1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

    [tex]\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}e^{ikx}e^{-ax^2}dx[/tex]

    "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

    [tex]\frac{1}{\sqrt{2\pi}}e^{\frac{-k^2}{4a}}\int^{+\infty}_{-\infty}e^{-a(x+\frac{ik}{2a})^2}dx[/tex]

    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:

    [tex]\frac{1}{\sqrt{2\pi}}e^{\frac{-k^2}{4a}}\int^{+\infty+\frac{ik}{2a}}_{-\infty+\frac{ik}{2a}}e^{ay^2}dy[/tex]

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

    And that leaves us with

    [itex]\frac{1}{\sqrt{2a}}e^{\frac{-k^2}{4a}}=g(k)[/itex]

    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

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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

    [tex]\int^{\infty}_{-\infty}e^{ikx}dx[/tex]

    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

    pasmith

    User Avatar
    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
    [/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted