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Im trying to solve the following integral using the residual theorem: $$ \int\limits_0^{2\pi} \frac{\mathrm{d}\phi}{3 + 2\cos \phi} $$ Using $2\cos \phi = z - \overline z$ and $\frac{\mathrm{d}z}{\mathrm{d}\varphi} = \mathrm{i} z$ I transformed the integral to $$ \int\limits_{|z|=1} -\frac{\mathrm{i}}{z(3 + z - \overline z)} \mathrm{d}z $$ Since the only pole in the unit circle is at $z=0$ (the others beeing of the form $z=-3/2 + \mathrm{i}b, \ b\in \mathbb{R}$) I concluded using the residual theorem that $$ \int\limits_0^{2\pi} \frac{\mathrm{d}\phi}{3 + 2\cos \phi} = 2\pi \mathrm{i} \lim_{z\to 0} \frac{-\mathrm{i}z}{z(3 + z - \overline z)} = \frac{2\pi}{3} $$

But Wolframalpha says it's $\frac{2\pi}{\sqrt{5}}$ and I don't think it is wrong since it even has a closed form for the undefined integral. So did I miss a pole or what is going on?

user44789
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  • I think Jack's answer below is right. But isn't $z-\overline{z}=2i\sin\phi$ and $z+\overline{z}=2\cos\phi$? – Demosthene May 15 '15 at 14:46
  • With $z = e^{i\phi}$, you have $2\cos\phi = z+\overline{z}$, but to get a holomorphic function, you must write $2\cos\phi = z + z^{-1}$. – Daniel Fischer May 15 '15 at 15:09

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You have to be careful since in order to apply the residue theorem you have to consider holomorphic functions, and $\frac{1}{z(z-\bar{z})}$ is not a holomorphic function. You may try:

$$\begin{eqnarray*}\int_{0}^{2\pi}\frac{d\theta}{3+2\cos\theta}&=&2\int_{0}^{\pi}\frac{d\theta}{3+2\frac{1-\tan^2\frac{\theta}{2}}{\tan^2\frac{\theta}{2}+1}}=4\int_{0}^{\pi/2}\frac{d\phi}{3+2\frac{1-\tan^2\phi}{1+\tan^2\phi}}\\&=&4\int_{0}^{+\infty}\frac{dt}{3(1+t^2)+2(1-t^2)}=2\int_{\mathbb{R}}\frac{dt}{5+t^2}\tag{1}\end{eqnarray*}$$ then: $$\int_{\mathbb{R}}\frac{dt}{5+t^2}=2\pi i\cdot\text{Res}\left(\frac{1}{5+t^2},t=i\sqrt{5}\right) = \frac{\pi}{\sqrt{5}}\tag{2}$$ from which: $$ \int_{0}^{2\pi}\frac{d\theta}{3+2\cos\theta}=\color{red}{\frac{2\pi}{\sqrt{5}}}\tag{3}$$ as claimed by WA.

Jack D'Aurizio
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