Re: FourierTransform of Sign^2 ?

*To*: mathgroup at smc.vnet.net*Subject*: [mg80140] Re: FourierTransform of Sign^2 ?*From*: "Dana DeLouis" <dana.del at gmail.com>*Date*: Mon, 13 Aug 2007 04:40:33 -0400 (EDT)

> Mathematica (v. 5.0, Mac) gives > -Sqrt[2 Pi]*DiracDelta[q] > with a minus sign. There was a discussion in another group recently about 5.2 with something similar. I'm not sure what the correct answer is, but I thought you might find this interesting. equ = Sech[x*a]; 5.2 takes about 1 minute to do this... (* 5.2 *) Timing[FourierTransform[equ, x, s]] {55.594, (Sqrt[Pi/2]*Sech[(Pi*s)/ (2*a)])/a} In Ver 6.0, we get a Minus sign. I thought it was ver 6.0 that has the bug, but I'm not sure of the correct answer. We do note that it is much faster in 6.0 though. (* 6.0.1 *) Timing[FourierTransform[equ, x, s]] {0.703, -((Sqrt[Pi/2]*Sech[(Pi*s)/ (2*a)])/a)} The real discussion focused on the fact that if you change 'a to a number like Log[2], then FourierTransform does not work. Ie FourierTransform[Sech[x*Log[2]], x, s] Unevaluated after a long long time. -- Dana Windows XP & Mathematica 5.2 "Jung-Tsung Shen" <jushen at gmail.com> wrote in message news:f9gv9e$b82$1 at smc.vnet.net... >I tried to evaluate the following Fourier Transform: > > FourierTransform[Sign[y] Sign[y], y, q] > > where Sign is the sign function. Instead of Sqrt[2 Pi]*DiracDelta[q], > Mathematica (v. 5.0, Mac) gives > > -Sqrt[2 Pi]*DiracDelta[q] > > with a minus sign. Is this a bug or I am doing something wrong here? > > Thanks. > > JT >