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MathGroup Archive 2007

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Re: Re: Definite Integration in Mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74489] Re: [mg74463] Re: Definite Integration in Mathematica
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Fri, 23 Mar 2007 19:02:14 -0500 (EST)
  • References: <etqo3f$10i$1@smc.vnet.net> <200703220612.BAA16861@smc.vnet.net>

Michael Weyrauch wrote:
> Hello,
> 
>   another nice example, where the result for the integral given by Mathematica just
> cannot be right. The indefinite integral of a continuuos function cannot have a jump.

This is simply untrue. 1/x is continuous away from the pole. But any 
antiderivative (a logarithm) of necessity will have a branch cut from 0 
to (complex) infinity. All one might do is control where to put it. 
Mathematica by and large makes no effort to do that, but instead opts to 
use whatever forms of antiderivatives it finds, with their branch cuts 
given by convention, and then to look for crossings on the integration 
path. In future versions (not the next one...) there may be some effort 
to formulate some antiderivatives more carefully so that branch 
crossings are avoided.


> That -- to my opinion -- is mathematical nonsens. We all learned that integration "smoothens",
> i.e. if the integrand is somewhat "ugly" the integral is less "ugly". (Never mind my English!).

This really applies to definite integration, as covered in "functions of 
a real variable" courses. Indefinite integration, that is, finding 
antiderivatives, is more in the realm of "foa complex variable". (I 
realize exceptions are made for e.g. distribution theory.) My point 
being that, in the complex plane, antiderivatives that are continuous 
almost everywhere may sometimes require branch cuts.


> So the result should actually be presented as
>  
> F[x_]=ArcTan[(1 + x)/(4 - x^2)]*UnitStep[2 - x] + 
> (Pi + ArcTan[(1 + x)/(4 - x^2)])*UnitStep[-2 + x]
> 
> This is a perfectly nice function without jumps and  it is the antiderivative of your integrand,
> and the fundamental theorem of calculus works with it...
> 
> So, despite my love for Mathematica, here it fools me....
> 
> Regards   Michael
> 

Not really. What you have is a nonanalytic "antiderivative" for which 
the FTOC will work on the real line. I do not dispute that this can be 
useful, when doing definite integration. Indeed, it might be viewed as a 
way of altering an antiderivative so as to avoid branch crossings. But 
it is not a valid antiderivative in the sense of complex analysis.


Daniel Lichtblau
Wolfram Research



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