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.
Answer
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.
No comments:
Post a Comment