# How to show that $\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx=3\pi$

Consider $$I=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx \qquad J=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx$$ I want to show that $I=3\pi$ and that $I=J$.

First, we noticed that $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$

So it gives us an idea to try and factorise $x^4-x^2+1$ but cannot find any factors.

Integrate $I$ (We try some substitutions to see where it will get us to)

$x=\sqrt{u}$ then $dx={2\over \sqrt{u}}du$

$$I=16\cdot{1\over 2}\int_{0}^{\infty}{u^{3/2}\over (1-u+u^2)^4}\mathrm du$$

$u=\tan(y)$ then $du=\sec^2(y)dy$

$$I=8\int_{0}^{\pi/2}{\tan^{3/2}(y)\over (1-\tan(y)+\tan^2(y))^4}{\mathrm dy\over \cos^2(y)}$$

then simplified down to

$$I=128\int_{0}^{\pi/2}{\cos^6(y)\tan^{3/2}(y)\over (2-\sin(2y))^4}\mathrm dy$$

we further simplified down to

$$I={128\over 2^{3/2}}\int_{0}^{\pi/2}{\cos^3(y)\sin^{3/2}(2y)\over (2-\sin(2y))^4}\mathrm dy$$

Not so sure what is the next step.

• Try the pattern of math.stackexchange.com/questions/2121087/… – lab bhattacharjee Jan 31 '17 at 8:38
• a sub $x\rightarrow 1/y$ should be really helpful here – tired Jan 31 '17 at 9:44
• Afterwards use Glasser's Master theorem – tired Jan 31 '17 at 10:31
• For your first approach: $$x^4-x^2+1=x^4+2x^2+1-3x^2=(x^2+1)^2-3x^2=(x^2+\sqrt{3}x+1)(x^2-\sqrt{3}x+1).$$ – mickep Jan 31 '17 at 15:52

Consider the transformation $x=1/y$. Then

$$J=16\int_0^{\infty} dy\frac{1}{y^2} \frac{1/y^4}{(1-1/y^2+1/y^4)^3}=8\int_{-\infty}^{\infty} dy\frac{1}{(y^2-1+1/y^2)^3}=\\8\int_{-\infty}^{\infty} dy\frac{1}{((y-1/y)^2+1)^3}\underbrace{=}_{(\star)}8\int_{-\infty}^{\infty}dz\frac{1}{(z^2+1)^3}=3\pi$$

and your proof is complete (for the next to last equality $(\star)$ we applied Glasser's Master theorem which is not difficult to proof for this special case)

Furthermore applying the same substitution $x=1/y$ again, it is a matter of straightforward algebra that

$$\Delta=J-I=-\Delta$$

so

$$\Delta=0 \,\,\text{or}\,\,I=J=3\pi$$

QED

\begin{align}I&=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx \qquad\\ &=\int_{0}^{\infty} \dfrac{16x^4(1+x^2)^4}{(1+x^6)^4} dx\\ &=\int_{0}^{\infty} \dfrac{16\cdot {{x}^{12}}+64\cdot {{x}^{10}}+96\cdot {{x}^{8}}+64\cdot {{x}^{6}}+16\cdot {{x}^{4}}}{(1+x^6)^4} dx\\ \end{align}

Perform the change of variable $y=x^6$,

\begin{align} I&=\int_{0}^{\infty} \dfrac{\tfrac{8}{3}x^{\tfrac{7}{6}}+\tfrac{32}{3}x^{\tfrac{5}{6}}+16x^{\tfrac{1}{2}}+\tfrac{32}{3}x^{\tfrac{1}{6}}+\tfrac{8}{3}x^{-\tfrac{1}{6}}}{(1+x)^4} dx\\ &=\dfrac{8}{3}\text{B}\left(\dfrac{13}{6},\dfrac{11}{6}\right)+\dfrac{32}{3}\text{B}\left(\dfrac{13}{6},\dfrac{11}{6}\right)+16\text{B}\left(\dfrac{3}{2},\dfrac{5}{2}\right)+\dfrac{32}{3}\text{B}\left(\dfrac{7}{6},\dfrac{17}{6}\right)+\dfrac{8}{3}\text{B}\left(\dfrac{5}{6},\dfrac{19}{6}\right)\\ &=\dfrac{8}{3}\times \dfrac{35\pi}{648}+\dfrac{32}{3}\times \dfrac{35\pi}{648}+16\times \dfrac{\pi}{16}+ \dfrac{32}{3}\times\dfrac{55\pi}{648}+\dfrac{8}{3}\times \dfrac{91\pi}{648}\\ &=\boxed{3\pi} \end{align}

$B$ is the Euler beta function.
\begin{align} \text{B}\left(\dfrac{13}{6},\dfrac{11}{6}\right)&=\dfrac{\Gamma\left(\dfrac{13}{6}\right)\Gamma\left(\dfrac{11}{6}\right)}{\Gamma\left(4\right)}\\ &=\dfrac{1}{6} \left(\dfrac{7}{6}\Gamma\left(\dfrac{7}{6}\right)\right)\times \left(\dfrac{5}{6}\Gamma\left(\dfrac{5}{6}\right)\right)\\ &=\dfrac{1}{6} \left(\dfrac{7}{6}\times\dfrac{1}{6}\Gamma\left(\dfrac{1}{6}\right)\right)\times \left(\dfrac{5}{6}\Gamma\left(\dfrac{5}{6}\right)\right)\\ &=\dfrac{35}{1296}\Gamma\left(\dfrac{1}{6}\right)\Gamma\left(\dfrac{5}{6}\right)\\ &=\dfrac{35}{1296}\times \dfrac{\pi}{\sin\left(\pi\times \tfrac{1}{6}\right)}\\ &=\boxed{\dfrac{35\pi}{648}} \end{align}
Note that the integrands are even, so $$I=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{4}}dx$$ and $$J=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{3}}dx.$$ Now let us consider the semicircular contour of radius $R$ centred in the origin on the upper plane of the complex plane. It is not difficult to note that the integral over the semicircumference vanish as $R\rightarrow\infty$ so $$\int_{-\infty}^{\infty}\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{4}}dx=2\pi i\left(\underset{x=\frac{\sqrt{3}}{2}+\frac{i}{2}}{\textrm{Res}}\left(\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{4}}\right)+\underset{x=-\frac{\sqrt{3}}{2}+\frac{i}{2}}{\textrm{Res}}\left(\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{4}}\right)\right)$$ $$=2\pi i\left(\frac{247+77i\sqrt{3}}{54\left(-\sqrt{3}+3i\right)}+\frac{247-77i\sqrt{3}}{54\left(\sqrt{3}+3i\right)}\right)=6\pi$$ hence $$I=\color{red}{3\pi}.$$ In a similar manner $$\int_{-\infty}^{\infty}\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{3}}dx=2\pi i\left(\underset{x=\frac{\sqrt{3}}{2}+\frac{i}{2}}{\textrm{Res}}\left(\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{3}}\right)+\underset{x=-\frac{\sqrt{3}}{2}+\frac{i}{2}}{\textrm{Res}}\left(\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{3}}\right)\right)$$ $$=2\pi i\left(\frac{13+i5\sqrt{3}}{3\left(-\sqrt{3}+3i\right)}+\frac{13-i5\sqrt{3}}{3\left(\sqrt{3}+3i\right)}\right)=6\pi$$ hence $$\color{blue}{I=J}.$$
The second integral $$J=I-\int_{0}^{\infty}{(2x)^4(x^2-x^4)\over (1-x^2+x^4)^4}\, dx$$
The integral on the right is zero by $x \to 1/x$.
• Tired: your latter integral is equal to $\text{B}\left(\dfrac{1}{2},\dfrac{5}{2}\right)$ – FDP Jan 31 '17 at 18:21