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Homework Help: Checking some work on a Fourier Transform

  1. Aug 18, 2014 #1
    1. The problem statement, all variables and given/known data

    OK, we're given to practice Fourier transforms. We are given

    [tex]f(x) = \int^{+\infty}_{-\infty} g(k) e^{ikx}dk[/tex]

    and told to get a Fourier transform of the following, and find g(k):

    [itex]f(x) = e^{-ax^2}[/itex] and [itex]f(x) = e^{-ax^2-bx}[/itex]

    2. Relevant equations

    3. The attempt at a solution

    For the first one I tried the following:

    The integral of [itex]e^{-ax^2}[/itex] is [itex]\sqrt{\frac{\pi}{a}}[/itex]. I saved that information for later.

    To get the Fourier transform I used:


    "completing the square" inside the integral I get [itex]e^{-a(x+\frac{ik}{2a})^2-\frac{k^2}{4a}}[/itex] and that leaves me with


    I can change the variable so that [itex]y^2=(x+\frac{ik}{2a})^2[/itex]
    and dx=dy in this case.

    After all that I can change the limits (but they are still infinity, right?) So that I get this:


    Which we know from before is [itex]\sqrt{\frac{\pi}{a}}[/itex].

    And that leaves us with


    Assuming all this is correct -- and I am not 100 percent sure it is -- then if I were to apply the same procedure to [itex]e^{ax^2-bx}[/itex] I should end up with the above multiplied by the integral of [itex]e^{-bx}[/itex]. But if I try to integrate that I get infinities.

    But I wasn't sure if that was a completely wrong approach, or if I even did the above bit correctly. The thing I have trouble getting my head around is whether e is periodic or not and what period to use when applying Fourier equations. I might be approaching this entirely wrongly.

    Thanks in advance for whatever assistance/ criticism/ telling me I am foolish.
    Last edited: Aug 18, 2014
  2. jcsd
  3. Aug 18, 2014 #2


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    If you define your Fourier transform from [itex]k[/itex] to [itex]x[/itex] space as
    [tex]f(x)=\int_{\mathbb{R}} \mathrm{d} k g(k) \exp(\mathrm{i} k x),[/tex]
    then the inverse is
    [tex]g(k)=\frac{1}{2 \pi} \int_{\mathbb{R}} \mathrm{d} x f(x) \exp(-\mathrm{i} k x).[/tex]
    Concerning the factors [itex]2 \pi[/itex] this is the opposite of what physicists (at least in the high-energy particle/nuclear physics community) use.

    In principle the way you do the integral is correct.
  4. Aug 18, 2014 #3
    thanks, that helps. One other question: say I want to integrate


    would I be better off using the Euler formula, which would get me [itex]\int^{\infty}_{-\infty}\cos x + i \sin x dx[/itex]? But then the cosine integral diverges...

    I could convert it to polar coordinates, but then I end up with

    [itex]x = r\cos \theta[/itex] and [itex]dx = -r\sin \theta + \cos \theta d\theta[/itex]

    which leads me to

    [tex]\int^{\infty}_{-\infty}e^{ikx}dx = \int^{\infty}_{0}e^{ikr\cos \theta}(-r \sin \theta d\theta) + \cos \theta dr= \int^{\infty}_{0}-re^{ikr\cos \theta}(\sin \theta) d\theta + \int^{\infty}_{0}e^{ikr\cos \theta}(\cos \theta) dr[/tex]

    where the first term would go to zero but I suspect I am approaching this the wrong way...
  5. Aug 18, 2014 #4


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    Homework Helper

    That integral diverges. There is a sense in which [tex]
    \int_{-\infty}^\infty e^{ikx}\,dx = 2\pi\delta(k)
    [/tex] where [itex]\delta[/itex] is the Dirac delta distribution, since then the inverse formula gives [tex]
    \frac{1}{2\pi}\int_{-\infty}^\infty 2\pi \delta(k)e^{-ikx}\,dk = \frac{1}{2\pi} 2\pi e^0 = 1
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