)}{1+x^3} dx<br />
= 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 <br />
J\left( \theta  \right) = \int_0^\infty  {\tfrac{{\ln \left( {1 + \theta ^3  \cdot x^3 } \right)}}<br />
{{1 + x^3 }}dx} <br />
differentiating by Leibniz rule: <br />
J'\left( \theta  \right) = 3 \cdot \theta ^{2}  \cdot \int_0^\infty  {\tfrac{{x^3 }}<br />
{{\left( {1 + \theta ^3\cdot x^3 } \right) \cdot \left( {1 + x^3 } \right)}}dx} <br />

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