integral $\int_0^a \frac{1}{\sqrt{1+x^6}} dx$ where $a=\frac{1}{\sqrt{\sqrt{3}-1}}$

Question

$I=\int_0^a \frac{1}{\sqrt{1+x^6}} dx$ where $a=\frac{1}{\sqrt{\sqrt{3}-1}}$ first, do binomial series on the integrand function, $\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n}x^{6n}$ after integration, we got $\sum_{n=0}^\infty \frac{(-1)^n (\frac{1}{2})_n}{n!}\cdot\frac{a^{6n}}{6n+1}$ further, write $6n+1=\frac{(\frac{7}{6})_n}{(\frac{1}{6})_n}$, so the integral equals $\int_0^a \frac{1}{\sqrt{1+x^6}} dx=a\cdot F(\frac{1}{6},\frac{1}{2},\frac{7}{6};-a^6)$, where $F$ is hypergeometric function.

Questions:

Q1: the binomial series converges for $|x|<1$, but here $a>1$, can we do the binomial series expansion here?

Q2: how to proceed, the answer is $\frac{1}{8}B(\frac{1}{3},\frac{1}{6})$, where $B$ is beta function.

Answer 1

Some days before i have answered a question (4463639) involving similar expressions. There is also some link to Wolfram on the Incomplete Beta Function, and the following relations were displayed: $$ \begin{aligned} B(z;a,b) &= \int_0^z u^{a-1}(1-u)^{b-1}\; du \\ &= \frac 1az^a\; {}_2F_1(a,1-b;\ a+1;\ z) \\ &\qquad\qquad\qquad= z^a\sum_{n\ge 0}\frac{(1-b)_n}{n!\;(a+n)}\; z^n\qquad\text{if convergent} \ , \\ \text{ and also }& \\ B(z;a,b) &= \int_0^z u^{a-1}(1-u)^{b-1}\; du \\ &= \int_0^{z/(1-z)}\frac{v^{a-1}}{(1+v)^{a+b}}\;dv \end{aligned} $$ Notations slightly collide, for this reason i will use $$ w = (\sqrt 3-1)^{-1/2}\ , $$ and integrate on $[0,w]$.

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(Q1) We are dealing with analytic functions, but we should avoid values for the (analitically extended) Gaussian hypergeometric function in the slit $(1, \infty)$, here is a link i found on this aspect: EM on hypergeometric functions.

(Q2) Let us compute the integral.

$$ \begin{aligned} 6\int_0^w\frac 1{(1+x^6)^{1/2}}\; dx &= 6\int_0^{w^6}\frac 1{(1+v)^{1/2}}\; \frac 16\; v^{-5/6}\; dv &&\text{ with } v = x^6 \\ &= \int_0^{w^6}\frac {v^{a-1}}{(1+v)^{a+b}}\; dv &&\text{ with } a=\frac 16\ ,\ b=\frac 13 \\ &= \int_0^{z/(1-z)}\frac {v^{a-1}}{(1+v)^{a+b}}\; dv&&\text{ with } z = \frac 19(3+2\sqrt 3) \\ % &= % B\left(z;a,b\right) % \\ &= B\left(z;\frac 16,\frac 13\right) \\ &= B\left(1;\frac 13,\frac 16\right) - \underbrace{B\left(1-z;\frac 13,\frac 16\right)} _{\color{red}{\text{Show it is }\frac 14B\left(\frac 13,\frac 16\right)\ !}} &&\text{ and let }Z = 1-z = \frac 29(3-\sqrt 3) \ . \end{aligned} $$ (Regarding (Q1) we can pass from $B(z,a,b)$ to an ${}_2F_1$-value for the argument $z$ with $z\approx 0.718233512793\in (0,1)$, also the last expression $B(1-z;b,a)$ has argument in $(0,1)$- so we have convergence if series are written. But we do not need this.)

Let us show the red marked equality: $$ \begin{aligned} B\left(1-z;\frac 13,\frac 16\right) &= B\left(Z;\frac 13,\frac 16\right) = \int_0^{V:=Z/(1-Z)}\frac{v^{1/3\ -\ 1}}{(1+v)^{1/3\ +\ 1/6}}\; dv \\[2mm] &\qquad\text{ and with $v=x^3$, $y=(1+v)^{1/2}$, } \\ &\qquad\text{ so that $y^2=1+v=1+x^3$, and $dv=3x^2\; dx$, and $v^{1/3\ -\ 1}=x^{-2}$} \\[2mm] &= 3\int _{(x,y)=(0,\ 1)} ^{(x,y)=\left(V^{1/3}\ ,\ (1+V)^{1/2}\right) =\left(\sqrt3-1\ ,\ \sqrt{6\sqrt 3-9}\right)} \ \frac{dx}y = 3\int_Q^P\frac{dx}y \\ &= 3\int_O^P\frac{dx}y - 3\int_O^Q\frac{dx}y \ , \end{aligned} $$ and the last integral is a path integral on $E(\Bbb R)$ where $E$ is the elliptic curve $y^2=x^3+1$. Let us use the notations $a=\sqrt 3$ and $b=\sqrt{6a-9}$

It turns out that the involved points $$ \begin{aligned} P &=(a-1\ ,\ b)=\left(\sqrt3-1\ ,\ \sqrt{6\sqrt 3-9}\right)\ ,\\ Q &=(0,1)\ , \end{aligned} $$ from $E(K)$ satisfy $4P=O=3Q$. Also $2P=(-1,0)$, and $2Q=-Q=(0,-1)$. Here is a piece of code giving this information in sage:

B = sqrt(6*sqrt(3) - 9)
K.<b> = NumberField( B.minpoly(), embedding=B.n() )
a = (b^2 + 9)/6
E = EllipticCurve(K, [0, 1])
P = E.point( (a - 1, b) )
P = E.point( (0    , 1) )
print(f'E = {E}')
print(f'The point P(a - 1, b) in E(K) has order {P.order()}.')
print(f'The point Q(    1, 0) in E(K) has order {Q.order()}.')

And we get (slightly manually rearranged to fit in the line):

E = Elliptic Curve defined by y^2 = x^3 + 1 over Number Field in b
    with defining polynomial x^4 + 18*x^2 - 27 with b = 1.179959679570986?
The point P(a - 1, b) in E(K) has order 4.
The point Q(    1, 0) in E(K) has order 3.

Because $dx/y$ is invariant w.r.t. the additive operation on $E$ we have: $$ \begin{aligned} \int_O^P\frac{dx}y &= \int_{P}^{2P}\frac{dx}y = \int_{2P}^{3P}\frac{dx}y = \int_{3P}^{O}\frac{dx}y \ , \\[2mm] \int_O^Q\frac{dx}y &= \int_Q^{2Q}\frac{dx}y = \int_{2Q}^O\frac{dx}y \ . \end{aligned} $$ Now use parametrizations of $E(\Bbb R)$ as follows. Let $x$ go from $\infty$ to $-1$, associate the path $\gamma_+(x)=(x, \ +\sqrt{1+x^3})\in E(\Bbb R)$, then let $x$ go from $-1$ to $+\infty$, and associate the path $\gamma_-(x)=(x, \ -\sqrt{1+x^3})\in E(\Bbb R)$. Let $\gamma$ be the full path, $\gamma =\gamma_+\cup\gamma_-$.

Let $J$ be the integral of $dx/y$ on $\gamma$, $J=-2\int_{-1}^\infty \frac{dx}{\sqrt{1+x^3}}$. Maybe it is best to have a picture, and to associate all integrals to fractions of an integral on the "whole contour $\gamma$ taken in trigonometric sense":

The two points on $Oy$ are $\pm Q$, the two points on the vertical $x=\sqrt3-1\approx 0.732$ are $\pm P$, the last marked point is $2P=(-1,0)$. We obtain $$ \begin{aligned} \int_{\sqrt 3-1}^\infty\frac{dx}{\sqrt{1+x^3}} &= \int_{\gamma_+\ ,\ x\in[\sqrt 3-1,\ \infty)}\frac{dx}{\sqrt{1+x^3}} = -\int_O^P\frac{dx}y = -\frac 14 J\ , \\ \int_{-1}^{\sqrt 3-1}\frac{dx}{\sqrt{1+x^3}} &= \int_{\gamma_+\ ,\ x\in[-1, \sqrt 3-1]}\frac{dx}{\sqrt{1+x^3}} = -\int_P^{2P}\frac{dx}y = -\frac 14J \ , \\ \int_0^\infty\frac{dx}{\sqrt{1+x^3}} &= \int_{\gamma_+\ ,\ x\in[0,\infty)}\frac{dx}{\sqrt{1+x^3}} = -\int_0^Q\frac{dx}y = -\frac 13 J\ , \\ 2\int_{-1}^0\frac{dx}{\sqrt{1+x^3}} &= \int_{\gamma_+\ ,\ x\in[-1,0]}\frac{dx}{\sqrt{1+x^3}} + \int_{\gamma_-\ ,\ x\in[-1,0]}\frac{dx}{\sqrt{1+x^3}} = -\int_Q^{2Q}\frac{dx}y = -\frac 13 J\ , \\ &\qquad\text{ and from here we can conclude} \\ \frac 13B\left(Z;\frac 13,\frac 16\right) &= \int_Q^P\frac {dx}y = \int_0^ {\sqrt3-1}\frac {dx}{\sqrt{1+x^3}} \\ &= \int_{-1}^ {\sqrt3-1}\frac {dx}{\sqrt{1+x^3}} - \int_{-1}^ 0\frac {dx}{\sqrt{1+x^3}} \\ &= -\frac 14J + \frac 16J = \left(\frac 14-\frac 16\right) \cdot 3 \int_0^\infty\frac {dx}{\sqrt{1+x^3}} = \frac 14 \int_0^\infty\frac {dx}{\sqrt{1+x^3}} \\ &= \frac 1{12} B\left(\frac 13,\frac 16\right) \end{aligned} $$

$\square$