>> 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.

>I hope you understand that the document is taylored to Mr. Bean.

>> 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.

>As a matter of fact, I'm curious ...

>> 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.

>OK, until now.

>> (iii) If a > - is fixed then

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

>What does 'a > -' mean?