Re: piecewise integration
- To: mathgroup at smc.vnet.net
- Subject: [mg66999] Re: piecewise integration
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Tue, 6 Jun 2006 06:28:49 -0400 (EDT)
- References: <20060605102611.774$dR_-_@newsreader.com>
- Sender: owner-wri-mathgroup at wolfram.com
On 5 Jun 2006, at 23:26, David W. Cantrell wrote:
>
>
> Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
>> On 4 Jun 2006, at 15:01, Chris Chiasson wrote:
>>
>>> The Integrate result seems pretty weak. Is there any way to obtain a
>>> more explicit exact answer besides manually converting the Piecewise
>>> function to two UnitStep functions? Can the same be done if the
>>> final
>>> limit of integration is a variable?
>>>
>>> in
>>>
>>> load[x_]=-9*10^3*DiracDelta[x]+Piecewise[{{x*10*(10^3/3),
>>> 0<=x<=3}}]-6*10^3*DiracDelta[x-5]//InputForm
>>>
>>> out
>>>
>>> -6000*DiracDelta[-5 + x] - 9000*DiracDelta[x] +
>>> Piecewise[{{(10000*x)/3, 0 <= x <= 3}}, 0]
>>>
>>> in
>>>
>>> Integrate[load[x],{x,0,5}]//InputForm
>>>
>>> out
>>>
>>> Integrate[InputForm[-6000*DiracDelta[-5 + x] - 9000*DiracDelta[x] +
>>> Piecewise[{{(10000*x)/3, 0 <= x <= 3}}, 0]], {x, 0, 5}]
>>>
>>> --
>>> http://chris.chiasson.name/
>>>
>>
>> I am sure Maxim does not need my help to advertise his package but
>> somehow people still keep posting such questions with surprising
>> frequency.
>
> It shouldn't surprise you. Such questions will continue to be posted
> frequently until Mathematica _itself_ can adequately handle
> integration of
> Piecewise functions.
No doubt you are right, but I was referring to the fact that I
mentioned the package in a related context only a few days earlier
and Maxim has posted several times in detail on this topic. So I
naturaly felt a little bit like a person trying to help another
person with a Sisyphean task.
>
>> Even if for some reason you do not want to use a third
>> party package you can always look at the Mathematica code inside,
>> which should answer questions such as these.
>>
>> load[x_] = -9*10^3*
>> DiracDelta[x] + Piecewise[{{x*10*(10^3/3), 0 <=
>> x <= 3}}] - 6*10^3*DiracDelta[x - 5];
>> << piecewise`
>>
>> In[3]:=
>> PiecewiseIntegrate[load[x],{x,0,5}]
>>
>> Out[3]=
>> 0
>
> That might or might not be what Chris wanted. It depends on whether he
> wanted integration of DiracDelta to be handled as Mathematica does
> or as
> PiecewiseIntegrate does.
>
> Using Mathematica, we have, for example,
>
> In[28]:= Integrate[DiracDelta[x],{x,0,2}]
> Out[28]= 1/2
>
> rather than 1, which I infer is what PiecewiseIntegrate would do.
I confess that I never noticed that Integrate and DiracDelta in
Mathematica behaved like this at end points. It seems to me that the
Piecewise approach, which assumes that boundary points are treated
the same as interior points, is the more natural. But Chris obviously
was not interested in the answer to this particular problem but in
more general matters. It is trivial to modify the behaviour of the
package in this respect (by adding ones own rules for handling
DiracDelta) to make it conform with what Mathematica does, if one
really wanted to. But my main point was that the package is
interesting in its own right and it seems to me that anyone seriously
interested in this topic would have already taken a look at it. Why,
even people not seriously interested in it, like myself, have done so
and found interesting and instructive things in it.
Andrzej
>
> If Chris really wanted the integration of the DiracDelta functions
> to be
> handled as Mathematica does, then I should think he'd want
> Integrate[load[x],{x,0,5}] to return 7500, rather than your Out[3]=0.
>
> David
>
>> In[4]:=
>> PiecewiseIntegrate[load[x], {x, 0, a}]
>>
>> Out[4]=
>> If[Inequality[0, Less, a, LessEqual, 3], (5000*a^2)/3, 0] + If[3 < a,
>> 15000, 0] + If[a < 0, 9000, 0] + If[0 <= a, -9000, 0] +
>> If[5 <= a, -6000, 0]
>>
>> Andrzej Kozlowski
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