# Prove: $\int_0^2 \frac{dx}{\sqrt{1+x^3}}=\frac{\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{3}\right)}{6\Gamma\left(\frac{1}{2}\right)}$

Prove: $$\int_{0}^{2}\frac{\mathrm{d}x}{\,\sqrt{\,{1 + x^{3}}\,}\,} = \frac{\Gamma\left(\,{1/6}\,\right) \Gamma\left(\,{1/3}\,\right)}{6\,\Gamma\left(\,{1/2}\,\right)}$$

First obvious sub is $$t = 1 + x^{3}$$: $$\frac{1}{3}\int_{1}^{9}{\left(\,{t - 1}\,\right)}^{-2/3}\, t^{-1/2}\, \mathrm{d}t$$ From here I tried many things like $$\frac{1}{t}$$, $$t-1$$, and more. The trickiest part is the bounds! Reversing it from the answer the integral should be like $$\frac{1}{6}\int_{0}^{1} x^{-2/3}\left(\,{1 - x}\,\right)^{-5/6}\,\mathrm{d}t$$ I'm not sure where the $$1/2$$ comes from and the $$0$$ to $$1$$ bounds. Any idea or tip please ?.

• Try the substitution $u=\frac{x}{2}$ Aug 5, 2020 at 21:30
• @TomHimler Now I get to $$\frac{2}{3} \int_0^1 t^{-\frac{2}{3}} {\left(1+8t\right)}^{-\frac{1}{2}} \; \mathrm{d}t$$
– user801111
Aug 5, 2020 at 21:33
• My bad, I made a dumb error. Aug 7, 2020 at 2:36
• May I ask where you found this integral? Aug 10, 2020 at 1:40
• @dxdydz tieba.baidu.com/p/4794735082
– user801111
Aug 10, 2020 at 1:54

An elementary solution: Consider the substitution $$t = \frac{{64 + 48{x^3} - 96{x^6} + {x^9}}}{{9{x^2}{{(4 + {x^3})}^2}}}$$ $$t$$ is monotonic decreasing on $$0, and $$\tag{1}\frac{{dx}}{{\sqrt {1 + {x^3}} }} = -\frac{{dt}}{{3\sqrt {1 + {t^3}} }}$$ this can be verified by explictly computing $$(dt/dx)^2$$ and compare it to $$9(1+t^3)/(1+x^3)$$. When $$x=2, t=-1$$, so $$\int_0^2 {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{1}{3}\int_{ - 1}^\infty {\frac{1}{{\sqrt {1 + {t^3}} }}dt}$$ I believe you now have no difficulty to solve last integral via Beta function.

A conceptual solution: Consider the elliptic curve $$E:y^2=x^3+1$$, $$P=(2,3),Q=(0,1)$$ on $$E$$, $$\omega = dx/y$$ is the invariant differential on $$E$$. For the multiplication-by-$$3$$ isogeny $$\phi:E\to E$$, we have $$3P=(-1,0), 3Q=O$$. So $$3\int_0^2 \omega \cong \int_{-1}^\infty \omega$$ up to an element of $$H_1(E,\mathbb{Z})$$.

$$t$$ given above is the $$x$$-component of $$\phi$$ and $$(1)$$ is equivalent to $$\phi^\ast \omega = 3\omega$$.

The $$P$$ above is $$6$$-torsion, if we consider $$4$$ or $$5$$-torsion instead, we obtain results like $$\int_0^\alpha {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{\Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{12 \sqrt{\pi }} \qquad \alpha = \sqrt[3]{2 \left(3 \sqrt{3}-5\right)} \approx 0.732$$ $$\int_0^\alpha {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{2 \Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{15 \sqrt{\pi }}\qquad \alpha = \left(9 \sqrt{5}+3 \sqrt{6 \left(13-\frac{29}{\sqrt{5}}\right)}-19\right)^{1/3}\approx 1.34$$

• how in the world did you pull out that substitution? Aug 6, 2020 at 4:48
• This is probably the most "obvious" substitution I ever saw. Back to serious, how did you find it ? Aug 6, 2020 at 4:54
• @CSquared How the substitution arises is explained in the "conceptual solution" below. For someone doesn't know much about elliptic curves, unfortunately he/she might simply need to accept the magical appearance of it. But once the substitution has been explicitly written down, every detail can be verified. Aug 6, 2020 at 4:59
• @adam I am not sure if there are contexts where the substitution arise more naturally than elliptic curve. But there is a wealth of theory behind such integral, such as theory of algebraic curves and elliptic functions. Aug 6, 2020 at 9:08
• @MoQabar Split it into $(-1,0)$ and $(0,\infty)$. The former can be converted into beta function via $t\mapsto -t$. Latter is of course also convertible to beta. Aug 6, 2020 at 13:59

A hypergeometric solution: Modulo Beta function $$I_0=\int_0^{\infty } \frac{1}{\sqrt{x^3+1}} \, dx=\frac{2 \Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{7}{6}\right)}{\sqrt{\pi }}$$ one may evaluate $$I_1=\int_2^{\infty } \frac{1}{\sqrt{x^3+1}} \, dx$$ instead. Substitute $$x\to\frac 1x$$ and binomial expansion gives $$I_1=\sqrt{2} \, _2F_1\left(\frac{1}{6},\frac{1}{2};\frac{7}{6};-\frac{1}{8}\right)=\frac{2 \sqrt{\frac{\pi }{3}} \Gamma \left(\frac{7}{6}\right)}{\Gamma \left(\frac{2}{3}\right)}$$ Where the last step has invoked the following formula $$\, _2F_1\left(a,a+\frac{1}{3};\frac{4}{3}-a;-\frac{1}{8}\right)=\frac{\left(\frac{2}{3}\right)^{3 a} \Gamma \left(\frac{2}{3}-a\right) \Gamma \left(\frac{4}{3}-a\right)}{\Gamma \left(\frac{2}{3}\right) \Gamma \left(\frac{4}{3}-2 a\right)}$$

Computing $$I_0-I_1$$ gives the desired result.

Update: Hypergeometric method can also establish @pisco's result (the case of $$4$$-torsion)

$$\int_0^\alpha {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{\Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{12 \sqrt{\pi }} \qquad \alpha = \sqrt[3]{2 \left(3 \sqrt{3}-5\right)} \approx 0.732$$

Since by binomial expansion again, it equals $$\left(\sqrt{3}-1\right) {_2F_1}\left(\frac{1}{3},\frac{1}{2},\frac{4}{3},10-6 \sqrt{3}\right)=\frac{\sqrt{\frac{1}{2} \left(6 \sqrt{3}-9\right) \pi } \Gamma \left(\frac{1}{3}\right)}{3\ 3^{3/4} \left(\sqrt{3}-1\right) \Gamma \left(\frac{5}{6}\right)}$$ due to certain transformation of hypergeometric series (see Special values of hypergeometric series by Akihito Ebisu). The rest are trivial.

This is a late solution, after the structural solution of pisco was already accepted, it also uses the elliptic curves intuition, and tries to give a "simpler substitution" and the way to obtain it. The substitution is $$X = \frac{x^3+4}{x^2}\ ,$$ but let us see how first it was obtained free of charge, since this is the main point. I will give full details of the computations, CAS support, and show the pictures of the involved elliptic curve paths.

(1) How to obtain the substitution?

(The reader considering sage code annoying may please completely skip (1) and extract only the information that an isogeny killing the $$3$$-torsion point $$(0,1)$$ is used.)

The given integral can be seen as the integral of the invariant differential $$dx/y$$ on the path $$\gamma$$ from the point $$P=(x_0,y_0)=(0,1)$$ to $$Q=(x_1,y_1)=(2,3)$$ on $$E(\Bbb R)$$, where $$E$$ is the elliptic curve given by the (affine) equation: $$E\ :\ y^2=x^3+1\ .$$ In a picture:

Just for the protocol: This was obtained in sage via:

sage: E = EllipticCurve(QQ, [0, 1])
sage: points = [E(P) for P in [ (-1,0), (0,1), (0,-1), (2,3), (2,-3) ]]

sage: pic = E.plot(xmin=-2, xmax=3)
sage: for P in points:
....:     pic += point(P.xy(), size=40, rgbcolor=hue(0.75))
....:
sage: pic


As the OP remarks, there would be no problem to compute the integral from $$P=(0,1)$$ to the infinity point using the substitution $$t=x^3+1$$ to introduce the beta function, but we integrate from $$P$$ to $$Q$$, and the same substitution leads to an "incomplete beta" value. So the problem is the upper integration limit $$2$$ corresponding to $$Q$$.

We would like to use an algebraic substitution and move $$Q$$ to a "simpler point" (possibly on some other elliptic curve). Note that the points that appear have finite order, using sage to print this information...

sage: for P in points:
....:     print(f'The point {P.xy()} has order {P.order()}')
....:
The point (-1, 0) has order 2
The point (0, 1) has order 3
The point (0, -1) has order 3
The point (2, 3) has order 6
The point (2, -3) has order 6


The idea to proceed is natural, we use the isogeny which "simplifies" the complicated torsion point $$Q=(2,3)$$ of order $$6$$. Note that $$2Q=P$$ on $$E$$:

sage: P, Q = E.point((0,1)), E.point((2,3))
sage: 2*Q == P
True


(So using the isogeny which "kills" $$P$$ will also "simplify" $$Q$$.)

We ask sage for this isogeny in the one-liner:

sage: phi = E.isogeny(kernel=P)


and let it give us its underlying information:

sage: phi
Isogeny of degree 3
from Elliptic Curve defined by y^2 = x^3 + 1  over Rational Field
to   Elliptic Curve defined by y^2 = x^3 - 27 over Rational Field
sage: phi.rational_maps()
((x^3 + 4)/x^2, (x^3*y - 8*y)/x^3)
sage: phi(P), phi(Q)
((0 : 1 : 0), (3 : 0 : 1))


This leads to a map $$\phi:(x,y)\to\phi(x,y)=(X,Y)$$ between the elliptic curves \begin{aligned} E\ :\qquad && y^2 &= x^3+1\text{ and}\\ E'\ :\qquad && Y^2 &= X^3-27\text{ given by the passage}\\ && X &=\frac{x^3+4}{x^2}\ ,\\ && Y &=y\cdot\frac{x^3-8}{x^3}\ . \end{aligned} and the image curve is $$Y^2=X^3-27$$, the path $$\gamma$$ from $$P$$ to $$Q$$ on $$E(\Bbb R)$$ becomes the path $$\gamma'$$ from $$P'=\infty$$ to $$Q'=(3,0)$$ on $$E'(\Bbb R)$$. I can plot only $$Q'$$ on $$E'(\Bbb R)$$...

So we integrate alternatively the invariant differential form $$dX/Y$$ on the "upper branch" starting from $$Q'=(3,0)$$ to the infinity point.

Indeed, the above expressions $$X,Y$$ satisfy: \begin{aligned} X^3-Y^2 &=\frac{(x^3+4)^3}{x^6}-\underbrace{y^2}_{x^3+1}\cdot\frac{(x^3-8)^2}{x^6}\\ &=\frac 1{x^6} \Big[\ (x^9+12x^6+48x^3+64) - (x^9-15x^6+48x^3+64) \ \Big] \\ &=27\ . \end{aligned}

(2) Using the substitution:

We formally use $$X=(x^3+4)/x^2$$, $$Y=y(x^3-8)/x^3$$, and compute formally: \begin{aligned} dX &= \left(x+\frac 4{x^2}\right)'\; dx = \left(1-\frac 8{x^3}\right)\; dx = \frac {x^3-8}{x^3}\; dx \ ,\text{ so as expected}\\ \frac{dX}Y %&=\frac {x^3-8}{x^3}\; dx\cdot\frac {x^3}{y(x^3-8)}\\ &=\frac{dx}y\ . \end{aligned} (The isogeny connects the invariant differentials.) This gives: \begin{aligned} \int_0^2\frac {dx}{\sqrt{x^3+1}} &= \int_{\gamma\text{ from }(0,1)\text{ to }(2,3)\text{ on }E(\Bbb R)} \frac{dx}y \\ &= \int_{\gamma'\text{ from }(3,0)\text{ to }\infty\text{ on }E'(\Bbb R)} \frac{dX}Y \\ &= \int_{Y=0}^{Y=\infty} \frac{d(Y^2+27)^{1/3}}Y =\int_0^\infty \frac 23(Y^2+27)^{1/3-1}\; dY \\ &\qquad\text{ (Substitution: 27t=Y^2+27, Y=27^{1/2}(t-1)^{1/2})} \\ &=\int_1^\infty \frac 23\cdot 27^{-2/3}\; t^{-2/3}\; 27^{1/2}\; \frac 12(t-1)^{-1/2}\; dt \\ &\qquad\text{ (Substitution: u=1/t)} \\ &=\int_0^1 \frac 13\cdot 3^{-2}\; u^{2/3}\; 3^{3/2}\; (1-u)^{-1/2}\; u^{1/2}\; \frac 1{u^2}\;du \\ &= 3^{-3/2}\int_0^1 u^{1/6-1}\; (1-u)^{1/2-1}\;du \\ &= 3^{-3/2}B\left(\frac 16,\frac 12\right) \ . \end{aligned} $$\square$$

(3) Note:

The above value involving the beta function $$B$$ is equal to $$\frac 16B\left(\frac 16,\frac 13\right)$$, use for this the multiplication theorem for the gamma function:

$$\Gamma\left(\frac 13\right) \Gamma\left(\frac 23\right) \Gamma\left(\frac 33\right) =(2\pi)\;3^{-1/2}\;\Gamma(1)\ .$$

• Do you have any sources or recommendations for learning more about elliptic curves, namely the types you use here?
– user879915
Apr 13, 2021 at 23:22

Consider $$y^2=4x^3+4,g_2=0,g_3=-4$$ The real period of $$\wp(z)$$ is $$\omega_1=\int_{-1}^{\infty} \frac{1}{\sqrt{1+x^3}} \text{d}x =\frac{\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{2\sqrt{\pi} }$$ And $$\wp\left ( \frac{\omega_1}{2} \right ) =-1$$

We have the addition formula: \begin{aligned} &\wp(3z) = \frac{1}{4}\left [ \frac{\wp'(z)-\wp'(2z)}{\wp(z)-\wp(2z)} \right ]^2 -\wp(z)-\wp(2z)\\ &\wp(2z)=\frac{1}{4} \left [ \frac{\wp''(z)}{\wp'(z)} \right ] ^2-2\wp(z)\\ &\wp(u+v) =\frac{1}{4} \left [ \frac{\wp'(u)-\wp'(v)}{\wp(u)-\wp(v)} \right ]^2 -\wp(u) -\wp(v) \end{aligned} where $$3z,2z,z$$ is not at the period lattices. Let $$x=\wp\left ( \frac{\omega_1}{6} \right ) ,y=\wp\left ( \frac{\omega_1}{3} \right )$$.And they have two relations: \begin{aligned} &\frac{1}{4} \left [ \frac{\sqrt{4x^3+4}-\sqrt{4y^3+4} }{x-y} \right ]^2-x-y = -1\\ &y=\frac{9x^4}{4x^3+4} -2x \end{aligned} Solve it and gives, $$\wp\left ( \frac{\omega_1}{6} \right )=2, \wp\left ( \frac{\omega_1}{3} \right )=0$$ So, \begin{aligned} I:&= \int_{0}^{2} \frac{1}{\sqrt{1+x^3} } \text{d}x\\ &=2\left(\frac{\omega_1}{3} -\frac{\omega_1}{6} \right)\\ &=\frac{\omega_1}{3} \end{aligned}

We have $$\wp\left ( \frac{\omega_1}{4} \right ) =\sqrt{3}-1$$,hence $$\int_{0}^{\sqrt{3} -1} \frac{1}{\sqrt{1+x^3} }\text{d}x =\frac{\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{12\sqrt{\pi} }$$

We have $$t=\wp\left ( \frac{\omega_1}{8} \right ) = -1+\sqrt{3} +\sqrt{3}\sqrt{2-\sqrt{3} } +\sqrt{3} \sqrt{2-\sqrt{3}+2\sqrt{2-\sqrt{3} } }$$,hence $$\int_{0}^{t} \frac{1}{\sqrt{1+x^3} } \text{d}x =\frac{5\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{24\sqrt{\pi} }$$

We have $$t=\wp\left ( \frac{\omega_1}{12} \right ) =2+\sqrt{3}+\sqrt{3} \sqrt{3+2\sqrt{3} }$$,hence $$\int_{0}^{t} \frac{1}{\sqrt{1+x^3} } \text{d}x =\frac{\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{4\sqrt{\pi} }$$

We have $$t=\wp\left ( \frac{\omega_1}{9} \right ) =2^{1/3} \left ( 16+3\cdot2^{2/3}\sqrt[3]{37+i\sqrt{3} }+ 6\cdot2^{-1/3}\sqrt[3]{37-i\sqrt{3} } \right )^{1/3}\approx4.5707391614928...$$,hence $$\int_{0}^{t} \frac{1}{\sqrt{1+x^3} } \text{d}x =\frac{2\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{9\sqrt{\pi} }$$

If $$t\ge0$$,and $$\int_{0}^{t} \frac{1}{\sqrt{1+x^3} } \text{d}x =R\cdot\frac{\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{\sqrt{\pi} }$$ where $$R$$ is rational with $$0\le R\le\frac{1}{3}$$.$$t$$ must be an algebraic number.

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{2}{\dd x \over \root{1 + x^{3}}}} = 2\int_{0}^{1}\pars{1 + 8x^{3}}^{-1/2}\,\dd x \\[5mm] = & {2 \over 3}\int_{0}^{1}x^{-2/3}\,\bracks{1 + 8x}^{-1/2}\,\dd x \\[5mm] & = {2 \over 3}\int_{0}^{1}x^{\color{red}{1/3}\ -\ 1}\, \pars{1 - x}^{\color{#0f0}{4/3}\ -\ \color{red}{1/3}\ -\ 1} \,\bracks{1 -\pars{\bf\color{red}{-8}}x}^{\,-{\bf 1/2}}\,\dd x \\[5mm] & = {2 \over 3}\,\mrm{B}\pars{\color{red}{1 \over 3}, \color{#0f0}{4 \over 3} - \color{red}{1 \over 3}} \mbox{}_{2}\mrm{F}_{1}\pars{\bf{1 \over 2}, \color{red}{1 \over 3};\color{#0f0}{4 \over 3};-8} \label{1}\tag{1} \\[5mm] & = {2 \over 3}\,{\Gamma\pars{1/3}\Gamma\pars{1} \over \Gamma\pars{4/3}} \,\mbox{}_{2}\mrm{F}_{1}\pars{\bf{1 \over 2}, \color{red}{1 \over 3};\color{#0f0}{4 \over 3};-8} \\[5mm] & = \bbx{\large {\Large 2}\,\,\mbox{}_{2}\mrm{F}_{1}\!\!\!\pars{\bf{1 \over 2}, \color{red}{1 \over 3};\color{#0f0}{4 \over 3};-8}} \approx 1.4022 \\ & \end{align} Step (\ref{1}): See the Euler Type Hypergeometric Function.