       Re: Fourier transform : failure with hyperbolic secant and log argument

• To: mathgroup at smc.vnet.net
• Subject: [mg79977] Re: Fourier transform : failure with hyperbolic secant and log argument
• From: dimitris <dimmechan at yahoo.com>
• Date: Fri, 10 Aug 2007 01:41:51 -0400 (EDT)
• References: <f9em4r\$kg2\$1@smc.vnet.net>

```On 9    , 12:18, lagaffe <spam... at neuf.fr> wrote:
> Hi all,
>
> Does someone know why Mathematica succeeds in findind the fourier transform of Sech[x*a] but not of Sech[x*Log], and why does it return a result in terms of gamma functions with Sech[x*Log[2.0]] ??
>
> I know this is not really an issue since I can replace Log by the symbol a, but it worries me about mathematica's reliability.
>
> Thanks.

Trust results from Mathematica (or other CAS) but verify them!
Mathematica is a reliable CAS but it is not perfect.

And as a second advice do not mix symbolic built in functions with
inexact arguments.
It is like you push Mathematica to do the things the wrong way!

BTW...

Here is something funnier!

In:=
TrigToExp[Sech[x*a]]
FourierTransform[%, x, s]

Out=
2/(E^((-a)*x) + E^(a*x))

Out=
(Sqrt[Pi/2]*Sech[(Pi*s)/(2*a)])/a

In:=
TrigToExp[Sech[x*Log]]
FourierTransform[%, x, s]

Out=
2/(2^(-x) + 2^x)

Out=
FourierTransform[2/(2^(-x) + 2^x), x, s]

Dimitris

```

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