# Conjecture ∫10dx3√x6√1−x√1−x(√6√12+7√3−3√3−6)2=π9(3+√24√27)\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})

Let

Note that $\alpha$ is the unique positive root of the polynomial equation

Now consider the following integral:

I have a conjectured elementary value for it

Actually, the integral $I$ can be evaluated in an exact form using Mathematica or manually, using formula DLMF 15.6.1:

but I could not find a way to simplify this result to $(4)$.

So, my conjecture can be restated in a different form:

or

The conjecture can also be given in terms of the incomplete beta function:

Question: Is the conjecture indeed true?

Note: It holds numerically up to at least $10^4$ decimal digits.

Conjecture 2

Let

Added later: We can simplify it to

I conjecture that:

I can imagine that if these conjectures are true, then there are generalizations for some other algebraic numbers.

## Answer

Let $t = 1 + y^3$, we can rewrite $B\left(\alpha^2; \frac12, \frac13 \right)$ as

Following the setup in my answer to a related question. Let
$\;\displaystyle\eta = \frac{\Gamma\left(\frac13\right)\Gamma\left(\frac16\right)}{\sqrt{3\pi}}\;$ and $\wp(z)$ be the Weierstrass elliptic $\wp$ function with fundamental periods $1$ and $e^{i\pi/3}$. $\wp(z)$ is known to satisfy an ODE of the form

If one perform variable substitution $\;\displaystyle y = -\frac{4}{\eta^2} \wp\left(\frac{iz}{\eta}\right)$, one has

Using this, we can express conjecture $(8)$ in terms of $y(\cdot)$ and/or $\wp(\cdot)$:

Let $u_0 = i\frac{\sqrt{3}}{3}$, $u_{-1} = i\frac{\sqrt{3}}{2}$ and $u = i\frac{5\sqrt{3}}{12} = \frac12(u_0 + u_{-1})$. Using the addition formula for $\wp$ function,
we have

Using the duplication formula of $\wp$ function, we get

Let $Y = \frac{4}{\eta^2}\wp(u)$ and $A^2 = 1 - Y^3$, above condition is equivalent to

• Since $\alpha$ is a root for one of the factors in $(*1d)$, $A = \alpha$ satisfies $(*1d)$ and hence $(*1c)$.
• Since $0 < \alpha < 1$ implies $1 - \alpha^2 > 0$, $(*1c) \implies (*1b)$ in this particular case.
i.e. $Y = \sqrt[3]{1-\alpha^2}$ satisfies $(*1b)$ and hence $(*1a)$.
• Since $u$ lies between $u_0$ and $u_{-1}$, $\frac{4}{\eta^2}\wp(u) > 0$. Using
the fact $(*1a)$ has only one positive root, we find $\frac{4}{\eta^2}\wp(u) = \sqrt[3]{1-\alpha^2}$. i.e. conjecture $(8)$ is true.

Attribution
Source : Link , Question Author : Vladimir Reshetnikov , Answer Author : Community