need help with integration
- To: mathgroup at smc.vnet.net
- Subject: [mg51997] need help with integration
- From: "symbio" <symbio at has.com>
- Date: Sat, 6 Nov 2004 02:08:53 -0500 (EST)
- Reply-to: "symbio" <symbio at sha.com>
- Sender: owner-wri-mathgroup at wolfram.com
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? 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)\)\)
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