On the integral $\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{1-x\,\gamma^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{2\gamma}}$

Question

V. Reshetnikov gave the interesting integral,

$$\int_0^1\frac{\mathrm dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{2-x\,\sqrt3}}=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi\tag1$$

After some experimentation, it turns out that more generally, given some integer/rational $N$, we are to find an algebraic number $\gamma$ that solves,

$$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{1-x\,\gamma^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{2\gamma}}\tag2$$

(Compare to the similar integral in this post.) Equivalently, to find $\gamma$ such that,

$$\begin{aligned} \frac{1}{N} &=I\left(\gamma^2;\ \tfrac14,\tfrac14\right)\\[1.8mm] &= \frac{B\left(\gamma^2;\ \tfrac14,\tfrac14\right)}{B\left(\tfrac14,\tfrac14\right)} \end{aligned} \tag3$$

with beta function $B(a,b)$, incomplete beta $B(z;a,b)$ and regularized beta $I(z;a,b)$, and $B\left(\tfrac14,\tfrac14\right)=\frac{\sqrt\pi}{\Gamma^2\left(\frac14\right)}$. Reshetnikov's example, after tweaking, was just the case $N=\frac{3}{2}$ and $\gamma=\frac{3^{1/4}}{\sqrt{2}}$.

Solutions for prime $N=2,3,5,7$ are known. Let $v=\gamma$, then, $$-1 + 2 v^2 = 0\quad\quad N=2\\ - 1 + 2 v + 2 v^2 = 0\quad\quad N=3\\ - 1 + 8 v - 4 v^2 - 8 v^3 + 4 v^4 = 0\quad\quad N=5$$ etc, with $N=7$ using a $12$-deg equation. I found these using Mathematica's FindRoot command but, unlike the other post, I couldn't find a nice common form for $\gamma$. (The pattern of this family is also different. I had expected $N=7$ to also involve a sextic only.)

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

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Update, Aug 16, 2019

In this comment, Reshetnikov gave the explicit solution to,

$$I\left(\gamma^2;\ \tfrac14,\tfrac14\right) = \tfrac17$$ as, $$\small\gamma = \frac16\left(5\cos x-\sqrt3\sin x-1-\sqrt3\sqrt{7+4\sqrt7-(11+2\sqrt7)\cos x+\sqrt3(5+2\sqrt7)\sin x}\right)$$ where $x = \tfrac13\arccos\big(\tfrac{13}{14}\big)$.

P.S. I forgot I also found $\gamma$ in this 2016 post as,

$$\gamma = \tfrac12\left(2\cos\tfrac{2\pi}7-\sqrt{2\cos\tfrac{4\pi}7+\sqrt2\csc\tfrac{9\pi}{28}}\right)$$

Answer 1

Taking $ \gamma $ to be the smallest positive root (the 4th root according to Mathematica) of the degree 30 polynomial \begin{align*} f(x) = -1 + 30 x + 170 x^2 + 672 x^3 - 6956 x^4 - 6808 x^5 + 42872 x^6 - 56576 x^7 - 241616 x^8 + 712800 x^9 + 1099296 x^{10} - 2718208 x^{11} - 3427264 x^{12} + 5028992 x^{13} + 8030592 x^{14} - 3956736 x^{15} - 14783232 x^{16} - 2065920 x^{17} + 20241920 x^{18} + 7954432 x^{19} - 19317760 x^{20} - 7817216 x^{21} + 12445696 x^{22} + 3342336 x^{23} - 5435392 x^{24} - 122880 x^{25} + 1662976 x^{26} - 393216 x^{27} - 344064 x^{28} + 98304 x^{29} + 32768 x^{30} \end{align*} seems to give a solution for $ N = 11 $.

Answer 2

It seems solving $I(\beta^2;\tfrac{1}{2},\tfrac{1}{4})=\frac1N$ also solves $I(\gamma^2;\tfrac{1}{4},\tfrac{1}{4})=\frac1N$ as they involve reciprocal polynomials. So again following Nemo's lead in this answer, we employ the duplication formula, $$\frac{1}{2}I(p^2;\tfrac{1}{2},\tfrac{1}{4})=I(1-q^2;\tfrac{1}{2},\tfrac{1}{4})$$ where $p,q$ are related by the $8$-deg, $$p^2 (-1 + 2 q + q^2)^4 = 16 q (-1 + q^2) (1 + q^2)^2\tag1$$ So given a solution to $\displaystyle I(\beta^2;\tfrac{1}{2},\tfrac{1}{4})= \frac{1}{2^n N}$ for $n=0$, the duplication formula enables us to find infinitely many for $n\geq1$.

For example, the solution to $I(p^2;\tfrac{1}{2},\tfrac{1}{4})=\frac{1}{3}$ is $p = 1-\sqrt3$. One can then solve for $I(\beta^2;\tfrac{1}{2},\tfrac{1}{4})=\frac{1}{6}$ by plugging the known $p$ into $(1)$.