On Galois groups and $\int_{-\infty}^{\infty} \frac{x^2}{x^{10} - x^9 + 5x^8 - 2x^7 + 16x^6 - 7x^5 + 20x^4 + x^3 + 12x^2 - 3x + 1}\,dx$

Given the solvable decic (among many in this database),

$$P(x) := x^{10} - x^9 + 5x^8 - 2x^7 + 16x^6 - 7x^5 + 20x^4 + x^3 + 12x^2 - 3x + 1$$

we have,

$$\int_{-\infty}^{\infty} \frac{x^2}{P(x)}\,dx = 2\pi\sqrt{\frac{y}{33}}$$

where $y\approx 0.005498$ is a root of the solvable $5$-real root quintic,

$$P(y):=410651^2 - 297369569963257 y + 64437688060325415 y^2 - 3213663132678906688 y^3 + 59485209442439490149 y^4 - (11^3\cdot67^2\cdot199^6) y^5 = 0$$

(Added later): A relation between the roots $x,y$ such that $P(x)=P(y)=0$ is,

$$12675353 + 84680609 x^3 + 55168143 x^6 - 6841070 x^9 - 1801451 x^{12} = (11\cdot67^2\cdot199^2) y$$

Q: In general, if, $$\int_{-\infty}^{\infty} \frac{x^2}{ P(x)}\,dx= 2\pi \sqrt{y}$$ and $P(x)$ has a solvable Galois group, is it true that $P(y)$ also has a solvable Galois group?

Solution 1:

Not an answer - just listing some obvious facts that seem to be relevant to get the ball rolling.

• The polynomials from that database surely only have simple roots, because they were selected with a suitable splitting field in mind.
• For the integral $\int_{-\infty}^\infty\dfrac{x^2}{P(x)}\,dx$ to converge it is necessary that $P(x)$ has no real roots.
• With $P(x)$ of high enough degree the usual business with sophomore level complex path integrals gives $$I=\int_{-\infty}^\infty\dfrac{x^2}{P(x)}\,dx=2\pi i\sum_{P(z)=0,z\in H}\operatorname{Res}(\frac{z^2}{P(z)},z),$$ where the summation ranges over the zeros of $P(x)$ in the upper half plane $H$.
• Those zeros are simple, so a single application of l'Hospital shows that at a zero $z_i\in H$ the residue is $$ \operatorname{Res}(\frac{z^2}{P(z)},z_i)=\frac{z_i^2}{P'(z_i)}. $$
• So if we write $I=2\pi U$, then $$U^2=-\left(\sum_i \frac{z_i^2}{P'(z_i)}\right)^2.$$
• If $K$ is the splitting field of $P(x)$ (inside $\Bbb{C}$), then $U^2\in K$. Furthermore, because $U^2$ is real, it belongs to the real subfield $L=K\cap \Bbb{R}$. So the minimal polynomial of $U^2$ is solvable, iff $Gal(K/\Bbb{Q})$ is. The number $U\in K(i)$, so it, too, has a solvable minimal polynomial.