Mathematica 9 is now available
Services & Resources / Wolfram Forums
MathGroup Archive
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2006

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

Search the Archive

Re: Re: piecewise integration

  • To: mathgroup at
  • Subject: [mg67054] Re: [mg66999] Re: piecewise integration
  • From: Andrzej Kozlowski <akoz at>
  • Date: Thu, 8 Jun 2006 04:53:33 -0400 (EDT)
  • References: <20060605102611.774$> <> <>
  • Sender: owner-wri-mathgroup at

On 8 Jun 2006, at 07:02, Chris Chiasson wrote:

>> 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
> I think Mathematica's Integrate does this to preserve the identity:
> Integrate[f[x],{x,a,c}]==Integrate[f[x],{x,a,b}]+Integrate[f[x], 
> {x,b,c}]
> There is a danger of violating this when using PiecewiseIntegrate:
> In[1]:=
> <<PiecewiseIntegrate.m
> load[x_]=-9*10^3*DiracDelta[x]-
>    Piecewise[{{x*10*(10^3/3),0\[LessEqual]x\[LessEqual]3}}]-6*10^3*
>      DiracDelta[x-5]
> Integrate[load[x],{x,-10,10}]
> PiecewiseIntegrate[load[x],{x,-10,10}]
> {PiecewiseIntegrate[load[x],{x,-10,0}],PiecewiseIntegrate[load[x], 
> {x,0,5}],
>  PiecewiseIntegrate[load[x],{x,5,10}]}
> Plus@@%
> Out[2]=
> -6000*DiracDelta[-5 + x] - 9000*DiracDelta[x] -
> Piecewise[{{(10000*x)/3, 0 <= x <= 3}}]
> Out[3]=
> -30000
> Out[4]=
> -30000
> Out[5]=
> {-9000,-30000,-6000}
> Out[6]=
> -45000
> -- 


I realised this. However,  I do not consider this identity correct in  
this context. Mathematica often treats generalised functions as if  
they were ordinary functions and this often lead to contradictions,  
or at least unpleasant behaviour. Even if we for the time being  
ignore the mathematical meaning of distributions you can get  
inconsistent results such as this:

{Integrate[DiracDelta[x], {x, -1, 0}],
   Integrate[DiracDelta[x], {x, -1, t}] /. t -> 0}

{1/2, 1}

Compare this with:


{PiecewiseIntegrate[DiracDelta[x], {x, -1, 0}],
   PiecewiseIntegrate[DiracDelta[x], {x, -1, t}] /. t -> 0}


Or looking at it form another angle:

Limit[PiecewiseIntegrate[DiracDelta[x], {x, -1, t}],
   t -> 0, Direction -> 1]


Limit[PiecewiseIntegrate[DiracDelta[x], {x, -1, t}],
   t -> 0, Direction -> -1]


So the Integral returned by Mathematica is not continuous either form  
the left or from the right. The integral returned by  
Piecewiseintegrate is continuous from the right, which is what is the  
usual assumption in mathematical texts. But there is something even  
worse about he Mathematica implementation.

In fact, if you write out the formula for the sum of integrals using  
limits you will see that it holds for PiecwiseIntegrate too.
In fact, generalised functions are not "functions" (they have no  
values at points) but are usually defined as functionals on a certain  
space of functions on real line (there is also another approach in  
terms of non-standard analysis, but that is not implemented in  
Mathematica).  These functionals are defined in terms of integrals  
over the entire real line so formally you cannot integrate the  
DiracDelta over a finite interval; you can only integrate it from - 
Infinity to +Infinity, because that is how the functional is defined.  
Integrals over finite intervals can then be defined as integrals over  
the whole real line of products of DiracDelta and the characteristic  
functions of these intervals. But look what happens when you use  

{Integrate[DiracDelta[x]*Boole[0 <= x <= 2],
    {x, -Infinity, Infinity}], Integrate[DiracDelta[x],
    {x, 0, 2}]}

{1, 1/2}

But these ought to be the same by definition! Again compare this with  

{PiecewiseIntegrate[DiracDelta[x]*Boole[0 <= x <= 2],
    {x, -Infinity, Infinity}], PiecewiseIntegrate[DiracDelta[x],
    {x, 0, 2}]}


 From the mathematical point of view I have no doubt here: Maxim got  
it right and Mathematica got it wrong. Of course one can always say  
that these are only different conventions and as is usual with  
various conventions, once you understand them and get used to them  
you usually find that you can use them equally successfully. Still,  
there are just too many departures form standard mathematics for me  
to be comfortable with the way Mathematica behaves in the above  

Andrzej Kozlowski 

  • Prev by Date: Re: Two questions (1) Sollve and (2) Precision
  • Next by Date: Re: Two questions (1) Sollve and (2) Precision
  • Previous by thread: Re: Re: piecewise integration
  • Next by thread: Re: Re: piecewise integration