Closed-form of $\int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx $

Question

I'm looking for a closed form of this integral.

$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx ,$$

where $\operatorname{Li}_2$ is the dilogarithm function.

A numerical approximation of it is

$$ I \approx 1.39130720750676668181096483812551383015419528634319581297153...$$

As Lucian said $I$ has the following equivalent forms:

$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx = \int_0^1 \frac{\operatorname{Li}_2\left( \sqrt{x} \right)}{2 \, \sqrt{x} \, \sqrt{1-x}} \,dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\sin x) \, dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\cos x) \, dx$$

According to Mathematica it has a closed-form in terms of generalized hypergeometric function, Claude Leibovici has given us this form.

With Maple using Anastasiya-Romanova's form I could get a closed-form in term of Meijer G function. It was similar to Juan Ospina's answer, but it wasn't exactly that form. I also don't know that his form is correct, or not, because the numerical approximation has just $6$ correct digits.

I'm looking for a closed form of $I$ without using generalized hypergeometric function, Meijer G function or $\operatorname{Li}_2$ or $\operatorname{Li}_3$.

I hope it exists. Similar integrals are the following.

$$\begin{align} J_1 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{1+x} \,dx = \frac{\pi^2}{6} \ln 2 - \frac58 \zeta(3) \\ J_2 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x}} \,dx = \pi^2 - 8 \end{align}$$

Related techniques are in this or in this paper. This one also could be useful.

Answer 1

We will go through a sequence of integrals, and, remarkably, we will see that at each step an integrand will have a continuous closed-form antiderivative in terms of elementary functions, dilogarithms and trilogarithms, so evaluation of an integral is then just a matter of calculating values (or limits) at end-points and taking a difference.

I used Mathematica to help me find some of those antiderivatives, but then I significantly simplified them manually. In each case correctness of the result was proved manually by direct differentiation, so we do not have to trust Mathematica on it. Maybe somebody will find a more elegant and enlightening way to evaluate them.

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First change the variable $x=\cos\theta$ and rewrite the integral as: $$I=\int_0^{\pi/2}\operatorname{Li}_2(\cos\theta)\,d\theta\tag{0}$$ Then we use a known integral representation of the dilogarithm: $$\operatorname{Li}_2(z)=-\int_0^1\frac{\ln(1-t\,z)}t\,dt.\tag1$$ Use it to rewrite $(0)$ and then change the order of integration: $$I=-\int_0^1\frac1t\int_0^{\pi/2}\ln(1-t\,\cos\theta)\,d\theta\,dt.\tag2$$

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Our first goal is to evaluate the inner integral in $(2)$. The integrand has a closed-form antiderivative in terms of elementary functions and dilogarithms that is continuous in the region of integration: $$\int\ln(1-t\,\cos\theta)\,d\theta=\theta\!\;\ln\!\left(\frac{1+\sqrt{1-t^2}}2\right)-2\,\Im\,\operatorname{Li}_2\!\left(\frac{1-\sqrt{1-t^2}}t\!\;e^{i\!\;\theta}\right).\tag3$$ (compare it with the raw Mathematica result)

Taking the difference of values of $(3)$ at the end-points $\pi/2$ and $0$, we obtain: $$\int_0^{\pi/2}\ln(1-t\,\cos\theta)\,d\theta=\frac\pi2\,\ln\!\left(\frac{1+\sqrt{1-t^2}}2\right)-2\,\Im\,\operatorname{Li}_2\!\left(i\,\frac{1-\sqrt{1-t^2}}t\right).\tag4$$ Recall that the imaginary part of the dilogarithm can be represented as the inverse tangent integral: $$\Im\,\operatorname{Li}_2(iz)=\operatorname{Ti}_2(z)=\int_0^z\frac{\arctan(v)}v dv.\tag{$4'$}$$ So, $$\int_0^{\pi/2}\ln(1-t\,\cos\theta)\,d\theta=\frac\pi2\,\ln\!\left(\frac{1+\sqrt{1-t^2}}2\right)-2\,\operatorname{Ti}_2\!\left(\frac{1-\sqrt{1-t^2}}t\right).\tag{$4''$}$$

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Now our goal is to evaluate the outer integral in $(2)$. Substituting $(4'')$ back into $(2)$ we get: $$I=-\frac\pi2\!\;I_1+2\!\;I_2,\tag5$$ where $$I_1=\int_0^1\frac1t\,\ln\!\left(\frac{1+\sqrt{1-t^2}}2\right)dt,\tag6$$ $$I_2=\int_0^1\frac1t\,\operatorname{Ti}_2\!\left(\frac{1-\sqrt{1-t^2}}t\right)dt.\tag7$$ The integrand in $(6)$ has a closed-form antiderivative in terms of elementary functions and dilogarithms. One way to find it is to change variable $t=2\sqrt{u-u^2}$ and integrate by parts. $$\int\frac1t\,\ln\!\left(\frac{1+\sqrt{1-t^2}}2\right)dt=\frac14\,\ln^2\!\left(\frac{1+\sqrt{1-t^2}}2\right)-\frac12\, \operatorname{Li}_2\!\left(\frac{1-\sqrt{1-t^2}}2\right).\tag8$$ (compare it with the raw Mathematica result)

Taking the difference of its values at the end-points, and using well-known values $$\operatorname{Li}_2(1)=\zeta(2)=\frac{\pi^2}6,\tag{$8'$}$$ $$\operatorname{Li}_2\left(\tfrac12\right)=\frac{\pi^2}{12}-\frac{\ln^22}2,\tag{$8''$}$$ we get: $$I_1=\frac{\ln^22}2-\frac{\pi^2}{24}.\tag9$$ To evaluate $I_2$ change the variable $t=\frac{2z}{1+z^2}$: $$I_2=\int_0^1\frac{1-z^2}{z\,(1+z^2)}\operatorname{Ti}_2(z)\,dz.\tag{10}$$ Again, the integrand has a closed-form antiderivative in terms of elementary functions, dilogarithms and trilogarithms. Before giving the result, we will try to split it into smaller parts. First, recall $(4')$ and a simple integral ${\large\int}\frac{1-z^2}{z\,(1+z^2)}dz=\ln\!\left(\frac z{1+z^2}\right)$, and integrate by parts: $$\int\frac{1-z^2}{z\,(1+z^2)}\operatorname{Ti}_2(z)\,dz=\ln\!\left(\frac z{1+z^2}\right)\operatorname{Ti}_2(z)\\-\underbrace{\int\frac{\ln z\cdot\arctan z}z\,dz}_{I_3}+\underbrace{\int\frac{\ln(1+z^2)\cdot\arctan z}z\,dz}_{I_4}.\tag{11}$$ The following results can be checked by direct differentiation: $$I_3=\operatorname{Ti}_2(z)\ln z-\Im\,\operatorname{Li}_3(iz),\tag{$11'$}$$ $$I_4=\left[\frac{\pi^2}3-\ln\left(1+z^2\right)\ln z-\frac12\,\operatorname{Li}_2\!\left(-z^2\right)\right]\arctan z\\-\frac\pi2\,\arctan^2z+\frac\pi8\,\ln^2\left(1+z^2\right)+\operatorname{Ti}_2(z)\ln\left(1+z^2\right)-2\,\Im\,\operatorname{Li}_3(1+iz).\tag{$11''$}$$ Plugging $(11')$ and $(11'')$ into $(11)$ we obtain: $$\int\frac{1-z^2}{z\,(1+z^2)}\operatorname{Ti}_2(z)\,dz=\left[\frac{\pi^2}3-\ln\left(1+z^2\right)\ln z-\frac12\,\operatorname{Li}_2\!\left(-z^2\right)\right]\arctan z\\-\frac\pi2\,\arctan^2z+\frac\pi8\,\ln^2\left(1+z^2\right)+\,\Im\,\operatorname{Li}_3(iz)-2\,\Im\,\operatorname{Li}_3(1+iz).\tag{$11'''$}$$ (compare it with the raw Mathematica result)

Taking the difference of its values at the end-points $1$ and $0$, we get: $$I_2=\frac{3\!\;\pi^3}{32}+\frac\pi8\!\;\ln^22-2\,\Im\,\operatorname{Li}_3(1+i).\tag{12}$$ Plugging $(9)$ and $(12)$ back into $(5)$ we get the final result:

$$\large\int_0^1\frac{\operatorname{Li}_2(x)}{\sqrt{1-x^2}}\,dx=\frac{5\!\;\pi^3}{24}-4\,\Im\,\operatorname{Li}_3(1+i).\tag{$\heartsuit$}$$

Answer 2

An alternative approach. As shown by nospoon here, \begin{equation}\label{shalev} \int_{0}^{1}\frac{\log^2(x)}{\sqrt{x(1- x\sin^2\theta )}}\,dx = \frac{8}{\sin\theta}\left[\frac{\theta^3}{3}-\text{Im}\,\text{Li}_3\left(1-e^{2i\theta}\right)\right]\tag{1}\end{equation} holds for any $\theta\in\left(0,\frac{\pi}{2}\right)$. This is an istance of a very nice principle, according to which every hypergeometric function of the $\phantom{}_{p+1}F_{p}\left(\frac{1}{2},\frac{1}{2},\ldots;\frac{3}{2},\frac{3}{2},\ldots;z\right)$ kind has a closed form in terms of polylogarithms. We need to evaluate $\phantom{}_4 F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)$, hence our arrival point is already pretty close to the statement of $(1)$.

The evaluation of $(1)$ at $\theta=\frac{\pi}{4}$ leads to \begin{equation}\label{shalev2} \int_{0}^{1}\frac{\log^2(x)}{\sqrt{x(2- x )}}\,dx = 8\left[\frac{\pi^3}{192}-\text{Im}\,\text{Li}_3\left(1-i\right)\right]\tag{2}\end{equation} and the functional relations for $\text{Li}_2$ reduce the original problem to $(2)$. In particular $$\begin{eqnarray*} \int_{0}^{\pi/2}\text{Li}_2(\sin\theta)\,d\theta &=& \int_{0}^{1}\frac{2}{1+t^2}\text{Li}_2\left(\frac{1-t^2}{1+t^2}\right)\,dt\\ \int_{0}^{\pi/2}\text{Li}_2(\cos\theta)\,d\theta&=&\int_{0}^{1}\frac{2}{1+t^2}\text{Li}_2\left(\frac{2t}{1+t^2}\right)\,dt\\ &=&\frac{5\pi^3}{24}+4\,\text{Im}\,\text{Li}_3(1-i)\tag{3}\end{eqnarray*}$$ as already shown by Reshetnikov.

Answer 3

My attempt. This is by no means closer to the answer, but I want to address several equivalent forms that might be helpful for future calculations.

First, from Landen's identity of the following form

$$ \mathrm{Li}_2(z) = -\mathrm{Li}_2\left(-\frac{z}{1-z}\right) - \frac{1}{2}\log^{2}(1-z), \quad z \notin [1, \infty)$$

we observe that

\begin{align*} I &= -\int_{0}^{1} \frac{1}{\sqrt{1-x^2}} \left\{ \mathrm{Li}_2\left(-\frac{x}{1-x}\right) - \frac{1}{2}\log^{2}(1-x) \right\} \, dx \\ &= -\int_{-\infty}^{0} \frac{2\mathrm{Li}_2 (t) + \log^{2}(1-t)}{2\sqrt{1-2t}(1-t)} \, dt \tag{1} \end{align*}

By noting that

$$ \frac{d}{dt} \arctan\left(\frac{1}{\sqrt{1-2t}}\right) = \frac{1}{2\sqrt{1-2t}(1-t)}, $$

integration by parts and the substitution $x = (1-2t)^{-1/2}$ shows that (1) is equal to

\begin{align*} I &= \int_{-\infty}^{0} \arctan\left(\frac{1}{\sqrt{1-2t}}\right) \frac{2\log(1-t)}{t(t-1)} \, dt \\ &= \int_{0}^{1} \frac{8x \arctan x}{1 - x^4} \log \left( \frac{1+x^2}{2x^2} \right) \, dx \tag{2} \end{align*}

The following observation

$$ \Re \log \left(\frac{1+ix}{\sqrt{2}} \right) = \frac{1}{2}\log \left( \frac{1+x^2}{2} \right) \quad \text{and} \quad \Im \log \left(\frac{1+ix}{\sqrt{2}} \right) = \arctan x $$

somehow seems to suggest complex-analytic approach, but I have not been successful with such approaches so far. Next, from the following simple formula

$$ \log \left( \frac{1+x^2}{2x^2} \right) \, dx = \int_{0}^{1} \frac{d}{dy} \log \left( \frac{y^2+x^2}{y^2 + 1} \right) \, dy $$

the integral (2) can be further decomposed into the following form

$$ I = \int_{0}^{1}\int_{0}^{1} \frac{16xy \arctan x}{(1+x^2)(1+y^2)(x^2+y^2)} \,dxdy. \tag{3} $$

Simple calculation shows that

$$ \int_{0}^{1}\int_{0}^{1} \frac{16xy}{(1+x^2)(1+y^2)(x^2+y^2)} \,dxdy = 2\zeta(2), $$

so I suspect that the situation in (3) is not that bad.

Answer 4

According to a CAS, $$I = \int_0^1 \frac{\operatorname{Li}_2\left( \sqrt{t} \right)}{2 \, \sqrt{t} \, \sqrt{1-t}} \,dt =\, _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right )+\frac{\pi ^3}{48}-\frac{1}{4} \pi \log ^2(2)$$

Enjoy !

Answer 5

From writing $$\operatorname{Li}_2(x)=-\int_0^1\frac{x\ln u}{1-xu}du$$

It follows that

$$-I=-\int_0^1\frac{\operatorname{Li}_2(x)}{\sqrt{1-x^2}}dx=\int_0^1\ln u\left[\int_0^1\frac{x}{(1-ux)\sqrt{1-x^2}}dx\right]du$$

$$=\int_0^1\ln u\left[\frac{\pi}{2}\cdot\left(\frac{1}{u\sqrt{1-u^2}}-\frac1u\right)+\frac{\sin^{-1}(u)}{u\sqrt{1-u^2}}\right]du$$ $$=\frac{\pi}2\int_0^1\frac{\ln u}{u}\left(\frac1{\sqrt{1-u^2}}-1\right)du+\int_0^1\frac{\ln u\sin^{-1}(u)}{u\sqrt{1-u^2}}du$$

For the first integral, let $u^2\to u$ first then apply integration by parts, we obtain

$$\frac{\pi}{2}\int_0^1\frac{\ln u}{u}\left(\frac{1}{\sqrt{1-u^2}}-1\right)\ du=\frac{\pi}{8}\int_0^1\ln^2u\ du\left(\frac{1}{\sqrt{1-u}}-1\right)du\\=-\frac{\pi}{32}\int_0^1\ln^2u (1-u)^{-3/2}du=-\frac{\pi}{32}\frac{\partial^2}{\partial\alpha^2}\lim_{\alpha\ \mapsto1}\text{B}\left(\alpha,-\frac12\right)\\=-\frac{\pi}{32}\left(\frac23\pi^2-8\ln^22\right)=\boxed{\frac{\pi}4\ln^2(2)-\frac{\pi^3}{48}}\, .$$

The second integral is already calculated here

$$\int_0^1\frac{\ln(x) \sin^{-1}(x)}{x\sqrt{1-x^2}}dx=\boxed{4 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) -\frac{3 \pi^3}{16} -\frac{\pi}{4} \ln^2(2)} \, .$$

Collecting the boxed results we get

$$I= \frac{5 \pi^3}{24}-4 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) \, .$$

Answer 6

Let's have a different perspective by exploiting the Fourier series.

A solution by Cornel Ioan Valean (in large steps)

In the book More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), at the end of section 6.49, two other Fourier series are announced, which can be derived, as explained, by using a strategy from (Almost) Impossible Integrals, Sums, and Series (2019). It is about

$$\tan(x)\log(1-\cos(x))=-\sum_{n=1}^{\infty}\left(\int_0^1 t^{n/2-1} \frac{1-t}{1+t}\textrm{d}t \right)\sin(n x), \ 0<x<\frac{\pi}{2};$$ $$\operatorname{Li}_2(\cos(x))=\frac{\pi^2}{24}-\frac{1}{2} \log^2(2)+\sum_{n=1}^{\infty}\frac{1}{n} \left(\int_0^1 t^{n/2-1}\frac{1-t}{1+t} \textrm{d}t\right)\cos(nx),\ 0<x<\frac{\pi}{2}.$$

Proof: The following result, $\displaystyle \int_0^{\infty} \tanh(x)e^{-n x}\textrm{d}x=\frac{1}{2}\left(\psi\left(\frac{n+2}{4}\right)-\psi\left(\frac{n}{4}\right)-\frac{2}{n}\right)$, can be extracted from (Almost) Impossible Integrals, Sums, and Series (2019), page $243$, and by the variable change $e^{-2y}=t$ it is immediately clear that $\displaystyle 2 \int_0^{\infty} \tanh(y)e^{-n y}\textrm{d}y=\int_0^1 t^{n/2-1}\frac{1-t}{1+t} \textrm{d}t$. Upon multiplying both sides by $\sin(n x)$ and making the summation from $n=1$ to $\infty$, we have

$$\sum_{n=1}^{\infty} \left(\int_0^1 t^{n/2-1}\frac{1-t}{1+t} \textrm{d}t\right) \sin(n x)=2 \sum_{n=1}^{\infty} \int_0^{\infty} \tanh(y) \sin(n x)e^{-n y}\textrm{d}y$$ $$=2 \int_0^{\infty} \tanh(y) \sum_{n=1}^{\infty} \sin(n x)e^{-n y}\textrm{d}y=\sin(x) \int_0^{\infty} \frac{\tanh(y)}{\cosh(y)-\cos(x)}\textrm{d}y$$ $$ \overset{1/\cosh(y)=t}{=}-\tan(x) \int_0^1 \frac{-\cos(x)}{1-\cos(x) t}\textrm{d}t=-\tan(x) \log(1-\cos(x) t)\biggr|_{t=0}^{t=1}$$ $$=-\tan(x) \log(1-\cos(x)),$$ which leads to the first Fourier series. In the calculations, the well-known Fourier series, $\displaystyle \sum_{n=1}^{\infty}p^n \sin(nx)=\frac{p\sin(x)}{1-2 p\cos(x)+p^2}, \ |p|<1$, is employed.

To get the second Fourier series, we exploit the first Fourier series, where we replace $x$ by $y$, and then integrate from $y=0$ to $y=x$, and further use that $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}\left(\int_0^1 t^{n/2-1} \frac{1-t}{1+t}\textrm{d}t \right)\overset{t=u^2}{=}2\sum_{n=1}^{\infty}\frac{1}{n}\left(\int_0^1 u^{n-1} \frac{1-u^2}{1+u^2}\textrm{d}u \right)=2\int_0^1 \frac{(1-u^2)\log(1-u)}{u(1+u^2)} \textrm{d}u=2\underbrace{\int_0^1 \frac{\log(1-u)}{u} \textrm{d}u}_{\displaystyle -\pi^2/6}-4 \int_0^1 \frac{u\log(1-u)}{1+u^2} \textrm{d}u=-\frac{1}{2}\log^2(2)-\frac{\pi^2}{8},$ where the second integral is $\displaystyle \int_0^1 \frac{u\log(1-u)}{1+u^2} \textrm{d}u=\frac{1}{8}\log^2(2)-\frac{5}{96}\pi^2$, given in (Almost) Impossible Integrals, Sums, and Series (2019), page $8$.

So, integrating from $x=0$ to $x=\pi/2$ using the second Fourier series, letting $t\mapsto t^2$, and further exploiting the parity, we get that

$$\int_0^{\pi/2}\operatorname{Li}_2(\cos(x))\textrm{d}x$$ $$=\frac{\pi^3}{48}-\frac{1}{4} \log^2(2)\pi+2\sum_{n=1}^{\infty}\frac{1}{n^2} \left(\int_0^1 t^{n-1}\frac{1-t^2}{1+t^2} \textrm{d}t\right)\sin\left(\frac{n \pi}{2}\right)$$ $$=\frac{\pi^3}{48}-\frac{1}{4} \log^2(2)\pi-2\sum_{n=1}^{\infty}(-1)^{n-1} \int_0^1\left(\int_0^1 t^{2n-2}\frac{1-t^2}{1+t^2} \textrm{d}t\right)u^{2n-2}\log(u)\textrm{d}u$$ $$=\frac{\pi^3}{48}-\frac{1}{4} \log^2(2)\pi+2\int_0^1\left(\int_0^1 \left(\frac{ (1+u^2) \log (u)}{(1-u^2) (1+ u^2 t^2)}-2\frac{ \log (u)}{(1+t^2)(1-u^2)}\right)\textrm{d}t\right)\textrm{d}u$$ $$=\pi \underbrace{\int_0^1\frac{\log(u)}{1-u^2}\textrm{d}u}_{\displaystyle \text{well-known}- \pi^2/8}-2\underbrace{\int_0^1\frac{\arctan(u)\log(u)}{u}\textrm{d}u}_{\displaystyle \text{well-known}- \pi^3/32}+2\int_0^1\frac{\arctan(u)\log(u)}{1+u}\textrm{d}u$$ $$-2\int_0^1\frac{\arctan(u)\log(u)}{1-u}\textrm{d}u$$ $$=\frac{\pi^3}{96}+\frac{3}{8}\log^2(2) \pi-4 \Im\biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr \},$$ where we also used that $\displaystyle \int_0^1\frac{\arctan(u)\log(u)}{1+u}\textrm{d}u=\frac{\log(2)}{2}G-\frac{\pi^3}{64}$, with solutions that may be found in Evaluating $\int_0^1 \frac{\arctan x \log x}{1+x}dx$ and (Almost) Impossible Integrals, Sums, and Series (2019), pages $140$-$142$, and $\displaystyle \int_0^1\frac{\arctan(t) \log(t)}{1-t}\textrm{d}t=\frac{1}{2}\log(2)G-\frac{\pi}{16}\log^2(2)-\frac{\pi^3}{16}+2\Im\biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\},$ which is given and derived in More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), pages $784$-$785$, that is obtained by using $\displaystyle \int_0^1\frac{\log(1+a t)\log(t)}{1-t}\textrm{d}t=-\zeta(2)\log(1+a)-\frac{1}{3}\log^3(1+a)-\log(1+a)\operatorname{Li}_2(-a)+\operatorname{Li}_3(-a)+2\operatorname{Li}_3\left(\frac{a}{1+a}\right),$ which is evaluated by differentiation.

End of story

Answer 7

Following Anastasiya-Romanova's approach, we have:

$$ I = \frac{1}{2}\sum_{n\geq 1}\frac{1}{n^2}\int_{0}^{\pi/2}\sin^n x\,dx =\frac{\pi}{16}\sum_{n\geq 1}\frac{\binom{2n}{n}}{n^2 4^n}+\frac{1}{4}\sum_{n\geq 1}\frac{4^n}{\binom{2n}{n}n(2n-1)^2}\tag{1}$$ where: $$ S_1 = \sum_{n\geq 1}\frac{\binom{2n}{n}}{n^2 4^n} = \zeta(2)-2\log^2 2 \tag{2}$$ and the second sum is the problematic one, leading to a value for a hypergeometric function $\phantom{}_4 F_3$:

$$\begin{eqnarray*}S_2 = \sum_{n\geq 1}\frac{4^n}{\binom{2n}{n}n(2n-1)^2} &=& -\int_{0}^{1}\frac{2\arcsin x}{x\sqrt{1-x^2}}\log x\,dx\\&=&-2\int_{0}^{\pi/2}\frac{\theta}{\sin\theta}\log\sin\theta\,d\theta. \tag{3}\end{eqnarray*}$$

However, since the Fourier cosine series of $\log\sin\theta$ is well-known: $$\log\sin\theta = -\log 2-\sum_{n\geq 1}\frac{\cos(2n\theta)}{n}\tag{4} $$ we just need to compute the Fourier cosine series of $\frac{\theta}{\sin\theta}$. Since the Fourier sine series of the triangle wave is given by: $$ \theta = \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\sin(2n\theta) \tag{5}$$ by exploiting $\frac{\sin(2n\theta)}{\sin\theta}=2\left(\cos\theta+\cos(3\theta)+\ldots+\cos((2n-1)\theta)\right)$ we have:

$$\frac{\theta}{\sin\theta}=2\sum_{k=1}^{+\infty}\left(\sum_{n\geq k}\frac{(-1)^{n+1}}{n}\right)\cos((2k-1)\theta)\tag{6}$$ so, at least in principle, $S_2$ is computable through a Fourier-analytic approach, by exploiting: $$ \int_{0}^{\pi/2}\cos((2n-1)\theta)\cos(2m\theta)\,d\theta = \frac{(2n-1)(-1)^m}{4m^2-(2n-1)^2}.\tag{7}$$

It is also interesting to notice that the last integral appearing in $(3)$ is very similar to the one appearing in this related question, but the latter is way easier to compute since in the Fourier cosine series of $\frac{\theta}{\sin\theta}\,\cos^2\theta$ there are only "even cosines".

Answer 8

Using Maple I am obtaining

$$1+\frac{\pi }{16}{\ _4F_3(1,1,1,3/2;\,2,2,2;\,1)}+\frac{\sqrt {\pi }}{8} G^{4, 1}_{4, 4}\left(-1\, \Big\vert\,^{1, 5/2, 5/2, 5/2}_{2, 3/2, 3/2, 1}\right) $$

and a numerical approximation is

$$1.3913063720392030337$$

Answer 9

Another one... replacing the Meijer G in Juan's answer $$ {\mbox{$_4$F$_3$}(1/2,1/2,1,1;\,3/2,3/2,3/2;\,1)}+ \frac{\pi \, {\mbox{$_4$F$_3$}(1,1,1,3/2;\,2,2,2;\,1)}}{16} \\ \approx 1.3913072075067666818109648381255138301541952863 $$ and with user's comment $$ {\mbox{$_4$F$_3$}(1/2,1/2,1,1;\,3/2,3/2,3/2;\,1)}+\frac{\pi^3}{48}-\frac{\pi (\log 2)^2}{4} $$ agreeing with Claude.