# 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?

• The next step is (usually) to complete the square. – mickep Jan 3 '15 at 20:47
• The integrals as written individually do not converge. You need to use finite intervals and evaluate the limit as the interval endpoint goes to infinity if you are to take this approach. – Ron Gordon Jan 3 '15 at 20:52
• Also, see math.stackexchange.com/questions/776812 for a nice way. (Maybe this is a duplicate of that question?) – mickep Jan 3 '15 at 20:53
• You can isolate the x terms away by forcing it to be derivative of the denominator so that the anti-derivative of terms with x at the numerator is log. The remaining terms at the numerator should contain no x and you ought to complete the square and use arctan as the anti-derivative. – Novice Jan 3 '15 at 20:57
• hint: use that $x^4+x^2+1=(x^2+1)^2-x^2=(x^2-x+1)(x^2+x+1)$ – Dr. Sonnhard Graubner Jan 3 '15 at 21:04

\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 $\;\displaystyle t=2\sinh z\quad\implies\quad \mathrm dt=2\cosh z\;\mathrm dz$

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.

$$\int_{\mathbb{R}}\frac{dx}{x^4+x^2+1}=2\int_{0}^{+\infty}\frac{dx}{x^4+x^2+1} = 2\int_{0}^{1}\frac{dx}{x^4+x^2+1}+2\int_{0}^{1}\frac{x^2\,dx}{x^4+x^2+1}$$ 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: $$I = \frac{\pi}{3}\tan\frac{\pi}{3} = \frac{\pi}{\sqrt{3}}.\tag{3}$$

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

\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}

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

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,

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

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.