Convert integral $\int_0^{\frac{\sqrt3}2} {\frac{1}{{{x^{3/4}}{{\left( {1 + x} \right)}^{3/4}}}}dx}$ to elliptic integral

Question

I found this integral from a friend of mine $$I = \int_0^{\frac{\sqrt3}2} {\frac{1}{{{x^{3/4}}{{\left( {1 + x} \right)}^{3/4}}}}dx} $$

Which its closed-form is :$$\frac{{2\sqrt 2 {\pi ^{3/2}}}}{{3{\Gamma ^2}( {\frac{3}{4}} )}}$$ Look at this closed form I have a feeling that the given integral can be involved with the complete elliptic integral of the first kind $K( {\frac{1}{2}} )$. But I don't know how to convert this integral to get that result.

I tried to use sub:$$\eqalign{ & x = \frac{{1 - t}}{{1 + t}} \Rightarrow t = \frac{{1 - x}}{{1 + x}} \Rightarrow dx = - \frac{2}{{{{\left( {1 + t} \right)}^2}}}dt \cr & \Rightarrow I = \frac{2}{{{2^{3/4}}}}\int\limits_{7 - 4\sqrt 3 }^1 {\frac{1}{{\sqrt {1 + t} {{\left( {1 - t} \right)}^{3/4}}}}} dt,{\text{ since }}7 - 4\sqrt 3 = {\left( {2 - \sqrt 3 } \right)^2}{\text{ then let}}:t = {u^2} \cr & \Rightarrow I = \frac{4}{{{2^{3/4}}}}\int\limits_{2 - \sqrt 3 }^1 {\frac{u}{{\sqrt {1 + {u^2}} {{\left( {1 - {u^2}} \right)}^{3/4}}}}} du \cr} $$

So I get stuck here. May I ask for help? Or give me a hint about substitution. Thank you very much.

Edit #1: After using generalized binomial theorem and changing order of summation and integration, with some manipulation with the last sum, I arrived at:$$I=2\ 2^{3/4} \sqrt[8]{3} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};-\frac{\sqrt{3}}{2}\right)$$

May be, there is a transformation with $_2F_1$ can link this result with the elliptic integral. I am still trying to figure out.

Answer 1

$\def\cn{\operatorname{cn}} \def\dn{\operatorname{dn}} \def\K{\operatorname K} \def\I{\operatorname I} \def\B{\operatorname B} \def\F{\text F} \def\sd{\operatorname {sd}} \def\sn{\operatorname {sn}} $

Here is how to find the OP’s closed form

Consider Jacobi $\sn(z,m)$, Jacobi $\sd(z,m)$, and the half period Jacobi $\cn(z,m)$ identity:

$$\cn\left(\frac23\K\left(\frac12\right),\frac12\right)= \cn\left(\frac13\K\left(\frac12\right)-\K\left(\frac12\right),\frac12\right)=\frac1{\sqrt2}\sd\left(\frac13\K\left(\frac12\right),\frac12\right)=\sqrt{\frac2{x^2+1}-1},x=\cn\left(\frac13\K\left(\frac12\right),\frac12\right)$$

and the Jacobi $\cn(z,m)$ double angle formula:

$$\cn\left(\frac23\K\left(\frac12\right),\frac12\right)=\frac{\cn^2\left(\frac13\K\left(\frac12\right),\frac12\right)-\sn^2\left(\frac13\K\left(\frac12\right),\frac12\right)\dn^2 \left(\frac13\K\left(\frac12\right),\frac12\right)}{1-\frac12 \sn^4\left(\frac13\K\left(\frac12\right),\frac12\right)}=-\frac{4x^2}{x^4-2x^2-1}-1$$

Solving for $x$:

$$-\frac{4x^2}{x^4-2x^2-1}-1= \sqrt{\frac2{x^2+1}-1}\implies x=\cn\left(\frac13\K\left(\frac12\right),\frac12\right)=\sqrt[4]{2\sqrt3-3}$$

the above method is from @Start Wearing Purple’s in this question. Now, apply a reciprocal type Jacobi $\dn(z,m)$ transformation:

$$\cn\left(\frac13\K\left(\frac12\right),\frac12\right)=\dn\left(\frac{1+i}6\K(2),2\right)\implies 1-\dn^4\left(\frac{1+i}6\K(2),2\right)=4-2\sqrt3$$

and notice inverse beta regularized in $1-\dn^4\left(\frac{1+i}2\K(2)x,2\right)=\I^{-1}_x\left(\frac12,\frac14\right)$ from section 4 of this post to get:

$$4-2\sqrt3=\I^{-1}_\frac13\left(\frac12,\frac14\right)$$

Taking the regularized beta function on both sides and converting it into an incomplete beta function gives us the closed form:

$$\I_{4-2\sqrt3}\left(\frac12,\frac14\right)=\frac13\implies \color{blue}{\B\left(\frac12,\frac14\right)-\B_{4-2\sqrt3}\left(\frac12,\frac14\right)}=\frac23\B\left(\frac12,\frac14\right)= \frac{{2\sqrt 2 {\pi ^{3/2}}}}{{3{\Gamma ^2}( {\frac{3}{4}} )}} $$

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The OP’s hypergeometric function converts to an incomplete beta function:

$$2^\frac74 \sqrt[8]3 \, _2\F_1\left(\frac14,\frac34;\frac54;-\frac{\sqrt{3}}{2}\right) =(-1)^{-\frac14}\B_{-\frac{\sqrt 3}2}\left(\frac14,\frac14\right)$$

Afterwards, we apply a complementary argument and reciprocal argument identity transformation:

$$(-1)^{-\frac14}\B_{-\frac{\sqrt 3}2}\left(\frac14,\frac14\right)=(-1)^{-\frac14}\left(\B\left(\frac14,\frac14\right)-\B_{1+\frac{\sqrt3}2}\left(\frac14,\frac14\right)\right)= \color{blue}{\B\left(\frac12,\frac14\right)-\B_{4-2\sqrt3}\left(\frac12,\frac14\right)}$$

The blue expressions matches the integral, so combine them and their equalities. Therefore the lemniscate arc length constant $\text s$ appears:

$$\bbox[5px,border: 3px solid #ADD8F7]{\int_0^{\frac{\sqrt3}2}x^{-\frac34}(x+1)^{-\frac34}dx =\frac13\sqrt{\frac2\pi}\Gamma^2\left(\frac14\right)=\frac23\text s}$$

Answer 2

This is to answer the OP how to convert it to the elliptic integral. Let $u=2x+1$

$$I=\int_1^{1+\sqrt3} \frac1{(u^2-1)^{3/4}}du$$

Let $u=\csc \theta$

$$I=\sqrt2\int_{\theta_0}^{\pi/2} \frac1{\sqrt{\sin\theta\cos\theta}}d\theta$$

where $\theta_0=\arcsin\frac{\sqrt3-1}2$, and note that

$$2\sin\theta\cos\theta=(\sin\theta+\cos\theta)^2-1=2\cos^2(\theta-\frac\pi4)-1=1-2\sin^2(\theta-\frac\pi4)$$

hence $$I=2\int_{\theta_0}^{\pi/2} \frac1{\sqrt{1-2\sin^2(\theta-\frac\pi4)}}d\theta$$

Let $\phi=\theta-\frac\pi4$

$$I=2\int_{\theta_0-\frac\pi4}^{\frac\pi4} \frac1{\sqrt{1-2\sin^2 \phi}}d\phi$$

Use the defintion of elliptic integral of the first kind

$$F(a,k)=\int_0^a \frac1{\sqrt{1-k^2\sin^2\phi}}d\phi$$

We get

$$\int_0^{\frac{\sqrt3}2} {\frac{1}{{{x^{3/4}}{{\left( {1 + x} \right)}^{3/4}}}}dx}=2F\left(\frac\pi4,\sqrt2\right)-2F\left(\arcsin\left(\frac{\sqrt3-1}2\right)-\frac\pi4,\sqrt2\right)\tag{1}$$

Using the Reciprocal-Modulus Transformation formula as suggested by @Tyma Gaidash, the first half of eq.(1) is converted to

$$2F\left(\frac\pi4,\sqrt2\right)=2\cdot \frac1{\sqrt2} F\left(\frac\pi2, \frac1{\sqrt2}\right)=\sqrt2 K\left(\frac1{\sqrt2}\right)$$

where $K(k)=\int_0^\frac\pi2 \frac1{\sqrt{1-k^2\sin^2\phi}}d\phi$ is complete elliptic integral of the first kind, hence

$$\boxed{\int_0^{\frac{\sqrt3}2} {\frac{1}{{{x^{3/4}}{{\left( {1 + x} \right)}^{3/4}}}}dx}=\sqrt2 K\left(\frac1{\sqrt2}\right)-2F\left(\arcsin\left(\frac{\sqrt3-1}2\right)-\frac\pi4,\sqrt2\right)}$$

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Remarks:

For the closed form, the integral limit $\sqrt3+1=\frac2{\sqrt3-1}$ reminds me maybe it is related to the elliptic curve, where I ever asked a question here. But I have very limited knowledge on this topic.

Answer 3

$$ \begin{align} I &:= \int_{0}^{\sqrt{3}/2}\frac{1}{x^{3/4}\left(1+x\right)^{3/4}}dx \\ &= \int_{0}^{\sqrt{3}/2}\frac{1}{x^{3/4}}\left(1-\left(-x\right)\right)^{5/4-5/4-3/4}\ _{2}F_{1}\left(\frac{5}{4}-\frac{5}{4},\frac{5}{4}-\frac{3}{4};\frac{5}{4};-x\right)dx \\ &= \int_{0}^{\sqrt{3}/2}\frac{1}{x^{3/4}}\ _{2}F_{1}\left(\frac{5}{4},\frac{3}{4};\frac{5}{4};-x\right)dx \tag{1}\\ &= \int_{0}^{\sqrt{3}/2}\frac{1}{x^{3/4}}\ _{2}F_{1}\left(\frac{1}{4}+1,\frac{3}{4};\frac{5}{4};-x\right)dx \\ &= \int_{0}^{\sqrt{3}/2}\frac{1}{x^{3/4}}\left(\frac{3/4 \cdot (-x)}{5/4}\ _{2}F_{1}\left(\frac{1}{4}+1,\frac{3}{4}+1;\frac{5}{4}+1;-x\right)+ \text{ } _{2}F_{1}\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};-x\right)\right)dx \tag{2}\\ &= \int_{0}^{\sqrt{3}/2}\left(\frac{1}{x^{3/4}}\cdot \text{ }_{2}F_{1}\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};-x\right)+4x^{1/4}\cdot\frac{1/4\cdot3/4}{-5/4}\ _{2}F_{1}\left(\frac{5}{4},\frac{7}{4};\frac{9}{4};-x\right)\right)dx \\ &= \int_{0}^{\sqrt{3}/2}\frac{d}{dx}4x^{1/4}\ _{2}F_{1}\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};-x\right)dx \tag{3}\\ &= 4x^{1/4}\ _{2}F_{1}\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};-x\right) \Bigg|_{0}^{\sqrt{3}/2} \\ &= 4\left(\frac{\sqrt{3}}{2}\right)^{1/4}\ _{2}F_{1}\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};-\frac{\sqrt{3}}{2}\right)-4\left(0\right)^{1/4}\ _{2}F_{1}\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};-0\right) \\ &= 2^{7/4}3^{1/8}\ _{2}F_{1}\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};-\frac{\sqrt{3}}{2}\right) \end{align} $$

$(1)$ $(2)$ $(3)$