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

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

Note that $\sqrt{9+4\sqrt5}=\phi^2$. 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. – Lucian Jul 27 '14 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. – user111187 Jul 27 '14 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 $\eta=5^{-1/6}\phi^2$. – Start wearing purple Jul 27 '14 at 23:34
• Using DLMF15.8.13 and 15.8.15, we have another equivalent identity: $${}_2F_1\left(\frac13;\frac12;\frac56;\frac45\right)\stackrel?=\frac{3}{\sqrt{5}}.$$ – Chen Wang Aug 7 '14 at 13:33
• Look like it's partially solved by the top answer in This question. – Chen Wang Oct 9 '14 at 8:08

## 3 Answers

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.

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 $dx \left/ \left(\sqrt{A+Bx} \, (1-x^2)^{2/3}\right)\right.$ 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 $${\large\int}_{-1}^1 \frac{dx}{\sqrt{c^2+1+2cx}\ \left(1-x^2\right)^{\small2/3}} = {\large\int}_{-1}^1 \frac{du}{\sqrt{c^2-u^2}\ \left(1-u^2\right)^{\small2/3}} .$$ 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 $\sqrt{9 + 4\sqrt{5}x}$ 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've found a generalization of your conjecture:

If $k>0$ real number, then

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

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

Another family of this type:

If $k>0$ real number, then

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