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Math Help - double improper integral

  1. #1
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    double improper integral

    find \int_{-\infty}^{\infty} \frac{e^{3 x}}{e^{6 x}+1}dx

    im lost with this one
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  2. #2
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    Quote Originally Posted by viet View Post
    find \int_{-\infty}^{\infty} \frac{e^{3 x}}{e^{6 x}+1}dx

    im lost with this one
    This is fun.

    First find,
    \int \frac{e^{3x}}{e^{6x}+1} dx
    Let t=e^{3x} \Rightarrow t' = (1/3)e^{3x}

    Thus,
    3\int \frac{1}{t^2+1} dt = 3\tan^{-1} t +C=3\tan^{-1} e^{3x} + C

    So now,
    \int_{-\infty}^{\infty} \frac{e^{3x}}{e^{6x}+1} = \int_{-\infty}^0 \frac{e^{3x}}{e^{6x}+1} + \int_0^{\infty} \frac{e^{3x}}{e^{6x}+1} dx = 3\tan^{-1} e^x \big|_{-\infty}^0 + 3\tan^{-1} e^x \big|_0^{\infty}  = \frac{3\pi}{4} - 0 + \frac{3\pi}{4} - 0 = \frac{3\pi}{2}
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    First find,
    \int \frac{e^{3x}}{e^{6x}+1} dx
    Let t=e^{3x} \Rightarrow t' = (1/3)e^{3x}
    Oops, you're not integrating there
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  4. #4
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    the answer is Pi/6

    Quote Originally Posted by ThePerfectHacker View Post
    This is fun.

    First find,
    \int \frac{e^{3x}}{e^{6x}+1} dx
    Let t=e^{3x} \Rightarrow t' = (1/3)e^{3x}

    Thus,
    3\int \frac{1}{t^2+1} dt = 3\tan^{-1} t +C=3\tan^{-1} e^{3x} + C

    So now,
    \int_{-\infty}^{\infty} \frac{e^{3x}}{e^{6x}+1} = \int_{-\infty}^0 \frac{e^{3x}}{e^{6x}+1} + \int_0^{\infty} \frac{e^{3x}}{e^{6x}+1} dx = 3\tan^{-1} e^x \big|_{-\infty}^0 + 3\tan^{-1} e^x \big|_0^{\infty}  = \frac{3\pi}{4} - 0 + \frac{3\pi}{4} - 0 = \frac{3\pi}{2}
    I think the answer is Pi/6.
    Attached Thumbnails Attached Thumbnails double improper integral-21june2007.gif  
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  5. #5
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    I think the answer is Pi/6.
    Yes.
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