how to prove identity for nth derivative of $(\text{arctanh}(x))^j$?

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I worked on integral problem and I got that $$ \int_0^1 \frac{x^n}{\ln \left(\frac{1-x}{1+x} \right) } dx=-\frac{2}{(n+1)!}\sum_{j=1}^{n+1}F(n,j) \eta'(-j)$$ where $\eta(x)$ is Dirichlet eta function and $$ F(n,j)=\sum_{p=1}^{n} \binom{n-1}{p-1} \sum_{m=j}^{n+1} \binom{m}{j} |s(n+1,m)| ((-p)^{m-j}+(1-p)^{m-j})$$ where $s(n,m)$ is Stirling numbers of first kind

and I searched for simple identity for $F(n,j)$ and I got this five identities

$$ F(n,j)=0 , n\equiv j \mod2$$ $$ F(n,n+1)=2^n$$ $$ F(n,n-1)=2^{n-1} \binom{n+1}{3}$$ $$ F(2n,1)=(2n)!$$ $$ F(2n+1,2)=2(2n+1)!\sum_{k=0}^n \frac{1}{2k+1}$$ and finally I found that $$ F(n,j)=\frac{2^{j-1}}{j!} \lim_{x\to 0} \frac{d^{n+1}}{dx^{n+1}} (\text{arctanh}(x))^j$$

which means that

$$ \lim_{x\to 0} \frac{d^{n+1}}{dx^{n+1}} (\text{arctanh}(x))^j=j! 2^{1-j}\sum_{p=1}^{n} \binom{n-1}{p-1} \sum_{m=j}^{n+1} \binom{m}{j} |s(n+1,m)| ((-p)^{m-j}+(1-p)^{m-j})$$

then I got explicit formula for A049214 , A049215 and A049216 now my questions

How to prove the last formula ?

is there simple formula for that double summation ?

How to prove five identities for $F(n,j)$ that I put ?

Answer 1

The derivative of the arctanh can be evaluated in terms of the Stirling numbers by following the suggestion of Tyma Gaidash.
Start from the generating function of the Stirling numbers of the first kind, $$\sum_{k=m}^\infty \frac{1}{k!}S_k^{(m)}\frac{(2x)^k}{(1-x)^k}=\frac{1}{m!}\ln^m\left(1+\frac{2x}{1-x}\right)=\frac{2^m}{m!}\operatorname{arctanh}^m x,$$ and differentiate using the identity $$\lim_{x\rightarrow 0}\frac{d^{n}}{dx^{n}}\frac{x^k}{(1-x)^k}=\begin{cases} n!{{n-1}\choose{k-1}}&\text{if}\;\;n\geq k,\\ 0&\text{if}\;\;n<k \end{cases}$$ to obtain (for $1\leq m\leq n+1$) $$\lim_{x\rightarrow 0}\frac{d^{n+1}}{dx^{n+1}}\operatorname{arctanh}^m x=\frac{m!(n+1)!}{2^m}\sum_{k=m}^{n+1}\frac{2^k}{k!}{{n}\choose{k-1}}S_k^{(m)}.$$ So this is a single finite sum, instead of the double sum of the OP.

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To reduce the double sum of the OP to my single sum, note that $$\sum_{p=1}^n{{n-1}\choose{p-1}}\bigl((-p)^q+(1-p)^q\bigr)=(-1)^q\sum_{p=1}^n{n\choose p}p^q,$$ so that $$F(n,m)= (-1)^{n+m+1}\sum_{k=m}^{n+1} \binom{k}{j} S_{n+1}^{(k)} \sum_{p=1}^n{n\choose p}p^{k-m}.$$ This needs more work to be reduced to a single sum.

Answer 2

Recall that $\mathrm{arctanh}(x) = \tfrac12 \log\tfrac{1+x}{1-x}$, which implies $$[x^{n+1}]\ (\mathrm{arctanh}(x))^j = j![x^{n+1}y^j]\ \exp\big(\frac12 y \log\frac{1+x}{1-x} \big) = j![x^{n+1}y^j]\ \big(\frac{1+x}{1-x}\big)^{y/2}$$ $$=j![x^{n+1}y^j]\ \big(1+\frac{2x}{1-x}\big)^{y/2} = j![y^j] \sum_{p=1}^{n+1} \binom{y/2}{p}2^p\binom{n}{p-1}$$ $$ = j!\sum_{p=1}^{n+1} s(p,j)\frac{2^{p-j}}{p!}\binom{n}{p-1}.$$