# 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),\$\$

(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)