Proof of a dilogarithm identity Through some experimentation, I've found:

$$\text{Li}_{2}(\sqrt{2}-1) \ + \text{Li}_{2}(1-\frac{1}{\sqrt{2}})\  =\frac{\pi^2}{8} - \frac{\ln^2(1 + \sqrt{2})}{2} - \frac{\ln^2{2}}{8}  $$

I'm sure a simple proof is evading me, but I have not been able to prove this so far.
I've been able to reduce it to proving the following integral equality:
$$\frac{\ln^2{\sqrt{2}+1}}{4}-\frac{\pi^2}{16}=\int_{\sqrt{2}-1}^1 \frac{\ln{t}}{1-t^2}$$
Which, again, agrees numerically but seems to be as hard as proving the original problem. How can I prove this dilogarithm identity? 
 A: We use the standard notation $\operatorname{Li}_2$ for the dilogarithm function and $\zeta$ for the Riemann zeta function. For $|z|<1$, we have
$$
\operatorname{Li}_2\left(2z-z^2\right) = -\ln^2\left(2-z\right)+\zeta(2)-2\operatorname{Li}_2\left(\frac{1}{2-z}\right)+2\operatorname{Li}_2(z).
$$
This result can be obtained from Abel's identity by using appropriate transformations and substitutions. For further details see Lewin, pp. 7$-$11.
Let $z:=1-\sqrt2$ and divide both side by $2$. We have
$$
\tfrac12\operatorname{Li}_2(-1) = -\tfrac12 \ln^2\left(1+\sqrt2\right)+\tfrac12\zeta(2) - \color{red}{\operatorname{Li}_2\left(\sqrt2 - 1\right)} + \operatorname{Li}_2\left(1-\sqrt2\right).\tag{$\spadesuit$}
$$
From the duplication formula and the definition of the dilogarithm function, we have
$$
\operatorname{Li}_2(-1) = -\tfrac12 \operatorname{Li}_2(1) = -\tfrac12 \zeta(2).\tag{$\clubsuit$}
$$
Now if we express $\operatorname{Li}_2\left(1-\sqrt2\right)$ from $\left(\spadesuit\right)$ and substitute $\left(\clubsuit\right)$, we arrive at
$$
\operatorname{Li}_2\left(1-\sqrt2\right) =  \color{red}{\operatorname{Li}_2\left(\sqrt2 - 1\right)} + \color{blue}{\tfrac12 \ln^2\left(1+\sqrt2\right)-\tfrac34\zeta(2)}.\tag{$\heartsuit$}
$$
Landen's identity states that for $z \notin \left(-\infty,0\right]$, we have
$$
\operatorname{Li}_2\left(1-z\right) + \operatorname{Li}_2\left(1-\frac{1}{z}\right)=-\tfrac12\ln^2(z).
$$
Let us substitute $z:=\sqrt2$ into Landen's identity. We arrive at
$$
\operatorname{Li}_2\left(1-\sqrt2\right) + \color{green}{\operatorname{Li}_2\left(1-\frac{1}{\sqrt2}\right)}=\color{blue}{-\tfrac18\ln^2(2)}.\tag{$\diamondsuit$}
$$
Now if we combine $\left(\heartsuit\right)$ and $\left(\diamondsuit\right)$ and use the fact that $\zeta(2) = \pi^2/6$, we arrive at
$$
\color{red}{\operatorname{Li}_2\left(\sqrt2 - 1\right)} + \color{green}{\operatorname{Li}_2\left(1-\frac{1}{\sqrt2}\right)} = \color{blue}{\frac{\pi^2}{8} -\frac18\ln^2(2) - \frac12\ln^2\left(1+\sqrt2\right)}.
$$

From this argument we can derive a general identity. For $|1-z|<1$ and $z \notin (-\infty,0]$, we have

$$
\operatorname{Li}_2\left(\frac{1}{z+1}\right)+\operatorname{Li}_2\left(1-\frac{1}{z}\right) = \tfrac12\left(\zeta(2)-\ln^2(z)-\ln^2(z+1)-\operatorname{Li}_2\left(1-z^2\right)\right).
$$

For $z:=\sqrt2$, we arrive at your particular problem. We give another example. For $z:=1/\sqrt{2}$, we have
$$
\operatorname{Li}_2\left(2-\sqrt2\right)+\operatorname{Li}_2\left(1-\sqrt{2}\right) = \frac{\pi^2}{24} - \frac38\ln^2(2)+\ln(2)\ln\left(\sqrt{2}+2\right)-\frac12\ln^2\left(\sqrt{2}+2\right).
$$
At this example we used the special value $\operatorname{Li}_2\left(\frac12\right)=\frac{\pi^2}{12}-\frac12\ln^2(2)$.
A: This identity was shown to be true in Theorem 4 of my paper published in 2012, see link to Elsevier site below.
https://www.sciencedirect.com/science/article/pii/S0019357711000425
Fabio M. S. Lima
