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Thread: Integral

  1. #1
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    Integral

    Hi I have another integral question that I wanted to ask.

    Question: Prove that $\displaystyle \displaystyle\int_{-\infty}^\infty\frac{e^{2x}}{(e^{3x}+1)^2}dx=\frac{ 2\pi}{9\sqrt{3}}$
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by nonsingular View Post
    Hi I have another integral question that I wanted to ask.

    Question: Prove that $\displaystyle \displaystyle\int_{-\infty}^\infty\frac{e^{2x}}{(e^{3x}+1)^2}dx=\frac{ 2\pi}{9\sqrt{3}}$

    $\displaystyle \displaystyle\int_{-\infty}^0\frac{e^{2x}}{(e^{3x}+1)^2}dx
    +\int_{0}^\infty\frac{e^{2x}}{(e^{3x}+1)^2}dx$

    the first one:
    $\displaystyle \displaystyle\int_{-\infty}^0\frac{e^{2x}}{(e^{3x}+1)^2}dx =$
    $\displaystyle \displaystyle\int_{-\infty}^0(\frac{e^x}{(e^{3x}+1)})^2dx =$
    $\displaystyle \displaystyle\int_{0+}^1\frac{u^2}{(u^3+1)^2}dx =$

    where $\displaystyle u=e^x$, and bounds change as well, and $\displaystyle dx= e^{-x}du =u^{-1} du $
    $\displaystyle \displaystyle\int_{0+}^1\frac{u^2}{(u^3+1)^2}u^{-1}du =$
    $\displaystyle \displaystyle\int_{0+}^1\frac{u}{(u^3+1)^2}du =$
    For the indefinite, Wolfram gives:

    integral x/(x^3+1)^2 dx = 1/18 (log(x^2-x+1)+(6 x^2)/(x^3+1)-2 log(x+1)+2 sqrt(3) tan^(-1)((2 x-1)/sqrt(3)))+constant

    Pretty messy, but you can use it for the remaining half.

    Hope this helps.
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  3. #3
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    complex analysis


    The integral $\displaystyle \int_{0}^{\infty} \frac{x}{(x^3+1)^2} ~dx $

    Substitute $\displaystyle x = t^{\frac{1}{3}}$

    The integral becomes $\displaystyle \frac{1}{3} \int_{0}^{\infty} \frac{t^{- \frac{1}{3}}}{(t+1)^2} ~ dt $

    Then use this formula :

    $\displaystyle \int_{0}^{\infty} x^{p-1} Q(x) ~dx = \frac{2\pi i}{1 - e^{2p \pi i}} \sum_{all~poles} Res[(z)^{p-1} Q(z)]$

    Now sub. p into $\displaystyle \frac{2}{3}$

    $\displaystyle = \frac{1}{3} \frac{2\pi i}{e^{\frac{2 \pi i}{3}}(e^{-\frac{2 \pi i}{3}} - e^{\frac{2 \pi i}{3}} )} ( (z)^{- \frac{1}{3}} )'_{z= -1} $

    $\displaystyle = \frac{1}{3} \frac{2\pi i}{e^{\frac{2 \pi i}{3}}(-2i sin{ \frac{2 \pi}{3}} )} ( -\frac{1}{3} ) (-1)^{-\frac{4}{3} }$

    $\displaystyle = \frac{ 2 \pi i}{ -6 i \sin{ \frac{ 2\pi }{3}} (-3) }$

    $\displaystyle = \frac{2 \pi}{9 \sqrt3 }$
    Last edited by simplependulum; Jun 15th 2009 at 01:22 AM.
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  4. #4
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    Quote Originally Posted by nonsingular View Post
    Hi I have another integral question that I wanted to ask.

    Question: Prove that $\displaystyle \displaystyle \int_{-\infty}^\infty\frac{e^{2x}}{(e^{3x}+1)^2}dx=\frac{ 2\pi}{9\sqrt{3}}$

    let $\displaystyle e^{3x}=\tan^2t.$ then $\displaystyle \int_{-\infty}^\infty\frac{e^{2x}}{(e^{3x}+1)^2}dx=\frac{ 2}{3}\int_0^{\frac{\pi}{2}}(\sin t)^{\frac{1}{3}} (\cos t)^{\frac{5}{3}} \ dt=\frac{1}{3}B \left(\frac{2}{3},\frac{4}{3} \right)=\frac{1}{3} \Gamma \left(\frac{2}{3} \right) \Gamma \left(\frac{4}{3} \right),$ and the result follows because $\displaystyle \Gamma \left(\frac{4}{3} \right)=\frac{1}{3}\Gamma \left(\frac{1}{3} \right)$ and by Euler's reflection formula:

    $\displaystyle \Gamma \left(\frac{2}{3} \right) \Gamma \left(\frac{1}{3} \right)=\frac{\pi}{\sin(\frac{\pi}{3})}=\frac{2\pi }{\sqrt{3}}.$
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