Re: Fourier transform of exponential function
- To: mathgroup at smc.vnet.net
- Subject: [mg108677] Re: Fourier transform of exponential function
- From: ALittleDog <leqia.he at gmail.com>
- Date: Sat, 27 Mar 2010 05:09:36 -0500 (EST)
- References: <hofg7f$f89$1@smc.vnet.net> <hofhko$g30$1@smc.vnet.net>
I got the following answers from Daniels: " A constant does not belong to L2(-Infinity,Infinity). However one can generalize the notion of function to define e.g. Fourier transforms of a constant. The generalized function only make sense inside an integral. The Fourier integral of const is zero with the expcetion of \omega==0. This is different from Exp[nt] that is unbounded. " I guess Daniels' answer also applies to Cos[t], which is also not integral summable and thus does not belong to L2. However, the Cos[t] function is square-integrable of [ - pi,pi]. And a constant can also be interpreted in this way to be square integrable of [-pi, pi]. In contrast to that, Exp[-t] is also square integrable of [-pi,pi]. Then, what makes the difference between a constant or Cos[t] and Exp[t], being not square integrable properties L2(- Infinity,Infinity) ? Is it due to the periodicity of Cos[t] and a constant?