Re: fourier transform of 1/|x|

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).
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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