To show $\lim_{\xi \to 1} (\xi -1) L(\xi,\chi_{0}) = \frac{\Phi(N)}{N}$

How to show if $\chi_{0}$ is the trivial $\text{Dirichlet Character}$ then $$\lim_{\xi \to 1} (\xi -1) L(\xi,\chi_{0}) = \frac{\Phi(N)}{N}$$

where $\Phi$ is the $\text{Euler's Totient}$.

-

Here are some hints:

• Use the product formula for $L(s, \chi) = \displaystyle\prod_{p} \biggl(1- \frac{\chi(p)}{p^s}\biggr)^{-1}$. Note $\chi_{0}(n) = \left\{\begin{array}{cc} 1 & p \nmid n \\\ 0 & p \mid n\end{array}\right.$

• Use the formula $\Phi(n) = \displaystyle n \cdot \prod_{p} \biggl(1-\frac{1}{p}\biggr)$

• Use the fact that $\displaystyle\lim_{z \to 1}\: (z-1)\:\zeta(z) =1$.

Complete solution:

See page 10 of the following link

-
Thanks a lot for the answer. – Sarah Apr 17 '12 at 19:09