Prove that $\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}$ 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?
 A: 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.
A: $$\int_{\mathbb{R}}\frac{dx}{x^4+x^2+1}=2\int_{0}^{+\infty}\frac{dx}{x^4+x^2+1} = 2\int_{0}^{1}\frac{dx}{x^4+x^2+1}+2\int_{0}^{1}\frac{x^2\,dx}{x^4+x^2+1}$$
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:
$$ I = \frac{\pi}{3}\tan\frac{\pi}{3} = \frac{\pi}{\sqrt{3}}.\tag{3} $$
A: \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 $\;\displaystyle t=2\sinh z\quad\implies\quad \mathrm dt=2\cosh z\;\mathrm dz$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\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}
A: $$x^4 + x^2 + 1 = 0 \implies x = \frac{1 \pm i\sqrt{3}}{2}, \frac{-1 \pm i\sqrt{3}}{2}$$
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,
$$\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{\infty} \frac{1}{x^4 + x^2 + 1} dx = \frac{\pi}{\sqrt{3}}$$
A: 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.
