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Re: FourierTransform and removable singularities (2nd workaround!)


As a second workaround, try

In[100]:=
FourierTransform[DiracDelta[t]*f[t], t, w]

Out[100]=
f[0]/Sqrt[2*Pi]

Then,

In[101]:=
% /. f[0] -> 1

Out[101]=
1/Sqrt[2*Pi]

since

In[102]:=
Normal[Sin[t]/t + O[t]]

Out[102]=
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.



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