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Re: Fourier Transform PDF Characteristic Function

  • To: mathgroup at
  • Subject: [mg14974] Re: [mg14955] Fourier Transform PDF Characteristic Function
  • From: BobHanlon at
  • Date: Wed, 2 Dec 1998 03:58:57 -0500
  • Sender: owner-wri-mathgroup at

In a message dated 11/27/98 8:47:13 PM, gauy at writes:

>>First, you must use Integrate vice Integral to integrate in Mathematica.



>I'm not very good in english could you explain above it bit more


A: In some of your lines you used the form

     Integral[ft Exp[I t x], {t, -Infinity, Infinity}] 

which should have been written

     Integrate[ft Exp[I t x], {t, -Infinity, Infinity}] 

Surprisingly, Mathematica does not give a spelling warning. Instead,
Mathematica just interprets Integral as a user-defined function for
which it has not been given a definition and leaves it unevaluated. If
Mathematica does not evaluate an expression which you expected to be
evaluated, then the first thing to check is the syntax of the

>I'm intriged by one of your remark about the Fourier Transform, 
>as I'm writing this down I have two papers in front of me stating that the >exponent is positive when your going from the characteristic
> function to the PDF and yet all of what you did came out correct?

A: The signs and constants used in the transform pair are a convention,
that is, they are arbitrary as long as the signs are different going in
the different directions and they are used consistently.  However,
since you were trying to use the built-in transforms from the add-on
package Calculus`FourierTransform` and compare results, then you must
use the same convention as the Mathematica package.

Different branches of science or engineering may use different
conventions and you need to make sure that you know which convention is
being used within a given context.

If you look in the on-line help for the standard package or in the
printed documentation for the add-on packages you will see the
convention which the package uses.  It uses a positive exponent for the
forward transform (to frequency domain) and a negative exponent for the
inverse transform.

> >This characteristic function is the truncated Levy distribution
and if = I >read the text correctly this time it's only known
analytically for Alpha = = 1 >=> Cauchy distribution and for Alpha = 2
=> Gaussian distribution.  >

Bob Hanlon

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