Re: Unexpected non-evaluation problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg63151] Re: Unexpected non-evaluation problem*From*: dh <dh at metrohm.ch>*Date*: Thu, 15 Dec 2005 07:14:11 -0500 (EST)*References*: <dnrch3$kie$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hallo Carl, go back to your analytic course notes and convince yourself that in a strict sense the FourierTransform and its reverse of x^2 do not exist. Only if the absolute value of a function is integrable will the FourierTransfrom and its reverse exist. Now Mathematica goes beyond this into the real of generalized functions, that is distributions. There a FourierTransform does exist, but not its reverse. I do not know enough about distributions to explain it properly, but here is an heuristic explanation for the FourierTransfrom of x^2: The FourierTransform is defined as: FT[x^2]¨= Integrate[ x^2 Exp[-I x y],{x,-Infinity,Infinity}] x^2 is even, therefore instead of Exp[-I x y] we may use Cos[x y]: FT[x^2]= Integrate[ x^2 Cos[ x y],{x,-Infinity,Infinity}] Clearly this integral does not exist. However, we may calculate the principal value where we take the limit: Limit[ FT[x^2]= Integrate[ x^2 Cos[ x y],{x,-x0,x0}] ,{x0->Infinity}] The indefinite integral gives: (2*x*Cos[x*y])/y^2 + ((-2 + x^2*y^2)*Sin[x*y])/y^3 convince yourself that both terms are antisymetric in x, therefore, the definite integral is zero for all y with the exception of y=0 where we get infinity (something like 1/0^3) have fun, Daniel Carl Cotner wrote: > I am totally baffled by the following Mathematica behavior: > > In[1]:= InverseFourierTransform[FourierTransform[x^2, x, y], y, x] > Out[1]= Sqrt[2 Pi] x^2 FourierTransform[DiracDelta[y], y, x] > > In[2]:= Sqrt[2 Pi] x^2 FourierTransform[DiracDelta[y], y, x] > Out[2]= x^2 > > I've already asked my local guru without success, so I'm hoping someone > here can help me. Does anyone know why the first expression doesn't > evaluate to x^2 all by itself? How I can force it to do so? > > Carl >