I have been given the integral $$\int_0^ {2\pi} \frac{sin^2\theta} {2 - cos\theta} d\theta $$ I have use the substitutions $z=e^{i\theta}$ |$d\theta = \frac{1}{iz}dz$ and a lot of algebra to transform the integral into this $$\frac{-i}{2} \oint \frac{1}{z^2}\frac{(z-1)^2}{z^2-4z+1}dz$$ In order to find the residues i further broke the integral into $$\frac{-i}{2} \oint \frac{1}{z^2}\frac{(z-1)^2}{(z+-r_1)(z-r_2)}dz$$ where $r_1 = 2+\sqrt{3}$ and $r_2 = 2-\sqrt{3}$ giving me three residues at $z=0|z=r_1|z=r_2 $

My question is where do I go from here? Thanks.

  • $\begingroup$ What I got is $$f=\frac{(z^2-1)^2}{2z(z^2-4z+1)}$$. And notice that you dont have to care about $r_1$. As I remember, residue for $f$ at $0$ is $\lim_{z\rightarrow 0} f z$. $\endgroup$
    – k99731
    Feb 19 '16 at 22:37
  • $\begingroup$ Let's say you chose to do a contour integral of that integrand, over a quarter circle in the first quadrant with a radius of $2\pi$. You would split that up into three separate contour integrals, one of which being the integral you're trying to find. You can use residue theorem to evaluate that quarter circle. $\endgroup$
    – Kaynex
    Feb 19 '16 at 22:40
  • $\begingroup$ Do we not care about $r_1$ because it is outside of the unit circle? $\endgroup$ Feb 19 '16 at 22:49

There was an error in the original post. We have

$$\int_0^{2\pi}\frac{\sin^2(\theta)}{2-\cos(\theta)}d\theta=-\frac i2\oint_{|z|=1}\frac{(z^2-1)^2}{z^2(z^2-4z+1)}\,dz$$

There are two poles inside $|z|=1$. The first is a second order pole at $z=0$ and the second is a first order pole at $z=r_2$.

To find the reside of the first pole we use the general expression for the residue of a pole of order $n$

$$\text{Res}\{f(z), z= z_0\}=\frac{1}{(n-1)!}\lim_{z\to z_0}\left(\frac{d^{n-1}}{dz^{n-1}}\left((z-z_0)^nf(z)\right)\right)$$

Here, we have

$$\begin{align} \text{Res}\left(\frac{-i(z^2-1)^2}{2z^2(z^2-4z+1)}, z= 0\right)&=\frac{1}{(2-1)!}\lim_{z\to 0}\left(\frac{d^{2-1}}{dz^{2-1}}\left((z-0)^2\frac{-i(z^2-1)^2}{2z^2(z^2-4z+1)}\right)\right)\\\\ \end{align}$$

To find the residue at $z=r_2$ we have simply

$$\text{Res}\left(\frac{-i(z^2-1)^2}{2z^2(z^2-4z+1)}, z= r_2\right)=\lim_{z\to r_2}\frac{-i(z^2-1)^2}{2z^2(z-r_1)}=-\frac{i}{2}\frac{(r_2^2-1)^2}{r_2^2(r_2-r_1)}$$

The rest is left as an exercise for the reader.


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