Results 1 to 4 of 4

Math Help - Evaluating Improper Integrals

  1. #1
    Member
    Joined
    Jun 2007
    Posts
    131

    Evaluating Improper Integrals

    Hi Again

    I have another question for you that I'm stuck on. I get an answer on ipi*e^(ipi/6) can anyone confirm this?

    Heres the question:
    Evaluating Improper Integrals-q3b.jpg


    Note that theorem 1.1 is as follows:

    Let p and q be polynomial functions such that:
    1. the degree of q exceeds the degree of p by at least 2
    2. any poles of p/q on the non-negative real axis are simple

    Then for 0<a<1,

    Integral (p(t)/q(t))t^a dt = -(pie^-piai cosec pia)S - (picot(pia)T

    Where S is the sum of the residues of the function
    f1(z) = p(z)/q(z) exp(alog2pi(z))

    in C2pi and T is the residues of the function

    f2(z)=p(z)/q(z) exp (alogz)

    on the positive real axis

    Thanks again
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    I am not familar with this theorem.

    Let x=t^{1/3} then,
    Thus,
    \int_0^{\infty} \frac{3x^3}{x^6+1}dx

    Consider a contour -R,+R in a semicircular fashion with f(z) = \frac{3z^3}{z^6+1}. So we get,
    \int_{-R}^R \frac{3x^2}{x^6+1} dx + \int_0^{\pi} \frac{3R^4 e^{3\theta i} i}{R^6 e^{6\theta i}+1} d\theta = 2\pi i \mbox{ RESIDUE }

    Now,
    \left| \frac{3R^4e^{3\theta i}i}{R^6e^{6\theta i}+1} \right| \leq \frac{3R^4|e^{3\theta i}i|}{|R^6e^{6\theta i}| - |1|} \leq \frac{3R^4}{R^6-1}\to 0 \mbox{ as }R\to \infty

    Now you need to find the poles and their residues and add them up to get your result.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by moolimanj View Post
    Hi Again

    I have another question for you that I'm stuck on. I get an answer on ipi*e^(ipi/6) can anyone confirm this?
    My only concern here (I'm not competent to help you with your theorem) is that you are apparently stating that the result of the integral is imaginary? This is (apparently) an integral over a real variable!

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by topsquark View Post
    This is (apparently) an integral over a real variable!
    I am sure the poster means the the real part of the number.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Evaluating integrals
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 29th 2010, 10:16 PM
  2. Evaluating improper integral for constant (C)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 1st 2010, 02:41 AM
  3. Evaluating improper integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 11th 2009, 07:57 AM
  4. Evaluating Improper integral
    Posted in the Calculus Forum
    Replies: 7
    Last Post: December 16th 2008, 09:11 AM
  5. Replies: 2
    Last Post: August 6th 2007, 06:10 AM

Search Tags


/mathhelpforum @mathhelpforum