Polylog integral $\int_{0}^{1}\frac{x\log x\operatorname{Li}_2(x)}{x^2+1}dx$

Question

I am trying to solve this: $\newcommand{\dilog}{\operatorname{Li}_2}$ $$\int_{0}^{1}\frac{x\log x\dilog(x)}{x^2+1}dx$$

My try: use the definition of $\dilog(x)$ as $$\dilog(x)=\sum_{k=1}^{\infty}\frac{x^k}{k^2}$$

Then change the order of integration and summation:$$\sum_{k=1}^{\infty}\frac{1}{k^2}\int_{0}^{1}\frac{x^{k+1}\log x}{x^2+1}dx$$

With some helps from CAS, the final form is:$$\frac{1}{16}\sum_{k=1}^{\infty}\frac{1}{k^2}\left(\Psi^{(1)}{\left(\frac{k}{4}+1\right)-\Psi^{(1)}\left(\frac{k}{4}+\frac{1}{2}\right)}\right)$$ where $\Psi^{(1)}{(s)}$ is the Trigamma function. I am stuck at this step, closed form for reference: $$\frac{G^2}{2}-\frac{41 \pi ^4}{7680}$$ where $G$ is the Catalan constant. My question is how to handle the last sum, and its alternating series means $\frac{1}{16}\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2}\left(\Psi^{(1)}{\left(\frac{k}{4}+1\right)-\Psi^{(1)}\left(\frac{k}{4}+\frac{1}{2}\right)}\right)$, or have any approach more effective? Thank you for your time.

Answer 1

Suppose $\sum_{n=1}^\infty|a_n|$ converges. Then $\left(\sum_{n=1}^\infty a_n\right)^2=\sum_{n=1}^\infty a_n^2+2\sum_{n=1}^\infty a_n\sum_{k=1}^\infty a_{n+k}$.

Replacing $a_n$ with $(-1)^n a_n$ and rearranging, we have $$\sum_{n=1}^\infty a_n\sum_{k=1}^\infty(-1)^k a_{n+k}=\frac12\left[\left(\sum_{n=1}^\infty(-1)^n a_n\right)^2-\sum_{n=1}^\infty a_n^2\right].\tag{*}\label{maineq}$$

Now, using the power series for $\operatorname{Li}_2(x)$ and $1/(1+x^2)$, we get \begin{align} \int_0^1\operatorname{Li}_2(x)\frac{x\log x}{1+x^2}dx&=\sum_{n=1}^\infty\frac1{n^2}\sum_{k=1}^\infty(-1)^{k-1}\int_0^1 x^{n+2k-1}\log x\,dx \\&=\sum_{n=1}^\infty\frac1{n^2}\sum_{k=1}^\infty\frac{(-1)^k}{(n+2k)^2}=S_\text{even}+S_\text{odd}, \end{align} where, using \eqref{maineq} for $a_n=1/n^2$ and $a_n=1/(2n-1)^2$, and known zeta-like sums,

\begin{align} S_\text{even}&=\sum_{n=1}^\infty\frac1{(2n)^2}\sum_{k=1}^\infty\frac{(-1)^k}{(2n+2k)^2}=\frac1{32}\left[\left(\frac{\pi^2}{12}\right)^2-\frac{\pi^4}{90}\right]; \\S_\text{odd}&=\sum_{n=1}^\infty\frac1{(2n-1)^2}\sum_{k=1}^\infty\frac{(-1)^k}{(2n+2k-1)^2}=\frac12\left(G^2-\frac{\pi^4}{96}\right).\end{align}

Answer 2

Another approach

Having seen in the answer Catalan constant and $\pi$, writing $$\frac{x \text{Li}_2(x)}{x^2+1}=\sum_{n=2}^\infty (a_n+b_n)\,x^n$$ with $$a_n=- \left(C \cos \left(\frac{\pi n}{2}\right)+\frac{\pi ^2}{48} \sin \left(\frac{\pi n}{2}\right)\right)\quad\text{and} \quad b_n=\frac{1}{2} (\Phi (-i,2,n+1)+\Phi (i,2,n+1))$$

$$\int_{0}^{1}\frac{x\log{(x)}\text {Li}_2{(x)}}{x^2+1}dx=-\sum_{n=2}^\infty\frac{a_n}{(n+1)^2}-\sum_{n=2}^\infty\frac{b_n}{(n+1)^2}$$

$$-\sum_{n=2}^\infty\frac{a_n}{(n+1)^2}=-C+C^2-\frac{\pi ^2}{192}+\frac{\pi ^4}{2304}$$ I am stuck with the second summation