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Math Help - Improper integrals

  1. #1
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    Improper integrals

    \int_{2}^{+\infty} \frac{dx}{x\sqrt{x^2-4}}

    i dunno what to do with this integral since both the limits are discontinuous:
    so
    \lim_{t \to 2} \frac{1}{2} \sec^{-1}(\frac{x}{2}) \big|_{t}^{2\sqrt{2}} + \lim_{s\to +\infty} \frac{1}{2} \sec^{-1}(\frac{x}{2}) \big|_{2\sqrt{2}}^{s}

    so in evaluating the integral:
    =\frac{1}{2}\sec^{-1}(\sqrt{2}) - \frac{1}{2}\sec^{-1}(1) + \frac{1}{2}\sec^{-1}(\frac{+\infty}{2}) - \frac{1}{2}\sec^{-1}(\sqrt{2})

    =\frac{1}{2}\sec^{-1}(1) which is infinite


    my answer seems to be divergent

    the answer at the back of the book seems to be \frac{\pi}{4}
    Last edited by ^_^Engineer_Adam^_^; September 12th 2007 at 04:46 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    \int_{2}^{+\infty} \frac{dx}{x\sqrt{x^2-4}}

    i dunno what to do with this integral since both the limits are discontinuous:
    so
    \lim_{t \to 2} \frac{1}{2} \sec^{-1}(\frac{x}{2}) \big|_{t}^{2\sqrt{2}} + \lim_{s\to +\infty} \frac{1}{2} \sec^{-1}(\frac{x}{2}) \big|_{2\sqrt{2}}^{s}

    so in evaluating the integral:
    =\frac{1}{2}\sec^{-1}(\sqrt{2}) - \frac{1}{2}\sec^{-1}(1) + \frac{1}{2}\sec^{-1}(\frac{+\infty}{2}) - \frac{1}{2}\sec^{-1}(\sqrt{2})

    =\frac{1}{2}\sec^{-1}(1) which is infinite


    my answer seems to be divergent

    the answer at the back of the book seems to be \frac{\pi}{4}
    Check two things.

    First
    sec^{-1}(\sqrt{2}) = \frac{\pi}{4}

    sec^{-1} ( \infty ) = \frac{\pi}{2}

    and (drum-roll please!)
    \theta = sec^{-1}(1) \implies sec(\theta) = 1

    sec(\theta) = 1 \implies cos(\theta) = \frac{1}{1} = 1

    So \theta = 0

    Thus
    sec^{-1}(1) = 0.

    So
    =\frac{1}{2}\sec^{-1}(\sqrt{2}) - \frac{1}{2}\sec^{-1}(1) + \frac{1}{2}\sec^{-1}(\frac{+\infty}{2}) - \frac{1}{2}\sec^{-1}(\sqrt{2})

    = \frac{1}{2} \cdot \frac{\pi}{4} - \frac{1}{2} \cdot 0 + \frac{1}{2} \cdot \frac{\pi}{2} - \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{4}

    And, of course, since the x = 2 limit isn't actually infinite we don't need to go through all that mess with the limit on it.

    -Dan
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  3. #3
    MHF Contributor red_dog's Avatar
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    I don't know if this is correct.
    Calculate \displaystyle\int_{a}^b\frac{dx}{x\sqrt{x^2-4}}.
    Let \displaystyle x=\frac{1}{t}\Rightarrow dx=-\frac{1}{t^2}dt.
    Now we have
    \displaystyle -\int_{\frac{1}{a}}^{\frac{1}{b}}\frac{dt}{\sqrt{1-4t^2}}=\left.-\frac{1}{2}\arcsin 2t\right|_{\frac{1}{a}}^{\frac{1}{b}}=-\frac{1}{2}\arcsin\frac{2}{b}+\frac{1}{2}\arcsin\f  rac{2}{a}.
    Now take the limits when b\to\infty and a\to 2 and we get \frac{\pi}{4}
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