How to evaluate $\int_{0}^{1}{\frac{{{\ln }^{2}}\left( 1-x \right){{\ln }^{2}}\left( 1+x \right)}{1+x}dx}$

I want to evaluate $$\int_{0}^{1}{\frac{{{\ln }^{2}}\left( 1-x \right){{\ln }^{2}}\left( 1+x \right)}{1+x}dx}$$ I run this integral on Maple, It does converge. How we get a closed form? Is that related to polylogs? $\operatorname{Li}_{5}\left(\frac{1}{2}\right)$

Solution 1:

$$\begin{eqnarray*} \int_{0}^{1}{\frac{{{\ln }^{2}}\left( 1-x \right){{\ln }^{2}}\left( 1+x \right)}{1+x}dx} &=& \left. \left(\frac{\partial^2}{\partial s^2}\frac{\partial^2}{\partial t^2} \int_0^1 dx\, (1-x)^s(1+x)^{t-1} \right) \right|_{s=t=0} \\ &=& \left. \left(\frac{\partial^2}{\partial s^2}\frac{\partial^2}{\partial t^2} \,\frac{{}_2F_1(1-t,1;s+2;-1)}{s+1} \right) \right|_{s=t=0} \end{eqnarray*}$$ Here we've used Euler's integral representation for the hypergeometric function.

Addendum: Using the series representation for the hypergeometric function we can take the derivatives. After a little work we find the integral can be written as $$\sum_{k=2}^\infty \frac{(-1)^k}{k+1} \left(H_k^2 - H_k^{(2)}\right) \left(H_{k+1}^2 + H_{k+1}^{(2)}\right),$$ where $H_k$ and $H_k^{(n)}$ are the harmonic and generalized harmonic numbers, respectively. This sum has bad convergence behavior, the terms go like $(-1)^k(\log k)^4/k$ $(k\to\infty)$. Since the sum is alternating we could accelerate it using the Euler transform, for example.

Solution 2:

The result is -4〖Li〗_5 (1/2)+4ζ(3) 〖Log〗^2 2-(2π^2)/9 〖Log〗^3 2- π^2/3 ζ(3)-π^4/20 Log2+ 7/30 〖Log〗^5 2+ 63/8 ζ(5) =0,418709830998418751408037243448…