Re: Apparent error integrating product of DiracDelta's
- To: mathgroup at smc.vnet.net
- Subject: [mg92073] Re: Apparent error integrating product of DiracDelta's
- From: magma <maderri2 at gmail.com>
- Date: Fri, 19 Sep 2008 05:16:32 -0400 (EDT)
- References: <gag2lg$39k$1@smc.vnet.net> <gal3ht$dv1$1@smc.vnet.net>
On Sep 18, 12:10 pm, 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 Very interesting answer! Yet, taking a naive (physicist's ?) point of view, the result seems intuitively correct. Mr. Delta himself (P.A.M. Dirac) in his "The principles of quantum mechanics" published in 1930 confirms the result (eq. 15.9 in the Italian 4th edition). He even "proves" it, a little bit later. The "proof" is a bit heuristics because he assumes the integral is defined in the first place. Yet the whole Delta function story was at the time just heuristics. Apparently the delta was described for the first time in this book (Wikipedia). I am also pretty sure he uses the result somewhere else later in the book (probably where double integrals appear). So, could he have been so wrong after having been so right before? More to the point: I found very intriguing your claim that " If different methods of approximation will lead to different results, then what you have cannot be defined". I agree with you of course, but could you "show" with Mathematica - maybe with a Manipulate program - that this is the case with this integral? Could you find 2 different approximation methods which give 2 different results? I look forward to seeing it. Concerning DiracDelta[x]^2, I would agree it is undefined at least because its integral, if it existed, would be DeltaDirac[0], which is undefined.
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- Re: Re: Apparent error integrating product of DiracDelta's
- From: Daniel Lichtblau <danl@wolfram.com>
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