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Integral of a rational function

  1. Jun 25, 2013 #1
    1. The problem statement, all variables and given/known data

    I came across this integral recently while tutoring:
    ##\displaystyle \frac{1}{5} \int \frac{-x^3+2x^2-3x+4}{x^4-x^3+x^2-x+1}~dx##

    2. Relevant equations

    3. The attempt at a solution

    I'm not sure how to approach this. At first I suspected partial fraction decomposition might be the way to go, but I don't know how to factor that denominator.

    I've checked with WolframAlpha, and I'm getting a result that I haven't come across yet and don't quite understand: http://www.wolframalpha.com/input/?i=Integrate%5B%28-x%5E3%2B2x%5E2-3x%2B4%29%2F%28x%5E4-x%5E3%2Bx%5E2-x%2B1%29%2Cx%5D

    ##\displaystyle \int \cdots=\frac{1}{5}\sum_{\left\{\omega~:~\omega^4-\omega^3+\omega^2-\omega+1=0\right\}}

    \frac{4\ln(x-\omega)-3\omega\ln(x-\omega)+2\omega^2\ln(x-\omega)-\omega^3\ln(x-\omega)}{4\omega^3-3\omega^2+2\omega-1}+C
    ##

    How did WA get that answer? Is there some specific name for the method? (I've tried looking it up, but I can't quite put the question into words.)

    From what I can tell, ##\omega## is an index indicating the roots of the polynomial in the denominator, or something like that.
     
  2. jcsd
  3. Jun 25, 2013 #2

    Bacle2

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    Have you tried using WA to factor the denominator?
     
  4. Jun 26, 2013 #3

    Ray Vickson

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    The ##\omega## will be related to the fifth roots of ##-1##, because the denominator is ##(x^5+1)/(x+1)##. WA likely did a partial-fraction expansion using these roots of the denominator.

    Maple 11 gets the integral as
    -3/10*5^(1/2)*ln(2*x^2-x-5^(1/2)*x+2)+5/(10-2*5^(1/2))^(1/2)*arctan((4*x-1-5^(1/2))/(10-2*5^(1/2))^(1/2))-7/5/(10-2*5^(1/2))^(1/2)*arctan((4*x-1-5^(1/2))/(10-2*5^(1/2))^(1/2))*5^(1/2)+3/10*5^(1/2)*ln(2*x^2-x+5^(1/2)*x+2)+7/5/(10+2*5^(1/2))^(1/2)*arctan((4*x-1+5^(1/2))/(10+2*5^(1/2))^(1/2))*5^(1/2)+5/(10+2*5^(1/2))^(1/2)*arctan((4*x-1+5^(1/2))/(10+2*5^(1/2))^(1/2))

    To parse this properly, you need to know that in ASCII, Maple prints a^b*c to mean (a^b)*c.

    When differentiated, the above produces a result that looks quite different from the original integrand, but when the difference between the two is plotted, it produces a plot of random roundoff noise. Probably, with more work one could demonstrate the equality of the two.
     
  5. Jun 26, 2013 #4

    ehild

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    Note that the numerator multiplied by x+1 is -x^4+x^3-x^2+x+4= - denominator +5, and the denominator multiplied by x+1 is x^5+1. You can find the complex roots of x^5+1=0 and factor it.

    ehild

    Ray beat me by 1 minute :)
     
  6. Jun 26, 2013 #5

    lurflurf

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    The wolfram alpha answer is the same as you would get from partial fractions.
     
  7. Jun 26, 2013 #6

    ehild

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    (x+1)(x^4-x^3+x^2-x+1)=x^5+1.

    X^5+1=0 has got the roots exp(i(2k+1)∏/5) for k=0,1,2,3,4.

    X^5+1 can be factored as

    x^5+1=(x-ei∏/5)(x-ei3∏/5)(x-ei5∏/5)(x-ei7∏/5)(x-ei9∏/5)

    The product of the first and last factor is
    (x-ei∏/5)(x-ei9∏/5)=(x-ei∏/5)(x-e-i∏/5)=x2-2cos(∏/5)+1.

    In the same way, (x-ei3∏/5)(x-ei7∏/5)=x2-2cos(3∏/5)+1.

    It is known (or you can prove ) that cos(∏/5)=(1+√5)/4 and cos(3∏/5)=(1-√5)/4.

    So the denominator in the integrand can be factored as
    (x2-x(1+√5)/2+1)(x2-x(1-√5)/2+1).

    ehild
     
  8. Jun 26, 2013 #7
    The Mathematica answer is a nicely-concise representation of the antiderivative. It's a rootsum: a sum over the roots of a polynomial. This is what I think you should do. Solve (via partial fractions):

    [tex]\int \frac{-x^3+2x^2-3x+4}{(x-a)(x-b)(x-c)(x-d)}dx[/tex]

    and then see if you can formulate your answer in terms of a rootsum.
     
    Last edited: Jun 26, 2013
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