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

Question

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 ?.

Answer 1

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 $0<x<2$, 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.

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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 $$\int_0^\alpha {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{\Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{12 \sqrt{\pi }} \qquad \alpha = \sqrt[3]{2 \left(3 \sqrt{3}-5\right)} \approx 0.732 $$ $$\int_0^\alpha {\frac{1}{{\sqrt {1 + {x^3}} }}dx} = \frac{2 \Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{15 \sqrt{\pi }}\qquad \alpha = \left(9 \sqrt{5}+3 \sqrt{6 \left(13-\frac{29}{\sqrt{5}}\right)}-19\right)^{1/3}\approx 1.34$$

Answer 2

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.

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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.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.

Answer 4

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.

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

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(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: $$ \begin{aligned} \int_0^2\frac {dx}{\sqrt{x^3+1}} &= \int_{\gamma\text{ from }(0,1)\text{ to }(2,3)\text{ on }E(\Bbb R)} \frac{dx}y \\ &= \int_{\gamma'\text{ from }(3,0)\text{ to }\infty\text{ on }E'(\Bbb R)} \frac{dX}Y \\ &= \int_{Y=0}^{Y=\infty} \frac{d(Y^2+27)^{1/3}}Y =\int_0^\infty \frac 23(Y^2+27)^{1/3-1}\; dY \\ &\qquad\text{ (Substitution: $27t=Y^2+27$, $Y=27^{1/2}(t-1)^{1/2}$)} \\ &=\int_1^\infty \frac 23\cdot 27^{-2/3}\; t^{-2/3}\; 27^{1/2}\; \frac 12(t-1)^{-1/2}\; dt \\ &\qquad\text{ (Substitution: $u=1/t$)} \\ &=\int_0^1 \frac 13\cdot 3^{-2}\; u^{2/3}\; 3^{3/2}\; (1-u)^{-1/2}\; u^{1/2}\; \frac 1{u^2}\;du \\ &= 3^{-3/2}\int_0^1 u^{1/6-1}\; (1-u)^{1/2-1}\;du \\ &= 3^{-3/2}B\left(\frac 16,\frac 12\right) \ . \end{aligned} $$ $\square$

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(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:

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

Answer 5

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 $$ \wp\left ( \frac{\omega_1}{2} \right ) =-1 $$

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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, $$ \begin{aligned} I:&= \int_{0}^{2} \frac{1}{\sqrt{1+x^3} } \text{d}x\\ &=2\left(\frac{\omega_1}{3} -\frac{\omega_1}{6} \right)\\ &=\frac{\omega_1}{3} \end{aligned} $$

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

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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 $$ \int_{0}^{t} \frac{1}{\sqrt{1+x^3} } \text{d}x =\frac{5\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{24\sqrt{\pi} } $$

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

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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 $$ \int_{0}^{t} \frac{1}{\sqrt{1+x^3} } \text{d}x =\frac{2\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{9\sqrt{\pi} } $$

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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.