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MathGroup Archive 2004

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Re: Re: Constant function Integrate Assumption

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47250] Re: [mg47218] Re: Constant function Integrate Assumption
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 1 Apr 2004 00:03:36 -0500 (EST)
  • References: <c4bdnf$6tr$1@smc.vnet.net> <200403310757.CAA20968@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 31 Mar 2004, at 16:57, David W. Cantrell wrote:

> dsr14 at Cornell.edu (Daryl Reece) wrote:
>> I am trying to use units within an Integration and would like
>> Mathematica to understand that the units (Month) are not a function of
>> the integration variable, but I have not had any luck with my
>> attempts.
>>
>> Specifically
>>
>> Integrate[UnitStep[t Months], t]
>>
>> should yield
>>
>> t UnitStep[t]
>
> Not exactly. I'll explain in a moment.
>
>> but I get the unevaluated form, because Mathematica does not know how 
>> to
>> assume that Month is not a function of t.
>
> Yes, you get an unevaluated form, but that's not due to Mathematica
> thinking that Month (or Months, whichever) might be a function of t.
> Consider, for example,
>
> In[1]:= Integrate[t Month, t]
>
> Out[1]= (Month t^2)/2
>
> showing that Mathematica assumes, just as you wish, that Month is
> independent of t.
>
>> I've tried Assumptions, SetAttribute, Replace...to no avail.  I've
>> also hunted extensively online (Wolfram, this group and MathGroup)
>> without any results, so any help would be greatly appreciated.
>
> The problem with something like Integrate[UnitStep[t Month], t] seems 
> to
> be that we cannot convince Mathematica that Month is positive unless we
> actually specify a numerical value for it. Note first that
>
> In[2]:= Integrate[UnitStep[Pi t], t]
>
> Out[2]= t UnitStep[t]
>
> works as desired because Mathematica knows that Pi>0. We also see that
>
> In[3]:= a=-3; Integrate[UnitStep[a t], t]
>
> Out[3]= t-t UnitStep[t]
>
> is correct because Mathematica knows that a<0.
>
> Prior to your posting, I would have thought that the following would 
> work:
>
> In[4]:= Clear[a]; Assuming[a>0,Integrate[UnitStep[a t], t]]
>
> Out[4]= Integrate[UnitStep[a t], t]
>
> but we see, alas, that it doesn't work. So can someone suggest a way to
> get Integrate[UnitStep[t Month], t] to evaluate to t UnitStep[t] 
> without
> having to assign a specific positive value to Month?
>
> David Cantrell
>
>

The closest I can get to it is:

Integrate[UnitStep[s Month],{s,-Infinity,t},Assumptions->Month>0]

t UnitStep[Month t]

(Note that since

D[Integrate[UnitStep[s*Month], {s, -Infinity, t}], t]

UnitStep[Month*t]

it is reasonable to view Integrate[UnitStep[s Month],{s,-Infinity,t}] 
as the same as Integrate[UnitStep[t Month], t].)



Andrzej Kozlowski
Chiba, Japan
http://www.mimuw.edu.pl/~akoz/


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