1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Evaluation of a tricky integral

  1. Jul 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Verify that [tex]\int_0^{\infty}\frac{x^8-4x^6+9x^4-5x^2+1}{x^{12}-10x^{10}+37x^8-42x^6+26x^4-8x^2+1}dx=\frac{\pi}{2}[/tex]

    2. Relevant equations

    Newton Leibniz formula.

    3. The attempt at a solution

    This is a particulary interesting integral in that one cannot directly use the partial fractions decomposition nor the methods of residues due to the irreducibility of both numerator and denominator. Advanced methods such as Feynman and Schwinger parametrizations also fail. On the other hand, Mathematica has no difficulty in evaluating this integral so I must be missing on something.

    Any help would be greatly appreciated.

  2. jcsd
  3. Jul 17, 2010 #2


    User Avatar
    Homework Helper

    I find that the denominator does factor, although I haven't seen how it will help.
  4. Jul 17, 2010 #3
    Yes, you're right, the denominator factors as the difference of squares of polynomials of the 6th degree. But, as you said - it doesn't really help.
  5. Jul 17, 2010 #4


    User Avatar
    Homework Helper

    "...due to the irreducibility of both numerator and denominator"

    It was this statement that pushed me to look.
  6. Aug 9, 2010 #5
    Playing around with the coefficients of the polynomial in the numerator as well as in the denominator I've found that the key to successful integration in Mathematica lies in the carefully chosen coefficients of the denominator. Is there any way I can use this fact even though I can't simplify / find roots?

    Also, a certain Risch algorithm is said to be helpful in integrals like this one. Can anybody give some info on the subject?

  7. Aug 9, 2010 #6

    Actually I believe you can use residues and Mathematica can represent the poles as "root objects". In Mathematica, these root objects are treated as "exact" symbolic expressions. Note in the code below, I treat them this way right up to the point that I obtain a numeric result. Not sure if I wait long enough with a Simplify if Mathematica can reduce it down to an exact symbolic answer of pi/2 however.

    Code (Text):

    myzeros = x /. Solve[x^12 - 10*x^10 +
           37*x^8 - 42*x^6 + 26*x^4 - 8*x^2 +
           1 == 0, x];

    mypoles = Select[myzeros, Im[#1] > 0 & ];

    myresidues =
       (Residue[(x^8 - 4*x^6 + 9*x^4 -
            5*x^2 + 1)/(x^12 - 10*x^10 +
            37*x^8 - 42*x^6 + 26*x^4 -
            8*x^2 + 1), {x, #1}] & ) /@

    N[Pi*I*Plus @@ myresidues,20]

    Out[82]= 1.5707963267948966192 + 0.*10^-20 I

    Last edited: Aug 9, 2010
  8. Aug 9, 2010 #7
    Yes, I was thinking the same thing - I thought I could express the sum of the residues via Viete formulas from the coefficients of the denominator. So far no success.
  9. Aug 9, 2010 #8
    You can definitely use residue theorem to do this, however, it is very tedious finding all 12 roots of the denominator and then using residues at poles.

    We know that if
    f(z) =
    [tex]\frac{\phi (z)}{(z- z_{0})^{m}}[/tex]

    [tex]Res_{z=z_{0}} f(z)[/tex] = [tex] \phi(z_{0})[/tex] if m =1


    [tex]Res_{z=z_{0}} f(z)[/tex] = [tex]\frac{\phi^{m-1}(z_{0})}{(m-1)!}[/tex] if [tex]m\geq 2[/tex]

    I don't see why this technique won't work the.

    OR we can use the fact that if p and q are analytic at a point [tex]z_{0}[/tex]. if p([tex]z_{0}[/tex]) [tex]\neq[/tex] 0, q([tex]z_{0}[/tex]) = 0 and q'([tex]z_{0}[/tex]) [tex]\neq[/tex] 0 then

    [tex]Res_{z=z_{0}} \frac{p(z)}{q(z)}[/tex] = [tex] \frac{p(z_{0})}{q'(z_{0})}[/tex]

    The amount of work required is too much!
    Last edited: Aug 9, 2010
  10. Aug 9, 2010 #9
    I'm affraid this formula applies only to the simple poles. Also, from the form of the denominator it is not clear that all the poles are simple so it can't be directly used. It just crossed my mind now.
  11. Aug 9, 2010 #10
    My mistake with the second formula... :(

    The first ,however, is applicable. It appears that all poles are simple except two ( according to wolfram alpha).

    Why don't you use the theorem of zero's and poles of orderm to evaluate the integral ?
  12. Aug 10, 2010 #11
    Do you mean the residue theorem? I can't use it explicitly because I can't factor out the denominator to find the poles. Maybe there's another way I'm not aware of?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook