Incomplete solution of integral

[freddyfish]

The task is to integrate dθ/(k-cosθ) from -pi to pi.

This is the case - I have solved this and also seen the teacher solve this on the board but I am not satisfied with the method and answer. We used the method of resdues.

This is how my instructor would solve it:

F(cosθ,sinθ)=1/(k-cosθ)

F(x,y)=1/(k-x)=P(x,y)/Q(x,y)

Q(x,y)=k-x

x^2+y^2=1 implies Q(x,y)≠0 iff |k| is greater than 1.

-1≤k≤1 implies that Q(x,y)=0 iff x=k.

Indeed this seems legitimate.

He then evaluates the residues for |k| greater than 1 only.

However, when k is not greater than one, the poles lie on the unit circle in the finite/complex plane. In this case the residue method cannot be used and the answer is therefore not complete. What is the solution of this integral without the assumption that |k| is greater than 1?

Thanks

 

[mfb]

Assuming real k: with $|k| \leq 1$, your function is not well-defined everywhere in the integration range, and you need a different way to write the problem.

 

[freddyfish]

Yeah, your assumtion is correct. Well, that is how the problem was stated. All you are given is the integral, which I have presented in words instead of using the integral sign. Thus, how would I write it in a different way to make it possible to solve?

 

[mfb]

It depends on the integral you want to calculate, and the reason why you want to do that.

Something like
$$\lim_{\epsilon \to 0} \left( \int_{-\pi}^{\arccos(k)-\epsilon} \frac{d\theta}{k-cos(\theta)} + \int_{\arccos(k)+\epsilon}^{\pi} \frac{d\theta}{k-cos(\theta)} \right)$$
might be possible to evaluate.

 

[freddyfish]

Ah, of course! That's Cauchy's principal value of the integral. :)