Integral through an algebraic expression

This integral is an absolute monster. There might be a faster way of doing it, but even with complex analysis it seems really hard. For this solution, I'll only use real analysis, paired with some complex algebra to shorten the partial fraction decompositions (PFD).

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Finding symmetry and reducing the denominator

Notice the rational part has an odd and an even part, while the cube root is odd. We expand by $x^2-1 \: (*)$ to shorten the denominator: $f(x) := \frac{1}{\pi}\frac{ x^3+x+1 }{ x^4+x^2+1 }\cdot x^{-\frac{1}{3}} \underset{(*)}{=} \frac{1}{\pi}\frac{ x^5+x^2-x-1 }{ x^6-1 }\cdot x^{-\frac{1}{3}} =\frac{1}{\pi}\frac{ (x^4-1)x^{\frac{2}{3}} }{ x^6-1 } + \frac{1}{\pi}\frac{ (x^2-1)x^{-\frac{1}{3}} }{ x^6-1 }=:e(x)+o(x)$ We factor the denominators to $x^6-1= (x^3-1)(x^3+1)$ . Comparing coefficients, we solve the PFDs: $\begin{aligned} e(x)&=\frac{1}{2\pi}\left[ \frac{x-1}{x^3-1} + \frac{x+1}{x^3+1} \right]x^{\frac{2}{3}} =:\frac{1}{2\pi}(g_1(x) + g_1(-x))x^{\frac{2}{3}}\\[1em] o(x)&=\frac{1}{2\pi}\left[ \frac{x^2-1}{x^3-1} + \frac{1-x^2}{x^3+1} \right]x^{-\frac{1}{3}} =:\frac{1}{2\pi}(g_2(x) + g_2(-x))x^{-\frac{1}{3}} \end{aligned}$ With a substitution $(**)\:u:=-x,\quad u\rightarrow x$ for the " $-x$ "-parts above we express the entire integral in terms of the reduced functions $g_{1,2}(x)$ : $\alpha=\int_0^\infty e(x)+o(x)\:dx\underset{(**)}{=}\frac{1}{2\pi}\int_{\mathbb{R}}g_1(x)x^{\frac{2}{3}}\:dx + \frac{1}{2\pi}\int_0^\infty g_2(x)x^{-\frac{1}{3}}\:dx + \frac{1}{2\pi}\int_0^{-\infty} g_2(x)x^{-\frac{1}{3}}\:dx=:\alpha_e+\alpha_o$ The even integral simplifies nicely (and could easily be evaluated by contour integration), while the odd integral does not. I couldn't come up with a nice complex curve to solve the odd integral, so let's just use real analysis, like before :)

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Solving the even part $\alpha_e$

Substitute $u:=x^{\frac{1}{3}},\: u^{(1)}(x)=\frac{1}{3u^2}$ to eliminate the cube root. The integration bounds stay the same: $\begin{aligned} \alpha_e&=\frac{1}{2\pi}\int_{\mathbb{R}} \frac{u^3-1}{u^9-1}\cdot u^2\cdot 3u^2\:du=\frac{3}{2\pi}\int_\mathbb{R}\frac{u^3-1}{u^9-1}\cdot u^4\:du, &&& \left|u_{k}\right.&:=w^{k}=:c_{k}+is_k, &k&\in\{1,\:2,\:4;\:5;\:7;\:8\},&&& u_{9-k}&=u_k^*,&w:=e^{\frac{2\pi i}{9}} \end{aligned}$ We define $u_k$ as the complex conjugate pole pairs and notice they all are powers of the fundamental ninth roots of unity $w$ . Their real- and imaginary parts are all sines and cosines we denote by $s_k,\:c_k$ . Even though the degree of the denominator suggests otherwise, we only have six of them: The remaining three cancel against the numerator (we introduced them ourselves when we expanded by $x^2-1$ to shorten the denominator). As all poles are distinct, we have a simple PFD: $\begin{aligned} \alpha_e&\underset{\text{PFD}}{=}\frac{3}{2\pi}\int_{\mathbb{R}}\sum_{k\in\{1;\:2;\:4\}}\frac{A_k}{u-u_k}+\frac{A_k^*}{u-u_k^*}\:du&&&\left|A_k=\lim_{u\rightarrow u_k}(u-u_k)\frac{u^3-1}{u^9-1}u^4\underset{\text{l'H.}}{=}\frac{u_k^3-1}{9u_k^8}u_k^4=\frac{1}{9}(u_k^{-1}-u_k^{-4})=:a_k+ib_k,\quad a_k,\:b_k\in\mathbb{R}\right. \end{aligned}$ The antiderivative we still need to find is listed at $(***)$ at the end. The log-parts cancel out when we let $x\rightarrow\pm\infty$ because $a_1+a_2+a_3=0$ : $\begin{aligned} \alpha_e&=\frac{3}{2\pi}\left[ \sum_{{k\in\{1;\:2;\:4\}}} a_k\log|(u-c_k)^2+s_k^2| - 2b_k\arctan\left(\frac{u-c_k}{s_k}\right) \right]_{-\infty}^\infty & s_{1,2,4}>0 \\ &= \frac{3}{2\pi}\cdot (-2\pi)(b_1+b_2+b_4)=-3\cdot\frac{1}{9}(-s_1+s_4-s_2+s_8-s_4+s_{16})=\frac{2}{3}(s_1+s_2) \end{aligned}$

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Solving the odd part $\alpha_o$

Using the exact same steps as in the even part, we can calculate the first part of the odd integral. The only change is the new definition of $A_k$ : $\begin{aligned} \frac{1}{2\pi}\int_0^\infty g_2(x)x^{-\frac{1}{3}}\:dx &= \frac{3}{2\pi}\int_0^\infty\frac{u^6-1}{u^9-1}\cdot u\:du=\frac{3}{2\pi}\int_0^\infty\sum_{k\in\{1;\:2;\:4\}}\frac{A_k}{u-u_k}+\frac{A_k^*}{u-u_k^*}\:du&&&\left|A_k=\frac{1}{9}(u_k^{-1}-u_k^{-7})=:a_k+ib_k,\quad a_k,\:b_k\in\mathbb{R}\right.\\ & =\frac{3}{2\pi}\left[ \sum_{{k\in\{1;\:2;\:4\}}} a_k\log|(u-c_k)^2+s_k^2| - 2b_k\arctan\left(\frac{u-c_k}{s_k}\right) \right]_{0}^\infty \\ &=\frac{3}{2\pi}\sum_{{k\in\{1;\:2;\:4\}}}-2b_k\left[ \frac{\pi}{2}-\left( \frac{2\pi k}{9}-\frac{\pi}{2} \right) \right] \end{aligned}$ Like before, the log-parts cancel at infinity because $a_1+a_2+a_3=0$ . At zero, they cancel because $s_k^2+c_k^2=1$ . The second part of the odd integral works just like the first one - only the upper border changes to $\red{-\infty}$ : $\begin{aligned} \frac{1}{2\pi}\int_0^{\red{-\infty}} g_2(x)x^{-\frac{1}{3}}\:dx &= \frac{3}{2\pi}\sum_{{k\in\{1;\:2;\:4\}}}-2b_k\left[ \red{-\frac{\pi}{2}}-\left( \frac{2\pi k}{9}-\frac{\pi}{2} \right) \right] \end{aligned}$ Adding both parts together: $\begin{aligned} \alpha_o&= \frac{3}{2\pi}\sum_{{k\in\{1;\:2;\:4\}}}-2b_k\left[ 0 - 2\left( \frac{2\pi k}{9}-\frac{\pi}{2} \right) \right]=\frac{3}{\pi}\cdot \frac{1}{9}\left( \frac{-5\pi}{9}(-s_1+s_7) + \frac{-\pi}{9}(-s_2+s_{14}) + \frac{7\pi}{9}(-s_4+s_{28}) \right)\\ &=\frac{1}{27}(12s_1+6s_2-6s_4)=\frac{2}{9}(2s_1+s_2-s_4) \end{aligned}$

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Solving the task

With solutions to both the odd and even part, we can finally calculate a solution to the integral. Remember the short-hand $s_k=\sin\left(\frac{2\pi k}{9}\right)$ : $\begin{aligned} \alpha&=\alpha_e+\alpha_0=\frac{2}{9}(5s_1+4s_2-s_4) & \Rightarrow && P(\alpha)&=27\alpha^2(7-3\alpha^2)^2\approx\boxed{1} \pm 10^{-8} \end{aligned}$ The last result I only checked numerically. I suspect you just have to use addition theorems repeatedly, but at this point, I'm a bit too tired to do that:)

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Standard anti-derivative of a complex pole pair

Let $u:=c+is\in\mathbb{C}$ with $|u|=1,\:s\neq 0$ and let $A=a+ib\in\mathbb{C}$ . Then $\begin{aligned} \int \frac{A}{x-u}+\frac{A^*}{x-u^*}\:dx&=a\log|(x-c)^2+s^2|-2b\arctan\left(\frac{x-c}{s}\right)+C,&C\in\mathbb{R}&&&&&(***) \end{aligned}$