Sum with reciprocal central binomial and harmonic number $\sum_{n=1}^{\infty}\frac{H_{2n}}{n^2 C_{2n}^{n}}$

Question

I am trying to calculate this sum,and below is my attempt:
$$\displaystyle{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{\mathrm{2}{n}} }{{n}^{\mathrm{2}} \left({C}_{\mathrm{2}{n}} ^{{n}} \right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{n}\left({C}_{\mathrm{2}{n}} ^{{n}} \right)}\left(\frac{{H}_{\mathrm{2}{n}} }{\mathrm{2}{n}}\right)=-\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({C}_{\mathrm{2}{n}} ^{{n}} \right)}\left(\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}−\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−{x}\right){dx}\right) \\=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−{x}\right){dx}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}−\mathrm{1}} }{{n}\left({C}_{\mathrm{2}{n}} ^{{n}} \right)}\right)=−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−{x}\right)\frac{\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{dx} \\{x}=\mathrm{2sin}\left({t}\right)\Rightarrow{t}=\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{2}}\right)\Rightarrow{dt}=\frac{{dx}}{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }} \\\Rightarrow{S}=−\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} {t}\mathrm{ln}\left(\mathrm{1}−\mathrm{2sin}\left({t}\right)\right){dt} \\ $$
But at this step, I don't know how to process further. Can I ask a hint or another approach? Thank you so much.

Answer 1

Hint

I did not check your steps, but, if the problem is $$I=\int t \,\log (1-2 \sin (t))\,dt$$ one integration by parts gives $$I=\frac{1}{2} t^2\, \log (1-2 \sin (t))+\int \frac{t^2 \cos (t)}{1-2 \sin (t)}\,dt$$ Using Euler representation, let $e^{it}=x$ $$\int\frac{t^2 \cos (t)}{1-2 \sin (t)}\,dt=\frac 12 \int\frac{\left(x^2+1\right) \log ^2(x)}{x \left(x^2-i x-1\right)}\,dx $$ Let $(a,b)$ be the roots of the quadratic $$\frac {x^2+1}{x(x-a)(x-b)}=\frac{1}{a b x}+\frac{a^2+1}{a (a-b) (x-a)}+\frac{b^2+1}{b (b-a) (x-b)}$$ The first integral is more than simple and what remains are two integrals $$I(c)=\int \frac{\log ^2(x)}{x-c}\,dx$$ Again,two integrations by parts $$I(c)=\log ^2(x) \log \left(1-\frac{x}{c}\right)+2 \log (x) \text{Li}_2\left(\frac{x}{c}\right)-2 \text{Li}_3\left(\frac{x}{c}\right)$$

I let you recombining everything, using $$a=-\frac{1}{2} \left(\sqrt{3}-i\right)\qquad \qquad b=\frac{1}{2} \left(\sqrt{3}+i\right)$$ go back to $t$ if you wish and use the bounds.

The result is nice.