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>*Reply-to*: kuska at informatik.uni-leipzig.de

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[1]:= FourierTransform[t^2,t,w] > > Out[1]= -(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. >