A closed form for a lot of integrals on the logarithm One problem that has been bugging me all this summer is as follows:
a) Calculate
$$I_3=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \ln{(1-x)} \ln{(1-xy)} \ln{(1-xyz)} \,\mathrm{d}x\, \mathrm{d}y\, \mathrm{d}z.$$
b) More generally, let $n \ge 1$ be an integer. Calculate, if possible, in terms of well known constants (find a closed form) this multiple logarithmic integral:
$$I_n=\int_{[0,1]^n} \ln{(1-x_1)}\ln{(1-x_1x_2)}\cdots\ln{(1-x_1x_2 \cdots x_n)}\,\mathrm{d}^nx.$$
My attempt so far is that I have got $I_1=-1$ and $I_2=3-2\zeta(3)$.
 A: Let $k$ be a positive integer, $s_1,\ldots,s_k$ positive integers with $s_1\geq 2$. The quantity
$$
\tag{$\star$}
\zeta(s_1,\ldots,s_k):=\sum_{n_1>\ldots>n_k\geq 1}\frac{1}{n_1^{s_1}\cdots n_k^{s_k}}\in\mathbb{R}
$$
is called a multizeta value. The weight of $(\star)$ is $s_1+\ldots+s_k$ and the depth is $k$. Depth one multizeta values are just values of the Riemann zeta function. In weight up to seven, every multizeta value can be expressed in terms of Riemann zeta values, but this is conjecturally not true in weight eight and higher. For instance, it is conjectured that $\zeta(5,3)$ cannot be expressed in closed form in terms of the Riemann zeta function.
Every iterated integral of the form
$$
\int_{\Delta}\frac{dt_1\ldots dt_n}{f_1(t_1)\ldots f_n(t_n)}
$$
is a multizeta value, where $\Delta$ is the region $0\leq t_1\leq\ldots\leq t_n\leq 1$ and $f_i(t_i)$ is either $t_i$ or $1-t_i$, and every multizeta value can be expressed as such an integral. Moreover the conversion between integrals of this form and multizeta values is easy to obtain by expanding the integrand as a power series and integrating term by term.
In your integral $I_n$, substitute $y_i=x_1\ldots x_i$, and write each term $\log(1-y_i)$ as $-\int_0^{y_i} 1/(1-t_i)dt_i$. The resulting multiple integral can be divided up into regions depending on the relative ordings of the $y_i$ and $t_i$, and the integral over each region has the form $(\star)$. This means $I_n$ can be written as a linear combination of multizeta values.
This computation can be performed by computer (I'm using the Maple package HyperInt, written by Erik Panzer). After obtaining an expression for $I_n$, the package simplifies using known relations among multizeta values. For $n=4$ the result is
$$
I_4=105-\frac{16}{7}\zeta(2)^3+6\zeta(3)^2-\frac{72}{5}\zeta(2)^2-30\zeta(3)-27\zeta(5)-\frac{65}{8}\zeta(7)+\frac{12}{5}\zeta(2)^2\zeta(3).
$$
This has weight up to $7$, so can be expressed in terms of ordinary zeta values. The next integral $I_5$ has weight $9$ terms:
$$
I_5=-945+{\frac {288}{7}}\,{\zeta(2)}^{3}-36\,{\zeta(3)}^{2}+108\,
{\zeta(2)}^{2}+2\,{\zeta(3)}^{3}+210\,\zeta (3)+255\,\zeta 
(5)+{\frac {963}{8}}\,\zeta(7)+{\frac {8112}{875}}\,{\zeta(2)}^{4}-{\frac {24}{5}}\,\zeta (5,3)+{\frac {3299}{72}}\,\zeta (9)-36\,{\zeta (2)}^{2}\zeta (3)-{\frac {16}{7}}\,{\zeta (2)}^
{3}\zeta(3)-{\frac {66}{5}}\,{\zeta (2)}^{2}\zeta (5)-54\,
\zeta (3)\zeta(5).
$$
It's a huge mess, but notice the term $\zeta(5,3)$. So conjecturally $I_5$ cannot be written in terms of ordinary zeta values.
To summarize: there is a finite algorithm to express $I_n$ in terms of multizeta values, and for $n\geq 5$, we expect that $I_n$ cannot be expressed in terms of ordinary zeta values.
A: This is not a solution, but it explains why $I_n$ for $n\geq 3$ is difficult. Indeed,
the change of variables $(y_1,\ldots,y_n)=(x_1,x_1x_2,\ldots,x_1x_2\ldots x_n)$, (i.e. $y_k=x_1x_2\ldots x_k$) shows that
$$\eqalign{
I_n&=\int_{1\geq y_1\geq y_2\geq \cdots\geq y_n\geq 0}\ln(1-y_1)
\ln(1-y_2)\cdots\ln(1-y_n)\frac{dy_1\cdots d y_n}{y_1\cdots y_{n-1}}\cr
&=\int_{\color{red}{y_n}=0}^1\left(\int_{1\geq y_1\geq y_2\geq \cdots\geq y_{n-1}\geq \color{red}{y_n}}
\prod_{k=1}^{n-1}\frac{\ln(1-y_k)}{y_k}dy_1\ldots dy_{n-1}\right)\ln(1-y_n)dy_n\cr
&=\frac{1}{(n-1)!}\int_{\color{red}{y_n}=0}^1\left(\int_{[\color{red}{y_n},1]^{n-1}}
\prod_{k=1}^{n-1}\frac{\ln(1-y_k)}{y_k}dy_1\ldots dy_{n-1}\right)\ln(1-y_n)dy_n
\cr
&=\frac{1}{(n-1)!}\int_{\color{red}{y_n}=0}^1\left(\int_{\color{red}{y_n}}^1
\frac{\ln(1-t)}{t}dt\right)^{n-1}\ln(1-y_n)dy_n
\cr
}
$$
So, our first equivalent expression for $I_n$ is
$$
I_n=\frac{1}{(n-1)!}\int_{0}^1\left(\int_{x}^1
\frac{\ln(1-t)}{t}dt\right)^{n-1}\ln(1-x)dx\tag 1
$$
Using integration by parts we have also
$$
I_n=\frac{1}{n!}\int_{0}^1\left(\int_{x}^1
\frac{\ln(1-t)}{t}dt\right)^{n} dx\tag 2
$$
Noting that $\int_x^1\frac{\ln(1-t)}{t}dt={\rm Li}_2(x)-\frac{\pi^2}{6}$
we get
$$\eqalign{
I_n
&=\frac{1}{(n-1)!}\int_{0}^1\left({\rm Li}_2(x)-\frac{\pi^2}{6}\right)^{n-1}\ln(1-x) dx\cr
&=\frac{1}{n!}\int_{0}^1\left({\rm Li}_2(x)-\frac{\pi^2}{6}\right)^{n} dx}
\tag 3
$$
So, the question is reduced to evaluating the integral of powers of the dilogarithm. To my knowledge this is not known for powers larger than $2$.
