Q:

What is the value of this expression involving the dilogarithm $\operatorname{Li}_2$ and $\sqrt{2}$?

I'm carrying out a calculation, and the end result is $$\operatorname{Li}_2(1/\sqrt2) - \operatorname{Li}_2(1 - 1/\sqrt2) +         
        \operatorname{Li}_2(2 - \sqrt2) - \operatorname{Li}_2(\sqrt{2} - 1).$$
The value appears to be $\pi^2/12$. How does one prove this?

A:

The following identity could be found in Russion version of article about dilogarithm:
$$\begin{aligned}\operatorname{Li}_2(xy)&=\operatorname{Li}_2(x)+\operatorname{Li}_2(y)-\operatorname{Li}_2\left(\frac{x(1-y)}{1-xy}\right)-\operatorname{Li}_2\left(\frac{y(1-x)}{1-xy}\right)-\\
&-\ln\left(\frac{1-x}{1-xy}\right)\ln\left(\frac{1-y}{1-xy}\right)\end{aligned}$$
Therefore
$$\begin{aligned}
\operatorname{Li}_2(x)+\operatorname{Li}_2(y)&=\operatorname{Li}_2(xy)+\operatorname{Li}_2\left(\frac{x(1-y)}{1-xy}\right)+\operatorname{Li}_2\left(\frac{y(1-x)}{1-xy}\right)+\\
&+\ln\left(\frac{1-x}{1-xy}\right)\ln\left(\frac{1-y}{1-xy}\right)
\end{aligned}$$
Now we can simplify the original expression (let's name it $E$):
$$\begin{aligned}E&=\left(\operatorname{Li}_2\frac1{\sqrt2}+\operatorname{Li}_2\left(2-\sqrt2\right)\right)-\operatorname{Li}_2\left(1-\frac1{\sqrt2}\right)-\operatorname{Li}_2(\sqrt2-1)=\\
&=\color{red}{\operatorname{Li}_2(\sqrt2-1)}+\operatorname{Li}_2\left(\frac{\frac1{\sqrt2}(1-(2-\sqrt2))}{1-(\sqrt2-1)}\right)+\operatorname{Li}_2\left(\frac{(2-\sqrt2)(1-\frac1{\sqrt2})}{1-(\sqrt2-1)}\right)+\\
&+\ln\left(\frac{1-(2-\sqrt2)}{1-(\sqrt2-1)}\right)\ln\left(\frac{1-\frac1{\sqrt2}}{1-(\sqrt2-1)}\right)-\operatorname{Li}_2\left(1-\frac1{\sqrt2}\right)\color{red}{-\operatorname{Li}_2(\sqrt2-1)}=\\
&=\operatorname{Li}_2\left(\frac12\right)+\color{red}{\operatorname{Li}_2\left(1-\frac1{\sqrt2}\right)}+\ln\left(\frac1{\sqrt2}\right)\ln\left(\frac12\right)\color{red}{-\operatorname{Li}_2\left(1-\frac1{\sqrt2}\right)}=\\
&=\frac{\pi^2}{12}-\frac12\ln^22+\frac12\ln^22=\frac{\pi^2}{12}
\end{aligned}$$
In the last line special value of $\operatorname{Li}_2\left(\frac12\right)$ was used which equals to $\frac{\pi^2}{12}-\frac12\ln^22$.

