80
$\begingroup$

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.

$\endgroup$
  • 22
    $\begingroup$ Goodness, do you mind sharing how you come up with those integrals in the first place, let alone a subsequent conjecture?? $\endgroup$ – imranfat Oct 24 '13 at 20:24
  • 10
    $\begingroup$ 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. $\endgroup$ – Vladimir Reshetnikov Oct 24 '13 at 20:33
  • 3
    $\begingroup$ Very cool indeed, especially for someone who claims not to be a math professional (Yes, I checked your profile...) $\endgroup$ – imranfat Oct 24 '13 at 20:37
  • $\begingroup$ Perhaps trying to express the integral in terms of the Beta function of arguments -$\frac13$ and -$\frac16$ might be a start ? $\endgroup$ – Lucian Oct 24 '13 at 20:40
  • 3
    $\begingroup$ Several related conjectures are published here. $\endgroup$ – Vladimir Reshetnikov Oct 26 '13 at 1:30
51
+100
$\begingroup$

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.
$\endgroup$
  • 5
    $\begingroup$ very nice solution +1 $\endgroup$ – Shobhit Bhatnagar Mar 1 '14 at 13:30
22
$\begingroup$

$\def\Beta{B}\def\tfrac#1#2{{\textstyle\frac{#1}{#2}}}$ 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 $f(z)=w$.

$\endgroup$
  • $\begingroup$ Looks like $f(1-z)=1-f(z)$. $\endgroup$ – Chen Wang Feb 11 '14 at 10:50
  • $\begingroup$ Maybe there will be an addition formula for $f^{-1}(z)$, similar to the addition formula for trigonometric and elliptic functions. $\endgroup$ – Chen Wang Feb 11 '14 at 14:20
  • $\begingroup$ Would you happen to know the solution for $p=11$? Kindly see this post. $\endgroup$ – Tito Piezas III Nov 30 '16 at 9:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.