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Re: FourierTransform and removable singularities
*To*: mathgroup at smc.vnet.net
*Subject*: [mg75345] Re: FourierTransform and removable singularities
*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>
*Date*: Thu, 26 Apr 2007 03:27:58 -0400 (EDT)
*References*: <f0kbmk$qvt$1@smc.vnet.net>
Roman <rschmied at gmail.com> wrote:
> 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)
I've got two suggestions, neither of which I like.
1) This is similar to what Jens-Peer suggested.
In[62]:=
Limit[FourierTransform[DiracDelta[t] Sin[t - c]/(t - c), t, w], c -> 0]
Out[62]=
1/Sqrt[2*Pi]
2)
In[63]:=
FourierTransform[DiracDelta[t](Sin[t] - t)/t + DiracDelta[t], t, w]
Out[63]=
1/Sqrt[2*Pi]
Note that, _formally_, the argument of the transform is indeed
your DiracDelta[t]*(Sin[t]/t), as the following (using diracDelta,
rather than DiracDelta) shows.
In[64]:=
Simplify[diracDelta[t] (Sin[t] - t)/t + diracDelta[t]]
Out[64]=
(diracDelta[t]*Sin[t])/t
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.
David W. Cantrell
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