Trigonometric integrals related to Gieseking's constant

Question

This question at MathOverflow conjectures certain relation between fast converging hypergeometric series and Gieseking's constant. The Gieseking constant ($G$) is the volume of the hyperbolic $3$-manifold, also called Gieseking's manifold.

The hypergeometric series under consideration has trigonometric integral representations, which give two different forms of the conjecture:

$$\int_0^{{\pi }/{3}} \frac{x \left({\sqrt{3}-{\sin x}}\right)\, dx}{\sin x \cdot \sqrt{3-2 \sqrt{3} \sin x}}\overset{?}{=}\frac{5}{2}\text{Im}\left(\text{Li}_2\left(e^{\frac{2 i \pi }{3}}\right)\right) = \frac{5}{3}G\tag{1}$$

$$ \int_0^{{\pi }/{3}} P(x)\, \frac{xdx}{\sin x \sqrt{\frac{3}{4}-\sin ^2x}}\overset{?}{=}5 \sqrt[4]{3}\, \rm{Im}\left(\rm{Li}_2\left(e^{\frac{2 i \pi }{3}}\right)\right) = \frac{10\sqrt[4]{3}}{3} G \tag{2} $$

where,

$$P(x)=\left(\frac{\sqrt{3}}{2}-\sqrt{\frac{3}{4}-\sin ^2x}\right)^{3/2}+\left(\frac{\sqrt{3}}{2}+\sqrt{\frac{3}{4}-\sin ^2x}\right)^{3/2}$$

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The 2nd representation is interesting because Lobachevsky has studied similar integrals in relation to volumes in hyperbolic geometry in his work "Application of imaginary geometry to certain integrals" (1836). Also see equation 3.842.7 in Gradshteyn and Ryzhik. For example he proves the following integration formula:

$$\int_0^\beta \frac{x\sin x\, dx}{(1-\sin^2 \alpha \sin^2 x)\sqrt{\sin^2 \beta -\sin^2 x}} = \frac{\pi\ln\frac{\cos \alpha +\sqrt{1-\sin^2\alpha \sin^2\beta}}{2\cos \beta \cos^2 \frac{\alpha}{2}}}{2\cos \alpha \sqrt{1-\sin^2\alpha \sin^2 \beta}}$$

This formula in the limit $\alpha\to\pm i\infty $ becomes $$ \int_0^\beta\frac{xdx}{\sin{x}\sqrt{\sin^2{\beta}-\sin^2{x}}}=\frac{\pi}{4\sin{\beta}}\ln{\frac{1+\sin{\beta}}{1-\sin{\beta}}}.\tag{3} $$ One can see that when $\beta=\pi/3$, then both (2) and (3) have the same factor $\frac{x}{\sin{x}\sqrt{\sin^2{\beta}-\sin^2{x}}}$, and if it wasn't for the convoluted factor $P(x)$ these integrals would be the same.

Lobachevsky also studied other integrals and in particular he shows that the integral with the variable upper limit $y<\beta$ $$ \int_0^y\frac{xdx}{\sin{x}\sqrt{\sin^2{\beta}-\sin^2{x}}} $$ can be expressed as a sum of elementary functions and Clausen functions.

Maybe (2) is related to hyperbolic geometry and there is certain substitution that can transform this integral into a sum of Clausen functions or maybe it has some geometric interpretation?

Q: How to prove conjectures (1) and (2)?

Answer 1

Both integrals are essentially some obscure variants of the Ahmed or Coxeter integral (as observed from a comment I left under the question several years ago). As such, the approach outlined here fits well - namely, bringing the integral to a form where we can utilize the following identity:

$$\int_a^b \frac{\arctan\left(f(x)\frac{\sqrt{g(x)}}{\sqrt{h(x)}} \right)}{p(x)\sqrt{g(x)}\sqrt{h(x)}}dx=\int_a^b \frac{1}{p(x)}\int_0^{f(x)} \frac{1}{h(x)+g(x)y^2}dydx$$

We will start by calculating the second integral, as it's smoother to deal with directly compared to the first integral. At the end of the post, we will also show how to connect the two integrals.

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Evaluation of the second integral

$$\int_0^{{\pi }/{3}} x \frac{\left(\frac{\sqrt{3}}{2}-\sqrt{\frac{3}{4}-\sin ^2x}\right)^{3/2}+\left(\frac{\sqrt{3}}{2}+\sqrt{\frac{3}{4}-\sin ^2x}\right)^{3/2}}{\sin x \sqrt{\frac{3}{4}-\sin ^2x}}dx$$

$$\overset{\sqrt{\frac34-\sin^2 x}\to \frac{\sqrt{3}}{2} x}=\ \sqrt 2\sqrt[4]{3}\int_0^1 \frac{{\color{blue}{(1-x)^{3/2}}}+(1+x)^{3/2}}{(1-x^2)\sqrt{1+3x^2}}\arcsin\left(\frac{\sqrt 3\sqrt{1-x^2}}{2}\right)dx$$

$$\overset{{\color{blue}{x\to -x}}}=\sqrt 2\sqrt[4]{3}\int_{-1}^1 \frac{(1+x)^{3/2} \arcsin\left(\frac{\sqrt 3\sqrt{1-x^2}}{2}\right)}{(1-x^2)\sqrt{1+3x^2}}dx \overset{x\to \frac{1-x}{1+x}} = \sqrt[4]{3}\int_0^\infty \frac{\arcsin\left(\frac{\sqrt{3x}}{1+x}\right)}{x\sqrt{1+x^3}}dx$$

$$=\sqrt[4]{3}\int_0^\infty \frac{\arctan \left(\frac{\color{red}{\sqrt{3x(1+x)}}}{\sqrt{1+x^3}}\right)}{x\sqrt{1+x^3}}dx=\sqrt[4]{3}\int_0^\infty \int_0^{\color{red}{f(x)}} \frac{1}{x(1+x^3+y^2)}dydx$$

$$\overset{y\to f(y)}=\sqrt[4]{3} \int_0^\infty \int_0^x \frac{f'(y)}{x(1+x^3+f^2(y))}dydx=\int_0^\infty \int_0^x (*)dydx = \int_0^\infty \int_y^\infty (*)dxdy$$

$$ = \frac{\sqrt 3 \sqrt[4]{3}}{2}\int_0^\infty \frac{(1+2y)\ln\left(\frac{1+y}{y}\right)}{\sqrt{y(1+y)}(1+3y+3y^2)}dy\overset{\frac{y}{1+y}\to y^2}=-2\sqrt 3 \sqrt[4]{3} \int_0^1 \frac{(1+y^2)\ln y}{1+y^2+y^4}dy$$

$$=2\sqrt[4]{3}\sum_{n=1}^\infty \frac{\sin\left(\frac{n\pi}{3}\right)+\sin\left(\frac{2n\pi}{3}\right)}{n^2} = 2\sqrt[4]{3}\left(\operatorname{Cl}_2\left(\frac{\pi}{3}\right)+\operatorname{Cl}_2\left(\frac{2\pi}{3}\right)\right)=\boxed{\frac{10\sqrt[4]{3}}{3}G}$$

Where $\operatorname{Cl}_2(x)$ is the Clausen Function and $G$ is the Gieseking's Constant. Also, the last integral was expanded into power series using $\frac{\sin t}{1-2x \cos t +x^2}=\sum\limits_{n=1}^\infty \sin (nt) x^{n-1}$ (after partial fractions).

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Connecting the first integral with the second integral

$$\small I_1=\frac{1}{\sqrt 3}\int_0^{{\pi }/{3}} \frac{x \left({\sqrt{3}-{\sin x}}\right)}{\sin x \sqrt{1-\frac{2}{ \sqrt{3}} \sin x}}dx\overset{\frac{2}{\sqrt 3}\sin x\to \cos x}=\frac{1}{\sqrt 2}\int_0^\frac{\pi}{2} \frac{(2-\cos x)\cos\frac{x}{2}}{\cos x}\frac{\arcsin \left(\frac{\sqrt 3}{2} \cos x\right)}{\sqrt{1-\frac{3}{4} \cos^2x}}dx$$

$$\overset{\frac{\pi}{2}-x\to \arcsin x}=\frac{1}{\sqrt 2}\int_0^1 \frac{(1-x)^{3/2}+(1+x)^{3/2}}{(1-x^2)\sqrt{1+3x^2}}\arcsin\left(\frac{\sqrt 3\sqrt{1-x^2}}{2}\right)dx= \frac{1}{2\sqrt[4]{3}}I_2$$

Note that above we used $\frac{(2-\cos x)\cos\frac{x}{2}}{\sqrt 2}=\sin^3\left(\frac{\pi}{4}-\frac{x}{2}\right)+\cos^3\left(\frac{\pi}{4}-\frac{x}{2}\right)$.