Generating function of series $\sum\limits_{n=1}^\infty H_n^{(m)}\binom{2n}{n}x^n$?

Question

I wish to find the generating function of infinite series $$\sum\limits_{n=1}^\infty H_n^{(m)}\binom{2n}{n}x^n= ?,$$ where $H_n^{(m)}$ is generalized harmonic number defined by $$H_n^{(m)}:=\sum\limits_{k=1}^n \frac{1}{k^m}$$ and $H_n=H_n^{(1)}$ is classical harmonic number. It is known that for $m=1$, $$\sum\limits_{n=1}^\infty H_n\binom{2n}{n}x^n=\frac{2}{\sqrt{1-4x}}\log\left(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\right),$$ see https://cs.uwaterloo.ca/journals/JIS/VOL19/Chen/chen21.pdf. But for any positive integer $m$, what is the generating function?

Answer 1

Well, you may exploit the fact that $\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}(2\cos\theta)^{2n}\,d\theta$. Since $$ \sum_{n\geq 1}H_n^{(m)}z^{2n} = \frac{\text{Li}_m(z^2)}{1-z^2} $$ for any $|x|<\frac{1}{4}$we have $$ \sum_{n\geq 1}H_n^{(m)}\binom{2n}{n}x^n =\frac{2}{\pi} \int_{0}^{\pi/2}\frac{\text{Li}_m(4x\cos^2\theta)}{1-4x\cos^2\theta}\,d\theta =\frac{2}{\pi}\int_{0}^{+\infty}\frac{\text{Li}_m\left(\frac{4x}{1+t^2}\right)}{(1-4x)+t^2}\,dt$$ which equals

$$\frac{2}{\pi\sqrt{1-4x}}\int_{0}^{\pi/2}\text{Li}_m\left(\tfrac{4x}{1+(1-4x)\tan^2\varphi}\right)\,d\varphi=\frac{2}{\pi\sqrt{1-4x}}\int_{0}^{\pi/2}\text{Li}_m\left(\tfrac{4x\cos^2\varphi}{1-4x\sin^2\varphi}\right)\,d\varphi.$$ This is not a trivial integral for $m\geq 2$, but it is approachable through integration by parts and known Fourier series.

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Addendum: in this recent article (section $3$, Twisted hypergeometric functions) Marco Cantarini and I have proved that both $\text{Li}_2$ and $\text{Li}_3$ have manageable (shifted) Fourier-Legendre expansions, and it is not a big deal to show that the same occurs for Fourier-Chebyshev expansions, so the cases $m=2$ and $m=3$ can be considered as known.