# Integral involving hypergeometric function $\int_0^1[{}_2F_1(\frac13,\frac23;1;x^3)]^2dx$

Question: How to prove $$I=\int_0^1\bigg[{}_2F_1\left(\frac13,\frac23;1;x^3\right)\bigg]^2dx=\frac{\sqrt3}{32\pi^5}\Gamma\left(\frac13\right)^9?$$

Source: An integral competition post of my country.

Attempt
Recall the series definition of hypergeometric function $$_2F_1(a,b,c,x)=\sum_{n=0}^\infty\frac{(a)_n(b)_n}{(c)_nn!}x^n,$$ we can transform $$I$$ into the series form $$I=\sum_{n,m=0}^\infty\frac{(a)_n(a)_m(b)_n(b)_m}{(c)_n(c)_mn!m!(3n+3m+1)}$$ But I can not handle this series.
I also thought of using complex method. $$I=\int_0^1\bigg[{}_2F_1\left(\frac13,\frac23;1;x\right)\bigg]^2\frac{x^{-2/3}}3dx$$ then let $$f(z)=\bigg[{}_2F_1\left(\frac13,\frac23;1;z\right)\bigg]^2(-z)^{-2/3}$$ and use keyhole contour, where $$(\cdot)^{-2/3}$$ is the principal branch of the multi-valued function. But the nature of the branch of the integrand in the inteval $$[1,\infty)$$ is too complex for me to handle. It involves another definite integral which is similar to $$I$$.

Before attacking the integral, I mention something about cubic theta function. The whole solution heavily exploits tools from modular forms. The "footnote" contains more information.

The three cubic theta functions are defined by \begin{aligned} a(q) &= \sum_{m,n} q^{m^2+mn+n^2}\\ b(q) &= \sum_{m,n} \zeta_3^{m-n} q^{m^2+mn+n^2}\\ c(q) &= \sum_{m,n} q^{{(m+\frac{1}{3})^2+(m+\frac{1}{3})(n+\frac{1}{3})+(n+\frac{1}{3})^2}} \end{aligned} where $$\zeta_3 = e^{2\pi i/3}$$, sum is over all $$m,n\in \mathbb{Z}$$. Then it can be shown$$^1$$ that $$a(q)^3 = b(q)^3+c(q)^3$$ $$a(q) = \frac{\eta^3(q) + 9 \eta^3(q^9)}{\eta (q^3)}\qquad b(q) = \frac{\eta^3(q)}{\eta(q^3)}\qquad c(q) = 3\frac{\eta^3(q^3)}{\eta(q)}$$ where $$\eta(q) = q^{1/24} \prod_{n\geq 1}(1-q^n)$$ is the Dedekind eta function.

Define $$K_3(m) = {_2F_1}(\frac{1}{3},\frac{2}{3};1;m)$$ Similar to elliptic integrals, denote $$K_3'(m) = K_3(1-m), m' = 1-m$$. Then one easily shows (I omit the subscript $$3$$): $$\frac{d}{dm}(\frac{K'}{K}) = -\frac{\sqrt{3}}{2\pi}\frac{1}{mm'K^2}$$

Moreover, letting $$q= \exp(-\frac{2\pi}{\sqrt{3}}\frac{K'(m)}{K(m)})$$, the following inversion formula holds$$^2$$ when $$0: $$a(q) = K(m)\qquad b(q)=(1-m)^{1/3} K(m)\qquad c(q) = m^{1/3} K(m)$$

Now we tackle the integral, $$I = \frac{1}{3}\int_0^1 {{m^{ - 2/3}}K{{(m)}^2}dm}$$ we make the substitution $$q = \exp ( - \frac{{2\pi }}{{\sqrt 3 }}\frac{{K'(m)}}{{K(m)}})$$, the above formulas imply $$dq = \frac{{q}}{{mm'{K^2}}}dm$$, as $$m$$ increases from $$0$$ to $$1$$, $$q$$ increases from $$0$$ to $$1$$. $$I = \frac{1}{3}\int_0^1 {\frac{{b{{(q)}^3}c(q)}}{{mm'{K^2}}}dm} = \frac{1}{3}\int_0^1 {\frac{{b{{(q)}^3}c(q)}}{q}dq} = \int_0^1 {\frac{{\eta {{(q)}^8}}}{q}dq}$$ Next, I will use notation $$\eta(q),\eta(\tau)$$ interchangably (the common notation in context of modular forms), where $$q = e^{2\pi i \tau}$$. Make $$q=e^{-2\pi x}$$, then $$I$$ becomes $$I = 2\pi \int_0^\infty {\eta {{(ix)}^8}dx} = 2\pi \int_0^\infty {{x^2}\eta {{(ix)}^8}dx}$$ where in last step, I used $$\eta(-1/\tau) = \sqrt{-i\tau} \eta(\tau)$$. Transform it back to $$q$$: $$\tag{1} I = \frac{1}{{4{\pi ^2}}}\int_0^1 {\frac{{{{\ln }^2}q}}{q}\eta {{(q)}^8}dq}$$ It can be shown that$$^3$$: $$\eta {(q)^8} = - \frac{1}{2}\sum\limits_{v \in S} {({v_0} - {v_1})({v_1} - {v_2})({v_0} - {v_2}){q^{{{\left\| v \right\|}^2}/6}}}$$ $$S = \left\{ {v \in {\mathbb{R}^3}|v = ({v_0},{v_1},{v_2}) = (3n,3m + 1,3r - 1),n + m + r = 0,n,m,r\in\mathbb{Z}} \right\}$$ with $$\|v\|$$ the norm of a vector. Plug this into (1): $$I = \frac{{ - 1}}{{{{(2\pi )}^2}}}{6^3}\sum\limits_{v \in S} {\frac{{({v_0} - {v_1})({v_1} - {v_2})({v_0} - {v_2})}}{{{{\left\| v \right\|}^6}}}}$$ Denote $$\rho = e^{\pi i/3}$$. Note that $$({v_0} - {v_1})({v_1} - {v_2})({v_0} - {v_2}) = 2\Re {({v_0} + \rho {v_1})^3}$$ and $${\left\| v \right\|^6} = 8{({v_0} + \rho {v_1})^3}{({v_0} + {\rho ^{ - 1}}{v_1})^3}$$ we obtain $$I = \frac{{ - 27}}{{2{\pi ^2}}}\Re \sum\limits_{v\in S} {\frac{1}{{{{({v_0} + {\rho ^{ - 1}}{v_1})}^3}}}} = - \frac{{27}}{{2{\pi ^2}}}\Re \sum\limits_{(m,n) \in {\mathbb{Z}^2}} {\frac{1}{{{{(3n + {\rho ^{ - 1}}(3m + 1))}^3}}}}$$ The latter can be recognized as an Eisenstein series of level $$3$$, but to calculate its value, it is best to use Weierstrass elliptic function. Let $$\wp_{1,\rho}$$ denote this elliptic function with periods $$\{1,\rho\}$$, then $${\wp _{1,\rho }}'(z) = - 2\sum\limits_{n,m} {\frac{1}{{{{(z + n + m\rho )}^3}}}}$$ gives $$I=\frac{1}{{4{\pi ^2}}}\Re \left[{\wp _{1,\rho }}'(\frac{{{\rho ^{ - 1}}}}{3})\right] = \frac{{{\omega ^3}}}{{4{\pi ^2}}}\Re\left[ {\wp _{\omega ,\omega \rho }}'(\frac{{{\omega\rho ^{ - 1}}}}{3})\right]$$ where $$\omega = \Gamma(1/3)^3/(2\pi)$$, then it is well-known that modular invariants associated to periods $$\{\omega,\omega\rho\}$$ are $$g_2 = 0, g_3 = 1$$. Therefore $${\wp _{\omega ,\omega \rho }}'(\frac{{\omega {\rho ^{ - 1}}}}{3})$$ is the $$y$$-coordinate of a $$3$$-torsion of the elliptic curve $$y^2 = 4x^3 - g_2 x - g_3 = 4x^3 -1$$, which can be readily calculated to be $$\sqrt{3}$$. Finally we finish the calculation:$$I = \omega^3\sqrt 3/(4\pi^2)$$.

$$^1$$: Proof outline: $$a(q^3),b(q^3),c(q^3)$$ are modular forms of weight $$1$$ and level $$27$$, therefore it suffices to verify their $$q$$-expansions to certain power of $$q$$. A self-contained approach can be found in the 1994 paper Cubic Analogues of the Jacobian Theta Function.

$$^2$$: Proof outline: $$f=c^3(\tau)/a^3(\tau)$$ is modular function of $$\Gamma_0(3)$$, by a fact in modular forms, $$b(\tau)$$ satisfies a 2nd order ODE in terms of $$f$$, its coefficients are rational functions of $$f$$ since modular curve $$X(3)$$ has genus $$0$$. Therefore, in certain region of $$\mathbb{H}$$, $$b(\tau) = (1-f)^{1/3} K_3(f)$$, we could replace $$\tau$$ by $$\gamma\tau$$ for $$\gamma\in \Gamma_0(3)$$, modularity of $$b$$ allows us to isolate the $$\tau$$. But doing this replacement might change it into another linear independent solution of the ODE, which explains why $$K'/K$$ arises. The details are more delicate.

$$^3$$: The exponent $$8$$ is special here, which is the dimension of semisimple Lie algebra $$A_2$$. There is a corresponding formula for $$\eta(q)^d$$ each semisimple Lie algebra with dimension $$d$$. See Affine Root Systems and Dedekind's eta-Function by I.G. Macdonald.

• +1. This actually proves a particular case of an old problem of mine: math.stackexchange.com/questions/1811490/… – nospoon Aug 18 '19 at 15:35
• (+1) amazing work. Could you take a look at this question? – clathratus Aug 18 '19 at 15:40
• @nospoon yeah, I was aware of your question. This proves your conjecture of $I(8)$, there are also similar expressions for $I(10), I(14)$. But I doubt this is also true for higher $n$. – pisco Aug 19 '19 at 5:17
• Yes, I already found the closed forms for $I(10)$ and $I(14)$, and currently working on the $n=26$ case. Regarding other $n$'s, I have the same doubts as you. – nospoon Aug 19 '19 at 5:23
• @nospoon Very interesting statement and paper! Using the expansion therein, I calculated the value fo $I(26)$, which involves both $\Gamma(1/4)$ and $\Gamma(1/3)$. – pisco Aug 20 '19 at 3:44

Not an answer, but an extended comment for now.

This hypergeometric function is a special case, and some complicated quadratic and cubic transformations apply to it. See this like for reference: https://dlmf.nist.gov/15.8.

The formulas 15.8.25 and 15.8.26 both apply here.

However, the most interesting one is so called Ramanujan’s Cubic Transformation (15.8.33):

$${_2 F_1} \left( \frac13, \frac23;1;x^3 \right)= \frac{1}{1+2 x} {_2 F_1} \left( \frac13, \frac23;1;1- \frac{(1-x)^3}{(1+2x)^3}\right)$$

Update:

The iteration:

$$x_{n+1}=\left( 1- \frac{(1-x_n)^3}{(1+2x_n)^3}\right)^{1/3}$$

Converges to $$x_{\infty}=1$$ for any $$x \in (0,1]$$. Not sure how to use this, because $${_2 F_1} \left( \frac13, \frac23;1;1 \right)= \infty$$.

This transformation is related to the cubic analogue of the arithmetic-geometric mean. See the references at DLMF and also these questions:

Integral identity related with cubic analogue of arithmetic-geometric mean

Evaluate the integral $\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$

Some formulas from the question above (and Nemo's answer) might be useful here, for example:

$$\int_0^\infty \frac{dt}{\sqrt{(t^3+1)(t^3+p)}}=\frac{2 \pi}{3 \sqrt{3}} {_2F_1} \left(\frac{1}{2},\frac{2}{3};1;1-p \right)= \\ =\frac{2 \pi}{3 \sqrt{3}p^{1/3}}{_2F_1} \left(\frac{1}{3},\frac{2}{3};1;\frac{(1-\sqrt{p})^2}{-4\sqrt{p}} \right)$$

This is just an application of already linked transformations, and can be applied backwards in this case.