fourier transform of 1/|x|
norbert9
The table of integrals says 1/|k|.
Mathematica gives something like log(|k|).
And I can't even figure out why the Fourier transform should exist.
David C. Ullrich
<norb...@gmail.com> wrote:
>What is the fourier transform of 1/|x|?
Are you talking about a one-dimensional FT? That is,
a function from R to R defined by 1/|x|, as opposed
to say a function defined on R^2?
If you're talking about the function 1/|x| on the line:
This function does not have a Fourier transform.
It's positive and not locally integrable so it simply
does not define even a tempered distribution.
(As opposed to, say, the function 1/x, which does
define a tempered distribution because of cancellation,
even though it's not locally integrable.)
Or an argument that applies to a hypothetical extension
of the Fourier transform to a larger space than tempered
distributions: Since 1/|x| has homogeneity -1 it follows
that the FT would have homogeneity 0. Ie the FT would
be a constant. But 1/x is not a delta function...
>The table of integrals says 1/|k|.
What table of integrals, and exactly what does it say?
(This is within a constant of being right _if_ we're
talking about the function on R^2 defined by the
formula 1/|x|. You can see this by homogeneity
again: In R^2 the function 1/|x| _is_ locally
integrable and _does_ define a tempered distribution.
Now since it has homogeneity -1 the FT must
also have homogeneity -1...)
>Mathematica gives something like log(|k|).
_Exactly_ what did you ask Mathematica, and exactly
what was the answer?
>And I can't even figure out why the Fourier transform should exist.
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.)
Herman Rubin
The sine transform is 0, as 1/|x| is even.
The cosine transform does not exist because of the
singularity at 0.
So it does not exist.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Musatov
> In article <ed455c34-8669-45f2-837d-bf16c039f...@e21g2000yqb.googlegroups.com>,
>
> norbert9 <norbe...@gmail.com> wrote:
> >What is the fourier transform of 1/|x|?
> >The table of integrals says 1/|k|.
> >Mathematica gives something like log(|k|).
> >And I can't even figure out why the Fourier transform should exist.
>
> The sine transform is 0, as 1/|x| is even.
>
> The cosine transform does not exist because of the
> singularity at 0.
>
> So it does not exist.
> --
> This address is for information only. I do not claim that these views
> are those of the Statistics Department or of Purdue University.
> Herman Rubin, Department of Statistics, Purdue University
> hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
fftshift(x[, axes]), Shift zero-frequency component to center of
spectrum. ... Because the discrete Fourier transform separates its
input into components that ... The routine np.fft.fftshift(A) shifts
transforms and their frequencies to put the ... Thus, n input points
produce n/2+1 complex output points. ..
--
Musatov
http://MeAmI.org
norbert9
Gradshteyn/Ryzhik (4th ed.) list the FT of 1/|x| as 1/|k| in 17.23.
I believe it is correct in some sense.
Mathematica gives
FourierTransform[1/Abs[x],x,k] = (-2 EulerGamma - 2 Log[Abs[k]])/Sqrt
[2 Pi]
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.
Thanks for any help.
David C. Ullrich
<norb...@gmail.com> 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?
>Thanks for any help.
G. A. Edgar
<18b30519-2b4e-4930...@c9g2000yqm.googlegroups.com>,
norbert9 <norb...@gmail.com> 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.
So the citation is SN 523, I guess page 523 of Sneddon _Fourier
Transforms_ 1951. So look there for more information, and also see if
G&R have it right.
>
> Mathematica gives
> FourierTransform[1/Abs[x],x,k] = (-2 EulerGamma - 2 Log[Abs[k]])/Sqrt
> [2 Pi]
>
> 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.
>
> Thanks for any help.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Rob Johnson
>I believe it is correct in some sense.
>
>Mathematica gives
>FourierTransform[1/Abs[x],x,k] = (-2 EulerGamma - 2 Log[Abs[k]])/Sqrt
>[2 Pi]
>
>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.
I can't speak for the 4th ed, but in the 5th ed, table 17.23 says
that the Fourier Transform of 1/|x| is divergent.
Rob Johnson <r...@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
Robert Israel
> On Tue, 30 Jun 2009 22:45:38 -0700 (PDT), norbert9
> <norb...@gmail.com> 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).
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
norbert9
>
> I can't speak for the 4th ed, but in the 5th ed, table 17.23 says
> that the Fourier Transform of 1/|x| is divergent.
Thanks. So this seems to have been an error in the 4th edition.
The book by Sneddon, which is referenced from Gradshteyn/Ryzhik, is
available on Google Books. On page 523 there is a table that lists the
FT of 1/|x| as 1/|k|, without derivation.
rancid moth
I don't have access to that book at the moment - and can't get page 523 off
the web....but it's more than likely that they mean 1/|x| =
1/sqrt(x^2+y^2+z^2) which transforms to 1/|k|^2 = 1/(k1^2+k2^2+k3^2)...at
least that's what i get. Here is how
consider
F = int(-oo,oo) exp(i*k#r)/(k_x^2+k_y^2+k_z^2) dk_x dk_y dk_z
where k#r is the dot product and k and r a vectors. convert to spherical
coords -->
F = int(0,oo) exp(i*k#r) dk_r int(0,pi) sin(k_theta) dk_theta int(0,2*pi)
dk_phi
since we are integrating over all k-space we can align the coordinate system
in any way we like. to make every thing easy choose to align k_z along the
r vector which implies r#k = k_r*|r|*cos(k_theta)
then
F = 4*pi/|r| int(0,oo) sin(t*|r|)/t dt
where t in the above integral is actually k_r but i have dropped it to make
it easier to read. This last integral can be evaluated in many ways - it's
a constant for all |r| !=0.
So F is like 1/|r| and hence 1/|r| has a transform of 1/|k|^2...so perhaps
it's a typo in the book?
I have read many physics books that make it hard to differentate between k
being just k, and k being (k_x,k_y,k_z).
David C. Ullrich
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".