Conjecture $\int_0^1\frac{dx}{\sqrt[3]x\,\sqrt[6]{1-x}\,\sqrt{1-x\left(\sqrt{6}\sqrt{12+7\sqrt3}-3\sqrt3-6\right)^2}}=\frac\pi9(3+\sqrt2\sqrt[4]{27})$

Let $$\alpha=\sqrt{6}\ \sqrt{12+7\,\sqrt3}-3\,\sqrt3-6\,=\,\big(2+\sqrt{3}\big) \big(\sqrt{2} \sqrt[4]{27}-3\big)\,=\,\frac{3\sqrt{3}}{3+\sqrt2\ \sqrt[4]{27}}.\tag1$$ Note that $\alpha$ is the unique positive root of the polynomial equation $$\alpha^4+24\,\alpha^3+18\,\alpha^2-27=0.\tag2$$ Now consider the following integral: $$I=\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}.\tag3$$ I have a conjectured elementary value for it $$I\stackrel?=\frac\pi9\Big(3+\sqrt2\ \sqrt[4]{27}\Big)=\color{blue}{\frac{\pi}{\sqrt{3}\,\alpha}}.\tag4$$

Actually, the integral $I$ can be evaluated in an exact form using Mathematica or manually, using formula DLMF 15.6.1: $$I=\frac{4\,\sqrt\pi}{\sqrt3}\cdot\frac{\Gamma\left(\frac56\right)}{\Gamma\left(\frac13\right)}\cdot{_2F_1}\left(\frac12,\frac23;\ \frac32;\ \alpha^2\right),\tag5$$ but I could not find a way to simplify this result to $(4)$.

So, my conjecture can be restated in a different form: $${_2F_1}\left(\frac12,\frac23;\ \frac32;\ \alpha^2\right)\stackrel?=\frac{\sqrt2+\sqrt[4]3}{4\,\sqrt[4]{27}}\cdot\frac{\Gamma\left(\frac13\right)}{\Gamma\left(\frac56\right)}\cdot\sqrt\pi\tag6$$ or $$\sum_{n=0}^\infty\frac{\Gamma\left(n+\frac23\right)}{(2\,n+1)\,\Gamma(n+1)}\alpha^{2\,n}\stackrel?=\frac{3+\sqrt2\,\sqrt[4]{27}}{18}\cdot\frac{\pi^{3/2}}{\Gamma\left(\frac56\right)}.\tag7$$

The conjecture can also be given in terms of the incomplete beta function: $$B\left(\alpha^2;\ \frac12,\frac13\right)\stackrel?=\frac{\sqrt\pi}2\cdot\frac{\Gamma\left(\frac13\right)}{\Gamma\left(\frac56\right)}.\tag8$$

Question: Is the conjecture indeed true?

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

Conjecture 2

Let $${\small \begin{multline} \beta=\frac{21}4+\frac{9\,\sqrt5}4-\frac{15}8\sqrt{750-330\,\sqrt5}-\frac{33}8\sqrt{150-66\,\sqrt5} \\ + \frac12\sqrt{3\left(165+75\,\sqrt5-46\,\sqrt{750-330\,\sqrt5}-103\,\sqrt{150-66\,\sqrt5}\right)}. \end{multline}}\tag9$$ Added later: We can simplify it to $$\small\beta=\frac 3 4 \left(7+3 \sqrt 5-\sqrt[4] 5 \sqrt{66+30 \sqrt 5}\right)+\frac 1 2 \sqrt{495+225 \sqrt 5-3 \sqrt{6 \big(8545+3821 \sqrt 5\big)}}.\tag{9'}$$

I conjecture that: $$\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\beta^2}}\stackrel?=\color{blue}{\frac{2\,\pi}{5\,\sqrt3\,\beta}}.\tag{10}$$

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

• Goodness, do you mind sharing how you come up with those integrals in the first place, let alone a subsequent conjecture?? Oct 24, 2013 at 20:24
• Initially, I tried to solve a reverse problem: determine $\alpha$ from a similar integral whose value I knew. I calculated $\alpha$'s approximate value using numerical methods, and when I found enough decimal digits, Mathematica's command RootApproximant quite unexpectedly returned a candidate polynomial of $4^{th}$ degree with moderate coefficients, that was supported by further numeric calculations. Then I made a conjecture of that. Oct 24, 2013 at 20:33
• Very cool indeed, especially for someone who claims not to be a math professional (Yes, I checked your profile...) Oct 24, 2013 at 20:37
• Perhaps trying to express the integral in terms of the Beta function of arguments -$\frac13$ and -$\frac16$ might be a start ? Oct 24, 2013 at 20:40
• Several related conjectures are published here. Oct 26, 2013 at 1:30

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

$$\int_0^{\alpha^2} t^{-1/2} (1-t)^{-2/3} dt = \int_{-1}^{-\sqrt[3]{1-\alpha^2}} \frac{3y^2dy}{\sqrt{1+y^3} y^2} = 3 \int_{-1}^{-\sqrt[3]{1-\alpha^2}} \frac{dy}{\sqrt{1+y^3}}$$

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 $$\wp'(z)^2 = 4 \wp(z)^3 - g_2 \wp(z) - g_3\quad\text{ where }\quad g_2 = 0 \;\text{ and }\;g_3 = \frac{\eta^6}{16}$$ If one perform variable substitution $\;\displaystyle y = -\frac{4}{\eta^2} \wp\left(\frac{iz}{\eta}\right)$, one has

$$\frac{dy}{\sqrt{1+y^3}} = -dz\quad\text{ and }\quad \begin{cases} y\left(\frac{\sqrt{3}\eta}{3}\right) = -\frac{4}{\eta^2}\wp\left(i\frac{\sqrt{3}}{3}\right) = 0\\ \\ y\left(\frac{\sqrt{3}\eta}{2}\right) = -\frac{4}{\eta^2}\wp\left(i\frac{\sqrt{3}}{2}\right) = -1 \end{cases}$$ Using this, we can express conjecture $(8)$ in terms of $y(\cdot)$ and/or $\wp(\cdot)$: \begin{align} & B\left(\alpha^2; \frac12, \frac13 \right) \stackrel{?}{=} \frac{\sqrt{\pi}}{2}\frac{\Gamma\left(\frac13\right)}{\Gamma\left(\frac56\right)} = \frac{\sqrt{3}}{4}\eta\\ \iff & 3\left[y^{-1}(-1) - y^{-1}(-\sqrt[3]{1-\alpha^2})\right] \stackrel{?}{=} \frac{\sqrt{3}}{4}\eta\\ \iff & y^{-1}(-\sqrt[3]{1-\alpha^2}) \stackrel{?}{=} \frac{5\sqrt{3}}{12}\eta\\ \iff & \frac{4}{\eta^2}\wp\left(i\frac{5\sqrt{3}}{12}\right) \stackrel{?}{=} \sqrt[3]{1-\alpha^2} \end{align} 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

$$\wp(2u) = \wp(u_0 + u_{-1}) = \frac14\left[\frac{\wp'(u_0)-\wp'(u_{-1})}{\wp(u_0)-\wp(u_{-1})}\right]^2 - \wp(u_0) - \wp(u_{-1})\\ =\frac14\left[\frac{-i\frac{\eta^3}{4} - 0}{0 - \frac{\eta^2}{4}}\right]^2 - 0 - \frac{\eta^2}{4} = -\frac{\eta^2}{2}$$ Using the duplication formula of $\wp$ function, we get $$-\frac{\eta^2}{2} = \wp(2u) = \frac14\left(\frac{(6\wp(u)^2-\frac12 g_2)^2}{4\wp(u)^3-g_2\wp(u)-g_3}\right) -2\wp(u) = \frac{9\wp(u)^4}{4\wp(u)^3-\frac{\eta^4}{16}} - 2\wp(u)$$ Let $Y = \frac{4}{\eta^2}\wp(u)$ and $A^2 = 1 - Y^3$, above condition is equivalent to

\begin{align} & Y^4 + 8 Y^3 + 8 Y - 8 = 0\tag{*1a}\\ \iff & Y(8+Y^3) = 8(1-Y^3)\tag{*1b}\\ \implies & (9-A^2)^3(1-A^2) = 512A^6\tag{*1c}\\ \iff & (A^4-24A^3+18A^2-27)(A^4+24A^3+18A^2-27) = 0\tag{*1d} \end{align}

• 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.
• very nice solution +1 Mar 1, 2014 at 13:30

$\def\Beta{B}\def\tfrac#1#2{{\textstyle\frac{#1}{#2}}}$ Perhaps this might be helpful to someone. The integral is equal to, as you note, $$J(y) = \int_0^1 x^{-1/3}(1-x)^{-1/6}(1-xy^2)^{-1/2}\,dx = \frac{2\pi}{y\sqrt{3}} \frac{\Beta(y^2,\frac12,\frac13)}{\Beta(\frac12,\frac13)},$$ where $\Beta(z,a,b)$ is the incomplete beta function. Consider the function $$I(y^2,a,b) = \frac{\Beta(y^2,a,b)}{\Beta(a,b)},$$ and rewrite using DLMF 8.17 it as $$I(y^2,\tfrac12,\tfrac13) = 1-2I(z,\tfrac13,\tfrac13), \qquad 4z(1-z) = 1-y^2.$$ Then the function $$f(z) = I(z,{\textstyle\frac13,\frac13}) = 3z^{1/3}\frac{{}_2F_1(\tfrac13,\tfrac23;\tfrac43;z)}{\Beta(\frac13,\frac13)}$$ is the one for which the conjectures are equivalent to (choosing the right root $z$): $$f(z) = \tfrac14 \quad\Leftrightarrow\quad z^4-14z^3+24z^2-14z+1=0,$$ $$f(z) = \tfrac25 \quad\Leftrightarrow\quad 1+17 z-107 z^2+164 z^3-155 z^4+164 z^5-107 z^6+17 z^7+z^8=0,$$

From my tests it appears (I don't know how to prove this) that the function $z(w)$ which solves the equation $f(z(w)) = w$ always has algebraic values when $w$ is a rational number. The original integral is $\frac{2\pi}{y\sqrt{3}}(1-2f(z))$, which is algebraic times $\pi$ when $f(z)$ is rational and $z$ algebraic, so I think the question is really about solving $f(z)=w$.

• Looks like $f(1-z)=1-f(z)$. Feb 11, 2014 at 10:50
• Maybe there will be an addition formula for $f^{-1}(z)$, similar to the addition formula for trigonometric and elliptic functions. Feb 11, 2014 at 14:20
• Would you happen to know the solution for $p=11$? Kindly see this post. Nov 30, 2016 at 9:05