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Interesting integral property

  1. Apr 3, 2007 #1
    so I'm having trouble with an integral. However I noticed an interesting property to it.


    with the integral being from -infinity to positive infinity.

    The problem arose while I was trying to solve the schroedinger equation in the presence of a potential, I'm trying to solve it through the use of a fourier transform, but in order to do that I need to know the fourier transform of the potential.

    can anybody lend a hand?
  2. jcsd
  3. Apr 4, 2007 #2


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    I don't understand what you mean by "d^2 u" inside the integral.
  4. Apr 4, 2007 #3
    I found the property from integration by parts, the d^2u represents the differential of du.

    techincally the c came out of the g function when I integrated it, however I don't know whether or not it can be used to develop other properties of the integral.

    if it helps at all g is equal to e^ikx and u is an unknown function which is the product of some known potential, and an unknown function psi.
  5. Apr 4, 2007 #4


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    The differential of du is zero. Did you mean something like u''(x) dx?
  6. Apr 5, 2007 #5
    yes exactly
  7. Apr 5, 2007 #6


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    I'd be interested in knowing HOW you got that.

    You are saying
    [tex]\int_{-\infty}^\infty g(x)u'(x)dx= c \int_{-\infty}^\infty g(x) u"(x)dx[/tex]
    for any g(x) and u(x)? I doubt that is true. In particular, if you take u= x, that turns into
    [tex]\int_{-\infty}^\infty g(x)dx= 0[/tex]
  8. Apr 6, 2007 #7
    no, as I said it was a property of a specific integral I was doing, where u is an unknown function, and g is equal to e^ikx, however running it through integration by parts I was able to deduce that that property exhisted, (because my function u goes to zero at -infinity and positive infinity, which is known because of other known properies of u)

    the integral came up as I was experimenting with fourier transforms of the schroedinger equation, I'm hoping that there may be some trick that would simplify the problem.
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