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 see that
\begin{align}
\frac{1}{2\pi i}\oint_{C_N}\frac{\pi\cot(\pi z)}{(3z+1)^2}{\rm d}z
&=\operatorname*{Res}_{z=-1/3}\frac{\pi\cot(\pi z)}{(3z+1)^2}+\sum^\infty_{n=-\infty}\frac{1}{(3n+1)^2}\\
&=-\frac{4\pi^2}{27}+\sum^\infty_{n=-\infty}\frac{1}{(3n+1)^2}\\
&=0
\end{align}
and
\begin{align}
\frac{1}{2\pi i}\oint_{C_N}\frac{\pi\cot(\pi z)}{(3z+2)^2}{\rm d}z
&=\operatorname*{Res}_{z=-2/3}\frac{\pi\cot(\pi z)}{(3z+2)^2}+\sum^\infty_{n=-\infty}\frac{1}{(3n+2)^2}\\
&=-\frac{4\pi^2}{27}+\sum^\infty_{n=-\infty}\frac{1}{(3n+2)^2}\\
&=0
\end{align}
This implies
$$L(2,\chi_1)=\frac{1}{2}\left(\frac{4\pi^2}{27}+\frac{4\pi^2}{27}\right)=\frac{4\pi^2}{27}$$
Similarly,
\begin{align}
\frac{1}{2\pi i}\oint_{C_N}\frac{\pi\cot(\pi z)}{3z+1}{\rm d}z
&=\operatorname*{Res}_{z=-1/3}\frac{\pi\cot(\pi z)}{3z+1}+\sum^\infty_{n=-\infty}\frac{1}{3n+1}\\
&=-\frac{\pi}{3\sqrt{3}}+\sum^\infty_{n=-\infty}\frac{1}{3n+1}\\
&=0
\end{align}
and
\begin{align}
\frac{1}{2\pi i}\oint_{C_N}\frac{\pi\cot(\pi z)}{3z+2}{\rm d}z
&=\operatorname*{Res}_{z=-2/3}\frac{\pi\cot(\pi z)}{3z+2}+\sum^\infty_{n=-\infty}\frac{1}{3n+2}\\
&=\frac{\pi}{3\sqrt{3}}+\sum^\infty_{n=-\infty}\frac{1}{3n+2}\\
&=0
\end{align}
Hence
$$L(1,\chi_2)=\frac{1}{2}\left(\frac{\pi}{3\sqrt{3}}-\left(-\frac{\pi}{3\sqrt{3}}\right)\right)=\frac{\pi}{3\sqrt{3}}$$
