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Let

$$\beta_n=\frac{\rho\sin\theta}{4+4\rho\cos\theta+\rho^2}+n\theta$$

where $0<\rho<2$. Could anyone give me some hints to prove analytically that

$$\frac1{2\pi\rho^n}\int_{-\pi}^{\pi}\exp\left(\frac{2+\rho\cos\theta}{4+4\rho\cos\theta+\rho^2}\right)\cos\beta_nd\theta$$

does not depend on $\rho$?

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    $\begingroup$ First instinct is take derivative with respect to $\rho$ $\endgroup$ – Simply Beautiful Art Jun 3 '17 at 15:44
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Indeed, notice that the integral can be written by

\begin{align*} \operatorname{Re}\int_{-\pi}^{\pi} \exp\left( \frac{1}{2+\rho e^{i\theta}} \right) e^{-in\theta} \, d\theta &= \rho^n \operatorname{Im}\int_{|z|=\rho} \exp\left( \frac{1}{2+z} \right) \frac{dz}{z^{n+1}} \\ &= 2\pi \rho^n \underset{z=0}{\operatorname{Res}} \left\{ \exp\left( \frac{1}{2+z} \right) \frac{1}{z^{n+1}} \right\}. \end{align*}

Therefore the integral does depend on $\rho$. (Well, it does not depend on $\rho$ if $n = 0$.)

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    $\begingroup$ Thanks. You are right and I've edited my question but your way is not analytically since you used calculus of residues! $\endgroup$ – user451471 Jun 5 '17 at 14:20
  • $\begingroup$ @user451471 I utilized the residue theorem as the integral is screaming out for it, but the role of residue here is not so crucial. Once you reached the first expression in my answer, you can also deduce the same conclusion by expanding the integrand as a Fourier series. If you want more elementary approaches, then I have no idea... $\endgroup$ – Sangchul Lee Jun 5 '17 at 15:10

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