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Is this integral possible to solve

  1. Aug 18, 2011 #1
    Hi

    I was wondering if anyone has seen this integral in a table, or indeed knows if it is possible to solve:

    [tex]
    \int^{\infty}_{-\infty} \frac{x^{2}}{ax^{4} + bx^{2} + c}
    [/tex]

    every table I look at seems to only go up to the first power of x in the numerator

    Thanks,
    Thrillhouse
     
  2. jcsd
  3. Aug 18, 2011 #2

    micromass

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    Your integral can be solved by a classical residue argument.

    If we assume that [itex]ax^4+bx^2+c[/itex] has no real roots, then the integral equals

    [tex]2 \pi i \sum{ Res(P/Q,z_k)}[/tex]

    where the [itex]z_k[/itex] are all the roots of Q in the upper half plane.
     
  4. Aug 18, 2011 #3
    Suppose it does have real roots. Can't we just go around them:

    [tex]\int_{-\infty}^{\infty}\frac{x^2}{-3x^4+2x^2+3}dx=-\frac{\pi}{2}\sqrt{\frac{1}{30}(\sqrt{10}-1)}[/tex]

    Or am I being a trouble-maker?
     
  5. Aug 18, 2011 #4
    Thanks for the help guys - can you briefly explain (or point me towards) why real roots are a problem ? is it something to do with branch points in the complex plane ?
     
  6. Aug 18, 2011 #5

    hunt_mat

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    Real roots are not a problem in the slightest, if the roots of the quadratic are real then it is possible to write the integral as:
    [tex]
    \int_{-\infty}^{\infty}\frac{x^{2}}{ax^{4}+bx^{2}+c}dx= \frac{1}{a}\int_{-\infty}^{\infty}\frac{x^{2}}{(x^{2}+\alpha )^{2}-\beta^{2}}= \frac{1}{a}\int_{-\infty}^{\infty}\frac{x^{2}}{(x^{2}+\alpha +\beta )(x^{2}+\alpha -\beta )}dx
    [/tex]
    You can now use standard partial fraction techniques to reduce it into a more manageable integral.

    The real roots will become a "problem" because the semi-circle used goes through the poles in question. To get around this you have a semi-circular indent of the contour (so the pole is now outside the contour). Now you have to look at the estimates of the complex integrand a little more carefully but it is not that hard to do, this technique is in all the standard textbooks on complex analysis.
     
  7. Aug 19, 2011 #6
    No, not branch-points, just singular points where the function becomes unbounded right since the denominator is going to zero. You can't directly integrate through those points. I was a little misleading in my post above. I conveniently neglected to explicitly state that I was taking the "Cauchy Principal Value" integral and really should have written it as:

    [tex]\text{P.V.}\int_{-\infty}^{\infty}\frac{x^2}{-3x^4+2x^2+3}dx=-\frac{\pi}{2}\sqrt{\frac{1}{30}(\sqrt{10}-1)}[/tex]

    In that case, rather than integrate it directly through the singular points, we take a simultaneous limit of the value of the intgral on either side of each singluar point or use techinques of Complex Analysis to "go around" them. If the limit exists, then the integral converges in the Cauchy principal sense. Like for example:

    [tex]\int_{-a}^a \frac{1}{x} dx[/tex]

    does not converge in the regular Riemann sense but:

    [tex]\text{P.V.} \int_{-a}^a \frac{1}{x} dx=0[/tex]

    Now here's a question: Does the principal-valued integral always exist no matter what the values of a, b, and c are? Because sometimes that coupled-limit across the singular point diverges. Does it ever in this case?
     
    Last edited: Aug 19, 2011
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