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MathGroup Archive 2004

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Re: need help with integration

  • To: mathgroup at smc.vnet.net
  • Subject: [mg52044] Re: need help with integration
  • From: "symbio" <symbio at has.com>
  • Date: Mon, 8 Nov 2004 03:13:30 -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

actually this is different, i used both assumptions and simplify but won't 
work.  Please try running the code, you'll see for yourself when you look at 
the plots, they are different even if you use assumptions.

"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!"
>
> <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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