1. Not finding help here? Sign up for a free 30min 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!

Evaluate Definite Integral with Complex Analysis

  1. Sep 28, 2014 #1
    1. The problem statement, all variables and given/known data

    [itex] I_1 = \int_0^{2\pi} \frac{sin\theta}{3+2cos\theta} d\theta [/itex]

    2. Relevant equations

    Using identities to change from cos, sin, to variables of z, I get:

    [itex] 2iz^2 + 6iz + 2i[/itex] in my denominator

    3. The attempt at a solution

    Looking for a singularity, will I use a quadratic for this denominator to find my two zeros and then use my positive result to calculate the residue?
     
  2. jcsd
  3. Sep 28, 2014 #2
    Also, is this a pole of second order? Yes, I think it is. But what this means is that I will have to take a derivative of this fraction with polynomials? I have tried to simplify it into products but to no avail.
     
  4. Sep 28, 2014 #3

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You forgot to write ##d\theta## in terms of ##dz##.

    Yes, you need to use the quadratic equation to factor the denominator. All of the poles in this problem are simple poles.
     
  5. Sep 28, 2014 #4
    Thanks Vela, I had that written but didn't include that. So, I found my residue for my singularity at [itex]z_0 = 0[/itex]
    which was [itex]-\frac{1}{2}[/itex]

    But when I find my residue for the [itex]z_0 = \frac{\sqrt5+3}{2}[/itex] (the value I got from my quadratic)
    I got a residue = 0.
    Quadratic being [itex](z^2)+3z+1[/itex]

    What do you think? I see from some examples I've done that this IS possible, but this is homework so I'd like to know what you think?
     
  6. Sep 28, 2014 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Show your calculation of the residue. It shouldn't be 0. The integral, however, turns out to be equal to 0. You can see this by changing the interval of integration to ##-\pi## to ##\pi## and noting the integrand is an odd function.
     
  7. Sep 28, 2014 #6
    Well for starters I am not getting my ([itex]z-z_0[/itex]) to cancel out, should I be using exponents instead of my z identities for the trig functions? The only reason I was able to get my residue at 0 was because [itex]\frac{z-0}{z}[/itex] occurred.
     
  8. Sep 28, 2014 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    From what you just said, it's not clear at all to me what you're doing to evaluate the residue.
     
  9. Sep 28, 2014 #8
    Using [itex] cos\theta = \frac{1}{2}(z+\frac{1}{z}) [/itex] and [itex] sin\theta = -\frac{1}{2i}(z-\frac{1}{z}) [/itex]

    my residue simplifies to [itex]-\frac{(z-z_0)(z^2-1)}{z(6z+2z^2+2)}[/itex]

    At [itex]z_0 = 0[/itex]

    I get

    [itex]Res[f,0] = 1/2[/itex]

    but my other two roots, of the polynomial in the denominator, are giving me problems when trying to compute the residue.

    [itex]\frac{1}{2}(\sqrt5-3)[/itex] and [itex]\frac{1}{2}(-3-\sqrt5)[/itex]

    I think both these points are in my region? it's a full revolution yes?
     
  10. Sep 28, 2014 #9
    I'm trying LHopital
     
  11. Sep 29, 2014 #10
    Seemed to have worked! With my root that is within the circle [itex]
    \frac{1}{2}(\sqrt5-3)
    [/itex] I get [itex]\frac{1}{2}[/itex]

    And so [itex]\frac{1}{2} + (-\frac{1}{2}) = 0 = Res[f,0] + Res[f,\frac{1}{2}(\sqrt5-3)] [/itex]
     
  12. Sep 29, 2014 #11

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I wouldn't call that the residue. You mean the residue for a simple pole at ##z=z_0## is given by
    $$b_1 = \lim_{z \to z_0} -\frac{(z-z_0)(z^2-1)}{z(6z+2z^2+2)}.$$

    I still don't know what problem you ran into.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Evaluate Definite Integral with Complex Analysis
Loading...