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

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.

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.

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: $$I=-\int_0^1\frac1t\int_0^{\pi/2}\ln(1-t\,\cos\theta)\,d\theta\,dt.\tag2$$

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, $$\int_0^{\pi/2}\ln(1-t\,\cos\theta)\,d\theta=\frac\pi2\,\ln\!\left(\frac{1+\sqrt{1-t^2}}2\right)-2\,\operatorname{Ti}_2\!\left(\frac{1-\sqrt{1-t^2}}t\right).\tag{$4''$}$$

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:

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

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.

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 !

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.

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".