Re: FourierTransform and removable singularities

*To*: mathgroup at smc.vnet.net*Subject*: [mg75344] Re: FourierTransform and removable singularities*From*: dimitris <dimmechan at yahoo.com>*Date*: Wed, 25 Apr 2007 05:46:47 -0400 (EDT)*References*: <f0kbmk$qvt$1@smc.vnet.net>

Hi. The only thing that comes in my mind is to replace the Dirac delta generalized function with a delta secuence. So for example, let In[64]:= delta[n_][x_] := n/(Sqrt[Pi]/E^((-n^2)*x^2)) Then In[66]:= FourierTransform[delta[n][t], t, w] Limit[%, n -> Infinity] Out[66]= n/(E^(w^2/(4*n^2))*(Sqrt[n^2]*Sqrt[2*Pi])) Out[67]= 1/Sqrt[2*Pi] which is correct. Also as you stated In[82]:= ({(1/Sqrt[2*Pi])*Integrate[Exp[I*w*t]*#1, {t, -Infinity, Infinity}], FourierTransform[#1, t, w]} & )[DiracDelta[t]] Out[82]= {1/Sqrt[2*Pi], 1/Sqrt[2*Pi]} Now, the example with the removable singularity In[83]:= FourierTransform[delta[n][t]*(Sin[t]/t), t, w] Limit[%, n -> Infinity] Out[83]= (n*((-(1/2))*Sqrt[Pi/2]*Erf[(-1 + w)/(2*Sqrt[n^2])] + (1/2)*Sqrt[Pi/ 2]*Erf[(1 + w)/(2*Sqrt[n^2])]))/Sqrt[Pi] Out[84]= 1/Sqrt[2*Pi] which is the desired (correct) result. On the contrary In[85]:= ({(1/Sqrt[2*Pi])*Integrate[Exp[I*w*t]*#1, {t, -Infinity, Infinity}], FourierTransform[#1, t, w]} & )[(DiracDelta[t]*Sin[t])/t] Out[85]= {0, 0} which is incorect as you have already said. I hope I helped you! Regards 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.