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Fourier Transform of this function

  1. Nov 6, 2013 #1
    Hi guys, I'm not sure how they got from first step to the second. Did they use integration by parts? I tried but I didn't arrive at the same result..

  2. jcsd
  3. Nov 6, 2013 #2
    It appears the power series of [itex]e^{-ikx}[/itex] was used.
  4. Nov 7, 2013 #3
    I've thought about that, but the integral is the product of two exponentials, not produt of two integrals:

    ∫eA.eB dx = ∫ e(A+B) dx

    ∫eA dx ∫eB dx = [Ʃ(1/r!)Ar] ∫eB dx

    I'm not sure why they simply moved the e-ikx out of the integral.
  5. Nov 7, 2013 #4
    It's just the Taylor expansion of the exponential. BTW that's not the easiest way to solve that integral. Try completing the square.
  6. Nov 7, 2013 #5
    You can't move it outside because of its dependence on x.
  7. Nov 10, 2013 #6
    My point exactly. So what magic did they perform at that step then?
  8. Nov 10, 2013 #7
    They performed no magic :) Actually, that looks like a rather clumsy approach to solve that integral.

    They did not bring exp(-ikx) outside the integral. They simply replaced exp(-ikx) with its Taylor expansion. What you want to bring outside the integral are just the Taylor coefficients of the xn terms.

    By the way, as you have been suggested by dauto, try instead to complete the square inside the brackets and you'll solve that integral in no time (no Taylor expansion involved). Interpret the terms inside the brackets as:


    [itex]2AB= ikx [/itex]

    [itex]B^2= \, ? [/itex]

    Try to determine [itex]B^2[/itex] and then write your integrand as e-(A2 + 2AB + B2-B2) which is equal to e-(A+B)2 eB2

    By the way, this thread would have fitted better in Mathematics/Calculus since your question is essentially about the evaluation of an integral.
    Last edited: Nov 10, 2013
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