# Closed form for ${\large\int}_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx$

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?

• Apparently, the integrals involving logarithm function have been trending topics here on MSE. – Tunk-Fey Sep 6 '14 at 18:38
• Somewhat related integrals people.sc.fsu.edu/~dduke/devoto-duke-84.pdf – ClassicStyle Sep 7 '14 at 1:17
• There is an identity I've found helpful in integrals involving the logarithm product $\ln{(1-z)}\ln{(1+z)}$. For $|z|<1$, this product equals: $\frac12\ln^2{(1+z)}+\frac12\ln^2{(1-z)}+2Li_2{\left(\frac{z}{z-1}\right)}+2Li_2{\left(\frac{z}{z-1}\right)}-Li_2{\left(\frac{z^2}{z^2-1}\right)}$. – David H Sep 7 '14 at 1:40
• Vladimir, IMHO, I doubt the given integral has a closed form. Well, I could certainly be wrong and I really hope I am wrong because I want to see its closed-form and how to prove it. I am able to derive the integral into three following forms and it turns out none of them have closed forms, at least it can be proven by using mathematics software packages. Using differentiation under integral sign method also gives me nothing. It looks like the term $1+2x$ makes thing so difficult since we cannot express it in form of series because of $0<x<1$. – Tunk-Fey Sep 14 '14 at 17:55
• When I looked at integral representations of multiple polylogarithms (as in Julian Rosen's answer), I found it is enough to evaluate the following integral, and perhaps there is a rational function identity that might simplify it (it looks symmetric): $$\int_{1,1,1,1}^{\{\infty\}^4} \frac{8(x^2 y^2 z^2w -1)}{x (x+2) y (x y+2) z \left(x^2 y^2 z^2-1\right) w \left(w^2 x^2 y^2 z^2-1\right)}$$ – Kirill Sep 18 '14 at 7:20

$\def\Li{\,\mathrm{Li}}$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:

(7*Pi^4)/540 + (5*Pi^2*Log[2]^2)/36 - (29*Log[2]^4)/144 - (Pi^2*Log[2]*Log[3])/2 +  (Log[2]^3*Log[3])/3 + (3*Pi^2*Log[3]^2)/8 - (9*Log[2]^2*Log[3]^2)/8 + (7*Log[2]*Log[3]^3)/4 -  (35*Log[3]^4)/48 - (Pi^2*PolyLog[2, 1/3])/12 + Log[2]*Log[3]*PolyLog[2, 1/3] - PolyLog[2, 1/3]^2/2 -  Log[2]*PolyLog[3, 1/3] + 3*Log[3]*PolyLog[3, 1/3] + 2*Log[3]*PolyLog[3, 2/3] - PolyLog[4, 1/4]/16 +  PolyLog[4, 1/3] + PolyLog[4, 1/2]/6 - (3*PolyLog[4, 2/3])/2 - PolyLog[4, 3/4]/4 +  (19*Log[2]*Zeta[3])/12 - (13*Log[3]*Zeta[3])/3


Edit. Here are the expressions for individual multiple polylogarithms above. The first two are rigorous, being the output of Integrate applied to integral representations:

MultiPolyLog[{2, 1}, {2/3, 3/4}] := -(1/4) \[Pi]^2 Log[2] - (8 Log[2]^3)/3 + 1/2 Log[2]^2 Log[3] + Log[2] Log[3]^2 - 2 Log[2] PolyLog[2, 1/4] + 3 Log[2] PolyLog[2, 2/3] - PolyLog[3, 1/4] - PolyLog[3, 1/3] + PolyLog[3, 2/3] + Zeta[3]/8
MultiPolyLog[{2, 1}, {2/3, 1}] := 1/6 (\[Pi]^2 Log[3/2] - 2 Log[3]^3 + Log[2]^2 Log[27/2] + 6 Log[3] PolyLog[2, -(1/2)]) + PolyLog[3, -(1/2)] + PolyLog[3, 2/3]


These two, of weight 4, come from an integer relation algorithm:

{MultiPolyLog[{2, 1, 1}, {2/3, 1, 3/4}] -> (11*Pi^4)/240 - (11*Pi^2*Log[2]^2)/240 - Log[2]^4/60 - (Pi^2*Log[2]*Log[3])/10 - (Log[2]^3*Log[3])/48 + (41*Pi^2*Log[3]^2)/480 - (7*Log[2]^2*Log[3]^2)/160 + (Log[2]*Log[3]^3)/48 - (9*Log[3]^4)/160 + (13*Pi^2*Log[2]*Log[4])/240 - (Log[2]^3*Log[4])/32 - (29*Pi^2*Log[3]*Log[4])/480 - (19*Log[2]^2*Log[3]*Log[4])/480 + (7*Log[2]*Log[3]^2*Log[4])/120 + (Log[3]^3*Log[4])/24 - (3*Pi^2*Log[4]^2)/80 - (Log[2]^2*Log[4]^2)/16 + (Log[2]*Log[3]*Log[4]^2)/16 - (Log[3]^2*Log[4]^2)/32 + (7*Log[2]*Log[4]^3)/480 - (7*Log[3]*Log[4]^3)/480 + Log[4]^4/32 - (13*Pi^2*PolyLog[2, 1/4])/480 + (19*Log[2]^2*PolyLog[2, 1/4])/240 - (Log[2]*Log[3]*PolyLog[2, 1/4])/48 - (41*Log[3]^2*PolyLog[2, 1/4])/480 + (Log[2]*Log[4]*PolyLog[2, 1/4])/80 + (17*Log[3]*Log[4]*PolyLog[2, 1/4])/160 + (Log[4]^2*PolyLog[2, 1/4])/40 + (11*PolyLog[2, 1/4]^2)/96 - (29*Pi^2*PolyLog[2, 1/3])/480 - (Log[2]^2*PolyLog[2, 1/3])/20 + (19*Log[2]*Log[3]*PolyLog[2, 1/3])/480 + (11*Log[3]^2*PolyLog[2, 1/3])/160 + (11*Log[2]*Log[4]*PolyLog[2, 1/3])/240 - (Log[3]*Log[4]*PolyLog[2, 1/3])/15 - (5*Log[4]^2*PolyLog[2, 1/3])/96 - (7*PolyLog[2, 1/4]*PolyLog[2, 1/3])/160 + PolyLog[2, 1/3]^2/120 - (7*Pi^2*PolyLog[2, 2/3])/96 + (Log[2]^2*PolyLog[2, 2/3])/60 + (11*Log[2]*Log[3]*PolyLog[2, 2/3])/480 + (Log[3]^2*PolyLog[2, 2/3])/48 + (17*Log[2]*Log[4]*PolyLog[2, 2/3])/480 + (11*Log[3]*Log[4]*PolyLog[2, 2/3])/240 + (11*Log[4]^2*PolyLog[2, 2/3])/160 + (49*PolyLog[2, 1/4]*PolyLog[2, 2/3])/480 - (7*PolyLog[2, 1/3]*PolyLog[2, 2/3])/240 + PolyLog[2, 2/3]^2/12 - (11*Pi^2*PolyLog[2, 3/4])/120 + (Log[2]^2*PolyLog[2, 3/4])/30 - (Log[2]*Log[3]*PolyLog[2, 3/4])/240 + (Log[3]^2*PolyLog[2, 3/4])/6 + (Log[2]*Log[4]*PolyLog[2, 3/4])/15 - (5*Log[3]*Log[4]*PolyLog[2, 3/4])/32 - (Log[4]^2*PolyLog[2, 3/4])/160 - (89*PolyLog[2, 1/4]*PolyLog[2, 3/4])/480 - (49*PolyLog[2, 1/3]*PolyLog[2, 3/4])/480 - (17*PolyLog[2, 2/3]*PolyLog[2, 3/4])/80 + PolyLog[2, 3/4]^2/24 - (37*Log[2]*PolyLog[3, 1/4])/240 - (Log[3]*PolyLog[3, 1/4])/40 - (77*Log[4]*PolyLog[3, 1/4])/480 + (3*Log[2]*PolyLog[3, 1/3])/80 - (Log[3]*PolyLog[3, 1/3])/20 - (11*Log[4]*PolyLog[3, 1/3])/160 - (Log[2]*PolyLog[3, 2/3])/240 - (71*Log[4]*PolyLog[3, 2/3])/480 - (Log[2]*PolyLog[3, 3/4])/48 - (Log[3]*PolyLog[3, 3/4])/40 - (91*Log[4]*PolyLog[3, 3/4])/480 - (21*PolyLog[4, 1/4])/16 - (7*PolyLog[4, 1/3])/4 + (5*PolyLog[4, 1/2])/2 - PolyLog[4, 2/3]/2 - (11*PolyLog[4, 3/4])/8 - (Log[2]*Zeta[3])/15 + (19*Log[3]*Zeta[3])/240 - (13*Log[4]*Zeta[3])/96
,MultiPolyLog[{2, 1, 1}, {2/3, 3/4, 4/3}] -> (-139*Pi^4)/1440 + (149*Pi^2*Log[2]^2)/1440 + Log[2]^4/30 + (347*Pi^2*Log[2]*Log[3])/1440 + (19*Log[2]^3*Log[3])/480 - (313*Pi^2*Log[3]^2)/1440 + (13*Log[2]^2*Log[3]^2)/120 - (7*Log[2]*Log[3]^3)/90 + (8*Log[3]^4)/45 - (19*Pi^2*Log[2]*Log[4])/180 + (97*Log[2]^3*Log[4])/1440 + (241*Pi^2*Log[3]*Log[4])/1440 + (23*Log[2]^2*Log[3]*Log[4])/288 - (47*Log[2]*Log[3]^2*Log[4])/480 - (37*Log[3]^3*Log[4])/240 + (37*Pi^2*Log[4]^2)/360 + (13*Log[2]^2*Log[4]^2)/96 - (5*Log[2]*Log[3]*Log[4]^2)/32 + (17*Log[3]^2*Log[4]^2)/144 - (31*Log[2]*Log[4]^3)/720 + (Log[3]*Log[4]^3)/360 - (7*Log[4]^4)/80 + (29*Pi^2*PolyLog[2, 1/4])/720 - (77*Log[2]^2*PolyLog[2, 1/4])/480 + (Log[2]*Log[3]*PolyLog[2, 1/4])/8 + (35*Log[3]^2*PolyLog[2, 1/4])/288 - (Log[2]*Log[4]*PolyLog[2, 1/4])/180 - (23*Log[3]*Log[4]*PolyLog[2, 1/4])/360 - (17*Log[4]^2*PolyLog[2, 1/4])/1440 + (11*PolyLog[2, 1/4]^2)/288 + (13*Pi^2*PolyLog[2, 1/3])/80 + (133*Log[2]^2*PolyLog[2, 1/3])/1440 - (133*Log[2]*Log[3]*PolyLog[2, 1/3])/1440 - (47*Log[3]^2*PolyLog[2, 1/3])/240 - (31*Log[2]*Log[4]*PolyLog[2, 1/3])/240 + (31*Log[3]*Log[4]*PolyLog[2, 1/3])/240 + (41*Log[4]^2*PolyLog[2, 1/3])/720 + (5*PolyLog[2, 1/4]*PolyLog[2, 1/3])/96 + (23*PolyLog[2, 1/3]^2)/720 + (247*Pi^2*PolyLog[2, 2/3])/1440 - (19*Log[2]^2*PolyLog[2, 2/3])/480 + (Log[2]*Log[3]*PolyLog[2, 2/3])/288 - (23*Log[3]^2*PolyLog[2, 2/3])/240 - (113*Log[2]*Log[4]*PolyLog[2, 2/3])/1440 + (Log[3]*Log[4]*PolyLog[2, 2/3])/144 - (113*Log[4]^2*PolyLog[2, 2/3])/720 - (59*PolyLog[2, 1/4]*PolyLog[2, 2/3])/1440 + (17*PolyLog[2, 1/3]*PolyLog[2, 2/3])/360 - (11*PolyLog[2, 2/3]^2)/144 + (103*Pi^2*PolyLog[2, 3/4])/480 - (127*Log[2]^2*PolyLog[2, 3/4])/1440 - (Log[2]*Log[3]*PolyLog[2, 3/4])/36 - (619*Log[3]^2*PolyLog[2, 3/4])/1440 - (127*Log[2]*Log[4]*PolyLog[2, 3/4])/720 + (187*Log[3]*Log[4]*PolyLog[2, 3/4])/720 - (7*Log[4]^2*PolyLog[2, 3/4])/180 + (331*PolyLog[2, 1/4]*PolyLog[2, 3/4])/1440 + (223*PolyLog[2, 1/3]*PolyLog[2, 3/4])/720 + (281*PolyLog[2, 2/3]*PolyLog[2, 3/4])/720 + (37*PolyLog[2, 3/4]^2)/720 + (49*Log[2]*PolyLog[3, 1/4])/360 + (191*Log[3]*PolyLog[3, 1/4])/240 - (59*Log[4]*PolyLog[3, 1/4])/1440 - (91*Log[2]*PolyLog[3, 1/3])/720 - (33*Log[3]*PolyLog[3, 1/3])/20 + (91*Log[4]*PolyLog[3, 1/3])/1440 - (61*Log[2]*PolyLog[3, 2/3])/360 + (31*Log[3]*PolyLog[3, 2/3])/60 - (17*Log[4]*PolyLog[3, 2/3])/720 - (5*Log[2]*PolyLog[3, 3/4])/48 + (11*Log[3]*PolyLog[3, 3/4])/12 + (17*Log[4]*PolyLog[3, 3/4])/160 + (23*PolyLog[4, 1/4])/16 - PolyLog[4, 1/3]/4 - (17*PolyLog[4, 1/2])/6 + (7*PolyLog[4, 2/3])/2 + (15*PolyLog[4, 3/4])/8 + (19*Log[2]*Zeta[3])/1440 - (203*Log[3]*Zeta[3])/288 + (Log[4]*Zeta[3])/40
}

• Very interesting! Have you used FindIntegerNullVector or some other implementation of PSLQ algorithm? Could you give separate expressions for multiple polylogarithms from the first formula in your answer? – Vladimir Reshetnikov Sep 20 '14 at 17:30
• @VladimirReshetnikov Sure, I put them in the answer. I used LatticeReduce. – Kirill Sep 20 '14 at 18:59
• @JulianRosen Yes, so am I. But then again, many, but not all, multiple zeta values can be written in terms of ordinary zeta values. – Kirill Sep 20 '14 at 21:02
• @Kirill Some polylogarithm terms in your result could be reduced to simpler arguments (using these identities), yielding a simpler overall result. – Vladimir Reshetnikov Sep 21 '14 at 0:28
• @Kirill It would be very nice to prove some kind of uniformity theorem for this class of expressions, so that a demonstration of a closed form of small enough syntactic complexity that matches the integral with sufficiently high precision would constitute a proof of its correctness (similar to how we can do it with algebraic numbers in certain cases). – Vladimir Reshetnikov Sep 21 '14 at 1:22

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.

• Very interesting: $$-{\frac{\log2\log3}2} {\mathrm{L}}_2\Big({\frac23}\Big) + {\frac{\log3}2}\mathrm{L}_{2,1}\Big({\frac23,\frac34}\Big) + {\frac{\log2}2} \mathrm{L}_{2,1}\Big({\frac23,1}\Big)-{\frac{1}{2}}\mathrm{L}_{2,1,1}\Big({{\frac{2}{3}},\frac{3}{4},\frac{4}{3}}\Big)-{\frac12}\mathrm{L}_{2,1,1}\Big({\frac{2}{3},1,\frac{3}{4}}\Big)$$ – Kirill Sep 16 '14 at 22:27
• parallel transport on algebraic vector bundles with nilpotent connection on open subsets of $\mathbb P^1$ - Took the words right out of my mouth! :-) – Lucian Sep 16 '14 at 22:31
• This is a really nice answer. I found Special values of multiple polylogarithms by Borwein, Bradley, Broadhurst, Lisonek, but the identities there are (I think) not enough to answer this question, or this one (but at least the integral representation I got looks imposing). Do you know more about the identities that multiple polylogarithms satisfy? – Kirill Sep 17 '14 at 19:13
• Thank you for your answer, the bounty is yours! I still would like to see an explicit result though. – Vladimir Reshetnikov Sep 18 '14 at 17:05
• I fail to see this the correct answer for this problem because you didn't give an explicit answer. Maybe I am just too dumb to understand it – Venus Sep 19 '14 at 17:11