Calculating $\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$ Do you see any fast way of calculating this one? $$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$$
Numerically, it's about 
$$\approx 111.024457130115028409990464833072173251135063166330638343951498907293$$
or in a predicted closed form 
$$\frac{4 }{3}\pi ^3+32 \pi  \log (2).$$
Ideas, suggestions, opinions are welcome, and the solutions are optionals.
Supplementary question for the integrals lovers: calculate in closed form
$$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^3 \, dx.$$ 
As a note, it would be remarkable to be able to find a solution for the generalization below
$$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^n \, dx.$$ 
 A: Another way to evaluate the integral $$I_{2} = \int_{0}^{\infty} \frac{\arctan (\frac{1}{x}) \log(1+x^{2})}{x} \, dx$$ in Jack D'Aurizio's answer (which according to Wolfram Alpha evaluates to $\frac{\pi^{3}}{12}$) is to consider the complex function $$f(z) = \frac{\arctan \left(\frac{1}{z} \right) \log(1-iz)}{z} = \frac{\text{arccot}(z) \log(1-iz)}{z}.$$
Using the principal branch of the logarithm, $\text{arccot}(z)$ has a branch cut on $[-i, i]$ and $\log(1-iz)$ has a branch cut on $(-i\infty, -i]$.
So integrating around a semicircle in the upper half-plane deformed around the branch cut and using the fact $\text{arccot}(z) \sim \frac{1}{z}$ when $z$ is large in magnitude, we get
$$\int_{-\infty}^{\infty} \frac{\arctan(\frac{1}{x}) \log(1-ix)}{x} \, dx + \int_{0}^{1} \frac{\frac{i}{2} (2 \pi i) \log(1+t)}{it} \, i \, dt =0.$$
Then equating the real parts on both sides of the equation, we get
$$\int_{0}^{\infty} \frac{\arctan(\frac{1}{x}) \log(1+x^{2})}{x} \, dx = \pi \int_{0}^{1} \frac{\log(1+t)}{t} \, dt = - \pi \, \text{Li}_{2}(-1) =  \frac{\pi^{3}}{12}.$$
A: Ok i want to tackle the problem from a different starting point:
Recall that the $\text{Li}_2(x)$ has the following integral representation
which is closely related to the Debye functions, well known in condensed matter physics:
$$
\text{Li}_2(z)=\frac{z}{\Gamma(2)}\int_{0}^{\infty}\frac{t}{e^t-z}dt  \quad \quad(1)
$$
This leads us to the following representation of our Integral
$$
I=\int_{0}^{\infty}\left(\text{Li}_2\left(\frac{1}{x^2}\right)\right)^2dx=\\
\int_{0}^{\infty}\mathrm{d}t_1t_1\int_{0}^{\infty}\mathrm{d}t_2t_2\int_{0}^{\infty}dx\frac{1}{x^2 e^{t_1}+1}\frac{1}{x^2 e^{t_2}+1}
$$
The inner integral is quite a straightforward exercise in residue calculus 
and yields (using parity)
$$
I=8\pi\int_{0}^{\infty}\mathrm{d}t_1t_1\int_{0}^{\infty}\mathrm{d}t_2t_2\frac{1}{e^{t_1}+e^{t_2}}
$$
Using (1) again to perform for example the $t_1$ integral, we get
$$
I=8\pi\int_{0}^{\infty}\mathrm{d}t_2t_2 e^{-t_2} \text{Li}_2\left(-e^{t_2}\right)
$$
Defining $p(t)=t e^{-t} $ and $ q(t)=\text{Li}_2\left(-e^{t}\right)$ (relabeling $t=t_2$)
we perform two integration by parts ending up with  ( $p^{(-n)}(x)$ denotes the $n$ primitive with respect to $x$)
$$
\frac{1}{8\pi}I=\underbrace{[p^{(-1)}(t)q(t)-p^{(-2)}(t)q'(t)]_0^{\infty}}_{=\frac{\pi ^2}{12}+2\log (2)}+\underbrace{\int_0^{\infty}\overbrace{\frac{t+2}{e^t+1}}^{p^{(-2)}(t)q"(t)}dt}_{J}
$$
Here we have used the special value $\text{Li}_2\left(-1\right)=-\frac{\pi^2}{12}\quad (2)$ (See also the appendix for a proof of this fact).
The $J$ is also straightforwardly calculated in the appendix and yields 
$$
J=\frac{\pi ^2}{12}+2\log (2)
$$
Ande therefore

$$
I=\frac{4\pi^3}{3}+32\pi \log(2)
$$

Q.E.D
Appendix
Calculation of $J$
$$
J=2\underbrace{\int_{0}^{\infty}\frac{1}{1+e^t}}_{J_1}+\underbrace{\int_{0}^{\infty}\frac{t}{1+e^t}}_{J_2}
$$
$J_1$ can be done by done by letting $e^t=z$ 
$$
J_1=\int_{1}^{\infty}\frac{1}{z(z+1)}=[\log (z)-\log (z+1)]_0^{\infty}=\log(2)
$$
$J_2$ may be done by different methods, but most straightforwardly i think by using geometric series and integrating term wise
$$
J_2 =\sum_{n=1}^{\infty}(-1)^n\int_{0}^{\infty}e^{-(n+1)x}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}=\eta(2)=\frac{\pi^2}{12} \quad (3)
$$
here we used a special value of the Dirichlet $\eta$-function
Note that by a change of variables $e^{-z}=q$ the last integral in (3) we also have $\sum_{n=1}^{\infty}\frac{(-1)^{(n+1)}}{n^2}=-\text{Li}_2(-1)$ which proves (2).
Using $J=2J_1+J_2$ the stated result follows
A: $$I=\int_0^\infty\operatorname{Li}_2^2\left(-\frac1{x^2}\right)\ dx\overset{IBP}{=}-4\int_0^\infty\operatorname{Li}_2\left(-\frac1{x^2}\right)\ln\left(1+\frac1{x^2}\right)\ dx$$
By using $$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$  setting $n=1$ and replacing $x$ with $-\frac1{x^2}$ we can write
$$\operatorname{Li}_2\left(-\frac1{x^2}\right)=\int_0^1\frac{\ln u}{u+x^2}\ du$$
Then
\begin{align}
I&=-4\int_0^1\ln u\left(\int_0^\infty\frac{\ln\left(1+
\frac1{x^2}\right)}{u+x^2}\ dx\right)\ du\\
&=-4\int_0^1\ln u\left(\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)\right)\ du, \quad \color{red}{\sqrt{u}=x}\\
&=-16\pi\int_0^1\ln x\left(\ln(1+x)-\ln x\right)\ dx\\
&=-16\pi\left(2-\frac{\pi^2}{12}-2\ln2-2\right)\\
&=\boxed{\frac43\pi^3+32\pi\ln2}
\end{align}


Proof of $\ \displaystyle\int_0^\infty\frac{\ln\left(1+
\frac1{x^2}\right)}{u+x^2}\ dx=\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)$

Let  $$J=\int_0^\infty\frac{\ln\left(1+\frac1{x^2}\right)}{u+x^2}\ dx\overset{x\mapsto\ 1/x}{=}\int_0^\infty\frac{\ln(1+x^2)}{1+ux^2}\ dx$$
and $$J(a)=\int_0^\infty\frac{\ln(1+ax^2)}{1+ux^2}\ dx, \quad J(0)=0,\quad J(1)=J$$
\begin{align}
J'(a)&=\int_0^\infty\frac{x^2}{(1+ux^2)(1+ax^2)}\ dx\\
&=\frac1{a-u}\int_0^\infty\left(\frac1{1+ux^2}-\frac1{1+ax^2}\right)\ dx\\
&=\frac1{a-u}*\frac{\pi}{2}\left(\frac1{\sqrt{u}}-\frac1{\sqrt{a}}\right)\\
&=\frac{\pi}{2}\frac{1}{\sqrt{u}a+u\sqrt{a}}
\end{align}
Then $$J=\int_0^1 J'(a)\ da=\frac{\pi}{2}\int_0^1\frac{da}{\sqrt{u}a+u\sqrt{a}}\\=\frac{\pi}{2}\left(\frac{2}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)\right)=\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)$$
A: With a substitution and a step of integration by parts, the problems boils down to computing:
$$ I = -4\int_{0}^{+\infty}\frac{\log(1+x^2)\,\text{Li}_2(-x^2)}{x^2}\,dx. \tag{1}$$
Integrating by parts again, the problem boils down to computing:
$$ I_1 = \int_{0}^{+\infty}\frac{\log(1+x^2)^2}{x^2}\,dx,\qquad I_2=\int_{0}^{+\infty}\arctan\left(\frac{1}{x}\right)\frac{\log(1+x^2)}{x}\,dx.\tag{2} $$
The first integral is straightforward:
$$ I_1 = \frac{1}{2}\left.\frac{d^2}{d\alpha^2}\int_{0}^{+\infty}\frac{(1+x)^{\alpha}-1}{x^{3/2}}\,dx\,\right|_{\alpha=0} \tag{3}$$
since the innermost integral can be evaluated in terms of the beta function. 
That leads to $I_1=4\pi\log 2$. Now we just need to compute:
$$ J = \int_{1}^{+\infty}\int_{0}^{+\infty}\frac{\log(1+x^2)}{1+t^2 x^2}\,dx\,dt \tag{4}$$
to recover the value of $I_2=\frac{\pi^3}{8}$. That proves the conjecture:

$$ I = \frac{4\pi^3}{3}+32\pi\log 2.\tag{5}$$

