       Re: FourierTransform

• To: mathgroup at smc.vnet.net
• Subject: [mg96048] Re: FourierTransform
• From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
• Date: Tue, 3 Feb 2009 06:31:58 -0500 (EST)
• Organization: Uni Leipzig
• References: <gm1dks\$3nk\$1@smc.vnet.net> <gm3r8h\$mev\$1@smc.vnet.net> <gm6kvu\$a39\$1@smc.vnet.net>

```Hi,

this is called a distribution or generalized function
and not a function and it is only defined
inside of an integral as my Vladimirov

http://www.amazon.de/Methods-Generalized-Functions-Analytical-Special/dp/0415273560/ref=sr_1_33?ie=UTF8&s=books-intl-de&qid=1233576829&sr=8-33

say.

Regards
Jens

John Doty wrote:
> Jens-Peer Kuska wrote:
>> Hi,
>>
>> the Fourier transform over the interval x in (-Infinity,Infinity)
>> converges only for quadratic integrable functions, i.e., functions
>> where Integrate[Conjugate[f[x]]*f[x],{x,-Infinity,Infinity}]< Infinity
>>
>> This is not the case for Cosh[x], and so no Fourier transform exist.
>
> Depends on what you mean by "function". Mathematica tries in its
> pragmatic way to do what you might want here:
>
> In:= FourierTransform[t^2,t,w]
>
> Out= -(Sqrt[2 Pi] DiracDelta''[w])
>
> t^2 is certainly not square integrable, but this is the kind of useful
> result scientists and engineers want.
>
> Mathematica's support for "generalized functions" still has room for
> improvement, but it has come a long way. The bizarre problems I saw in
> the past trying Fourier methods to perform fractional differentiation
> and integration
> (http://forums.wolfram.com/mathgroup/archive/2000/Apr/msg00043.html)
> seem no longer to be with us in Mathematica 7.
>

```

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