
Re: dirac delta function is not a TRUE function??
Posted:
Jul 18, 2011 1:53 PM


In article <gerryE6603D.09354915072011@news.eternalseptember.org>, Gerry Myerson <gerry@maths.mq.edi.ai.i2u4email> writes: >In article <ivn53v$7e1$1@dontemail.me>, mstemper@walkabout.empros.com (Michael Stemper) wrote: >> In article <9f2192a597764c559a2deeb93f0559ab@g3g2000prf.googlegroups.com>, MBALOVER <mbalover9@gmail.com> writes: >> >On Jul 11, 7:36=A0pm, Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote:
>> >> The Dirac delta function does not assign a real number to zero, >> >> and is thus not a true function. >> >> >Thanks Gerry. But I am still confused. I think at x =3D 0, delta(x) = >> >infinity. >> >> That's one way of looking at it  in fact it's the one that I was >> taught while studying electical engineering. However, look more >> carefully at what was said: "... does not assign a real number to >> zero ...". >> >> There is no real number called "infinity". > >Quite. > >You could try to think of delta as a function >from the reals to the extended reals, the extended >reals being "the reals union infinity", but that >won't help you to understand why >integral from minus infinity to infinity of f(x) delta(x) dx >is f(0).
It's been a long time since I studied this, but isn't that true for the integral from minus epsilon to epsilon (for any positive epsilon), as well?
 Michael F. Stemper #include <Standard_Disclaimer> Life's too important to take seriously.

