Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$

Let

$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$ $$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$

Prove that $I=J={\pi \over 2\sqrt3}$

Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$

$x=\infty \rightarrow u={\pi\over 2}$, $x=0\rightarrow u=0$

Rewrite $(1)$ as

$$I=\int_{0}^{\infty}{1\over (1+x^2)^2-x^2}dx$$ then

$$\int_{0}^{\pi/2}{\sec^2{u}\over \sec^4{u}-\tan^2{u}}du\tag3$$

Simplified to

$$I=\int_{0}^{\pi/2}{1\over \sec^2{u}-\sin^2{u}}du\tag4$$

Then to

$$I=2\int_{0}^{\pi/2}{1+\cos{2u}\over (2+\sin{2u})(2-\sin{2u})}du\tag5$$

Any hints on what to do next?

Re-edit (Hint from Marco)

$${1\over x^8+x^4+1}={1\over 2}\left({x^2+1\over x^4+x^2+1}-{x^2-1\over x^4-x^2+1}\right)$$

$$M=\int_{0}^{\infty}{x^2+1\over x^4+x^2+1}dx=\int_{0}^{\infty}{x^2\over x^4+x^2+1}dx+\int_{0}^{\infty}{1\over x^4+x^2+1}dx={\pi\over \sqrt3}$$

$$N=\int_{0}^{\infty}{x^2-1\over x^4-x^2+1}dx=0$$

$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx={1\over 2}\left({\pi\over \sqrt3}-0\right)={\pi\over 2\sqrt3}.$$

• Sorry I re-edit for another question. – gymbvghjkgkjkhgfkl Jun 18 '16 at 11:18
• Please don't repurpose questions like that. If you have a new question, you can post it in a new thread. – joriki Jun 18 '16 at 11:30
• Are you aware of the residue theorem? – joriki Jun 18 '16 at 11:32
• Yes I am aware of it and have see users applying them but not sure how to use it. – gymbvghjkgkjkhgfkl Jun 18 '16 at 11:34
• As you've been posting a lot of definite integral questions, I think the residue theorem is one of the first things you should learn to apply. – joriki Jun 18 '16 at 12:07

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{I \equiv \int_{0}^{\infty}{\dd x \over x^{4} + x^{2} + 1}\,,\qquad J \equiv \int_{0}^{\infty}{\dd x \over x^{8} + x^{4} + 1}}$

Note that $\ds{x^{8} + x^{4} + 1 = \pars{x^{4} + x^{2} + 1}\pars{x^{4} - x^{2} + 1}}$ such that \begin{align} I - J & = \int_{0}^{\infty}{x^{4} - x^{2} \over x^{8} + x^{4} + 1}\,\dd x\ \stackrel{x\ \to\ 1/x}{=}\ \int_{\infty}^{0}{1/x^{4} - 1/x^{2} \over 1/x^{8} + 1/x^{4} + 1} \,{\dd x \over -x^{2}} = \int_{0}^{\infty}{x^{2} - x^{4} \over x^{8} + x^{4} + 1}\,\dd x \\[3mm] & = J - I\quad\imp\quad \fbox{$\ds{\quad\color{#f00}{I} = \color{#f00}{J}\quad}$} \end{align}

The problem is reduced to evaluate $\ds{\underline{just\ one}}$ of the above integrals: For example, $\ds{\color{#f00}{I}}$. \begin{align} \color{#f00}{I} & = \int_{0}^{\infty}{\dd x \over x^{4} + x^{2} + 1}\ \stackrel{x\ \to\ 1/x}{=}\ \int_{0}^{\infty}{\dd x \over 1/x^{2} + 1 + x^{2}} = \int_{0}^{\infty}{\dd x \over \pars{x - 1/x}^{2} + 3}\tag{1} \\[3mm] & \mbox{Similarly,} \\[3mm] \color{#f00}{I} & = \int_{0}^{\infty}{\dd x \over x^{4} + x^{2} + 1} = \int_{0}^{\infty}{1 \over x^{2} + 1 + 1/x^{2}}\,{\dd x \over x^{2}} = \int_{0}^{\infty}{1 \over \pars{x - 1/x}^{2} + 3} \,\dd\pars{-\,{ 1\over x}}\tag{2} \end{align}
With $\pars{1}$ and $\pars{2}$: \begin{align} \color{#f00}{I} & = \color{#f00}{J} = \half\int_{x = 0}^{x \to \infty}{1 \over \pars{x - 1/x}^{2} + 3} \,\dd\pars{x - {1 \over x}}\ \stackrel{\pars{x - 1/x}\ \to x}{=}\ \half\int_{-\infty}^{\infty}{\dd x \over x^{2} + 3} \\[3mm] \stackrel{x/\root{3}\ \to\ x}{=}\ &\ {1 \over \root{3}}\ \underbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}}_{\ds{=\ {\pi \over 2}}}\ =\ \color{#f00}{\pi \over 2\root{3}} \end{align}

For $\theta$ an arbitrary constant, we have $$x^4+2x^2\cos2\theta+1=(x^2-2x\sin\theta+1)(x^2+2x\sin\theta+1)\tag1$$ Now, let’s evaluate $$I=\int_0^\infty\frac{1}{x^4+2x^2\cos2\theta+1}\ dx\tag2$$ By making substitution $x\mapsto\frac1x$, then $$I=\int_{0}^\infty\frac{x^2}{x^4+2x^2\cos2\theta+1}\ dx\tag3$$ Adding $(2)$ and $(3)$, we get $$I=\frac12\int_{0}^\infty\frac{1+x^2}{x^4+2x^2\cos2\theta+1}\ dx=\frac14\int_{-\infty}^\infty\frac{1+x^2}{x^4+2x^2\cos2\theta+1}\ dx\tag4$$ Since the integrand is even, an odd function like $2x\sin\theta$ doesn't change the value of the integral, so by using $(1)$ we have \begin{align} I&=\frac14\int_{-\infty}^\infty\frac{x^2+2x\sin\theta+1}{x^4+2x^2\cos2\theta+1}\ dx\\[10pt] &=\frac14\int_{-\infty}^\infty\frac{1}{x^2-2x\sin\theta+1}\ dx\\[10pt] &=\frac14\int_{-\infty}^\infty\frac{1}{(x-\sin\theta)^2+\cos^2\theta}\ dx\\[10pt] &=\frac1{4\cos\theta}\int_{-\infty}^\infty\frac{1}{y^2+1}\ dy\quad\longrightarrow\quad y={x-\sin\theta\over\cos\theta}\\[10pt] &=\frac\pi{4\cos\theta} \end{align} For $\theta=\frac\pi3$, then $$\int_0^\infty\frac{1}{x^4+x^2+1}\ dx=\frac{\pi}{2\sqrt3}$$ as announced.

For $J$ we use $$x^8+2x^4\cos2\theta+1=(x^4-2x^2\sin\theta+1)(x^4-2x^2\sin\theta+1)$$ Then apply the same methods for $$J=\int_0^\infty\frac{1}{x^8+2x^4\cos2\theta+1}\ dx$$

• A little note: the first integral we want to calculate is $$\int_{0}^{\infty}\frac{1}{x^{4}+x^{2}+1}dx$$ not $$\int_{0}^{\infty}\frac{1}{x^{4}+x+1}dx.$$ – Marco Cantarini Jun 19 '16 at 8:02
• +1, very nice! Where it says "Since the integrand is even", that's not really the reason -- it's more like "Since the denominator is even and the integration region is symmetrical, adding an odd function in the numerator doesn't change the integral". – joriki Jun 19 '16 at 10:48

Note that $$\frac{1}{x^{4}+x^{2}+1}=\frac{1}{2}\left(\frac{x+1}{x^{2}+x+1}-\frac{x-1}{x^{2}-x+1}\right)$$ and \begin{align} \int\frac{x+1}{x^{2}+x+1}dx= & \frac{\log\left(x^{2}+x+1\right)}{2}+\frac{1}{2}\int\frac{1}{x^{2}+x+1}dx \\ = & \frac{\log\left(x^{2}+x+1\right)}{2}+\frac{1}{2}\int\frac{1}{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}dx \\ = & \frac{\log\left(x^{2}+x+1\right)}{2}+\frac{2}{3}\int\frac{1}{\left(\frac{2x+1}{\sqrt{3}}\right)^{2}+1}dx \\ = & \frac{\log\left(x^{2}+x+1\right)}{2}+\frac{\arctan\left(\frac{2x+1}{\sqrt{3}}\right)}{\sqrt{3}} \end{align} and in a similar way we can compute the other integral, hence \begin{align} \int_{0}^{\infty}\frac{1}{x^{4}+x^{2}+1}dx= & \frac{1}{2}\left(\log\left(\frac{\sqrt{x^{2}+x+1}}{\sqrt{x^{2}-x+1}}\right)+\frac{\arctan\left(\frac{2x+1}{\sqrt{3}}\right)+\arctan\left(\frac{2x-1}{\sqrt{3}}\right)}{\sqrt{3}}\right)_{0}^{\infty} \\ = & \frac{\pi}{2\sqrt{3}}. \end{align} For the second note that $$\frac{1}{x^{8}+x^{4}+1}=\frac{1}{4}\left(\frac{1}{x^{2}-x+1}+\frac{1}{x^{2}+x+1}\right)+\frac{1}{2}\left(\frac{x^{2}-1}{x^{4}-x^{2}+1}\right)$$ the first two integral are part of the previous calculation, so we have only to analyze $$I=\int_{0}^{\infty}\frac{x^{2}-1}{x^{4}-x^{2}+1}dx$$ and $$I=\int_{0}^{1}\frac{x^{2}-1}{x^{4}-x^{2}+1}dx+\int_{1}^{\infty}\frac{x^{2}-1}{x^{4}-x^{2}+1}dx \tag{1}$$ and note that if we put $x=1/y$ in the second integral of $(1)$ we get $$\int_{1}^{\infty}\frac{x^{2}-1}{x^{4}-x^{2}+1}dx=-\int_{0}^{1}\frac{y^{2}-1}{y^{4}-y^{2}+1}dy$$ then $$I=0.$$

• @ Marco what is the decomposition of ${1\over x^4-x^2+1}$? I can't work it out. – gymbvghjkgkjkhgfkl Jun 20 '16 at 19:37
• @Chinacat Do you mean this? $$\frac{1}{x^{4}-x^{2}+1}=\frac{x-\sqrt{3}}{2\sqrt{3}(-x^{2}+\sqrt{3}x+1)}$$ $$+\frac{x+\sqrt{3}}{2\sqrt{3}(x^{2}+\sqrt{3}x+1)}$$ – Marco Cantarini Jun 20 '16 at 20:02
• Yes thank you @Marco Cantarini it looks much complicated than the other one. – gymbvghjkgkjkhgfkl Jun 20 '16 at 20:06

\begin{align} \int_0^{\infty}\frac{\mathrm dx}{a^2x^4+bx^2+c^2} &= \int_0^{\infty}\frac{\mathrm dx}{\left(ax-\frac{c}{x}\right)^2+b+2ac}\cdot \frac{1}{x^2}\\[9pt] &=\frac{c}{a} \underbrace{\int_0^{\infty}\frac{\mathrm dy}{\left(ay-\frac{c}{y}\right)^2+b+2ac}}_{\large\color{blue}{x=\frac{c}{ay}}}\\[9pt] &=\frac{c}{a} \underbrace{\int_0^{\infty}\frac{\mathrm dy}{y^2+b+2ac}}_{\large\color{blue}{y=z\sqrt{b+2ac}}}\tag{$\spadesuit$}\\[9pt] &=\frac{c}{a\sqrt{b+2ac}} \int_0^{\infty}\frac{\mathrm dz}{z^2+1}\\[9pt] &=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{c\pi}{2a\sqrt{b+2ac}}}} \end{align}

Setting $a=b=c=1$, then

$$I=\int_0^{\infty}\frac{\mathrm dx}{x^4+x^2+1}=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{\pi}{2\sqrt{3}}}}$$

$(\spadesuit)$ Cauchy-Schlomilch transformation for $f(\cdot)$ is a continuous function and $a,c>0$

$$\int_0^\infty f\left(\left(ay-\frac{c}{y}\right)^2\right)\ \mathrm dy=\frac{1}{a}\int_0^\infty f\left(y^2\right)\ \mathrm dy$$

Both integrands are even, so the integrals can be extended to negative infinity. The integrands decay quickly enough for the contour to be closed with a semicircle at infinity in the upper half plane without chaning the integral. Then the integrals can be determined with the residue theorem.

Multiplying the denominator of the first integrand by $x^2-1$ yields $x^6-1$, so there are simple poles at $x=\omega_k=\exp\left(\frac{2\pi\mathrm i k}6\right)$ for $k=1,2,4,5$. The ones for $k=1,2$ lie in the upper half plane, and the corresponding residues are

$$r_k=\lim_{x\to\omega_k}\frac{\left(x^2-1\right)\left(x-\omega_k\right)}{x^6-1}=\lim_{x\to\omega_k}\frac{(x-\omega_0)(x-\omega_3)(x-\omega_k)}{\prod_{j=0}^5(x-\omega_j)}\;,$$

with

$$r_1=\frac1{(\omega_1-\omega_2)(\omega_1-\omega_4)(\omega_1-\omega_5)}=-\frac14-\frac1{4\sqrt3}\mathrm i$$

and

$$r_2=\frac1{(\omega_2-\omega_1)(\omega_2-\omega_4)(\omega_2-\omega_5)}=\frac14-\frac1{4\sqrt3}\mathrm i$$

Thus the contour integral is

$$2\pi\mathrm i\left(\left(-\frac14-\frac1{4\sqrt3}\mathrm i\right)+\left(\frac14-\frac1{4\sqrt3}\mathrm i\right)\right)=\frac\pi{\sqrt3}\;,$$

and your real integral is half of that.

For the second integral, multiplying the denominator by $x^4-1$ yields $x^{12}-1$, so there are simple poles at $\nu_k=\exp\left(\frac{2\pi\mathrm ik}{12}\right)$ for $k=1,2,4,5,7,8,10,11$, of which the ones for $k=1,2,4,5$ lie in the upper half plane. I'll leave it to you to find and add their residues as above.

Solution for lazy people who want to trade as much work as possible for effortless formal reasoning. Let's consider a more general integral of the form:

$$I = \int_0^{\infty}\frac{dx}{\sum_{k=0}^{N}x^{r k}}$$

Summing the geometric series in the denominator yields:

$$I = \int_0^{\infty}\frac{(1-x^r) dx}{1-x^{r(N+1)}}$$

Put $t = x^{r(N+1)}$ and define $s = \frac{1}{r(N+1)}$

$$I =s\int_0^{\infty}\frac{(1-t^{rs})t^{s-1} dt}{1-t}\tag{1}$$

Let's try to evaluate this using the standard integral:

$$\int_0^{\infty}\frac{x^{-p}dx}{1+x}=\frac{\pi}{\sin{\pi p}}\tag{2}$$

for $0<p<1$. Proving this is an easy contour integration exercise, there are lot of explanations on how to do this on this site. To evaluate (1) we need to generalize this to:

$$\int_0^{\infty}\frac{x^{-p}dx}{1+ax}=a^{p-1}\frac{\pi}{\sin{\pi p}}\tag{3}$$

Deriving (3) from (2) involves just a rescaling of the integration variable, but we'll use the r.h.s. of (3) for $a=-1$ when the integral on the l.h.s. is not defined. We'll circumvent this problem using an analytic continuation argument.

Consider generalizing the integral (1) to:

$$I(a) =s\int_0^{\infty}\frac{(1-t^{rs})t^{s-1} dt}{1+at}\tag{4}$$

Then this is well defined for $a = -1$, it is an analytic function of $a$. There is clearly no problem splitting $I(a)$ up as:

$$I(a) =s\int_0^{\infty}\frac{t^{s-1} dt}{1+at} - s\int_0^{\infty}\frac{t^{(r+1)s-1} dt}{1+at}\tag{5}$$

as long as $a\neq 1$. But note that if we substitute the expressions for these integrals using (3) then the resulting expression for $I(a)$ will be analytic in $a$, therefore by the principle of analytic continuation it will yield the correct expression also for $a = -1$. The only thing to be careful about is to use a consistent definition of the fraction power of $a$ for the two integrals.

Using the r.h.s. of (3) for $a = \exp(i\pi)$ yields the result:

$$I = \pi s \left[\cot(\pi s) - \cot(\pi(r+1)s)\right]$$

• +1, nice (but not what I'd call "effortless formal reasoning" :-) – joriki Jun 19 '16 at 10:53

Here is an alternative approch using complex analysis. Since both integrands are even one can start with integrating around the singularities in the positive half-plane using the residue theorem which yields

\begin{align*} \frac{1}{2}\int_{-\infty}^\infty \frac{1}{x^4+x^2+1}\,\mathrm dx &= \pi\mathrm i\left(\operatorname{Res}_{z=\sqrt[3]{-1}}\left(\frac{1}{z^4+z^2+1}\right) + \operatorname{Res}_{z=(-1)^{2/3}}\left(\frac{1}{z^4+z^2+1}\right)\right)\\ &= \pi\mathrm i\left(\frac{1}{4\sqrt[3]{-1}^3+2\sqrt[3]{-1}} + \frac{1}{4((-1)^{2/3})^3+2(-1)^{2/3}}\right)\\ &= \pi\mathrm i\left(-\frac{\mathrm i}{2\sqrt{3}}\right)\\ &= \frac{\pi}{2\sqrt{3}} \end{align*}

and similiarly the second integral \begin{align*} \frac{1}{2}\int_{-\infty}^\infty \frac{1}{x^8+x^4+1}\,\mathrm dx &= \pi\mathrm i\left(\operatorname{Res}_{z=\sqrt[6]{-1}}\left(\frac{1}{z^8+z^4+1}\right) + \operatorname{Res}_{z=\sqrt[3]{-1}}\left(\frac{1}{z^8+z^4+1}\right)\right.\\ &\qquad \left.+ \operatorname{Res}_{z=(-1)^{2/3}}\left(\frac{1}{z^8+z^4+1}\right) + \operatorname{Res}_{z=(-1)^{5/6}}\left(\frac{1}{z^8+z^4+1}\right)\right)\\ &= \pi\mathrm i\left(\frac{1}{8(\sqrt[6]{-1})^7+4(\sqrt[6]{-1})^3} + \frac{1}{8(\sqrt[3]{-1})^7+4(\sqrt[3]{-1})^3}\right.\\ &\qquad \left. + \frac{1}{8((-1)^{2/3})^7+4((-1)^{2/3})^3} +\frac{1}{8((-1)^{5/6})^7+4((-1)^{5/6})^3} \right)\\ &= \pi\mathrm i\left(-\frac{\mathrm i}{2\sqrt{3}}\right)\\ &= \frac{\pi}{2\sqrt{3}} \end{align*}

thus both are equal indeed.