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MathGroup Archive 2002

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Re: Re: partial fraction

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32917] Re: [mg32881] Re: [mg32865] partial fraction
  • From: Murray Eisenberg <murraye at attbi.com>
  • Date: Tue, 19 Feb 2002 02:29:57 -0500 (EST)
  • Organization: Mathematics & Statistics, Univ. of Mass./Amherst
  • References: <200202181022.FAA14858@smc.vnet.net>
  • Reply-to: murray at math.umass.edu
  • Sender: owner-wri-mathgroup at wolfram.com

What's wrong with the following:

   Integrate[1/(1 - x^2)^2, x]
-x/(2*(-1 + x^2)) - Log[-1 + x]/4 + Log[1 + x]/4

Or, if you want to see the partial fractions, using the built-in Apart:

   Apart[1/(1 - x^2)^2]
1/(4*(-1 + x)^2) - 1/(4*(-1 + x)) + 1/(4*(1 + x)^2) + 1/(4*(1 + x))
   Integrate[%, x]
-1/(4*(-1 + x)) - 1/(4*(1 + x)) - Log[-1 + x]/4 +  Log[1 + x]/4

(Or, as you show, then use Map[Integrate[#, x]&, %] to make sure
Mathematica is really doing the integration term-by-term rather than
first somehow simplifying the partial fraction expression.)

BobHanlon at aol.com wrote:
> 
> In a message dated 2/16/02 4:45:52 AM, kyguan at hotmail.com writes:
> 
> >Do you have any ideas on how to integrate the following function by partial
> >
> >fraction?
> >
> >1/(1-x^2)^2 (Sorry it was supposed to be a minus instead of plus)
> >
> >I have found the answer by using 'the integrator'. However, I have no single
> >
> >clue on how the answer were calculated.
> >
> 
> Although it does not appear to be documented, the standard add-on
> package DiscreteMath`RSolve` defines the function PartialFractions.
> 
> Needs["DiscreteMath`RSolve`"];
> 
> f[x_] := 1/(1-x^2)^2;
> 
> pfe1 = PartialFractions[f[x], x]
> 
> 1/(4*(-1 + x)^2) - 1/(4*(-1 + x)) + 1/(4*(1 + x)^2) +
>  1/(4*(1 + x))
> 
> Integrating term-by-term
> 
> Integrate[#,x]& /@ pfe1
> 
> -1/(4*(-1 + x)) - 1/(4*(1 + x)) - Log[-1 + x]/4 +
>  Log[1 + x]/4
> 
> %//Simplify
> 
> ((-2*x)/(-1 + x^2) - Log[-1 + x] + Log[1 + x])/4
> 
> You could also do the partial fraction expansion yourself
> 
> Factor[f[x]]
> 
> 1/((-1 + x)^2*(1 + x)^2)
> 
> The partial fraction is then of the form
> 
> pfe2 = {a, b, c, d}.(1/{x+1, (x+1)^2 , (x-1), (x-1)^2})
> 
> d/(-1 + x)^2 + c/(-1 + x) + b/(1 + x)^2 + a/(1 + x)
> 
> eqn = Thread[
>     (cl =
> 
>           CoefficientList[Numerator[Together[pfe2]],x])==
>       PadRight[
>         CoefficientList[Numerator[f[x]],x],
>         Length[cl]]]
> 
> {a + b - c + d == 1, -a - 2*b - c + 2*d == 0,
> 
>  -a + b + c + d == 0, a + c == 0}
> 
> soln = Solve[eqn, {a,b,c,d}][[1]]
> 
> {a -> 1/4, b -> 1/4, c -> -1/4, d -> 1/4}
> 
> (pfe2 /. soln) == pfe1
> 
> True
> 
> Bob Hanlon
> Chantilly, VA  USA

-- 
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.       
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street
Amherst, MA 01375


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