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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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