Prove ${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{2/3}}=\frac{3^{3/2}}{2^{4/3}5^{5/6}\pi }\Gamma^3\left(\frac13\right)$

Here is one more conjecture I discovered numerically: $${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{\small2/3}}\stackrel{\color{#808080}?}=\frac{3^{\small3/2}}{2^{\small4/3}\,5^{\small5/6}\,\pi }\Gamma^3\!\!\left(\tfrac13\right)$$ How can we prove it?

Note that $\sqrt[3]{9+4\sqrt5}=\phi^2$. Mathematica can evaluate this integral, but gives a large expression in terms of Gauss and Appel hypergeometric functions of irrational arguments.

• Perhaps it would be better to ask directly for a general approach to evaluating $\displaystyle\int_{0\text{ or }-1}^1(x^n+a)^m(1-x^p)^q~dx$. For $a=0$, the connection to the beta function is obvious. Jul 27, 2014 at 19:38
• Perhaps the formula $\int_0^1 x^{a-1} (1-x)^{b-1} (p + x)^{-a - b} \,dx= \frac{ B(a, b) }{ p^b (1+p)^a }$ can be of use here. I made an attempt but with no success so far. Jul 27, 2014 at 21:49
• The integral clearly gives a hypergeometric function (set $x=2t-1$). The equivalent identity is $$_2F_1\left(\frac13,\frac13;\frac23; -40\eta^3\right)= \frac{3}{5\eta},$$ where $\eta=5^{-1/6}\phi^2$. Jul 27, 2014 at 23:34
• Using DLMF15.8.13 and 15.8.15, we have another equivalent identity: $${}_2F_1\left(\frac13;\frac12;\frac56;\frac45\right)\stackrel?=\frac{3}{\sqrt{5}}.$$ Aug 7, 2014 at 13:33
• Look like it's partially solved by the top answer in This question. Oct 9, 2014 at 8:08

As suggested by Chen Wang, this integral is related to an integral of the form $$\int_0^1 \frac{dx}{ \sqrt{1-x} \; x^{2/3} (1-zx)^{1/3} }$$ which appears in this forum's Question 879089 by the same author.

Here's a more direct route than given by Kirill, without conversions between definite integrals and hypergeometric functions. A differential $dx \left/ \left(\sqrt[3]{A+Bx} \, (1-x^2)^{2/3}\right)\right.$ has four poles of fractional order, two at $x = \pm 1$ of order $2/3$, and two at $x = \infty$ and $x = -B/A$ of order $1/3$. Hence the differential is invariant under an involution in the form of a fractional linear transformation that switches $-1 \leftrightarrow +1$ and $\infty \leftrightarrow -B/A$. To exploit this symmetry it is convenient to use a coordinate $u = (cx+1)/(x+c)$: for any $c$ we have $u=\pm 1$ at $x = \pm 1$ and $u = c$ at $x = \infty$, and we choose $c>1$ so that $u=-c$ at $x=-B/A$. Explicitly, this change of variable gives $${\large\int}_{-1}^1 \frac{dx}{\sqrt[3]{c^2+1+2cx}\ \left(1-x^2\right)^{\small2/3}} = {\large\int}_{-1}^1 \frac{du}{\sqrt[3]{c^2-u^2}\ \left(1-u^2\right)^{\small2/3}} .$$ By symmetry the second $\int_{-1}^1$ is $2\int_0^1$ of the same integrand, and then the change of variable $u^2 = t$ gives $${\large\int}_0^1 \frac{dt}{\sqrt{t} \, (c^2-t)^{\small1/3} (1-t)^{\small2/3}}.$$ Now integrating with respect to $X=1-t$ instead of $t$ gives $${\large\int}_0^1 \frac{dX}{\sqrt{1-X} \; X^{2/3} \left(c^2-1+X\right)^{\small1/3}},$$ and we have reached an integral of the desired form.

In the present case we take $c = \frac12 \! \sqrt 5$. Then $c^2+1 + 2cx = (9 + 4\sqrt{5}x)/4$, so our integral is $4^{1/3}$ times the one with $\sqrt[3]{9 + 4\sqrt{5}x}$ in the denominator. Hence $c^2-1 = 1/4$, so we gain another factor of $4^{1/3}$ and we've reached the integral in the first display with $z=-4$. Vladimir Reshetnikov already noted in a comment to his question that the integral seemed to have an exact value at $z = -4$, and the analysis I gave in my answer there (or Kirill's answer here) leads to the period of a CM elliptic curve with $j$-invariant $146329141248 \sqrt{5} - 327201914880$; this curve is related by $5$-isogeny with a curve of $j$-invariant zero, and thus has a period proportional to a Beta integral.

I will start with and prove Chen Wang's equivalent formulation: $$F\left({\tfrac13,\tfrac12\atop \tfrac56}\middle| \tfrac45 \right) = \frac{3}{\sqrt{5}}.$$

By the integral representation of hypergeometric functions (DLMF 15.6.E1), this is equal to $$\frac1{B(\frac13,\frac12)}\int_0^1 \frac{dx}{x^{2/3}(1-x)^{1/2}(1-A^6x)^{1/2}},$$ where $A = (4/5)^{1/6}$ is easier to use than $\frac45$. Let the integral be denoted by $I$. Introducing two changes of variables, $x\mapsto 1/u^3$ and later $u=A^2/v$, we see that $$I = \int_1^\infty \frac{3u du}{\sqrt{(u^3-1)(u^3-A^6)}} = \int_0^{A^2} \frac{3A\,dv}{\sqrt{(1-v^3)(A^6-v^3)}}.$$

The hyperelliptic curve $$y^2 = (x^3-1)(x^3-A^6), \qquad \frac{1}{3A}I = \int_0^{A^2}\frac{dx}{y} = \int_1^\infty \frac{x}{A}\frac{dx}{y}$$ admits an involution $x\mapsto A^2/x$, and, as demonstrated very clearly by Jyrki Lahtonen here, there is a rational change of variables that maps this curve onto the curve $$s^2 = t^3 + 9A^2t^2 + 6A(A^3+1)^2t+(A^3+1)^4.$$

In particular, first by writing $$u = x+A^2/x, \qquad v = y\left(\frac1x + \frac A{x^2}\right), \qquad \frac{v/y}{du/dx} = \frac1{x-A},$$ we get $$\frac{2}{3A} I = \int_0^{A^2}\frac{dx}{y} + \int_1^\infty \frac{x}{A}\frac{dx}{y} = \int_{1+A^2}^\infty \frac{du}{v}\frac{v/y}{du/dx}\left(\frac xA-1\right) = \frac{{\color{red}6}}{A}\int_{1+A^2}^\infty \frac{du}{v}.$$ (I lost a factor of $6$ somewhere in my notes; I'll edit this once I find it.) And transforming to $$t = -\frac{(A^3+1)^2}{u+2A}, \qquad s = \frac{(A^3+1)^2v}{(u+2A)^2},$$ gives $$I = 9\int_{t_1}^{0}\frac{dt}{s}, \qquad t_1 = -(1-A+A^2)^2.$$

Finally, the curve $(s,t)$ is elliptic, and sage's function isogenies_prime_degree tells us that there exists a rational map given by $$\begin{eqnarray}z &=& \Big(9000 A^2 \left(754+843 A^3\right) t+63000 \left(94+105 A^3\right) t^2+67500 A \left(34+35 A^3\right) t^3\\&&+112500 A^2 \left(4+3 A^3\right) t^4+45000 t^5\Big)\Big/\\&&\Big(60508 A^2+67650 A^5+100 \left(754+843 A^3\right) t+75 A \left(514+575 A^3\right) t^2\\&&+625 A^2\left(14+15 A^3\right) t^3+1250 t^4\Big),\end{eqnarray}$$ $$w/s = \left(345600 \left(51841+57960 A^3\right)+7776000 A \left(2889+3230 A^3\right) t+1620000 A^2 \left(8278+9255 A^3\right) t^2+1080000 \left(4136+4635 A^3\right) t^3+48600000 A \left(21+25 A^3\right) t^4+10125000 A^2 \left(14+15 A^3\right) t^5+13500000 t^6\right)/\left(32 \left(832040+930249 A^3\right)+1200 A \left(46368+51841 A^3\right) t+300 A^2 \left(159454+178275 A^3\right) t^2+5000 \left(3872+4329 A^3\right) t^3+7500 A \left(648+725 A^3\right) t^4+46875 A^2 \left(14+15 A^3\right) t^5+62500 t^6\right)$$ with $$\frac{w/s}{dz/dt} = 6,$$ that maps the curve $(s,t)$ to the curve $$w^2 = z^3+180^3.$$

This means that the integral is given by $$I = 9\times 6\times \int_{-180}^0 \frac{dz}{\sqrt{z^3+180^3}} = \frac{3}{\sqrt{5}}B(\tfrac12,\tfrac13),$$ where the last integral is elementary in terms of beta functions. Putting things together gives the desired result.

I've found a generalization of your conjecture:

If $k>0$ real number, then

$${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{k\pm\tfrac{4\sqrt5k}{9}\,x}\ \left(1-x^2\right)^{\small2/3}}\stackrel{?}{=} \frac{1}{\sqrt[3]{k}}\cdot\frac{3^{\small13/6}}{2^{\small4/3}5^{\small5/6}\pi}\Gamma^{3}\left(\tfrac{1}{3}\right).$$

For $k=9$ with $+$ sign it gives back your integral.

Another family of this type:

If $k>0$ real number, then

$${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{k\pm\tfrac{4k}{5}\,x}\ \left(1-x^2\right)^{\small2/3}}\stackrel{?}{=} \frac{1}{\sqrt[3]{k}}\cdot\frac{\sqrt[3]{5}}{\sqrt[3]{2}\sqrt{3}\pi}\Gamma^3\left(\tfrac{1}{3}\right).$$