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.