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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 $|x|/2$.

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  • $\begingroup$ Numerically i found $f(m)\sim 2\log(m)+\text{Const}$ where $\text{Const}\approx 3.23$. i see what i can do by hand.. $\endgroup$
    – tired
    Nov 4 '17 at 14:15
  • $\begingroup$ the leading term can be found by some more or less straightforward local analysis around the origin, but i am too busy at the moment to work it out completly $\endgroup$
    – tired
    Nov 4 '17 at 14:49
  • $\begingroup$ ok, and how comes that the following mathematica code Num=Table[{5i, NIntegrate[((1- Cos[5 i x])/(Sqrt[(2-Cos[x])^2-1])),{x,-Pi,Pi}]-2Log[5 i]},{i,5,40}] ListPlot[Num,PlotRange->Full] Gives us a Plot which brutally goes to the aforementioned constant 3.23387? $\endgroup$
    – tired
    Nov 4 '17 at 17:08
  • $\begingroup$ ok, have you tried what happens if $i \in [50,100]$? right, the same behaviour...this is not a real proof, but it gives us very good confidence that $2\log(m)+C+o(1)$ is the correct asymptotic behaviour ... $\endgroup$
    – tired
    Nov 4 '17 at 18:37
  • $\begingroup$ The constant hidden in the $O(1)$ term is $\color{red}{2\gamma+3\log 2}\approx 3.2338728714829$, in agreement with tired's simulation. $\endgroup$ Nov 7 '17 at 21:21
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$$f(m) = \int_{-\pi}^{\pi}\frac{1-\cos(mx)}{\sqrt{(1-\cos x)(3-\cos x)}}\,dx=\int_{-\pi}^{\pi}m F_m(x)\sqrt{\frac{1-\cos x}{3-\cos x}}\,dx $$ 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:

$$\boxed{ f(m) = 2\log m+\underbrace{\color{red}{(2\gamma+3\log 2)}}_{3.2338728714829}+O\left(\frac{1}{m}\right).}$$

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    $\begingroup$ @EzWin: If we replace $\sqrt{\frac{1-\cos x}{3-\cos x}}$ with $\frac{|x|}{2}$ we have an explicit constant which depends on $\gamma,\pi^2$ and $\log 2$. The explicit constant for $\sqrt{\frac{1-\cos x}{3-\cos x}}$ only depends on the previous one and on the integral $\int_{0}^{\pi}\frac{\pi-x}{2}\cdot\frac{d^2}{dx^2}\sqrt{\frac{1-\cos x}{3-\cos x}}\,dx$, which introduces an extra contribution related to $\sqrt{2}$. $\endgroup$ Nov 4 '17 at 20:53
  • $\begingroup$ nice teamwork (+1) $\endgroup$
    – tired
    Nov 5 '17 at 3:00
  • $\begingroup$ @EzWin: I have expanded my answer. The explicit computation of the wanted constant now depends only on the computation of a horrible integral. Edit: not so horrible. $\endgroup$ Nov 7 '17 at 21:12

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