Contour integral $\int_{-\infty}^{\infty}e^{-iax}/(-b+\cos(x))\mathrm dx$ with $a>0$ and $0The integral is 
$$\text{PV}\int_{-\infty}^{\infty}\frac{e^{-iax}}{(-b+\cos(x))}\, dx$$
 with $a>0$ and $0<b<1$.
This integral stems from the Fourier transform of a Green's function in energy-momentum domain to Energy-position domain. The poles in the integrand are the eigenvalues of the system (tight binding model). How to deal with these poles? I know that since $a>0$, so we can choose the contour which goes to the lower half complex plane. But I have no idea how to deal with this kind of poles which are periodic on the real axis. If anybody has any idea, please help. Thanks.
 A: Let $f(x)$ be any function that is periodic with period $L$, so that $f(x+L) = f(x)$. Then the continuous Fourier transform of $f(x)$ is:
\begin{align}
\hat{f}(a)
&= \int_{-\infty}^{+\infty} dx\, f(x)\, e^{-i a x}\\
&= \sum_{n = -\infty}^{+\infty}\int_{nL}^{(n+1)L}dx\, f(x)\, e^{-i a x}\\
&= \int_0^L dx\, f(x)\, e^{-ia x}\sum_{n = -\infty}^{+\infty} e^{-i n a L}\\
&= \sum_{n = -\infty}^{+\infty}\delta\left(a - \frac{2\pi n}{L}\right)
\; \frac{2\pi}{L} \int_0^L dx\, f(x)\, e^{-i\frac{2\pi n x}{L}}\, .
\end{align}
In going from the second to the third line above, I have made a change of variables $x \rightarrow x+nL$, and used the fact that $f(x+nL) = f(x)$. In going from the third to fourth line, I've made use of the $\delta$-function identity
$$
\sum_{n = -\infty}^{+\infty}e^{\pm i n x} = 2\pi\sum_{n = -\infty}^{+\infty}\delta(x - 2\pi n)\, .
$$
For this problem, $L = 2\pi$ and $f(x) = 1/(\cos(x) - b)$, and so the integral in question becomes:
$$
I(a,b) = \sum_{n = -\infty}^{+\infty} C_n(b)\, \delta(a - n)\, ,
$$
where
$$
C_n(b) = PV \int_0^{2\pi}dx\, \frac{e^{-i n x}}{\cos(x) - b}\, .
$$
This expression for the coefficients $C_n(b)$ can be evaluated by contour methods. Make a change of variables $z = e^{-i x}$, so that $dz = -i z\, dx$ and $\cos(x) = (z + 1/z)/2$. Then:
$$
C_n(b) = 2i\, PV\oint dz\, \frac{z^n}{z^2 - 2 b z + 1}
$$
Here the contour runs around the unit circle in the complex plane in the clockwise direction (because we defined $z = e^{-i x}$). The integrand has two poles, located at
$$
z_{\pm} = b \pm i\sqrt{1 - b^2}\, .
$$
Note that $|z_{\pm}| = 1$, so the contour passes through both of these poles. Since this is a principal value integration, we can use standard contour integration techniques and each pole on the contour will contribute half its residue to the integral. (Perhaps a real mathematician, which I am not, can correct me on the generality of that assertion.) Then:
\begin{align}
C_n(b)
&= 2i\times (-2\pi i) \times \frac{1}{2} \times
\left[
\frac{{(z_+)}^n}{z_+ - z_{-}} + 
\frac{{(z_{-})}^n}{z_{-} - z_{+}}
\right]\\
&= \frac{2\pi}{\sqrt{1 - b^2}}\;\mathrm{Im}\left\{{\left(b + i\sqrt{1-b^2}\right)}^n\right\}
\end{align}
Finally, by using the fact that $b + i\sqrt{1-b^2}$ has modulus unity, this becomes:
$$
C_n(b) = \frac{2\pi}{\sqrt{1 - b^2}}\, \sin(n\phi)\, ,
$$
where
$$
\phi = \tan^{-1}\left(\frac{\sqrt{1-b^2}}{b}\right)\, .
$$
Together these expressions for $I(a,b)$ and $C_n(b)$ should answer the question.
