MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Re: Constant function Integrate Assumption

  • To: mathgroup at
  • Subject: [mg47250] Re: [mg47218] Re: Constant function Integrate Assumption
  • From: Andrzej Kozlowski <akoz at>
  • Date: Thu, 1 Apr 2004 00:03:36 -0500 (EST)
  • References: <c4bdnf$6tr$> <>
  • Sender: owner-wri-mathgroup at

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

> dsr14 at (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, 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]


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

  • Prev by Date: RE: exporting animations to powerpoint
  • Next by Date: Re: Defining anti-symmetric operation. New ideas requested.
  • Previous by thread: Re: exporting animations to powerpoint
  • Next by thread: RE: Re: Constant function Integrate Assumption