Re: InverseFourierTransform strange behaviour

*To*: mathgroup at smc.vnet.net*Subject*: [mg82335] Re: InverseFourierTransform strange behaviour*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>*Date*: Thu, 18 Oct 2007 04:44:53 -0400 (EDT)*Organization*: The Open University, Milton Keynes, UK*References*: <fesjid$ks8$1@smc.vnet.net><ff1qf0$9ao$1@smc.vnet.net> <ff4hrl$k1c$1@smc.vnet.net>

roby.nowak at gmail.com wrote: > Hi Frank, > > sorry but you are not correct. > > 2 Sqrt[2/Pi] Exp[(w/2)^2] Sin[w]/w > is pefectly even (as it is without the negative sign as you suggested) > > > Exp[(w/2)^2] is even > Sin[w] is odd > 1/w is odd > Sin[w]/w is even > > a simple numerical test too shows that the function is even ( f[x] == > f[-x] ) And one can even (no pun intended) use Mathematica's symbolic capabilities to check that f(x) is even, indeed. In[1]:= f[x_] := 2 Sqrt[2/Pi] Exp[(w/2)^2] Sin[w]/w In[2]:= f[x] == f[-x] Out[2]= True Just my two cents, Jean-Marc > @everybody > > I am in doubt now if the IFT exists for my function. any suggestions > to this ? > > regards robert > > > On 16 Okt., 09:49, Frank Iannarilli <frank... at cox.net> wrote: >> On Oct 14, 4:21 am, roby.no... at gmail.com wrote: >> >>> In[1]:= 2 Sqrt[2/Pi] Exp[(w/2)^2] Sin[w]/w; >>> InverseFourierTransform[%, w, x] >> Your function is actually not even, but I believe you intended it to >> be: >> 2 Sqrt[2/Pi] Exp[-(w/2)^2] Sin[w]/w >> ^^^^ >> (note inserted negative sign) >> >> In this case, everything will work as you expected. >> >> Regards, > > >