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MathGroup Archive 1998

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Re: Fourier Transform

  • To: mathgroup at smc.vnet.net
  • Subject: [mg13774] Re: Fourier Transform
  • From: David Withoff <withoff>
  • Date: Mon, 24 Aug 1998 05:07:23 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

> Let f(x) = 1/x. If f is regarded as the generalized function, then its
> Fourier transform is:
>                                   -Pi*I*Sign[t] (see, e.g., G.B.
> Folland, "Fourier Analysis and Its Applications," p. 337).
> Using Mathematica 3.0 we get:
> In[1]:=
> << "Calculus`FourierTransform`"
>
> In[2]:=
> FourierTransform[1/x, x, t]
>
> Out[2]=
> 2*I*Pi*(-(1/2) + UnitStep[t, ZeroValue -> 1/2]).
>
> This agrees with the above result only if t = 0. Bug?
>
> Edward Neuman

You can use the FourierFrequencyConstant option to get the definition of
Fourier transform from the reference that you quoted.  The result is
then equivalent to -Pi*I*Sign[t].

In[20]:= FourierTransform[1/x, x, t, FourierFrequencyConstant -> -1]

                   1                              1 Out[20]= 2 I Pi
(-(-) + UnitStep[-t, ZeroValue -> -])
                   2                              2

Different authors use difference choices for these constants.

Dave



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