Re: FourierTransform and removable singularities

• To: mathgroup at smc.vnet.net
• Subject: [mg75393] Re: FourierTransform and removable singularities
• From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
• Date: Sat, 28 Apr 2007 05:54:51 -0400 (EDT)
• References: <f0kbmk\$qvt\$1@smc.vnet.net> <f0pk9v\$1t0\$1@smc.vnet.net> <f0sfan\$n4v\$1@smc.vnet.net>

```Peter Pein <petsie at dordos.net> wrote:
> David W.Cantrell schrieb:
> ...
>
> > 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.
>
> Hi David,
>
> sorry, I did not believe this.

Thanks, Peter!
Although rather embarrassed, I'm _glad_ you didn't believe it!

> And indeed:
>
> In[1]:= Off[General::spell];
>         Sinc[t_] := Piecewise[{{1, t == 0}}, Sin[t]/t]
>
> In[3]:= FourierTransform[DiracDelta[t]*Sinc[t], t, w]
> Out[3]= 1/Sqrt[2*Pi]

Yes, that's the desired result.  <blush>

Despite my having defined sinc above in a piecewise manner, when I defined
it _in Mathematica_, I did not do it in that manner. Rather, I had used

Sinc[t_] := Limit[Sin[u]/u, u -> t]

and then FourierTransform[DiracDelta[t]*Sinc[t], t, w] had given 0.

Regards,
David

```

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