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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

cactus314
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    Although there are elementary proofs, the most standard derivation of the value of $L(1,\chi_2)$ is through Dirichlet's "class number formula". – Greg Martin Aug 22 '14 at 23:22

3 Answers3

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

Jack D'Aurizio
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After thinking about it for a while:

$$ L(\chi_1,2) = \tfrac{1}{1^2} + \tfrac{1}{2^2} + 0 + \tfrac{1}{4^2} + \tfrac{1}{5^2} + 0 + \dots = \sum \frac{1}{n^2} - \sum \frac{1}{(3n)^2} = \left( 1 - \frac{1}{9}\right) \sum \frac{1}{n^2} = \frac{8}{9} \zeta(2)$$

cactus314
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  • Lol, so much easier than the way I took :) – Jack D'Aurizio Aug 22 '14 at 22:30
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    Did you find a clever way to prove that $L(1,\chi_2)=\frac{\pi}{3\sqrt{3}}$? I though it is possible to use the Kronecker theorem, relating $L(1,\chi)$ (where $\chi$ is the Legendre symbol) to the number of reduced quadratic form. – Jack D'Aurizio Aug 22 '14 at 22:45
  • @JackD'Aurizio No I haven't. My proof extends to $L(2k,\chi_1)= (1 - \tfrac{1}{3^{2k}})\zeta(2k)$ but in the odd case, I am trying to add $$ L(1, \chi_2)= \sum \frac{1}{9n^2 + 9n + 2} $$ This type of problem has appeard on Math.SE I think. – cactus314 Aug 22 '14 at 22:54
  • @johnmangual That contradicts what you read about $L(\chi_1,2)\in\pi^2\sqrt 3\mathbb Q$. Where did you read that? – Thomas Andrews Aug 22 '14 at 22:56
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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 $$L(1,\chi_2)=\frac{1}{2}\left(\frac{\pi}{3\sqrt{3}}-\left(-\frac{\pi}{3\sqrt{3}}\right)\right)=\frac{\pi}{3\sqrt{3}}$$

SuperAbound
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