Asymptotic expansion of $\int_{-\pi}^{\pi} dk \frac{1-\cos(km)}{\sqrt{(2-\cos(k))^2 -1}}$ I'm looking for asymptotic behaviour of 
$$f(m)=\int_{-\pi}^{\pi} dk \frac{1-\cos(km)}{\sqrt{(2-\cos(k))^2 -1}}$$
as $m\to\infty$. 
UPD:
Looking for constant term (since $f(m) = 2\ln m + O(1)$. Following attempts from the present answer I found "first approximation" of it, getting something like $2(1+\gamma+ln2)$. Is it correct? How to evaluate error from representing $\sqrt{\frac{1-\cos x}{3-\cos x}}$ as $|x|/2$.
 A: $$f(m) = \int_{-\pi}^{\pi}\frac{1-\cos(mx)}{\sqrt{(1-\cos x)(3-\cos x)}}\,dx=\int_{-\pi}^{\pi}m F_m(x)\sqrt{\frac{1-\cos x}{3-\cos x}}\,dx $$
where $F_m(x)$ is Fejér kernel. In particular, assuming $\sqrt{\frac{1-\cos x}{3-\cos x}}$ has the following Fourier cosine series
$$\sqrt{\frac{1-\cos x}{3-\cos x}}=\frac{1}{2}+\sum_{n\geq 1} c_n \cos(nx)$$
$$ mF_m(x) = \sum_{|j|\leq m}(m-|j|)e^{ijx}=m+\sum_{n=1}^{m}2(m-n)\cos(nx) $$
we get
$$ f(m) = \pi m+\pi \sum_{n=1}^{m}2c_n (m-n)=-2\pi m\sum_{n>m}c_n-2\pi\sum_{n=1}^{m}nc_n. $$
The behaviour of the Fourier coefficients $c_n$ depends on the regularity of $\sqrt{\frac{1-\cos x}{3-\cos x}}$ at the origin. Such function is given by $\frac{1}{2}|x|+g(x)$ where $g(x)$ is a function of class $C^2$, hence the first term of the asymptotic behaviour of $f(m)$ can by simply computed by considering the Fourier cosine series of $|x|$, which is well-known. This proves tired's conjecture
$$ f(m) = 2\log(m) +O(1). $$

Let us find the implicit constant in the $O(1)$ term now. It is practical to decompose $\sqrt{\frac{1-\cos x}{3-\cos x}}$ as $\frac{1}{2}|x|+g(x)$ and to recall that over $(-\pi,\pi)$
$$ |x| = \frac{\pi}{2}-\frac{4}{\pi}\sum_{n\geq 0}\frac{\cos((2n+1)x)}{(2n+1)^2} $$
hence
$$\begin{eqnarray*} \int_{-\pi}^{\pi}2m F_{2m}(x)\frac{|x|}{2}\,dx &=& m \pi^2-4\sum_{n=0}^{m-1}\frac{2m-(2n+1)}{(2n+1)^2}\\&=& 2(1+\gamma+2\log 2)+2\log m+O\left(\frac{1}{m^2}\right).\end{eqnarray*}$$
On the other hand
$$ g(x) = \frac{2-\pi}{4}+\sum_{n\geq 1}c_n \cos(nx),\qquad c_n=O\left(\frac{1}{n^2}\right),\sum_{n\geq 1}c_n=\frac{\pi-2}{4} $$
hence
$$\begin{eqnarray*} \int_{-\pi}^{\pi}mF_m(x)g(x)\,dx &=& \frac{2-\pi}{2}\pi m + \pi\sum_{n=1}^{m}2(m-n)c_n\\&=&\underbrace{-2\pi m\sum_{n>m}c_n}_{\to 0}-2\pi\sum_{n=1}^{m}nc_n\end{eqnarray*} $$
where
$$ -2\pi\sum_{n\geq 1}n c_n = 4\int_{0}^{\pi}g''(x)\sum_{n\geq 1}\frac{\cos(nx)}{n}\,dx = 16 \int_{0}^{\pi}\frac{(3+\cos x)\sin^4\frac{x}{2}\log\sin\frac{x}{2}}{(1-\cos x)^{3/2} (3-\cos x)^{5/2}}\,dx$$
gives the explicit correction we have to apply to $2(1+\gamma+2\log 2)$.
By integration by parts and the help of Mathematica we get:
$$\boxed{ f(m) = 2\log m+\underbrace{\color{red}{(2\gamma+3\log 2)}}_{3.2338728714829}+O\left(\frac{1}{m}\right).}$$
