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Thread: Improrper Integrals

  1. #1
    Sep 2009

    Improrper Integrals

    I have a midterm tomorrow morning and was looking at example problems and got stuck on this one. Please Help!

    Determine if the following integral converges or diverges. If it converges find its value
    It is the integral from 8 to +infinity of
    [(64-x^2)/(64+x^2)^2]*arctan(x/8) dx
    It tells us that the integral of [(64-x^2)/(64+x^2)^2] = x/(64+x^2)dx

    I know that since they give us the integral for half of the function to use integration by parts. I have gotten to the point where you plug the infinities back into the integral.

    for my integral I got arctan(x/8)*x/(64+x^2)+4/(64+x^2) evaluated from 8 to +inf

    I have no problem plugging 8 into the integral but I am having trouble plugging the +inf into the integral and knowing whether or not it is determinate or indeterminate

    please help
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  2. #2
    A Plied Mathematician
    Jun 2010
    CT, USA
    You have

    $\displaystyle \displaystyle\int_{8}^{\infty}\frac{64-x^{2}}{64+x^{2}}\,\tan^{-1}(x/8)\,dx.$

    I believe your intuition about by-parts is accurate. Let

    $\displaystyle dv=\frac{64-x^{2}}{64+x^{2}}\,dx,$ and let

    $\displaystyle u=\tan^{-1}(x/8).$ Then the integral becomes

    $\displaystyle \displaystyle\int_{8}^{\infty}\frac{64-x^{2}}{64+x^{2}}\,\tan^{-1}(x/8)\,dx=\left(\tan^{-1}(x/8)\,\frac{x}{64+x^{2}}\right)\Bigg|_{8}^{\infty}-\int_{8}^{\infty}\frac{x}{64+x^{2}}\,\frac{8}{64+x ^{2}}\,dx$

    $\displaystyle \displaystyle=\left(\tan^{-1}(x/8)\,\frac{x}{64+x^{2}}\right)\Bigg|_{8}^{\infty}-8\int_{8}^{\infty}\frac{x}{(64+x^{2})^{2}}\,dx.$

    For the boundary term (the first one on the RHS), the arctan function is bounded, and the denominator beats the numerator, so it should be zero at the upper end. The lower end presents no problems. As for the integral, I would do a u substitution to finish it off. What do you get?
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