Evaluating $\int_{-\infty}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^n \ \mathrm{d}x$

Question

I am trying to find a general form of $$I_n = \int_{-\infty}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^n \ \mathrm{d}x$$ as well as methods of solving it (preferably elementary). This is easy enough to solve for individual $n$, which leads me to believe that there is likely a $\frac{\pi}{\sqrt{3}}$ term in the general expression, but I am having trouble finding a general solution. I have tried constructing a recurrence via IBP, but I always have an extra $\frac{(1 - x^2)^m}{(1 + x^2 + x^4)^n}$ term with $m \neq n$. I also tried creating a $2D$ recurrence, but that doesn't seem solvable, nor can I figure out how to incorporate the requirement of $m \leq 2n - 1$ (for the integral to be finite).

Additionally, I tried using a semicircular contour in the upper half-plane, but computing the residues at $z = e^{\frac{i\pi}{3}}, e^{\frac{2i\pi}{3}}$ seems very bashy.

I have tried using Mathematica, and that just outputs the input.

Any help would be great.

Answer 1

A systemic procedure is to utilize the reduction formula

$$\int_0^\infty \frac{x^{2m}}{(1+x^2+x^4)^n}dx =I_{m,n}=-2I_{m-1,n}+ \frac{4n-2m-3}{2(n-1)}I_{m-1, n-1}$$

Then, given the initial values $$I_{0,1}=I_{1,1}=\frac\pi{2\sqrt3}, \>\>\>I_{0,2}=\frac\pi{3\sqrt3}, \>\>\> I_{0,3}=\frac{13\pi}{48\sqrt3}, \>\>\>\cdots$$ we evaluate recursively \begin{align} &I_{1,2}=-2 I_{0,2}+\frac32 I_{0,1}=\frac\pi{12\sqrt3}\\ &I_{2,2}=-2 I_{1,2}+\frac12 I_{1,1}=\frac\pi{12\sqrt3}\\ &I_{1,3}=-2 I_{0,3}+\frac74 I_{0,2}=\frac\pi{24\sqrt3}\\ &I_{2,3}=-2 I_{1,3}+\frac54 I_{1,2}=\frac\pi{48\sqrt3}\\ &I_{3,3}=-2 I_{2,3}+\frac34 I_{2,2}=\frac\pi{48\sqrt3}\\ &\ \cdots \end{align} As a result \begin{align} &\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )dx=I_{0,1}- I_{1,1}=0\\ &\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^2dx=I_{0,2}- 2I_{1,2}+ I_{2,2} =\frac\pi{4\sqrt3}\\ &\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^3 dx= I_{0,3}- 3I_{1,3}+ 3I_{2,3} -I_{3,3}=\frac{3\pi}{16\sqrt3}\\ &\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^4 dx=\cdots=\frac{\pi}{6\sqrt3}\\ &\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^5 dx=\cdots\\ \end{align}

Answer 2

comment
Some starting values:

$$I_{{1}}=0,\\I_{{2}}={\frac {\pi\,\sqrt {3}}{6}}={\frac {\pi\,\sqrt {3}}{ \left( 2 \right) \left( 3 \right) }},\\I_{{3}}={\frac {\pi\, \sqrt {3}}{8}}={\frac {\pi\,\sqrt {3}}{ ( 2 ) ^{3}}},\\I_{{4}}={\frac {\pi\,\sqrt {3}}{9}}={\frac {\pi\,\sqrt {3}}{ ( 3 ) ^{2}}},\\I_{{5}}={\frac { 115\,\pi\,\sqrt {3}}{1152}}={\frac { \left( 5 \right) \left( 23 \right) \pi\,\sqrt {3}}{ ( 2 ) ^{7} ( 3 ) ^{2}}},\\I_{{6}}={\frac {211\,\pi\,\sqrt {3}}{2304} }={\frac { \left( 211 \right) \pi\,\sqrt {3}}{ ( 2 ) ^{8} ( 3 ) ^{2}}},\\I_{{7}}={\frac {49\,\pi\,\sqrt {3}}{576}}={\frac { ( 7 ) ^{2}\pi\,\sqrt {3}}{ ( 2 ) ^{6} ( 3 ) ^{2}}},\\I_{{8}}={\frac {13235\,\pi \,\sqrt {3}}{165888}}={\frac { \left( 5 \right) \left( 2647 \right) \pi\,\sqrt {3}}{ ( 2 ) ^{11} ( 3)^{4}}},\\I_{{9}}={\frac {22229\,\pi\,\sqrt {3}}{294912}}={\frac { \left( 22229 \right) \pi\,\sqrt {3}}{ ( 2 ) ^{15} ( 3 ) ^{2}}},\\I_{{10}}={\frac {142585\,\pi\,\sqrt {3}}{1990656}}={\frac { \left( 5 \right) \left( 28517 \right) \pi\,\sqrt {3}}{ ( 2 ) ^{13} ( 3 ) ^{5}}},\\I_{{11}}={\frac { 4356451\,\pi\,\sqrt {3}}{63700992}}={\frac { \left( 11 \right) \left( 396041 \right) \pi\,\sqrt {3}}{ ( 2 ) ^{18} ( 3 ) ^{5}}},\\I_{{12}}={\frac {927977\,\pi\, \sqrt {3}}{14155776}}={\frac { \left( 53 \right) \left( 17509 \right) \pi\,\sqrt {3}}{ ( 2 ) ^{19} ( 3 ) ^{3}}},\\I_{{13}}={\frac {1004029\,\pi\,\sqrt {3}}{ 15925248}}={\frac { ( 13 ) ^ {3} \left( 457 \right) \pi\,\sqrt {3}}{ ( 2 ) ^{16} ( 3 ) ^{5}}},\\I_{{14}}={\frac {371846825\,\pi\,\sqrt {3}}{6115295232}}={ \frac { ( 5 ) ^{2} \left( 7 \right) \left( 2124839 \right) \pi\,\sqrt {3}}{ ( 2 ) ^{23} ( 3 ) ^{6}}},\\I_ {{15}}={\frac {479349895\,\pi\,\sqrt {3}}{8153726976}}={ \frac { \left( 5 \right) \left( 61 \right) \left( 787 \right) \left( 1997 \right) \pi\,\sqrt {3}}{ ( 2 ) ^{25} ( 3 ) ^{5}}},\\I_{{16}}={ \frac {130624067\,\pi\,\sqrt {3}}{2293235712}}={ \frac { \left( 7 \right) \left( 43 \right) \left( 433967 \right) \pi \,\sqrt {3}}{ ( 2 ) ^{20} ( 3 ) ^{7}}},\\I_{{17}}={\frac { 259686035873\,\pi\,\sqrt {3}}{4696546738176}}={\frac { \left( 11 \right) \left( 17 \right) \left( 41 \right) \left( 389 \right) \left( 87071 \right) \pi\,\sqrt {3}}{ ( 2 ) ^{31} ( 3 ) ^{7}}},\\I_{{18}}={\frac { 505009314013\,\pi\,\sqrt {3}}{9393093476352}}={\frac { \left( 1689893 \right) \left( 298841 \right) \pi\,\sqrt {3}}{ ( 2 ) ^{32} ( 3 ) ^{7 }}},\\I_{{19}}={\frac { 40981417091\,\pi\,\sqrt {3}}{782757789696}}={\frac { ( 19 ) ^{2} \left( 113521931 \right) \pi\,\sqrt {3}}{ ( 2 ) ^ {30} ( 3 ) ^{6}}},\\I_{{20}}={\frac { 103579369887065\,\pi\,\sqrt {3}}{2028908190892032}}={\frac { \left( 5 \right) \left( 23 \right) \left( 61 \right) \left( 73 \right) \left( 10181 \right) \left( 19867 \right) \pi\,\sqrt {3}}{ ( 2 ) ^{35} ( 3 ) ^{10}}}$$