Compute $\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx$

How to prove that

$$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx$$

$$=2\text{Li}_4\left(\frac12\right)-2\zeta(4)+\frac{15}8\ln(2)\zeta(3)-\frac12\ln^2(2)\zeta(2)$$

where $$\text{Li}_r$$ is the polylogarithm function and $$\zeta$$ is Riemann zeta function.

I managed to prove the equality above using the following harmonic series,

$$\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n}, \ \sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}H_n}{n},\ \sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^2}\ \text{and }\ \sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$$

so definitely this approach is pretty boring. Is it possible to solve it in a different way? Thank you.

Set $$x=2t-1$$ \begin{align} & =\int_{\frac{1}{2}}^{1}{\frac{\ln \left( t \right)\ln \left( 2t \right)}{t}\ln \left( 2-2t \right)dt} \\ & =\int_{\frac{1}{2}}^{1}{\frac{\ln \left( t \right)\ln \left( 2t \right)}{t}\left( \ln \left( 2 \right)-\sum\nolimits_{n=1}^{\infty }{\frac{{{t}^{n}}}{n}} \right)dt} \\ & =\int_{\frac{1}{2}}^{1}{\left\{ \frac{\ln \left( t \right)\ln \left( 2t \right)\ln \left( 2 \right)}{t}-\sum\nolimits_{n=1}^{\infty }{\frac{{{t}^{n-1}}\ln \left( t \right)\ln \left( 2t \right)}{n}} \right\}dt} \\ & =\int_{\frac{1}{2}}^{1}{\frac{\ln \left( t \right)\ln \left( 2t \right)\ln \left( 2 \right)}{t}dt-}\sum\nolimits_{n=1}^{\infty }{\frac{1}{n}\int_{\frac{1}{2}}^{1}{{{t}^{n-1}}\ln \left( t \right)\ln \left( 2t \right)}dt} \\ & =-\frac{1}{6}{{\ln }^{4}}\left( 2 \right)-\sum\nolimits_{n=1}^{\infty }{\left( \frac{2}{{{n}^{4}}}-\frac{2}{{{2}^{n}}{{n}^{4}}}-\frac{\ln \left( 2 \right)}{{{n}^{3}}}-\frac{\ln \left( 2 \right)}{{{2}^{n}}{{n}^{3}}} \right)} \\ & \vdots \\ & \vdots \\ \end{align}

• Really nicely done! (+1) Commented Mar 19, 2020 at 6:17
• Very nice approach. Simple and short. +1 Commented Mar 19, 2020 at 16:07

Here is another proof based on the integral

$$i(z) =\int_0^z\frac{\log(1-x)\log(1+x)}{1+x}\log\left(\frac{1+x}{2}\right)\ dx\tag{1}$$

from which the proof follows as a special case.

The advantage of this approach is that we can check the validity of the final expression by just forming the derivative with respec to to $$z$$, and this is independent of the way we have have found it, may it be by strict derivation or by any heuristic tool, like e.g. Mathematica. With this method we have obviously a lot more information available than the proof of the intial assertion.

Let us first look at the indefinite integral (the antiderivative) and later adjust it to $$\int_0^z\,dx$$.

Writing $$i = i_1 + i_2$$ with

$$i_1 = \int \frac{\log(1-x)\log(1+x)^2}{1+x})\ dx\tag{2}$$ $$i_2 = -\log(2) \int \frac{\log(1-x)\log(1+x)}{1+x}\ dx\tag{3}$$

Mathematica finds

$$i_1 = -2 \text{Li}_4\left(\frac{x+1}{2}\right)-\text{Li}_2\left(\frac{x+1}{2}\right) \log ^2(x+1)+2 \text{Li}_3\left(\frac{x+1}{2}\right) \log (x+1)+\frac{1}{3} \log (2) \log ^3(x+1)$$

subtracting the values at $$x=0$$ which is $$-2 \text{Li}_4\left(\frac{1}{2}\right)$$ and renaming the $$x\to z$$ we find

$$i_1(z) = -2 \text{Li}_4\left(\frac{z+1}{2}\right)-\text{Li}_2\left(\frac{z+1}{2}\right) \log ^2(z+1)+2 \text{Li}_3\left(\frac{z+1}{2}\right) \log (z+1)+2 \text{Li}_4\left(\frac{1}{2}\right)+\frac{1}{3} \log (2) \log ^3(z+1)\tag{2a}$$

Similarly we get

$$i_2(z) = \frac{1}{24} \log (2) \left(21 \zeta (3)+4 \log ^3(2)-\pi ^2 \log (4)\right)-\log (2) \left(\text{Li}_3\left(\frac{z+1}{2}\right)-\text{Li}_2\left(\frac{z+1}{2}\right) \log (z+1)+\frac{1}{2} \log (2) \log ^2(z+1)\right)\tag{3a}$$

Notice that both $$i_1(z)$$ and $$i_2(z)$$ contain only real terms, vanish at $$z=0$$ and have a meaning also for $$-1\lt z \le 1$$.

The complete integral is then

$$i(z) =-2 \text{Li}_4\left(\frac{z+1}{2}\right)-\text{Li}_2\left(\frac{z+1}{2}\right) \log ^2(z+1)-\log (2) \left(\text{Li}_3\left(\frac{z+1}{2}\right)-\text{Li}_2\left(\frac{z+1}{2}\right) \log (z+1)+\frac{1}{2} \log (2) \log ^2(z+1)\right)+2 \text{Li}_3\left(\frac{z+1}{2}\right) \log (z+1)+2 \text{Li}_4\left(\frac{1}{2}\right)+\frac{1}{3} \log (2) \log ^3(z+1)+\frac{1}{24} \log (2) \left(21 \zeta (3)+4 \log ^3(2)-\pi ^2 \log (4)\right)\tag{4}$$

Here is a plot of the integrals as a function of $$z$$

Now we can look at specific values of $$z$$.

For $$z\to1$$ we get

$$i_1(1) = 2 \text{Li}_4\left(\frac{1}{2}\right)+2 \zeta (3) \log (2)-\frac{\pi ^4}{45}+\frac{\log ^4(2)}{3}-\frac{1}{6} \pi ^2 \log ^2(2)\tag{2b}$$

$$i_2(1) = \frac{1}{24} \log (2) \left(21 \zeta (3)+4 \log ^3(2)-\pi ^2 \log (4)\right)-\log (2) \left(\zeta (3)+\frac{\log ^3(2)}{2}-\frac{1}{6} \pi ^2 \log (2)\right)\tag{3b}$$

and

$$i(1) = 2 \text{Li}_4\left(\frac{1}{2}\right)+\frac{15}{8} \zeta (3) \log (2)-\frac{\pi ^4}{45}-\frac{1}{12} \pi ^2 \log ^2(2)\tag{4a}$$

in agreement with the result announced in the OP.

As a second example we let $$z\to -\frac{1}{2}$$

$$i_1(-\frac{1}{2})=2 \text{Li}_4\left(\frac{1}{2}\right)-2 \text{Li}_4\left(\frac{1}{4}\right)-\text{Li}_2\left(\frac{1}{4}\right) \log ^2(2)-2 \text{Li}_3\left(\frac{1}{4}\right) \log (2)-\frac{1}{3} \log ^4(2)\tag{2c}$$

$$i_2(-\frac{1}{2}) = \frac{1}{24} \log (2) \left(21 \zeta (3)+4 \log ^3(2)-\pi ^2 \log (4)\right)-\log (2) \left(\text{Li}_3\left(\frac{1}{4}\right)+\text{Li}_2\left(\frac{1}{4}\right) \log (2)+\frac{\log ^3(2)}{2}\right)\tag{3c}$$

$$i(-\frac{1}{2}) = 2 \text{Li}_4\left(\frac{1}{2}\right)-2 \text{Li}_4\left(\frac{1}{4}\right)-2 \text{Li}_2\left(\frac{1}{4}\right) \log ^2(2)-3 \text{Li}_3\left(\frac{1}{4}\right) \log (2)+\frac{7}{8} \zeta (3) \log (2)-\frac{1}{3} 2 \log ^4(2)-\frac{1}{12} \pi ^2 \log ^2(2)\tag{4b}$$

• Thanks Wolf another good idea. My solution is somewhat similar. Commented Mar 19, 2020 at 16:09
• Ali, if we adopt the term "mma-integral" for an integral whose antiderivative is immediately calculated by mathematica this one here is an example of an "mma" integral. Hence it's an easy one if we are only interested in the result and its proof. Commented Mar 19, 2020 at 18:16

I proved here

$$\small{\int_0^a\frac{\ln(1-x)\ln(1+x)}{1+x} \ dx=\text{Li}_3\left(\frac{1+a}{2}\right)-\text{Li}_3\left(\frac{1}{2}\right)-\ln(1+a)\text{Li}_2\left(\frac{1+a}{2}\right)+\frac12\ln2\ln^2(1+a)}$$

Divide both sides by $$1+a$$ the integrate from $$a=0$$ to $$a=1$$ we get

$$\int_0^1\int_0^a\frac{\ln(1-x)\ln(1+x)}{(1+x)(1+a)} \ dxda=\int_0^1\frac{\text{Li}_3\left(\frac{1+a}{2}\right)}{1+a}\ da-\text{Li}_3\left(\frac{1}{2}\right)\underbrace{\int_0^1\frac{da}{1+a}}_{\ln(2)}$$ $$-\int_0^1\frac{\ln(1+a)\text{Li}_2\left(\frac{1+a}{2}\right)}{1+a}\ da+\frac12\ln2\underbrace{\int_0^1\frac{\ln^2(1+a)}{1+a}\ da}_{1/3 \ln^3(2)}$$

where

$$\int_0^1\int_0^a\frac{\ln(1-x)\ln(1+x)}{(1+x)(1+a)} \ dxda=\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x} \left(\int_x^1\frac{da}{1+a}\right)\ dx$$

$$=-\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx=-I$$

$$\int_0^1\frac{\text{Li}_3\left(\frac{1+a}{2}\right)}{1+a}\ da=\text{Li}_4\left(\frac{1+a}{2}\right)\bigg|_0^1=\zeta(4)-\text{Li}_4\left(\frac{1}{2}\right)$$

$$\int_0^1\frac{\ln(1+a)\text{Li}_2\left(\frac{1+a}{2}\right)}{1+a}\ da\overset{IBP}{=}\ln(1+a)\text{Li}_3\left(\frac{1+a}{2}\right)\bigg|_0^1-\int_0^1\frac{\text{Li}_3\left(\frac{1+a}{2}\right)}{1+a}\ da$$

$$=\ln(2)\zeta(3)-\zeta(4)+\text{Li}_4\left(\frac{1}{2}\right)$$

Combine all results and use $$\text{Li}_3\left(\frac{1}{2}\right)=\frac78\zeta(3)-\frac12\ln(2)\zeta(2)+\frac16\ln^3(2)$$, we get the closed form of $$I$$.

Substitute $$t=\frac{1+x}2$$ \begin{align} I=&\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\frac{1+x}{2}\ dx\\ =&\int_{\frac12}^1 \frac{\ln(2(1-t))(\ln^2t+\ln2\ln t)}t \>dt \overset{ibp}=\frac13 \int_{\frac12}^1 \frac{\ln^3t}{1-t}dt+ \frac12\ln2 \int_{\frac12}^1 \frac{\ln^2t}{1-t}dt \end{align} where $$\int_{0}^1 \frac{\ln^3t}{1-t}dt=-6Li_4(1)$$, $$\int_{0}^1 \frac{\ln^2t}{1-t}dt=2Li_3(1)$$ \begin{align} &\int^{\frac12}_0\frac{\ln^2t}{1-t}dt =2Li_3(\frac12)+2\ln2 Li_2(\frac12)+\ln^3 2\\ &\int^{\frac12}_0\frac{\ln^3t}{1-t}dt =-6Li_4(\frac12)-6\ln2 Li_3(\frac12)-3\ln^22Li_2(\frac12)-\ln^42 \end{align} Thus \begin{align} I=&\>2\>[ Li_4(\frac12)-Li_4(1) ]+\ln2 \>[Li_3(\frac12)+Li_3(1)] -\frac16\ln^42\\ \end{align}