# Analytic Continuation of Dirichlet L-Function

## Theorem

Let $\chi : G := \left({\Z / q \Z}\right)^\times \to \C^\times$ be a Dirichlet character modulo $q$.

Let $L \left({s, \chi}\right)$ be the Dirichlet $L$-function for $\chi$.

Let $\chi$ be the trivial character.

Then $L \left({s, \chi}\right)$ has an analytic continuation to $\C$ except for a simple pole at $s = 1$.

Let $\chi$ be non-trivial.

Then $L \left({s, \chi}\right)$ is analytic on $\Re \left({s}\right) > 0$.

## Proof

Let $\chi$ be the trivial character.

$\displaystyle L \left({s, \chi}\right) = \zeta \left({s}\right) \cdot \prod_{p \mathop \backslash q} \left(1 - p^{-s}\right)$

where $\backslash$ denotes divisibility.

Also, by Poles of Riemann Zeta Function, $\zeta$ is analytic on $\C$ except for a simple pole at $s = 1$.

Since $L \left({s, \chi}\right)$ is just $\zeta$ times some constant, the same holds for this function.

If $\chi$ is nontrivial, then by the Orthogonality Relations for Characters:

$\displaystyle \sum_{x \mathop \in G} \chi \left({x}\right) = 0$

By definition, $\chi$ is $q$-periodic, and zero on integers not coprime to $q$.

So for any $M \in \N$:

$\displaystyle \sum_{n \mathop = M + 1}^{M + Q} \chi \left({n}\right) = 0$

Let $M, N \in \N$ be arbitrary.

Let $d$ be such that $M + q d \le N \le M + q \left({d + 1}\right)$.

Then:

 $\displaystyle \sum_{n \mathop = M}^N \chi \left({n}\right)$ $=$ $\displaystyle \sum_{k \mathop = 0}^{d-1} \sum_{n \mathop = 0}^{q-1} \chi \left({M + kq + n}\right) + \sum_{n \mathop = M + q d}^N \chi \left({n}\right)$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = M \mathop + q d}^N \chi \left({n}\right)$ because $\chi$ is $q$-periodic, and zero on integers not coprime to $q$ $\displaystyle$ $\le$ $\displaystyle q$ because $\left\vert{N - M + q d}\right\vert \le q$

So the coefficients $\chi \left({n}\right)$ of $L \left({s, \chi}\right)$ have bounded partial sums.

Therefore, by Convergence of Dirichlet Series with Bounded Partial Sums, $L \left({s, \chi}\right)$ converges locally uniformly to an analytic function on $\Re \left({s}\right) > 0$.

$\blacksquare$