       Re: need help with integration

• To: mathgroup at smc.vnet.net
• Subject: [mg52024] Re: [mg51997] need help with integration
• From: DrBob <drbob at bigfoot.com>
• Date: Sun, 7 Nov 2004 01:04:18 -0500 (EST)
• References: <200411060708.CAA26072@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```I assume you mean

eq2 + eq3 - eq1 // FullSimplify

(3*(-2 + t))/Sqrt[(-2 + t)^2] -
(3*t)/Sqrt[t^2] -
6*UnitStep[-2 + t] +
(6*(E*UnitStep[-2 + t] -
UnitStep[t]))/E^(t/2) +
6*UnitStep[t]

should be zero for all t?

If we reduce x[t] to just x[t_]=3Exp[-.5t]UnitStep[t] or x[t_]=DiracDelta[t + 3], it works properly, so I think it should work for your x function as well. But it doesn't.

Perhaps, when we add the terms together, there's some ambiguity about the value at -3, and that leads to the problem? Hopefully the experts can tell you more.

Bobby

On Sat, 6 Nov 2004 02:08:53 -0500 (EST), symbio <symbio at has.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?
>
> Try this in the input cell
>
> In:=
> 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=
> \!\(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=
> \!\(\((\(-6.`\) +
> 16.30969097075427`\ \[ExponentialE]\^\(\(-0.5`\)\ t\))\)\ UnitStep[\
> \(-2\) + t] + \((\(\(6.`\)\(\[InvisibleSpace]\)\) -
> 6.`\ \[ExponentialE]\^\(\(-0.5`\)\ t\))\)\ UnitStep[t]\)
>
>
>
> Out=
> \!\(1\/2\ \((\(-\(\(1 +
> t\)\/\@\((1 + t)\)\^2\)\) + \(3 + t\)\/\@\((3 + t)\)\^2)\)\)
>
>
>
>

--
DrBob at bigfoot.com
www.eclecticdreams.net

```

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