Conjectured closed form for $\operatorname{Li}_2\!\left(\sqrt{2-\sqrt3}\cdot e^{i\pi/12}\right)$

Reduction of dilogarithm. Known identities for the dilogarithm allow us the following simple transform:

$$ \operatorname{Li}_2\left(\tfrac{1}{2} + \tfrac{i}{2} \tan\theta\right) = - \operatorname{Li}_2(e^{i(\pi+2\theta)}) -\tfrac{1}{2}\log^2\left(\tfrac{1}{2} - \tfrac{i}{2} \tan\theta \right). $$

In this case, we have $\theta = \tfrac{1}{12}\pi$ and hence the dilogarithm in question is written as

$$ \operatorname{Li}_2\left(\tfrac{1}{2} + \tfrac{i}{2}\tan\tfrac{\pi}{12} \right) = - \operatorname{Li}_2(e^{7\pi i/6}) -\tfrac{1}{2}\log^2\left( \tfrac{1}{2} - \tfrac{i}{2}\tan\tfrac{\pi}{12} \right). \tag{1} $$

Now utilizing the Fourier series of the Bernoulli polynomial, we know that $\operatorname{Li}_2(e^{i\theta})$ reduces to

$$ \operatorname{Li}_2(e^{i\theta}) = \sum_{k=1}^{\infty} \tfrac{1}{k^2}(\cos (k\theta) + i\sin(k\theta)) = \tfrac{1}{6}\pi^2 - \tfrac{1}{2}\pi\theta + \tfrac{1}{4}\theta^2 + i \operatorname{Cl}_2(\theta) \tag{2} $$

for $0 \leq \theta \leq 2\pi$, where $\operatorname{Cl}_2(\theta) = \sum_{k=1}^{\infty} k^{-2}\sin(k\theta)$ is the Clausen function. So it remains to simplify $\operatorname{Cl}_2 \left( \tfrac{7}{6}\pi \right)$.

Reduction of Clausen function. To this end, we group the terms

$$ \operatorname{Cl}_2 \left(\tfrac{7}{6}\pi\right) = \sum_{k=1}^{\infty} \frac{1}{k^2} \sin \left(\tfrac{7}{6}k\pi\right) $$

according the value of sine and simplifying each group as in this proof, we find that

$$ \operatorname{Cl}_2 \left(\tfrac{7}{6}\pi\right) = \tfrac{1}{144} \sum_{j=1}^{6} \sin \left(\tfrac{7}{6}j\pi\right) \left( \psi^{(1)}\left(\tfrac{j}{12}\right) - \psi^{(1)}\left(\tfrac{1}{2}+\tfrac{j}{12}\right) \right). \tag{3} $$

Now utilizing the reflection formula and the duplication and triplication formula extensively, we can simplify the above sum as

$$ \operatorname{Cl}_2 \left(\tfrac{7}{6}\pi\right) = -\tfrac{2}{3}G - \tfrac{1}{12\sqrt{3}}\pi^2 + \tfrac{1}{8\sqrt{3}} \psi ^{(1)}\left(\tfrac{1}{3}\right). \tag{4} $$

Indeed, we expand the summation in (3) and utilize the triplication formula to the green-colored groups and the duplication formula to the blue-colored groups to obtain

\begin{align*} \operatorname{Cl}_2 \left(\tfrac{7}{6}\pi\right) & = - \tfrac{1}{288} \psi^{(1)}\left(\tfrac{1}{12}\right) + \tfrac{1}{96 \sqrt{3}} \psi^{(1)}\left(\tfrac{1}{6}\right) - \tfrac{1}{144} \psi^{(1)}\left(\tfrac{1}{4}\right) + \tfrac{1}{96 \sqrt{3}} \psi^{(1)}\left(\tfrac{1}{3}\right) - \tfrac{1}{288} \psi^{(1)}\left(\tfrac{5}{12}\right) \\ &\quad + \tfrac{1}{288} \psi^{(1)}\left(\tfrac{7}{12}\right) - \tfrac{1}{96\sqrt{3}} \psi^{(1)}\left(\tfrac{2}{3}\right) + \tfrac{1}{144} \psi^{(1)}\left(\tfrac{3}{4}\right) - \tfrac{1}{96 \sqrt{3}} \psi^{(1)}\left(\tfrac{5}{6}\right) + \tfrac{1}{288} \psi^{(1)}\left(\tfrac{11}{12}\right) \\ & = - \tfrac{1}{288} \color{green}{\left( \psi^{(1)}\left(\tfrac{1}{12}\right) + \psi^{(1)}\left(\tfrac{5}{12}\right) + \psi^{(1)}\left(\tfrac{9}{12}\right) \right)} - \tfrac{3}{288} \psi^{(1)}\left(\tfrac{1}{4}\right) \\ &\quad + \tfrac{1}{288} \color{green}{\left( \psi^{(1)}\left(\tfrac{3}{12}\right) + \psi^{(1)}\left(\tfrac{7}{12}\right) + \psi^{(1)}\left(\tfrac{11}{12}\right) \right)} + \tfrac{3}{288} \psi^{(1)}\left(\tfrac{3}{4}\right) \\ &\quad + \tfrac{1}{96 \sqrt{3}} \color{blue}{\left( \psi^{(1)}\left(\tfrac{1}{6}\right) + \psi^{(1)}\left(\tfrac{2}{3}\right) \right)} + \tfrac{1}{48 \sqrt{3}} \psi^{(1)}\left(\tfrac{1}{3}\right) \\ &\quad - \tfrac{1}{96 \sqrt{3}} \color{blue}{\left( \psi^{(1)}\left(\tfrac{1}{3}\right) + \psi^{(1)}\left(\tfrac{5}{6}\right) \right)} - \tfrac{1}{48\sqrt{3}} \psi^{(1)}\left(\tfrac{2}{3}\right) \\ & = \tfrac{1}{16 \sqrt{3}} \left( \psi^{(1)}\left(\tfrac{1}{3}\right) - \psi^{(1)}\left(\tfrac{2}{3}\right) \right) - \tfrac{1}{24} \left( \psi^{(1)}\left(\tfrac{1}{4}\right) - \psi^{(1)}\left(\tfrac{3}{4}\right) \right). \end{align*}

Now we focus on the last line. Applying the reflection formula to the first term and comparing the definition of the Catalan constant $G$ with the second term, we obtain (4) as claimed.

Finally, plugging this back gives the desired result.

Addendum: Fourier series of the Bernoulli polynomial. Taking imaginary part of

$$ \log(1-e^{2\pi i x}) = - \sum_{k=1}^{\infty} \frac{e^{2\pi i k x}}{k} $$

for $x \in (0, 1)$, we find that the Bernoulli polynomial $B_1(x)$ of degree 1 is written as

$$ B_1( x ) = x - \tfrac{1}{2} = - \frac{1}{2\pi i} \sum_{k \neq 0} \frac{e^{2\pi i k x}}{k}, \quad 0 < x < 1.$$

Integrating both sides repeatedly and using the relation $B_n'(x) = nB_{n-1}(x)$, we find that for any $n \geq 1$,

$$ B_n( x ) = - \frac{n!}{(2\pi i)^n} \sum_{k \neq 0} \frac{e^{2\pi i k x}}{k^n}, \quad 0 < x < 1.$$

Notice that depending on whether $n$ is even or odd, this reduces to either cosine series or sine series. For example, when $n = 2$ we have

$$ x^2 - x + \tfrac{1}{6} = B_2(x) = \frac{1}{\pi^2} \sum_{k=1}^{\infty} \frac{\cos(2\pi k x)}{k^2} $$

and hence we obtain the formula which was used in our solution:

$$ \sum_{k=1}^{\infty} \frac{\cos(k x)}{k^2} = \pi^2 B_2\left(\tfrac{1}{2\pi} x\right) = \tfrac{1}{4} x^2 - \tfrac{1}{2}\pi x + \tfrac{1}{6}\pi^2. $$

We use the main three steps from my answer to one of your previous questions.

$$\operatorname{Li}_2(z) = -\operatorname{Li}_2\left(\frac{z}{z-1}\right)-\frac{1}{2}\ln^2(1-z), \quad z \notin (1,\infty).\tag{$\diamondsuit$}$$ $$\operatorname{Li}_2\left(e^{i\theta}\right) = \operatorname{Sl}_2(\theta)+i\operatorname{Cl}_2(\theta), \quad \theta \in [0,2\pi).\tag{$\heartsuit$} $$ $$ \operatorname{Sl}_2(\theta) = \frac{\pi^2}{6}-\frac{\pi\theta}{2}+\frac{\theta^2}{4}, \quad \theta \in [0,2\pi).\tag{$\spadesuit$} $$ For definitions and notations see my previous answer.

Let $z:=\tfrac{1}{2}+i\left(1-\tfrac{\sqrt3}{2}\right).$ Note that $$ \frac{z}{z-1} = -\frac{\sqrt 3}{2} - \frac{i}{2}, $$ and therefore $\left|\frac{z}{z-1}\right|=1$.

The equation $$ e^{i\theta} = \frac{z}{z-1} = -\frac{\sqrt 3}{2} - \frac{i}{2} $$ has the only solution $\theta=\tfrac76 \pi$ in $[0,2\pi)$.

Because of $(\diamondsuit)$ and $(\heartsuit)$ we have $$ \operatorname{Li}_2(z) = -\color{red}{\operatorname{Sl}_2(\theta)} - i \color{green}{\operatorname{Cl}_2(\theta)} - \color{blue}{\frac{1}{2}\ln^2(1-z)}, $$ for $z=\tfrac{1}{2}+i\left(1-\tfrac{\sqrt3}{2}\right)$ and $\theta =\tfrac76 \pi$.

For the logarithm term we get $$ \Re{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = \frac{1}{8}\ln^2\left(2-\sqrt{3}\right)-\frac{\pi^2}{288} $$ and $$ \Im{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = -\frac{\pi}{24}\ln\left(2-\sqrt{3}\right). $$ We know that $\color{red}{\operatorname{Sl}_2(\theta)}$ and $\color{green}{\operatorname{Cl}_2(\theta)}$ are real quantities. By using $(\spadesuit)$ for the SL-type Clausen term we get $$ \color{red}{\operatorname{Sl}_2(\theta)} =-\frac{11\pi^2}{144}. $$ Now we could obtain your conjectured closed-form for the real part: $$\Re\left[\operatorname{Li}_2(z)\right] = -\color{red}{\operatorname{Sl}_2(\theta)} - \Re{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = \frac{23\pi^2}{288}-\frac{1}{8}\ln^2\left(2-\sqrt{3}\right).$$

For the imaginary part we have $$\begin{align}\Im\left[\operatorname{Li}_2(z)\right] &= -\color{green}{\operatorname{Cl}_2(\theta)} - \Im{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} \\ &= -\operatorname{Cl}_2\left(\tfrac{7}{6}\pi\right)+\frac{\pi}{24}\ln\left(2-\sqrt{3}\right).\end{align}$$ Now by using the relationship between Clausen function and polygamma function and by using some polygamma tricks, we can get that $$\operatorname{Cl}_2 \left(\frac{7}{6}\pi\right) = -\frac{2}{3}G - \frac{\pi^2}{12\sqrt{3}} + \frac{1}{8\sqrt{3}} \psi ^{(1)}\left(\frac{1}{3}\right).$$ This completes the proof.