Re: fourier transform of 1/|x|
- From: David C. Ullrich <dullrich@xxxxxxxxxxx>
- Date: Thu, 02 Jul 2009 07:21:33 -0500
On Wed, 01 Jul 2009 11:51:22 -0500, Robert Israel
<israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
David C. Ullrich <dullrich@xxxxxxxxxxx> writes:
On Tue, 30 Jun 2009 22:45:38 -0700 (PDT), norbert9
<norbert9@xxxxxxxxx> wrote:
Indeed, I meant a one-dimensional FT.
Gradshteyn/Ryzhik (4th ed.) list the FT of 1/|x| as 1/|k| in 17.23.
I believe it is correct in some sense.
In what sense? The most general notion of FT I know is the
FT of a tempered distribution, and this is certainly not right
in that sense. (Unless, again, they're talking about R^2.)
Mathematica gives
FourierTransform[1/Abs[x],x,k] = (-2 EulerGamma - 2 Log[Abs[k]])/Sqrt
[2 Pi]
Do you also believe that
1/|k| = (-2 EulerGamma - 2 Log[Abs[k]])/Sqrt[2 Pi]
in some sense?
I'm well aware that \int\cos(kx)/|x| diverges around x=0, but I
suspect there is a way to restrict the support of the function and
define a meaningful FT.
How?
Perhaps something like this. For eps > 0,
2 int_eps^infty cos(|k| x)/x dx = -2 Ci(|k| eps)
= -2 ln(eps) - 2 gamma - 2 ln |k| + O(eps^2)
To get the Mathematica result as eps -> 0, you have to remove the -2 ln(eps),
which would correspond to the Fourier transform of - 2 ln(eps) Dirac(x).
Of course one's first reaction is something like "fine, but given
that we have to 'remove' those things it seems a little wacky
to call this a calculation of a Fourier transform".
Second reaction: Hmm, how can we make this into a calculation
of an FT?
Say S is the space of Schwarz functions, with dual S', the space
of tempered distributions. Say S_0 is the space of Schwarz
functions that vanish at the origin.
The dual of S_0 would be (S_0)^* = S'/D, where D is the
one-dimensional space of multiples of a delta function.
The Fourier transform induces a quotient map from S'/D
onto S'/C, where C is the space of constant functions;
this is a topological-vector-space isomorphism.
Now 1/|x| _does_ define an element of (S_0)^*, and those
truncations converge to 1/|x| in the appropriate weak
topology, hence the calculation above does show that
the "Fourier transform" of 1/|x| is "1/|k| mod constants".
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
.
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- Re: fourier transform of 1/|x|
- From: norbert9
- Re: fourier transform of 1/|x|
- From: David C . Ullrich
- Re: fourier transform of 1/|x|
- From: Robert Israel
- Re: fourier transform of 1/|x|
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