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Help with a tricky improper integration problem

  1. Aug 23, 2012 #1
    Show that:

    ∫(x^2)e^(-x^2)=(1/2)∫(e^(-x^2)

    (both integrals are from 0 to ∞)



    I don't even know if it's possible to integrate these functions, so it makes me think that what's interesting about the problem is that it only works with this integration limits.

    I know improper integration, but don't know if prooving this "numerically" is the right path.

    Thanks, I would appreciate any advice
     
  2. jcsd
  3. Aug 23, 2012 #2

    Curious3141

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    Hint:

    [tex]\int_0^{\infty} x^2e^{-x^2}dx = \int_0^{\infty} x.(xe^{-x^2})dx[/tex]

    Integrate that by parts (use [itex]u = x, dv = xe^{-x^2}dx[/itex]), then use L'Hopital's Rule to calculate the limit of one of the expressions.
     
    Last edited: Aug 23, 2012
  4. Aug 23, 2012 #3
    is this correct?

    I integrated by pats (differentiating x and integrating x^2e^{-x^2} )

    until I got to the point where (e^{-x^2})/2 (evaluated from o to infinity)=0

    Using L'Hopital's Rule, I effectively got 0=0




    Is this the way?
     
  5. Aug 23, 2012 #4

    Curious3141

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    No, I think you messed up the integration by parts. You should end up with:

    [tex]\int_0^{\infty} x^2e^{-x^2}dx = xe^{-x^2}|_0^{\infty} +\frac{1}{2}\int_0^{\infty}e^{-x^2}dx[/tex]

    and the first expression on the RHS vanishes when you take the limit by L' Hopital's.
     
  6. Aug 23, 2012 #5
    It's the same thing (maybe my english isn't that good, I don't know) but Thanks!!!
     
  7. Aug 23, 2012 #6

    Curious3141

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    How is it the same thing? You said you got "0 = 0" which means nothing. You also said you got [itex]\frac{1}{2}\int_0^{\infty}e^{-x^2}dx = 0[/itex] which is clearly wrong.

    Anyway, if you're confident you have the right solution, then it's fine.
     
  8. Aug 23, 2012 #7
    What about this one? Show:

    ∫e-x2=∫√(-lny)
    (from 0 to ∞) (from 0 to 1)



    I tried this one by solving for y in the right side of the equation, getting y=-ex2

    However, when checking the solution graphically, it's not right.

    So what the hell am I supposed to do? I also wonder if the integration limits changed when expressed in y or x.


    Thanks
     
  9. Aug 24, 2012 #8
    You should do a substitution to change the limits and get the log into play, then integrate by parts. Your previous result might help here.
     
    Last edited: Aug 24, 2012
  10. Aug 24, 2012 #9

    Curious3141

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    Use this (your previous result):

    [tex]\int_0^{\infty} x^2e^{-x^2}dx = \frac{1}{2}\int_0^{\infty}e^{-x^2}dx[/tex]

    and make the substitution [itex]e^{-x^2} = y[/itex] on the LHS. The result follows very quickly.
     
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