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MathGroup Archive 2004

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An Integration bug?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg51677] An Integration bug?
  • From: "symbio" <symbio at s.dn.com>
  • Date: Fri, 29 Oct 2004 03:38:53 -0400 (EDT)
  • Reply-to: "symbio" <symbio at s.dn.com>
  • Sender: owner-wri-mathgroup at wolfram.com

Is this another integration bug in Mathematica 5.0?  Or am I doing something 
wrong?  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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