Given algebraic $a$, find the closed form of $\int_0^a \dfrac{dx}{\sqrt{1-x^4}}$

Question

Let $$A=\int_0^1 \dfrac{dx}{\sqrt{1-x^4}}.$$ Given an algebraic number $0\le a\le 1$, can we determine if there exists a rational number $b$ such that $$\int_0^a \dfrac{dx}{\sqrt{1-x^4}}=Ab?$$ If so, can we find the rational number $b$?

Examples: $$\int_0^{\sqrt{\sqrt{2}-1}}\dfrac{dx}{\sqrt{1-x^4}}=\dfrac{1}{2}A,$$ $$\int_0^{\frac{1}{2}(\sqrt{3}+1-\sqrt[4]{12})}\dfrac{dx}{\sqrt{1-x^4}}=\dfrac{1}{3}A.$$

Note: The reverse procedure, however, is easy and I know how to do it (i.e. given $Ab$, find the algebraic number $a\in [0,1]$ such that $\int_0^a dx/\sqrt{1-x^4}=Ab$). The key is to use the addition formula of the elliptic function corresponding to the elliptic integral.

This is a follow-up question to Given an algebraic number $a$, find the closed form of $\arctan (a)$, so I tried to build on analogy. The problem in that question is finding the rational number $b$ such that $$\arctan (a)=\int_0^a \dfrac{dx}{1+x^2}=\pi b$$ for a given algebraic $a\ge 0$. The key to that problem was expressing $e^{i\theta}$ as an algebraic function of $a=\tan (\theta)$ where $\theta$ is the division point, checking if $(e^{i\theta})^n=1$ for some positive integers $n$ and finding a non-trivial lower bound for the degree of the $n$th cyclotomic polynomial, so that there are only finitely many cases to check and this gives an algorithm.

However, in the elliptic integral case, I'm not aware of an elliptic analog of the exponential function. I know that inverting the elliptic integral on $[0,1]$, $$\int_0^{f(u)} \dfrac{dx}{\sqrt{1-x^4}}=u,$$ gives an elliptic function $f$ with addition formula $$f(x+y)=\dfrac{f(x)f'(y)+f(y)f'(x)}{1+f(x)^2f(y)^2}$$ where $f'$ is an algebraic function of $f$. This should be somehow analogous to the "tangent case" in my previous question because $$\tan (x+y)=\dfrac{\tan (x)+\tan (y)}{1-\tan (x)\tan (y)},$$ but I really don't know what to do next.

Note that $A$ can be expressed by the gamma function as $$A=\dfrac{\Gamma (1/4)^2}{4\sqrt{2}\Gamma (1/2)}.$$ It is possible that an answer to my question is not as elementary and is at least as hard as proving, for example, $$\int_0^2 \dfrac{dx}{\sqrt{1+x^3}}=\dfrac{\Gamma (1/6)\Gamma (1/3)}{6\Gamma (1/2)},$$ which is another example of an elliptic integral. Answers (Prove: $\int_0^2 \frac{dx}{\sqrt{1+x^3}}=\frac{\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{3}\right)}{6\Gamma\left(\frac{1}{2}\right)}$) to that problem use sophisticated tools from the theory of elliptic curves and hypergeometric functions.

Answer 1

$\DeclareMathOperator{\cl}{cl}$ $\DeclareMathOperator{\sl}{sl}$

Let $C/\mathbb{Q}$ be the plane curve $$c^2 + s^2 + c^2 s^2 = 1$$ with distinguished point $O:(c_O,s_O)=(1,0)$. Then $C$ is an elliptic curve, birational by some $\phi: C\rightsquigarrow E$ to the curve $E/\mathbb{Q}$ with Weierstrass form $$y^2 = x^3 + 4x\text{,}$$ and therefore admits the usual group law $(+,-,O)$. Let $K$ be an algebraic number field, let $\mathfrak{p}$ be a real place of $K$, and let $P:(\gamma,\sigma)$ be a point of $C/K$. Then the question has two parts:

• Can we decide whether $P$ an $N$-torsion point of $C/K$ for some $N$?
• Suppose $P$ is an $N$-torsion point. Let $\Omega$ be the period lattice and $D$ the fundamental parallelogram of the lemniscate elliptic functions. Can we find the unique $t\in\tfrac{1}{N}\Omega \cap D$ such that $$(\gamma_{\mathfrak{p}},\sigma_{\mathfrak{p}})= (\cl t, \sl t)\text{?}$$

And the answer to both parts is "yes". Sketch of an algorithm:

1. Compute the torsion subgroup $E(K)_{\text{tors}}$ of $E/K$. This the "hi-tech" part, but it has been possible for a while, basically because we have computable bounds for the size of the torsion subgroup in terms of the properties of $K$, and is implemented in Magma.
2. Test membership of $\phi(P)\in E(K)_{\text{tors}}$. If false, then $P$ is not an $N$-torsion point for any $N$, and we are done.
3. Otherwise, we know $P$ is an $N$-torsion point for some $N$. But we have a real place $\mathfrak{p}$, so we know that $-1 < \sigma_{\mathfrak{p}} < 1$. And $\sl$ and $\cl$, as elliptic functions, take every value away from ramification points exactly twice. So writing $2\varpi$ for the real period, we know that $$(\gamma_{\mathfrak{p}},\sigma_{\mathfrak{p}})=(\cl \tfrac{2\varpi M}{N},\sl \tfrac{2\varpi M}{N})$$ for a unique value of $0<M<N$. But the left side, and each of the $N-2$ possible values for the right side, are effectively computable; consequently, our estimate for the left side is inconsistent with all but one of the possible values of the right side after computing to sufficient precision. Therefore we can compute the value of $M$, and so have solved the second part of the question.

Examples

$\mathbb{Z}/(8)$-torsion from $\mathbb{Q}(\sqrt{\sqrt{2}-1})$

Let $K=\mathbb{Q}(\alpha)$, where $\alpha=\sqrt{\sqrt{2}-1}$. Then, even if we didn't know already, we could establish that $(\sqrt{\sqrt{2}-1},\sqrt{\sqrt{2}-1})=(\cl \tfrac{M\varpi}{4},\sl\tfrac{M\varpi}{4})$, i.e., $\int_0^{\alpha}\frac{\mathrm{d}s}{\sqrt{1-s^4}}=\tfrac{M\varpi}{4}$, for some integer $M$. That's because we can compute the torsion subgroup to be isomorphic to $\mathbb{Z}/(8)$. Using the Magma calculator

R<z> := PolynomialRing(Integers());
K<a> := NumberField(z^4 + 2*z^2 - 1);
P<c, s, l> := ProjectiveSpace(K, 2);
C := Curve(P, c^2*s^2 + c^2*l^2 + s^2*l^2 - l^4);
O := C ! [0, 1, 1];
E, phi := EllipticCurve(C, O);
EKtors<e>, map := TorsionSubgroup(E);
EKtors;
RationalPoints(Pullback(phi, map(EKtors ! 5)));

yields

Abelian Group isomorphic to Z/8
Defined on 1 generator
Relations:
    8*e = 0
{@ (0 : 1 : 1), (a : a : 1), (0 : 1 : 0), (1 : 0 : 0) @}

$\mathbb{Z}/(12)$-torsion from $\mathbb{Q}(\sqrt[4]{\sqrt{12}-3})$

Let $K=\mathbb{Q}(\alpha)$, where $\alpha=\sqrt[4]{\sqrt{12}-3}$. Once again, even if we didn't know already, we could establish that $(\sqrt[4]{\sqrt{12}-3},\tfrac{1}{2}(1+\sqrt{3}-\sqrt[4]{12}))=(\cl \tfrac{M\varpi}{6},\sl\tfrac{M\varpi}{6})$, i.e., $\int_0^{\tfrac{1}{2}(1+\sqrt{3}-\sqrt[4]{12})}\frac{\mathrm{d}s}{\sqrt{1-s^4}}=\tfrac{M\varpi}{6}$, for some integer $M$. That's because we can compute the torsion subgroup to be isomorphic to $\mathbb{Z}/(12)$. Using the Magma calculator again

R<z> := PolynomialRing(Integers());
K<a> := NumberField(z^8 + 6*z^4 - 3);
P<c, s, l> := ProjectiveSpace(K, 2);
C := Curve(P, c^2*s^2 + c^2*l^2 + s^2*l^2 - l^4);
O := C ! [0, 1, 1];
E, phi := EllipticCurve(C, O);
EKtors<e>, map := TorsionSubgroup(E);
EKtors;
RationalPoints(Pullback(phi, map(EKtors ! 10)));

yields

Abelian Group isomorphic to Z/12
Defined on 1 generator
Relations:
    12*e = 0
{@ (0 : 1 : 1), (a : 1/4*(-a^6 + a^4 - 5*a^2 + 5) : 1), (0 : 1 : 0), (1 : 0 : 0)
@}

Answer 2

Too long for a comment, What I am thinking is, as we know that

$$ \sum_{n=0}^\infty \binom{2n}{n} x^n = \frac{1}{\sqrt{1-4x}} \, \, \text{for} \, \, |x| < 1/4 $$

now replace $x$ by $x^4/4$ and put in $A$

\begin{align*} A &= \int_0^a \frac{1}{\sqrt{1-x^4}} dx \\ &= \int_0^a \sum_{n=0}^\infty \binom{2n}{n} \frac{x^{4n}}{4^n} dx \\ &= \sum_{n=0}^\infty \binom{2n}{n} \frac{1}{4^n} \int_0^a x^{4n} dx \\ &= \sum_{n=0}^\infty \binom{2n}{n} \frac{a^{4n+1}}{4^n (4n+1)} \end{align*}

Here I assumed that $a < 1/4$.