Re: Re: need help with integration

*To*: mathgroup at smc.vnet.net*Subject*: [mg52060] Re: [mg52044] Re: need help with integration*From*: DrBob <drbob at bigfoot.com>*Date*: Tue, 9 Nov 2004 01:36:50 -0500 (EST)*References*: <cmhv6l$pqv$1@smc.vnet.net> <cmklju$jk2$1@smc.vnet.net> <200411080813.DAA07936@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

Symbio seems to be right, but a solution CAN be salvaged. Here's Wolfgang's code from David's link, executed in Mathematica 5.0.1: x[t_] = 3*Exp[-t/2]*UnitStep[t] + DiracDelta[t + 3]; h[t_] = UnitStep[t] - UnitStep[t - 2]; eq1 = Integrate[x[a]*h[t - a], {a, -Infinity, Infinity}]; eq2 = Integrate[x[a]*h[t - a], {a, 0, Infinity}]; eq3 = Integrate[x[a]*h[t - a], {a, -Infinity, 0}]; Simplify[eq1, t \[Element] Reals] FullSimplify[eq2 + eq3, t \[Element] Reals] (1/2)*(-((6*(-2 + t))/ Abs[-2 + t]) + (6*t)/Abs[t] - (1 + t)/Abs[1 + t] + (3 + t)/Abs[3 + t]) -((1 + t)/(2*Abs[1 + t])) + (3 + t)/(2*Abs[3 + t]) + 6*((-1 + E^(1 - t/2))*UnitStep[-2 + t] + UnitStep[t] - UnitStep[t]/ E^(t/2)) Plot[{(1/2)*(-((6*(-2 + t))/Abs[-2 + t]) + (6*t)/Abs[t] - (1 + t)/Abs[1 + t] + (3 + t)/Abs[3 + t]), (1/2)*(-Sign[1 + t] + Sign[3 + t] + 12*(-UnitStep[-2 + t] + (E*UnitStep[-2 + t] - UnitStep[t])/ E^(t/2) + UnitStep[t]))}, {t, -10, 10}, PlotRange -> All] The Plots are NOT the same, so what worked for Wolfgang in Mathematica 4.0 doesn't work in 5.0.1. His Integrate with Assumptions gives yet a third expression (after a LOOOONG calculation): eq1 = Integrate[x[a]*h[t - a], {a, -Infinity, Infinity}, Assumptions -> t \[Element] Reals] (1 + t + Abs[1 + t])/(-2 - 2*t) + (3 + t + Abs[3 + t])/(6 + 2*t) - 3*((2 - 2*E^(1 - t/2))* UnitStep[-2 + t] + 2*(-1 + E^(-t/2))*UnitStep[t]) This one plots the same as eq2+eq3, so that solves the problem.... ...IF we're convinced the original eq1 was wrong and the other two answers are right. I can't muster much confidence in that. Bobby On Mon, 8 Nov 2004 03:13:30 -0500 (EST), symbio <symbio at has.com> wrote: > actually this is different, i used both assumptions and simplify but won't > work. Please try running the code, you'll see for yourself when you look at > the plots, they are different even if you use assumptions. > > "David W. Cantrell" <DWCantrell at sigmaxi.org> wrote in message > news:cmklju$jk2$1 at smc.vnet.net... >> "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)\)\) >> > > > > -- DrBob at bigfoot.com www.eclecticdreams.net

**References**:**Re: need help with integration***From:*"symbio" <symbio@has.com>