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?
On
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.
$\def\Li{\,\mathrm{Li}}$I followed the technique suggested by Julian Rosen in his answer, and decomposed your integral (and your other integral) as a linear combination of multiple polylogarithms: $$\textstyle -\frac12\log2\log3 \Li_2({\frac23}) + \frac12\log3\Li_{2,1}({\frac23,\frac34}) + \frac12\log2\Li_{2,1}({\frac23,1}) \\\textstyle - \frac12\Li_{2,1,1}({\frac23,\frac34,\frac43}) - \frac12\Li_{2,1,1}({\frac23,1,\frac34}) $$ There is a paper by Borwein, Bradley, Broadhurst, Lisonek, that explains what few families of identities apply to multiple polylogarithms, and it mentions a conjecture that those mentioned there are all the identities that apply at all.
Multiple polylogarithms are generalizations of zeta functions (and polylogarithms, and multiple zeta functions) in that the important thing is not the depth $k$ of $\mathrm{Li}_{s_1,\ldots,s_k}(z_1,\ldots,z_k)$, but the weight $\sum_{i=1}^{k}s_i$. My Mathematica was able, with some amount of hand-holding, to compute directly the (rational) integral representations involved in multiple polylogarithms of weight $1$, $2$, and $3$, but couldn't handle weight $4$. Your integral has three logs, so it's weight 4.
It seems multiple polylogarithms of weight 4 with small rational arguments are still algebraically related to ordinary polylogarithms. By analogy with multiple zeta values, I suspect the same won't necessarily be true of higher weights at least in general.
I made some guesses based on exact weight-3 values about what terms weight-4 values might involve, and used an integer relation algorithm to try and find an expression for your integral. I found this one, which matches the integral to $3000$ digits, and when I looked for an integer relation I used a tolerance of only $10^{-200}$.
Here you go: $$\textstyle\def\Li{\mathrm{Li}} -\frac{1}{2} \Li_2(\frac{1}{3}) \zeta (2)-\frac{1}{4}\Li_4(\frac{3}{4})-\frac{3}{2} \Li_4(\frac{2}{3})+\frac{1}{6}\Li_4(\frac{1}{2})+\Li_4(\frac{1}{3}) -\frac{1}{16}\Li_4(\frac{1}{4})-\frac{1}{2}\Li_2(\frac{1}{3}){}^2+2 \Li_3(\frac{2}{3}) \log3+3 \Li_3(\frac{1}{3}) \log3-\Li_3(\frac{1}{3}) \log2 +\Li_2(\frac{1}{3}) \log2 \log3-\frac{13}{3} \zeta (3) \log3+\frac{19}{12} \zeta (3) \log2+\frac{7}{6} \zeta (4)+\frac{9}{4} \zeta (2) \log^23 +\frac{5}{6} \zeta (2) \log^22-3 \zeta (2) \log2 \log3-\frac{35}{48} \log^43-\frac{29}{144} \log^42+\frac{7}{4} \log2 \log^33 +\frac{1}{3} \log^32 \log3-\frac{9}{8} \log^22 \log^23 $$
Here is the Mathematica expression verbatim, to save people some typing:
Edit. Here are the expressions for individual multiple polylogarithms above. The first two are rigorous, being the output of
Integrateapplied to integral representations:These two, of weight 4, come from an integer relation algorithm: