Evaluate $\lim\limits_{\eta \to +0} \int_{-\infty}^{\infty}dx\frac{\cos^4{x}}{2+\cos{x}}\cdot \frac{\eta}{(x-\pi/3)^2+\eta^2}$

Question

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)

Answer 1

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 $$ \lim_{a\to 0^+}\int_{-\infty}^{+\infty}\frac{\cos^4(x)}{2+\cos(x)}\cdot\frac{a}{(x-\pi/3)^2+a^2}\,dx = \pi\cdot\frac{\cos^4(\pi/3)}{2+\cos(\pi/3)}=\frac{\pi}{40}.\tag{4}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}