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Problem Understanding Divergence in Improper Integrals

  1. Mar 31, 2010 #1
    I am having a problem with the definition of divergence in improper integrals. My understanding of the logic behind convergence and divergence is that, for example, as a improper integral approaches infinity the area under the function will be approaching but never reach zero. This implies that the area may be unbounded (the area under the curve is approaching infinity as the function approaches infinity) or the area may be bounded (the area under the curve - although never actually reaching a finite number - is approaching a finite number as the function approaches infinity).

    However, the definition seems to be at odds with this conception in certain instances. The definition states that a function is divergent if any part of that function is divergent. This would include improper integral problems that introduce limits of this form: lim x-->a f(x) + lim x-->b g(x), where plugging a and b into their respective equations results in the term (infinity)-(infinity). (In some cases a=-infinity and b=infinity and in others a is a number approached from the negative side and b is that same number approached from the positive side.)

    My question is, why is it that when dealing with improper integrals the term (infinity)-(infinity) is not an indeterminate form, which then may or may not be divergent or convergent? Take as an example the improper integral of 1/x from -1 to 1. Intuition would tell you that the negative and positive space under the curve on either side of zero would cancel each other out, so that the integral would converge to 0.

    Is there an logical definition for why the divergence is defined the way it is, and can this be found in any of the literature? Thank you.
  2. jcsd
  3. Mar 31, 2010 #2


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    The problem with your intuition there is that it's only correct if in a certain instance. Let's define the improper integral of 1/x from -1 to 1 by

    [tex]\lim_{a \rightarrow 0^-} \int_{-1}^a \frac{dx}{x} + \lim_{b \rightarrow 0^+} \int_b^1 \frac{dx}{x} = \lim_{a \rightarrow 0^-} \ln(|a|) - \lim_{b \rightarrow 0^+} \ln b[/tex]

    You see that if a and b are not equal then the two terms don't exactly cancel. If b = |a|/2, for example, the result is [itex]\ln(2)[/itex]. Hence, a consistent value cannot be assigned to the integral, and thus it is considered to not exist (perhaps this is better to say than divergent?). The symmetric case, however, does have some uses, and is know as a principal value integral.
  4. Mar 31, 2010 #3
    Thanks for the reply. I can see how ln|a| and ln(b) are not exactly equal, but shouldn't their limits be as they approach zero?
  5. Apr 1, 2010 #4

    Gib Z

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    In Mute's formulation, the result in the calculations is two seperate limits, both of which fail to tend to any real number. Subtraction isn't even well defined. You can't conclude "infinity - infinity = 0" . If you go on to do real analysis and learn how limits, and sequences and the like are all defined, a lot of this will become clearer to you. One fact you will learn is that if a sequence approaches a limit, all subsequences must also tend to the same limit. Or in terms of limits, we can vaguely say "if the limit exists, it must be the same no matter how we choose to approach it". As Mute showed, we can easily adjust the limit to be any value we like.
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