Re: FourierTransform and removable singularities

*To*: mathgroup at smc.vnet.net*Subject*: [mg75393] Re: FourierTransform and removable singularities*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Sat, 28 Apr 2007 05:54:51 -0400 (EDT)*References*: <f0kbmk$qvt$1@smc.vnet.net> <f0pk9v$1t0$1@smc.vnet.net> <f0sfan$n4v$1@smc.vnet.net>

Peter Pein <petsie at dordos.net> wrote: > David W.Cantrell schrieb: > ... > > > 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. > > Hi David, > > sorry, I did not believe this. Thanks, Peter! Although rather embarrassed, I'm _glad_ you didn't believe it! > And indeed: > > In[1]:= Off[General::spell]; > Sinc[t_] := Piecewise[{{1, t == 0}}, Sin[t]/t] > > In[3]:= FourierTransform[DiracDelta[t]*Sinc[t], t, w] > Out[3]= 1/Sqrt[2*Pi] Yes, that's the desired result. <blush> Despite my having defined sinc above in a piecewise manner, when I defined it _in Mathematica_, I did not do it in that manner. Rather, I had used Sinc[t_] := Limit[Sin[u]/u, u -> t] and then FourierTransform[DiracDelta[t]*Sinc[t], t, w] had given 0. Regards, David