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Homework Help: Complex Analysis of a trigonometric function integral

  1. Jul 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Find [tex]I = \int_0^{2\pi} \frac{1}{cos\phi+b} d\phi[/tex]

    2. Relevant equations
    Given above..

    3. The attempt at a solution
    This problem is an introductory problem to trigonometric functions and here is how the answer is obtained - but I have a question about it. First, here is the solution.

    Use complex analysis. Make the substitution [tex]z=e^{i\phi}[/tex] and [tex]cos\phi=\frac{e^{i\phi}+e^{-i\phi}}{2}[/tex] to get [tex]\frac{1}{i}\oint_C {\frac{1}{z^2+2bz+1}dz}[/tex] integrated over the unit circle contour.
    We now find the poles: [tex]z_\pm=-b\pm \sqrt{b^2-1}[/tex].

    However, as b goes to infinity, you see that the [tex]z_-[/tex] goes to -infinity and [tex]z_+[/tex] goes to 0. Obviously, only [tex]z_+[/tex] stays inside the unit circle for |b|>1 so the solution is:

    This is fine and I have no problem with this. This only applies for |b|>1 though. What about for |b|<=1? Well, obviously the integral diverges if you look back at the original integral. Here's my problem - if you try to use contour integration for |b|<=1, you will have to include the residues of both poles because both poles will be on the unit circle as b decreases. Wouldn't the residues cancel giving an answer of zero? What if b is actually a function and you want to do further integrations afterward with respect to a different variable? Thanks for any help on this.
  2. jcsd
  3. Jul 27, 2008 #2
    I think you have to consider the cases [tex]b>1[/tex] and [tex]b<-1[/tex] separately.

    (1) When [tex]b>1[/tex] :

    The pole at [tex]z=-b- \sqrt{b^2-1}<-1[/tex] lies outside the unit circle, while the pole at [tex]z=-b+ \sqrt{b^2-1}= \frac{1}{-b- \sqrt{b^2-1}}[/tex], since [tex]\left| \frac{1}{-b-\sqrt{b^2-1}} \right|<1[/tex], is enclosed by the unit circle, where we have used the fact that the product of the roots of the equation [tex]z^2+2bz+1=0[/tex] is 1.

    Thus the required pole in this case is [tex]z=-b+ \sqrt{b^2-1}[/tex].

    (2) When [tex]b<-1[/tex] :

    The pole at [tex]z=-b+ \sqrt{b^2-1}>1[/tex] lies outside the unit circle, while a similar argument to the above shows that the pole at [tex]z=-b- \sqrt{b^2-1}[/tex] lies within the unit circle.

    Thus there would be two general solutions, corresponding to the cases b > 1 and b < -1.

    As for extending the integral to the case [tex]\left| b \right| \leq 1[/tex], I think Cauchy's Residue Theorem cannot be applied at all. If, indeed, [tex]\left| b \right| \leq 1[/tex], then the roots of the equation [tex]z^2+2bz+1=0[/tex] must be a conjugate pair, and since their product is equal to 1, the modulii of each of the roots is 1. The expression [tex]\frac{1}{z^2+2bz+1}dz}[/tex] is therefore not analytic on the contour of integration, and so we cannot apply the residue theorem in this case.

    The last question seems a little open-ended. There are probably very many possibilities, and the most suitable methods of integration in each case may vary considerably. Complex integration may not always be the best choice - I guess it all depends on what specific function you are given for b.
  4. Sep 5, 2008 #3
    I would like to know how to solve integral with complex analysis:

    Integrate[Sin[5*x]/Sin[x], {x,0,2*Pi}] !

    I made substitution, get poles +1 and -1, but residue came 0.

    Please help!
    Last edited: Sep 5, 2008
  5. Sep 7, 2008 #4
    I have an exam in one week and I really have to know this!
  6. Oct 13, 2008 #5
    Complex Analysis of Branch Cut

    I am trying to figure out the following hw questions:

    1 a. Say what it means to define a branch cut for log z. Define a branch for log z and give the location of the branch cut. Note make your definition appropriate for part b.

    1b. Find integral [0,inf] {(logx)^4}/{1+x^2 } dx =

    Do I have to divide the countour to 3 different path and integrate from there?
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