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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?

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    $\begingroup$ See my answer under user name "mathematics" for the series you already had an answer for it. I'll try to give an integral representation for the one you are asking for. $\endgroup$ – Mhenni Benghorbal Dec 25 '14 at 20:06
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    $\begingroup$ I don't know if you already know the answer, but anyway the closed form turns out to be: $$ S = \frac{151 \pi^4}{2880} + \frac{\pi^2}{6} \ln^2 2 - \frac{ \ln^4 2}{6} - \frac{7}{2} \zeta(3) \ln2 - 4 \text{Li}_4\left( \frac12 \right).$$ $\endgroup$ – nospoon Jun 26 '16 at 20:43
  • $\begingroup$ @nospoon I would be very interested in seeing the calculation you used in order to arrive at the value for this sum. Particularly so if you managed to avoid the $\operatorname{Re} \operatorname{Li}_4 (1 + i)$ term I struck in my solution for this sum given below. $\endgroup$ – omegadot Jul 14 '19 at 4:49
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Noting that $$\sum_{k = 1}^n \frac{1}{2k - 1} = H_{2n} - \frac{1}{2} H_n,$$ where $H_n$ denotes the $n$th harmonic number, your sum $S$ can be re-written in terms of the following two Euler sums: $$S = \sum_{n = 1}^\infty \frac{(-1)^{n - 1}}{n^3} \left (H_{2n} - \frac{1}{2} H_n \right ) = -\sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n^3} + \frac{1}{2} \frac{(-1)^n H_n}{n^3}.$$

Dealing with these two Euler sums, their values can be found from the following generating function \begin{align} \sum^\infty_{n=1}\frac{H_n}{n^3}z^n &=2{\rm Li}_4(x)+{\rm Li}_4\left(\frac{x}{x-1}\right)-{\rm Li}_4(1-x)-{\rm Li}_3(x)\ln(1-z)-\frac{1}{2}{\rm Li}_2^2\left(\frac{x}{x-1}\right)\\ &+\frac{1}{2}{\rm Li}_2(x)\ln^2(1-x)+\frac{1}{2}{\rm Li}_2^2(x)+\frac{1}{6}\ln^4(1-x)-\frac{1}{6}\ln{x}\ln^3(1-x)\\ &+\frac{\pi^2}{12}\ln^2(1-x)+\zeta(3)\ln(1-x)+\frac{\pi^4}{90},\tag1 \end{align} which is proved in this answer here.

Setting $x = -1$ in (1) gives \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3}=2{\rm Li}_4\left(\tfrac{1}{2}\right)-\frac{11\pi^4}{360}+\frac{7}{4}\zeta(3)\ln{2}-\frac{\pi^2}{12}\ln^2{2}+\frac{1}{12}\ln^4{2}, \end{align} while setting $x = i$ in (1) gives \begin{align} \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n^3} &= 8 \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{(2n)^3}\\ &= 8 \operatorname{Re} \sum_{n = 1}^\infty \frac{H_n}{n^3} i^n\\ &= -16 \operatorname{Re} \operatorname{Li}_4(1 + i) + \frac{29 \pi^4}{288} + \frac{35}{8} \zeta (3) \ln 2 + \frac{\pi^2}{8} \ln^2 2. \end{align}

Substituting these two values for the Euler sums back into the expression for the sum $S$ gives $$S = 16 \operatorname{Re} \operatorname{Li}_4 (1 + i) - \frac{167}{1440} \pi^4 - \frac{7}{2} \zeta (3) \ln 2 - \frac{\pi^2}{6} \ln^2 2 + \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{1}{24} \ln^4 2.\tag2$$

The above result can be further simplified since recently a closed form for $\operatorname{Re} \operatorname{Li}_4 (1 + i)$ has been found. As (for a proof of this, see here):

$$\operatorname{Re} \operatorname{Li}_4 (1 + i) = -\frac{5}{16} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{97 \pi^4}{9216} + \frac{\pi^2}{48} \ln^2 2 - \frac{5}{384} \ln^4 2,$$

the value for the required sum becomes

$$\sum_{n = 1}^\infty \frac{(-1)^{n - 1}}{n^3} \left (H_{2n} - \frac{1}{2} H_n \right ) = -4 \operatorname{Li}_4 \left (\frac{1}{2} \right ) - \frac{7}{2} \zeta (3) \ln 2 + \frac{151 \pi^4}{2880} - \frac{1}{6} \ln^4 2 + \frac{\pi^2}{6} \ln^2 2.$$

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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.

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