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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?


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.


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.)

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  • $\begingroup$ Meta note: I've been struggling with this problem for the latter part of this decade. I finally broke down and decided to ask here at MO, in the hope that this gets resolved before this year/decade ends, so any good leads would be very much appreciated. $\endgroup$ Commented Dec 18, 2019 at 11:48
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    $\begingroup$ can you give some background/motivation for this quest? $\endgroup$ Commented Dec 18, 2019 at 14:29
  • $\begingroup$ Your integral is an Abelian integral on your elliptic curve with 6 poles (they are obtained by solving $y^2=\alpha x$ and $x^2+y^3-3\alpha xy=1$ simultaneously. You can bring your elliptic curve to the standard Weierstrass form, and then the integral will be the sum of 6 standard elliptic integrals of the third kind. $\endgroup$ Commented Dec 18, 2019 at 16:52
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    $\begingroup$ @Alexandre, "Your integral is an Abelian integral" - Dixon explicitly states that this is the Abelian integral of the first kind for this cubic, yes. "You can bring your elliptic curve to the standard Weierstrass form" - I have already (implicitly) done this in the second and third formulae: $$v^2=4u^3-\frac{\alpha}{12}\left(\alpha^3-8\right)u-\frac{8-20\alpha^3-\alpha^6}{216}$$ What to do after is what is giving me trouble, so a further push would be appreciated. $\endgroup$ Commented Dec 18, 2019 at 22:29
  • $\begingroup$ @Carlo: two things: I want to see a plot of the inverse function in the complex plane, and I am aiming to eventually write a numerical method to evaluate them. As Alexandre says, since it might be possible to express in terms of elliptic integrals, evaluating this can then be done through known efficient algorithms for elliptic integrals, but I would of course need the closed form first. $\endgroup$ Commented Dec 18, 2019 at 22:32

1 Answer 1

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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


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 $(r,s)$.

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  • $\begingroup$ If $\alpha=0,2$ or a root of $8-20 \alpha^3-\alpha^6$, the inverse of $\text{sm}(z,\alpha)$ is expressible in terms of an incomplete beta function $\endgroup$ Commented Nov 27, 2022 at 18:51

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