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Improper integral of 1/x from –1 to 1

  1. Feb 25, 2006 #1
    I am a physicist, not a mathematician.
    This problem has bothered me for 40 years.

    All introductory Calculus texts would consider this integral divergent.
    An example found in many texts is the integral of 1/(x-2) from 0 to 3, which is just a variant to the question I am asking. What I find interesting is the accompanying statement that says if you find the integral equal to -ln|2| you would be making a terrible mistake????.

    Now back to my problem of the area under the curve of 1/x from –1 to 1.
    Using a symmetry argument I would state to however close to the Y axis you want to come the area is exactly equal to 0.0. Even a bigger computer would yield the same sum, exactly 0.0.

    Now I know I’m using the word area under the curve and not integral, but I do have an infinite number of Riemann Sums missing only the last sum at X=0,
    Not sure what to do with this last sum? But again I can make a strong argument that last Sum is also 0.0 based on symmetry.

    I’ll gladly accept any help on this as I am now teaching math, and I want to cover this in my calculus class.

    Dr Bob
  2. jcsd
  3. Feb 25, 2006 #2


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    It is quite similar to how the value of a conditionally convergent series is dependent upon the manner in which you sum up your terms:

    Let [itex]\epsilon[/itex] be a small parameter, and let [itex]f(\epsilon),g(\epsilon)[/itex] be positive functions vanishing as epsilon goes to zero.

    Consider the integral:

    As you can see, the limiting value of [itex]I(\epsilon)[/itex] as epsilon goes to zero is entirely dependent upon how we choose to approach our singularity!

    But, in general, we want our integral value to be INDEPENDENT of how we partition our interval of integration.
    (And for Riemann integrable functions, independence of a particular partition scheme is a requirement that must be met).

    Why should "symmetric" integration be preferred over other ways?

    As a contrasting example, consider the improper integral:
    By a similar stratagem, we can set:
    This will tend to the value 4 in the limit, no matter what continuous functions f and g are.
    Last edited: Feb 25, 2006
  4. Feb 26, 2006 #3


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    Consider that in oder for the integral to converge, we require that

    [tex]\int_{-1}^{1}\frac{dx}{x} =\lim_{\epsilon_1,\epsilon_2\rightarrow 0^+ }\left( \int_{-1}^{-\epsilon_1} \frac{dx}{x} +\int_{\epsilon_2}^{1} \frac{dx}{x}\right) = \lim_{\epsilon_1,\epsilon_2\rightarrow 0^+ }\left( \ln (\epsilon_1) -\ln (\epsilon_2) \right) = \lim_{\epsilon_1,\epsilon_2\rightarrow 0^+ }\ln \left( \frac{\epsilon_1}{\epsilon_2} \right)[/tex]

    exist independent of the manner in which [tex] \epsilon_1,\epsilon_2\rightarrow 0^+ [/tex]. Clearly this is not the case, since, for example, were we to let [tex] \epsilon_1\rightarrow 0^+ [/tex] first, then the value of the limit is [tex]-\infty,[/tex] but if instead we let [tex] \epsilon_2\rightarrow 0^+ [/tex] first, then the value of the limit is [tex]+\infty,[/tex] and if, rather, we put [tex] \epsilon_1=\epsilon_2=\epsilon[/tex], and then let [tex]\epsilon\rightarrow 0^+ [/tex] then the limit is 0 -- this is the case you have built up in your mind, this particular value of the limit is called the Cauchy principal value (P.V. for short). If you evaluate an integral along the real axis by using contour integrals and residue theorem, the value of the integral thereby obtained is the Cauchy PV.
    Last edited: Feb 26, 2006
  5. Feb 26, 2006 #4

    The Integral exists only if limit of the Reimann sums exists under *any* partition You are right, you can form a sequence of partitions such that the corresponding Reimann sums are all zero, but you can also partition [-1,1] in such a way that they are not, and in fact, do not converge to a finite value at all. As arildno said, this is very simmilar to conditionally convergent series.
  6. Feb 26, 2006 #5


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    As for the computer result, any numerical approximation will use some specific partition scheme.
    Although many cases exist where a non-uniform partition would yield a good approximation earlier than a uniform partition scheme, I would think that the default option for any integration routine uses a uniform partition scheme. That would explain why you get 0.0 as the answer on your computer.
  7. Feb 26, 2006 #6


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    To cover this example in your class is an excellent idea, in my opinion.
    Note that the requirement of partition independence is actually an element of a precise formulation of our intuitive notion concerning areas:
    However we choose to subdivide a particular region into simpler pieces, we always want the area of our region to equal the sum of the areas of the simpler pieces, however we choose to add them together.
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