TITLE: Proving $\Im\operatorname{Li}_2(\sqrt i(\sqrt 2-1))=\frac34G+\frac18\pi\ln(\sqrt2-1)$
QUESTION [5 upvotes]: $\newcommand{\Li}{\operatorname{Li}_2}$

I found, numerically, that $$\Im\Li(\sqrt i(\sqrt 2-1))=\frac34G+\frac18\pi\ln(\sqrt2-1).$$
  How can we prove it?

My attempt of proving this equation:
Using identity $$\Li(x)=\int_0^1\frac{x}{xt-1}\ln tdt,$$
we can deduce$$\begin{align}\Im\Li(\sqrt i(\sqrt 2-1))&=\frac1{2i}\int_0^1\left(\frac{\sqrt i(\sqrt2-1)}{\sqrt i(\sqrt2-1)t-1}-\frac{\sqrt {-i}(\sqrt2-1)}{\sqrt {-i}(\sqrt2-1)t-1}\right)\ln tdt\\
&=\int_0^1\frac{2-\sqrt{2}}{\left(4 \sqrt{2}-6\right) t^2-2 \left(\sqrt{2}-2\right) t-2}\ln tdt\\
&=\int_0^{2-\sqrt2}-\frac{1}{u^2-2u+2}\ln\frac u{2-\sqrt 2}du\\
&=\frac18\pi\ln(2-\sqrt2)-\int_{-1}^{1-\sqrt2}\frac{\ln(v+1)}{v^2+1}dv\\
&=\frac18\pi\ln(2-\sqrt2)-\int_{\pi/8}^{\pi/4}\ln(1-\tan x)dx\\
&=\frac18\pi\ln(\sqrt2-1)-\int_{\pi/8}^{\pi/4}\ln\sec x+\ln\sin\left(\frac\pi4-x\right)dx\\
\end{align}$$
I have no idea how to deal with the log-trig integral.

REPLY [2 votes]: \begin{align}
I&=-\int_{\pi/8}^{\pi/4}\log(\sec x)+\log\sin \left(\frac\pi4-x\right)\ dx\\
&=\int_{\pi/8}^{\pi/4}\log(\cos x)\ dx-\int_{\pi/8}^{\pi/4}\log\sin \left(\frac\pi4-x\right)\ dx\\
&\{\text{break the interval for the first integral and let } \frac{\pi}{4}-x\mapsto x \text{ for the second integral}\}\\
&=\int_{0}^{\pi/4}\log(\cos x)\ dx-\int_{0}^{\pi/8}\log(\cos x)\ dx-\int_{0}^{\pi/8}\log(\sin x)\ dx\\
&=\int_{0}^{\pi/4}\log(\cos x)\ dx-\int_{0}^{\pi/8}\log(\cos x\ \sin x)\ dx\\
&=\int_{0}^{\pi/4}\log(\cos x)\ dx-\int_{0}^{\pi/8}\log\left(\frac{\sin(2x)}{2}\right)\ dx\\
&=\int_{0}^{\pi/4}\log(\cos x)\ dx-\frac12\int_{0}^{\pi/4}\log\left(\frac{\sin x}{2}\right)\ dx,\quad \{\ln(\cos x)=\ln(\sin x)-\ln(\tan x)\}\\
&=\int_{0}^{\pi/4}\log(\sin x)\ dx-\int_{0}^{\pi/4}\log(\tan x)\ dx-\frac12\int_{0}^{\pi/4}\log(\sin x)\ dx+\frac12\int_{0}^{\pi/4}\ln2\ dx\\
&=\frac12\int_{0}^{\pi/4}\log(\sin x)\ dx-\int_{0}^{\pi/4}\log(\tan x)\ dx+\frac12\int_{0}^{\pi/4}\ln2\ dx\\
&=\frac12\left(-\frac{G}{2}-\frac{\pi}{4}\ln2\right)-(-G)+\frac12\ln2\left(\frac{\pi}{4}\right)\\
&=\frac34G
\end{align}