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Improper integral, infinite limits of integration

  1. Mar 28, 2013 #1

    ∫ x/(x^2+1) dx

    I basicaly evaluated the integral and Ln (x^2+1) as the antiderivative and when taking the limits I get ∞-∞

    (ln |1| -ln|b+1|) + (ln|n+1|- ln|1|)

    lim b-> neg. infinity lim n-> infinity

    does this function converge or diverge? this was a question on the quiz and everyone in class after said the function converges to zero. I'm a littled puzzled by this.
  2. jcsd
  3. Mar 28, 2013 #2


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    ##\int_{-\infty}^\infty \frac{x}{x^2+1} \, dx = \lim_{a \to -\infty} \int_{a}^{0} \frac{x}{x^2+1} \, dx + \lim_{b\to\infty} \int_{0}^{b} \frac{x}{x^2+1} \, dx##

    The limits are separate, and you need convergence in both. Your friends are thinking of the Cauchy principle value, but unless the problem explicitly asks for it, you should not give that as the answer.
  4. Mar 28, 2013 #3
    so does it converge or diverge?
  5. Mar 28, 2013 #4
    Also I looked on harvard, and university of portlands website, and it says that the function diverges.

    http://www.math.harvard.edu/~keerthi/files/1b_fall_2011/worksheet11_solns.pdf [Broken]
    end of pg. 1 and beginning of page 2
    Last edited by a moderator: May 6, 2017
  6. Mar 28, 2013 #5
    Also, when looking at the antiderivative of Ln|x^2+1| i see that it is an even function does this mean you can subtract ∞-∞ since the function is approaching infinity at the same rate? If not, when is this method applied, if it is at all? ha? I really wanna understand this problem. Apparently my teacher loves surprising us on tests and quizzes.
  7. Mar 29, 2013 #6


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    For large positive values of ##x##, the integrand is approximately ##1/x##. What can you conclude?
  8. Mar 29, 2013 #7
    it would have to diverge because taking the limit as Ln|x| as x-> infinity diverges, correct?
  9. Mar 29, 2013 #8


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    Yes, that's right.
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