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how evaluate $\int_0^{\pi}\frac{1}{(a+\cos{\theta})^2}, a>1 $, using residues theorem? This problem is an exercise book Complex Analysis of Conway.

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  • $\begingroup$ Is complex analysis an obligation? $\endgroup$ – mvggz Oct 18 '14 at 15:02
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Firstly $\int_0^{\pi}\dfrac{1}{(a+\cos{\theta})^2}d\theta = \dfrac{1}{2}\int_0^{2\pi}\dfrac{1}{(a+\cos{\theta})^2}d\theta$.

Write $\cos \theta = \dfrac{e^{i\theta} + e^{-i\theta}}{2} = \dfrac{z+ z^{-1}}{2}$ with $z = e^{i\theta}$, then

$$\int_0^{2\pi}\dfrac{1}{(a+\cos{\theta})^2}d\theta = \int_{|z|= 1} \dfrac{1}{(a+ \dfrac{z + z^{-1}}{2})^2} \dfrac{dz}{iz} = -4i\int_{|z|= 1} \dfrac{z}{(z^2 + 2az + 1)^2}dz$$

Denote the two roots of $z^2 + 2az + 1 = 0$ by $z_1$ and $z_2$ we have $z_1 = \dfrac{-2a + \sqrt{4a^2 - 4}}{2}$ and $z_2 = \dfrac{-2a - \sqrt{4a^2 - 4}}{2}$ and $z_1z_2 = 1$

Since $a>1$, only $z_1 = \sqrt{a^2 -1 } - a$ is inside $|z|=1$, so the residue of $f(z) = \dfrac{z}{(z^2 + 2az + 1)^2}$ at $z_1$ is the value of $\dfrac{d}{dz}\dfrac{z}{(z-z_2)^2}$ taken at $z = z_1$. Then it's easy to get the result by residue theorem

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  • $\begingroup$ How can the first integral equation came from? $\endgroup$ – 최선웅 May 28 '18 at 1:13

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