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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?

Tito Piezas III
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  • It's interesting that the new equation has all the roots real. (I think you meant to give it a different name from $P$.) Btw , have you tried with other rational functions? – orangeskid Dec 25 '15 at 04:39
  • @orangeskid: I tried it with solvable sextics and octics, and the new equation (as $P(y)$) has deg $< 5$. So solvable decics was the way to test the observation. It was tricky to find those that converged, and those that did, yielded solvable $P(y)$. However, a general result is required. – Tito Piezas III Dec 25 '15 at 04:46
  • If it can be show that, in general, the root $y$ is a polynomial function of the root $x$, then that should answer the question. – Tito Piezas III Dec 25 '15 at 08:03
  • Why tricky to find those that converged? If your numerator is $x^2$ then take the denominator with degree $\ge 4$ and without real zeros, that guarantees convergence. – orangeskid Dec 25 '15 at 12:55

1 Answers1

5

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.
Jyrki Lahtonen
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  • I would like to be able to say more about $U^2$. After all, the sum is highly symmetric in the zeros $z_i$. However, those are only one half of the zeros of $P(x)$. More can probably be said, if we know how the Galois group of $P(x)$ acts on the conjugate pairs. – Jyrki Lahtonen Dec 25 '15 at 08:53
  • If I understood it correctly, the example degree ten $P(x)$ has Galois group $D_{5}$. Because $U^2$ is stable under complex conjugation it has at most five distinct conjugates. I think that because complex conjugation is not in the center of the Galois group, we can say more... – Jyrki Lahtonen Dec 25 '15 at 10:28
  • This is already informative enough. :) – Tito Piezas III Dec 26 '15 at 02:56
  • Can you look at Q2 of this post? It's about integrals and Galois groups again, but a different angle. – Tito Piezas III Nov 29 '16 at 09:01