1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

1/x^2 integral

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

    2. Relevant equations
    power rule


    3. The attempt at a solution
    1/x^2
    x^-2
    x^-1
    (-1/x)

    plugging in values:
    (-1/1) - (-1/-1)
    -1 - 1
    -2

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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
    INFINITY - 1 = INFINITY

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

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: 1/x^2 integral
Loading...