Very tricky complex integral, with poles on both sides of the real line,

Question

I am trying to evaluate$$\int_{-\infty}^{\infty} \frac {x^2 -x^4}{1-x^6}\,dx,$$

which is an old exam problem. There is a special note on this problem that reads:

Note:

Your answer need not be a simple expression; it suffices to give the answer as a finite sum of terms of the form $\large \frac{z_1…z_m}{w_1…w_n}$ where the $z_i$'s and $w_i$'s are nonzero.

Some ideas:

Simplifying the numerator, using difference of squares, getting $(x-x^2)(x+x^2)$ doesn't appear to help.

The integrand has 6 simple poles; I have found all of them. The problem? There is a pole at $-1$ and at $+1$ -- on both sides of the real line. This makes choosing a contour problematic. I may need an upper-semicircular contour, and use semi-circular indents to avoid the poles on the real line, but perhaps there is a better approach.

Another idea is to first make a change of variable $y=x^2$. This also doesn't appear to make things any simpler or more insightful.

A last idea is to try to use a wedge contour to minimize computation of residues, i.e., choose a wedge that encloses as few poles as possible. However this wedge will still have to include at least the positive real axis - and then perhaps integrate and use the evenness of the integrand to account for computation along all of $R$. But still the problem remains: there is a pole on the positive real axis.

Any hints, comments or suggestions are welcome.

Thanks,

Answer 1

It turns out that $x^2-1$ divides $x^6-1$, so there aren't actually poles at $\pm 1$. So you really want to evaluate something like $$\int_{\mathbb R} \frac{x^2\, dx}{x^4+x^2+1}.$$

Now you can just sum the residues in the upper half-plane as usual. (They're at the $6$th roots of unity that aren't $\pm1$, so there are two of them you need to get at.) I got $\frac{\pi}{\sqrt{3}}$ as the answer, which a quick WolframAlpha query confirms.

Answer 2

Notice, $$\int_{-\infty}^{\infty}\frac{x^2-x^4}{1-x^6}\ dx=\lim_{b\to \infty}\int_{-b}^{b}\frac{x^2-x^4}{1-x^6}\ dx$$ $$=2\lim_{b\to \infty}\int_{0}^{b}\frac{x^2(1-x^2)}{(1-x^2)(x^4+x^2+1)}\ dx$$ $$=2\lim_{b\to \infty}\int_{0}^{b}\frac{x^2}{x^4+x^2+1}\ dx$$ $$=\lim_{b\to \infty}\int_{0}^{b}\frac{2}{x^2+\frac{1}{x^2}+1}\ dx$$ $$=\lim_{b\to \infty}\int_{0}^{b}\frac{\left(1+\frac{1}{x^2}\right)+\left(1-\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}+1}\ dx$$

$$=\lim_{b\to \infty}\int_{0}^{b}\frac{\left(1+\frac{1}{x^2}\right)\ dx}{\left(x-\frac{1}{x}\right)^2+3}+\lim_{b\to \infty}\int_{0}^{b}\frac{\left(1-\frac{1}{x^2}\right)\ dx}{\left(x+\frac{1}{x}\right)^2-1}$$

$$=\lim_{b\to \infty}\int_{0}^{b}\frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+(\sqrt 3)^2}+\lim_{b\to \infty}\int_{0}^{b}\frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^2-1^2}$$

$$=\lim_{b\to \infty}\frac{1}{\sqrt 3}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt 3}\right)\biggm|_0^{b}+\lim_{b\to \infty}\frac{1}{2}\ln\left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|\biggm|_{0}^b$$ $$=\frac{1}{\sqrt3}(\pi)+0=\color{red}{\frac{\pi}{\sqrt 3}}$$

Answer 3

the two non-removable singularities in the upper half-plane are the simple poles located at $\frac{\pm 1 +i\sqrt{3}}2$ which we may write as $a,a^2$. we require the residues $r(a),r(a^2)$ of $-\frac{x^2-x^4}{x^6-1}=\frac{-x^2(1-x^2)}{\prod_{k=0}^5 (x-a^k)}$

so $$ -r(a) = \frac{a^2(1-a^2)}{(a-1)(a-a^2)(a-a^3)(a-a^4)(a-a^5)} $$ since $a^3=-1$ and $a(a-a^5)=a^2-1$ this gives $-r(a)=\frac1{2(1-a)^2(1-a^2)}$ and $$ -r(a^2) = \frac{a^4(1-a^4)}{(a^2-1)(a^2-a)(a^2-a^3)(a^2-a^4)(a^2-a^5)} \\ =-\frac{1+a^2}{2(1-a)^2(1-a^2)} $$ thus: $$ -r(a)-r(a^2) = \frac{-a^2}{2(1-a)^2(1-a^2)}= \frac12\frac{-a}{(1-a)^2}\frac1{a^{-1}-a}\\ $$ now $a^{-1}-a=\frac{1-i\sqrt{3}}2 -\frac{1+i\sqrt{3}}2 = -i\sqrt{3}$, and $(1-a)^2=1-2a+a^2= (1-a+a^2)-a=-a$ (since $a$ is a primitive sixth root of unity) so $$ 2\pi i\left(r(a)+r(a^2)\right) = \frac{-2\pi i}{-2i\sqrt{3}} \\ =\frac{\pi}{\sqrt{3}} $$

Answer 4

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[2]{\,\mathrm{Li}_{#1}\left(\,{#2}\,\right)} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\color{#f00}{\int_{-\infty}^{\infty}{x^{2} - x^{4} \over 1 - x^{6}}\,\dd x} = 2\int_{0}^{\infty}{x^{2} - x^{4} \over 1 - x^{6}}\,\dd x = 2\int_{0}^{1}{x^{2} - x^{4} \over 1 - x^{6}}\,\dd x + 2\int_{1}^{\infty}{x^{2} - x^{4} \over 1 - x^{6}}\,\dd x \\[3mm] = &\ 2\int_{0}^{1}{x^{2} - x^{4} \over 1 - x^{6}}\,\dd x - 2\int_{1}^{0}{x^{2} - 1 \over x^{6} - 1}\,\dd x = 2\int_{0}^{1}{1 - x^{4} \over 1 - x^{6}}\,\dd x = 2\int_{0}^{1}{1 - x^{2/3} \over 1 - x}\,{1 \over 6}\,x^{-5/6}\,\dd x \\[3mm] = &\ {1 \over 3}\int_{0}^{1}{x^{-5/6} - x^{-1/6} \over 1 - x}\,\dd x = {1 \over 3}\pars{\int_{0}^{1}{1 - x^{-1/6} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{-5/6} \over 1 - x}\,\dd x} \\[3mm] = &\ {1 \over 3}\bracks{\Psi\pars{5 \over 6} - \Psi\pars{1 \over 6}} = {1 \over 3}\,\pi\ \underbrace{\cot\pars{\pi\,{1 \over 6}}}_{\ds{\root{3}}} \end{align}

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$$ \color{#f00}{\int_{-\infty}^{\infty}{x^{2} - x^{4} \over 1 - x^{6}}\,\dd x} = \color{#f00}{\pi \over \root{3}} $$