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Re: Unexpected non-evaluation problem

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

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