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Integral of an even function

  1. Dec 20, 2013 #1
    I am getting two different answers with two different methods so can someone point out the error?
    [itex]\int\limits_{-\infty}^{\infty} \frac{1}{x^2}dx[/itex] = 2 [itex]\int\limits_{0}^{\infty} \frac{1}{x^2}dx[/itex] = [itex] - \frac{2}{x} |_{0}^{\infty} = \infty [/itex]


    [itex]\int\limits_{-\infty}^{\infty} \frac{1}{x^2}dx = - \frac{1}{x} |_{- \infty}^{\infty} [/itex] = 0
     
  2. jcsd
  3. Dec 20, 2013 #2

    vanhees71

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    The answer is simple: The integral is undefined, because the integrand has a singularity at [itex]x=0[/itex], and you must define, what you mean by your integral. Often, one likes to calculate the socalled "Cauch principle value", which is defined as
    [tex]\text{PV} \int_{-\infty}^{\infty} \mathrm{d} x \frac{1}{x^2} = \lim_{\epsilon \rightarrow 0^+} \left [\int_{-\infty}^{-\epsilon} \mathrm{d} x \frac{1}{x^2} + \int_{\epsilon}^{\infty} \mathrm{d} x \frac{1}{x^2} \right ].[/tex]
    This means, you take out a little interval of the integration region that is symmetric around the singularity and then (i.e., after taking the integrals) make this interval arbitrarily small. Now, you can evalute this Cauchy principle value easily yourself :-).

    BTW: This is only one possibility to make sense to a previously meaningless expression. Whether the principle value is what you really want or not has to be decided by the application you have in mind when solving this integral.
     
  4. Dec 20, 2013 #3

    HallsofIvy

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    Strictly speaking neither of those is correct because an "ordinary" integral cannot be applied here.

    What you need to do is take limits as x goes to 0. But there are two different ways to do that:

    1) The way the standard integral is defined: take the limit as x goes to 0 from above and below independently:
    [tex]\lim_{\alpha\to 0^-}\int_{-\infty}^\alpha \frac{1}{x^2}dx+ \lim_{\beta\to 0^+}= \lim_{\alpha\to 0^-}\left[-\frac{1}{x}\right]_{-\infty}^{\alpha}\left[-\frac{1}{x}\right]_{\beta}^\infty[/tex]. Here, NEITHER of those limits exist. The integral does not exist.

    2) The "Cauchy Principle Value" (NOT the integral you learn in Calculus)
    [tex]\lim_{\alpha\to 0}\int_{-\infty}^{-\alpha}\frac{1}{x^2}dx+ \int_{\alpha}^\infty \frac{1}{x^2}dx[/tex]
    Here the two, with the same "[itex]\alpha[/itex]" cancel before the limit so the limit is 0.

    Any time the integral itself exists, the "Cauchy Principle Value" is equal to the limit. When the integral does NOT exist (I would not say "equals infinity" but "does not exist"), the Cauchy Principal value might exist. But it is still NOT the standard integral.
     
    Last edited: Dec 20, 2013
  5. Dec 20, 2013 #4

    Ray Vickson

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    No, the limit does not exist: the sum equals ##2/\alpha##, which → ∞ as ##\alpha \to 0+##. Basically, we are adding up the areas under y = 1/x^2 to the left of ##x = -\alpha## and the right of ##x = \alpha##, and these two areas are positive and equal.
     
  6. Dec 20, 2013 #5
    Thanks for all the replies...I think I understand now. I had thought it had something to do with the singularity at 0 but was unsure of how to deal with it. As I understand it, in this case even the Principle value diverges does it not? I often hear mathematicians say that an integral does not exist as opposed to it being infinite or it diverges, how do you distinguish between these?
     
  7. Dec 20, 2013 #6
    Also is there some criteria for an integral not to exist if the integrand is is infinite at some point? For example a delta function has a formal infinity in it, but the integral over a delta function is still well defined. Also what is the criteria for a point to be called a singularity? I have done some complex analysis and I know simple poles, poles of higher order and essential singularities. I know that you can distinguish them from their corresponding Laurent expansions. But I have also encountered singularities that are finite valued which was not so clear to me. For example the series

    [itex]\sum\limits_{n=1}^{\infty} \dfrac{z^n}{n^2}[/itex]

    converges on the unit circle in the complex plane. It is finite on all points on the circle but is said to be singular at z=1. I don't know why this is singular or what kind of singularity this is...
     
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