Background: I'm computing harmonic series using FL expansion. For instance, the following

  • $\frac{\log (1-x)}{x}=\sum _{n=0}^{\infty } 2 (-1)^{n-1} (2 n+1) P_n(2 x-1) \left(\sum _{k=n+1}^{\infty } \frac{(-1)^{k-1}}{k^2}\right)$

Can be used to compute $\sum_{n=1}^\infty\frac{H_n}{n}\left(\frac{(2n)!}{4^n(n!)^2}\right)^2$ (see here). This expansion (and the solution linked) is given by On the interplay between hypergeometric series, Fourier-Legendre expansions and Euler sums by M. Cantarini, J. D’Aurizio.

Problem: As I confront higher weight sums FL expansion of the following are needed:

  • $\large\frac{\text{Li}_2(x)}{x},\frac{\log ^2(1-x)}{x},\frac{\log (x) \log (1-x)}{x}$

I haven't figured out how to compute them based on known results. Any help will be appreciated.

Update: I summarize Jack's result here for sake of others' convenience.

If $f(x)\sim\sum_{n=0}^\infty c_n P_n(2x-1)$, then $$\frac{f(x)}x\sim\sum_{n=0}^\infty(-1)^n (2n+1)\left(\int_0^1 \frac{f(x)}x dx+2\sum_{m=1}^{n}\frac{1}{m}\sum_{k=m}^\infty (-1)^k c_k\right)P_n(2x-1)$$

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    $\begingroup$ With a lot of patience you may understand the structure of the $n$-th derivative and then recall that $\int_{0}^{1}f(x)P_n(2x-1)\,dx$ is related to $\int_{0}^{1}f^{(n)}(x) x^n(1-x)^n\,dx$ via Rodrigues' formula and integration by parts. $\endgroup$ – Jack D'Aurizio Jun 23 '20 at 9:32
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    $\begingroup$ This approach leads to $$ \log(x)\log(1-x)=\left(2-\frac{\pi^2}{6}\right)-\sum_{n\geq 1}\frac{4n+1}{2n^2(2n+1)^2} P_{2n}(2x-1), $$ for instance. $\endgroup$ – Jack D'Aurizio Jun 23 '20 at 9:40

Here a possible alternative to the approach outlined in the comments. Let us assume we start with the FL-expansion of $\text{Li}_2(x),\log^2(1-x)$ or $\log(x)\log(1-x)$ (they all are reasonably simple and related to each other via the dilogarithm reflection formula), for instance

$$\text{Li}_2(x) = (\zeta(2)-1)+\sum_{n\geq 1}\frac{2n+1}{n^2(n+1)^2}\,P_n(2x-1).$$

We have $$ \frac{\text{Li}_2(x)}{x}= \zeta(3)+\sum_{n\geq 1} c_n\,P_n(2x-1) $$ and by Bonnet's recursion formula $$ x P_n(2x-1) = \frac{n+1}{4n+2}P_{n+1}(2x-1)+\frac{1}{2}P_n(2x-1)+ \frac{n}{4n+2}P_{n-1}(2x-1) $$ so the coefficients of the FL-expansion of $\frac{\text{Li}_2(x)}{x}$ are given by the solution of the recurrence

$$ \frac{n}{4n-2} c_{n-1} + \frac{1}{2} c_n + \frac{n+1}{4n+6}c_{n+1} = \frac{2n+1}{n^2(n+1)^2}\tag{R} $$ with the initial conditions $c_0=\zeta(3)$ and $c_1=\pi^2-6-3\zeta(3)$. My version of Mathematica is not able to directly crack this through $\text{RSolve}$, but we may establish once again the superiority of humans over machines through insights, i.e. our chaotic thought process.

By eye-balling it is not difficult to get $$ c_n=(2n+1)\int_{0}^{1}\frac{\text{Li}_2(x)}{x}P_n(2x-1)\,dx = (-1)^n(2n+1)\left[\zeta(3)-H_n\frac{\pi^2}{3}+r(n)\right]$$ with $r(n)\in\mathbb{Q}$, and since $c_n\to 0$ we may expect that the structure of $r(n)$ is close to the one of $2 H_n H_n^{(2)}-H_n^{(3)}$. If we let $c_n=(-1)^n (2n+1) d_n$ in $(R)$ we end up with

$$ -n d_{n-1} + (2n+1) d_n - (n+1) d_{n+1} = 2(-1)^n \left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\tag{R'} $$ and by letting $e_n=d_n-d_{n-1}$, then $f_n = ne_n$, we get $$ n e_n - (n+1) e_{n+1} = 2(-1)^n \left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\tag{R''} $$ $$ f_n - f_{n+1} = 2(-1)^n \left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\tag{R'''} $$ so

$$ f_n = \sum_{k\geq n}2(-1)^k \left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right) $$ $$ e_n = \frac{1}{n}\sum_{k\geq n}2(-1)^k \left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right) $$ $$ d_n = \sum_{m=1}^{n}\frac{1}{m}\sum_{k\geq m}2(-1)^k \left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right) $$ and by dropping the $\zeta(3)$-related part

$$\boxed{ c_n = (-1)^n(2n+1)\sum_{m=1}^{n}\frac{1}{m}\sum_{k\geq m}2(-1)^k \left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right). }$$

A similar approach can be used for finding the FL-expansions of $\text{Li}_n(x)$, since $\text{Li}_{n+1}(x)=\int\text{Li}_n(x)\frac{dx}{x}$ and $\int P_n(x)\,dx=\frac{1}{2n+1}(P_{n+1}-P_n)$. According to my knowledge the FL-expansion of $\text{Li}_4$ is yet unpublished, but I think it is not a big deal to directly share this technique with the community (I hope Marco agrees :/)

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    $\begingroup$ I agree :D (+1) $\endgroup$ – Marco Cantarini Jun 23 '20 at 11:41
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    $\begingroup$ Disregarding the initial terms we have that if $c_n$ is the coefficient of $P_n(2x-1)$ in the FL-expansion of $f(x)$, then the coefficient of $P_n(2x-1)$ in the FL-expansion of $f(x)/x$ is given by $$(-1)^n (2n+1)\sum_{m=1}^{n}\frac{1}{m}\sum_{k\geq m}2(-1)^k c_k $$ $\endgroup$ – Jack D'Aurizio Jun 23 '20 at 13:20
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    $\begingroup$ The difference equation is always the same, since it is just given by Bonnet's recursion formula, what changes is the non-homogenous part (the RHS). $\endgroup$ – Jack D'Aurizio Jun 23 '20 at 13:21

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