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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)

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

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

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

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

Marco Cantarini
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gymbvghjkgkjkhgfkl
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7 Answers7

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

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

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}
Felix Marin
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For $\theta$ an arbitrary constant, we have \begin{equation} x^4+2x^2\cos2\theta+1=(x^2-2x\sin\theta+1)(x^2+2x\sin\theta+1)\tag1 \end{equation} Now, let’s evaluate \begin{equation} I=\int_0^\infty\frac{1}{x^4+2x^2\cos2\theta+1}\ dx\tag2 \end{equation} By making substitution $x\mapsto\frac1x$, then \begin{equation} I=\int_{0}^\infty\frac{x^2}{x^4+2x^2\cos2\theta+1}\ dx\tag3 \end{equation} Adding $(2)$ and $(3)$, we get \begin{equation} 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 \end{equation} 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 \begin{equation} \int_0^\infty\frac{1}{x^4+x^2+1}\ dx=\frac{\pi}{2\sqrt3} \end{equation} as announced.


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

Sophie Agnesi
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  • 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 08:02
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    +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
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\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 $a,c>0$

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

Venus
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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 Cantarini
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5

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.

joriki
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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:

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

Count Iblis
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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.

Christian Ivicevic
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