Improper integral, infinite limits of integration

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Discussion Overview

The discussion revolves around the evaluation of the improper integral of the function x/(x^2+1) over the interval from negative infinity to positive infinity. Participants explore whether this integral converges or diverges, referencing various approaches and interpretations related to limits and antiderivatives.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant evaluates the integral and suggests that it leads to an indeterminate form of ∞-∞, expressing confusion over the convergence.
  • Another participant clarifies that the integral should be split into two separate limits and emphasizes the need for convergence in both parts, suggesting that classmates may be considering the Cauchy principal value.
  • A participant cites external sources indicating that the integral diverges, referencing specific educational materials from Harvard and the University of Portland.
  • Further discussion includes questioning whether the even nature of the antiderivative allows for the subtraction of ∞-∞, seeking clarity on when such methods are applicable.
  • Another participant notes that for large positive values of x, the integrand approximates 1/x, prompting further conclusions about divergence.
  • One participant asserts that the limit of the natural logarithm diverges as x approaches infinity, suggesting that this indicates divergence.
  • A later reply confirms the assertion about divergence, indicating some agreement on this point.

Areas of Agreement / Disagreement

Participants express differing views on the convergence of the integral, with some suggesting it diverges based on external sources and reasoning, while others remain uncertain or seek clarification on the evaluation process. The discussion does not reach a consensus.

Contextual Notes

Participants reference the Cauchy principal value and the behavior of the integrand at infinity, indicating that assumptions about convergence may depend on specific interpretations of improper integrals.

nick.martinez
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∫ 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.
 
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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.
 
so does it converge or diverge?
 
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
end of pg. 1 and beginning of page 2
 
Last edited by a moderator:
pwsnafu said:
##\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.

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 want to understand this problem. Apparently my teacher loves surprising us on tests and quizzes.
 
nick.martinez said:
so does it converge or diverge?
For large positive values of ##x##, the integrand is approximately ##1/x##. What can you conclude?
 
it would have to diverge because taking the limit as Ln|x| as x-> infinity diverges, correct?
 
nick.martinez said:
it would have to diverge because taking the limit as Ln|x| as x-> infinity diverges, correct?
Yes, that's right.
 

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