Evaluating $\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$

Question

In this thread

a friend posted the following integral

$$I=\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$$

The best we could do is expressing it in terms of Euler sums

$$I=-\frac{\zeta^2(2)}{2}+ \sum_{n\geq 1}\frac{(-1)^{n-1}}{n^2} H_{n}^{(2)}+\sum_{n\geq 1}\frac{(-1)^{n-1}}{n^3}H_{n}$$

I am wondering if the approach I followed made the integral complicated ? What approach would you follow to solve the integral?, can we find a better solution ?

Answer 1

The values of the two Euler Sums are

$$\displaystyle \sum_{n=1}^\infty (-1)^{n-1} \frac{H_n}{n^{3}} = \frac{11\pi^4}{360}-2\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7}{4}\log(2) \zeta(3)+\frac{\pi^2}{12}\log^2(2)-\frac{1}{12}\log^4(2)$$ $$\displaystyle \sum_{n=1}^\infty (-1)^{n-1} \frac{H_n^{(2)}}{n^{2}} =-\frac{17}{480}\pi^4 +4 \text{Li}_4 \left(\frac{1}{2} \right)+\frac{7}{2}\log(2) \zeta(3)-\frac{\pi^2 \log^2(2)}{6}+\frac{\log^4(2)}{6}$$

Therefore the integral evaluates to

$$\begin{align*} \int_0^1 \frac{\log(1-x)\log(x)\log(1+x)}{x}dx &=-\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12} \\ &\quad+ 2 \text{Li}_4 \left(\frac{1}{2} \right) \sim 0.290721 \end{align*}$$

Refer to this page for the evaluation of Euler Sums.

Answer 2

$$-\frac{\partial^2}{\partial s\partial t}\left[B(s+1,t+1)\;_3 F_2(1,1,s+1;2,s+t+2;-1)\right]_{s=t=0}$$

It may be that the Hypergeometric function is summable. In this case, the differentiation is trivial. (B denotes Euler's beta function.)

Answer 3

I would do the following variable change.

$$x=e^{-t}$$ Then we can represent the integral as follows:

$$I=-\int_{0}^{\infty}t\ln(1+e^{-t})\ln(1-e^{-t})\;dt$$ Now, apply the Taylor expansion of the logarithm:

$$\ln(1+x)=\sum_{i=1}^{\infty}(-1)^{i-1}\frac{x^i}{i}$$

$$I=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\frac{(-1)^{i-1}}{ij}\int_{0}^{\infty}te^{-(i+j)t}dt=$$

$$=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\frac{(-1)^{i-1}}{ij(i+j)^2}$$

Answer 4

Related problems: (I). You can have the following solution

$$ \frac{3\gamma}{4}\,\zeta( 3 )+{\frac {7\pi^4}{360}}+\sum _{m=1}^{\infty }{\frac { \left( -1 \right) ^{m-1}\psi \left( m \right) }{{m}^{3}}}+\sum _{m=1}^{\infty }-{\frac { \left( -1 \right) ^{m-1}\psi' \left( m \right) }{{m}^{2}}}\sim 0.2907212779,$$

which you might be able to simplify it further.

Note: If you use the identity

$$ \frac{\pi^4}{90}=\zeta(4), $$

in the above expression, then you will have the form

$$ \frac{3\gamma}{4}\,\zeta( 3 )+{\frac {7}{4}}\zeta(4)+\sum _{m=1}^{\infty }{\frac { \left( -1 \right) ^{m-1}\psi \left( m \right) }{{m}^{3}}}+\sum _{m=1}^{\infty }-{\frac { \left( -1 \right) ^{m-1}\psi' \left( m \right) }{{m}^{2}}}.$$