From: sparapan@flash.ra.anl.gov (RA Sparapani RA/208/xxx gwing)
Subject: Thanks for Fourier Transform Help
Organization: Argonne National Laboratory, Argonne IL
Date: Thu, 1 Oct 1992 19:48:18 GMT


Thanks for all who responded to my post.  Some people answered this very
quickly.  I include those responses for all who might be interested.  Also,
there is a response of my own that bounced so it is also included.

In article <71795714322538@flash.ra.anl.gov> you write:
>
>sci.math.stat:
>
>I am having great difficulty computing the Fourier transform of the sinc
>function, i.e. (sinx)/x.  If anyone knows of the solution or a reference
>to a solution, I would be thankful.  The problem is that I need to 
>integrate functions of the following forms.
>
>sin(1+constant)x        cos(1+constant)x
>----------------  and   ----------------
>        x                       x
>
>If anyone has information on the integrals of either of these two forms, I
>would appreciate that, also.  Thanks.
>
>Rodney

Hopefully someone has already sent you the "exact" answer.  I'll just point you
in two useful directions:
	a) since  sin(x)/x is NOT Lebesgue integrable on (-oo, oo) (it's
              absolute value goes to 0 like 1/x, not fast enough)
	      the Fourier transform must be taken in some "distributional"
	      or limiting sense.
	b) an integration by parts converts  the integral of sinc(x)exp(-iwx)
           over [-a,a] into a numeric piece with 1/x factor and an
           integral part with 1/x^2  as a factor.  The numerical part -> 0
           as a -> oo, the integral part (in the limit) can be calculated 
	   using contour integral techniques (you need some complex variable
           theory to do this - and you need to know how to compute "residues").

	c) The upshot is that transform of sinc(x) is some multiple of the
	   function which is 1 on [-1,1] and 0 elsewhere.  (If you believe
           that the inverse Fourier transform gives back the original, its
           easy to verify this fact, since  
		
		integral from -1 to 1 of  exp(iwx)dw = (exp(ix) - exp(-ix))/ix
				= 2 sinc(x)

Now all you need to do is get the constant right.

Prof. David Ragozin,        Dept. of Mathematics, GN-50
Univ. of Washington, Seattle, WA. 98195. (206-543-1148)
rag@math.washington.edu  or   ...!uw-coco!rag
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The Fourier transform is quite simple.  We start with the fact that
the integral of (sin tx)/x is sign(t)*pi.  Since the function is even,
the FT is the integral of 

	(sin x)(cos tx) dx/x.

But the trigonometric product is (sin(t+1)x + sin(1-t)x)/2.  So the
FT is

	pi			if |t| < 1

	pi/2			if |t| = 1

	0			if |t| > 1

I have given you the sin integral you ask for.  The cos integral is
only defined as a Cauchy principle value, and since cos is even, the
integral is always 0.
-- 
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
Phone: (317)494-6054
hrubin@pop.stat.purdue.edu (Internet, bitnet)  
{purdue,pur-ee}!pop.stat!hrubin(UUCP)
--------------------------------------------------------------------------------
Well, I think that I can respond to my own post.  There is an example in 
Schaum's outline of Advanced Mathematics that I borrowed from a co-worker.
As some of you might have guessed, the Fourier Transform of the sinc function
is 1/(2pi)**.5.  In other words, it is the 1 function times the symmetric
constant 1/(2pi)**.5, a rather uninteresting result(or interesting depending on 
your point of view).

Rodney