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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)

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  • There is always the possibility to compute $L(0,\chi)$ http://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 09 '16 at 16:49
  • I see Eric Naslund explains a general method for computing $L(1,\chi)$ http://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 09 '16 at 16:52
  • Take a look at http://math.stackexchange.com/questions/426325/evaluate-int-01-frac-log-left-1x2-sqrt3-right1x-mathrm-dx/428709#428709 – B. S. Jun 09 '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 09 '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.... – B. S. Jun 09 '16 at 17:31
  • Then we restate the sum as a sum of representatives mod the group of units, then induce a character which is +-1, depending on the sign of the norm of a number.Since the number ring is a PID, we have that the sums can be restated as sums over ideals.Using the class number formula and the euler product, David Speyer calculates the first of the two sums.The second sum comes down to calculating an L-function, which is the product $\sum \chi_4(n).\sum \chi_3(n)$,which is where my question is relevant. – B. S. Jun 09 '16 at 17:36
  • All of this was just a restatement of what is in the answer and won't probably illuminate anything to you, but I tried my best :) – B. S. Jun 09 '16 at 17:37
  • so your second question is only about $N_{\mathbb{Z}(\sqrt{3})/\mathbb{Z}}(2 +\sqrt{3})$ https://en.wikipedia.org/wiki/Ideal_norm why is known to be the same as the field norm $ N_{\mathbb{Q}(\sqrt{3})/\mathbb{Q}}(2 +\sqrt{3})$ which is the determinant of the $\mathbb{Q}$ linear operator representing $x \mapsto (2+\sqrt{3}) x$ in $\mathbb{Q}(\sqrt{3})$. And it is related to $L(s,\chi)$ where $\chi$ is the quadratic character modulo $3$ by a general property of quadratic fields (non trivial, but David tried to explain it by elementary means) https://en.wikipedia.org/wiki/Dedekind_zeta_function – reuns Jun 09 '16 at 17:46
  • and I don't get what you mean with $\chi_4$, this is why you should re-write your question, for making it self-contained (and when writing it everything will become much clearer at least for you) – reuns Jun 09 '16 at 17:53
  • i'm afraid in order to make it self contained ill have to rewrite the whole post.Also, chi_4 is a character sending 1 mod 4 to 1, 2 mod 4 and 4 mod 4 to 0 and 3 mod 4 to -1, so the sum i mentioned is one minus a third plus a fifth and so on, which is arctan 1. – B. S. Jun 09 '16 at 17:58
  • right, his $\sigma_2(m+ \sqrt{3}n) = (-1)^{n+m}$ takes us to a $L(s,\chi_{\bmod 12}) =L(s,\chi_3 \chi_4)$ – reuns Jun 09 '16 at 18:13

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I have found a very easy proof from Jack D'Aurizio's answer here:

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

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