From: ullrich@math.okstate.edu (David C. Ullrich)
Subject: Re: Q: Fourier Transform Weirdness
Date: Mon, 27 Nov 2000 15:47:01 GMT
Newsgroups: sci.math
Summary: [missing]

On Mon, 27 Nov 2000 13:10:24 GMT, Dan Greenfield <dan@eAware.com>
wrote:

>Hi,
>
>  I was hoping someone out there could clarify something for me:
>
>why is Integral(exp(iwx)dx) from -infinity to infinity = zero everywhere
>but w = 0?

	It isn't. There are lies in those books...

>I don't understand how an integral of say w=1 yielding
>Integral((cos(x)+isin(x))dx) could be zero when taking the 'boundaries'
>-infinity and infinity. Surely it is absurd to give exp(i*infinity) any
>definite value let alone presume to say it cancels with exp(-i*infinity)
>?
	Yes, it's absurd.

>And yet the (continuous) fourier transform of a constant is merely a
>multiple of delta(w). Is there something I am missing?

	But nonetheless the Fourier Transform of a constant
_is_ a multiple of that "delta function" (the delta function is
_not_ a function, it's something else!).

	The Fourier Transform of 1 is delta. But the Fourier
Transform is _not_ defined by that integral that your book
says it's defined by! If f is a "nice" function then the Fourier
Transform F(f) is defined by that integral (here saying that
the integral of |f| from -ininifty to infinity is finite is nice
enough.) A person uses that integral to define the FT of nice 
functions. Then a person notices that

(*) int(f*F(g)) = int(F(f)*g)

if f and g are both nice functions (here "int" means
"integral from -infinity to infinity", F(f) is the FT of F
and F(g) is the FT of g, and * denotes multiplication
(not convolution).

	At this point the definition of the FT gets
extended from just nice functions to a much larger
class of objects, called distributions. The FT
of a distribution is _not_ defined as the integral
of exp(-itx) times the distribution. Exactly how it
is defined is a long story. But it's defined in
such a way that (*) remains valid if f is a distribution
and g is an extremely nice function (extremely
nice is a little more than nice...) In fact (*) is more
or less the definition of the FT of a distribution,
although it takes a little while to say exactly what
I mean by that.

	Now what should the FT of the constant
1 be? It's delta. I'm not going to say exactly what
delta _is_, but we know what delta is supposed to
_do_: delta is supposed to have the property that

(**) int(g*delta) = g(0)

if g is nice enough. In fact (**) is more or less the
definition of the "delta function", actually. 
	So verifying that the F(1) = delta
amounts to verifying that (*) is correct if we
set f=1 and F(f) = delta. And that works:
Using (**) we see

int(f*F(g)) = int(F(g)) = g(0) = int(delta*g),

and if F(f) = delta then this says 

int(f*F(g)) = int(F(f)*g)

as desired.

	This is sort of a "top-down" approach
to the whole thing. There's a more "bottom-up"
approach in ZK's reply. If you really want to
know how the math works you need to study
"distributions". (These are not the
"distributions" that come up in probability, btw.)

>- DG
>