Re: FourierTransform and removable singularities

*To*: mathgroup at smc.vnet.net*Subject*: [mg75345] Re: FourierTransform and removable singularities*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Thu, 26 Apr 2007 03:27:58 -0400 (EDT)*References*: <f0kbmk$qvt$1@smc.vnet.net>

Roman <rschmied at gmail.com> wrote: > It seems to me that Mathematica 5.2 is not careful enough when doing > Fourier transforms of functions with delta functions at removable > singularities: if you call > > FourierTransform[DiracDelta[t], t, w] > > you get the right answer, > > 1/Sqrt[2*Pi] > > But if you call something of the sort of > > FourierTransform[DiracDelta[t]*(Sin[t]/t), t, w] > > which has a removable singularity at the point where the Dirac delta > function acts, the answer is zero, which is wrong. > > Does anyone know how to resolve this by reformulating the problem? (a > workaround) I've got two suggestions, neither of which I like. 1) This is similar to what Jens-Peer suggested. In[62]:= Limit[FourierTransform[DiracDelta[t] Sin[t - c]/(t - c), t, w], c -> 0] Out[62]= 1/Sqrt[2*Pi] 2) In[63]:= FourierTransform[DiracDelta[t](Sin[t] - t)/t + DiracDelta[t], t, w] Out[63]= 1/Sqrt[2*Pi] Note that, _formally_, the argument of the transform is indeed your DiracDelta[t]*(Sin[t]/t), as the following (using diracDelta, rather than DiracDelta) shows. In[64]:= Simplify[diracDelta[t] (Sin[t] - t)/t + diracDelta[t]] Out[64]= (diracDelta[t]*Sin[t])/t Perhaps the best way to handle you problem would be to have the sine cardinal function | { 1 if x = 0, | sinc(x) = { | { sin(x)/x otherwise implemented in Mathematica. But defining that function yourself, it does not work as desired with FourierTransform. David W. Cantrell