$$\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}\stackrel?=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi\tag1$$ 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 $$I\left(\frac{\sqrt3}2;\ \frac14,\frac14\right)\stackrel?=\frac23,\tag2$$ where $I\left(z;\ a,b\right)$ is the regularized beta function.

Even simpler case: $$\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[6]{9-x}\ \sqrt[3]x}\stackrel?=\frac\pi{\sqrt3},\tag3$$ which is equivalent to $$I\left(\frac19;\ \frac16,\frac13\right)\stackrel?=\frac12.\tag4$$

A related question.

  • 52
    $\begingroup$ Is it a routine of yours to randomly choose equations in the morning, approximate them by the afternoon, and conjecture them in the night? $\endgroup$
    – chubakueno
    Oct 27, 2013 at 5:36
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    $\begingroup$ Where is this coming from ? $\endgroup$ Oct 27, 2013 at 5:40
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    $\begingroup$ @chubakueno I would say it's a hobby. $\endgroup$ Oct 27, 2013 at 5:40
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    $\begingroup$ I am sure you know that this is a hypergeometric function$$ \int_{0}^1 \frac{dx }{\sqrt{1-x} \sqrt[4]{x}\sqrt[4]{ 2-x \sqrt{3}}}=\frac{2^{9/4} \pi^{3/2} }{\Gamma^2\left(\frac14\right)}{}_2F_1\left( \frac14, \frac34, \frac54,\frac{\sqrt{3}}{2} \right) $$ Your answer almost certainly follows from some combination of its quadratic (or higher-order) transformations. But to be honest, I don't see what can one learn in the course of such calculation. $\endgroup$ Oct 27, 2013 at 5:45
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    $\begingroup$ Vladimir, you've been posting a lot of conjectures about integrals of this form. Given as @O.L. says that they seem to be related to Hypergeometric functions, could your questions be more systematically written down as - "How can we simplify the hypergeometric function for various inputs?" $\endgroup$ Oct 27, 2013 at 5:51

2 Answers 2


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

$$ F_{\alpha\beta}(\mu) = \int_0^1\frac{dx}{x^\alpha(1-x)^\beta(1-\mu x)^\gamma} \quad\text{ and }\quad \Delta = \frac{\Gamma(1-\alpha)\Gamma(1-\beta)}{\Gamma(1+\gamma)} $$ When $|\mu| < 1$, we can rewrite the integral $F_{\alpha\beta}(\mu)$ as

$$\begin{align} F_{\alpha\beta}(\mu) = & \int_0^1 \frac{1}{x^\alpha(1-x)^{\beta}}\left(\sum_{n=0}^{\infty}\frac{(\gamma)_n}{n!}\mu^n x^n\right) dx = \sum_{n=0}^{\infty}\frac{(\gamma)_n}{n!}\frac{\Gamma(n+1-\alpha)\Gamma(1-\beta)}{\Gamma(n+1+\gamma)}\mu^n\\ = & \Delta\sum_{n=0}^{\infty}\frac{(\gamma)_n (1-\alpha)_n}{n!(\gamma+1)_n}\mu^n = \Delta\gamma \sum_{n=0}^{\infty}\frac{(1-\alpha)_n}{n!(\gamma+n)}\mu^n \end{align}$$ This implies $$ \mu^{-\gamma} \left(\mu\frac{\partial}{\partial \mu}\right) \mu^{\gamma} F_{\alpha\beta}(\mu) = \Delta\gamma \sum_{n=0}^{\infty}\frac{(1-\alpha)_n}{n!}\mu^n = \Delta\gamma\frac{1}{(1-\mu)^{1-\alpha}} $$ and hence $$F_{\alpha\beta}(\mu) = \Delta\gamma \mu^{-\gamma} \int_0^\mu \frac{\nu^{\gamma-1}d\nu}{(1-\nu)^{1-\alpha}} = \Delta\gamma \int_0^1 \frac{t^{\gamma-1} dt}{(1-\mu t)^{1-\alpha}} = \Delta \int_0^1 \frac{dt}{(1 - \mu t^{1/\gamma})^{1-\alpha}}$$

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

$$F_{\alpha\beta}(\mu) = \int_0^1 \frac{dy}{y^\beta(1-y)^\alpha(1-\mu - \mu y)^{\gamma}} = \frac{1}{(1-\mu)^\gamma} F_{\beta\alpha}(-\frac{\mu}{1-\mu})$$

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

$$F_{\alpha\beta}(\mu) = \frac{\Delta}{(1-\mu)^{\gamma}}\int_0^1 \frac{dt}{( 1 + \omega^{1/\gamma} t^{1/\gamma})^{1-\beta}} = \frac{\Delta}{\mu^\gamma}\int_0^\omega \frac{dt}{(1 + t^{1/\gamma})^{1-\beta}}$$

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

$$\frac{\Gamma(\frac34)\Gamma(\frac12)}{\Gamma(\frac54) (\sqrt{3})^{1/4}}\int_0^\omega \frac{dt}{\sqrt{1+t^4}} \stackrel{?}{=} \frac{2\sqrt{2}}{3\sqrt[8]{3}} \pi\tag{*1}$$

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

$$K(m) = \int_0^1 \frac{dx}{\sqrt{(1-x^2)(1-mx^2)}}$$ 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

$$\int_0^\omega \frac{dt}{\sqrt{1+t^4}} \stackrel{?}{=} \frac23 K(\frac12)\tag{*2}$$

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:

$$u = \int_0^{\varphi(u)} \frac{dt}{\sqrt{1+t^4}}$$

Let $\psi(u)$ be $\frac{1}{\sqrt{2}}(\varphi(u) + \varphi(u)^{-1})$. It is easy to check/verify $$ \varphi'(u)^2 = 1 + \varphi(u)^4 \implies \psi'(u)^2 = 4 (1 - \psi(u)^2)(1 - \frac12 \psi(u)^2) $$

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

$$\psi(u) = \text{sn}(2u + \text{constant} | \frac12 )\tag{*3}$$

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.

$$\psi(u) = \text{sn}(2u + iK )$$

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

$$\frac{1}{\sqrt{2}} (\omega + \omega^{-1}) \stackrel{?}{=} \text{sn}( \frac43 K + i K)\tag{*4}$$

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 $\text{sn}(u+v)$

$$\text{sn}(u+v) = \frac{\text{sn}(u)\text{cn}(v)\text{dn}(v)+\text{sn}(v)\text{cn}(u)\text{dn}(u)}{1-m\,\text{sn}(u)^2 \text{sn}(v)^2}$$

One find in order for $\text{sn}(3u)$ to blow up, $\text{sn}(u)$ will be a root of following polynomial equation: $$3 m^2 s^8-4 m^2 s^6-4 m s^6+6 m s^4-1 = 0$$ Substitute $m = \frac12$ and $s = \frac{1}{\sqrt{2}}(t+\frac{1}{t})$ into this, the equation $\omega$ need to satisfy is given by:

$$(t^8 - 6 t^4 - 3)(3 t^8 + 6 t^4 - 1 ) = 0$$

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.

  • 8
    $\begingroup$ A nice solution! (+1) Can we verify the other integrals in the same way? $\endgroup$ Oct 31, 2013 at 9:35
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    $\begingroup$ @Integral, I don't know but I fear the chance is slim. The radicals in the other question has degree > 4! So the functions involved should be hyper-elliptic instead of elliptic functions. Without an elliptic function around, it is sort of hard (at least for me) to impose non-trivial polynomial constraints on the parameters/values in the corresponding integrals. $\endgroup$ Oct 31, 2013 at 9:52
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    $\begingroup$ @achillehui This is brilliant! What about the integral $(3)$ from my question above? $\endgroup$ Oct 31, 2013 at 16:24
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    $\begingroup$ @VladimirReshetnikov Let $I(s) = \int_0^{s} \frac{dt}{\sqrt{1+t^6}}$, it is easy to verify (3) is equivalent to $I(\frac{1}{\sqrt{2}}) = \frac12 I(\infty)$. If I didn't make any mistake, then $$\frac{(1-\sqrt{3})s^2+1}{(1+\sqrt{3})s^2+1} = \text{cn}\left( 2\sqrt[4]{3} I(s)\,\Big|\,\frac{2+\sqrt{3}}{4} \right)$$ and (3) can be verified using the double angle formula for $\text{cn}(\cdot)$. $\endgroup$ Nov 1, 2013 at 2:59

Here is another approach for evaluating the integral (3).

Using @achille hui's transformation, we can write

$$ I=\int_0^1\frac{dx}{\sqrt[3]{x}\sqrt{1-x}\sqrt[6]{9-x}}= \frac{4\pi^2 2^{\frac{1}{3}}}{\Gamma\left(\frac{1}{3}\right)^3}\int_0^\frac{1}{\sqrt{2}}\frac{1}{\sqrt{1+x^6}}dx $$ Using the substitution $x=\sqrt{t}$, we get $$ I=\frac{\pi^2 2^{\frac{4}{3}}}{\Gamma\left(\frac{1}{3}\right)^3}\int_0^\frac{1}{2}\frac{1}{\sqrt{t+t^4}}dt $$ We can now transform this integral into a beta integral using the substitution $t=\frac{1-y}{2+y}$. This gives us

\begin{align*} I &= \frac{\pi^2 2^{\frac{4}{3}}}{\Gamma\left(\frac{1}{3}\right)^3} \int_0^1\frac{1}{\sqrt{1-y^3}}dy \\ &= \frac{\pi^2 2^{\frac{4}{3}}}{3\Gamma\left(\frac{1}{3}\right)^3}B\left(\frac{1}{3},\frac{1}{2} \right) \\ &= \frac{\pi^2 2^{\frac{4}{3}}}{3\Gamma\left(\frac{1}{3}\right)^3}\left( \frac{\sqrt{3}\Gamma\left(\frac{1}{3}\right)^3}{\pi 2^{\frac{4}{3}}}\right) \\ &= \frac{\pi}{\sqrt{3}} \end{align*}


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