How do we evaluate this Dirichlet L-series

In this answer, David Speyer, whose answer is magnificent, states that "The sum $\sum \chi_3(n)/n$ is only slightly less well known; it is $\pi/(3 \sqrt{3})$.", where $\chi_3(n)$ is the character sending 1 mod 3 to 1 , 2 mod 3 to -1 and 0 mod 3 to 0.

How does one calculate this Dirichlet L-function?

Bonus question:Also, is there a way to generalise the methods in David Speyer's answer, at least for when the number alpha is a fundamental unit in a quadratic number ring that is a PID?Can someone explain why the number in the question (namely $2+\sqrt3$) has these miraculous properties (for instance, the region D becomes a fundamental one mod $\Gamma$).All this seems a bit serendipitous to me ( but then again, I'm no expert)

• There is always the possibility to compute $L(0,\chi)$ math.stackexchange.com/a/1683657/276986 $L(0,\chi) = - \frac{1}{3} \sum_{n=1}^2 \chi(n) n = \frac{1}{3}$ and then using the functional equation – reuns Jun 9 '16 at 16:49
• I see Eric Naslund explains a general method for computing $L(1,\chi)$ math.stackexchange.com/a/112925/276986 using the discrete Fourier transform of the character so that it reduces to computing $\displaystyle \sum_{n=1}^\infty \frac{e^{2 i \pi n a / q}}{n}$ for $a = 1 \ldots q-1$ (here $q=3$) – reuns Jun 9 '16 at 16:52
• Take a look at math.stackexchange.com/questions/426325/… – Bogdan Simeonov Jun 9 '16 at 17:03
• Come on, your second question assumes we studied in details David Speyer's 100 lines answer, so try to make it self-contained in a few lines. – reuns Jun 9 '16 at 17:12
• That's exactly why it's difficult to contain in a few lines, but i will try: So firstly, we tried to calculate the integral in the question, which Jim Belk restated as a double sum.Then, we notice that the quadratic form in the denominator actually arose from a particular kind of set of the numbers in $Z(\sqrt3),$ which turns out to be a fundamental domain modulo the set of units, which the number alpha is a generator of.... – Bogdan Simeonov Jun 9 '16 at 17:31

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: $$\color{red}{L(1,\chi_2)}=\int_{0}^{1/2}\frac{dx}{x^2+3/4}=\frac{1}{\sqrt{3}}\arctan\sqrt{3}=\color{red}{\frac{\pi}{3\sqrt{3}}.}$$