On the atypical harmonic series $\sum _{k=1}^{\infty }\left(\frac{\overline{O}_k}{k}\right)^2$

Question

The following series is proposed by Cornel Ioan Vălean: $$\sum _{k=1}^{\infty }\left(\frac{\overline{O}_k}{k}\right)^2=\frac{\pi ^4}{32}-2G^2,$$ where $\overline{O}_k=\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{2n-1}$ represents the $k$-th skew-harmonic number of the second kind and $G$ denotes Catalan's constant.

The main components used to establish a solution are the following key identities: $$\frac{\operatorname{arctanh} \left(x\right)}{1+x^2}=\sum _{k=1}^{\infty }\left(-1\right)^{k+1}\overline{O}_kx^{2k-1},\quad\left|x\right|<1;$$ $$\int _0^1x^{2k-1}\arctan \left(x\right)\:dx=\frac{1}{2k}\left(\frac{\pi }{4}+\frac{\pi }{4}\left(-1\right)^{k+1}-\left(-1\right)^{k+1}\overline{O}_k\right),\quad k\in \mathbb{Z}^+.$$ Their proofs are straightforward and may be found here and here. In my approach below, by applying these identities, we are able to first obtain complicated expressions which are reduced through a range of manipulations, leading to the desired closed form.

Given this, can we obtain such closed form in a more direct way? Are there other ways to proceed with these identities?

My approach (in large steps) is the following:

Through straightforward use of the main components, we have $$\frac{\operatorname{arctanh} \left(x\right)}{1+x^2}=\sum _{k=1}^{\infty }\left(-1\right)^{k+1}\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{2n-1}x^{2k-1}$$ $$\frac{1}{x}\int _0^x\frac{\operatorname{arctanh} \left(t\right)}{1+t^2}\:dt=\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k}\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{2n-1}x^{2k-1}$$ $$\int _0^1\frac{\arctan \left(x\right)}{x}\int _0^x\frac{\operatorname{arctanh} \left(t\right)}{1+t^2}\:dt\:dx=\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k}\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{2n-1}\int _0^1x^{2k-1}\arctan \left(x\right)\:dx$$ $$\int _0^1\frac{\operatorname{arctanh} \left(t\right)}{1+t^2}\left(G-\operatorname{Ti}_2\left(t\right)\right)\:dt=\frac{\pi }{16}\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k^2}\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{2n-1}+\frac{\pi }{16}\sum _{k=1}^{\infty }\frac{1}{k^2}\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{2n-1}$$ $$-\frac{1}{4}\sum _{k=1}^{\infty }\frac{1}{k^2}\left(\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{2n-1}\right)^2$$ $$\sum _{k=1}^{\infty }\frac{1}{k^2}\left(\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{2n-1}\right)^2=\frac{\pi }{4}\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k^2}\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{2n-1}+\frac{\pi }{4}\sum _{k=1}^{\infty }\frac{1}{k^2}\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{2n-1}$$ $$+4\int _0^1\frac{\operatorname{arctanh} \left(t\right)\operatorname{Ti}_2\left(t\right)}{1+t^2}\:dt-2G^2.\tag{1}$$ Moreover, by interchanging the order of summation and performing simple manipulations, we obtain $$\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k^2}\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{2n-1}=-4\int _0^1\frac{\ln \left(t\right)\operatorname{arctanh} \left(t\right)}{1+t^2}\:dt\tag{2}$$ and $$\sum _{k=1}^{\infty }\frac{1}{k^2}\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{2n-1}=-4\int _0^1\frac{\ln \left(t\right)\arctan \left(t\right)}{1-t^2}\:dt.\tag{3}$$ Furthermore, if we use that $\operatorname{arctanh} \left(\frac{1}{t}\right)-\operatorname{arctanh} \left(t\right)=\frac{i\pi }{2},\:t>1$ and $\operatorname{Ti}_2\left(t\right)-\operatorname{Ti}_2\left(\frac{1}{t}\right)=\frac{\pi }{2}\ln \left(t\right),\:t>0$, it follows that $$\int _0^1\frac{\operatorname{arctanh} \left(t\right)\operatorname{Ti}_2\left(t\right)}{1+t^2}\:dt=\frac{1}{2}\operatorname{\mathfrak{R}} \left\{\int _0^{\infty }\frac{\operatorname{arctanh} \left(t\right)\operatorname{Ti}_2\left(t\right)}{1+t^2}\:dt\right\}+\frac{\pi }{4}\int _0^1\frac{\ln \left(t\right)\operatorname{arctanh} \left(t\right)}{1+t^2}\:dt,$$ and since $\int _0^1\frac{t}{1-t^2u^2}\:du=\operatorname{arctanh} \left(t\right)$ and $\operatorname{Ti}_2\left(t\right)=-\int _0^1\frac{t\ln \left(x\right)}{1+t^2x^2}\:dx$, it follows that $$\operatorname{\mathfrak{R}} \left\{\int _0^{\infty }\frac{\operatorname{arctanh} \left(t\right)\operatorname{Ti}_2\left(t\right)}{1+t^2}\:dt\right\}$$ $$=-\int _0^1\ln \left(x\right)\int _0^1\operatorname{\mathfrak{R}} \left\{\int _0^{\infty }\frac{t^2}{\left(1+t^2\right)\left(1+x^2t^2\right)\left(1-u^2t^2\right)}\:dt\right\}\:du\:dx$$ $$\overset{x=t}=-\frac{\pi }{2}\int _0^1\ln \left(t\right)\int _0^1\left(\frac{t}{\left(1+u^2\right)\left(t^2+u^2\right)}-\frac{1}{\left(1+u^2\right)\left(1+t\right)}\right)\:du\:dt,$$ and thus $$\operatorname{\mathfrak{R}} \left\{\int _0^{\infty }\frac{\operatorname{arctanh} \left(t\right)\operatorname{Ti}_2\left(t\right)}{1+t^2}\:dt\right\}=-\frac{\pi ^2}{8}\int _0^1\frac{\ln \left(t\right)}{1-t^2}\:dt+\frac{\pi }{2}\int _0^1\frac{\ln \left(t\right)\arctan \left(t\right)}{1-t^2}\:dt.$$ This means that $$\int _0^1\frac{\operatorname{arctanh} \left(t\right)\operatorname{Ti}_2\left(t\right)}{1+t^2}\:dt=-\frac{\pi ^2}{16}\int _0^1\frac{\ln \left(t\right)}{1-t^2}\:dt+\frac{\pi }{4}\int _0^1\frac{\ln \left(t\right)\arctan \left(t\right)}{1-t^2}\:dt$$ $$+\frac{\pi }{4}\int _0^1\frac{\ln \left(t\right)\operatorname{arctanh} \left(t\right)}{1+t^2}\:dt\tag{4}.$$ Therefore, if we apply $\left(2\right)$, $\left(3\right)$, and $\left(4\right)$ to $\left(1\right)$, we arrive at $$\sum _{k=1}^{\infty }\frac{1}{k^2}\left(\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{2n-1}\right)^2=-\frac{\pi ^2}{4}\int _0^1\frac{\ln \left(t\right)}{1-t^2}\:dt-2G^2,$$ and by using that $\int _0^1\frac{\ln \left(t\right)}{1-t^2}\:dt=-\frac{\pi ^2}{8}$, the closed form follows.

---

A first observation:

By independent calculations, we can also find that $$4\int _0^1\frac{\operatorname{Ti}_3\left(x\right)}{1+x^2}\:dx$$ $$=2\pi \int _0^1\frac{\arctan \left(x\right)\operatorname{arctanh}\left(x^2\right)}{x}\:dx-4\int _0^1\frac{\arctan ^2\left(x\right)\operatorname{arctanh}\left(x^2\right)}{x}\:dx$$ $$=\frac{\pi ^4}{32}-2G^2.$$ Perhaps this could be exploited in order to answer my first question.

A second observation:

By applying summation by parts to the main series, we obtain the following: $$\sum _{k=1}^{\infty }\left(\frac{\overline{O}_k}{k}\right)^2=\lim _{n\to \infty }\left(H_n^{\left(2\right)}\overline{O}_n^2\right)-\sum _{k=1}^{\infty }H_k^{\left(2\right)}\left(\overline{O}_{k+1}^2-\overline{O}_k^2\right)$$ $$=\frac{\pi ^4}{96}-\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{\left(2k+1\right)^2}+2\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}\overline{O}_kH_k^{\left(2\right)}}{2k+1},$$ and by using the result obtained in this post along with the result of the remaining series, we find $$\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}\overline{O}_kH_k^{\left(2\right)}}{2k+1}$$ $$=-G^2-\frac{91}{32}\zeta \left(4\right)+4\operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{7}{2}\ln \left(2\right)\zeta \left(3\right)-\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{6}\ln ^4\left(2\right).$$

Answer 1

A (Miraculous) Solution by C. I. Valean

The present solution is also meant to answer Jorge's questions in the main post Given this, can we obtain such a closed form more directly? Are there other ways to proceed with these identities? We start with the same two identities (it's natural to consider them as soon as we are aware of them), but then the flow of calculations goes completely differently, leading to amazing connections with other results, turning everything into a dream-like solution.

So, we start by combining $\displaystyle \frac{1}{x}\int_0^x\frac{\operatorname{arctanh}(y)}{1+y^2}\textrm{d}y=\frac{1}{2}\sum_{n=1}^{\infty} (-1)^{n-1}x^{2n-1}\frac{\overline{O}_n}{n}$ and $\displaystyle \int_0^1 x^{2n-1}\arctan(x)\textrm{d}x=\frac{1}{2n}\left(\frac{\pi}{4}+\frac{\pi}{4}(-1)^{n-1}-(-1)^{n-1} \overline{O}_n\right)$, which leads to $$\int_0^1 \frac{\arctan(x)}{x}\left(\int_0^x \frac{\operatorname{arctanh}(y)}{1+y^2}\textrm{d}y \right) \textrm{d}x=\int_0^1 \frac{\operatorname{arctanh}(y)}{1+y^2} \left(\int_y^1 \frac{\arctan(x)}{x}\textrm{d}x \right) \textrm{d}y $$ $$ =\int_0^1 \frac{\operatorname{arctanh}(y)(\operatorname{Ti}_2(1)-\operatorname{Ti}_2(y))}{1+y^2} \textrm{d}y=\int_0^1 \frac{\operatorname{arctanh}(y)(G-\operatorname{Ti}_2(y))}{1+y^2} \textrm{d}y$$ $$ \small =\frac{\pi}{16}\left(\sum_{n=1}^{\infty} \frac{\overline{O}_n}{n^2}+\sum_{n=1}^{\infty} (-1)^{n-1}\frac{\overline{O}_n}{n^2}\right)-\frac{1}{4}\sum_{n=1}^{\infty} \left(\frac{\overline{O}_n}{n}\right)^2=\frac{\pi}{8}\sum_{n=1}^{\infty} \frac{\overline{O}_{2n-1}}{(2n-1)^2}-\frac{1}{4}\sum_{n=1}^{\infty} \left(\frac{\overline{O}_n}{n}\right)^2.$$

Now, it's straightforward to get that $\displaystyle \int_0^1 \frac{\operatorname{arctanh}(x)}{1+x^2} \textrm{d}x=\frac{1}{2}G$, by letting $(1-x)/(1+x)=y$ and then integrating by parts. Also, we recall that $\displaystyle G=\operatorname{Ti}_2(1)=\int_0^1 \frac{\arctan(x)}{x}\textrm{d}x=\sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{(2n-1)^2}$. Moreover, we may also easily observe that based on the Cauchy product, $\displaystyle \arctan(x)\operatorname{arctanh}(x)=\sum_{n=1}^{\infty}x^{4n-2}\frac{\overline{O}_{2n-1}}{2n-1} , \ |x|<1$, which is derived in More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), pages $492$ - $493$, we have $\displaystyle \int_0^1 \frac{ \arctan(x)\operatorname{arctanh}(x)}{x}\textrm{d}x=\frac{1}{2}\sum_{n=1}^{\infty} \frac{\overline{O}_{2n-1}}{(2n-1)^2}$, and then the result written above reduces to $$\sum_{n=1}^{\infty} \left(\frac{\overline{O}_n}{n}\right)^2$$ $$=-2G^2+\pi \int_0^1 \frac{ \arctan(x)\operatorname{arctanh}(x)}{x}\textrm{d}x+ 4\underbrace{\int_0^1 \frac{ \operatorname{arctanh}(x)\operatorname{Ti}_2(x)}{1+x^2}\textrm{d}x}_{\displaystyle x\mapsto \arctan(x)}$$ $$\small =-2G^2+\pi \int_0^1 \frac{ \arctan(x)\operatorname{arctanh}(x)}{x}\textrm{d}x+4 \int_0^{\pi/4}\operatorname{arctanh}(\tan(x))\operatorname{Ti}_2(\tan(x))\textrm{d}x. \tag1$$

Next, combining the powerful Fourier series, $$\operatorname{Ti}_2(\tan(x))$$ $$=\sum _{n=1}^{\infty } \left((-1)^{n-1}\frac{ H_n}{n}-\frac{\overline{H}_n}{n}+\log (2)\frac{1}{n}+\log (2) (-1)^{n-1} \frac{1}{n}\right)\sin (2 n x)$$ $$=2\sum _{n=1}^{\infty } (-1)^{n-1} \left(\int_0^1 t^{n-1}\operatorname{arctanh}(t)\textrm{d}t\right)\sin (2 n x), \ 0<x<\frac{\pi}{2},$$ where $H_n=\sum_{k=1}^n \frac{1}{k}$ is the $n$th harmonic number and $\overline{H}_n=\sum_{k=1}^n(-1)^{k-1}\frac{1}{k}$ represents the $n$th skew-harmonic number, and the version with the first equality is given in the paper Deriving Special Fourier Series Involving the Inverse Tangent Integrals of Order Two and Three, and Other Curious Functions by C.I. Valean, and further we easily observe that the second equality is obtained by using that $\displaystyle \int_0^1 t^{n-1}\operatorname{arctanh}(t)\textrm{d}t=\frac{1}{2}\log(2)\frac{1}{n}+\frac{1}{2}\log(2)(-1)^{n-1}\frac{1}{n}+\frac{1}{2}\frac{H_n}{n}-\frac{1}{2}(-1)^{n-1}\frac{\overline{H}_n}{n}$, also appearing in the same book mentioned above, page $28$, together with the useful trigonometric integral \begin{equation*} \small \ \int_0^{\pi/4} \sin(2 n x)\operatorname{arctanh}(\tan(x))\textrm{d}x= \begin{cases} \displaystyle \quad \dfrac{\pi}{8} (-1)^{m-1}\dfrac{1}{2m-1}, \ \ \quad \qquad \qquad \quad n=2m-1, \ m \in \mathbb{N}; \\[10pt] \displaystyle \quad \dfrac{1}{4}(-1)^{m-1}\dfrac{ H_{2m}}{m}-\frac{1}{8}(-1)^{m-1}\dfrac{ H_m}{m}\\[10pt] \displaystyle =\dfrac{1}{4} (-1)^{m-1}\dfrac{\overline{H}_{2m}}{m}+\frac{1}{8}(-1)^{m-1}\dfrac{ H_m}{m}, \ \ n=2m, \ m \in \mathbb{N}, \end{cases} \end{equation*} where $H_n=\sum_{k=1}^n \frac{1}{k}$ is the $n$th harmonic number and $\overline{H}_n=\sum_{k=1}^n(-1)^{k-1}\frac{1}{k}$ represents the $n$th skew-harmonic number, which is found and evaluated in A Valuable Elementary Generalized Trigonometric Integral by C.I. Valean, then the second integral in $(1)$ can be written as $$\int_0^{\pi/4}\operatorname{arctanh}(\tan(x))\operatorname{Ti}_2(\tan(x))\textrm{d}x$$ $$=2\int_0^{\pi/4}\operatorname{arctanh}(\tan(x))\sum _{n=1}^{\infty } (-1)^{n-1} \left(\int_0^1 t^{n-1}\operatorname{arctanh}(t)\textrm{d}t\right)\sin (2 n x) \textrm{d}x$$ $$=2\sum _{n=1}^{\infty } (-1)^{n-1} \left(\int_0^1 t^{n-1}\operatorname{arctanh}(t)\textrm{d}t\right)\int_0^{\pi/4}\operatorname{arctanh}(\tan(x))\sin (2 n x) \textrm{d}x$$ $$\text{\{split the series according to parity\}}$$ $$=2\sum _{n=1}^{\infty } \left(\int_0^1 t^{2n-2}\operatorname{arctanh}(t)\textrm{d}t\right)\int_0^{\pi/4}\operatorname{arctanh}(\tan(x))\sin (2 (2n-1) x) \textrm{d}x$$ $$-2\sum _{n=1}^{\infty }\left(\int_0^1 t^{2n-1}\operatorname{arctanh}(t)\textrm{d}t\right)\int_0^{\pi/4}\operatorname{arctanh}(\tan(x))\sin (2 (2n) x) \textrm{d}x$$ $$=\frac{\pi}{4}\sum _{n=1}^{\infty } \left(\int_0^1 t^{2n-2}\operatorname{arctanh}(t)\textrm{d}t\right) \frac{ (-1)^{n-1}}{2n-1}$$ $$ -\frac{1}{4}\sum _{n=1}^{\infty }\left(\int_0^1 t^{2n-1}\operatorname{arctanh}(t)\textrm{d}t\right)\left((-1)^{n-1}\frac{2 H_{2n}- H_n}{n}\right)$$ $$=\frac{\pi}{4} \int_0^1 \frac{\operatorname{arctanh}(t)}{t} \left( \sum _{n=1}^{\infty } (-1)^{n-1} \frac{t^{2n-1}}{2n-1}\right)\textrm{d}t$$ $$-\frac{1}{4}\int_0^1\frac{\operatorname{arctanh}(t)}{t} \left(\sum _{n=1}^{\infty }(-1)^{n-1}t^{2n}\frac{2 H_{2n}- H_n}{n} \right)\textrm{d}t$$ $$=\frac{\pi}{4}\int_0^1\frac{\arctan(t) \operatorname{arctanh}(t)}{t} \textrm{d}t-\frac{1}{2}\int_0^1\frac{\arctan^2(t) \operatorname{arctanh}(t)}{t} \textrm{d}t, \tag2$$ where above I also used the known Cauchy product, $\displaystyle \arctan^2(x)=\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}x^{2n} \frac{2H_{2n}-H_n}{n}, \ |x|\le1$, derived in the mention book, too.

Combining $(1)$ and $(2)$, we conclude that $$\sum_{n=1}^{\infty} \left(\frac{\overline{O}_n}{n}\right)^2$$ $$=-2G^2+2\left(\color{blue}{\underbrace{\pi \int_0^1 \frac{ \arctan(x)\operatorname{arctanh}(x)}{x}\textrm{d}x-\int_0^1\frac{\arctan^2(x) \operatorname{arctanh}(x)}{x} \textrm{d}x}_{\displaystyle \pi^4/64}}\right)$$ $$=\frac{\pi^4}{32}-2 G^2=\frac{45}{16}\zeta(4)-2G^2,$$ which is an identity immediately obtained from this solution, since we have that $$\int_0^1 \frac{\displaystyle\left(\pi/2-\arctan(x)\right)^2 \operatorname{arctanh}(x)}{x}\textrm{d}x=\frac{\pi^4}{64},$$ where upon expanding and using the trivial result $\displaystyle \int_0^1\frac{\operatorname{arctanh}(x)}{x}\textrm{d}x=\frac{\pi^2}{8}$, the needed auxiliary result follows.

The key identity in blue is derived in two different ways in More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), the sequel of (Almost) Impossible Integrals, Sums, and Series (2019), as you may see on page $783$ and then on pages $786$-$788$.

End of story (it's stunning to see any bit of ugly calculations has been completely avoided)

Answer 2

(Supplementary work) A solution by C. I. Valean to the alternating version of the main series

The main series and some auxiliary tools employed for its calculation will immediately allow us to extract the value of the alternating version.

Let's get started ... and first we want to consider the following Cauchy product of two series, $\displaystyle \arctan(x)\operatorname{arctanh}(x)=\sum_{n=1}^{\infty}x^{4n-2}\frac{\overline{O}_{2n-1}}{2n-1}, \ |x|<1$, which is derived in More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), pages $492$ - $493$, that may be written in a useful way as $$\frac{\arctan(x)\operatorname{arctanh}(x)}{x}=\frac{1}{2}\sum_{n=1}^{\infty}x^{2n-1}\frac{\overline{O}_{n}}{n}+\frac{1}{2}\sum_{n=1}^{\infty}x^{2n-1}(-1)^{n-1}\frac{\overline{O}_{n}}{n}, \ |x|<1,$$ which if we combine it with the integral identity from A Simple Useful Generalized Integral Involving the Arctan Function from the Table of I. S. Gradshteyn and I. M. Ryzhik (8th Edition), 4.532.1 by C. I. Valean, $$\int_0^1 x^{2n-1}\arctan(x)\textrm{d}x=\frac{1}{2n}\left(\frac{\pi}{4}+\frac{\pi}{4}(-1)^{n-1}-(-1)^{n-1} \overline{O}_n\right),$$ leads to $$\int_0^1\frac{\arctan^2(x)\operatorname{arctanh}(x)}{x}\textrm{d}x$$ $$=\frac{\pi}{8}\sum_{n=1}^{\infty}\frac{\overline{O}_{n}}{n^2}+\frac{\pi}{8}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\overline{O}_{n}}{n^2}-\frac{1}{4}\sum_{n=1}^{\infty}\left(\frac{\overline{O}_{n}}{n}\right)^2 -\frac{1}{4}\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{\overline{O}_{n}}{n}\right)^2$$ $$=\frac{\pi}{4}\sum_{n=1}^{\infty}\frac{\overline{O}_{2n-1}}{(2n-1)^2}-\frac{1}{4}\sum_{n=1}^{\infty}\left(\frac{\overline{O}_{n}}{n}\right)^2 -\frac{1}{4}\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{\overline{O}_{n}}{n}\right)^2,$$

and since from my first post, we can collect the results: $\displaystyle \sum_{n=1}^{\infty} \left(\frac{\overline{O}_n}{n}\right)^2=\frac{\pi^4}{32}-2 G^2$, further $\displaystyle \int_0^1 \frac{ \arctan(x)\operatorname{arctanh}(x)}{x}\textrm{d}x=\frac{1}{2}\sum_{n=1}^{\infty} \frac{\overline{O}_{2n-1}}{(2n-1)^2}$, then $\displaystyle \color{blue}{\pi \int_0^1 \frac{ \arctan(x)\operatorname{arctanh}(x)}{x}\textrm{d}x-\int_0^1\frac{\arctan^2(x) \operatorname{arctanh}(x)}{x} \textrm{d}x=\frac{\pi^4}{64}}$, all together with the trivial integral (more ways of evaluating the integral are possible here, of course), $\displaystyle \int_0^1\frac{\arctan(x)\operatorname{arctanh}(x)}{x}\textrm{d}x=\frac{1}{2}\log(2)G-\frac{\pi}{32}\log^2(2)-\frac{\pi^3}{128}+ \Im\biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\},$ appearing with a solution in the mentioned book pages $783$-$785$ we arrive at the desired result

$$\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{\overline{O}_{n}}{n}\right)^2$$ $$=2 G^2-\log(2)\pi G+\frac{3}{8}\log^2(2)\zeta(2)+\frac{135}{32}\zeta(4)-2\pi \Im\biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\}.$$

End of story

Other ideas to develop and attack both series

$\color{brown}{\text{We might combine (cleverly)} }$

A useful identity

$$\overline{O}_n^2=\sum_{i=1}^n \sum_{j=1}^n \frac{(-1)^{i-1}}{2i-1 }\frac{(-1)^{j-1}}{2j-1 }=\sum_{i=1}^n\left(\sum_{j=1}^i+\sum_{j=i}^n\right)\frac{(-1)^{i-1}}{2i-1 }\frac{(-1)^{j-1}}{2j-1 }-\sum_{k=1}^n \frac{1}{(2k-1)^2} $$ $$ =2\sum_{i=1}^n \frac{(-1)^{i-1}}{2i-1 }\sum_{j=1}^i\frac{(-1)^{j-1}}{2j-1 }-O_n^{(2)}=2 \sum_{i=1}^n (-1)^{i-1}\frac{\overline{O}_i}{2i-1 }-O_n^{(2)},$$ whence we obtain that $$ \color{brown}{\sum_{k=1}^n (-1)^{k-1}\frac{\overline{O}_k}{2k-1 }=\frac{\overline{O}_n^2+O_n^{(2)}}{2}},$$ where $O_k=\sum _{n=1}^k\frac{1}{2n-1}$ and $\overline{O}_k=\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{2n-1}$,

$\color{brown}{\text{and} }$ $$ \color{brown}{\int_0^1 x^{2n-1}\arctan(x)\textrm{d}x=\frac{1}{2n}\left(\frac{\pi}{4}+\frac{\pi}{4}(-1)^{n-1}-(-1)^{n-1} \overline{O}_n\right)}.$$

Together, they may lead to the auxiliary integral,

$$\color{brown}{\int_0^1 \frac{(1-x^{2n}) \log(1+x^2)}{1-x^2} \textrm{d}x= \log (2) O_n+\frac{\pi}{2} \overline{O}_n-O_n^{(2)}-\overline{O}_n^2}.$$

Cauchy products of two series with $\chi_n(x)$ and $\operatorname{Ti}_n(x)$

We simply view the result $\displaystyle \frac{\operatorname{arctanh} \left(x\right)}{1+x^2}=\sum _{k=1}^{\infty }(-1)^{k+1}\overline{O}_kx^{2k-1},\ |x|<1$ more generally, and using that Legendre chi function of order $n$ is defined by $\displaystyle \chi_n(x)=\sum_{k=1}^{\infty} \frac{x^{2k-1}}{(2k-1)^n}, |x|<1$, we obtain that $$\frac{\chi_n(x)}{1+x^2}=\sum _{k=1}^{\infty }(-1)^{k-1}\overline{O}_k^{(n)}x^{2k-1},\ |x|<1.$$

Similarly, for the Inverse tangent integral of order $n$, $\operatorname{Ti}_n(x)=\sum_{k=1}^{\infty} (-1)^{k-1}\frac{x^{2k-1}}{(2k-1)^n}, \ |x|<1$, we have that $$\frac{\operatorname{Ti}_n(x)}{1+x^2}=\sum _{k=1}^{\infty } (-1)^{k-1} O_k^{(n)}x^{2k-1},\ |x|<1.$$

Further, it's easy to note we also have that $$\frac{\chi_n(x)}{1-x^2}=\sum _{k=1}^{\infty } O_k^{(n)}x^{2k-1},\ |x|<1,$$

and

$$\frac{\operatorname{Ti}_n(x)}{1-x^2}=\sum _{k=1}^{\infty} \overline{O}_k^{(n)}x^{2k-1},\ |x|<1,$$

where $\displaystyle O_n^{(m)}=\sum_{k=1}^n \frac{1}{(2k-1)^m}$ is the $n$th harmonic number of the second kind, of order $m$ and $\displaystyle \overline{O}_n^{(m)}=\sum_{k=1}^n (-1)^{k-1}\frac{1}{(2k-1)^m}$ represents the $n$th skew-harmonic number of the second kind, of order $m$.