How to Prove $\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$

Question

Question:- Prove that $\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$

On factoring the denominator we get,

$\int_{0}^{\infty}\frac {1}{(x^4+x^2+1)(x^4-x^2+1)}dx$

Partial fraction of the integrand contains big terms with their long integral.So i didn't proceed with partial fraction.I'm unable figure out any other method.I think that there might be some other method for evaluation of this definite integral since its value is $\frac{π}{2\sqrt{3}}$.

Does anyone have nice way to solve it!

Answer 1

Since your function is even, your integral is$$\require{cancel}\frac12\int_{-\infty}^\infty\frac{\mathrm dx}{x^8+x^4+1}.$$On the other hand, $x^8+x^4+1=\dfrac{x^{12}-1}{x^4-1}$ and therefore the roots of $x^8+x^4+1$ are the roots of order $12$ of $1$ which are not fourth roots of $1$. Among these, those with imaginary part greater than $0$ are $e^{\pi i/6}$, $e^{\pi i/3}$, $e^{2\pi i/3}$ and $e^{5\pi i/6}$. The residue of $\dfrac1{z^8+z^4+1}$ at these points is, respectively, $-\dfrac1{4\sqrt3}$, $-\dfrac i{4\sqrt3}$, $-\dfrac i{4\sqrt3}$, and $\dfrac1{4\sqrt3}$. Therefore, if $R>1$, then, by the residue theorem\begin{multline}\int_{-R}^R\frac{\mathrm dx}{x^8+x^4+1}+\int_{|z|=R,\ \operatorname{Im}z\geqslant0}\frac{\mathrm dz}{z^8+z^4+1}=\\=2\pi i\left(\cancel{-\frac1{4\sqrt3}}-\frac i{4\sqrt3}-\frac i{4\sqrt3}+\cancel{\frac1{4\sqrt3}}\right)=\frac\pi{\sqrt3}\end{multline}and so, since$$\lim_{R\to\infty}\int_{|z|=R,\ \operatorname{Im}z\geqslant0}\frac{\mathrm dz}{z^8+z^4+1}=0,$$we have\begin{align}\int_{-\infty}^\infty\frac{\mathrm dx}{x^8+x^4+1}&=\lim_{R\to\infty}\int_{-R}^R\frac{\mathrm dx}{x^8+x^4+1}\\&=\frac\pi{\sqrt3}.\end{align}

Answer 2

Let $(a,b,c,d)$ to be the complex roots of $x^8+x^4+1=0$. So, after partial fraction decomposition, the integrand write $$\frac{1}{(a-b) (a-c) (a-d) \left(x^2-a\right)}+\frac{1}{(b-a) (b-c) (b-d) \left(x^2-b\right)}+$$ $$\frac{1}{(c-a) (c-b) (c-d) \left(x^2-c\right)}+\frac{1}{(d-a) (d-b) (d-c) \left(x^2-d\right)}$$ This makes the definite integral to be $$\frac \pi 2\frac{ \left(\sqrt{a} \left(\sqrt{b}+\sqrt{c}+\sqrt{d}\right)^2+a \left(\sqrt{b}+\sqrt{c}+\sqrt{d}\right)+\left(\sqrt{b}+\sqrt{c}\right) \left(\sqrt{b}+\sqrt{d}\right) \left(\sqrt{c}+\sqrt{d}\right)\right)}{ \sqrt{a} \sqrt{b} \sqrt{c} \sqrt{d} \left(\sqrt{a}+\sqrt{b}\right) \left(\sqrt{a}+\sqrt{c}\right) \left(\sqrt{a}+\sqrt{d}\right) \left(\sqrt{b}+\sqrt{c}\right) \left(\sqrt{b}+\sqrt{d}\right) \left(\sqrt{c}+\sqrt{d}\right)}$$ Now, using $$a=\frac{1+i \sqrt{3}}{2}\qquad b=\frac{1-i \sqrt{3}}{2}\qquad c=-\frac{1+i \sqrt{3}}{2}\qquad d=-\frac{1-i \sqrt{3}}{2}$$ leads to the result.

Answer 3

Use $x^8+x^4+1=(x^4+x^2+1)(x^4-x^2+1)$ to partial decompose \begin{align} \int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx &=\frac12\int_{0}^{\infty}\frac {x^2+1}{x^4+x^2+1}dx - \frac12\int_{0}^{\infty}\frac {x^2-1}{x^4-x^2+1}dx\\ &= \frac12\int_{0}^{\infty}\frac {d(1-\frac1x)}{(x-\frac1x)^2+3} - \frac12\int_{0}^{\infty}\frac {d(1+\frac1x)}{(x+\frac1x)^2-3}\\ &=\frac12\cdot \frac{π}{\sqrt{3}}+\frac12\cdot0= \frac{π}{2\sqrt{3}} \end{align}

Answer 4

Taking $x\mapsto \frac{1}{x}$ transforms the integral into $I= \int_0^{\infty} \frac{x^6}{1+x^4+x^8} d x$ and then taking the average of them gives $$ \begin{aligned} I & =\frac{1}{2} \int_0^{\infty} \frac{x^6+1}{x^8+x^4+1} d x \\ & =\frac{1}{2} \int_0^{\infty} \frac{\left(x^2+1\right)\left(x^4-x+1\right)}{\left(x^4+x^2+1\right)\left(x^4-x+1\right)} d x \\ & =\frac{1}{2} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1} d x \\ & =\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+3} \\ & =\frac{1}{2 \sqrt{3}}\left[\tan ^{-1}\left(\frac{\left.x-\frac{1}{x}\right)}{\sqrt{3}}\right)\right]_0^{\infty} \\ & =\frac{\pi}{2 \sqrt{3}} \end{aligned} $$