Area enclosed by the closed loop of an elliptic curve $y^2 = x^3 + a x + b$

Question

Let $E:y^2=x^3+ax+b$ be a real elliptic curve such that $0=x^3+ax+b$ has $3$ distinct real roots $w_0<w_1<w_2$ so that $x^3+ax+b=(x-w_0)(x-w_1)(x-w_2)$ and thus $w_0+w_1+w_2=0$. My question then is

Can we express the area $\text{Ar}(E):=2\int_{w_0}^{w_1}\sqrt{x^3+ax+b}\ dx$ in terms of already know special functions?

I have obtained the following infinite series that converges whenever $w_1<0$ $$\boxed{\text{Ar}(E)=\frac{\pi}{2}\sqrt{w_2-w_1}(w_1-w_0)^2\sum_{n=0}^\infty\frac{(-1)^{n+1}(2n+1)}{2n^3+5n^2+n-2}{2n\choose n}^2\frac{1}{16^n}\left(\frac{w_1-w_0}{w_2-w_1}\right)^n}$$ but I haven't been able to link it to other already known special functions. So for example, if we consider the elliptic curve $E_0:y^2=x^3-13x-12=(x+3)(x+1)(x-4)$, we have $$\text{Ar}(E_0)=2\int_{-3}^{-1}\sqrt{x^3-13x-12}dx=2\pi\sqrt{5}\sum_{n=0}^\infty\frac{(-1)^{n+1}(2n+1)}{2n^3+5n^2+n-2}{2n\choose n}^2\frac{1}{40^n}=7.68858977...$$ I think this will be possible via elliptic integrals, but I don't know how exactly. Sorry if this question has already been asked here; I've searched and found nothing.

Answer 1

The affine substitution $u := \frac{x - w_0}{w_1 - w_0}$ transforms the integral into $$(w_1 - w_0)^{5 / 2} \int_0^1 \sqrt{u (u - 1) (u - \lambda)} \,du , $$ where $$\lambda := \frac{w_2 - w_0}{w_1 - w_0} > 1 ,$$ which can be expressed in terms of the complete elliptic integral functions of the First and Second Kinds, $K$ and $E$, respectively: $$\int_0^1 \sqrt{u (u - 1) (u - \lambda)} \,du = \frac{4 \sqrt\lambda}{15} \left[(\lambda^2 - \lambda + 1) E\left(\frac1{\sqrt \lambda}\right) - \left(\lambda - \frac12\right) (\lambda - 1) K\left(\frac1{\sqrt \lambda}\right)\right] .$$ Both $E$ and $K$ are expressible in terms of the Gauss hypergeometric function, ${}_2 F_1$, hence so is the integral: \begin{multline}\int_0^1 \sqrt{u (u - 1) (u - \lambda)} \,du \\ = \frac{\pi \sqrt\lambda}{15} \left[2 (\lambda^2 - \lambda + 1) \cdot {}_2 F_1\left(-\frac12, \frac12; 1; \frac1\lambda\right) - (2 \lambda - 1) (\lambda - 1) \cdot {}_2 F_1\left(\frac12, \frac12; 1; \frac1\lambda\right)\right] .\end{multline}

Answer 2

Computing such an area, based on the series representation that you've arrived at, appears to involve integrals of elliptic integrals. Recall the generating function for the squared binomial coefficient,

$$\sum_{n\ge0} \binom{2n}n^2 x^n = \frac4\pi \frac1{1+\sqrt{1-16x}} K\left(\frac{1-\sqrt{1-16x}}{1+\sqrt{1-16x}}\right)$$

In the example with $\operatorname{Ar}(E_0)$, replace $\dfrac1{40}$ with $x$ and denote the power series by $F(x)$. We have by partial fraction expansion $(1)$ and the fundamental theorem of calculus that

$$\begin{align*} F(x) &= \sum_{n\ge0} \frac{(-1)^{n+1}(2n+1)}{(n+1)(n+2)(2n-1)} \binom{2n}n^2 x^n \\ &= \frac13 f(x) - \frac35 g(x) + \frac8{15} h(x) \tag1 \\ &= - \frac4{3\pi x} \int_0^x \frac{1}{1 + \sqrt{1+16t}} K \left(\frac{1-\sqrt{1+16t}}{1+\sqrt{1+16t}}\right) \, dt \\ & \qquad + \frac{12}{5\pi x^2} \int_0^x \frac{t}{1 + \sqrt{1+16t}} K \left(\frac{1-\sqrt{1+16t}}{1+\sqrt{1+16t}}\right) \, dt \\ & \qquad - \frac{32x}{15\pi} \int_0^x \frac1{t^2 \left(1 + \sqrt{1+16t^2}\right)} K\left(\frac{1 - \sqrt{1+16t^2}}{1 + \sqrt{1+16t^2}}\right) \, dt \tag2 \end{align*}$$

Each of $f,g,h$ are obtained $(2)$ by solving linear ODEs. As an example, $f$ is derived below:

$$\begin{align*} f(x) &= \sum_{n\ge0} \binom{2n}n^2 \frac{(-1)^{n+1} x^n}{n+1} \\ x f(x) &= \sum_{n\ge0} \binom{2n}n^2 \frac{(-1)^{n+1} x^{n+1}}{n+1} \\ \left[x f(x)\right]' = x f'(x) + f(x) &= - \sum_{n\ge0} \binom{2n}n^2 x^n \\ f(x) &= - \frac4{\pi x} \int_0^x \frac{1}{1 + \sqrt{1+16t}} K \left(\frac{1-\sqrt{1+16t}}{1+\sqrt{1+16t}}\right) \, dt \end{align*}$$

I do not know if any more can be done here to get a simpler expression for $F(x)$...

Answer 3

As @Travis Willse wrote, the summation

$$\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n+1)}{2n^3+5n^2+n-2}\,\,{2n\choose n}^2\,\left(\frac{x}{16}\right)^n$$ is, in terms of the Gaussian hypergeometric function, $$\frac{1}{2} \,\, _2F_1\left(-\frac{1}{2},\frac{1}{2};3;-x\right)+\frac{x}{12} \,\, _2F_1\left(\frac{1}{2},\frac{3}{2};4;-x\right)$$

In terms of elliptic integrals, it is $$\frac{8}{15 \pi x^2}\Big(2 \left(x^2+x+1\right) E(-x)-\left(x^2+3 x+2\right) K(-x) \Big)$$

For $x=\frac 25$, this is $$\frac{4}{5 \pi }\left(13 E\left(-\frac{2}{5}\right)-14 K\left(-\frac{2}{5}\right)\right)=0.5472450212\cdots$$ and then the result