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Integrating the function 1/x^2. Something I don't understand!

  1. Aug 20, 2005 #1

    VietDao29

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    While integrating the function [tex]f(x) = \frac{1}{x ^ 2}[/tex], I came across something I don't understand:
    [tex]\int \frac{1}{x ^ 2}dx = - \frac{1}{x} + C[/tex]
    Let [tex]f(x) := \frac{1}{x ^ 2}[/tex]
    [tex]f(x) > 0, \forall x \in \mathbb{R}[/tex]
    [tex]\int_{-1}^{1}f(x)dx = -1 - (-(-1)) = -2[/tex]:confused:
    Why this happened??? :confused: It's obvious that [tex]f(x) > 0, \forall x \in \mathbb{R}[/tex] and -1 < 1, but why [tex]\int_{-1}^{1}f(x)dx < 0[/tex]
    :confused:
    I think it should be : [tex]\int_{-1}^{1}f(x)dx = + \infty[/tex]
    What have I done wrong???? :cry:
    Viet Dao,
     
  2. jcsd
  3. Aug 20, 2005 #2
    It's an improper integral.
     
    Last edited: Aug 20, 2005
  4. Aug 20, 2005 #3

    Fermat

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    There's a discontinuity in f(x) at x = 0.

    You can't integrate over a range that includes a discontituity.
     
  5. Aug 20, 2005 #4

    lurflurf

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    The fundamental theorem does not hold as 1/x^2 is not continuous at x=0. It is also an improper integral that does not converge in the traditional sense.
     
    Last edited: Aug 20, 2005
  6. Aug 20, 2005 #5

    VietDao29

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    Yup, thanks for the help. I see what I missed now. :smile:
    Viet Dao,
     
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