Evaluate $\frac{1}{2\pi}\int _{-\pi}^\pi \frac{e^{-i\varphi}}{1-k\cos (\varphi)} \, \mathrm{d}\varphi$ I couldn't evaluate this integral.  Could you please help me?
$$\frac{1}{2\pi}\int _{-\pi}^\pi \frac{e^{-i\varphi}}{1-k\cos (\varphi)} \, \mathrm{d}\varphi,\text{ where $k$ is a constant}$$
 A: $$|k|<1:$$
$$\begin{align*}\int_{-\pi}^{\pi}\frac{e^{-i\phi}}{1-k\cos \phi}d\phi &=\int_{-\pi}^{\pi}\frac{\cos \phi}{1-k\cos\phi}d\phi-i\underbrace{\int_{-\pi}^{\pi}\frac{\sin\phi}{1-k\cos\phi}d\phi}_{=\,0}\\&=2\int_{-\infty}^{\infty}\frac{1-t^2}{(1+t^2)(1+t^2-k(1-t^2))}dt\\[7pt]&=\frac{2}{k}\left(\int_{-\infty}^{\infty}\frac{1}{(k+1)t^2-k+1}dt-\int_{-\infty}^{\infty}\frac{1}{t^2+1}dt\right)\\[7pt]&=\frac{2}{k}\left(\left[\frac{1}{\sqrt{1-k^2}}\arctan\left(t\cdot\sqrt{\frac{1+k}{1-k}}\right)\right]_{-\infty}^{\infty}-\pi\right)\\[7pt]&=\frac{2\pi}{k}\left(\frac{1}{\sqrt{1-k^2}}-1\right)\end{align*}$$
Hence
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{e^{-i\phi}}{1-k\cos \phi}d\phi=\frac{1}{k}\left(\frac{1}{\sqrt{1-k^2}}-1\right)$$
A: Let $z=e^{-i \phi}$ and use the residue theorem.  The integral becomes, assuming $|k|<1$:
$$-\frac{1}{i 2 \pi} \frac{2}{k}\oint_{|z|=1} dz \frac{z}{z^2-\frac{2}{k}z+1}$$
when you use the fact that $\cos{\phi} = \frac{1}{2}(e^{i \phi}+e^{-i \phi})$.  
This may be evaluated with the residue theorem.  The poles of the integrand are at
$$z_{\pm} = \frac{1}{2} \left ( \frac{1}{k} \pm \sqrt{\frac{1}{k^2}-1}\right )$$
Note that only $z_-$ is inside the unit circle.  By the residue theorem, the integral is equal to 
$$-\frac{2}{k} \lim_{z \rightarrow z_-} \left[(z-z_-) \frac{z}{z^2-\frac{2}{k}z+1}\right]= -\frac{2}{k} \frac{z_-}{z_--z_+} = \frac{1}{k} \left ( \frac{1}{\sqrt{1-k^2}}-1 \right )$$
A: Following up on Yimin's suggestion:
$$
\int_{-\pi}^\pi \frac{e^{-i\varphi}}{1-k\cos\varphi} \, d\varphi = \int_{-\pi}^\pi \frac{\cos\varphi}{1-k\cos\varphi} \, d\varphi  - i\int_{-\pi}^\pi \frac{\sin\varphi}{1-k\cos\varphi} \, d\varphi.
$$
Since the second integral is that of an odd function over an interval symmetric about $0$, it is $0$, and we only need to work on the first integral.
The Weierstrass substitution is
\begin{align}
x & = \tan\frac\varphi2 \tag{1}\\[8pt]
\frac{1-x^2}{1+x^2} & = \cos\varphi \tag{2}\\[8pt]
\frac{2\,dx}{1+x^2} & = d\varphi\tag{3}
\end{align}
How to get $(2)$ and $(3)$ from $(1)$ using trigonometric identities could be discussed here (I wonder if anyone's ever posted that as a question?).  There is also the identity $\dfrac{2x}{1+x^2}=\sin\varphi$, but we won't need that here.
We get
$$
\int_{-\pi}^\pi \frac{\cos\varphi}{1-k\cos\varphi} \, d\varphi = \int_{-\infty}^{\infty}\frac{\frac{1-x^2}{1+x^2}}{1-k\frac{1-x^2}{1+x^2}} \cdot\frac{2\,dx}{1+x^2}
$$
$$
= \int_{-\infty}^\infty \frac{1-x^2}{(1+x^2)-k(1-x^2)}\cdot\frac{2\,dx}{1+x^2}
$$
$$
= \int_{-\infty}^\infty \frac{1-x^2}{(1-k)+(1+k)x^2}\cdot\frac{2\,dx}{1+x^2}
$$
$$
= \int_{-\infty}^\infty \left( \frac{Ax+B}{(1-k)+(1+k)x^2} + \frac{Cx+D}{1+x^2} \right) \, dx
$$
etc.  The antiderivative should involve logarithms and/or arctangents.
