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Re: FourierTransform

  • To: mathgroup at
  • Subject: [mg57623] Re: FourierTransform
  • From: dh <dh at>
  • Date: Thu, 2 Jun 2005 05:17:02 -0400 (EDT)
  • References: <d7k36h$oi1$>
  • Sender: owner-wri-mathgroup at

$Version --> "5.1 for Microsoft Windows (October 25, 2004)"

Hi Marcin,
The Fouriertransfrom of a function exists if the integral of its 
absolute value exists (plus some additional requirements). Therefore, 
the Fouriertransform of the Unitstep does not exist as a function 
(however, it is possible to give it some meaning in the context of 
distributions that only make sense inside an integral).  Therefore,
what is
or more accurate:
Integrate[Exp[-I*w*t], {t, 0, Infinity}, Assumptions ->Element[w,Reals]]
should fail to converge because the integrand is a unit vector whose 
phase changes with constant velocity (with the exception w=0). However, 
Mathematica happily gives the answer:
what is an ordinary function and wrong.

I will notify Wolfram and hope that Daniel Lichtblau will not be angry 
if I bother him again.
Sincerely, Daniel

Marcin Rak wrote:
> Hi,
> I was wondering what the exact mathematica equation of the following 
> Mathematica Command was:
> FourierTransform[UnitStep[t],t,w,FourierParameters->{1,-1}]
> This gives -i/w + pie*DiracDelta[w]
> which is correct.  However, when I substitute my UnitStep[t] function 
> into the direct definition employed by FourierTransform, I don't get the 
> same result.  ie
> Integrate[UnitStep[t]*Exp[-i*w*t],{t,-infinity,infinity}] Doesn't give 
> the same result, despite the fact that in section 3.5.11 of the 
> Mathematica book it is defined as such?
> Thanks
> MR

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