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Is there a simple way to prove the following identity?

$$\int_0^1\sqrt{\frac{u^2-2-2 \sqrt{u^4-u^2+1}}{4 u^6-8 u^4+8 u^2-4}}\mathrm du=\frac{\sqrt{3+2 \sqrt{3}}}{2^{10/3}\pi}\Gamma\left(\frac13\right)^3$$


Context:

This integral came up in trying to evaluate the complete elliptic integral of the first kind $K(m)$ ($m$ is the parameter),

$$K\left(\exp\left(\frac{i\pi}{3}\right)\right)$$

in terms of simpler functions. In particular,

$$K\left(\exp\left(\frac{i\pi}{3}\right)\right)=C\left(1+i \left(2-\sqrt{3}\right)\right)$$

and $C$ is the integral mentioned in the first part.

I was able to show this through an indirect route, but I am hoping my messy method can be easily outdone.

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Here is another approach that doesn’t requires Hypergeometric Functions.

$$I=\int_{0}^{1}\sqrt{\frac{u^2-2-2\sqrt{u^4-u^2+1}}{4u^6-8u^4+8u^2-4}}du=\int_{0}^{1}\underbrace{\sqrt{\frac{2-u^2+2\sqrt{{u^4-u}^2+1}}{4\left(1-u^2\right)\left({u^4-u}^2+1\right)}}}_{u\rightarrow\sqrt{x}}du$$

$$I=\frac{1}{4}\int_{0}^{1}\underbrace{\sqrt{\frac{2-x+2\sqrt{x^2-x+1}}{x\left(1-x\right)\left(x^2-x+1\right)}}}_{x\rightarrow y+\frac{1}{2}}dx=\frac{1}{4}\int_{-\frac{1}{2}}^{\frac{1}{2}}\underbrace{\sqrt{\frac{\frac{3}{2}-y+2\sqrt{y^2+\frac{3}{4}}}{\left(\frac{1}{4}-y^2\right)\left(y^2+\frac{3}{4}\right)}}}_{f(y)}dy$$

$$I=\frac{1}{4}\int_{0}^{\frac{1}{2}}\left(f\left(y\right)+f\left(-y\right)\right)dy=\frac{\sqrt{2+\sqrt3}}{2\sqrt2}\int_{0}^{\frac{1}{2}}\underbrace{\sqrt{\frac{\frac{\sqrt3}{2}+\sqrt{y^2+\frac{3}{4}}}{\left(\frac{1}{4}-y^2\right)\left(y^2+\frac{3}{4}\right)}}}_{y=\sqrt{z^2-\frac{3}{4}}}dy$$

$$I=\frac{\sqrt{2+\sqrt3}}{2\sqrt2}\int_{\frac{\sqrt3}{2}}^{1}\underbrace{\frac{dz}{\sqrt{\left(1-z^2\right)\left(z-\frac{\sqrt3}{2}\right)}}}_{z=\cos{\left(\theta\right)}}=\frac{\sqrt{2+\sqrt3}}{2\sqrt2}\int_{0}^{\frac{\pi}{6}}\underbrace{\frac{dz}{\sqrt{\cos{\left(\theta\right)}-\cos{\left(\frac{\pi}{6}\right)}}}}_{\sin{\left(\frac{\theta}{2}\right)}=\sin{\left(\phi\right)}\sin{\left(\frac{\pi}{12}\right)}}$$

$$I=\frac{\sqrt{2+\sqrt3}}{2}\int_{0}^{\frac{\pi}{2}}\frac{d\phi}{\sqrt{1-\sin^2{\left(\frac{\pi}{12}\right)\sin^2{\left(\phi\right)}}}}=\frac{\sqrt{2+\sqrt3}}{2}K\left(\sin{\left(\frac{\pi}{12}\right)}\right)$$

$$I=\frac{\sqrt{2+\sqrt3}}{2}\left(\frac{1}{2\ 3^\frac{3}{4}}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{4}{6}\right)}\right)=\frac{\sqrt{3+2\sqrt3}}{2^\frac{10}{3}\pi}{\Gamma\left(\frac{1}{3}\right)}^3$$

Some explanation:

1) I’ve started factorizing the denominator to check if there was any common term that could be canceled.

2) Then, I applied a couple of substitutions. The first one had the objective of decreasing the degree of the variable, and the second was aiming to eliminate the first-degree term of the polynomial inside the inner root.

3) Then I rewrote the integral exploiting the symmetry of its integration limits, which result in a sum of two square roots ($f(y)+f(-y)$) that could be rewritten after some algebra as a single square root.

4) After that, some other substitutions were applied to make calculations easier. The last one may look kind trick, but it’s intuitive after rewriting the expression using sines of half angle.

5) Finally, a well-known representation of the Complete Elliptic Integral of the First Kind was found, and its value was taken from a table and then the result was simplified using both duplication and reflection formulas of Gamma.

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    $\begingroup$ Ah, I wish you'd posted earlier; you would have otherwise gotten the bounty. Nice work! $\endgroup$ – J. M. isn't a mathematician Apr 12 '20 at 0:46
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Let $\mathcal{I}$ denote the value of the definite integral

$$\mathcal{I}:=\int_{0}^{1}\mathrm{d}u\,\sqrt{\frac{u^{2}-2-2\sqrt{u^{4}-u^{2}+1}}{4u^{6}-8u^{4}+8u^{2}-4}}\approx1.5436866339.$$

Note: the denominator of the radicand of the outer square root has the factorization

$$\begin{align} 4u^{6}-8u^{4}+8u^{2}-4 &=4t^{3}-8t^{2}+8t-4;~~~\small{\left[u^{2}=t\right]}\\ &=4\left(t^{3}-2t^{2}+2t-1\right)\\ &=4\left(t-1\right)\left(t^{2}-t+1\right).\\ \end{align}$$


Using the substitution $u^{2}=t$, the integral $\mathcal{I}$ can be rewritten as

$$\begin{align} \mathcal{I} &=\int_{0}^{1}\mathrm{d}u\,\sqrt{\frac{u^{2}-2-2\sqrt{u^{4}-u^{2}+1}}{4u^{6}-8u^{4}+8u^{2}-4}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{1}{2\sqrt{t}}\sqrt{\frac{t-2-2\sqrt{t^{2}-t+1}}{4t^{3}-8t^{2}+8t-4}};~~~\small{\left[u=\sqrt{t}\right]}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{1}{2\sqrt{t}}\sqrt{\frac{t-2-2\sqrt{t^{2}-t+1}}{4\left(t-1\right)\left(t^{2}-t+1\right)}}\\ &=\frac14\int_{0}^{1}\mathrm{d}t\,\sqrt{\frac{2-t+2\sqrt{t^{2}-t+1}}{t\left(1-t\right)\left(t^{2}-t+1\right)}}.\\ \end{align}$$

Now the inner radical is just a square root of a quadratic function, which suggests that the integral might be further simplified using an appropriate Euler substitution.

Consider a substitution given implicitly by the relation

$$\sqrt{t^{2}-t+1}=t+x.$$

Solving for $t$, we obtain $t=\frac{1-x^{2}}{1+2x}$. The integral $\mathcal{I}$ is then transformed to

$$\begin{align} \mathcal{I} &=\frac14\int_{0}^{1}\mathrm{d}t\,\sqrt{\frac{2-t+2\sqrt{t^{2}-t+1}}{t\left(1-t\right)\left(t^{2}-t+1\right)}}\\ &=\frac14\int_{1}^{0}\mathrm{d}x\,\frac{(-2)\left(1+x+x^{2}\right)}{\left(1+2x\right)^{2}}\sqrt{\frac{3\left(1+x\right)}{\left(1-x\right)}\cdot\frac{\left(1+2x\right)}{x\left(2+x\right)}\cdot\frac{\left(1+2x\right)^{2}}{\left(1+x+x^{2}\right)^{2}}};~~~\small{\left[t=\frac{1-x^{2}}{1+2x}\right]}\\ &=\frac12\int_{0}^{1}\mathrm{d}x\,\sqrt{\frac{3\left(1+x\right)}{x\left(1-x\right)\left(2+x\right)\left(1+2x\right)}}\\ &=\frac{\sqrt{3}}{2}\int_{0}^{1}\mathrm{d}x\,\frac{\left(1+x\right)}{\sqrt{x\left(1-x\right)\left(1+x\right)\left(1+2x\right)\left(2+x\right)}}.\\ \end{align}$$


Next, look what happens when we transform the integral using the linear fractional transformation $x=\frac{1-y}{1+y}$:

$$\begin{align} \mathcal{I} &=\frac{\sqrt{3}}{2}\int_{0}^{1}\mathrm{d}x\,\frac{\left(1+x\right)}{\sqrt{x\left(1-x\right)\left(1+x\right)\left(1+2x\right)\left(2+x\right)}}\\ &=\frac{\sqrt{3}}{2}\int_{0}^{1}\mathrm{d}x\,\frac{\left(1+x\right)}{\sqrt{\left(1+x\right)^{6}\left(\frac{1-x}{1+x}\right)\left(\frac{x}{1+x}\right)\left(\frac{1}{1+x}\right)\left(\frac{1+2x}{1+x}\right)\left(\frac{2+x}{1+x}\right)}}\\ &=\frac{\sqrt{3}}{2}\int_{0}^{1}\mathrm{d}x\,\frac{\left(\frac{1}{1+x}\right)^{2}}{\sqrt{\left(\frac{1-x}{1+x}\right)\left(\frac{x}{1+x}\right)\left(\frac{1}{1+x}\right)\left(\frac{1+2x}{1+x}\right)\left(\frac{2+x}{1+x}\right)}}\\ &=\frac{\sqrt{3}}{2}\int_{1}^{0}\mathrm{d}y\,\frac{(-2)}{\left(1+y\right)^{2}}\cdot\frac{\left(\frac{1+y}{2}\right)^{2}}{\sqrt{y\left(\frac{1-y}{2}\right)\left(\frac{1+y}{2}\right)\left(\frac{3-y}{2}\right)\left(\frac{3+y}{2}\right)}};~~~\small{\left[x=\frac{1-y}{1+y}\right]}\\ &=\sqrt{3}\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{y\left(1-y\right)\left(1+y\right)\left(3-y\right)\left(3+y\right)}}\\ &=\sqrt{3}\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{y\left(1-y^{2}\right)\left(9-y^{2}\right)}}.\\ \end{align}$$


Recalling Euler's integral formula for the Gauss hypergeometric function: for real argument and parameters,

$$\int_{0}^{1}\mathrm{d}t\,\frac{t^{b-1}\left(1-t\right)^{c-b-1}}{\left(1-zt\right)^{a}}=\operatorname{B}{\left(b,c-b\right)}\,{_2F_1}{\left(a,b;c;z\right)};~~~\small{\left(a,b,c,z\right)\in\mathbb{R}^{4}\land0<b<c\land z<1},$$

(where $\operatorname{B}$ here denotes the usual beta function), we arrive at the following representation for $\mathcal{I}$ as a particular value of ${_2F_1}$:

$$\begin{align} \mathcal{I} &=\sqrt{3}\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{y\left(1-y^{2}\right)\left(9-y^{2}\right)}}\\ &=\frac{1}{\sqrt{3}}\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{y}\sqrt{1-y^{2}}\sqrt{1-\frac19y^{2}}}\\ &=\frac{1}{\sqrt{3}}\int_{0}^{1}\mathrm{d}t\,\frac{1}{2\sqrt{t}}\cdot\frac{1}{\sqrt[4]{t}\sqrt{1-t}\sqrt{1-\frac19t}};~~~\small{\left[y=\sqrt{t}\right]}\\ &=\frac{1}{2\sqrt{3}}\int_{0}^{1}\mathrm{d}t\,\frac{t^{b-1}\left(1-t\right)^{c-b-1}}{\left(1-zt\right)^{a}};~~~\small{\left[a:=\frac12,b:=\frac14,c:=\frac34,z:=\frac19\right]}\\ &=\frac{1}{2\sqrt{3}}\operatorname{B}{\left(b,c-b\right)}\,{_2F_1}{\left(a,b;c;z\right)}\\ &=\frac{1}{2\sqrt{3}}\operatorname{B}{\left(\frac14,\frac12\right)}\,{_2F_1}{\left(\frac12,\frac14;\frac34;\frac19\right)}.\\ \end{align}$$


Given $\left(a,b,z\right)\in\mathbb{R}_{>0}\times\mathbb{R}_{>0}\times\left(0,1\right)$, the Gauss hypergeometric function obeys the following two functional relations:

$${_2F_1}{\left(a,b;2b;z\right)}=\left(\frac{1+\sqrt{1-z}}{2}\right)^{-2a}\,{_2F_1}{\left(a,a-b+\frac12;b+\frac12;\left(\frac{1-\sqrt{1-z}}{1+\sqrt{1-z}}\right)^{2}\right)};~~~\small{b<a+\frac12},$$

and

$$\begin{align} {_2F_1}{\left(a,b;\frac12;z\right)} &=\frac{\Gamma{\left(a+\frac12\right)}\,\Gamma{\left(b+\frac12\right)}}{2\,\Gamma{\left(\frac12\right)}\,\Gamma{\left(a+b+\frac12\right)}}\bigg{[}{_2F_1}{\left(2a,2b;a+b+\frac12;\frac{1-\sqrt{z}}{2}\right)}\\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+{_2F_1}{\left(2a,2b;a+b+\frac12;\frac{1+\sqrt{z}}{2}\right)}\bigg{]}.\\ \end{align}$$

The following pair of identities are then immediate corollaries of the pair above by setting $b=a$: for $0<a\land0<z<1$,

$${_2F_1}{\left(a,\frac12;a+\frac12;z^{2}\right)}=\left(1+z\right)^{-2a}\,{_2F_1}{\left(a,a;2a;\frac{4z}{\left(1+z\right)^{2}}\right)},$$

and

$$\begin{align} {_2F_1}{\left(a,a;\frac12;z\right)} &=\frac{\left[\Gamma{\left(a+\frac12\right)}\right]^{2}}{2\,\Gamma{\left(\frac12\right)}\,\Gamma{\left(2a+\frac12\right)}}\bigg{[}{_2F_1}{\left(2a,2a;2a+\frac12;\frac{1-\sqrt{z}}{2}\right)}\\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+{_2F_1}{\left(2a,2a;2a+\frac12;\frac{1+\sqrt{z}}{2}\right)}\bigg{]}.\\ \end{align}$$


Continuing with our main evaluation of the integral $\mathcal{I}$, the quadratic transformations of ${_2F_1}$ given above allow us to reduce our integral to standard complete elliptic integrals:

$$\begin{align} \mathcal{I} &=\frac{1}{2\sqrt{3}}\operatorname{B}{\left(\frac14,\frac12\right)}\,{_2F_1}{\left(\frac12,\frac14;\frac34;\frac19\right)}\\ &=\frac{1}{2\sqrt{3}}\operatorname{B}{\left(\frac14,\frac12\right)}\,{_2F_1}{\left(\frac14,\frac12;\frac34;\frac19\right)}\\ &=\frac{1}{2\sqrt{3}}\operatorname{B}{\left(\frac14,\frac12\right)}\,\frac{\sqrt{3}}{2}\,{_2F_1}{\left(\frac14,\frac14;\frac12;\frac34\right)}\\ &=\frac{1}{2\sqrt{3}}\operatorname{B}{\left(\frac14,\frac12\right)}\,\frac{\sqrt{3}}{2}\cdot\frac{\left[\Gamma{\left(\frac34\right)}\right]^{2}}{2\,\Gamma{\left(\frac12\right)}}\bigg{[}{_2F_1}{\left(\frac12,\frac12;1;\frac{2-\sqrt{3}}{4}\right)}+{_2F_1}{\left(\frac12,\frac12;1;\frac{2+\sqrt{3}}{4}\right)}\bigg{]}\\ &=\frac{\Gamma{\left(\frac14\right)}\,\Gamma{\left(\frac34\right)}}{8}\bigg{[}{_2F_1}{\left(\frac12,\frac12;1;\frac{2-\sqrt{3}}{4}\right)}+{_2F_1}{\left(\frac12,\frac12;1;\frac{2+\sqrt{3}}{4}\right)}\bigg{]}\\ &=\frac{\sqrt{2}\,\pi}{8}\bigg{[}{_2F_1}{\left(\frac12,\frac12;1;\frac{2-\sqrt{3}}{4}\right)}+{_2F_1}{\left(\frac12,\frac12;1;\frac{2+\sqrt{3}}{4}\right)}\bigg{]}\\ &=\frac{1}{2\sqrt{2}}\bigg{[}K{\left(\frac{\sqrt{2-\sqrt{3}}}{2}\right)}+K{\left(\frac{\sqrt{2+\sqrt{3}}}{2}\right)}\bigg{]}\\ &=\frac{1}{2\sqrt{2}}\bigg{[}K{\left(\frac{\sqrt{2-\sqrt{3}}}{2}\right)}+K^{\prime}{\left(\frac{\sqrt{2-\sqrt{3}}}{2}\right)}\bigg{]}\\ &=\frac{1}{2\sqrt{2}}\bigg{[}K{\left(\sin{\frac{\pi}{12}}\right)}+K^{\prime}{\left(\sin{\frac{\pi}{12}}\right)}\bigg{]}.\\ \end{align}$$

where here $K{(k)}$ is the complete elliptic integral of the first kind defined as a function of elliptic modulus $k$ by

$$K{(k)}:=\int_{0}^{1}\mathrm{d}x\,\frac{1}{\sqrt{(1-x^{2})(1-k^{2}x^{2})}};~~~\small{-1<k<1},$$

and $K^{\prime}{(k)}$ is the complementary complete elliptic integral of the first kind and is defined in terms of $K$ by

$$K^{\prime}{(k)}:=K{\left(\sqrt{1-k^{2}}\right)}.$$

We can complete our calculation by recognizing that the modulus $k=\sin{\frac{\pi}{12}}$ is in fact the third elliptic integral singular value, $k_{3}$.

Finally, we obtain:

$$\begin{align} \mathcal{I} &=\frac{1}{2\sqrt{2}}\bigg{[}K{\left(\sin{\frac{\pi}{12}}\right)}+K^{\prime}{\left(\sin{\frac{\pi}{12}}\right)}\bigg{]}\\ &=\frac{1}{2\sqrt{2}}\bigg{[}K{\left(k_{3}\right)}+K^{\prime}{\left(k_{3}\right)}\bigg{]}\\ &=\frac{1}{2\sqrt{2}}\bigg{[}1+\frac{K^{\prime}{\left(k_{3}\right)}}{K{\left(k_{3}\right)}}\bigg{]}K{\left(k_{3}\right)}\\ &=\frac{1+\sqrt{3}}{2\sqrt{2}}\,K{\left(k_{3}\right)}\\ &=\frac{1+\sqrt{3}}{2\sqrt{2}}\cdot\frac{\sqrt[4]{3}}{6}\operatorname{B}{\left(\frac12,\frac16\right)}\\ &=\frac{1+\sqrt{3}}{2^{5/2}\,3^{3/4}}\cdot\frac{\Gamma{\left(\frac12\right)}\,\Gamma{\left(\frac16\right)}}{\Gamma{\left(\frac23\right)}}\\ &=\frac{1+\sqrt{3}}{2^{11/6}\,3^{3/4}}\cdot\frac{\pi\,\Gamma{\left(\frac13\right)}}{\left[\Gamma{\left(\frac23\right)}\right]^{2}}\\ &=\frac{\sqrt{3+2\sqrt{3}}}{2^{10/3}}\cdot\frac{\left[\Gamma{\left(\frac13\right)}\right]^{3}}{\pi}.\blacksquare\\ \end{align}$$


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    $\begingroup$ Impressive, for sure ! $\endgroup$ – Claude Leibovici Apr 5 '20 at 14:47

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