# A difficult logarithmic integral ${\Large\int}_0^1\log(x)\,\log(2+x)\,\log(1+x)\,\log\left(1+x^{-1}\right)dx$

A friend of mine shared this problem with me. As he was told, this integral can be evaluated in a closed form (the result may involve polylogarithms). Despite all our efforts, so far we have not achieved anything, so I decided to ask for your advice. $$\int_0^1\log(x)\,\log(2+x)\,\log(1+x)\,\log\left(1+x^{-1}\right)dx$$

I found some similar questions here on MSE: (1), (2), (3), (4), (5), (6), (7), (8), (9), (10), (11), (12), (13), (14).

• It can be represented as a difference of 3-factor integrals (with powers): ${\Large\int}_0^1\ln(x)\,\ln(2+x)\,\ln^2(1+x)\,dx - {\Large\int}_0^1\ln^2(x)\,\ln(2+x)\,\ln(1+x)\,dx$, that makes it more similar to other integrals you mentioned. – Vladimir Reshetnikov Jul 28 '15 at 0:48
• $$I=4\log^2(2)-2\log^3(2)-\frac74 \zeta(3)\log(2)-2\sum_{n,m=1}^{\infty}\frac{(-1)^{m+n}H_m}{2^nn(m+1)(m+n+2)^2}-2\sum_{n,m=1}^{\infty}\frac{(-1)^{m+n}}{2^nnm(m+n+1)^3}$$ These double series has me stumped... – nospoon Jul 29 '15 at 18:46

The main ingredient here is the integral representation $$\operatorname{Li}_n(z)=\frac{(-1)^{n-1}}{(n-2)!}\int_0^1 \frac{\ln\left(1-zx\right)\ln^{n-2}x\,dx}{x},\tag{\spadesuit}$$ valid for $|z|<1,n\in\mathbb{N}_{\ge 2}$.

The derivation goes as follows:

1. Rewrite the initial integral as \begin{align*} \mathcal{I}&=\int_0^1\ln(x+2)\underbrace{\left[\ln x\ln^2(1+x)-\ln^2x\ln(1+x)\right]}_{=\frac13\left(\ln x-\ln(x+1)\right)^3-\frac13\ln^3 x+\frac13\ln^3 (x+1)} dx=\\ &=\frac13\biggl[\underbrace{\int_0^1\ln(x+2)\ln^3(x+1)dx}_{\mathcal{I}_1}-\underbrace{\int_0^1\ln(x+2)\ln^3x\,dx}_{\mathcal{I}_2}-\underbrace{\int_0^1\ln(x+2)\ln^3\frac{x+1}{x}dx}_{\mathcal{I}_3}\biggr]. \end{align*}

2. The integrals $\mathcal{I}_{1,2}$ have antiderivatives that can be expressed in terms of polylogarithms (say, with Mathematica), therefore we concentrate on $\mathcal{I}_3$. After the change of variables $t=\frac{2x}{x+1}$, we obtain \begin{align*}\mathcal{I_3}&=-2\int_0^1\frac{\ln\frac{4-t}{2-t}\ln^3\frac t2\,dt}{(2-t)^2}=\\&=-2\int_0^1\frac{\ln\frac{4-t}{2-t}\ln^3 t\,dt}{(2-t)^2}+ 6\ln 2 \int_0^1\frac{\ln\frac{4-t}{2-t}\ln^2 t\,dt}{(2-t)^2} \tag{$\clubsuit$}\\&\quad -6\ln^22\int_0^1\frac{\ln\frac{4-t}{2-t}\ln t\,dt}{(2-t)^2}+ 2\ln^32\int_0^1\frac{\ln\frac{4-t}{2-t}dt}{(2-t)^2}. \end{align*}

3. Now let me explain how these integrals can be computed. Consider, for instance, the first term in ($\clubsuit$): \begin{align*} 2\int_0^1\frac{\ln\frac{4-t}{2-t}\ln^3 t\,dt}{(2-t)^2}&=\int_0^1\ln\frac{4-t}{2-t}\ln^3 t\,d\left(\frac{t}{2-t}\right)=- \int_0^1\frac{t}{2-t}d\left(\ln\frac{4-t}{2-t}\ln^3 t\right)=\\ &=- \int_0^1\frac{t}{2-t}\left[\color{red}{-\frac{\ln^3 t}{4-t}+\frac{\ln^3 t}{2-t}}+\frac{3}{t}\ln\frac{4-t}{2-t}\ln^2 t\right]dt \end{align*} The terms shown in red lead to integrals computable with the help of ($\spadesuit$) (e.g. differentiate it with respect to $z$ and see what happens). The remaining nontrivial piece is thus $$\int_0^1\frac{\ln\frac{4-t}{2-t}\ln^2 t}{2-t}dt= \int_0^1\frac{\ln(4-t)\ln^2 t}{2-t}dt-\int_0^1\frac{\ln(2-t)\ln^2 t}{2-t}dt$$ The second part again has again a polylogarithmic antiderivative computable with Mathematica, so it remains to compute $$\mathcal{I}_4=\int_0^1\frac{\ln(4-t)\ln^2 t}{2-t}dt.$$ Note that the same procedure applied to the other three terms in ($\clubsuit$) leads to easily computable integrals (as instead of $\ln^2t$ in the analog of $\mathcal{I}_4$ we have $\ln t$ or $1$).

4. Thus it remains to compute $\mathcal{I}_4$. And this is the only place where a certain miracle takes place, which indicates that there should be an easier way to do the initial integral. Making the change of variables $s=2-t$, we get \begin{align*} \mathcal{I}_4&=\int_1^2\frac{\ln(2+s)\ln^2(2-s)\,ds}{s}=\\ &=\frac16\int_1^2\frac{\left[\ln(2+s)+\ln(2-s)\right]^3+\left[\ln(2+s)-\ln(2-s)\right]^3-2\ln^3(2+s)}{s}ds=\\ &=\frac16\int_1^2\frac{\ln^3(4-s^2)}{s}ds+\frac16\int_1^2\frac{\ln^3\frac{2+s}{2-s}}{s}ds-\frac13\int_1^2\frac{\ln^3(s+2)}{s}ds. \end{align*} Each of these three pieces again has polylogarithmic antiderivatives that can be computed by Mathematica. This becomes obvious after change of variables $u=s^2$ in the first integral (the miracle is here: due to special parameter values we don't have an additional linear term under logarithm which would spoil the things) and $u=\frac{2+s}{2-s}$ in the second.

So the conclusion is that indeed, the integral $\mathcal{I}$ can be expressed in terms of polylogarithms (up to $\operatorname{Li}_4$), but I was too lazy to type the answer. Fortunately, for that we have Cleo.

Added: As suggested by Vladimir Reshetnikov, the integration bounds $(0,1)$ are not really important: the above approach yields an explicit antiderivative which I posted at https://gist.github.com/anonymous/4c35e5617cf846e8f517

• So, it looks like the integrand in the question has a closed-form antiderivative in terms of elementary functions and polylogarithms, right? – Vladimir Reshetnikov Aug 4 '15 at 19:30
• @VladimirReshetnikov Maybe, but not necessarily - see for instance my red terms where the integration bounds look essential. – Start wearing purple Aug 4 '15 at 19:33
• Mathematica can find antiderivatives for the red terms. – Vladimir Reshetnikov Aug 4 '15 at 19:35
• @VladimirReshetnikov You were right, now I have the antiderivative explicitly, but it takes the whole screen and does not seem to simplify. – Start wearing purple Aug 4 '15 at 22:41
• Simplified from 1924 down to 720 nodes: goo.gl/rNPLj9 – Vladimir Reshetnikov Aug 6 '15 at 21:49

• More than $1500$ decimal digits match. – Vladimir Reshetnikov Aug 4 '15 at 23:27