Why Does Integral 1/x dx from -1 to 1 Diverge?

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Discussion Overview

The discussion centers around the improper integral of 1/x from -1 to 1, specifically why it diverges. Participants explore the nature of the integral, its symmetry, and the implications of breaking it into separate limits.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why the integral diverges, noting that breaking it into two parts (from -1 to 0 and 0 to 1) leads to divergent results of negative infinity and infinity, suggesting a symmetry that implies it "should be zero."
  • Another participant reflects on their research, stating that the integral is ill-defined due to the infinite ways it can be approached, leading to different limits and concluding that it diverges.
  • A third participant provides a definition of improper integrals, explaining that the limits do not exist when evaluating at a point where the function is undefined (in this case, at x=0).
  • This participant also describes the Cauchy Principal Value and how it is calculated, indicating that it can yield a different result than the integral itself.
  • A later reply reiterates the ill-defined nature of the integral and agrees with the conclusion that it diverges, while also acknowledging the possibility of different interpretations.

Areas of Agreement / Disagreement

Participants express a lack of consensus on the interpretation of the integral, with some agreeing on its ill-defined nature and divergence, while others explore the implications of the Cauchy Principal Value and its potential to yield different results.

Contextual Notes

The discussion highlights the complexities involved in evaluating improper integrals, particularly in terms of limits and the behavior of the function near points of discontinuity.

aceetobee
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Can someone explain why the following improper integral diverges?

Integral 1/x dx from -1 to 1

I know if you break it up the individual integrals (from -1 to 0 and 0 to 1) diverge to negative infinity and infinity, whose sum is indeterminant in general, but the symmetry of the integral suggests it "should be zero".

Thanks!
 
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OK... well, a little research and I think I answered my own question.

It really is an ill-definied integral, because when broken up into the limit of two separate integrals, there are an infinite number of ways this can be done, with one side approaching zero at a different speed than the other.

I guess the Cauchy Principle Value would be zero, but there are other possiblilities, so it diverges.

Please correct me if I'm wrong on this...
 
If [itex]\int_b^c f(x)dx[/itex] is "improper" because f(a) is not defined, with b< a< c, then the integral is DEFINED as:
[tex]lim_{x_1->a^-}\int_b^{x_1}f(x)dx+ lim_{x_2->a^+}\int_{x_2}^cf(x)dx[/tex].

Since the anti-derivative of 1/x is ln|x|, neither of those limits exists when a= 0.

The Cauchy Principal Value, on the other hand is:
[tex]lim_{x_1->a}\left(\int_b^{x_1}f(x)dx+ \int_{x_1}^cf(x)dx\right)[/tex].
Since the limit is taken after both integrals are done, we can cancel the "ln|x1|" terms before the limit and just have ln|c|-ln|b|.
 
Last edited by a moderator:
aceetobee said:
It really is an ill-definied integral, because when broken up into the limit of two separate integrals, there are an infinite number of ways this can be done, with one side approaching zero at a different speed than the other.

I guess the Cauchy Principle Value would be zero, but there are other possiblilities, so it diverges.

Please correct me if I'm wrong on this...
That's totally correct.
 

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