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Re: Re: need help with integration
*To*: mathgroup at smc.vnet.net
*Subject*: [mg52086] Re: Re: need help with integration
*From*: "symbio" <symbio at has.com>
*Date*: Wed, 10 Nov 2004 04:45:28 -0500 (EST)
*References*: <cmhv6l$pqv$1@smc.vnet.net> <cmklju$jk2$1@smc.vnet.net> <200411080813.DAA07936@smc.vnet.net> <cmppj3$mj4$1@smc.vnet.net>
*Reply-to*: "symbio" <symbio at sha.com>
*Sender*: owner-wri-mathgroup at wolfram.com
Equation eq2+eq3 is the correct value for sure, I've verified through
Frequency domain (remember this is just a convolution integral). eq1 is not
being calculated correctly by Mathematica. It seems to be a bug , I wonder
what Wolfram has to say about it.
"DrBob" <drbob at bigfoot.com> wrote in message
news:cmppj3$mj4$1 at smc.vnet.net...
> Symbio seems to be right, but a solution CAN be salvaged.
>
> Here's Wolfgang's code from David's link, executed in Mathematica 5.0.1:
>
> 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]
> FullSimplify[eq2 + eq3, t \[Element] Reals]
>
> (1/2)*(-((6*(-2 + t))/ Abs[-2 + t]) +
> (6*t)/Abs[t] - (1 + t)/Abs[1 + t] + (3 + t)/Abs[3 + t])
> -((1 + t)/(2*Abs[1 + t])) + (3 + t)/(2*Abs[3 + t]) +
> 6*((-1 + E^(1 - t/2))*UnitStep[-2 + t] + UnitStep[t] - UnitStep[t]/
> E^(t/2))
>
> Plot[{(1/2)*(-((6*(-2 + t))/Abs[-2 + t]) +
> (6*t)/Abs[t] - (1 + t)/Abs[1 + t] +
> (3 + t)/Abs[3 + t]),
> (1/2)*(-Sign[1 + t] + Sign[3 + t] +
> 12*(-UnitStep[-2 + t] +
> (E*UnitStep[-2 + t] - UnitStep[t])/
> E^(t/2) + UnitStep[t]))},
> {t, -10, 10}, PlotRange -> All]
>
> The Plots are NOT the same, so what worked for Wolfgang in Mathematica 4.0
> doesn't work in 5.0.1.
>
> His Integrate with Assumptions gives yet a third expression (after a
> LOOOONG calculation):
>
> eq1 = Integrate[x[a]*h[t - a],
> {a, -Infinity, Infinity},
> Assumptions -> t \[Element] Reals]
>
> (1 + t + Abs[1 + t])/(-2 - 2*t) +
> (3 + t + Abs[3 + t])/(6 + 2*t) -
> 3*((2 - 2*E^(1 - t/2))* UnitStep[-2 + t] +
> 2*(-1 + E^(-t/2))*UnitStep[t])
>
> This one plots the same as eq2+eq3, so that solves the problem....
>
> ...IF we're convinced the original eq1 was wrong and the other two answers
> are right.
>
> I can't muster much confidence in that.
>
> Bobby
>
> On Mon, 8 Nov 2004 03:13:30 -0500 (EST), symbio <symbio at has.com> wrote:
>
>> 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)\)\)
>>>
>>
>>
>>
>>
>
>
>
> --
> DrBob at bigfoot.com
> www.eclecticdreams.net
>
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