Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2000
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2000

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

Search the Archive

Re: Re: Simple integral wrong

  • To: mathgroup at smc.vnet.net
  • Subject: [mg25165] Re: [mg25073] Re: Simple integral wrong
  • From: David Withoff <withoff at wolfram.com>
  • Date: Tue, 12 Sep 2000 02:58:51 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

> A couple of people told me that
> 
>             Plot[Integrate[Abs[Cos[u]], {u, 0, x  Pi]}], {x, 0, 3}]
> 
> works fine. The result is monotonic increasing as expected.
> 
> But try
> 
>             Plot[Evaluate[Integrate[Abs[Cos[u]],{u,0,Pi*x}]],{x,0,3}]
> 
> and see what happens! The evaluate forces Mathematica to do the
> integral symbolically. It was doing it numerically without the Evaluate.

These examples behave differently because the limits of integration are
numerical in the first example and non-numerical in the second.  All of
the integrals here are done symbolically.  (You could use NIntegrate to
see what happens if the integrals are done numerically.)  With numerical
limits for the integral Mathematica can invoke a difficult but
implementable procedure to correct for the discontinuities of the
corresponding indefinite integral that fall within the range of
integration.  Without numerical values for the limits it is not possible
in general to know which discontinuities fall within the range of integration.

One conceivable approach to this problem would be to add some sort of
symbolic correction for the discontinuities, as suggested in:

> Or just type
> 
>              Integrate[Abs[Cos[u]],{u,0,Pi x}]
>  
> Mathematica 4 returns
>  
>                                    2
> Out[1]= Sqrt[Cos[Pi x] ] Tan[Pi x]
>  
> This plots as a saw-tooth. The true solution should be 
> 
> Sqrt[Cos[Pi x]^2] Tan[Pi x] + 2 Floor[x + 1/2]
> 
> Mathematica misses the step functions necessary to make the solution
> continuous.

One problem with this is that the suggested result isn't correct for all
values of x.   Apparently there is an implicit assumption that x is real.
Even if one makes that assumption (which exposes another set of difficulties)
there is the problem that there is no known computer algorithm for
constructing the necessary correction.  If you try constructing
the necessary correction for a non-trivial examples you can quickly
come to appreciate why no one has worked out a way to construct
such corrections automatically (and are not likely to do so any
time soon).

Examples like this are very common.  If you look up the following
integral in a book, for example, you will probably find a result
similar to the result given by Mathematica:

In[1]:= Integrate[1/(3 + Cos[2 x]), x]

Out[1]= ArcTan[Tan[x]/Sqrt[2]]/(2 Sqrt[2])

This result has discontinuities along the real axis.  Tables of
integrals do not normally include piecewise constant functions to
correct for those discontinuities, and neither does Mathematica.

Changing this to a definite integral with symbolic limits of integration
doesn't really change the situation.  A definite integral with
symbolic limits of integration is still essentially handled as
an indefinite integral:

In[2]:= Integrate[1/(3 + Cos[2 x]), {x, 0, t}]

Out[2]= ArcTan[Tan[t]/Sqrt[2]]/(2 Sqrt[2])

There isn't a single formula that will give a correct result for this
integral for any value of x and for any contour of integration.  If
someone figures out a better approach for handling some subset of
examples like this one, that would probably make a nice addition to
some future version of Mathmematica. 

Dave Withoff
Wolfram Research


  • Prev by Date: Re: Displaying Mixed Numbers
  • Next by Date: Re: Re: Simple integral wrong
  • Previous by thread: Re: Re: Simple integral wrong
  • Next by thread: Re: Re: Simple integral wrong