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Math Help - [SOLVED] Improper integrals

  1. #1
    Junior Member superman69's Avatar
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    [SOLVED] Improper integrals

    Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as "divergent."


    I don't get what to do next, but turns out the integral is

    arctan(x) I don't know where to continue from there.



    Last edited by superman69; October 21st 2009 at 10:33 PM. Reason: error
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  2. #2
    MHF Contributor chisigma's Avatar
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    Because \frac{1}{1+x^{2}} is an 'even function' we can swap x with -x and obtain...

    \int_{-\infty}^{1}\frac{dx}{1+x^{2}} = \int_{-1}^{\infty}\frac{dx}{1+x^{2}}= \int_{-1}^{0}\frac{dx}{1+x^{2}} + \int_{0}^{\infty}\frac{dx}{1+x^{2}} (1)

    Because for n-1<x<n, n\ge 1 integer is...

    \frac{1}{1+x^{2}} < \frac{1}{n^{2}} (2)

    ... is...

    \int_{0}^{\infty}\frac{dx}{1+x^{2}} < \sum_{n=1}^{\infty}\frac{1}{n^{2}} = \frac{\pi^{2}}{6} (3)

    ... so that the integral (1) converges. Its value can be computed in 'standard form'...

    \int_{-1}^{\infty} \frac{1}{1+x^{2}} = \lim_{x \rightarrow \infty} (tan^{-1} x + \frac{\pi}{4}) = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4} (4)

    Kind regards

    \chi \sigma
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