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$$\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$$

I have difficulty to evaluating above integrals.

First I try the substitution $x^4 =t$ or $x^4 +x^2+1 =t$ but it makes integral worse. Using Mathematica I found the result $\dfrac{\pi}{48\sqrt{3}}$ I want to know the procedure of evaluating this integral.

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    $\begingroup$ Are you familiar with methods from complex analysis? Residues and such? $\endgroup$ May 20, 2015 at 6:04
  • $\begingroup$ @Jyki Lahtonen, Okay i will try Residue integration $\endgroup$
    – phy_math
    May 20, 2015 at 6:13
  • $\begingroup$ @phy_math See my answer below avoiding complex integration and avoiding a lenghty computation with partial decompositions. $\endgroup$ May 20, 2015 at 21:33
  • $\begingroup$ @Olivier Oloa, Wow great! Is there any books or papers related to such a nice integral method? $\endgroup$
    – phy_math
    May 21, 2015 at 5:26

5 Answers 5

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Here is an approach.

You may write

$$\begin{align} \int_0^{\infty}\frac{x^4}{\left(x^4+x^2+1\right)^3}dx &=\int_0^{\infty}\frac{x^4}{\left(x^2+\dfrac1{x^2}+1\right)^3\,x^6}dx\\\\ &=\int_0^{\infty}\frac{1}{\left(x^2+\dfrac1{x^2}+1\right)^3}\frac{dx}{x^2} \\\\ &=\int_0^{\infty}\frac{1}{\left(x^2+\dfrac1{x^2}+1\right)^3}dx\\\\ &=\frac12\int_0^{\infty}\frac{1}{\left(x^2+\dfrac1{x^2}+1\right)^3}\left(1+\dfrac1{x^2}\right)dx\\\\ &=\frac12\int_0^{\infty}\frac{1}{\left(\left(x-\dfrac1{x}\right)^2+3\right)^3}d\left(x-\dfrac1{x}\right)\\\\ &=\frac12\int_{-\infty}^{+\infty}\frac{1}{\left(u^2+3\right)^3}du\\\\ &=\frac14\:\partial_a^2\left(\left.\int_{-\infty}^{+\infty}\frac{1}{\left(u^2+a\right)}du\right)\right|_{a=3}\\\\ &=\frac14\:\partial_a^2\left.\left(\frac{\pi}{\sqrt{a}}\right)\right|_{a=3}\\\\ &=\color{blue}{\frac{\pi }{48 \sqrt{3}}} \end{align}$$

as desired.

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    $\begingroup$ thanks for the "artful" approach! +1 $\endgroup$
    – Math-fun
    May 20, 2015 at 21:33
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    $\begingroup$ This is a very elegant solution indeed ! $\endgroup$ May 21, 2015 at 6:16
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For the antiderivative, you could also "simplify" the problem using partial fraction decomposition since $$\frac{x^4}{\left(x^4+x^2+1\right)^3}=-\frac{3 (x-1)}{16 \left(x^2-x+1\right)}+\frac{3 (x+1)}{16 \left(x^2+x+1\right)}-\frac{3}{16 \left(x^2-x+1\right)^2}-\frac{3}{16 \left(x^2+x+1\right)^2}+\frac{x}{8 \left(x^2-x+1\right)^3}-\frac{x}{8 \left(x^2+x+1\right)^3}$$ which leads to the result. But, I suspect that using residues will make the problem easier.

Could you prove that $$\int_0^1 \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{1}{288} \left(-28+\sqrt{3} \pi +27 \log (3)\right)$$

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  • $\begingroup$ Plug in Mathematica, it gives $\int_0^1 \frac{x^4}{(x^4+x^2+1)^3}dx =\frac{1}{288} \left(-28+\sqrt{3} \pi +27 \text{Log}[3]\right)$ which you mention above. $\endgroup$
    – phy_math
    May 20, 2015 at 15:44
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    $\begingroup$ And without Mathematica ? Cheers :-) $\endgroup$ May 20, 2015 at 17:55
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A yet another approach:

\begin{align} I&=\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx\\ &=\int_0^1 \frac{x^4}{(x^4+ x^2 +1)^3} dx+\int_1^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx\\ &=\int_0^1\frac{x^4}{(x^4+ x^2 +1)^3} dx+\int_0^1\frac{x^6}{(x^4+ x^2 +1)^3} dx\\ \end{align} where I used $x \to 1/x$.

Now use $x\to \tan x$ to obtain \begin{align} I&=\int_0^{\pi/4} \frac{32\sin^4 2x}{(7+\cos 4x)^3} dx \end{align} which is manageable.

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  • $\begingroup$ Math-fun, Wow, it's great!. Oh i see, now i can handle the integral supposed by Claude Leibovic $\endgroup$
    – phy_math
    May 21, 2015 at 5:29
  • $\begingroup$ This makes the antiderivative much much simpler ! $\endgroup$ May 21, 2015 at 6:14
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Letting $x\mapsto\frac{1}{x}$ transforms $I$ into $$ I=\int_0^{\infty} \frac{x^6}{\left(1+x^2+x^4\right)^3} d x $$ Averaging them yields $$ \begin{aligned} I & =\frac{1}{2} \int_0^{\infty} \frac{x^4+x^6}{\left(1+x^2+x^4\right)^3} d x \\ & =\frac{1}{2} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{\left(x^2+\frac{1}{x^2}+1\right)^3} d x \\ & =\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left[\left(x-\frac{1}{x}\right)^2+3\right]^3} \cdot \\ & =\frac{1}{2} \int_{-\infty}^{\infty} \frac{d y}{\left(y^2+3\right)^3} \quad ( \textrm{ via } y=x-\frac{1}{x}) \\&=\int_0^{\infty} \frac{d y}{\left(y^2+3\right)^3} \end{aligned} $$ Using $y=\sqrt 3\tan \theta$ gives us $$ I=\frac{1}{9 \sqrt{3}} \int_0^{\frac{\pi}{2}} \cos ^4 \theta d \theta =\frac{1}{36\sqrt{3}}\left[\theta+\sin 2 \theta+\frac{1}{2} \theta+\frac{\sin 4 \theta}{8}\right]_0^{\frac{\pi}{2}} =\frac{\pi}{48 \sqrt{3}} \quad \blacksquare $$

In general, for any real number $a> \frac{1}{2}$, we can prove similarly that $$I_a=\int_0^{\infty} \frac{x^{2 a-2}}{\left(x^4+x^2+1\right)^a} d x = \frac{1}{3^{a-\frac{1}{2}}} \int_0^{\frac{\pi}{2}} \cos ^{2 a-2} \theta d \theta \\ \quad = \frac{1}{2 \cdot 3^{a-\frac{1}{2}}} B\left(a-\frac{1}{2}, \frac{1}{2}\right) = \boxed{\frac{\sqrt \pi\Gamma\left(a-\frac{1}{2}\right) }{2 \cdot 3^{a-\frac{1}{2}}\Gamma(a)} }$$

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Not so elegant a solution, but one can substitute $x\mapsto\dfrac{1-x}{1+x}$, then simplify the quotient and expand into partial fractions. The subsequent integral is taken over a symmetric interval so some of the terms will drop out right away.

$$\begin{align*} I &= \int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} \, dx \\[1ex] &= \int_{-1}^1 \frac{\frac{(1-x)^4}{(1+x)^4}}{\left(\frac{(1-x)^4}{(1+x)^4}+ \frac{(1-x)^2}{(1+x)^2} +1\right)^3} \, \frac{2}{(x+1)^2} \, dx \\[1ex] &= 2 \int_{-1}^1 \frac{(1-x)^4(1+x)^{10}}{(x^2+3)^3 (3x^2+1)^3} \, dx \\[1ex] &= \frac2{81} \int_{-1}^1 \left(6x^2+36x+6 - \frac{162(2x+3)}{x^2+3} + \frac{648(2x+5)}{(x^2+3)^2} - \frac{5184}{(x^2+3)^3} \right. \\ &\qquad\qquad\qquad\qquad \left. - \frac{6(30x+13)}{3x^2+1} + \frac{8(18x+17)}{(3x^2+1)^2} - \frac{64}{(3x^2+1)^3}\right) \, dx \\[1ex] &= \frac4{81} \int_{-1}^1 \left(3x^2+3 - \frac{243}{x^2+3} + \frac{1620}{(x^2+3)^2} - \frac{2592}{(x^2+3)^3} \right. \\ &\qquad\qquad \qquad\qquad \left. - \frac{39}{3x^2+1} + \frac{68}{(3x^2+1)^2} - \frac{32}{(3x^2+1)^3}\right) \, dx \end{align*}$$

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