Analogue of $\zeta(2) = \frac{\pi^2}{6}$ for Dirichlet L-series of $\mathbb{Z}/3\mathbb{Z}$? Consider the two Dirichlet characters of $\mathbb{Z}/3\mathbb{Z}$.
$$ 
\begin{array}{c|ccr}
& 0 & 1 & 2 \\ \hline
\chi_1 & 0 & 1 & 1 \\
\chi_2  & 0 & 1 & -1
\end{array}
$$
I read the L-functions for these series have special values


*

*$ L(2,\chi_1) \in \pi^2 \sqrt{3}\;\mathbb{Q} $

*$ L(1,\chi_2) \in \pi \sqrt{3}\;\mathbb{Q} $


In other words, these numbers are $\pi^k \times \sqrt{3} \times \text{(rational number)}$.  Is there a way to derive this similar to to the famous $\zeta(2) = \tfrac{\pi^2}{6}$ formula?

Here are 14 proofs of $\zeta(2) = \tfrac{\pi^2}{6}$ for reference

We have:
$$L(2,\chi_1)=\sum_{j=0}^{+\infty}\left(\frac{1}{(3j+1)^2}+\frac{1}{(3j+2)^2}\right)=-\int_{0}^{1}\frac{(1+x)\log x}{1-x^3}\,dx$$
and integration by parts gives:
$$\begin{eqnarray*}\color{red}{L(2,\chi_1)}&=&-\int_{0}^{1}\frac{\log(1-x)}{x}\,dx+\frac{1}{3}\int_{0}^{1}\frac{\log(1-x^3)}{x}\,dx\\&=&-\frac{8}{9}\int_{0}^{1}\frac{\log x}{1-x}=\frac{8}{9}\zeta(2)=\color{red}{\frac{4\pi^2}{27}.}\end{eqnarray*}$$
With a similar technique:
$$L(1,\chi_2)=\sum_{j=0}^{+\infty}\left(\frac{1}{3j+1}-\frac{1}{3j+2}\right)=\int_{0}^{1}\frac{1-x}{1-x^3}\,dx=\int_{0}^{1}\frac{dx}{1+x+x^2}$$
so:
$$\color{red}{L(1,\chi_2)}=\int_{0}^{1/2}\frac{dx}{x^2+3/4}=\frac{1}{\sqrt{3}}\arctan\sqrt{3}=\color{red}{\frac{\pi}{3\sqrt{3}}.}$$
