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I Dirac Delta using Fourier Transformation

  1. Dec 25, 2016 #1
    We know,
    $$\delta(x) = \begin{cases}
    \infty & \text{if } x = 0 \\
    0 & \text{if } x \neq 0
    \end{cases}$$
    And, also,
    $$\int_{-\infty}^{\infty}\delta(x)\,dx=1$$
    Using Fourier Transformation, it can be shown that,
    $$\delta(x)=\lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega x)}}{\pi x}$$
    Let's look at the graph of ##y=\frac{\sin{(\Omega x)}}{\pi x}##, for large ##\Omega##.

    Untitled.png

    So, from this graph, it seems,
    $$\lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega x)}}{\pi x}\neq 0 \text{ , for } x\neq 0 $$

    But, this is actually the fourier transformation of the function,
    $$f(t)=1=\exp{(i\cdot 0\cdot t)}$$

    So, intution tells me that the frequency spectrum of ##f(t)## should have infinite amplitude at ##x=0## and ##0## amplitude otherwise, just like the Dirac-delta function.

    Now, how can I show that,
    $$\lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega x)}}{\pi x} = 0 \text{ when, } x\neq 0$$
     
    Last edited: Dec 25, 2016
  2. jcsd
  3. Dec 25, 2016 #2

    Delta²

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    Gold Member

    The rigorous definition of dirac delta function is not the one you write at the beginning of your post. Even if we accept that definition (because intuitively that's what the dirac delta is) then I don't know how you come up with ##\delta(x)=\lim_{\Omega \rightarrow \infty} \frac{sin(\Omega x)}{\pi x}## cause that limit simply does NOT exist. (it does not exist for ##x\neq 0##, if we first take the limit x->0 then it exists and is infinite).
     
    Last edited: Dec 25, 2016
  4. Dec 25, 2016 #3
    Using fourier transformation, we have,
    AS.PNG

    Comparing with the equation,
    $$f(t)=\int_{-\infty}^{\infty}\delta(t-u)f(u)\,du$$
    we have,
    ASS.PNG

    Thus,
    $$\begin{align}
    \delta(t)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega t}\,d\omega\\
    &=\frac{1}{2\pi}\lim_{\Omega\rightarrow \infty}\left[\frac{e^{i\omega t}}{i\omega t}\right]^{\Omega}_{\omega=-\Omega}\\
    &=\lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega t)}}{\pi t}
    \end{align}$$

    Reference:
    Mathematical Methods for Physics and Engineering: A Comprehensive Guide, 3rd Edition by K. F. Riley, M. P. Hobson, S. J. Bence, Page/442
     
  5. Dec 25, 2016 #4

    Delta²

    User Avatar
    Gold Member

    The last two equalities would hold only if the limit ##\frac{1}{2\pi}\lim_{\Omega\rightarrow \infty}\left[\frac{e^{i\omega t}}{it}\right]^{\Omega}_{\omega=-\Omega}=
    \lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega t)}}{\pi t}

    ## existed but it does not exist, as you can easily prove for any x<>0 the limit ##\lim_{\Omega \rightarrow \infty}\sin{(\Omega x)}## does not exist.
     
    Last edited: Dec 25, 2016
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