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Math Help - Improper Integrals

  1. #1
    Member zangestu888's Avatar
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    Improper Integrals

    The integral of

    e^x / e^2x + 1 from upper limit:infinite and lower limit:0

    So for I have the antderirive as:

    this after changin the integration limits to a limit

    limit t--> (arctan(e^x)) from b=infinte and a=0

    and also this one

    the integral of

    ln(x)/x^2 from upper limit=b=infinte and lower limit=a=1 not sure how to start this one at all!
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  2. #2
    o_O
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    (1) Recall that: \lim_{x \to \infty} \arctan x = \frac{\pi}{2}

    So: \lim_{t \to \infty} \arctan x \Big|_{0}^{t} \ = \ \lim_{t \to \infty} \left( \arctan t - \arctan 0\right) \ = \ \cdots

    (2) To integrate, use integration by parts: \begin{array}{lll} u = \ln x & \quad & dv = \frac{1}{x^2} \ dx \\ du = \frac{1}{x}dx & \quad & v = -\frac{1}{x} \end{array}

    See if you can carry on from there.
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  3. #3
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    \frac{{{e}^{x}}}{{{e}^{2x}}+1}\le \frac{{{e}^{x}}}{{{e}^{2x}}}={{e}^{-x}}, hence \int_{0}^{\infty }{\frac{{{e}^{x}}}{{{e}^{2x}}+1}\,dx} converges.
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  4. #4
    Member zangestu888's Avatar
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    ..hmm

    Okay so am guessing my first part was correct, now I have to evaulate the integral as it approces infinite, I have one more question this is my weakness. am not sure how to exactly evaulte limits so for in my case:

    lim
    t-->arctan(e^x)=limt --> infinite (arctan(e^t) - arctan(o))
    =lim
    t->infinte (arctan(e^t) -0))

    =lim t-->infinte (arctan(e^t))=pi/2 ?? i think am write lol that ws great help am gonna work on the other one now ! thanks!BANKAI
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  5. #5
    o_O
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    Yep, as t \rightarrow \infty then e^t \rightarrow \infty as well. So \lim_{t \to \infty} \arctan e^t = \frac{\pi}{2}

    Good job, Kurosaki Ichigo.
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    You forgot that it's  \tan^{-1} (e^0) not  \tan^{-1} (0)

     \lim_{t \to \infty} \tan^{-1} (t) - \tan^{-1} (1) = \frac{\pi}{2}-\frac{\pi}{4} = \frac{\pi}{4}
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