Re: need help with integration

• To: mathgroup at smc.vnet.net
• Subject: [mg52045] Re: need help with integration
• From: "symbio" <symbio at has.com>
• Date: Mon, 8 Nov 2004 03:13:33 -0500 (EST)
• References: <cmhv6l\$pqv\$1@smc.vnet.net> <cmklju\$jk2\$1@smc.vnet.net>
• Reply-to: "symbio" <symbio at sha.com>
• Sender: owner-wri-mathgroup at wolfram.com

```Also, please note, that FullSimplify with assumptions also does not work,
see below:
\!\(x[t_] =
3*\(E\^\(-\(t\/2\)\)\) UnitStep[t] + DiracDelta[t + 3]\[IndentingNewLine]
h[t_] = UnitStep[t] - UnitStep[t - 2]\[IndentingNewLine]
Integrate[
x[\[Tau]]
h[t - \[Tau]], {\[Tau], \(-\[Infinity]\), \[Infinity]}]\
\[IndentingNewLine]
eq1 = FullSimplify[%,
t \[Element] Reals\ && \ \[Tau]\ \[Element] Reals]\[IndentingNewLine]
Plot[eq1, {t, \(-10\), 10}, PlotRange -> All]\[IndentingNewLine]
Integrate[
x[\[Tau]] h[t - \[Tau]], {\[Tau], 0, \[Infinity]}]\[IndentingNewLine]
eq2 = FullSimplify[%,
t \[Element] Reals\ && \ \[Tau]\ \[Element] Reals]\[IndentingNewLine]
Plot[eq2, {t, \(-10\), 10}, PlotRange -> All]\[IndentingNewLine]
Integrate[
x[\[Tau]] h[t - \[Tau]], {\[Tau], \(-\[Infinity]\),
0}]\[IndentingNewLine]
eq3 = FullSimplify[%,
t \[Element] Reals\ && \ \[Tau]\ \[Element] Reals]\[IndentingNewLine]
Plot[eq3, {t, \(-10\), 0}, PlotRange -> All]\)

"David W. Cantrell" <DWCantrell at sigmaxi.org> wrote in message
news:cmklju\$jk2\$1 at smc.vnet.net...
> "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!"
>
>
> 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,
>
> 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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