Evaluate $\lim\limits_{\eta \to +0} \int_{-\infty}^{\infty}dx\frac{\cos^4{x}}{2+\cos{x}}\cdot \frac{\eta}{(x-\pi/3)^2+\eta^2}$ I want to evaluate the value of $\displaystyle\lim_{\eta \to +0} \int_{-\infty}^{\infty}dx\frac{\cos^4{x}}{2+\cos{x}}\cdot \frac{\eta}{(x-\pi/3)^2+\eta^2}$  . But I am not sure how.
Things I noticed:

*

*Looking at the graph, the integrand does not seem to be uniformly convergent.(it has "a needle" around $x=1$.)

*$\int \frac{\eta dx}{(x-\pi/3)^2+\eta^2} = \arctan{\frac{x-\pi /3}{\eta}}+const.$

*$\int \frac{\cos^4{x}}{2+\cos{x}}dx=\frac{1}{12}\left\{-108x+57\sin{x}-6\sin{2x}+\sin{3x}+128\sqrt{3}\arctan{\frac{\tan{\frac{x}{2}}}{\sqrt{3}}}\right\}+const.$ (according to Wolfram Alpha)

*Maybe we can use Fourier transform?(The question set had a Fourier analysis question)

 A: For any $a>0$ we have that $f_a(x)=\frac{a}{x^2+a^2}$ is an even function with an absolute maximum at the origin such that $\int_{-\infty}^{+\infty}f_a(x)\,dx=\pi$. In terms of the Laplace transform
$$(\mathcal{L}^{-1} f_a)(s)=\sin(as)$$
holds, and if we assume that
$$ g(x) \stackrel{L^2}{=} c_0 + \sum_{n\geq 1}\left( c_n \cos(nx) + s_n \sin(nx)\right) $$
has a uniformly convergent Fourier series (which is granted, for instance, by $\max(s_n,c_n)=O\left(\frac{1}{n^2}\right)$) we have
$$\begin{eqnarray*} \int_{-\infty}^{+\infty}g(x)f_a(x)\,dx &=& \int_{0}^{+\infty}(g(x)+g(-x))f_a(x)\,dx\\&=&\pi c_0+2\sum_{n\geq 1}c_n\int_{0}^{+\infty}\cos(nx)\frac{a}{a^2+x^2}\,dx\\
&=&\pi c_0+2\sum_{n\geq 1}c_n \int_{0}^{+\infty}\frac{s}{n^2+s^2}\sin(as)\,ds
\\&=&\pi c_0+2\sum_{n\geq 1}c_n \int_{0}^{+\infty}\frac{s}{1+s^2}\sin(nas)\,ds\\&=&\pi c_0+\pi\sum_{n\geq 1}c_n e^{-na}\end{eqnarray*} $$
by the self-adjointness of the Laplace transform. Since $c_n=O\left(\frac{1}{n^2}\right)$, by the dominated convergence theorem
$$ \lim_{a\to 0^+}\int_{-\infty}^{+\infty}g(x)f_a(x)\,dx = \pi\sum_{n\geq 0}c_n = \pi g(0)$$
and by translation
$$ \lim_{a\to 0^+}\int_{-\infty}^{+\infty}g(x-\pi/3)f_a(x-\pi/3)\,dx = \pi g(\pi/3).\tag{1}$$
In order to apply $(1)$, it only remains to prove that $g(x)=\frac{\cos^4(x+\pi/3)}{2+\cos(x+\pi/3)}$ meets the wanted constraints. In order to do that, it is sufficient to estimate the Fourier coefficients of $\frac{1}{2+\cos(x)}$ or $\frac{1}{2-\cos(x)}$. For any $R>1$ we have
$$ \frac{1}{1-\frac{1}{R}e^{ix}}\cdot \frac{1}{1-\frac{1}{R}e^{-ix}}=\frac{R^2}{(1+R^2)-2R\cos(x)}=\sum_{n\geq 0}\frac{e^{nix}}{R^n}\sum_{m\geq 0}\frac{e^{-mix}}{R^m} \tag{2}$$
hence by picking $R=2+\sqrt{3}$ we have
$$\frac{2+\sqrt{3}}{2}\cdot\frac{1}{2-\cos(x)}=\sum_{n\geq 0}\frac{e^{nix}}{(2+\sqrt{3})^n}\sum_{m\geq 0}\frac{e^{-mix}}{(2+\sqrt{3})^m} \tag{3}$$
and the coefficients of the Fourier series of $\frac{1}{2\pm\cos(x)}$ are given by explicit convolutions. They decay like $\frac{1}{(2+\sqrt{3})^n}$, i.e. way faster than $\frac{1}{n^2}$, and this proves that
$$ \lim_{a\to 0^+}\int_{-\infty}^{+\infty}\frac{\cos^4(x)}{2+\cos(x)}\cdot\frac{a}{(x-\pi/3)^2+a^2}\,dx = \pi\cdot\frac{\cos^4(\pi/3)}{2+\cos(\pi/3)}=\frac{\pi}{40}.\tag{4}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[10px,#ffd]{\lim_{\eta \to 0^{+}}\int_{-\infty}^{\infty}{\cos^{4}\pars{x} \over 2 + \cos\pars{x}}
\,{\eta \over \pars{x - \pi/3}^{2} + \eta^{2}}\,\dd x}
\\[5mm] = &\
\pi\lim_{\eta \to 0^{+}}\int_{-\infty}^{\infty}{\cos^{4}\pars{x} \over
2 + \cos\pars{x}}\,\,
\underbrace{{\eta/\pi \over \pars{x - \pi/3}^{2} + \eta^{2}}}
_{\ds{\begin{array}{l}\delta\pars{x - \pi/3}
\\ \mbox{representation}
\\ \mbox{as}\ \eta \to 0^{+}
\end{array}}}\,\,\dd x
\\[5mm] = &\
\left. \pi\,{\cos^{4}\pars{x} \over 2 + \cos\pars{x}}
\right\vert_{\ x\ =\ \pi/3} = \bbx{\large{\pi \over 40}} \approx 0.0785
\\ &
\end{align}
