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Rational integral

  1. Aug 6, 2007 #1
    can anyone show me step by step of how to evaluate the integral of
    [x^m/(x^n+a^n)^p](dx) from negative infinity to positive infinity. all i know is that contour integration is required to solve this problem.
  2. jcsd
  3. Aug 7, 2007 #2
    You'll have to use Residue Theorem. In order to do that, consider the complex integral

    [tex]\int_\gamma \frac{z^m}{(z^n+a^n)^p}dz=\int_{\gamma_1}\frac{z^m}{(z^n+a^n)^p}dz +\int_{\gamma_{2}}\frac{z^m}{(z^n+a^n)^p}dz,[/tex]

    where [itex]\gamma_1=(x,0)[/itex] and [itex]\gamma_2=Re^{i\theta}[/itex], [itex]x\in(-R,R)[/itex] and [itex]\theta\in(0,\pi)[/itex] (i.e. [itex]\gamma[/itex] is the semicircle of radius [itex]R[/itex] enclosing the upper semiplamen [itex]\mathbb{H}^+[/itex]). Then

    [tex]\int_\gamma \frac{z^m}{(z^n+a^n)^p}dz=\int_{-R}^R \frac{x^m}{(x^n+a^n)^p}dx+\int_0^\pi \frac{R^m e^{im\theta}}{(R^n e^{in\theta}+a^n)^p}iRe^{i\theta}d\theta.[/tex]

    Now you have to prove that as [itex]R\rightarrow \infty[/itex] the second integral goes to zero, so

    [tex]\left|\int_0^\pi \frac{R^m e^{im\theta}}{(R^n e^{in\theta}+a^n)^p}iRe^{i\theta}d\theta\right|\le \frac{\pi R^{m+1}}{(R^n-a^n)^p}, \qquad R\gg |a| [/tex]

    and therefore [itex]m,n[/itex] and [itex]p[/itex] cannot be arbitrary (doh!). So assuming [itex]m+1<np[/itex], then

    [tex]\int_\gamma \frac{z^m}{(z^n+a^n)^p}dz=\int_{-\infty}^\infty \frac{x^m}{(x^n+a^n)^p}dx.[/tex]

    Now, using the Residue Theorem

    [tex]\int_{-\infty}^\infty \frac{x^m}{(x^n+a^n)^p}dx=\int_\gamma \frac{z^m}{(z^n+a^n)^p}dz=2\pi i \sum_i Res\left(\frac{z^m}{(z^n+a^n)^p},a_i\right),[/tex]

    where the [itex]a_i[/itex] are roots of the equation [itex]z^n+a^n=0[/itex] in [itex]\mathbb{H}^+[/itex].

    It's all straightforward from here.


    I'm also assuming that [itex]m,n,p \in \mathbb{N}[/itex]. If not, then you'll have to pick branches of the logarithm and the problem is a lot different.
    Last edited: Aug 7, 2007
  4. Aug 7, 2007 #3
    AIRAVATA did a good job showing the remainder goes to zero. This happens to be a special case of a more general case.

    Theorem: Consider a complex function [tex]f(z)[/tex] having the following properties:
    1)[tex]f[/tex] is meromorphic in upper half plane having finitely many poles [tex](c_j)\ 1\leq j\leq n[/tex].
    2)[tex]c_j\not \in \mathbb{R}[/tex], i.e. not on real line.
    3)[tex]\lim_{|z|\to \infty}zf(z) = 0[/tex] with [tex]z[/tex] in upper half plane.
    [tex]\mbox{PV}\int_{-\infty}^{\infty} f(x) dx = 2\pi i \sum_{j=1}^n \mbox{res}(f,c_j)[/tex]

    EDIT: Not if [tex]m,n,p\not \in \mathbb{N}[/tex] then the theorem does not apply since it is no longer meromorphic.
  5. Aug 7, 2007 #4

    how would you arrive at your final answer. i understand how you have done it so far but i have no idea how arrive at the final answer which has the gamma function and sin in it. I actually have the final answer right here: http://www.sosmath.com/tables/integral/integ41/integ41.html it is the last integral at the bottom of the page (or #8).
  6. Aug 8, 2007 #5
    Well, I can assure you is not an easy task but it is possible. Although it migh have nothing to do with what I posted before.

    I present to you the beta function:

    [tex]B(\xi,\eta)=\int_0^\infty \frac{u^{\xi-1}}{(1+u)^{\xi+\eta}}du,\qquad \hbox{Re }\xi>0,\quad \hbox{Re }\eta>0,[/tex]

    wich can be writen as*


    Also, the Gamma function satisfies the following identity*

    [tex]\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z}.[/tex]

    So, taking the change of variable [itex]x^n/a^n[/tex] in the Beta function, we have

    [tex]B(\xi,\eta)=na^{n\eta}\int_0^\infty \frac{x^{n\xi-1}}{(a^n+x^n)^{\xi+\eta}}dx.[/tex]

    Then, taking [itex]\xi=(m+1)/n[/itex] and [itex]\eta=p-(m+1)/n[/itex], you have the following relation between your integral and the Beta function:

    [tex]\int_0^\infty \frac{x^m}{(a^n+x^n)^p}dx=\frac{a^{m+1-np}}{n}B\left(\frac{m+1}{n}, p-\frac{m+1}{n}\right),[/tex]

    and I assume is straight sustitution and Gamma function manipulation from here to the answer you want.

    *The proof of this relations can be found in
    N.N. LEBEDEV, Special Functions and their Applications, Dover Publications Inc, New York, 1972.
    Last edited: Aug 8, 2007
  7. Aug 8, 2007 #6
    thaks man i was having trouble with that problem for quite a while now
  8. Aug 8, 2007 #7
    The identity can also be established by computing,
    [tex]\int_{-\infty}^{\infty} \frac{e^{ax}}{e^x+1} dx[/tex] on the rectangle [tex]\pm R \pm \pi i, \pm R[/tex]. And then using a Beta substitution this converts into the Gamma integral.
  9. Aug 8, 2007 #8

    do you have link to access that book as a pdf file?
  10. Aug 9, 2007 #9
    Nope, I have the book right here in my right. There must be one in your local library. There is a lot of information regarding the Gamma and Beta functions in the wikipedia. I'm sure it will be of some help.
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