I'm trying to find a closed form (in terms of simpler functions) for the following hypergeometric function with a complex argument: $$\mathcal{Q}=\,_2F_1\left(\frac12,\frac23;\,\frac32;\,\frac{8\,\sqrt{11}\,i-5}{27}\right).\tag1$$ I have a guess (supported by thousands of digits from numerical calculations) about its argument (phase), but no ideas about its absolute value yet: $$\arg\mathcal{Q}\approx0.168669236010871306727578153...\stackrel?=\arccos\left(\frac{12+\sqrt{33}}{18}\right)\tag2$$ $$|\mathcal{Q}|\approx0.915170225773196416688677425...\tag3$$

Can you suggest any ideas how to prove the conjecture $(2)$? Is there a closed form for the absolute value?

As suggested by gammatester in a comment, the conjecture $(2)$ is equivalent to $$\arg\,B\left(\frac{8\,\sqrt{11}\,i-5}{27};\,\frac12,\frac13\right)\stackrel?=\frac\pi3,\tag4$$ where $B$ denotes the incomplete beta function.

Also, it can be shown that another equivalent form of the conjecture $(2)$ is $$B\left(\frac19;\,\frac16,\frac13\right)\stackrel?=\frac{\Gamma\left(\frac16\right)\,\Gamma\left(\frac13\right)}{2\,\sqrt\pi}.\tag5$$

  • $\begingroup$ Let $x=\frac{-5+i8\sqrt{11}}{27}$ (assuming that the positive square-root is chosen). Then one can show that $27 x^2 + 10 x + 27=0$ and $|x|=1$. You can try using the formula for the hypergeometric function as a series and then using the quadratic formula to reduce every higher power of $x$ to one that is at most linear in $x$. $\endgroup$
    – suresh
    Feb 24 '14 at 1:43
  • 3
    $\begingroup$ I don't know if this really helps, but your hypergeometric function is related to the incomplete Beta function with $a=\frac{1}{2}, \; b=\frac{1}{3}$ $$B_x(a,b)= \frac{x^a}{a}F(a,1-b,a+1,x), $$ see functions.wolfram.com/ $\endgroup$ Feb 24 '14 at 13:47
  • $\begingroup$ @gammatester Thanks, it makes the conjecture much more nicely-looking. $\endgroup$ Feb 24 '14 at 18:09
  • 1
    $\begingroup$ The right hand side of (5) can be rewritten as $\frac12 B(\frac16,\frac13)$. $\endgroup$
    – Kirill
    Feb 25 '14 at 19:21
  • 1
    $\begingroup$ @VladimirReshetnikov: Care to shed some light regarding this question? :) $\endgroup$ Dec 11 '16 at 6:19

Not sure how to transform conjecture $(2)$ to $(5)$. However, $(5)$ is true.

Let $\;\displaystyle t = \frac{1}{1+y^3}\;$, we can rewrite the integral $\;\displaystyle B\left(\frac19;\frac13,\frac16\right)\;$ as $$ \int_0^{1/9} t^{-5/6} (1-t)^{-2/3} dt = \int_\infty^2 (1+y^3)^{5/6}\left(\frac{1+y^3}{y^3}\right)^{2/3}\frac{-3y^2 dy}{(1+y^3)^2} = 3 \int_2^\infty \frac{dy}{\sqrt{1+y^3}} $$ Let $\omega = e^{\pi i/3}$ and $\mathbb{T} = \big\{\; m+n\omega : m, n \in \mathbb{Z}\;\big\}$ be the triangular lattice span by $1$ and $\omega$. Let $\wp(z)$ be the Weierstrass elliptic $\wp$ function with double poles on lattice $\mathbb{T}$:

$$\wp(z) = \frac{1}{z^2} + \sum_{\lambda \in \mathbb{T} \setminus \{ 0 \}} \left(\frac{1}{(z-\lambda)^2} - \frac{1}{\lambda^2}\right)$$

Let $\;\displaystyle\eta = \frac{\Gamma\left(\frac13\right)\Gamma\left(\frac16\right)}{\sqrt{3\pi}}\;$, it is known that $\wp(z)$ satisfies a differential equation of the form:

$$\wp'(z)^2 = 4 \wp(z)^3 - g_2 \wp(z) - g_3\quad\text{ where }\quad g_2 = 0 \;\text{ and }\;g_3 = \frac{\eta^6}{16}$$

Let $\;\displaystyle y(z) = -\frac{4}{\eta^2} \wp\left(\frac{iz}{\eta}\right)$. Using symmetry, it is not hard to see as $z$ varies from $0$ to $\frac{\eta}{\sqrt{3}}$ along the real axis, $y(z)$ remains real and positive, decreases from $\infty$ at $z = 0$ to $0$ at $z = \frac{\eta}{\sqrt{3}}$.

In terms of $z$, we have:

$$\frac{dy}{\sqrt{1+y^3}} = \frac{-\frac{4i}{\eta^3}\wp'\left(\frac{iz}{\eta}\right)dz}{ \sqrt{1-\frac{64}{\eta^6}\wp\left(\frac{iz}{\eta}\right)^3}} = \frac{-4i\wp'\left(\frac{iz}{\eta}\right)dz}{ \sqrt{16g_3-64\wp\left(\frac{iz}{\eta}\right)^3}} = - dz $$ From this, we get

$$B\left(\frac19;\frac13,\frac16\right) = -3 \left[ y^{-1}(\infty) - y^{-1}(2)\right] = 3y^{-1}(2)$$

This allow us to simplify conjecture $(5)$

$$ B\left(\frac19;\frac13,\frac16\right) \stackrel{?}{=} \frac{\Gamma\left(\frac13\right)\Gamma\left(\frac16\right)}{2\sqrt{\pi}} = \frac{\sqrt{3}}{2}\eta \iff y^{-1}(2) \stackrel{?}{=} \frac{\eta}{2\sqrt{3}} \iff \wp(\frac{i}{2\sqrt{3}}) \stackrel{?}{=} -\frac{\eta^2}{2} $$ Let $u = \frac{i}{2\sqrt{3}}$. Since $\wp(2u) = \wp\left(\frac{i}{\sqrt{3}}\right) = 0$, we can use the duplication formula for Weierstrass elliptic $\wp$ function to obtain

$$\begin{align} & 0 = \wp(2u) = \frac14\left(\frac{(6\wp(u)^2-\frac12 g_2)^2}{4\wp(u)^3-g_2\wp(u)-g_3}\right) -2\wp(u) \\ \iff & \wp(u)\left(\frac{\wp(u)^3 + 2g_3}{4\wp(u)^3-g_3}\right) = 0\\ \implies & \wp(u) = (-2g_3)^{1/3} = \left(-\frac{\eta^6}{8}\right)^{1/3} = -\frac{\eta^2}{2} \end{align}$$

i.e. the conjecture $(5)$ is true.

  • $\begingroup$ Could you give a reference to understand how the value $g_3 = \eta^6/16$ is found? Thanks. $\endgroup$ Mar 1 '14 at 10:06
  • 2
    $\begingroup$ @EstebanCrespi Writing $g_3$ as $\frac{\eta^6}{16}$. In our case, it corresponds to half period $\frac12$. This implies the half period for the case $g_3 = 1$ will be equal to $\frac12 \left(\frac{\eta^6}{16}\right)^{1/6} = 2^{-5/3}\eta$. The last number is known as the $\omega_2$ constant and has value equal to $\frac{\Gamma(1/3)^3}{4\pi}$, you can use that to deduce the expression of $\eta$ I use. $\endgroup$ Mar 1 '14 at 18:42
  • 1
    $\begingroup$ Two questions: 1.) How do we know that we need the lattice with $\omega=\exp(i \pi/3)$, despite of the fact that it works out in the end? For example, the choice of $\wp(z)$ with $g_2=0, g_3=-1$ seems to be more natural at first glance 2.) How do we know that $\wp(i\eta/\sqrt3)=0$? Away from the case of square lattices finding solutions to $\wp(z)=0$ is highly non-trivial (See for example people.mpim-bonn.mpg.de/zagier/files/doi/10.1007/BF01453974/…) $\endgroup$
    – asgeige
    Sep 10 '20 at 16:38
  • $\begingroup$ @achillehui push $\endgroup$
    – asgeige
    Oct 9 '20 at 18:33

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.