Dirichlet $L$ functions at $s=2$

Question

Let $\chi$ be a Dirichlet character and let $L(\chi,s)$ denotes its Dirichlet $L$-function.

What is the value of $L(2,\chi)$ ? Or simply, is $L(2,\chi)/\pi^2$ rational ?

Many thanks for your answer !

Answer 1

It depends on which character you are considering.

For instance, let $\chi$ be the non-principal real character $\!\!\pmod{3}$. Then: $$ L(2,\chi)=\sum_{n\geq 0}\left(\frac{1}{(3n+1)^2}-\frac{1}{(3n+2)^2}\right)\tag{1}$$ and since $\int_{0}^{1} x^{n}\log x\,dx = -\frac{1}{(n+1)^2}$, we have: $$ L(2,\chi) = -\int_{0}^{1}\frac{1-x}{1-x^3}\log x\,dx=-\int_{0}^{1}\frac{\log x}{1+x+x^2}\,dx =\int_{0}^{+\infty}\frac{t\,dt}{1+2\cosh t}\tag{2}$$ or: $$ L(2,\chi) = \frac{1}{9}\left(\psi'\left(\frac{1}{3}\right)-\psi'\left(\frac{2}{3}\right)\right)\tag{3}$$ where $\psi$ is the digamma function $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$. However, $$\begin{eqnarray*}\frac{L(2,\chi)}{\pi^2}&=&0.079162485257286498865527381554913\ldots\\&=&[0; 12, 1, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 14, 14, 1, 33,\ldots]\end{eqnarray*}$$ does not look as a rational number. By Dirichlet convolution: $$\frac{L(2,\chi)}{\pi^2}=\frac{1}{6}\sum_{n\geq 1}\frac{u(n)}{n^2},\qquad u(n)=\sum_{d\mid n}\chi(d)\mu(n/d)\tag{4}$$ and by Euler's product: $$\frac{L(2,\chi)}{\pi^2}=\frac{1}{6}\prod_{p\equiv -1\!\!\pmod{\!\!3}}\frac{p^2-1}{p^2+1}.\tag{5}$$

Answer 2

I finally find a partial answer, I leave it here in addition to Jack's answer.

Take $\chi$ an even Dirichlet character (say modulo $N$) of conductor $N_0$ associated to the primitive character $\chi_0$. We know (see Apostol's Introduction to Analytic Number Theory for instance) that $$L(2,\chi)=L(2,\chi_0)\prod_{p|N}{\left(1-\frac{\chi_0(p)}{p^2}\right)}$$ Now, $\chi_0$ is even and primitive so we can apply the functional equation to find $$\frac{\sqrt{\pi}}{G(\bar{\chi_0})}\Gamma(-1/2)L(-1,\bar{\chi_0})=\frac{N_0}{\pi}L(2,\chi_0)$$ whith $\Gamma(-1/2)=-2\sqrt{\pi}$, $G(\chi_0)$ the Gauss' sum of $\chi_0$ and \begin{align} L(-1,\chi_0) &= N_0\sum_{n=1}^{N_0}{\chi_0(n)\zeta(-1,\frac{n}{N_0})} \\ &= -\frac{N_0}{2}\sum_{n=1}^{N_0}{\chi_0(n)\left(\frac{n^2}{N_0^2}-\frac{n}{N_0}+\frac{1}{6}\right)} \\ &=-\frac{1}{2N_0}\sum_{n=1}^{N_0}{n^2\chi_0(n)} \end{align} (where $\zeta(s,a)$ is the Hurwitz zeta function) since one has (again $\chi_0$ is even, see Apostol) $$\sum_{n=1}^{N_0}{\chi_0(n)}=\sum_{n=1}^{N_0}{n\chi_0(n)}=0.$$ From that, we easily find $$L(2,\chi)=\frac{1}{G(\bar{\chi_0})}\left(\frac{\pi}{N_0}\right)^2\left(\sum_{n=1}^{N_0}{n^2\chi_0(n)}\right)\prod_{p|N}{\left(1-\frac{\chi_0(p)}{p^2}\right)}$$ In particular, when $\chi$ is real and even, $\sqrt{N_0} L(2,\chi)/\pi^2$ is rational.