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Homework Help: 1/x^2 integral

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data
    INTEGRAL(-1, 1): 1^x2

    2. Relevant equations
    power rule

    3. The attempt at a solution

    plugging in values:
    (-1/1) - (-1/-1)
    -1 - 1

    Now i KNOW that -2 is an incorrect answer, the solution is infinity according to maple, nut im not 100% why. I would guess its something to do with the fact that the integral involves the value x = 0 which is undefined for f(x) = -1/x as it asymptotically approaches 0.

    Could i get a nice concise, clarification. thx
  2. jcsd
  3. Nov 24, 2009 #2


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    That needs to be done as an improper integral since the original function is not defined at x=0. You can't just integrate over 0, you have to approach it as a limit. Take the integral of 1/x^2 from a to 1 and let a->0 (similarly on the negative side).
  4. Nov 24, 2009 #3
    ok so could i do it like this.

    INT (-1, 1) : 1/x^2
    INT (-1, A)(Lim A--> 0, from the left): 1/x^2
    INT(A , 1)(Lim A -->0, from the right): 1/x^2

    (-1/-A) -(-1/-1)
    as A approaches 0 from the left we get

    (-1/1) - (-1/A)
    as A approaches 0 from the right we get

    Now by sum of the 2 limits
    inf + inf = inf
    Last edited: Nov 24, 2009
  5. Nov 24, 2009 #4


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    Yes. That's the idea. Don't be too casual about doing arithmetic with 'infinity' though. E.g. infinity-infinity is not necessarily 0.
  6. Nov 24, 2009 #5


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    The "Cauchy principal value" for such an integral of a function not defined at 0 is
    [tex]\lim_{\epsilon\to 0}\left[\int_{-1}^{-\epsilon} f(x) dx+ \int_{\epsilon}^1 f(x)dx[/itex]

    The difference between that and the "regular" integral is that we do not take the limits independently. With f(x)= 1/x2, we have
    [tex]\lim_{\epsilon\to 0}\left[-x^{-1}\right]_{-1}^{-\epsilon}+ \left[-x^{-1}\right]_\epsilon^1[/tex]
    [tex]= \lim_{\epsilon\to 0} \left(\frac{1}{\epsilon}- 1\right)+ \left(1- \frac{1}{\epsilon}\right)= 0[/tex].
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