Re: need help with integration

*To*: mathgroup at smc.vnet.net*Subject*: [mg52049] Re: need help with integration*From*: "David W. Cantrell" <DWCantrell at sigmaxi.org>*Date*: Mon, 8 Nov 2004 03:13:47 -0500 (EST)*References*: <cmhv6l$pqv$1@smc.vnet.net> <cmklju$jk2$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

"David W. Cantrell" <DWCantrell at sigmaxi.org> wrote: > "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 That comment of mine was made too hastily, and I apologize. Although your question does resemble what you'd asked about earlier, I see now that it is not the same question at all. In fact, I think that there is a bug of some sort in giving the result for the integral from -Infinity to +Infinity. One work-around seems to be just to add the integrals from -Infinity to 0 and from 0 to +Infinity. 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)\)\)