Relationship between Fourier and Mellin transforms

[mnb96]

Hello,

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

$$\mathcal{M}\{f(x)\}(s) = \mathcal{F}\{f(e^{-x}\}(-is)$$

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?

 

[jackmell]

If you let:

$$F(f)=\int_{-\infty}^{\infty} e^{-i x z} f(x)dx$$

$$M(f)=\int_0^{\infty} x^{s-1}f(x)dx$$

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

$$\int_{-\infty}^{\infty} e^{-izv}f(e^{-v})dv\biggr|_{z=-is}=F(f(e^{-v})\biggr|_{z=-is}$$

 

[mnb96]

Yes.
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 e^sx: 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 :-|

 

[jackmell]

Yes.
But you plugged -is as argument of the Fourier transform.
I thought the Fourier transform was defined only for real arguments.
|

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

f[x_] := Exp[-x^2]

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

Integrate[Exp[(-I)*s*x]*f[Exp[-x]], 
  {x, -Infinity, Infinity}]

Out[6]=
ConditionalExpression[(1/2)*Gamma[s/2], Re[s] > 0]

Out[7]=
ConditionalExpression[(1/2)*Gamma[(I*s)/2], 
  Im[s] < 0]