# Calculating in closed form $\int_0^1 \log(x)\left(\frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}}\right)^2 \,dx$

What real tools excepting the ones provided here Closed-form of $\int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx$ would you like to recommend? I'm not against them, they might be great, but it seems they didn't lead anywhere for the version $\displaystyle \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx$. Perhaps we can find an approach that covers both cases, also

$$\int_0^1 \log(x) \left(\frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}}\right)^2 \,dx$$ that I would like to calculate.

Might we possibly expect a nice closed form as in the previous case? What do you propose?

EDIT: Thanks David, I had to modify it a bit to fix the convergence issue. Also, for the previous question there is already a 300 points bounty offered for a full solution with all steps clearly explained.

• We have $~I=J'(0),~$ where $~J(k)=\displaystyle\int_0^1\frac{1-x^k}{1-x^2}\cdot\text{Li}_2^2(x)~dx,~$ which is reminiscent of Euler's integral formula for generalized harmonic numbers. – Lucian Jul 24 '15 at 23:17

## 1 Answer

We have: $$\sum_{k=1}^{n-1}\frac{1}{k^2(n-k)^2}=\frac{1}{n^2}\sum_{k=1}^{n-1}\left(\frac{1}{k}+\frac{1}{n-k}\right)=\frac{2H_{n-1}^{(2)}}{n^2}+\frac{4H_{n-1}}{n^3}$$ so: $$\text{Li}_2(x)^2 = \sum_{n\geq 2}\left(\frac{2H_{n-1}^{(2)}}{n^2}+\frac{4H_{n-1}}{n^3}\right) x^n\tag{1}$$ and since: $$\int_{0}^{1}\frac{x^n \log x}{1-x^2}\,dx = -\sum_{m\geq 0}\frac{1}{(n+2m+1)^2}\tag{2}$$ we have: $$\int_{0}^{1}\log(x)\left(\frac{\text{Li}_2(x)}{\sqrt{1-x^2}}\right)^2\,dx = -\sum_{n\geq 2}\left(\frac{2H_{n-1}^{(2)}}{n^2}+\frac{4H_{n-1}}{n^3}\right)\sum_{m\geq 0}\frac{1}{(n+2m+1)^2}\tag{3}$$ and the problem boils down to the computation of a complicated Euler sum.

In order to perform partial summation, it is useful to recall that: $$\sum_{n=1}^{N}\frac{2H_{n-1}^{(2)}}{n^2}=\left(H_{N}^{(2)}\right)^2-H_{N}^{(4)},$$ $$\sum_{n=1}^{N}\frac{H_{n-1}}{n^3} = H_N^{(3)}H_{N-1}-\sum_{n=1}^{N-1}\frac{H_{n}^{(3)}}{n}.\tag{4}$$

• I think it's a good start. (+1) – user 1357113 Jul 24 '15 at 22:46
• @Chris'ssistheartist: about the bounty, are you so confident that a nice closed form for the $\phantom{}_4 F_3$ value shown by Claude Leibovici really exists, to waste $300$ reputation points? Sometimes, the most cunning approach does not work simply because it cannot work. Maybe that is one of these cases. – Jack D'Aurizio Jul 24 '15 at 23:14
• Yes, it definitely exists, and that form can be expressed in terms of trilogarithm. However, I would have liked to see a complete solution there. – user 1357113 Jul 24 '15 at 23:17
• @Chris'ssistheartist: why not just provide yours? – Jack D'Aurizio Jul 24 '15 at 23:18
• It's a known fact the connection with the trilogarithm, here math.stackexchange.com/questions/918680/…, but I would have liked to see a complete approach of the problem. – user 1357113 Jul 24 '15 at 23:22