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Fourier transform of a constant

  1. Feb 24, 2007 #1
    1. The problem statement, all variables and given/known data

    Fourier transform of a constant

    2. Relevant equations

    3. The attempt at a solution

    I am trying to prove that Fourier Transform of a constant is a Dirac delta function.I have fed f(x)=1 in the formula of Forward Fourier transform and got F(k)=int{exp[-ik*pi*x]}dx
    I know that this is delta function with arguement k.But cannot prove it.It's sure to give infinity at k=0.But,for other values of k I am not getting the integral=0.So,an attempt through the way of definition fails.Can anyone help me?
  2. jcsd
  3. Feb 24, 2007 #2

    Meir Achuz

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    Start with \delta(x) and Fourier transform it. You get a constant.
  4. Feb 25, 2007 #3
    Here is a method how to do it.It took some time,however.

    First, you find the F.T. of delta(x) and it turns out as a constant F(k)=1
    Then,you did inverse transform of 1 which yields a delta function.
    Like this: int{exp[ik'x']dk'}=2*pi*delta(x')
    Putting k'=x and x'=-k,we have

    Now consider the F.T. of a constant function g(x)=c
    G(k)=int{c exp[-ikx]dx}=c[2*pi*delta(-k)]=2*pi*c delta(k)
  5. Oct 17, 2011 #4
    If you compute Int( exp(-ikx), x=-Infinity..Infinity) you get Infinity for k=0 and 2/k sin(k Infinity) that is undefined for k=/=0. To remove this ambiguity use the Residue theorem giving Int( exp(-ikz), z on C ) = 0 where C is the contour of a half-disk of radius R. Then using the parametrisation z = R exp (i alpha) and letting R->Infinity you will find Int( exp(-ikx), x=-Infinity..Infinity) = 0 for k =/=0. Using this reasoning you get that Int( exp(-ikx), x=-Infinity..Infinity) is "proportional" to delta(k).

    To understand the factor 2*pi you have to work in the context of the definitions of the Fourier transform and of his inverse. These definitions results from the Fourier serie as T->Infinity, see http://en.wikipedia.org/wiki/Fourier_Transform.
    Last edited: Oct 17, 2011
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