I need to perform the following complex integral analytically, $$ I_n(\nu) := \int_{-\pi}^{+\pi} \frac{\mathrm{d} k}{2\pi} \; \frac{ e^{i k n} }{\nu + cos(k) + i \eta^+} ~, \tag{*} $$ where $i = \sqrt{-1}$, $ n \in \mathbb{N} > 0 $, $\nu \in \mathbb{R}$, $| \nu | > 1$, and $\eta^+ \rightarrow 0^+$. The latter limit should be taken after obtaining the value of the integral.
I have tried to perform the integral by the standard methods used in the calculus of residues; namely, by mapping $I_n$ to a contour integral via a variable change, $$ z := e^{i k} ~, $$ so that $|z| = 1$ and $\mathrm{d} k = \frac{\mathrm{d} z}{i z}$ for $ z \neq 0$, and $$ cos(k) = \frac{e^{i k} + e^{-i k}}{2} = \frac{1}{2} (z + \frac{1}{z}) ~. $$ Then, $$ I_n(\nu) = \frac{1}{2\pi i} \, \oint_{C_1} \frac{\mathrm{d} z}{z} \; \frac{z^n}{\nu + \frac{1}{2}(z + \frac{1}{z}) + i \eta^+} , $$ where $C_1$ is the unit circle centred at the origin, $z = 0$. The result of the integral will be then the residue of the (simple) poles inside the unit circle.
For $|\nu| > 1$, I find the poles inside $C_1$ as $$ z^- = - \nu - \sqrt{\nu^2 - 1} + i\eta^+ \quad \text{for} \quad \nu < 0, \\ z^+ = - \nu + \sqrt{\nu^2 - 1} + i\eta^+ \quad \text{for} \quad \nu > 0 ~, $$ with the residues, $$ \mathrm{Res}(z^-) = \frac{ (-\nu - \sqrt{\nu^2 - 1})^n }{-2 \sqrt{\nu^2 - 1}} , \\ \mathrm{Res}(z^+) = \frac{ (-\nu + \sqrt{\nu^2 - 1})^n }{2 \sqrt{\nu^2 - 1}} ~. $$ Therefore, the integral yields (for $ |\nu| > 1 $) $$ I_n(\nu) = \begin{cases} \frac{ (-\nu - \sqrt{\nu^2 - 1})^n }{- \sqrt{\nu^2 - 1}} &, \nu < 0 \\ \frac{ (-\nu + \sqrt{\nu^2 - 1})^n }{\sqrt{\nu^2 - 1}} &, \nu > 0 \end{cases} ~. \tag{**} $$ The problem is that the asymptotic behaviour does not match what I expect from the original expression (*); that is, $$ I_n(\nu) \sim \frac{1}{\nu} \quad \text{for} \quad |\nu| \gg 1 ~, $$ since this does not conform to the asymptotic behaviour of the results (**), $$ I_n(|\nu| \gg 1) = \begin{cases} -\frac{1}{ |\nu|^{n-1} } &, \nu < 0 \\ (-1)^n \, \frac{1}{ |\nu|^{n-1} } &, \nu > 0 \end{cases} ~, $$ especially for positive $\nu$.
I'd like to understand which part of my calculation or reasoning is incorrect.