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MathGroup Archive 2004

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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)\)\)


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