Complex Analysis of a trigonometric function integral

[badphysicist]

Homework Statement

Find $$I = \int_0^{2\pi} \frac{1}{cos\phi+b} d\phi$$

Homework Equations

Given above..

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 $$z=e^{i\phi}$$ and $$cos\phi=\frac{e^{i\phi}+e^{-i\phi}}{2}$$ to get $$\frac{1}{i}\oint_C {\frac{1}{z^2+2bz+1}dz}$$ integrated over the unit circle contour.
We now find the poles: $$z_\pm=-b\pm \sqrt{b^2-1}$$.

However, as b goes to infinity, you see that the $$z_-$$ goes to -infinity and $$z_+$$ goes to 0. Obviously, only $$z_+$$ stays inside the unit circle for |b|>1 so the solution is:
$$\frac{2\pi}{\sqrt{b^2-1}}$$

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.

 

[xalvyn]

I think you have to consider the cases $$b>1$$ and $$b<-1$$ separately.

(1) When $$b>1$$ :

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

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

(2) When $$b<-1$$ :

The pole at $$z=-b+ \sqrt{b^2-1}>1$$ lies outside the unit circle, while a similar argument to the above shows that the pole at $$z=-b- \sqrt{b^2-1}$$ 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 $$\left| b \right| \leq 1$$, I think Cauchy's Residue Theorem cannot be applied at all. If, indeed, $$\left| b \right| \leq 1$$, then the roots of the equation $$z^2+2bz+1=0$$ must be a conjugate pair, and since their product is equal to 1, the modulii of each of the roots is 1. The expression $$\frac{1}{z^2+2bz+1}dz}$$ 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.

 

[aljazek]

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!

 

[aljazek]

I have an exam in one week and I really have to know this!

 

[logmarie]

Complex Analysis of Branch Cut

I am trying to figure out the following homework 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?