Prove that $\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}$

Question

Good evening everyone,

how can I prove that

$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}\;?$$

Well, I know that $\displaystyle\frac{1}{x^4+x^2+1} $ is an even function and the interval $(-\infty,+\infty)$ is symmetric about zero, so

$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = 2\int_0^\infty \frac{1}{x^4+x^2+1}dx.$$ Then I use the partial fraction: $$2\int_0^\infty \frac{1}{x^4+x^2+1}dx= 2\int_0^\infty \left( \frac{1-x}{2(x^2-x+1)} + \frac{x+1}{2(x^2+x+1)} \right)dx.$$ So that's all. What's next step?

Answer 1

The next step is to write, for example, $$\frac{x+1}{x^2+x+1} = \frac{1}{2} \cdot \frac{2x+1}{x^2+x+1} + \frac{1}{2} \cdot \frac{1}{x^2+x+1}$$ from which we then have $$\int \frac{x+1}{x^2+x+1} \, dx = \frac{1}{2} \log\left|x^2+x+1\right| + \frac{1}{2} \int \frac{dx}{(x+1/2)^2+(\sqrt{3}/2)^2},$$ and the second integral is, after an appropriate substitution, expressible as an inverse tangent. A analogous approach applies to the other term in your original expression.

Answer 2

$$\int_{\mathbb{R}}\frac{dx}{x^4+x^2+1}=2\int_{0}^{+\infty}\frac{dx}{x^4+x^2+1} = 2\int_{0}^{1}\frac{dx}{x^4+x^2+1}+2\int_{0}^{1}\frac{x^2\,dx}{x^4+x^2+1}$$ so we just have to find: $$ I=2\int_{0}^{1}\frac{1+x^2}{1+x^2+x^4}\,dx = 2\int_{0}^{1}\frac{1-x^4}{1-x^6}\,dx.$$ By expanding the integrand function as a geometric series we have: $$ I = 2\sum_{n=0}^{+\infty}\left(\frac{1}{6n+1}-\frac{1}{6n+5}\right)=2\sum_{n=1}^{+\infty}\frac{\chi(n)}{n} $$ where $\chi(n)$ is the primitive non-principal Dirichlet character $\!\!\pmod{6}$. Since, by the residue theorem:

$$\frac{x^2+1}{x^4+x^2+1}=-\frac{i}{2\sqrt{3}}\left(\frac{1}{x-\omega}+\frac{1}{x-\omega^2}-\frac{1}{x-\omega^4}-\frac{1}{x-\omega^5}\right)$$ where $\omega=\exp\frac{2\pi i}{6}$, it follows that: $$ I = \int_{0}^{1}\left(\frac{1}{1-x+x^2}+\frac{1}{1+x+x^2}\right)\,dx=\frac{2\pi}{3\sqrt{3}}+\frac{\pi}{3\sqrt{3}}=\color{red}{\frac{\pi}{\sqrt{3}}}. $$

As an alternative approach, we can just manipulate the series representation: $$ I = 2\sum_{n\geq 0}\frac{4}{(6n+3)^2-4}=\frac{1}{9}\sum_{n\geq 0}\frac{8}{(2n+1)^2-\frac{4}{9}}\tag{1}$$ through the logarithmic derivative of the Weierstrass product for the cosine function: $$ \cos z = \prod_{n\geq 0}\left(1-\frac{4z^2}{(2n+1)^2 \pi^2}\right), $$ $$ \tan z = \sum_{n\geq 0}\frac{8z}{(2n+1)^2 \pi^2 - 4z^2}$$ $$ \pi\tan(\pi z) = \sum_{n\geq 0}\frac{8z}{(2n+1)^2 - 4z^2}\tag{2}$$ from which it follows that: $$ I = \frac{\pi}{3}\tan\frac{\pi}{3} = \frac{\pi}{\sqrt{3}}.\tag{3} $$

Answer 3

\begin{align} \int_{-\infty}^{\infty} \frac{1}{x^4+x^2+1}\,\mathrm dx&=2\int_{0}^{\infty} \frac{1}{x^4+x^2+1}\,\mathrm dx\\[7pt] &=2\int_{0}^{\infty} \frac{1}{\left(x-\frac{1}{x}\right)^2+3}\cdot\frac{\mathrm dx}{x^2}\\[7pt] &=2\int_{0}^{\infty} \frac{1}{\left(y-\frac{1}{y}\right)^2+3}\,\mathrm dy\tag1\\[7pt] &=\int_{-\infty}^{\infty}\frac{1}{\left(y-\frac{1}{y}\right)^2+3}\,\mathrm dy\\[7pt] &=\int_{-\infty}^{0}\frac{1}{\left(y-\frac{1}{y}\right)^2+3}\,\mathrm dy+\int_{0}^{\infty}\frac{1}{\left(y-\frac{1}{y}\right)^2+3}\,\mathrm dy\\[7pt] &=\int_{-\infty}^{\infty}\frac{e^{z}}{\left(e^{z}-e^{-z}\right)^2+3}\,\mathrm dz+\int_{-\infty}^{\infty}\frac{e^{-z}}{\left(e^{-z}-e^{z}\right)^2+3}\,\mathrm dz\tag2\\[7pt] &=\int_{-\infty}^{\infty}\frac{2\cosh z}{\left(2\sinh z\right)^2+3}\,\mathrm dz\tag3\\[7pt] &=\int_{-\infty}^{\infty}\frac{1}{t^2+3}\,\mathrm dt\tag4\\[7pt] &=\left.\frac{\arctan\left(\frac{t}{\sqrt{3}}\right)}{\sqrt{3}}\right|_{-\infty}^{\infty}\\[7pt] &=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{\pi}{\sqrt{3}}}}\tag{$\color{red}{❤}$} \end{align}

---

Explanation :

$(1)\;$ Use substitution $\;\displaystyle y=\frac{1}{x}\quad\implies\quad \mathrm dy=-\frac{\mathrm dx}{x^2}$

$(2)\;$ Use substitution $\;\displaystyle y=e^{z}\,$ for the left integral and $\;\displaystyle y=e^{-z}\,$ for the right integral

$(3)\;$ Adding both integrals then using the fact that $\;\displaystyle \cosh z=\frac{e^{z}+e^{-z}}{2}\,$ and $\;\displaystyle \sinh z=\frac{e^{z}-e^{-z}}{2}\,$

$(4)\;$ Use substitution $\;\displaystyle t=2\sinh z\quad\implies\quad \mathrm dt=2\cosh z\;\mathrm dz$

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{-\infty}^{\infty}{1 \over x^{4} + x^{2} + 1}\,\dd x ={\pi \over \root{3}}\;?}$

---

\begin{align}&\color{#66f}{\large% \int_{-\infty}^{\infty}{1 \over x^{4} + x^{2} + 1}\,\dd x} =2\int_{0}^{\infty}\ {1 \over x^{2} + 1 + 1/x^{2}}\,{1 \over x^{2}}\,\dd x\ =\ \overbrace{% 2\int_{0}^{\infty}\ {1 \over \pars{x - 1/x}^{2} + 3}\,{1 \over x^{2}}\,\dd x} ^{\dsc{I_{1}}} \\[5mm]&\stackrel{\dsc{x}\ \mapsto\ \dsc{1 \over x}}{=}\ \overbrace{% 2\int_{0}^{\infty}\ {1 \over \pars{x - 1/x}^{2} + 3}\,\dd x} ^{\dsc{I_{2}}} \end{align}

---

Then, \begin{align}&\color{#66f}{\large% \int_{-\infty}^{\infty}{1 \over x^{4} + x^{2} + 1}\,\dd x} ={\dsc{I_{1}} + \dsc{I_{2}} \over 2}\ =\ \overbrace{% \int_{0}^{\infty}\ {1 \over \pars{x - 1/x}^{2} + 3}\pars{{1 \over x^{2}} + 1} \,\dd x}^{\ds{\dsc{x - {1 \over x}}\ \mapsto\ \dsc{x}}} \\[5mm]&=\int_{-\infty}^{\infty}\ {1 \over x^{2} + 3}\,\dd x ={1 \over \root{3}}\ \overbrace{\int_{-\infty}^{\infty}\ {1 \over x^{2} + 1}\,\dd x}^{\ds{=\ \dsc{\pi}}} \ =\ \color{#66f}{\large{\pi \over \root{3}}} \end{align}

Answer 5

$$x^4 + x^2 + 1 = 0 \implies x = \frac{1 \pm i\sqrt{3}}{2}, \frac{-1 \pm i\sqrt{3}}{2}$$

Only consider the two positive roots,

$$a, b = \frac{1 + i\sqrt{3}}{2}, \frac{-1 + i\sqrt{3}}{2}$$

Consider a semi circle contour $C$, with radius $R$ and upper semi-circle $\Gamma$

In other words, $C = \text{line} + \Gamma$

The integral around the whole contour $C$ is given by:

$$\oint_{C} f(z) dz = (2\pi i)(\sum Res)$$

The sum of the residues is as follows: (ask if you dont understand)

$$\sum \text{Res}f(z) = \frac{-i}{2\sqrt{3}} $$

Then,

$$\oint_{C} f(z) dz = (2\pi i)(\sum Res) = (2\pi i)\cdot \frac{-i}{2\sqrt{3}} = \frac{\pi}{\sqrt{3}} $$

But realize that:

$$\oint_{C} f(z) dz = \int_{-R}^{R} f(x) dx + \int_{\Gamma} f(z) dz = \frac{\pi}{\sqrt{3}}$$

Using the M-L estimation lemma

You see that:

$$\left| \int_{\Gamma} \frac{1}{z^4 + z^2 + 1} dz \right| \le \int_{\Gamma} \left| \frac{1}{z^4 + z^2 + 1} \right|$$

You see that:

The point in polar representation is $z = Re^{i\theta}$

$$\left| \frac{1}{z^4 + z^2 + 1} \right| = \left| \frac{1}{R^4e^{4i\theta} + (R^2e^{2i\theta}) + 1} \right| $$

Since $\theta > 0$ we have:

$$\left| \frac{1}{R^4e^{4i\theta} + (R^2e^{2i\theta}) + 1} \right| = \left| \frac{1}{R^4(1) + R^2(1)) + 1} \right| = M$$

The perimeter along the semi-circle is $L(\Gamma) = \frac{1}{2} (2\pi R) = \pi R$

The Ml inequality states:

$$\left| \int_{\Gamma} \frac{1}{z^4 + z^2 + 1} dz \right| \le \int_{\Gamma} \left| \frac{1}{z^4 + z^2 + 1} \right| \le ML(\Gamma) = \frac{\pi R}{R^4(1) + R^2(1)) + 1}$$

$$\lim_{R \to \infty} \frac{\pi R}{R^4(1) + R^2(1)) + 1} = 0$$

Back to:

$$\oint_{C} f(z) dz = \int_{-R}^{R} f(x) dx + \int_{\Gamma} f(z) dz = \frac{\pi}{\sqrt{3}}$$

Take the limit as $R \to \infty$

$$\frac{\pi}{\sqrt{3}} = \int_{-\infty}^{\infty} f(x) dx + \lim_{R \to \infty} \int_{\Gamma} f(z) dz = \int_{-\infty}^{\infty} f(x) dx + 0$$

Thus,

$$\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{\infty} \frac{1}{x^4 + x^2 + 1} dx = \frac{\pi}{\sqrt{3}}$$

Answer 6

Another way to prove is to use series: \begin{eqnarray} \int_{-\infty}^{\infty} \frac{1}{x^4 + x^2 + 1} dx&=&2\int_{0}^{\infty} \frac{1}{x^4 + x^2 + 1} dx\\ &=&2\int_{0}^{1} \frac{1}{x^4 + x^2 + 1} dx+2\int_{1}^{\infty} \frac{1}{x^4 + x^2 + 1} dx\\ &=&2\int_{0}^{1} \frac{1}{x^4 + x^2 + 1} dx+2\int_{0}^{1} \frac{x^2}{x^4 + x^2 + 1} dx\\ &=&2\int_{0}^{1} \frac{1+x^2}{x^4 + x^2 + 1} dx\\ &=&2\int_{0}^{1} \frac{1-x^4}{1-x^6} dx\\ &=&2\int_{0}^{1}\sum_{n=0}^\infty x^{6n}(1-x^4)dx\\ &=&2\sum_{n=0}^\infty\left(\frac{1}{6n+1}-\frac{1}{6n+5}\right)\\ &=&8\sum_{n=0}^\infty\frac{1}{(6n+1)(6n+5)}\\ &=&\frac{1}{9}\sum_{n=-\infty}^\infty\frac{1}{(n+\frac{1}{2})^2+(\frac{i}{3})^2}\\ &=&\frac{1}{9}\frac{\pi\sinh(2\pi b)}{b\left(\cosh(2\pi b)-\cos(2\pi a)\right)}\bigg|_{a=-\frac12,b=\frac i3}\\ &=&\frac{\pi}{\sqrt3}. \end{eqnarray} Here we use a result from this post.