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MathGroup Archive 2004

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Re: Integration of UnitStep has bugs!? help!

  • To: mathgroup at smc.vnet.net
  • Subject: [mg51616] Re: Integration of UnitStep has bugs!? help!
  • From: "Dr. Wolfgang Hintze" <weh at snafu.de>
  • Date: Wed, 27 Oct 2004 01:54:38 -0400 (EDT)
  • References: <clcnlr$q41$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

It's not a bug. Mathematic is simply careful. It needs to know more 
about t to carry on.

You need simply to state that t is real. Here we go:

Either during Simplify

In[13]:=
x[t_] = 3*Exp[-t/2]*UnitStep[t] + DiracDelta[t + 3];
   h[t_] = UnitStep[t] - UnitStep[t - 2];
   eq1 = Integrate[x[a]*h[t - a], {a, -Infinity,
      Infinity}]; eq2 = Integrate[x[a]*h[t - a],
     {a, 0, Infinity}];
eq3 = Integrate[x[a]*h[t - a], {a, -Infinity, 0}];
Simplify[eq1, t \[Element] Reals]
Simplify[eq2 + eq3, t \[Element] Reals]

Out[15]=
(-6 + 6*E^(1 - t/2))*UnitStep[-2 + t] +
   (6 - 6/E^(t/2))*UnitStep[t] - UnitStep[1 + t] +
   UnitStep[3 + t]

Out[16]=
(-6 + 6*E^(1 - t/2))*UnitStep[-2 + t] +
   (6 - 6/E^(t/2))*UnitStep[t] - UnitStep[1 + t] +
   UnitStep[3 + t]

or even in doing Integrate (using Assumptions)

In[18]:=
eq1 = Integrate[x[a]*h[t - a], {a, -Infinity, Infinity},
    Assumptions -> t \[Element] Reals]

Out[18]=
(-6 + 6*E^(1 - t/2))*UnitStep[-2 + t] +
   (6 - 6/E^(t/2))*UnitStep[t] - UnitStep[1 + t] +
   UnitStep[3 + t]

Hence it works fine (in Mathematica 4.0. I hope it does the same in 5).

Wolfgang

news wrote:

> Can anyone explain why Integrate does not work propery with UnitStep 
> function?  When you integrate from -inf to inf you get one answer, but when 
> you split it to two sections (from -inf to 0, and then from 0 to +inf) you 
> get a different answer.  Which is correct?  And why is Mathematica version 5 
> has not fixed this problem yet?  See below.
> 
> In[1]:=
> \!\(\(x[t_] = \((3 Exp[\(\(-1\)\/2\) t] UnitStep[t])\) +
> DiracDelta[t + 3];\)\[IndentingNewLine]
> \(h[t_] = UnitStep[t] - UnitStep[t - 2];\)\[IndentingNewLine]
> \(eq1 =
> Integrate[
> x[a] h[t -
> a], {a, \(-\[Infinity]\), \[Infinity]}];\)\[IndentingNewLine]
> \(eq2 = Integrate[x[a] h[t - a], {a, 0, \[Infinity]}];\)\[IndentingNewLine]
> \(eq3 =
> Integrate[
> x[a] h[t - a], {a, \(-\[Infinity]\), 0}];\)\[IndentingNewLine]
> eq1\ // FullSimplify\[IndentingNewLine]
> eq2 + eq3 // FullSimplify\)
> Out[6]=
> \!\(1\/2\ \((\(-\(\(6\ \((\(-2\) +
> t)\)\)\/\@\((\(-2\) + t)\)\^2\)\) + \(6\ t\)\/\@t\^2 - \
> \(1 + t\)\/\@\((1 + t)\)\^2 + \(3 + t\)\/\@\((3 + t)\)\^2)\)\)
> Out[7]=
> \!\(1\/2\ \((\(-\(\(1 +
> t\)\/\@\((1 + t)\)\^2\)\) + \(3 + t\)\/\@\((3 + t)\)\^2)\) \
> + 6\ \((\(-1\) + \[ExponentialE]\^\(1 - t\/2\))\)\ UnitStep[\(-2\) +
> t] + \((6 - 6\ \[ExponentialE]\^\(\(-t\)/2\))\)\ UnitStep[t]\)
> 
> 


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