On Fri, 10 Dec 2004 09:29:05 +0100, Han de Bruijn

Curious document. At the start you define delta to be that gaussian -
that's simply redefining what the term means, which really makes
it kind of easy to prove whatever you want...

But later you say we have to let sigma -> 0. So you really meant
that delta is the limit (in some sense - the correct sense is
"in the sense of distributions") of that gaussian as sigma -> 0.
_That_ is of course exactly correct, and it's no surprise that
if you do that then the Laplace transform comes out to exactly
what it does using the much simpler argument from the
actual standard definition of delta.

In case you're curious, there's a standard way to prove what
you prove in that pdf, which is both simpler and much more
powerful. Let's say g_sigma(t-T) is that gaussian function.
The result follows immediately from the following three
properties of g_sigma (the proof is more powerful because
it works for _any_ family of functions satisfying these
three properties):

(i) g_sigma >= 0.

(ii) int g_sigma = 1.

(iii) If a > - is fixed then

  int_{|t-T| > a} g_sigma(t-T) dt -> 0 as s -> 0.

************************

David C. Ullrich