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Homework Help: Property fourier transform

  1. Dec 26, 2008 #1
    1. The problem statement, all variables and given/known data
    If [tex]F(k)=TF\{f(x)\},k\neq 0[/tex] where TF is the Fourier transform ,and

    [tex]F(0)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(u)du\neq 0[/tex] ,

    show that

    [tex]TF\{\int_{-\infty}^{x}f(u)du\}=-i \frac{F(k)}{k} +\pi F(0)\delta(k)[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I attempt the following:

    [tex]TF\{\int_{-\infty}^{x}f(u)du\}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \{\int_{-\infty}^{x}f(u)du\}e^{-ikx}dx[/tex]

    and , integration by parts, with

    [tex]w=\int_{-\infty}^{x}f(u)du[/tex]

    [tex]dw=f(x)dx[/tex]

    give

    [tex]\frac{1}{\sqrt{2\pi}}\{\int_{-\infty}^{x}f(u)du \frac{i}{k} e^{-ikx}\}|_{-\infty}^{\infty} -\frac{1}{\sqrt{2\pi}} \frac{i}{k}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx[/tex]

    The last term is

    [tex] -\frac{i}{k}F(k)[/tex]

    What should i do with the first?. I do this, but it probably be bad:

    [tex]\frac{i}{k} e^{-ikx}\}|_{-\infty}^{\infty}=\int_{-\infty}^{\infty}e^{-ikx}dx = 2\pi\delta(k)[/tex]

    and

    [tex]\frac{1}{\sqrt{2\pi}}\left[\lim_{x\rightarrow\infty}\int_{-\infty}^xf(z)dz-\lim_{x\rightarrow -\infty}\int_{-\infty}^xf(z)dz\right] = F(0) [/tex]

    Then,

    [tex]
    \int_{-\infty}^{\infty}\frac{dx}{\sqrt{2\pi}}\int_{-\infty}^xf(z)dze^{-ikx}=-i\frac{F(k)}{k}+2\pi F(0)\delta(k)
    [/tex]

    where is an extra 2.

    I tried to do this problem using convolution, but again i cant achieve the desired result:

    [tex](f*g)(x)=\int_{-\infty}^{\infty}f(u)g(x-u)du[/tex]

    If [tex]g(x-u)[/tex] is the Heaviside function, and with [tex]x-u>0[/tex] , the convolution is

    [tex](f*g)(x)=\int_{-\infty}^{\infty}f(u)g(x-u)du=\int_{-\infty}^{x}f(u)du[/tex]

    Then,

    [tex]TF\{\int_{-\infty}^{x}f(u)du\}=TF\{(f*g)(x)\}=\sqrt{2\pi}F(k)G(k)[/tex]

    But, the fourier transform of the Heaviside function is:

    [tex]TF\{g(x)\}=\frac{-i}{k\sqrt{2\pi}}+\sqrt{\frac{\pi}{2}}\delta(k)[/tex]

    Hence,


    [tex]TF\{\int_{-\infty}^{x}f(u)du\}=\frac{-iF(k)}{k}+\pi F(k)\delta(k)[/tex]


    But, in the last term appears [tex]F(k)[/tex] , and i want to get [tex]F(0)[/tex]. What is wrong??


    Thanks
     
  2. jcsd
  3. Dec 26, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    There's nothing wrong with your convolution argument. Think what happens if you integrate a test function against F(k)*delta(k). You get the same thing as if you integrate it against F(0)*delta(k) if F is continuous.F(k)*delta(k) and F(0)*delta(k) are the same distribution.
     
  4. Dec 27, 2008 #3
    Ok,i understand it now, thanks for the help :smile:
     
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