2

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.

Community
  • 1
user 1591719
  • 44,987

1 Answers1

2

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: $$ \sum_{n=1}^{N}\frac{2H_{n-1}^{(2)}}{n^2}=\left(H_{N}^{(2)}\right)^2-H_{N}^{(4)},$$ $$\sum_{n=1}^{N}\frac{H_{n-1}}{n^3} = H_N^{(3)}H_{N-1}-\sum_{n=1}^{N-1}\frac{H_{n}^{(3)}}{n}.\tag{4}$$

Jack D'Aurizio
  • 361,689
  • I think it's a good start. (+1) – user 1591719 Jul 24 '15 at 22:46
  • 1
    @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 1591719 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 http://math.stackexchange.com/questions/918680/closed-form-for-the-imaginary-part-of-textli-3-big-frac1i2-big/920995#920995, but I would have liked to see a complete approach of the problem. – user 1591719 Jul 24 '15 at 23:22
  • All right, but I still do not get it. If it is just a matter of collecting pieces from there and there, why don't you do it by yourself? – Jack D'Aurizio Jul 24 '15 at 23:30
  • From what I saw in that post it doesn't seem to be anything about collecting pieces from there and there at that stage, but I think one needs a clever strategy to get the answer. I'm just curious to see the full solution and I just offered 300 points bounty for that. – user 1591719 Jul 24 '15 at 23:33
  • Actually I've construated the integral to get a deeper knowledge about the connection between that hypergeom series and that trilogarithm. On the other hand, if someone could find a closed-form of that hypergeom series, that would be the best. I've worked on this problem a lot. I've posted another related integral problem. – user153012 Jul 26 '15 at 09:11