Evaluating the challenging sum $\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}$.

Question

I managed to evaluate the sum, my approach can be found $\underline{\operatorname{below as an answer}}$, I'd truly appreciate if any of you could share new methods to evaluate this series, thank you.

The following are the short proofs of the other series I encountered during my approach.

$$\sum _{k=1}^{\infty }\frac{x^k}{4^k}\binom{2k}{k}=\frac{1}{\sqrt{1-x}}-1\tag{$\ast$}$$ $$\sum _{k=1}^{\infty }\frac{x^k}{k\:4^k}\binom{2k}{k}=\:-2\ln \left(1+\sqrt{1-x}\right)+2\ln \left(2\right)$$

$$\sum _{k=1}^{\infty }\frac{1}{k\:4^k}\binom{2k}{k}=\:2\ln \left(2\right)$$

On $\left(\ast\right)$ perform the following manipulations. $$-\sum _{k=1}^{\infty }\frac{1}{4^k}\binom{2k}{k}\int _0^1x^{k-1}\ln \left(x\right)\:dx=-\int _0^1\frac{\ln \left(x\right)\left(1-\sqrt{1-x}\right)}{x\sqrt{1-x}}\:dx$$ $$\sum _{k=1}^{\infty }\frac{1}{k^2\:4^k}\binom{2k}{k}=-2\int _0^1\frac{\ln \left(1-x^2\right)}{1+x}\:dx$$ $$=-2\int _0^1\frac{\ln \left(1-x\right)}{1+x}\:dx-2\int _0^1\frac{\ln \left(1+x\right)}{1+x}\:dx$$ $$=2\operatorname{Li}_2\left(\frac{1}{2}\right)-\ln ^2\left(2\right)$$

$$\sum _{k=1}^{\infty }\frac{1}{k^2\:4^k}\binom{2k}{k}=\zeta \left(2\right)-2\ln ^2\left(2\right)$$

Answer 1

First part.

Evaluating $$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx$$

$$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx=\frac{1}{2}\int _0^1\ln ^2\left(t\right)\left(\int _0^1\frac{1}{\left(1-tx\right)\sqrt{1-x^2}}\:dx\right)dt$$ $$=\frac{\pi }{4}\int _0^1\frac{\ln ^2\left(t\right)}{\sqrt{1-t^2}}\:dt+\frac{1}{2}\int _0^1\frac{\ln ^2\left(t\right)\arcsin \left(t\right)}{\sqrt{1-t^2}}\:dt$$ $$=\frac{\pi }{4}\int _0^{\frac{\pi }{2}}\ln ^2\left(\sin \left(x\right)\right)\:dx+\frac{1}{2}\int _0^{\frac{\pi }{2}}x\ln ^2\left(\sin \left(x\right)\right)\:dx$$ Using the Beta function for the first integral and using this result for the $2$nd one proved there we conclude that: $$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx=\frac{41}{64}\zeta \left(4\right)+\frac{1}{2}\operatorname{Li}_4\left(\frac{1}{2}\right)+\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{48}\ln ^4\left(2\right)$$

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Second part.

Finding $$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}$$

By employing $1+\sum _{k=1}^{\infty }\frac{x^{2k}}{4^k}\binom{2k}{k}=\frac{1}{\sqrt{1-x^2}}$ on the previous integral we have: $$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx=\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x}\:dx+\sum _{k=1}^{\infty }\frac{1}{4^k}\binom{2k}{k}\int _0^1x^{2k-1}\operatorname{Li}_3\left(x\right)\:dx$$ $$=\zeta \left(4\right)+\sum _{k=1}^{\infty }\frac{1}{4^k}\binom{2k}{k}\left(\frac{1}{2k}\zeta \left(3\right)-\frac{1}{2k}\int _0^1x^{2k-1}\operatorname{Li}_2\left(x\right)\:dx\right)$$ $$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx=\zeta \left(4\right)+\frac{1}{2}\zeta \left(3\right)\sum _{k=1}^{\infty }\frac{1}{k\:4^k}\binom{2k}{k}-\frac{1}{4}\zeta \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^2\:4^k}\binom{2k}{k}$$ $$+\frac{1}{8}\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}$$ $$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}=8\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx-3\zeta \left(4\right)-8\ln \left(2\right)\zeta \left(3\right)-4\ln ^2\left(2\right)\zeta \left(2\right)$$ Using the result found on the first part we finalize. $$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}=\frac{17}{8}\zeta \left(4\right)+4\operatorname{Li}_4\left(\frac{1}{2}\right)-8\ln \left(2\right)\zeta \left(3\right)+4\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{6}\ln ^4\left(2\right)$$

Answer 2

\begin{align*}J&= \sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}\\ &=\frac{2}{\pi}\sum_{k=1}^\infty \frac{1}{k^3}\int_0^1 \frac{1-x^{2k}}{1-x}\left(\int_0^{\frac{\pi}{2}} \sin^{2k} t\,dt\right)dx\\ &=\frac{2}{\pi}\int_0^1 \frac{1}{1-x} \left(\int_0^{\frac{\pi}{2}}\left( \text{Li}_3\left(\sin^2 t\right)-\text{Li}_3\left(x^2\sin^2 t\right)\right)dt\right)dx\\ &\overset{\text{IBP}}=-\frac{2}{\pi}\int_0^{\frac{\pi}{2}}\Big[\ln(1-x)\left( \text{Li}_3\left(\sin^2 t\right)-\text{Li}_3\left(x^2\sin^2 t\right)\right)\Big]_0^1 dt-\\& \frac{4}{\pi}\int_0^1 \frac{\ln(1-x)}{x}\left(\int_0^{\frac{\pi}{2}} \text{Li}_2\left(x^2\sin^2 t\right) dt\right)dx\\ &=-\frac{4}{\pi}\int_0^1 \frac{\ln(1-x)}{x}\left(\int_0^{\frac{\pi}{2}} \text{Li}_2\left(x^2\sin^2 t\right) dt\right)dx\\ &\overset{\text{IBP}}=\frac{4}{\pi}\text{Li}_2(1)\int_0^{\frac{\pi}{2}}\text{Li}_2\left(\sin^2 t\right) dt+\frac{8}{\pi}\int_0^1 \frac{\text{Li}_2(x)}{x}\left(\int_0^{\frac{\pi}{2}} \ln(1-x^2 \sin^2 t)\right)dx\\ &\overset{\text{IBP}}=\frac{4}{\pi}\text{Li}_2(1)\int_0^{\frac{\pi}{2}}\text{Li}_2\left(\sin^2 t\right) dt+\frac{8}{\pi}\text{Li}_3(1)\times 2\int_0^{\frac{\pi}{2}} \ln\left(\cos t\right)dt+\frac{16}{\pi} \int_0^1 \text{Li}_3(x)\left(\int_0^{\frac{\pi}{2}} \frac{x\sin^2 t}{1-x^2\sin^2 t}\right)dx\\ &=\frac{2\pi}{3}\int_0^{\frac{\pi}{2}}\text{Li}_2\left(\sin^2 t\right) dt-8\zeta(3)\ln 2+\frac{16}{\pi}\int_0^1 \text{Li}_3(x)\left(\int_0^{\frac{\pi}{2}} \frac{x\sin^2 t}{1-x^2\sin^2 t}\right)dx\\ &=\frac{2\pi}{3}\int_0^{\frac{\pi}{2}}\text{Li}_2\left(\sin^2 t\right) dt-8\zeta(3)\ln 2+\frac{16}{\pi}\int_0^1 \text{Li}_3(x)\left[\frac{\arctan\left(\tan t\sqrt{1-x^2}\right)}{x\sqrt{1-x^2}}-\frac{t}{x}\right]_0^{\frac{\pi}{2}} dx\\ &=\frac{2\pi}{3}\int_0^{\frac{\pi}{2}}\text{Li}_2\left(\sin^2 t\right) dt-8\zeta(3)\ln 2-8\int_0^1 \frac{x\left(\text{Li}_4(x)-\text{Li}_4(1)\right)}{(1-x^2)^{\frac{3}{2} }}dx\\ \end{align*}

\begin{align*} K&=\int_0^{\frac{\pi}{2}} \text{Li}_2\left(\sin^2 t\right)dt\\ &\overset{\text{IPP}}=\Big[t\text{Li}_2\left(\sin^2 t\right)\Big]_0^{\frac{\pi}{2}}+4\int_0^{\frac{\pi}{2}}\frac{t\ln\left(\cos t\right)}{\tan t}dt\\ &\overset{\text{IPP}}=\frac{\pi^3}{12}+4\Big[t\ln\left(\sin t\right)\ln\left(\cos t\right)\Big]_0^{\frac{\pi}{2}}-\\& 4\int_0^{\frac{\pi}{2}} \ln\left(\sin t\right)\ln\left(\cos t\right)dt+4\int_0^{\frac{\pi}{2}} u\tan u\ln\left(\sin u\right)du\\ &\overset{t=\frac{\pi}{2}-u}=\frac{\pi^3}{12}-4\int_0^{\frac{\pi}{2}} \ln\left(\sin t\right)\ln\left(\cos t\right)dt+4\int_0^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-t\right)\ln\left(\cos t\right)}{\tan t}dt\\ &=\frac{\pi^3}{12}-4\int_0^{\frac{\pi}{2}} \ln\left(\sin t\right)\ln\left(\cos t\right)dt+2\pi\int_0^{\frac{\pi}{2}} \frac{\ln\left(\cos t\right)}{\tan t}dt+\left(\frac{\pi^3}{12}-K\right)\\ K&=\frac{\pi^3}{12}-2\int_0^{\frac{\pi}{2}} \ln\left(\sin t\right)\ln\left(\cos t\right)dt+\pi\int_0^{\frac{\pi}{2}} \frac{\ln\left(\cos u\right)}{\tan u}du\\ &\overset{t=\cos^2 u}=\frac{\pi^3}{12}-2\int_0^{\frac{\pi}{2}} \ln\left(\sin t\right)\ln\left(\cos t\right)dt+\frac{\pi}{4}\int_0^1 \frac{\ln t}{1-t}dt\\ &=\frac{\pi^3}{24}-2\int_0^{\frac{\pi}{2}} \ln\left(\sin t\right)\ln\left(\cos t\right)dt\\ &=\frac{\pi^3}{24}+\frac{1}{2}\int_0^{\frac{\pi}{2}} \ln^2\left(\tan t \right)dt-\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln^2\left(\sin u\cos u\right)du\\ &\overset{x=\tan t,x=2u}=\frac{\pi^3}{24}+\frac{1}{2}\int_0^{\infty}\frac{\ln^2 x}{1+x^2}dx-\frac{1}{4}\int_0^\pi \ln^2\left(\frac{\sin x}{2}\right)dx\\ &=\frac{5\pi^3}{48}-\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{\sin x}{2}\right)dx-\frac{1}{4}\int_{\frac{\pi}{2}}^{\pi}\ln^2\left(\frac{\sin u}{2}\right)du\\ &\overset{x=\pi-u}=\frac{5\pi^3}{48}-\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{\sin x}{2}\right)dx\\ &=\frac{5\pi^3}{48}-\frac{\pi\ln^2 2}{4}+\ln 2\int_0^{\frac{\pi}{2}} \ln\left(\sin x\right)dx-\frac{1}{2}\int_0^{\frac{\pi}{2}} \ln^2\left(\sin x\right)dx \end{align*} \begin{align*} &=\frac{5\pi^3}{48}-\frac{3\pi\ln^2 2}{4}-\frac{1}{2}\int_0^{\frac{\pi}{2}} \ln^2\left(\sin x\right)dx\\ &=\frac{5\pi^3}{48}-\frac{3\pi\ln^2 2}{4}-\frac{1}{4}\left(\int_0^{\frac{\pi}{2}} \ln^2\left(\sin x\right)dx+\int_0^{\frac{\pi}{2}} \ln^2\left(\sin u\right)du\right)\\ &\overset{x=\frac{\pi}{2}-x}=\frac{5\pi^3}{48}-\frac{3\pi\ln^2 2}{4}-\frac{1}{4}\left(\int_0^{\frac{\pi}{2}} \ln^2\left(\sin x\right)dx+\int_0^{\frac{\pi}{2}} \ln^2\left(\cos x\right)dx\right)\\ &=\frac{5\pi^3}{48}-\frac{3\pi\ln^2 2}{4}-\frac{1}{4}\left(\int_0^{\frac{\pi}{2}} \ln^2\left(\tan x\right)dx+2\int_0^{\frac{\pi}{2}}\ln\left(\sin x\right)\ln\left(\cos x\right)dx\right)\\ &=\frac{7\pi^3}{96}-\frac{3\pi\ln^2 2}{4}-\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln\left(\sin x\right)\ln\left(\cos x\right)dx\\ &=\frac{7\pi^3}{96}-\frac{3\pi\ln^2 2}{4}-\frac{1}{2}\left(\frac{\pi^3}{48}-\frac{J}{2}\right)\\ &=\frac{\pi^3}{16}-\frac{3\pi\ln^2 2}{4}+\frac{J}{4}\\ K&=\boxed{\frac{\pi^3}{12}-\pi\ln^2 2} \end{align*}

\begin{align} L&=\int_0^1 \frac{x\left(\text{Li}_4(x)-\text{Li}_4(1)\right)}{(1-x^2)^{\frac{3}{2} }}dx\\ &\overset{x=\sin t}=\int_0^{\frac{\pi}{2}}\frac{\sin t\left(\text{Li}_4(\sin t)-\text{Li}_4(1)\right)}{\cos^2 t}dt\\ &\overset{\text{IPP}}=\Big[\frac{1}{\cos x}\big(\text{Li}_4(\sin t)-\text{Li}_4(1)\big)\Big]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \frac{\text{Li}_3(\sin t)}{\sin t}dt\\ &=\frac{\pi^4}{90}-\int_0^{\frac{\pi}{2}} \frac{\text{Li}_3(\sin t)}{\sin t}dt\\ &=\frac{\pi^4}{90}-\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_0^1 \frac{\ln^2 a}{1-a\sin t}da\right)dt\\ &=\frac{\pi^4}{90}-\int_0^1 \left[\frac{\arctan \left(\frac{\tan\left(\frac{t}{2}\right)-a}{\sqrt{1-a^2}}\right)}{\sqrt{1-a^2}}\right]_0^{\frac{\pi}{2}}da\\ &=\frac{\pi^4}{90}-\int_0^1 \frac{\arctan\left(\sqrt{\frac{1-a}{1+a}}\right)\ln^2 a}{\sqrt{1-a^2}}da-\int_0^1 \frac{\arctan\left(\frac{b}{\sqrt{1-b^2}}\right)\ln^2 b}{\sqrt{1-b^2}}db\\ &\overset{a=\frac{1-\tan^2\left(\frac{u}{2}\right)}{1+\tan^2\left(\frac{u}{2}\right)},b=\sin t}=\frac{\pi^4}{90}-\frac{1}{2}\int_0^{\frac{\pi}{2}}u\ln^2\left(\cos u\right)du-\int_0^{\frac{\pi}{2}}t\ln^2\left(\sin t\right)dt\\ &\overset{t=\frac{\pi}{2}-u}=\frac{\pi^4}{90}-\frac{1}{2}\int_0^{\frac{\pi}{2}}t\ln^2\left(\sin t\right)dt-\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\sin t\right)dt\\ &=\frac{\pi^4}{90}-\frac{1}{2}\int_0^{\frac{\pi}{2}}t\ln^2\left(\sin t\right)dt-\frac{\pi}{8}\left(\int_0^{\frac{\pi}{2}}\ln^2\left(\tan t\right)dt+2\int_0^{\frac{\pi}{2}}\ln\left(\sin t\right)\ln\left(\cos t\right)dt\right)\\ &=-\frac{13\pi^4}{2880}-\frac{1}{2}\int_0^{\frac{\pi}{2}}t\ln^2\left(\sin t\right)dt+\frac{\pi}{8}\left(K-\frac{\pi^3}{24}\right)\\ &=-\frac{13\pi^4}{2880}-\frac{1}{2}\int_0^{\frac{\pi}{2}}t\ln^2\left(\sin t\right)dt+\frac{\pi}{8}\left(\frac{\pi^3}{24}-\pi\ln^2 2\right)\\ &=\boxed{\frac{\pi^4}{1440}-\frac{1}{2}\int_0^{\frac{\pi}{2}}t\ln^2\left(\sin t\right)dt-\frac{\pi^2\ln^2 2}{8}} \end{align}

From tough definite integral: $\int_0^\frac{\pi}{2}x\ln^2(\sin x)~dx$

\begin{align}\int_0^{\frac{\pi}{2}}t\ln^2\left(\sin t\right)dt&=\operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{1}{24}\ln^42+\frac{\pi^2}{12}\ln^22-\frac{{19}\pi^4}{2880}\end{align}

Therefore,

\begin{align}L&=\frac{23\pi^4}{5760}-\frac{\pi^2\ln^22}{6}-\frac{\ln^4 2}{48}-\frac{1}{2}\text{Li}_4\left(\frac{1}{2}\right)\\ J=&\boxed{\frac{17\pi^4}{720}+\frac{2\pi^2\ln^2 2}{3}+\frac{\ln^4 2}{6}-8\zeta(3)\ln 2+4\text{Li}_4\left(\frac{1}{2}\right)} \end{align}