# On the integral $\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt[4]{1-x}\ \sqrt{1-x\,\beta^2}}=\frac{2\pi}{7\sqrt{2}\,\beta}$ and $\cos\frac{2\pi}{7}$

(This continues the post on integrals that use roots of reciprocal polynomials.)

Given $N=7$. First, how do we show that the algebraic number $\beta$ that solves,

$$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt[4]{1-x}\ \sqrt{1-x\,\beta^2}}=\frac{1}{7}\,\frac{2\pi}{\sqrt{2}\;|\beta|}\tag1$$

is a root of,

$$\beta^2-\frac{2}{1+2\cos\tfrac{\pi}{14}}\beta+\frac{4}{2-\sqrt{2}\,\sec\tfrac{13\pi}{28}}=0$$

Second, the algebraic number $\gamma$ for the related integral,

$$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{1-x\,\gamma^2}}=\frac{1}{7}\,\frac{2\pi}{\sqrt{2\gamma}}\tag2$$

is a root of,

$$\gamma^2-2\cos\tfrac{2\pi}{7}\,\gamma+\frac{2-\sqrt{2}\,\csc\tfrac{9\pi}{28}}{4}=0$$

Equivalently, these roots $\beta$ and $\gamma$ satisfy,

$$I\left(\beta^2;\ \tfrac12,\tfrac14\right)=I\left(\gamma^2;\ \tfrac14,\tfrac14\right)=\frac{1}{7}$$

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

Lastly, why is the minpoly of $\beta$ and $\gamma$ which are $12$-deg equations actually reciprocal polynomials?

P.S. The cases $N=2,3,5$ can also be expressed by trigonometric functions, such as for $N=5$, $$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt[4]{1-x}\ \sqrt{1-x\big(1-\tan\tfrac{3\pi}{20}\big)^2}}=\frac{\sqrt{2}\,\pi}{5-5\tan\tfrac{3\pi}{20}}$$ though to do so for $N=11$ given in this post would be difficult.

• For $N=7$ : $\gamma=\frac{\beta^2}{2-\beta^2}$. I think the relation $\gamma^2=\frac{\beta^4}{(2-\beta^2)^2}$ suggested by ${I\left(z,\frac{1}{2},\frac{1}{4}\right)}={I\left({\frac{z^2}{(2-z)^2}},\frac{1}{4},\frac{1}{4}\right)}$ holds for all rational N, since the function $B\left(z,\frac{1}{4},\frac{1}{4}\right)$ is monotonic. – Nemo Dec 3 '16 at 20:56
• BTW, I have got an explicit expression in radicals and trig functions for the 30-degree root arising in case $N=11$. It is huge, but I am working on simplification. – Vladimir Reshetnikov Aug 16 at 4:35
• @VladimirReshetnikov: Ok. Also, that 30-deg minpoly of $\gamma$ for $$I\left(\gamma^2;\ \tfrac14,\tfrac14\right) = \tfrac1{11}$$ is here. We didn't express it in radicals. – Tito Piezas III Aug 16 at 8:19
• I finally did express it (see the same link). – Vladimir Reshetnikov Sep 9 at 5:39