Simpler proof of identity $ \sum_{n=1}^{\infty} \frac{H_n^2}{n2^n} = \frac{7}{8} \zeta(3) $

Question

The identity in question can be obtained by first proving $$ \sum_{n=1}^{\infty} \frac{H_n^2}{n} z^n = - \frac{1}{3} \log^3(1-z) -\log(1-z) \text{Li}_2(z) + \text{Li}_3(z), \hspace{0.5cm} |z|<1, $$ which can be established by differentiating the expression and applying some algebra coupled with the observation that $ H_n^2 = 2\sum_{k=1}^n \frac{H_k}{k} - \sum_{k=1}^n \frac{1}{k^2} $ for all $n$. Letting $z=\frac{1}{2}$ and using the well-known values $$ \text{Li}_2(\frac{1}{2}) = \frac{\pi^2}{12}-\frac{1}{2}\log^22, \hspace{1cm} \text{Li}_3(\frac{1}{2}) = \frac{1}{6}\log^32 -\frac{\pi^2}{12} \log2+\frac{7}{8} \zeta(3) $$ we obtain $ \sum_{n=1}^{\infty} \frac{H_n^2}{n2^n} = \frac{7}{8} \zeta(3) $ after lots of cancellation. This approach is quite convoluted so I'm wondering if there is a more direct way of proving this identity, something that will bypass the need to use the values of $\text{Li}_2(\frac{1}{2})$ and $\text{Li}_3(\frac{1}{2})$. Thanks

Answer 1

\begin{align} \sum_{n=1}^\infty\frac{H_n^2}{n2^n}&=\sum_{n=1}^\infty\frac{H_n}{2^n}\left(-\int_0^1 x^{n-1} \ln(1-x)dx\right)\\ &=-\int_0^1 \frac{\ln(1-x)}{x}\sum_{n=1}^\infty(x/2)^n H_ndx\\ &=\int_0^1\frac{\ln(1-x)\ln(1-x/2)}{x(1-x/2)}dx\qquad x\to 1-x\\ &=2\int_0^1\frac{\ln(x)\ln\left(\frac{1+x}{2}\right)}{1-x^2}dx\qquad x\to (1-x)/(1+x)\\ &=\int_0^1\frac{\ln\left(\frac{1+x}{1-x}\right)\ln(1+x)}{x}dx\\ &=\frac14\underbrace{\int_0^1\frac{\ln^2(1-x^2)}{x}dx}_{1-x^2\to x}-\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}dx}_{1-x\to x}+\frac34\underbrace{\int_0^1\frac{\ln^2\left(\frac{1-x}{1+x}\right)}{x}dx}_{(1-x)/(1+x)\to x}\\ &=\frac18\int_0^1\frac{\ln^2(x)}{1-x}dx-\int_0^1\frac{\ln^2(x)}{1-x}dx+\frac32\int_0^1\frac{\ln^2(x)}{1-x^2}dx\\ &=-\frac78\int_0^1\frac{\ln^2(x)}{1-x}dx+\frac32\int_0^1 \ln^2(x)\left(\frac{1}{1-x}-\frac{x}{1-x^2}\right)dx\\ &=\frac58\int_0^1\frac{\ln^2(x)}{1-x}dx-\frac32\underbrace{\int_0^1 \frac{x\ln^2(x)}{1-x^2}dx}_{\sqrt{x}\to x}\\ &=\frac58\int_0^1\frac{\ln^2(x)}{1-x}dx-\frac{3}{16}\int_0^1\frac{\ln^2(x)}{1-x}dx\\ &=\frac{7}{16}\int_0^1\frac{\ln^2(x)}{1-x}dx\\ &=\frac{7}{8}\zeta(3) \end{align}