A nasty integral of a rational function

The integral over $mathbb{R}$ of a meromorphic function $f(z)$, $O(|z|^{-2})$ at infinity, non-vanishing over $mathbb{R}$, is equal to $2pi i$ times the sum of residues in the poles located in the complex upper-half plane. Since:

$$p(y) = y^6-2y^5-2y^4+4y^3+3y^2-4y+1 = p_{+}(y)cdot p_{-}(y),$$
$$p_{+}(y)= y^3-(i+1)y^2+(i-2)x+1,qquad p_{-}(y)=y^3+(i-1)y^2-(2+i)y+1,$$

(I got this through a numerical calculation of the roots of $p(y)$, followed by a separation of the roots with positive and negative imaginary part, say $zeta_1,zeta_2,zeta_3$ and $bar{zeta_1},bar{zeta_2},bar{zeta_3}$ – so $p_{+}(z)$ is just $prod_{j=1}^3 (z-zeta_j)$) we have:

$$ I = int_{mathbb{R}}frac{y^2 dy}{p(y)} = 2pi isum_{j=1}^{3}operatorname{Res}_{z=zeta_j}left(frac{z^2}{p_{+}(z)cdot p_{-}(z)}right),$$

but $p_{-}(x)-p_{+}(x)=2i(x^2-x)$, so:

$$ I = int_{mathbb{R}}frac{y^2 dy}{p(y)} = 2pi isum_{j=1}^{3}operatorname{Res}_{z=zeta_j}left(frac{z^2}{p_{+}^2(z)+2i(z^2-z)p_{+}(z)}right).$$

By De l’Hopital theorem, and since $zeta_j$ is a double zero of $p_{+}^2(x)$:

$$lim_{ztozeta_j}frac{z^2(z-zeta_j)}{p_{+}^2(z)+2i(z^2-z)p_{+}(z)}=frac{zeta_j^2}{2i(zeta_j^2-zeta_j)p_{+}'(zeta_j)},$$

so:

$$ I = int_{mathbb{R}}frac{y^2 dy}{p(y)} = pisum_{j=1}^{3}frac{zeta_j}{(zeta_j-1)p_{+}'(zeta_j)}.$$

Now we compute the remainder between $(z-1)p_{+}'(z)$ and $p_{+}(z)$, in order to have:

$$ I = int_{mathbb{R}}frac{y^2 dy}{p(y)} = pisum_{j=1}^{3}frac{zeta_j}{-(1+i)+6zeta_j-(2-i)zeta_j^2}.$$

If now we take $alpha=frac{3+sqrt{6-i}}{2-i}$ and $beta=frac{3-sqrt{6-i}}{2-i}$ we can re-write the last line as:

$$ I = frac{pi}{(i-2)(alpha-beta)}left(sum_{j=1}^{3}frac{alpha}{zeta_j-alpha}-sum_{j=1}^{3}frac{beta}{zeta_j-beta}right)=-frac{pi}{2sqrt{6-i}}left(Sigma_1-Sigma_2right).$$

Now $Sigma_1$ is the sum of the reciprocal of the roots of the polynomial $p_{+}(alpha(z+1))$, and $Sigma_2$ is the sum of the reciprocal of the roots of the polynomial $p_{+}(beta(z+1))$. This quantities can be computed through the coefficients of $p_{+}$, since the sum of the reciprocal of the roots of a polynomial $q(z)$ is just $-frac{q'(0)}{q(0)}$. This gives:

$$Sigma_1 = -alphafrac{p_{+}'(alpha)}{p_{+}(alpha)},qquad Sigma_2 = -betafrac{p_{+}'(beta)}{p_{+}(beta)}.$$

Up to a massive amount of long but straightforward computations, we get:

$$Sigma_1 = (i-2)-sqrt{6-i},qquad Sigma_2 = (i-2)+sqrt{6-i}, $$

from which $color{red}{I=pi}$ finally follows.

I am really grateful to Jon Haussmann for the proof that

$$int_0^{infty} frac{x^8 – 4x^6 + 9x^4 – 5x^2 + 1}{x^{12} – 10 x^{10} + 37x^8 – 42x^6 + 26x^4 – 8x^2 + 1}dx = frac{1}{2}int_{mathbb{R}}frac{y^2 dy}{p(y)},$$

where only the second integral is treated here.

IMPORTANT UPDATE:
In fact, there is no need to compute the coefficients of $p_{+}(x)$ and $p_{-}(x)$ (we only need the identity $p_{-}(x)-p_{+}(x)=2i(x^2-x)$), or introduce $alpha$ and $beta$. Since $p_{+}(x)$ is a third-degree polynomial with roots in the upper half-plane,
$$0=int_{mathbb{R}}frac{dz}{p_{+}(z)}=sum_{j=1}^{3}frac{1}{p_{+}'(zeta_j)}.$$
This gives:
$$ I = pisum_{j=1}^{3}frac{zeta_j}{(zeta_j-1)p_{+}'(z)} = pisum_{j=1}^{3}frac{1}{(zeta_j-1)p_{+}'(zeta_j)}, $$
but if we decompose $frac{1}{p_{+}(z)}$ in simple fractions, we get:
$$frac{1}{p_{+}(z)}=sum_{j=1}^{3}frac{1}{p_{+}'(zeta_j)(z-zeta_j)}, $$
so the magic gives:
$$ I = -frac{pi}{p_{+}(1)}.$$
Since $p(x)=p_{+}(x)cdot p_{-}(x)$, $p(1)=1$, $Iinmathbb{R}^+$ and $p_{-}(1)$ is the conjugate of $p_{+}(1)$, $p_{+}(1)$ can be only $+1$ or $-1$, so $I=pi$.

Some progess: The integrand actually decomposes as
$$frac{1}{2} left( frac{x^2 + 2x + 1}{x^6 + 4x^5 + 3x^4 – 4x^3 – 2x^2 + 2x + 1} + frac{x^2 – 2x + 1}{x^6 – 4x^5 + 3x^4 + 4x^3 – 2x^2 – 2x + 1} right).$$
Note that the second term is the same as the first term, except with $-x$ instead of $x$. Thus, with some substitutions, the integral becomes
$$frac{1}{2} int_{-infty}^infty frac{y^2}{y^6 – 2y^5 – 2y^4 + 4y^3 + 3y^2 – 4y + 1} ; dy.$$

Let,

$$I=int_0^{infty} frac{x^8 – 4x^6 + 9x^4 – 5x^2 + 1}{x^{12} – 10 x^{10} + 37x^8 – 42x^6 + 26x^4 – 8x^2 + 1} , dx$$

As noted by Jon Haussmann,

$$2I=int_0^{infty} frac{(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 – 2x^2 -2x + 1}dx + int_0^{infty}frac{(x+1)^2}{x^6 +4x^5 + 3x^4 – 4x^3 – 2x^2 + 2x + 1}dx$$

Perform the change of variable $x=dfrac{1}{u-1}$ in the second integral,

$begin{align}2I&=int_0^{infty} frac{(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 – 2x^2 -2x + 1}dx+int_1^{infty} frac{x^2(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 – 2x^2 -2x + 1}dx\
&=left(int_0^1 frac{(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 – 2x^2 -2x + 1}dx+\
int_1^{infty} frac{(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 – 2x^2 -2x + 1}dxright)+\&int_1^{infty} frac{x^2(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 – 2x^2 -2x + 1}dxend{align}$

Perform the change of variable $x=dfrac{u-1}{u}$ in the first integral of the latter equality,

$begin{align}2I&=int_1^{infty} frac{x^2}{x^6-4x^5 + 3x^4 +4x^3 – 2x^2 -2x + 1}dx+int_1^{infty} frac{(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 – 2x^2 -2x + 1}dx+int_1^{infty} frac{x^2(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 – 2x^2 -2x + 1}dx\
&=int_1^{infty} frac{x^2+(x-1)^2+x^2(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 – 2x^2 -2x + 1}dx\
&=Big[arctanleft(dfrac{x^3-2x^2-x+1}{x(x-1)}right)Big]_0^{+infty}\
&=pi
end{align}$

(Problem found in American Mathematical Monthly, vol. 112, april 2005.

Solution found in vol. 114)

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