Let $J$ be the integral to be computed. The following solution was written so that the parallel "clean" formula or a related integral $J_1$ (with no minus sign under the blue dilogarithm) is also mentioned:
$$
\tag{$*$}
J_1 =
\int_0^1\log y\; \color{blue}{\operatorname{Li}_2(y)}\;\frac1{1-y^2}\; dy
=\bbox[yellow]{\ -\frac 1{192}\pi^4\ }\ .
$$
We record formulas for its "pieces":
$$
\begin{aligned}
J_{11} &=
\int_0^1\log y\; \color{blue}{\operatorname{Li}_2(y)}\;\frac1{1-y}\; dy
=\bbox[yellow]{\ -\frac 1{120}\pi^4\ }\ ,
\\
J_{12} &=
\int_0^1\log y\; \color{blue}{\operatorname{Li}_2(y)}\;\frac1{1+y}\; dy
=\bbox[yellow]{\ -\frac 1{480}\pi^4\ }\ ,
\end{aligned}
$$
----------
Up to some point, integrals are manipulated using the "usual" properties of
integrals. The solution is given up to referenced results. However, afterwards, also a symbolic point of view is taken, key words are multiple zeta values (MZV) and generalized polylogarithms.
----------
We will use first relations from the [wolfram dilog page][1] to split the wanted $J$ into pieces.
For instance the relation $(6)$ is useful:
$$
\tag{6}
\begin{aligned}
\operatorname{Li}_2(-y)
- \operatorname{Li}_2(1-y)
+\frac 12 \operatorname{Li}_2(1-y^2)
&=-\frac {\pi^2}{12} - \log(+y)\log(1+y)\ ,
\\
\operatorname{Li}_2(+y)
- \operatorname{Li}_2(1+y)
+\frac 12 \operatorname{Li}_2(1-y^2)
&=-\frac {\pi^2}{12} - \log(-y)\log(1-y)\ .
\end{aligned}
$$
Now subtract to get rid of the dilog computed in $(1-y^2)$,
we obtain a way to express the dilog of $-y$ in terms of the dilog
in $y$, $1\pm y$, and arguably simpler products of logarithms.
Our $y$ runs in $[0,1]$, so we replace tacitly $\log(-y)$ by $\log y$, and pass to the real part of all expressions involved.
(Equalities in the real part only, so modulo $i\Bbb R$ are denoted by $\equiv$.)
From the above we obtain:
$$
\tag{$\dagger$}
\operatorname{Li}_2(-y)
\equiv
\color{blue}{\operatorname{Li}_2(y)}
+
\color{brown}{\operatorname{Li}_2(1-y)
-
\operatorname{Li}_2(1+y)}
+
\color{green}{\log(y)\log\frac{1-y}{1+y}}\ .
$$
----------
Note that we may have used $(5)$ from *loc. cit.* instead,
which apparently should be simpler, since it involves only two dilogs, so we replace in integral involving $\operatorname{Li}_2(-y)$
with an other one involving $\operatorname{Li}_2(1+y)$, and products of
$\log$'s. Using $(6)$ there are three dilogs appearing! Why? It turns out that the integral corresponding to $\operatorname{Li}_2(y)$ is "*clean*", and the difference $\operatorname{Li}_2(1+y)-\operatorname{Li}_2(1-y)$ also leads to a "*clean*" integral. Details follow.
----------
We will use $(5)$ from *loc. cit.* to isolate the real part from
$$
\operatorname{Li}_2(1+y)
=
\underbrace{
-\operatorname{Li}_2(-y)
+\frac{\pi^2}6
-\log (-y)\log(1+y)}_{\Re \operatorname{Li}_2(1+y)}
\pm i\pi\log(1+y)\ ,\qquad y\in[0,1]\ ,
$$
where the monodromy is hidden in the imaginary part,
and we try to avoid powers of $\operatorname{Li}_2(1+y)$
in calculus, to mix different branches.
Below, taking the real part is pointing tacitly to this dilog of $(1+y)$, $y\in[0,1]$, where $1+y$ leaves the disk of convergence of the dilog.
In particular,
$\Re \operatorname{Li}_2(2)
=\Re \operatorname{Li}_2(1+1)
=-\operatorname{Li}_2(-1)+\frac {\pi^2}6
=\frac{\pi^2}4$.
----------
For these simpler products, for the integrals that
are obtained from it is convenient to also introduce the notation:
$$
\Bbb J_{abc,def}=
\Bbb J\binom{a\ b\ c}{d\ e\ f}
=
\int_0^1
\frac {\log^a(1-y)\log^b(y)\log^c(1+y)}{(1-y)^d(y)^e(1+y)^f}\; dy\ .
$$
A similar notation is introduced in
the arXiv the article [Au1], page 1, of Kam Cheong Au,
as mentioned in the other answer of . We use two references, to his articles, the information can be also extracted from the Book [Z], Chapter 13:
- [Au1][2] ,
- [Au2][3] ,
- [Z] Multiple Zeta Functions, Multiple Polylogarithms And Their Special Values, Jianqiang Zhao.
In the notations from [Au1] we have
- $i_{abc0}=\Bbb J_{abc,100}$,
- $i_{abc1}=\Bbb J_{abc,010}$,
- $i_{abc2}=\Bbb J_{abc,001}$.
Considering the results from [Au1], uur stand point is
that for powers $d,e,f$ between $0,1$,
and small values of the "weight" $w=a+b+c+1$ ($w\le 4$) we can compute / look up in a table for $\Bbb J_{abc,def}$, and take the value as a present.
Note that a partial fraction decomposition reduces the cases for $(d,e,f)$
among $110$, $101$, $011$, $111$ to the simpler $100$, $010$, $001$.
In fact, we will soon do a similar split for
$\frac 1{1-y^2}=\frac 12\left(\frac 1{1-y}+\frac 1{1+y}\right)$.
Then we have from $(\dagger)$ a linear dependence between the **real part** of the following integrals $J, J_1,J_2,J_3$:
$$
\begin{aligned}
J &=
\int_0^1\log y\; \operatorname{Li}_2(-y)\;\frac1{1-y^2}\; dy\ ,\text{ the integral of interest,}
\\
J_1 &=
\int_0^1\log y\; \color{blue}{\operatorname{Li}_2(y)}\;\frac1{1-y^2}\; dy
=\bbox[yellow]{\ -\frac 1{192}\pi^4\ }\ ,\text{ a clean value to be shown below,}
\\[3mm]
J_2
&=
\Re
\int_0^1\log y\;
\color{brown}{\Big(
\operatorname{Li}_2(1+y)
-
\operatorname{Li}_2(1-y)
\Big)}
\;\frac1{1-y^2}\; dy
\\
&=
\frac 12
\Re\int_0^1\log y\;
\left(
\frac{\operatorname{Li}_2(1+y)}{1+y}
+
\frac{\operatorname{Li}_2(1+y)}{1-y}
-
\frac{\operatorname{Li}_2(1-y)}{1+y}
-
\frac{\operatorname{Li}_2(1-y)}{1-y}
\right)\; dy
\\
&\equiv
-\frac 12
\int_0^1
\Big(\Re\operatorname{Li}_2(1+y)\Big)'\Re\operatorname{Li}_2(1+y)\; dy
-
\frac 12
\int_0^1
\Big(\operatorname{Li}_2(1-y)\Big)'\operatorname{Li}_2(1-y)\; dy
\\
&\qquad\qquad
+
\frac 12
\int_0^1
\underbrace{\frac{\log y}{1-y}}
_{=(\ +\operatorname{Li}_2(1-y)\ )'}
\Re\operatorname{Li}_2(1+y)\; dy
-
\frac 12
\underbrace{\int_0^1\frac{\log y}{1+y}}
_{\sim (\ -\Re\operatorname{Li}_2(1+y)\ )'}
\operatorname{Li}_2(1-y)\; dy
\\
&=
-\frac 14\Bigg[\ (\Re\operatorname{Li}_2(1+y))^2\ \Bigg]_0^1
-\frac 14\Bigg[\ \operatorname{Li}_2(1-y)^2\ \Bigg]_0^1
\\
&\qquad\qquad
+\frac 12\Bigg[\ \operatorname{Li}_2(1-y)\Re\operatorname{Li}_2(1+y)\ \Bigg]_0^1
\\
&=
-\frac 14\Big[\
(\Re \operatorname{Li}_2(2))^2 - \operatorname{Li}_2(1)^2\ \Big]
-\frac 14\Big[\
0-\operatorname{Li}_2(1)^2 \ \Big]
+\frac 12\Big[ \ 0-\operatorname{Li}_2(1))^2 \ \Big]
\\
&=-\frac 14\cdot
(\Re \operatorname{Li}_2(2))^2
=\bbox[yellow]{\ -\frac {\pi^4}{64}\ }\ ,
\\[3mm]
J_3
&=\int_0^1\log y\cdot
\color{green}{\log y\log\frac{1-y}{1+y}}\cdot \frac1{1-y^2}\; dy
\\
&=\frac 12\Bbb J_{120-021,100+001}\qquad\text{(with a linear split
w.r.t. the symbolic indices)}
\\
&=
\frac{53}{1440}\pi^4
+\frac 16\pi^2\log^2 2
-4\operatorname{Li}_2\left(\frac 12\right)
-\frac 72\log(2)\,\zeta(3)
-\frac 16\log^42\ .
\end{aligned}
$$
Then we obtain a formula for $J$, from $J=J_1-J_2+J_3$,
$$
\bbox[lightyellow]{\qquad
J=
\frac{17}{360}\pi^4
+\frac 16\pi^2\log^2 2
-4\operatorname{Li}_2\left(\frac 12\right)
-\frac 72\log(2)\,\zeta(3)
-\frac 16\log^42\ .
\qquad}
$$
----------
----------
**MZV and symbolic computations.** I will try to collect some facts, see the above references. The problem with MZV's (*M*ultiple *Z*eta *Values) is that all articles go in the direction of using the (Hopf-)algebraic part of them,
there are not so many examples - so that the many MSE integrals that appear still look like each one is a world for itself. Dictionary and formulas follow.
We fix a ("chromatic") level $N$. For our needs, $N=1$ and $N=2$ are enough.
We deal with iterated integrals, computed for functions
$f,g,\dots,h$,
- identified in notation with the right objects to be taken, which are forms
$f(t)\; dt$,
$g(t)\; dt$, ... , $h(t)\; dt$ -
which are defined on some interval (or manifold parametrized interval) $\gamma$. In our case (the manifold is $\Bbb R$ and) $\gamma=[0,1]$. So i will write the integral limits $0,1$ below.
$$
\operatorname{It}(f,g,\dots,h)
=\int_0^1 f(t)\; dt\int_0^t g(u)\; du\dots \int_0^? h(v)\; dv\ ,
$$
where the question mark stays for the previously used integration variable.
So we integrate $f\otimes g\otimes\dots\otimes h$ on the simplex
$1\ge t\ge u\ge\dots\ge v\ge 0$.
Write
$$
\operatorname{It}_\gamma(f,g,\dots,h)
$$
if needed in order to integrate on an other path, e.g. $[0,t]$ with a dynamic $t$.
We are mainly interested in functions $f,g,\dots,h$ in a very small world of functions. Here are they:
- level $N=1$,
$\frac 1t$ and $\frac 1{1-t}$ - written as forms $\frac{dt}t$, $\frac{dt}{1-t}$,
- level $N=2$,
$\frac 1t$ and $\frac 1{1-t}$ and $\frac 1{-1-t}$ - written as forms $\frac{dt}t$, $\frac{dt}{1-t}$, $\frac{dt}{-1-t}$,
- and with a general $N$ we associate $\mu$ the primitive root of unity which is $\exp$ of $2\pi i/N$, and the alphabet contains now $\frac 1t$ and all fractions $\frac 1{\mu^{-k}-t}=\frac{\mu^k}{1-\mu^k t}$.
Iterated integrals look then like (non-commutative) words in this alphabet.
Since the alphabet is so small, and the the characters in it so complicated,
we really use established "simple letters".
For level $1$, in a first decade the convention was to take $0$ for $1/t$ and $1$ for $1/(1-t)$. Then further levels came in, and an other notation was with $a$ and $b_0$ instead, as in the references [Au1], [Au2], [Z]. The index zero of $b_0$ points to the power $k=0$ of the unit $\mu$ to be taken. So for a general $N$ we have $b_0$, $b_1$, ... till $b_{N-1}$. My convention below is to use
- $A$ for $0$, and $a$, and explicitly $1/t$,
- $B$ for $1$, and $b_0$ in level $N$, and explicitly $1/(1-t)$,
- $C$ for $b_1$ in level two $N=2$, and explicitly $-1/(1+t)$.
$$
\begin{array}{|c|c|}
\hline
\text{Words in few symbols} & \text{Tuples of few functions}\\\hline
A &\frac{dt}t\\\hline
B=B_0 &\frac{dt}{1-t}\\\hline
B'=B_1 &-\frac{dt}{1+t}\\\hline
X=\text{Alphabet }\{A,B,C\}^* & \Omega=\text{Alphabet in }\frac{dt}t,\ \frac{dt}{1-t},\ -\frac{dt}{1+t}\\\hline
X\text{-word} & \Omega\text{-word}\\\hline
X\text{-word }w & \Omega\text{-word often also written }w\\\hline
\int_0^1 w\in\Bbb R & \operatorname{It}(w)=\operatorname{It}_{[0,1]}(w)\in\Bbb R\\\hline
\operatorname{Li}_w(t)=\operatorname{Li}(w,t)=
\int_0^t w & \operatorname{It}_{[0,t]}(w)\\\hline
\int_\gamma w & \operatorname{It}_\gamma(w)\\\hline
\bbox[lightyellow]{\qquad
\int_\gamma w\cdot \int_\gamma w'
=
\int_\gamma w\ ш\ w'\qquad} &
\operatorname{It}_\gamma(w)\operatorname{It}_\gamma(w')
=
\operatorname{It}_\gamma(w \ ш\ w')\\\hline
ш\text{ is the shuffle product of words} &\\\hline
\int_t^1 A^n & \frac 1{n!}(-1)^n\log^n t\\\hline
\int_0^t B^n & \frac 1{n!}(-1)^n\log^n (1-t)=\frac 1{n!}\operatorname{Li}_1(t)^n\\\hline
\int_0^t C^n & \frac 1{n!}(-1)^n\log^n (1+t)=\frac 1{n!}\operatorname{Li}_1(-t)^n\\\hline
\operatorname{Li}(B,t) & \operatorname{Li}_1(t) =-\log(1-t)=\int_0^tB=\int_0^t\frac{dt}{1-t}\\\hline
\operatorname{Li}(AB,t) & \operatorname{Li}_2(t)
=\int_0^t\frac {\operatorname{Li}_1(t)}{t}\;dt
=\int_0^t\frac {dt}t\int_0^t\frac {du}{1-u}
\operatorname{It}_{[0,t]}\left(\frac 1t,\frac1{1-t}\right)
\\\hline
\operatorname{Li}(AAB,t) & \operatorname{Li}_3(t)
=\int_0^t\frac {\operatorname{Li}_2(t)}{t}\;dt
\operatorname{It}_{[0,t]}\left(\frac 1t,\frac 1t,\frac1{1-t}\right)
\\\hline
\operatorname{Li}(\underbrace{AA\dots AB}_{n\text{ letters}},t) & \operatorname{Li}_n(t)
\\\hline
Z(s) = Z_0(s)=\underbrace{AA\dots AB}_{s\text{ letters}} &\\
Z'(s) = Z_1(s) = \underbrace{AA\dots AB'}_{s\text{ letters}} &\\\hline
Z_j(s)= \underbrace{AA\dots AB_j}_{s\text{ letters}} &\\\hline
s =(s_1,s_2,\dots,s_d)\text{ poly-index} & s =(s_1,s_2,\dots,s_d)\in\Bbb N_{>0}^d\\\hline
j =(j_1,j_2,\dots,j_d)\text{ poly-index} & j =(j_1,j_2,\dots,j_d)\in(\Bbb Z/N)^d\\\hline
\Delta j=(j_1-0, j_2-j_1,\dots,j_d-j_{d-1}) & \\\hline
x =(x_1,x_2,\dots,x_d)\text{ poly-argument} & x =(x_1,x_2,\dots,x_d)\\\hline
\mu =(\mu,\mu,\dots,\mu)\text{ as poly-argument} & \mu =(\mu,\mu,\dots,\mu)\\\hline
\prod Z(s,j) =\prod_j Z_j(s)\\
=Z_{j_1}(s_1)\dots Z_{j_d}(s_d) &\\\hline
&
\displaystyle
\zeta(s) = L(s)=L(s,1)\\
&
\displaystyle
=\sum_{n_1>\dots >n_d\ge 1}\frac 1{n_1^{s_1}\dots n_d^{s_d}}\\\hline
&
\displaystyle
L(s,x)=\sum_{n_1>\dots >n_d\ge 1}\frac {x_1^{n_1}\dots x_d^{n_d}}{n_1^{s_1}\dots n_d^{s_d}}\\\hline
\int_0^1\prod Z(s,j) & L(s, \mu^{\Delta j})\\\hline
\end{array}
$$
We are computing integrals in this answer. Some part of the manipulations is purely algebraic, e.g. hidden in the shuffle algebra. Making the computations on the symbolic side is easier. Often, a MZV is obtained, and relations in the MZV-algebra are finally used to obtain the "usual" (even) powers of $\pi$, and zeta-values of odd arguments.
There are some convergence issues, for some words, but we neglect them here.
----------
----------
Let us illustrate the above with some computations.
Words will occur, corresponding to functions/forms to be considered inside
an iterated integral.
- We hate any $B=B_0$ at the beginning of a word, and in this case, we replace
$B^m w_1 w'$ by $-\frac 1m$ times $B^{m-1}w_1(B\ ш\ w')$.
In other words, we "formally take" the last $B$ in the prefix $B^m$ and move it through the word
$w_1w'$ at any place after each letter, first time after the first character $w_1$, then after the first two, and so on, the last entry time we see $B$ at the last place. The factor $-1/m$ correspond to the fact that there is a full $B$-prefix $B^m$.
- We hate any $A$ at the end of a word, and in this case, we replace
$w'w_dA^m$ by $-\frac 1m$ times $(w'\ ш\ A)w_dA^{m-1}$.
In other words, we "formally take" the first $A$ in the suffix $A^m$ and move it through the word
$w'w_d$ at any place beyond $w_d$.
The iterated integrals and the $L$-values are extended to follow this rule.
----------
The computation of $J_1$. The following part $J_{11}$ of $J_1$ is simpler,
because only $y$ and $1-y$ appear, so the symbolic part involve only $A,B$:
$$
\begin{aligned}
J_{11}
&=
\int_0^1\log y\; \operatorname{Li}_2(y)\;\frac1{1-y}\; dy
\\
&\overset{\text{up to }\pm}
=
\int_0^1\frac {dy}{1-y}
\left(\int_y^1\frac{du}{u}\right)
\left(\int_0^y\frac{du}{u}\int_u^1\frac {dv}{1-v}\right)
\\
&=
\int_0^1 B\int_0^t A\ ш\ AB
\\
&=
\int_0^1 B( \color{blue}{A}\ ш\ AB )
\\
&=
\int_0^1
(
B\color{blue}{A}AB
+
BA\color{blue}{A}B
+
BAB\color{blue}{A}
)
=2\int_0^1\color{brown}{B}AAB+\int_0^1 \color{brown}{B}ABA
\\
&\qquad\text{ move $\color{brown}{B}$ from the first place behind}\\
&=
-2\int_0^1
(A\color{brown}{B}AB + AA\color{brown}{B}B
+ AAB\color{brown}{B})
-\int_0^1
(A\color{brown}{B}BA + AB\color{brown}{B}A+ ABA\color{brown}{B})
\\
&=
-3\int_0^1ABAB
-4\int_0^1AABB
-2\int_0^1ABB\color{blue}{A}
\\
&\qquad\text{ move $\color{blue}{A}$ from the last place to the left}\\
\\
&=-3\int_0^1ABAB
-4\int_0^1AABB
+2\int_0^1(AB\color{blue}{A}B + A\color{blue}{A}BB + \color{blue}{A}ABB)
\\
&=-\int_0^1ABAB=-\int_0^1Z(2)Z(2)
\\
&\overset{\text{up to }\pm}=\int_0^1 Z(2,2)=\zeta(2,2)
\\
&\qquad\text{ recall $\zeta(k)\zeta(n)=\zeta(k,n)+\zeta(n,k)+\zeta(n+k)$, use
$k=n=2$}
\\
&=\frac 12(\zeta(2)^2-\zeta(4)) = \frac 1{120}\pi^4\ .
\end{aligned}
$$
Numerical check, pari/gp:
? intnum(y=0, 1, log(y) * dilog(y) / (1 - y))
%20 = -0.8117424252833536436370027724058759270810632139390451807622321615830904621402266349176822269562749379
? Pi^4/120
%21 = 0.8117424252833536436370027724058759270810632139390451807622321615830904621402266349176822269562749379
----------
The computation of the remained part $J_{12}$ from $J_1$ follows,
we have algebraically a similar situation, but together with $y$ and $1-y$ there appears also $1+y$, so the symbolic part involve $A,B$ and also the $2$-chromatic piece $B'$:
$$
\begin{aligned}
J_{12}
&=
\int_0^1\log y\; \operatorname{Li}_2(y)\;\frac1{1+y}\; dy
\\
&\overset{\text{up to }\pm}
=
\int_0^1\frac {dy}{1+y}
\left(\int_y^1\frac{du}{u}\right)
\left(\int_0^y\frac{du}{u}\int_u^1\frac {dv}{1-v}\right)
\\
&=
\int_0^1 B'\int_0^t A\ ш\ AB
\\
&=
\int_0^1 B'( \color{blue}{A}\ ш\ AB )
\\
&=
\int_0^1
(
B'\color{blue}{A}AB
+
B'A\color{blue}{A}B
+
B'AB\color{blue}{A}
)
=2\int_0^1\color{brown}{B'}AAB+\int_0^1 \color{brown}{B'}ABA
\\
&\qquad\text{ move $\color{brown}{B'}$ from the first place behind}\\
&=
-2\int_0^1
(A\color{brown}{B'}AB + AA\color{brown}{B'}B
+ AAB\color{brown}{B'})
-\int_0^1
(A\color{brown}{B'}BA + AB\color{brown}{B'}A+ ABA\color{brown}{B'})
\\
&=
-2\int AB'AB-2\int_0^1AA(BB'+B'B)
-\int_0^1(AB'B+ABB')\color{blue}{A}
-\int_0^1ABAB'
\\
&\qquad\text{ move $\color{blue}{A}$ from the last place to the left}
\\
&=
-2\int AB'AB-2\int_0^1AA(BB'+B'B)
+\int_0^1\color{blue}{A}A(B'B+BB')
\\
&\qquad
+\int_0^1A\color{blue}{A}(B'B+BB')
+\int_0^1AB'\color{blue}{A}B
+\int_0^1AB\color{blue}{A}B'
\\
&=-\int_0^1AB'AB=-\int_0^1Z_1(2)Z_0(2)
\\
&\overset{\text{up to }\pm}=\int_0^1 Z_{(1,0)}(2,2)=L_{(1,1)}(2,2)
\\
&=\zeta(\bar 2,\bar 2)
\\
&\qquad\text{ recall $\zeta(\bar k)\zeta(\bar n)=\zeta(\bar k,\bar n)+\zeta(\bar n,\bar k)+\zeta(n+k)$, use
$k=n=2$,}
\\
&\qquad\text{ recall $\zeta(\bar 2)=-\frac 1{1^2}+\frac 1{2^2}-\frac 1{3^2}+\frac 1{4^2}\pm \dots=-\left(1-\frac 2{2^2}\right)\zeta(2)=-\frac1{12}\pi^2$,
}
\\
&=\frac 12(\zeta(\bar 2)^2-\zeta(4))
= \frac 12\pi^4\left(\frac 1{12^2}-\frac 1{90}\right)
= -\frac 1{480}\pi^4\ .
\end{aligned}
$$
Numerical check, pari/gp:
? intnum(y=0, 1, log(y) * dilog(y) / (1 + y))
%25 = -0.2029356063208384109092506931014689817702658034847612951905580403957726155350566587294205567390687345
? lindep([% , Pi^4])
%26 = [480, 1]~
? -Pi^4/480
%27 = -0.2029356063208384109092506931014689817702658034847612951905580403957726155350566587294205567390687345
----------
I have to stop here, typing kills my energy, and the MO-web-interpreter hangs. It remains to show the formula for
$J_3$, again the same symbolic computations apply. Similar examples are in [Au1], [Au2]. Initially, i considered the case to be simpler to put down on the paper, but - as the result shows, there must be a mess in weight four words that show up. If a particular aspect is of interest, i will fill in. The notation $\Bbb J$ was introduced to conquer better the situation, but...
[1]: https://mathworld.wolfram.com/Dilogarithm.html
[2]: https://arxiv.org/pdf/1910.12113.pdf
[3]: https://arxiv.org/pdf/2007.03957.pdf