       Re: need help with integration

• To: mathgroup at smc.vnet.net
• Subject: [mg52049] Re: need help with integration
• From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
• Date: Mon, 8 Nov 2004 03:13:47 -0500 (EST)
• References: <cmhv6l\$pqv\$1@smc.vnet.net> <cmklju\$jk2\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```"David W. Cantrell" <DWCantrell at sigmaxi.org> wrote:
> "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

That comment of mine was made too hastily, and I apologize. Although your
question does resemble what you'd asked about earlier, I see now that it is
not the same question at all. In fact, I think that there is a bug of
some sort in giving the result for the integral from -Infinity to
+Infinity. One work-around seems to be just to add the integrals from
-Infinity to 0 and from 0 to +Infinity.

David

> > 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)\)\)

```

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