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Hellp with fourier integral via completing the square

  1. Sep 28, 2003 #1
    This comes from quantum mechanics but it's basically a Fourier integral I can't quite do...

    F(k) = 1/sqrt(2a*[pi]) * [inte] exp( -(ax^2+ikx) dx over infinite limits. i is sqrt(-1)

    to do this, I complete the square getting

    exp( -(sqrt(a)*(x +ik/(2a))^2 * exp(k^2 / (4a))

    sticking this in the integral and integrating over x I get

    F(k) = 1/sqrt(2a*[pi]) * exp(k^2 / 4a )

    I like the k^2 / a part but the factor of 4 seems wrong as well as the sign of the exponent possibly.

    exp(-a x^2) should transform to exp(k^2/a)?

    Help with using completing the square to do this integral would be greatly appreciated.

    I used [inte] exp(-y^2) = sqrt([pi]) from a table.
     
  2. jcsd
  3. Sep 28, 2003 #2

    Hurkyl

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    Your completion of the square looks wrong. Try completing this square (and testing your answer) as a seperate problem... working subproblems seperately tend to make it easier to see what's wrong.

    (Incidentally, Mathematica agrees with the factor of 4)

    Oh, I also presume that a is supposed to be positive? (It makes a big difference)
     
  4. Sep 28, 2003 #3
    Ok,

    x^2 + ikx/a = (x + ik/2a)^2 + k^2/4a^2

    so my exponential becomes

    exp(-a * (k/2a)^2) * exp(-a * (x + ik/2a))

    the integral then is exp(-a * (k/2a)^2) * sqrt([pi]/a)

    (i'm ignoring the constant in front of the integral for now)

    and a is positive constant

    the final answer becomes 1/sqrt(2*[pi]) * sqrt([pi]/a) * exp(-k^2/4a)

    I like that better, exp(k^2) is not absolutely integralable.
     
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