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Improper Integral with Infinity in Limits

  1. Oct 11, 2012 #1

    ∫xe^[-x^2] dx
    -∞

    So basically I've solved for everything in this problem and it looks like it should be an indeterminate form and thus divergent. My book and Wolfram both say it's 0 and convergent though.

    I get it down into:
    lim [[e^(-t^2)] - e^0]/2 + lim [e^0 - [e^(-v^2)]]/2
    t->-∞__________________v->∞

    When I plug stuff in I get:

    [e^∞ - e^∞ - e^0 + e^0]/2

    I can see why it might be 0 from the stuff above, but e^∞ - e^∞ should be indeterminate rather than 0. Can someone please explain what I'm not getting?


    Wolfram:
    http://www.wolframalpha.com/input/?...=DefiniteIntegralCalculator.rangeend_infinity
     
  2. jcsd
  3. Oct 11, 2012 #2

    jbunniii

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    The integrand is an odd function, so if you can show that [itex]\int_{0}^{\infty} x e^{-x^2} dx[/itex] is finite, then the integral from [itex]-\infty[/itex] to [itex]\infty[/itex] will be zero. Seems like a simple substitution will do the trick.
     
  4. Oct 11, 2012 #3
    I'm afraid I don't understand what you're saying.
     
    Last edited: Oct 11, 2012
  5. Oct 11, 2012 #4
    Well you know that sin(x) is an odd function right? An odd function means that f(-x)=-f(x) and the function just flips over when it crosses the origin like sin(x) does. So that if f(-x)=-f(x) then surely:

    [tex]\int_{-1}^{1} f(x)dx=0[/tex]

    if f(x) is odd. Same diff for any symmetric interval like -2 to 2, -100 to 100 even -infty to infty.

    So what about the function [itex]x e^{-x^2}[/itex]? Is that one f(-x)=-f(x)? Then it would be an odd function then so that:

    [tex]\int_{-a}^{a} f(x)dx=0[/tex]

    And would be likewise zero if a were infinity and the integral is finite in the interval

    [tex]\int_0^{\infty} f(x)dx[/tex]

    And I think you can compute:

    [tex]\int_0^{\infty}x e^{-x^2} dx[/tex]

    right?
     
  6. Oct 11, 2012 #5

    HallsofIvy

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    You are trying to evaluate "[itex]e^{\infty}[/itex]" when you should be looking at "[itex]e^{-\infty}[/itex].
     
    Last edited: Oct 12, 2012
  7. Oct 12, 2012 #6
    I see what you're talking about. Thanks. I messed up on my math anyway : D
     
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