Closed form and limit of the integral of a rational function

Question

While trying to answer this question, I wondered whether there could be a way to:

(A) Find the closed form of the generalization of integrals $I$ and $J$, that is

$$I_n=\int_{-\infty}^{+\infty}\frac{dx}{x^{4n}+x^{2n}+1}$$

(B) Prove my own conjecture that

$$\lim_{n \to \infty}I_n = 2$$

Is there a clean way to do these?

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EDIT: I have managed to find a closed form for the poles of the integrand that lie in the upper half-plane, which would be the ones lying within the usual semicircular contour:

$$\omega^+_n(j)=\exp\left(i\frac{\pi}{4n}\big(2j-1-(-1)^n\big)\right) \qquad 1 \leq j \leq 2n, \quad j \in \mathbb{N}$$

My integral would become

$$\def\res{\mathop{\text{Res}}} \begin{align} \oint_\Gamma \frac{dz}{z^{4n}+z^{2n}+1}&=\lim_{R\to\infty} \int_{-R}^{+R}\frac{dx}{x^{4n}+x^{2n}+1}+\lim_{R\to\infty}iR\int_{0}^{\pi}\frac{e^{i\theta}d\theta}{R^{4n}e^{i\theta4n}+R^{2n}e^{i\theta2n}+1} \\ &=2\pi i \sum_{j=1}^{2n} \res_{z=\omega^+_n(j)} \left(\frac{1}{z^{4n}+z^{2n}+1}\right) \end{align}$$ The limit of the integral in $\theta$ is $0$, as one would expect, so my question reduces to:

Is there a way to express $$\sum_{j=1}^{2n}\res_{z=\omega^+_n(j)} \left(\frac{1}{z^{4n}+z^{2n}+1}\right)$$ in terms of $n$ and $j$?

The answers to (A) and (B) above should follow from this in a relatively straightforward way.

Answer 1

This exercise is so much easier when you simply get rid of the $n$'s in the denominator, i.e., through a substitution $x=u^{1/(2 n)}$.

Write

$$I_n = 2 \int_0^{\infty} \frac{dx}{x^{4 n}+x^{2 n}+1} = \frac1n \int_0^{\infty} du \,\frac{ u^{-(2 n-1)/(2 n)}}{u^2+u+1} $$

Now consider the following contour integral

$$\oint_C dz \frac{z^a}{z^2+z+1} $$

where $a=-(2 n-1)/(2 n)$ and $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$ about the positive real axis. Letting $R \to \infty$ and $\epsilon \to 0$, the contour integral is equal to

$$\left (1-e^{i 2 \pi a} \right ) \int_0^{\infty} dx \frac{x^a}{x^2+x+1} $$

By the residue theorem, the contour integral is also equal to

$$i 2 \pi \left (\frac{e^{i 2 \pi a/3}}{i \sqrt{3}} + \frac{e^{i 4 \pi a/3}}{-i \sqrt{3}} \right ) $$

Thus,

$$\int_0^{\infty} dx \frac{x^a}{x^2+x+1} = \frac{2 \pi}{\sqrt{3}} \frac{\displaystyle\sin{\left (\frac{\pi a}{3} \right )}}{\sin{(\pi a)}} = \frac{\displaystyle\pi \sin{\left (\frac{a \pi}{3} \right )}}{\sin{(a \pi)} \sin{\left (\frac{\pi}{3} \right )}} $$

Now, letting $a=1/(2 n)-1$, we find that

$$I_n = \frac{2 \pi}{\sqrt{3}} \frac{\displaystyle \sqrt{3}\cos{\left (\frac{\pi }{6 n} \right )} - \sin{\left (\frac{\pi }{6 n} \right )}}{2 n \sin{\left (\frac{\pi }{2 n} \right )}} $$

and, indeed,

$$\lim_{n \to \infty} I_n = 2$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With $\ds{x \equiv y^{1/6n}}$: \begin{align} \color{#f00}{I_{n}} & = \int_{-\infty}^{\infty}{\dd x \over x^{4n} + x^{2n} + 1} = 2\int_{0}^{\infty}{\dd x \over x^{4n} + x^{2n} + 1} = {1 \over 3n} \int_{0}^{\infty}{y^{1/\pars{6n} - 1} \over y^{2/3} + y^{1/3} + 1}\,\dd y \\[3mm] & = {1 \over 3n} \int_{0}^{\infty}{y^{1/\pars{6n} - 2/3} - y^{1/\pars{6n} - 1}\over y + 1}\,\dd y = {1 \over 3n}\mathrm{f}\pars{{1 \over 6n} - {2 \over 3},{1 \over 6n} - 1} \\[3mm] &\mbox{where}\quad\mathrm{f}\pars{a,b} \equiv \lim_{\Lambda \to\ \infty}\bracks{% \int_{0}^{\Lambda}{y^{a} - 1 \over y - 1}\,\dd y - \int_{0}^{\Lambda}{y^{b} - 1 \over y - 1}\,\dd y} \end{align}

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Note that \begin{align} \int_{0}^{\Lambda}{y^{\mu} - 1 \over y - 1}\,\dd y & = \int_{0}^{1}{y^{\mu} - 1 \over y - 1}\,\dd y + \int_{1}^{\Lambda}{y^{\mu} - 1 \over y - 1}\,\dd y \\[3mm] & = \int_{0}^{1}{1 - y^{\mu} \over 1 - y}\,\dd y + \int_{1}^{1/\Lambda}{y^{-\mu} - 1 \over 1/y - 1}\,\pars{-1 \over y^{2}}\,\dd y \\[3mm] & = \int_{0}^{1}{1 - y^{\mu} \over 1 - y}\,\dd y + \int_{1/\Lambda}^{1}{y^{-\mu - 1} - y^{-1} \over 1 - y}\,\dd y \end{align} Then, \begin{align} \mathrm{f}\pars{a,b} & = \int_{0}^{1}{1 - y^{a} \over 1 - y}\,\dd y - \int_{0}^{1}{1 - y^{b} \over 1 - y}\,\dd y + \int_{0}^{1}{1 - y^{-b - 1} \over 1 - y}\,\dd y - \int_{0}^{1}{1 - y^{-a - 1} \over 1 - y}\,\dd y \\[3mm] & = \Psi\pars{a + 1} - \Psi\pars{b + 1} + \Psi\pars{-b} - \Psi\pars{-a} \\[3mm] & = -\pi\cot\pars{\pi a} + \pi\cot\pars{\pi b} \end{align}

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$$ \color{#f00}{I_{n}} = \color{#f00}{{\pi \over 3n}\braces{\cot\pars{\pi \over 6n} - \cot\pars{{\pi \over 6n} - {2\pi \over 3}}}} \quad\imp\quad\color{#f00}{\lim_{n \to \infty}I_{n}} = \color{#f00}{2} $$

The 'limiting situation' is clearly seen by rewriting $\color{#f00}{I_{n}}$ in the following way: $$ I_{n} = \color{#f00}{\large 2}\ \underbrace{{\pi/\pars{6n} \over \sin\pars{\pi/\bracks{6n}}}}_{\ds{\to 1}}\ \underbrace{\cos\pars{\pi \over 6n}}_{\ds{\to\ 1}}\ -\ \underbrace{{\pi \over 3n}\,\cot\pars{{\pi \over 6n} - {2\pi \over 3}}} _{\ds{\to 0}} $$