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).

  • $\begingroup$ 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. $\endgroup$ – Vladimir Reshetnikov Jul 28 '15 at 0:48
  • $\begingroup$ $$I=4\log^2(2)-2\log^3(2)-\frac74 \zeta(3)\log(2)-2\sum_{n,m=1}^{\infty}\frac{(-1)^{m+n}H_m}{2^nn(m+1)(m+n+2)^2}-2\sum_{n,m=1}^{\infty}\frac{(-1)^{m+n}}{2^nnm(m+n+1)^3}$$ These double series has me stumped... $\endgroup$ – 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

  • $\begingroup$ So, it looks like the integrand in the question has a closed-form antiderivative in terms of elementary functions and polylogarithms, right? $\endgroup$ – Vladimir Reshetnikov Aug 4 '15 at 19:30
  • $\begingroup$ @VladimirReshetnikov Maybe, but not necessarily - see for instance my red terms where the integration bounds look essential. $\endgroup$ – Start wearing purple Aug 4 '15 at 19:33
  • $\begingroup$ Mathematica can find antiderivatives for the red terms. $\endgroup$ – Vladimir Reshetnikov Aug 4 '15 at 19:35
  • $\begingroup$ @VladimirReshetnikov You were right, now I have the antiderivative explicitly, but it takes the whole screen and does not seem to simplify. $\endgroup$ – Start wearing purple Aug 4 '15 at 22:41
  • 2
    $\begingroup$ Simplified from 1924 down to 720 nodes: goo.gl/rNPLj9 $\endgroup$ – Vladimir Reshetnikov Aug 6 '15 at 21:49

\begin{align} & \int_0^1\ln(2+x)\,\ln(1+x)\,\ln\left(1+x^{-1}\right)\ln x\,dx\\ & \quad=\frac{71}{36}\,\ln^42+2\ln^32\cdot\ln3+4\ln2\cdot\ln^33-7\ln^22\cdot\ln^23-\frac23\,\ln^32-\frac23\,\ln^33-\ln^22\cdot\ln3\\ & \quad \quad +6\ln^22+3\ln^23-12\ln2-\frac{\pi^4}{216}+\pi^2\!\left(\frac{49}{36}\,\ln^22-2\ln2\cdot\ln3-\frac{\ln2}3+\frac{\ln3}3-\frac16\right)\\ & \quad \quad+\left(6-2\ln2-2\ln^22\right)\operatorname{Li}_2\!\left(\tfrac13\right)+(2-12\ln2)\left[\operatorname{Li}_3\!\left(\tfrac13\right)+\operatorname{Li}_3\!\left(\tfrac23\right)\right]-\frac23\,\operatorname{Li}_4\!\left(\tfrac12\right)+3\operatorname{Li}_4\!\left(\tfrac14\right)\\ & \quad \quad +\left(\frac54+\frac{221}{12}\ln2\right)\zeta(3). \end{align}

  • 6
    $\begingroup$ any details on how you derived this? $\endgroup$ – user190080 Aug 4 '15 at 17:22
  • $\begingroup$ @VladimirReshetnikov could you maybe share your in-depth testing espacially the result for the right hand side? If this is indeed the solution - which I cannot conform - it would be really impressive $\endgroup$ – user190080 Aug 4 '15 at 18:02
  • 7
    $\begingroup$ More than $1500$ decimal digits match. $\endgroup$ – Vladimir Reshetnikov Aug 4 '15 at 23:27
  • 5
    $\begingroup$ Is @Cleo the next Ramanujan? $\endgroup$ – vs_292 Dec 31 '17 at 7:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.