Expressing the inverse Dixon function in terms of more familiar functions

Question

If $x^3+y^3-3\alpha xy=1$, is there an expression for the integral $$\int_0^z \frac{\mathrm dx}{y^2-\alpha x}$$ in terms of more familiar functions?

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A.C. Dixon introduced the elliptic functions $\operatorname{sm}(u,\alpha)$ and $\operatorname{cm}(u,\alpha)$ now named after him in this article. It can be shown (see e.g. my writeup here) that these functions can be expressed in terms of the more conventional Weierstrass elliptic functions, e.g.

$$\operatorname{sm}(u,\alpha)=-\frac{2\wp\left(u;g_2,g_3\right)+\frac{\alpha^2}{2}}{\wp^\prime\left(u;g_2,g_3\right)+\alpha\wp\left(u;g_2,g_3\right)-\frac{\alpha^3+4}{12}}$$

where

$$\begin{align*} g_2&=\frac{\alpha}{12}\left(\alpha^3-8\right)\\ g_3&=\frac{8-20\alpha^3-\alpha^6}{216} \end{align*}$$

The problem stated above is then equivalent to asking for a (hopefully simpler) explicit expression for the inverse Dixon elliptic function, $\operatorname{sm}^{(-1)}(z,\alpha)$.

Lagrangian inversion of the Maclaurin series for $\operatorname{sm}(u,\alpha)$ yields the series

$$z-\frac{\alpha z^2}{2}+\frac{5 \alpha^3+2}{12} z^4-\frac{\alpha\left(7 \alpha ^3+4\right)}{15} z^5+\frac{44 \alpha^6+40 \alpha^3+5}{63} z^7+c_8 z^8+\cdots$$

where the Maclaurin coefficients $c_n$ satisfy the recurrence

$$n(n+1)(n+2)c_n-(n+3)\left(18 \alpha ^3+4 \alpha ^3 n^2+n^2+18 \alpha ^3 n-3 n-24\right)c_{n+3}-(n+6)\left(180 \alpha ^3+\left(4 \alpha ^3+1\right) n^2+3 \left(18 \alpha ^3+7\right) n+84\right)c_{n+6}+(n+7)(n+8)(n+9) c_{n+9}=0$$

and with a little more work, one can derive the differential equation satisfied by $w=\operatorname{sm}^{(-1)}(z,\alpha)$:

$$(z+1)\left(z^2-z+1\right)\left(z^6-2\left(2 \alpha^3+1\right) z^3+1\right)w^{(3)}(z)+6 z^2\left(z^6+\left(1-\alpha^3\right) z^3-3\alpha^3-2\right)w^{\prime\prime}(z)+2z\left(3z^6+\left(\alpha^3+7\right)z^3-5\alpha^3-2\right) w^\prime(z)=0$$

Unfortunately, I have not succeeded in making further headway. I have reason to suspect that a (generalized) hypergeometric function (e.g. Appell's $F_1$) is involved, considering that the special value $\operatorname{sm}^{(-1)}(1,\alpha)$ is expressible in terms of the Gaussian hypergeometric function:

$$\operatorname{sm}^{(-1)}(1,\alpha)=\frac13 B\left(\frac13,\frac13\right) {}_2F_1\left({{\frac13,\frac13}\atop{\frac23}}\middle|-\alpha^3\right)+\frac{\alpha}{3} B\left(\frac23,\frac23\right) {}_2F_1\left({{\frac23,\frac23}\atop{\frac43}}\middle|-\alpha^3\right)$$

where $B(x,y)=\dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ is the usual beta function.

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This has been out for a while, but I have not yet gotten anything that is computationally useful (or even something I can numerically experiment with).

So, to spell things out very explicitly:

Given the standard Weierstrass reduction outlined above for $x^3+y^3-3\alpha xy=1$, can someone please write out the explicit representation of the inverse Dixon function $\operatorname{sm}^{(-1)}(z,\alpha)=\int_0^z \frac{\mathrm dx}{y^2-\alpha x}$, perhaps in terms of standard elliptic integrals or multivariate hypergeometric functions?

(I can worry about implementation details later; I just need a starting point.)

Answer 1

Here is a solution using a slightly different notation for mathjax convenience assuming $0\le a\le1$ although a similar solution for $0\not\le a\not\le 1$ exists:

$$\frac13 \text B\left(\frac13,\frac13\right) \,_2\text F_1\left(\frac13,\frac13;\frac23;-a^3\right)+\frac a3 \text B\left(\frac23,\frac23\right) \,_2\text F_1\left(\frac23,\frac23;\frac43;-a^3\right) $$

Now convert to an Associated Legendre Q function of the third kind $Q_m^n(z)$:

$$\,_2\text F_1\left(\frac13,\frac13;\frac23,A\right)=\frac{2^\frac23\Gamma\left(\frac56\right)Q^0_{-\frac23}\left(\frac2A-1\right)}{\sqrt \pi\sqrt[3]A\Gamma\left(\frac13\right)}$$

and the second conversion:

$$\,_2\text F_1\left(\frac23,\frac23;\frac43,A\right)=\frac{2\sqrt[3]2\Gamma\left(\frac76\right)Q_{-\frac13}^0\left(\frac2A-1\right)}{\sqrt\pi A^\frac23\Gamma\left(\frac23\right)}$$

Now we simplify to a Legendre P and Legendre Q expression:

$$Q_{-\frac13}^0\left(\frac2A-1\right) =Q_{-\frac13}\left(\frac2A-1\right)-\frac{i\pi}2 P_{-\frac 13}\left(\frac2A-1\right)$$

and

$$Q_{-\frac23}^0\left(\frac2A-1\right) =Q_{-\frac23}\left(\frac2A-1\right)-\frac{i\pi}2 P_{-\frac 23}\left(\frac2A-1\right) $$

Now bring everything together with $(-a^3)^\frac13=-a$

$$\begin{align}\text{sm}^{-1}(1,a)= \frac13 \text B\left(\frac13,\frac13\right) \,_2\text F_1\left(\frac13,\frac13;\frac23;-a^3\right)+\frac a3 \text B\left(\frac23,\frac23\right) \,_2\text F_1\left(\frac23,\frac23;\frac43;-a^3\right) = \frac2{3a}\left(Q_{-\frac13}^0\left(-\frac2{a^3}-1\right)-Q^0_{-\frac23}\left(-\frac2{a^3}-1\right) \right)= \frac2{3a}\left(Q_{-\frac13}\left(-\frac2{a^3}-1\right)-\frac{i\pi}2 P_{-\frac 13}\left(-\frac2{a^3}-1\right)-Q_{-\frac23}\left(-\frac2{a^3}-1\right)+\frac{i\pi}2 P_{-\frac 23}\left(-\frac2{a^3}-1\right) \right)\end{align}$$

However, this identity is just for a special case

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Experimentally, one gets:

$$\operatorname{sm}(u,\alpha)=-\frac{2\wp\left(u;g_2,g_3\right)+\frac{\alpha^2}{2}}{\wp^\prime\left(u;g_2,g_3\right)+\alpha\wp\left(u;g_2,g_3\right)-\frac{\alpha^3+4}{12}}=-\frac{2p+\frac{\alpha^2}{2}}{-\sqrt{4p^3-g_2 p-g_3} +\alpha p-\frac{\alpha^3+4}{12}},p=\wp(u;g_2,g_3)$$

Using $f(z)$ to invert $-\frac{2z+\frac{\alpha^2}{2}}{-\sqrt{4z^3-g_2 z-g_3} +\alpha z-\frac{\alpha^3+4}{12}}$, Weierstrass half periods $\omega_1(g_2,g_3),\omega_3(g_2,g_3)$, and inverse Weierstrass P $\wp^{-1}(x;a,b)$:

$$\text{sm}(y,\alpha)=z\implies y=\wp^{-1}\left(f(z); \frac{\alpha^4}{12}-\frac{2\alpha}3, \frac1{27}-\frac{5\alpha^3}{54}-\frac{\alpha^6}{216}\right)+2\left(\omega_1m+\omega_3n\right);m,n\in\Bbb Z$$

Except for possible $r,s$ value and $\wp(z)$ in terms of $\wp’(z)$ issues, simplifying $f(z)$ gives:

$$\boxed{\begin{align}\text{sm}(y,\alpha)=z\implies y=\wp^{-1}\left(\frac{2r((2a+a^2z-z^2)(az+1)z+1)+\sqrt[3]2s\left(2 + 6 \alpha z + 12 \alpha^2 z^2 - 3 z^3 + 11 \alpha^3 z^3 - 6 \alpha z^4 + 6 \alpha^4 z^4 - 6 \alpha^2 z^5 + z^6 - \alpha^3 z^6 + \sqrt{z^6 + 2 \alpha^3 z^6 + \alpha^6 z^6 - 2 z^9 - 8 \alpha^3 z^9 - 10 \alpha^6 z^9 - 4 \alpha^9 z^9 + z^{12} + 2 \alpha^3 z^{12} + \alpha^6 z^{12}}\right)^\frac23}{3\cdot2^\frac23z^2 \sqrt[3]{2 + 6 \alpha z + 12 \alpha^2 z^2 - 3 z^3 + 11 \alpha^3 z^3 - 6 \alpha z^4 + 6 \alpha^4 z^4 - 6 \alpha^2 z^5 + z^6 - \alpha^3 z^6 + \sqrt{z^6 + 2 \alpha^3 z^6 + \alpha^6 z^6 - 2 z^9 - 8 \alpha^3 z^9 - 10 \alpha^6 z^9 - 4 \alpha^9 z^9 + z^{12} + 2 \alpha^3 z^{12} + \alpha^6 z^{12}}}}+ \frac{\left(\frac2z+a\right)^2}{12};\frac{\alpha^4}{12}-\frac{2\alpha}3, \frac1{27}-\frac{5\alpha^3}{54}-\frac{\alpha^6}{216}\right)+2\left(m\omega_1\left(\frac{\alpha^4}{12}-\frac{2\alpha}3, \frac1{27}-\frac{5\alpha^3}{54}-\frac{\alpha^6}{216}\right)+ n\omega_3\left(\frac{\alpha^4}{12}-\frac{2\alpha}3, \frac1{27}-\frac{5\alpha^3}{54}-\frac{\alpha^6}{216}\right)\right);(r,s)=(1,1), \left(\frac{\pm \sqrt3 i-1}2, \frac{\mp \sqrt3 i-1}2\right);m,n\in\Bbb Z \end{align}}$$

where one of each sign is taken for $(r,s)$.