# Conjecture ∫10dx√1−x 4√x 4√2−x√3?=2√238√3π\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}\stackrel?=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi

The equality numerically holds up to at least $10^4$ decimal digits.

Can we prove that the equality is exact?

An equivalent form of this conjecture is

where $I\left(z;\ a,b\right)$ is the regularized beta function.

Even simpler case:

which is equivalent to

For $\alpha, \beta, \gamma \in (0,1)$ satisfying $\alpha+\beta+\gamma = 1$ and
$\mu \in \mathbb{C} \setminus [1,\infty)$, define

When $|\mu| < 1$, we can rewrite the integral $F_{\alpha\beta}(\mu)$ as

This implies

and hence

Notice if we substitute $x$ by $y = 1-x$, we have

Combine these two representations of $F_{\alpha\beta}(\mu)$ and let $\omega = \left(\frac{\mu}{1-\mu}\right)^{\gamma}$, we obtain

Let $(\alpha,\beta,\gamma) = (\frac14,\frac12,\frac14)$ and $\mu = \frac{\sqrt{3}}{2}$, the identity we want to check becomes

Let $K(m)$ be the complete elliptic integral of the first kind associated with modulus $m$. i.e.

It is known that $\displaystyle K(\frac12) = \frac{8\pi^{3/2}}{\Gamma(-\frac14)^2}$. In term of $K(\frac12)$, it is easy to check $(*1)$ is equivalent to

To see whether this is the case, let $\varphi(u)$ be the inverse function of above integral.
More precisely, define $\varphi(u)$ by following relation:

Let $\psi(u)$ be $\frac{1}{\sqrt{2}}(\varphi(u) + \varphi(u)^{-1})$. It is easy to check/verify

Compare the ODE of $\psi(u)$ with that of a Jacobi elliptic functions with modulus $m = \frac12$, we find

Since we are going to deal with elliptic functions/integrals with $m = \frac12$ only,
we will simplify our notations and drop all reference to modulus, i.e
$\text{sn}(u)$ now means $\text{sn}(u|m=\frac12)$ and $K$ means $K(m = \frac12)$.

Over the complex plane, it is known that $\text{sn}(u)$ is doubly periodic with
fundamental period $4 K$ and $2i K$. It has two poles at $i K$ and $(2 + i)K$ in the fundamental domain.
When $u = 0$, we want $\varphi(u) = 0$ and hence $\psi(u) = \infty$. So the constant
in $(*3)$ has to be one of the pole. For small and positive $u$, we want $\varphi(u)$ and hence $\psi(u)$ to be positive. This fixes the constant to $i K$. i.e.

and the condition $(*2)$ becomes whether following equality is true or not.

Notice $3( \frac43 K + i K) = 4 K + 3 i K$ is a pole of $\text{sn}(u)$. if one repeat
apply the addition formula for $\text{sn}(u+v)$

One find in order for $\text{sn}(3u)$ to blow up, $\text{sn}(u)$ will be a root of
following polynomial equation:

Substitute $m = \frac12$ and $s = \frac{1}{\sqrt{2}}(t+\frac{1}{t})$ into this, the equation $\omega$ need to satisfy is given by:

One can check that $\omega = \sqrt[4]{\frac{\sqrt{3}}{2-\sqrt{3}}}$ is indeed a root of this polynomial. As a result, the original equality is valid.