how to evaluate this integral $\int_0^1 \frac{\ln x \, \text{Li}_2(1-x)}{2+x} \, dx$

Question

Question statement: how to evaluate this integral $$\int_0^1 \frac{\ln x \, \text{Li}_2(1-x)}{2+x} \, dx$$

I don't know if there is a closed form for this integral or not.

Here is my attempt to solve the integral

\begin{align*} \Omega &= \int_0^1 \frac{\ln x \, \text{Li}_2(1-x)}{2+x} \, dx \\ &= \sum_{n=1}^{\infty} \frac{1}{n^2} \int_0^1 \frac{(1-x)^n \ln x}{2+x} \, dx \\ &= \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{k=0}^{n} \binom{n}{k} (-1)^k \int_0^1 \frac{x^k \ln x}{2+x} \, dx \\ &= \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{k=0}^{n} \binom{n}{k} (-1)^k \sum_{j=k+1}^{\infty} \frac{(-1)^j}{2^j j^2} \\ &= \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{k=0}^{n} \binom{n}{k} \sum_{j=k+1}^{\infty} \frac{(-1)^j}{2^j j^2} \\ &= \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{k=0}^{n} \binom{n}{k} 2^k f(k) \\ &= \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{k=1}^{n} \binom{n}{k} 2^k f(k) + \text{Li}_2 \left(-\frac{1}{2}\right) \frac{\pi^2}{6} \end{align*}

I don't know how to proceed. Is this a possible way to evaluate the integral? I'm not sure if it will work.

Answer 1

Too long for a comment:

Combine the Dilogarithm function reflection formula, $$\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\zeta(2)-\log(x)\log(1-x)$$ and $$\int_0^1 \frac{\displaystyle \log(x)\operatorname{Li}_2(x) }{1-a x}\textrm{d}x=\frac{1}{2}\frac{(\operatorname{Li}_2(a))^2}{a}-2\zeta(2)\frac{\operatorname{Li}_2(a)}{a}+3\frac{\operatorname{Li}_4(a)}{a}, |a|\le 1, \ a\neq0.$$

presented in More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), page 61.

End of story {here is a small note: given the questions you ask, I might think that expecting from you to derive and exploit the related generating functions (you need) constructed on the skeleton of the integral identity, $$\displaystyle \int_{0}^{1}x^{n-1}\log(x)\log^2(1-x)\textrm{d}x=2\zeta(3)\frac{1}{n}+\frac{\pi^2}{3}\frac{H_n}{n}-\frac{H_n^2}{n^2}-\frac{H_n^{(2)}}{n^2}-2\frac{H_n^{(3)}}{n}-2\frac{H_n H_n^{(2)}}{n}$$ is not a difficulty for you - the identity is found on page 33 of the same book}