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Re: FourierTransform and removable singularities (2nd workaround!)
As a second workaround, try In:= FourierTransform[DiracDelta[t]*f[t], t, w] Out= f/Sqrt[2*Pi] Then, In:= % /. f -> 1 Out= 1/Sqrt[2*Pi] since In:= Normal[Sin[t]/t + O[t]] Out= 1 Cheers Dimitris =CF/=C7 Roman =DD=E3=F1=E1=F8=E5: > 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) > > Cheers! > Roman.