Re: FourierTransform

• To: mathgroup at smc.vnet.net
• Subject: [mg57623] Re: FourierTransform
• From: dh <dh at metrohm.ch>
• Date: Thu, 2 Jun 2005 05:17:02 -0400 (EDT)
• References: <d7k36h\$oi1\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```\$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,
Integrate[UnitStep[t]*Exp[-I*w*t],{t,-infinity,infinity}]
what is
Integrate[Exp[-I*w*t],{t,0,infinity}]
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,
-I/w
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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