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I want to evaluate $\displaystyle \frac{1}{2\pi}\int_{0}^{2\pi} \frac{1}{1-2rcos\theta + r^2} d\theta$ for $0 < r< 1$.

I was thinking or replacing $2cos\theta = (e^{i\theta} + e^{-i\theta}) $ and trying to reduce this to an integral over the unit circle but I get stuck early on.

I would very much appreciate it if anyone would either use my proposed methodology to solve it or maybe come up with a better one if mine does not seem to work.

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See my answer. –  Mhenni Benghorbal Nov 20 '12 at 4:39

2 Answers 2

up vote 1 down vote accepted

Your method seems okay, a lot of times solving these types of trig integrals using complex methods reduces to factoring a polynomial, and thats basically what happens here.

Knowing that $2\cos(\theta) = e^{i\theta} + e^{-i\theta}$ and substituing will make our integral

$$ \frac{1}{2i\pi} \int_{\lvert z\rvert = 1} \frac{1}{z} \frac{1}{1 - rz - \frac{r}{z} + r^2}\mathrm{d}z =\frac{1}{2i\pi} \int_{\lvert z \rvert =1} \frac{1}{z - rz^2 - r + r^2z}\mathrm{d}z$$$$ = \frac{1}{2i\pi} \int_{\lvert z \rvert =1} \frac{1}{(r-z)(rz - 1)}\mathrm{d}z$$

Now we can evaluate this using Cauchy's integral formula, because since $0 < r < 1$, there is only one singularity in our domain. So write this in a revealing way to exploit that fact $$ = \frac{1}{2 i \pi} \int_{\lvert z \rvert = 1} \frac{(1 - rz)^{-1}}{(z-r)}\mathrm{d}z$$ So using Cauchy's integral formula we get $$\frac{1}{2\pi}\int_0^{2\pi}\frac{1}{1 - 2r\cos(\theta) + r^2} = \frac{1}{1-r^2}$$

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$$z=e^{i\theta}\,\,,\,\theta\in [0,2\pi]\Longrightarrow -\frac{i\,dz}{z}=d\theta\;\;,\;\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}\Longrightarrow$$

$$\frac{1}{2\pi}\int_0^{2\pi}\frac{d\theta}{1-2r\cos\theta+r^2}=-\frac{i}{2\pi }\oint_{S^1}\frac{dz}{z\left(1-r\left(z+\frac{1}{z}\right)+r^2\right)}=$$

$$=-\frac{i}{2\pi }\oint_{S^1}\frac{dz}{\left(z-rz^2-r+zr^2\right)}=\frac{i}{2\pi r}\oint_{S^1}\frac{dz}{(z-r)\left(z-\frac{1}{r}\right)}=$$

$$=\frac{i}{2\pi r}\oint_{S^1}\frac{\left(z-\frac{1}{r}\right)^{-1}dz}{z-r}\stackrel{\text{Cauchy's Formula}}=-\frac{1}{r}\left.\frac{1}{z-\frac{1}{r}}\right|_{z=r}=\frac{1}{1-r^2}$$

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