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V. Reshetnikov gave the remarkable integral, $$\int_0^1\frac{dx}{\sqrt[3]x\,\sqrt[6]{1-x}\,\sqrt{1-x\left(\sqrt{6}\sqrt{12+7\sqrt3}-3\sqrt3-6\right)^2}}=\frac\pi9(3+\sqrt2\sqrt[4]{27})\tag1$$ More generally, given some integer/rational $N$, we are to find an algebraic number $\alpha$ that solves,

$$\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{3}\,|\alpha|}\tag2$$

and absolute value $|\alpha|$. (Compare to the similar integral in this post.) Equivalently, to find $\alpha$ such that,

$$\begin{aligned} \frac{1}{N} &=I\left(\alpha^2;\ \tfrac12,\tfrac13\right)\\[1.8mm] &= \frac{B\left(\alpha^2;\ \tfrac12,\tfrac13\right)}{B\left(\tfrac12,\tfrac13\right)}\\ &=B\left(\alpha^2;\ \tfrac12,\tfrac13\right)\frac{\Gamma\left(\frac56\right)}{\sqrt{\pi}\,\Gamma\left(\frac13\right)}\end{aligned} \tag3$$

with beta function $\beta(a,b)$, incomplete beta $\beta(z;a,b)$ and regularized beta $I(z;a,b)$. Solutions $\alpha$ for $N=2,3,4,5,7$ are known. Let, $$\alpha=\frac{-3^{1/2}+v^{1/2}}{3^{-1/2}+v^{1/2}}\tag4$$ Then, $$ - 3 + 6 v + v^2 = 0, \quad N = 2\\ - 3 + 27 v - 33v^2 + v^3 = 0, \quad N = 3\\ 3^2 - 150 v^2 + 120 v^3 + 5 v^4 = 0, \quad N = 5\\ - 3^3 - 54 v + 1719 v^2 - 3492v^3 - 957 v^4 + 186 v^5 + v^6 = 0, \quad N = 7$$

and (added later),

$$3^4 - 648 v + 1836 v^2 + 1512 v^3 - 13770 v^4 + 12168 v^5 - 7476 v^6 + 408 v^7 + v^8 = 0,\quad N=4$$

using the largest positive root, respectively. The example was just $N=2$, while $N=4$ leads to,

$$I\left(\tfrac{1-\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\right)=\tfrac{3}{8},\quad\quad I\left(\tfrac{1+\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\right)=\tfrac{5}{8}$$

I found these using Mathematica's FindRoot command, and some hints from Reshetnikov's and other's works, but as much as I tried, I couldn't find prime $N=11$.

Q: Is it true one can find algebraic number $\alpha$ for all $N$? What is it for $N=11$?

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  • $\begingroup$ Do you know about the RootApproximant function? It tries to find the simplest algebraic number matching a numeric approximation. It likes to be given quite a lot of correct decimal digits, though. From my experience, if the degree $>8$, then you need to provide $>200$ digits. $\endgroup$ Dec 1 '16 at 0:14
  • $\begingroup$ I used RootApproximant[N[InverseBetaRegularized[1/7, 1/2, 1/3], 250]] to find my conjectures. Had no luck with 1/11. I got results for dozens of other fractions though. $\endgroup$ Dec 1 '16 at 0:41
  • $\begingroup$ @VladimirReshetnikov: These "dozens" of fractions, did you get unit fractions other than $p=2,3,5,7$? $\endgroup$ Dec 1 '16 at 2:35
  • $\begingroup$ With InverseBetaRegularized[1/n, 1/2, 1/3] -- no. But InverseBetaRegularized[1/6,1/3,1/3], InverseBetaRegularized[1/4,1/4,1/2] and others seem to be algebraic. No fraction with a denominator $>7$ produced an apparently algebraic number. $\endgroup$ Dec 1 '16 at 3:11
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    $\begingroup$ Hasn't achille hui in his answer already showed that $\alpha$ is algebraic for all rational $N$? In his notations we have $\sqrt[3]{1-\alpha^2}=\frac{4}{\eta^2}\wp\left(i\frac{\sqrt{3}}{2N}\right)$. We also know that $\wp\left(i\frac{\sqrt{3}}{2N}\right)=\wp\left(\frac{2\omega_2-\omega_1}{N}\right)$. Now addition theorem for $\wp$ shows that $\wp\left(\frac{2\omega_2-\omega_1}{N}\right)/\eta^2$ is algebraic for all rational $N$, which means that $\alpha$ is also algebraic. $\endgroup$
    – Nemo
    Dec 1 '16 at 6:34
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(Too long for a comment. And courtesy of V. Reshetnikov's result here, though as he points out it is tentative.)

The algebraic number $\alpha$ that solves,

$$\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}=\frac{1}{11}\,\frac{2\pi}{\sqrt{3}\;\alpha}$$

seems to have a $40$-deg minpoly. However, it turns out we can also reduce its degree and express it using the common form above. Let,

$$\alpha=\frac{3^{1/2}-v^{1/2}}{3^{-1/2}+v^{1/2}}$$

where $v$ is a the second largest positive root ($r_9$ in Mathematica syntax) of,

$$\small P(v)=-3^{10} + 23816430 v^2 - 323903448 v^3 + 2177615583 v^4 - 9297934272 v^5 + 25869358152 v^6 - 37475802144 v^7 - 16459141842 v^8 + 180065426112 v^9 - 338100745356 v^{10} + 329418595440 v^{11} - 211367836746 v^{12} + 102243404736 v^{13} - 8162926200 v^{14} - 9999738144 v^{15} + 1006439643 v^{16} - 134177472 v^{17} - 2246706 v^{18} + 30888 v^{19} + 11 v^{20} = 0$$

Also,

$$I\big(\alpha^2;\tfrac{1}{2},\tfrac{1}{3}\big)=\tfrac{1}{5}\,I\big(\tfrac{1-\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\big)=\tfrac{1}{6}\,I\big(\tfrac{1+\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\big)=\frac{1}{11}$$

with regularized beta function $I(z;a,b)$. Furthermore, if

$$y =\frac{r_1+r_9+r_{13}+r_{14}}{12}$$

then $y$ is a root of the solvable quintic,

$$67 - 1748 y - 7033 y^2 - 1378 y^3 + 234 y^4 + y^5=0$$

with discriminant divisible by $11^4\times23^4$. Using the other quartic symmetric polynomials show that the $20$-deg is just a quartic in disguise, hence is solvable. All these suggest that $P(v)$ is the correct polynomial for $N=11$.

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I. Duplication

Following Nemo's lead in this answer, we find the formula, $$\frac{1}{2}I(p^2;\tfrac{1}{2},\tfrac{1}{3})=I(1+q^3;\tfrac{1}{2},\tfrac{1}{3})$$ where $p,q$ are related by the $12$-deg, $$p^2(-2 + 2 q + q^2)^6 = 36(1 + q^3) (4 + 4 q + 6 q^2 - 2 q^3 + q^4)^2$$ This then enables us to find infinitely many $\displaystyle\frac{1}{2^n N}$.

For example, since $I(p^2;\tfrac{1}{2},\tfrac{1}{3})=\frac{1}{3}$ is known, then solving for $I(\alpha^2;\tfrac{1}{2},\tfrac{1}{3})=\frac{1}{6}$ turns out to involve a $36$-deg equation.

II. Triplication

(Courtesy of Nemo.) Starting with, $$B\left(z;\frac{1}{2},\frac{1}{3}\right)=2 \sqrt{z} \, _2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};z\right). $$ The transformation $$ \, _2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};-\frac{3 z \left(1-\frac{z}{9}\right)^2}{(1-z)^2}\right)=\frac{(1-z) \, }{1-\frac{z}{9}}{}_2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};z\right) $$ applied two times gives $$ \frac{1}{3} B\left({\frac{(9-z)^2 z \left(z^3+225 z^2-405 z+243\right)^2}{729 (1-z)^2 (z+3)^6}};\frac{1}{2},\frac{1}{3}\right)=B\left(z;\frac{1}{2},\frac{1}{3}\right). $$

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