Re: Apparent error integrating product of DiracDelta's
- To: mathgroup at smc.vnet.net
- Subject: [mg92080] Re: Apparent error integrating product of DiracDelta's
- From: Michael Mandelberg <mmandelberg at comcast.net>
- Date: Fri, 19 Sep 2008 05:17:50 -0400 (EDT)
- References: <gag2lg$39k$1@smc.vnet.net> <gal3ht$dv1$1@smc.vnet.net>
Hmmm. I appreciate (mostly) the niceties of this argument, but what about the practical matter that as an intermediate step in some problem I may want to have some DiracDelta's floating around prior to a final integration? Michael On Sep 18, 6:10 am, Daniel Lichtblau <d... at wolfram.com> wrote: > magma wrote: > > On Sep 15, 9:40 am, "Nasser Abbasi" <n... at 12000.org> wrote: > >> "Michael Mandelberg" <mmandelb... at comcast.net> wrote in message > > >>news:gag2lg$39k$1 at smc.vnet.net... > > >>> How do I get: > >>> Integrate[DiracDelta[z- x] DiracDelta[z- y], {z-Infinity, Infinit= y}= > > ] > >>> to give DiracDelta[x-y] as the result? Currently it gives 0. I = ha= > > ve > >>> all three variable assumed to be Reals. I am using 6.0.0. > >>> Thanks, > >>> Michael Mandelberg > >> I think you have synatx error in the limit part. I assume you mean to = wri= > > te > >> {z, -Infinity,Infinity} > > >> Given that, I think zero is the correct answer. When you multiply 2= de= > > ltas > >> at different positions, you get zero. Integral of zero is zero. > > >> Nasser > > > No Nasser, the correct value of the integral should be DiracDelta[x- > > y], as Michael said. > > This value is indeed 0 if x != y but it is not 0 if x==y. > > It is not 0 at x==y, but neither is it DiracDelta[x-y]. The value the= re > is undefined. > > > Mathematica correctly calculates: > > > Integrate[f[z - x] DiracDelta[z - y], {z, -Infinity, Infinity}, > > Assumptions -> y \[Element] Reals] > > > as > > > f[-x + y] > > This is making a tacit assumption that f is a "nice" function. Nice, in > this context, means it is an element of Schwartz space S: C^infinity and > vanishing faster than any polynomial at +-infinity. DiracDelta, suffice > it to say, is not an element of S (it's not even a function). > > > However it fails to recognize that if f[z-x] is replaced by > > DiracDelta[z-x], the result should be > > > DiracDelta[-x + y] > > > or the equivalent > > > DiracDelta[x - y] > > This is not a failure but rather an active intervention. > > > In the help file, under "possible issues" it is mentioned that > > "Products of distributions with coinciding singular support cannot be > > defined:" > > This is a statement of mathematics and not specific to Mathematica. > > > So perhaps at the moment the only way to do the integral is: > > > Integrate[f[z - x] DiracDelta[z - y], {z, -Infinity, Infinity}, > > Assumptions -> y \[Element] Reals] /. f -> DiracDelta > > > hth > > Here is a general rule of thumb. If you are working with DiracDelta > function(al)s, instead approximate them as ordinary functions. If > different methods of approximation will lead to different results, then > what you have cannot be defined. One can use this notion to see that, > for example, DiracDelta[x]^2 is not defined. > > Daniel Lichtblau > Wolfram Research- Hide quoted text - > > - Show quoted text -