Re: need help with integration

*To*: mathgroup at smc.vnet.net*Subject*: [mg52019] Re: need help with integration*From*: "David W. Cantrell" <DWCantrell at sigmaxi.org>*Date*: Sun, 7 Nov 2004 01:03:50 -0500 (EST)*References*: <cmhv6l$pqv$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

"symbio" <symbio at sha.com> wrote: > Below I define two functions x[t] and h[t], then in eq1 I integrate > the integrand ( x[tau] h[t-tau] ) from -inf to +inf and get one set of > results and plots, then I integrate the same integrand as before but this > time in two steps, once from -inf to 0 and once from 0 to +inf, but the > results and plots from the integration performed in two steps are NOT the > same as the results and plots from integration in one step. Is this a > bug and if so what's the work around? This looks essentially like what you asked here about two weeks ago in "Integration of UnitStep has bugs!? help!" <http://groups.google.com/groups?threadm=clneor%24obq%241%40smc.vnet.net> I would have answered your question in the previous thread had you not already gotten two good answers. Look at them carefully. I suspect that the trouble you're having now is the same as you were having previously; if so, then your question has already been answered, twice. David > Try this in the input cell > > In[19]:= > x[t_] = (3Exp[-.5t]UnitStep[t]) + DiracDelta[t + 3] > h[t_] = UnitStep[t] - UnitStep[t - 2] > eq1 = Integrate[x[\[Tau]]h[t - \[Tau]], {\[Tau], -\[Infinity], > \[Infinity]}] Plot[eq1, {t, -10, 10}, PlotRange -> All] > eq2 = Integrate[x[\[Tau]]h[t - \[Tau]], {\[Tau], 0, \[Infinity]}] > Plot[eq2, {t, -10, 10}, PlotRange -> All] > eq3 = Integrate[x[\[Tau]]h[t - \[Tau]], {\[Tau], -\[Infinity], 0}] > Plot[eq3, {t, -10, 0}, PlotRange -> All] > > You will get the following output expressions (I couldn't paste the plots > here, but if you run the above input cells you should get the plots too, > then it will be more obvious what the problem is) > > Out[21]= > \!\(1\/2\ \((1 + \(6.`\ \@\((\(-2\) + t)\)\^2\)\/\(\(\(2.`\)\(\ > \[InvisibleSpace]\)\) - 1.`\ t\) + \(6.`\ \@t\^2\)\/t + \@\((3 + > t)\)\^2\/\(3 \ > + t\) - \(1 + t + \@\((1 + t)\)\^2\)\/\(1 + t\))\)\) > > Out[23]= > \!\(\((\(-6.`\) + > 16.30969097075427`\ \[ExponentialE]\^\(\(-0.5`\)\ t\))\)\ UnitStep[\ > \(-2\) + t] + \((\(\(6.`\)\(\[InvisibleSpace]\)\) - > 6.`\ \[ExponentialE]\^\(\(-0.5`\)\ t\))\)\ UnitStep[t]\) > > Out[25]= > \!\(1\/2\ \((\(-\(\(1 + > t\)\/\@\((1 + t)\)\^2\)\) + \(3 + t\)\/\@\((3 + t)\)\^2)\)\)