How to calculate $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}h_n}{n^3}$ where $h_n=\sum_{k=1}^{n}\frac{1}{2k-1}$?

Question

Well known is the relationship $$\displaystyle\sum_{n=1}^{\infty}\frac{h_n}{n^3}=\operatorname{Li}_4(\frac{1}{2})+\frac{1}{24}\ln^42-\frac{\pi^2}{24}\ln^22+\frac{7}{8}\zeta(3)\ln2-\frac{53\pi^4}{5760}$$ Where$$h_n=\displaystyle\sum_{k=1}^{n}\frac{1}{2k-1}$$ Just consider: $$\displaystyle\int_0^1\frac{\log\,x}{x}\left[\log\frac{1-x}{1+x}\right]^2dx=-8\sum_{n=1}^{\infty}\frac{h_n}{n^3} \text{ put }\,x=\frac{1-t}{1+t}$$ But how to calculate $$S=\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}h_n}{n^3}$$ We have $$\displaystyle\int_0^{\pi/4}{x}[\log\,\tan\,x]^2dx=\frac{1}{4}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}h_n}{n^3}$$ But how deduct the integral?

Answer 1

I already gave an integral representation for a more general case for your first series yet here is a more general case which makes the evaluation of both series easier to evaluate

$$\sum_{n=1}^{\infty} \frac{t^n}{n^q}\sum_{k=1}^{n}\frac{x^n}{(2k-1)^p} = \frac{(-1)^{q-1}\sqrt{x t}}{2\Gamma(q)} \int_{0}^{1} \frac{(\ln(u))^{q-1} ( \rm Li_p( \sqrt{x t u}) - \rm Li_p(-\sqrt{x t u})) }{\sqrt{u}(1-xu)}du,$$

where $\rm Li_p(z)$ is the polylogarithm function. In your case, subs $q=3, p=1, x=1$ and $t=-1$ in the above formula and try to evaluate the resulted integral. See related techniques and problems I, II.