Re: Bug in Calculus`DiracDelta`

*To*: mathgroup at smc.vnet.net*Subject*: [mg14842] Re: Bug in Calculus`DiracDelta`*From*: "Kevin J. McCann" <kevinmccann at Home.com>*Date*: Fri, 20 Nov 1998 02:16:51 -0500*Organization*: @Home Network*References*: <72je71$3f9@smc.vnet.net> <72tscl$iqi@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

I don't believe that the change of variables suggested by Bill "gives the correct result for all limits of integration." Try 0->2Pi and you get zero which is not correct. To get a handle on DiracDelta manipulations, I like to go to one of the limiting forms. Try the following: delta[n_, x_] := n/Sqrt[Pi]/E^(-(-n^2*x^2)) Integrate[delta[n, x], {x, -Infinity, Infinity}] NIntegrate[delta[10, x], {x, -5, 5}] Plot[delta[10, x], {x, -2, 2}, PlotRange -> All] NIntegrate[delta[10, Cos[x]], {x, 0, 2*Pi}] Plot[delta[10, Cos[x]], {x, 0, 2*Pi}, PlotRange -> All]; (If you copy this into a NB and then highlight the right notebook bracket and type Ctl-Shift-N, it will prettify the input) . Anyway, you can see that the correct result for Integrate[DiracDelta[Cos[x]],{x,0,2Pi}] is 2, not zero. Kevin W. K. Bertram wrote in message <72tscl$iqi at smc.vnet.net>... > >M. Rommel wrote: > >> I know normally "it's the user, stupid!" but this one seems real. >> >> In[1]:= <<Calculus`DiracDelta` >> >> In[2]:= Integrate[DiracDelta[Cos[x]],{x,0,\[Pi]}] >> >> Out[2]= 1 >> >> That's what I agree with but the next line I cannot: >> >> In[3]:= Integrate[DiracDelta[Cos[x]],{x,\[Pi],2\[Pi]}] >> >> Out[3]= 0 >> >> Any comments/insights/etc. pp.? >> > > Martin, >The second result is certainly wrong! I have always been told to be very >careful >with expressions containing delta functions. As a result I would always >transform >an integral such as the one above by making the argument of the delta >function the >variableof integration. Therefore, making the substitution, > > u = Cos[x] > >you would instead evaluate > > Integrate[Sqrt[1-u^2] DiracDelta[u],{u, Cos[Pi], Cos[2Pi]}] > >and this gives the correct result for all limits of integration. >Cheers, > Bill > >