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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,
> 
> 
>