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From: ALittleDog on 25 Mar 2010 07:08
I am trying to do a Fourier transform of a simple exponential
function, by using the following code:
FourierTransform[Exp[-t], t, =F9]
By Mahtematica 5.0, I obtain
\sqrt(2 Pi) DiracDelta[ I + =F9 ]
By Mathematica 7.0, I obtain a repetition of the original input as
FourierTransform[Exp[-t], t, =F9].

I search to look for a classical solution, but most of them are
talking about FourierTransform[Exp[-Abs[t]], t, =F9]. Indeed, what is
the classical solution of this problem? Why 5.0 version and 7.0
version try to figure it out in a different manner? Thanks!



From: dh on 25 Mar 2010 07:21
Hi,
there is no Fourier transform of Exp[-t]. For t->-Infinity the function
is not bounded.
Daniel

On 25.03.2010 12:08, ALittleDog wrote:
> I am trying to do a Fourier transform of a simple exponential
> function, by using the following code:
> FourierTransform[Exp[-t], t, =F9]
> By Mahtematica 5.0, I obtain
> \sqrt(2 Pi) DiracDelta[ I + =F9 ]
> By Mathematica 7.0, I obtain a repetition of the original input as
> FourierTransform[Exp[-t], t, =F9].
>
> I search to look for a classical solution, but most of them are
> talking about FourierTransform[Exp[-Abs[t]], t, =F9]. Indeed, what is
> the classical solution of this problem? Why 5.0 version and 7.0
> version try to figure it out in a different manner? Thanks!
>
>
>


--

Daniel Huber
Metrohm Ltd.
Oberdorfstr. 68
CH-9100 Herisau
Tel. +41 71 353 8585, Fax +41 71 353 8907
E-Mail:<mailto:dh(a)metrohm.com>
Internet:<http://www.metrohm.com>


From: Kevin J. McCann on 25 Mar 2010 07:32
I suggest that you look at the integral you are trying to do. It is not
clear from your post whether you mean to have t run from zero to
infinity or from minus infinity to infinity. In the latter case, the
integral does not converge, and this is what corresponds to your
example. I suspect that you mean to take the FT of exp(-t) for t>0, i.e.
the function is zero for negative t. If that is so, you could try

FourierTransform[UnitStep[t] Exp[-t], t, w]

with result:

I/(Sqrt[2 \[Pi]] (I + w))

I am not sure why V5 and V7 give different results. Perhaps it is a
feature ;).

Kevin

ALittleDog wrote:
> I am trying to do a Fourier transform of a simple exponential
> function, by using the following code:
> FourierTransform[Exp[-t], t, =F9]
> By Mahtematica 5.0, I obtain
> \sqrt(2 Pi) DiracDelta[ I + =F9 ]
> By Mathematica 7.0, I obtain a repetition of the original input as
> FourierTransform[Exp[-t], t, =F9].
>
> I search to look for a classical solution, but most of them are
> talking about FourierTransform[Exp[-Abs[t]], t, =F9]. Indeed, what is
> the classical solution of this problem? Why 5.0 version and 7.0
> version try to figure it out in a different manner? Thanks!
>
>
>

From: ALittleDog on 26 Mar 2010 06:33
In fact, I want that t runs from minus infinity to infinity.
But why Mathematica 5.0 gives \sqrt(2 Pi) DiracDelta[ I + \omega ] as
the answer, when I run the code FourierTransform[Exp[-t], t, \omega],
if the integral doesn't converge? Is there a different interpretation?
Moreover, Exp[I n t] (n from minus infinity to positive infinity) are
function basis of infinite number in Hilbert space L2. Is Exp[nt] also
function basis of L2?


From: ALittleDog on 26 Mar 2010 06:35
So the function that has a Fourier transform must be square summable,
which is a condition that exp[-t] (t from -\inf to \inf) does not satisfy.
By the way, what is mathematical interpretation of \sqrt(2 Pi)
DiracDelta[ I + \omega ], that is given by Mathematica 5.0 for
FourierTransform[Exp[-t], t, \omega] ?

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