Analytic Continuation of Dirichlet L-Function

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Theorem

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



Let $\map L {s, \chi}$ be the Dirichlet $L$-function for $\chi$.


Let $\chi$ be the trivial character.

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


Let $\chi$ be non-trivial.

Then $\map L {s, \chi}$ is analytic on $\map \Re s > 0$.


Proof

Let $\chi$ be the trivial character.

Then by Dirichlet L-Function from Trivial Character:

$\ds \map L {s, \chi} = \map \zeta s \cdot \prod_{p \mathop \divides q} \paren {1 - p^{-s} }$

where $\divides$ denotes divisibility.

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

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


If $\chi$ is non-trivial, then by the Orthogonality Relations for Characters:

$\ds \sum_{x \mathop \in G} \map \chi x = 0$

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

So for any $M \in \N$:

$\ds \sum_{n \mathop = M + 1}^{M + Q} \map \chi n = 0$


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

Let $d$ be such that $M + q d \le N \le M + q \paren {d + 1}$.


Then:

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

So the coefficients $\map \chi n$ of $\map L {s, \chi}$ have bounded partial sums.

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

$\blacksquare$