Here is another integral I'm trying to evaluate:
$$I=\int_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx.\tag1$$
A numeric approximation is:
$$I\approx-0.19902842515384155925817158058508204141843184171999583129…\tag2$$
(click here to see more digits).
Unfortunately, so far I have made no progress in finding a closed form for it. Could you please suggest any ideas how to do that?
Best Answer
The value of $I$ is a $\mathbb{Q}$-linear combination of values of the multiple polylogarithm at rational arguments. I'll explain how to compute this.
Expanding each logarithm in the integrand as an integral, multiplying out, and dividing into regions, and making the substitution $x\leftrightarrow 1-x$, we get that $I$ is a $\mathbb{Q}$-linear combination of iterated integrals of the form $$ \int_{1\geq t_1\geq t_2\geq t_3\geq t_4\geq 0} \frac{dt_4}{f_4(t_4)}\frac{dt_3}{f_3(t_3)}\frac{dt_2}{f_2(t_2)}\frac{dt_1}{f_1(t_1)}, $$ where each $f_i(t)$ is either $t$ or $1-wt$ for some $w\in\{1/2,2/3\}$.
Claim: each iterated integral of this form is a value of the multiple polylogarithm, defined by $$ Li_{s_1,\ldots,s_k}(z_1,\ldots,z_k):=\sum_{n_1>\ldots>n_k\geq 1}\frac{z_1^{n_1}\ldots z_k^{n_k}}{n_1^{s_1}\ldots n_k^{s_k}}. $$ For $k=1$ this is the ordinary polylogarithm, and $Li_{s_1,\ldots,s_k}(1,\ldots,1)=\zeta(s_1,\ldots,s_k)$ is the multiple zeta value.
The claim isn't too hard to see by induction on the number of terms in the iterated integral: we have $$ \int_0^{z_1} \frac{Li_{s_1,\ldots,s_k}(t,z_2,\ldots,z_k)}{t}\,dt=Li_{s_1+1,\ldots,s_k}(z_1,z_2,\ldots,z_k), $$ $$ \int_0^{z_1} \frac{Li_{s_1,\ldots,s_k}(t,z_2,\ldots,z_k)}{1-wt}\,dt=\frac{1}{w}Li_{1,s_1,\ldots,s_k}(wz_1,1/w,z_2,z_3,\ldots,z_k). $$ (I hope I wrote this all out correctly)
Values of multiple polylogarithms satisfy many relations, so it's possible the expression one gets can be simplified.
Iterated integrals like this show up when computing the action of parallel transport on algebraic vector bundles with nilpotent connection on open subsets of $\mathbb{P}^1$. It's not so hard to write down such a thing on $\mathbb{P}^1\backslash\{1,-1,-1/2,\infty\}$ giving $I$ as a matrix coefficient for transport along $[0,1]$.
There's a thing called the unipotent fundamental group of a variety, which has the structure of a motive. Without getting into what exactly this is, I'll just say that the observation about parallel transport essentially amounts to $I$ being a period of $\pi_{1,\cdot}(\mathbb{P}^1\backslash\{1,-1,-1/2,\infty\})$. One doesn't get a good model of $X:=\mathbb{P}^1\backslash\{1,-1,-1/2,\infty\}$ over $\mathbb{Z}$ because the removed points collide mod 2 and mod 3, but there is a good model over $\mathbb{Z}[1/6]$. It is known that the fundamental group of a rational curve has the structure of a mixed Tate motive, so $I$ is the period of a mixed Tate motive over $\mathbb{Z}[1/6]$. I don't really understand the construction of mixed Tate motives, so I'm just viewing it as a black box. Probably someone who understood them better than I could see directly that $I$ is the period of a mixed Tate motive without thinking about $\pi_1$.
For comparison: if the only denominators appearing in the iterated integral were $t$ and $1-t$, then the value of the integral is a multiple zeta value. These numbers are periods of the fundamental group of $\mathbb{P}^1\backslash\{0,1,\infty\}$, which is a mixed Tate motive over $\mathbb{Z}$. It's a theorem that the space of all periods of mixed Tate motives over $\mathbb{Z}$ is the $\mathbb{Q}[(2\pi i)^{-1}]$ span of the multiple zeta values. I think in the case of $I$ the mixed Tate motive we need is only defined over $\mathbb{Z}[1/6]$, so that $I$ can't necessarily be written in terms of multiple zeta values.
There's a conjecture that all periods of mixed Tate motives over any ring $\mathbb{Z}[1/N]$ are linear combinations of values of multiple polylogarithms.