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MathGroup Archive 2002

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Re: partial fraction

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32881] Re: [mg32865] partial fraction
  • From: BobHanlon at aol.com
  • Date: Mon, 18 Feb 2002 05:22:00 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

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


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