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Relationship between Fourier and Mellin transforms

  1. Aug 16, 2011 #1

    it is possible to prove that the Mellin transform of a function f(x) can be expressed in terms of Fourier transform, namely:

    [tex]\mathcal{M}\{f(x)\}(s) = \mathcal{F}\{f(e^{-x}\}(-is)[/tex]

    I am not convinced of that imaginary unit i as argument of the Fourier transform. In fact, since the argument (-is) is imaginary, that is not a Fourier transform anymore.

    I don't see I could compute a Mellin transform using a Fourier transform. Am I missing something?
  2. jcsd
  3. Aug 16, 2011 #2
    If you let:

    [tex]F(f)=\int_{-\infty}^{\infty} e^{-i x z} f(x)dx[/tex]

    [tex]M(f)=\int_0^{\infty} x^{s-1}f(x)dx[/tex]

    can you not let v=-ln(x) in the Mellin integral and come out with:

    [tex]\int_{-\infty}^{\infty} e^{-izv}f(e^{-v})dv\biggr|_{z=-is}=F(f(e^{-v})\biggr|_{z=-is}[/tex]
  4. Aug 17, 2011 #3
    But you plugged -is as argument of the Fourier transform.
    I thought the fourier transform was defined only for real arguments.

    In fact, if the argument is real the Fourier bases are complex sinusoids e-isx=cos(x)-isin(x), while if you use imaginary arguments you get real exponential functions esx: sin and cos disappear, the bases are not orthogonal anymore, Fourier integrals might not converge, and some theorems of Fourier theory might not be valid anymore :-|
    Last edited: Aug 17, 2011
  5. Aug 17, 2011 #4
    The arguments for Fourier and Laplace transforms are in general, complex. Also, this is what I did in Mathematica to check this. Notice in th second calculation, you need to substitute s=-iz to obtain the Mellin result

    Code (Text):
    f[x_] := Exp[-x^2]

    Integrate[x^(s - 1)*f[x], {x, 0, Infinity}]

      {x, -Infinity, Infinity}]

    ConditionalExpression[(1/2)*Gamma[s/2], Re[s] > 0]

      Im[s] < 0]
    Last edited: Aug 17, 2011
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