# Tag Info

## New answers tagged dirichlet-series

2

By Kronecker's formula, the value at $s=1$ of $L(s,\chi)$ functions associated with quadratic characters depends on a class number, so there actually is an algebraic counterpart dealing with reduced binary quadratic forms. The basic problem is to find the sign of a Gauss sum, where $G(\chi)^2 = p$ is almost trivial. However, I am not so sure that "the ...

2

Well, by the same approach shown in my previous answer, if $\chi$ is a non-principal character $\!\!\pmod{p}$ and $s\in\mathbb{N}^*$, we have that: $$L(s,\chi)= \sum_{n\geq 1}\frac{\chi(n)}{n^s} = \int_{0}^{1}\frac{q_{\chi}(x)\log(x)^{s-1}}{1-x^{p}}\,dx$$ can be evaluated through partial fraction decomposition in terms of $\text{Li}_{s}$ evaluated at the $... 1 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\...

Top 50 recent answers are included