Re: Question: DiracDelta simplifies/integrates incorrectly?

*To*: mathgroup at smc.vnet.net*Subject*: [mg65163] Re: [mg65132] Question: DiracDelta simplifies/integrates incorrectly?*From*: John Harker <harker at me.rochester.edu>*Date*: Wed, 15 Mar 2006 23:59:40 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

On Wed, 15 Mar 2006, gardyloo wrote: >John Harker wrote: >> Hello, >> >> I have a question regarding the behavior of the DiracDelta function in >> Mathematica 5.2. >> >> The following two inputs produce the following results: >> >> In[61]:= >> Simplify[Pi*DiracDelta[Pi*x]] >> >> Out[61]= >> DiracDelta[x] >> >> In[60]:= >> Simplify[Pi*DiracDelta[Pi*(x-3)]] >> >> Out[60]= >> \[Pi] DiracDelta[\[Pi] (-3+x)] >> >> As you can see, although the Pi is correctly simplified out in the first >> case, it is not simplified in the second case. This is a problem because >> of the following result: >> >> In[68]:= >> Clear[f]; >> >> In[69]:= >> Integrate[f[x]*Pi*DiracDelta[Pi*x],{x,-Infinity,Infinity}] >> >> Out[69]= >> f[0] >> >> In[70]:= >> Integrate[f[x]*Pi*DiracDelta[Pi*(x-3)],{x,-Infinity,Infinity}] >> >> Out[70]= >> 0 >> >> As you can see, the output [69] is correct, but the output [70] should >> correctly be f[3], and instead it returns 0. >> >> Is there a flaw in my understanding of the DiracDelta function, or is this >> a bug? >> >> All of the above poses a problem because Mathematica will return results >> such as the following: >> >> In[78]:= >> Simplify[ >> FourierTransform[Exp[I*2*Pi*3*x],x,f, >> FourierParameters\[Rule]{0,-2*\[Pi]}] >> ] >> >> Out[78]= >> \[Pi] DiracDelta[(-3+f) \[Pi]] >> >> So you see that just by asking for a simple Fourier transform, I can get >> an output result which Mathematica cannot integrate correctly. >> >> Does anyone have any ideas about a better way to perform this math >> in order to get around the problem? Or something illuminating about how >> the DiracDelta function works? >> >> Many thanks! >> >> John >> >> >> > Perhaps this is overkill (more things get evaluated symbolically when >they don't need to be?), but is it possible to use an unassigned symbol, >like "n", in place of Pi, evaluate whatever you want, and then do a >replacement with result/.{n->Pi} ? > > >-- >========================================================== >Curtis Osterhoudt >PGP Key ID: 0x088E6D7A >Please avoid sending me Word or PowerPoint attachments >See http://www.gnu.org/philosophy/no-word-attachments.html >========================================================== > Thanks, Curtis, you're absolutely correct! For some reason, evaluating with a general variable in place of Pi DOES work. How odd. For my purposes, I need to carry through the definition of pi from the Fourier transform beforehand, so the workaround (which is more complete than my previously mentioned workaround) looks like this: (edited for clarity) In[17]:= tmp=FourierTransform[Exp[I*2*jpi*n*x],x,f,FourierParameters\[Rule]{0,-2*jpi}] Out[17]= Sqrt[\[Pi]] * Sqrt[Abs[jpi]] * DiracDelta[f*jpi - jpi*n] In[19]:= Integrate[g[f]*tmp,{f,-Infinity,Infinity}] /. jpi->\[Pi] Out[19]= g[n] So it works almost effortlessly. You're right, there's a little more complexity involved, but I'll take the tradeoff happily. You can see that the Fourier transform gives the correct answer: In[18]:= tmp/.jpi->\[Pi] Out[18]= \[Pi] * DiracDelta[f*\[Pi] - n*\[Pi]] ...but somehow the use of the general variable jpi to replace Pi causes everything to evaluate correctly when it's integrated. Thanks for your advice! John