Prove ${\large\int}_0^\infty\left({_2F_1}\left(\frac16,\frac12;\frac13;-x\right)\right)^{12}dx\stackrel{\color{#808080}?}=\frac{80663}{153090}$

I discovered the following conjectured identity numerically (it holds with at least $1000$ digits of precision). How can I prove it? $${\large\int}_0^\infty\left({_2F_1}\left(\frac16,\frac12;\frac13;-x\right)\right)^{12}dx\stackrel{\color{#808080}?}=\frac{80663}{153090}$$

Update: It looks like this hypergeometric function assumes algebraic values at algebraic points (it's only a guess because I have only approximations to those algebraic numbers). Looking at those values, I was able to further conjecture that the hypergeometric function for $x<0$ is actually the following elementary function:

$${_2F_1}\left(\frac16,\frac12;\frac13;x\right)\stackrel{\color{#808080}?}= \\ \frac1{\sqrt[4]2\sqrt3}\cdot\sqrt{\frac{\alpha}{1-x}+\frac{1}{\alpha} \sqrt{\frac{4\left(\alpha\sqrt{2}+2\right)+x\left(\sqrt[3]{4\beta}-2\left(\alpha\sqrt{2}+4\right)\right)+2\sqrt[3]{2\beta^2}}{1-x}}}~,$$

where

$$\alpha=\sqrt{2-2x+\sqrt[3]{2\beta^2}}~,\qquad\beta=x(x-1)~.$$

• I suspect that the integrand might be an elementary function. It seems to have algebraic values at algebraic points. Jul 26, 2014 at 2:33
• The integrand can be expressed in terms of Legendre functions; see WRA. Jul 26, 2014 at 5:15
• Let $~f(x)~=~_2F_1\bigg(\dfrac16,\dfrac12;\dfrac13;-x\bigg),~$ and $~I_n=\displaystyle\int_0^\infty f^n(x)~dx.~$ Then $I_1=-\dfrac85$ , $I_8=\dfrac{481}{405}$ , $I_{10}=\dfrac{1231}{1701}$ , $I_{12}=\dfrac{80663}{153090}$ , $I_{14}=\dfrac{59975}{144342}$ , $I_{16}=\dfrac{6765377}{19702683}$ , $I_{18}=\dfrac{28837643}{98513415}$ , $I_{20}=\dfrac{3076840073}{12058041996}$ , etc. Jul 26, 2014 at 14:02
• @Lucian Actually, $I_1$ does not converge. Also, it cannot have a negative value, because the integrand is positive on the integration region. Jul 26, 2014 at 23:00
• @Lucian It depends on what do you mean by $\sum\limits_{n=1}^\infty$. If it is its most direct meaning, the limit of partial sums, then yes, I am denying that. I know that it's possible to assign a different meaning to this symbol, that would make this identities true. :-) Jul 26, 2014 at 23:27

Consider the hypergeometric equation with parameters $(a,b,c)=\left(\frac16,\frac12,\frac13\right)$, and build from its two canonical solutions near $z=0$ the vector $$\vec{y}(z)=\left(\begin{array}{c} y_1 \\ y_2 \end{array}\right)=\left(\begin{array}{c} _2F_1(a,b;c;z) \\ z^{1-c}{}_2F_1(a-c+1,b-c+1;2-c;z) \end{array}\right).\tag{1}$$ This is a single-valued vector function on $\mathbb{C}\backslash\{(-\infty,0]\cup[1,\infty)\}$. Its analytic continuation along a closed loop $\gamma$ gives rise to monodromy representation of $\pi_1(\mathbb{C}\backslash\{0,1\})$: $$\gamma\mapsto M_{[\gamma]},\qquad y(\gamma z)=M_{[\gamma]}y(z).$$ The monodromy group $G\subset GL(2,\mathbb{C})$ of the hypergeometric equation is generated by two matrices corresponding to simple loops around $0$ and $1$. In the case we are interested in these matrices are explicitly given by $$M_0=\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{-2\pi i /3}\end{array}\right),\qquad M_1=C\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{2\pi i /3}\end{array}\right)C^{-1},\tag{2}$$ where the connection matrix $C=\left(\begin{array}{cc} 1 & 2^{\frac43} \\ -2^{\frac83} & 8\end{array}\right)$. If $G$ is finite, then $\vec{y}(z)$ has a finite number of branches, and moreover (Schwarz 1872), is algebraic.

It is not difficult to check that the monodromy group $G$ generated by $M_0$, $M_1$ from (2) is indeed finite. In particular, note that $$M_0^3=M_1^3=I,\qquad M_1^2=-M_0M_1M_0,$$ $$M_1M_0M_1=-M_0^2,\qquad M_1M_0^2M_1=M_0^2M_1M_0^2.$$ It turns out that $G$ has order $24$ and is isomorphic to the binary tetrahedral group: $$G\cong 2T=\langle s,t\,|\,(st)^2=s^3=t^3\rangle,$$ where the generators can be identified as $s=M_0M_1M_0M_1M_0$, $t=M_1M_0M_1M_0M_1$.

Corollary: The hypergeometric functions in (1) are algebraic.

Algebraic solutions of the hypergeometric equations are classified by the so-called Schwarz table, and have been studied by many mathematicians, see e.g. the bibliography in this paper by R.Vidunas. Their explicit construction is somewhat involved but relatively straightforward - at least when the corresponding algebraic curve has genus $0$ (the genus can be determined independently from the Riemann-Hurwitz formula).

In our case the task simplifies even more as our parameter values can be obtained from the genus $0$ tetrahedral formula (2.4) of the above mentioned paper by a combination of a linear trasformation (sending $\frac56$ to $\frac43-\frac56=\frac12$) and differentiation (transforming $\frac43$ into $\frac13$). The result is $$_2F_1\left(\frac16,\frac12;\frac13;-\frac{r(r+2)^3}{(r+1)(1-r)^3}\right)=\frac{\sqrt{1-r^2}}{2r+1}.$$

Corollary: The antiderivative $\displaystyle\int \mathcal{R}\left(x,y(x)\right)dx$, where $y(x)={}_2F_1\left(\frac16,\frac12;\frac13;-x\right)$ and $\mathcal{R}(x,y)$ is rational in both arguments, can be expressed in terms of elementary functions.

Example: The transformation $r\mapsto x(r)=\frac{r(r+2)^3}{(r+1)(1-r)^3}$ bijectively maps $(0,1)$ to $(0,\infty)$, and therefore the initial integral becomes \begin{align}\mathcal{I}&=\int_0^1 \left(\frac{\sqrt{1-r^2}}{2r+1}\right)^{12}\left(\frac{r(r+2)^3}{(r+1)(1-r)^3}\right)'dr=\\&= 2\int_0^1 \frac{(1+r)^4(1-r)^2(r+2)^2}{(2r+1)^{10}} dr=\\&=\frac{80\,663}{153\,090}. \end{align}

• +1) I've always wondered what on earth was so "geometric" about hypergeometric functions that someone decided to call them that. The ideas glimpsed in this answer now make me confident there's a very good reason. @_@ Jul 26, 2014 at 19:08
• The only thing I find a bit disappointing is that, while it renders the generalization discussed by Lucian's comments tractable, it doesn't lend itself to any "nice" formula for $\mathcal{I}_n$. But I'm quibbling. Jul 26, 2014 at 19:50
• @Semiclassical Yes, one obtains more and more complicated rational function to integrate. By the way one can compute any integral of any $R({}_2F_1,x)$ (rational in the two arguments) provided the integral converges. Jul 26, 2014 at 19:59
• The only way I could see a 'nice' formula emerging is to find some parametric differentiation approach (which would also be nice since it'd make Lucian's answer the coefficients of some generating function.) But that's probably too much to wish for. Jul 26, 2014 at 20:02
• @Semiclassical My guess is $\frac{2(n(8(n-8)n+113)-21)\ {_2F_1}\left(1,\frac{n-1}2;\frac{n+4}2;-3\right)-\frac{6(n-1)(4n(4n-31)+195)\ {_2F_1}\left(2,\frac{n+1}2;\frac{n+6}2;-3\right)}{n+4}}{(n-6)(n-4)(n-2)(n+2)}$ Jul 26, 2014 at 23:48