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, Mathematica happily gives the answer: -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 > >