[proofplan] We first prove the formula in the absolute convergence region $\operatorname{Re}(s)>1$ by comparing Euler products: the principal character deletes exactly the Euler factors corresponding to primes dividing $q$. The finite product over $p \mid q$ is entire, so multiplying the meromorphic continuation of $\zeta(s)$ gives the meromorphic continuation of $L(s,\chi_0)$ to $\operatorname{Re}(s)>0$. Finally, the pole and residue at $s=1$ are obtained by multiplying the simple pole of $\zeta(s)$ by the nonzero finite factor evaluated at $s=1$. [/proofplan] [step:Compare Euler products in the half-plane of absolute convergence] Define the finite Euler factor \begin{align*} P_q: \mathbb{C} &\to \mathbb{C} \\ s &\mapsto \prod_{p \mid q}(1-p^{-s}), \end{align*} where the product is over the prime divisors of $q$, and where the empty product is $1$ when $q=1$. For $\operatorname{Re}(s)>1$, the Dirichlet series defining $L(s,\chi_0)$ is absolutely convergent because $|\chi_0(n)| \leq 1$ for all $n \in \mathbb{N}$ and $\sum_{n=1}^{\infty} n^{-\operatorname{Re}(s)}$ converges. In this region, the Euler product for Dirichlet series attached to completely multiplicative characters gives \begin{align*} L(s,\chi_0) = \prod_p (1-\chi_0(p)p^{-s})^{-1}. \end{align*} For a prime $p$, one has $\chi_0(p)=0$ if $p \mid q$ and $\chi_0(p)=1$ if $p \nmid q$. Hence \begin{align*} L(s,\chi_0) &= \prod_{p \nmid q}(1-p^{-s})^{-1} \\ &= \left(\prod_p (1-p^{-s})^{-1}\right)\prod_{p \mid q}(1-p^{-s}) \\ &= \zeta(s)P_q(s). \end{align*} Here the last equality uses the Euler product for the Riemann zeta function in $\operatorname{Re}(s)>1$. These Euler product facts are being used as prerequisites not yet resolved to wiki theorem IDs: Euler product for Dirichlet $L$-functions and Euler product for $\zeta(s)$. [guided] The purpose of this step is to identify exactly how the principal character changes the Euler product for $\zeta(s)$. Define \begin{align*} P_q: \mathbb{C} &\to \mathbb{C} \\ s &\mapsto \prod_{p \mid q}(1-p^{-s}). \end{align*} This is a finite product over the prime divisors of $q$; if $q=1$, there are no such primes, and the product is defined to be $1$. Let $s \in \mathbb{C}$ satisfy $\operatorname{Re}(s)>1$. Since $|\chi_0(n)| \leq 1$ for every $n \in \mathbb{N}$, we have \begin{align*} \sum_{n=1}^{\infty}\left|\frac{\chi_0(n)}{n^s}\right| \leq \sum_{n=1}^{\infty}\frac{1}{n^{\operatorname{Re}(s)}}<\infty. \end{align*} Thus the Dirichlet series defining $L(s,\chi_0)$ is absolutely convergent in this half-plane, so its Euler product is valid: \begin{align*} L(s,\chi_0) = \prod_p (1-\chi_0(p)p^{-s})^{-1}. \end{align*} The principal character satisfies $\chi_0(p)=0$ exactly for primes $p$ dividing $q$, and $\chi_0(p)=1$ for primes $p$ not dividing $q$. Therefore the Euler factor at a prime divisor of $q$ is \begin{align*} (1-\chi_0(p)p^{-s})^{-1}=(1-0)^{-1}=1, \end{align*} while the Euler factor at a prime not dividing $q$ is \begin{align*} (1-\chi_0(p)p^{-s})^{-1}=(1-p^{-s})^{-1}. \end{align*} Hence \begin{align*} L(s,\chi_0) = \prod_{p \nmid q}(1-p^{-s})^{-1}. \end{align*} The zeta Euler product in the same half-plane is \begin{align*} \zeta(s)=\prod_p (1-p^{-s})^{-1}. \end{align*} Multiplying this product by $\prod_{p \mid q}(1-p^{-s})$ cancels precisely the Euler factors with $p \mid q$, so \begin{align*} \zeta(s)\prod_{p \mid q}(1-p^{-s}) = \prod_{p \nmid q}(1-p^{-s})^{-1} = L(s,\chi_0). \end{align*} The Euler product facts used here are prerequisites not yet resolved to wiki theorem IDs: Euler product for Dirichlet $L$-functions and Euler product for $\zeta(s)$. [/guided] [/step] [step:Continue the identity meromorphically to $\operatorname{Re}(s)>0$] For each prime $p \mid q$, the function \begin{align*} s \mapsto 1-p^{-s}=1-e^{-s\log p} \end{align*} is entire, so $P_q$ is entire as a finite product of entire functions. The Riemann zeta function has a meromorphic continuation to $\operatorname{Re}(s)>0$ with its only pole in this half-plane at $s=1$ (citing a result not yet in the wiki: meromorphic continuation of the Riemann zeta function). Therefore \begin{align*} s \mapsto \zeta(s)P_q(s) \end{align*} is meromorphic on $\operatorname{Re}(s)>0$, with possible pole only at $s=1$. Since this [meromorphic function](/page/Meromorphic%20Function) agrees with $L(s,\chi_0)$ on the nonempty [open set](/page/Open%20Set) $\operatorname{Re}(s)>1$, it gives the meromorphic continuation of $L(s,\chi_0)$ to $\operatorname{Re}(s)>0$. Thus, for every $s$ with $\operatorname{Re}(s)>0$ and $s \neq 1$, \begin{align*} L(s,\chi_0)=\zeta(s)\prod_{p \mid q}(1-p^{-s}). \end{align*} [guided] The identity has already been proved where the original Dirichlet series converges absolutely, namely in $\operatorname{Re}(s)>1$. To move it into the larger half-plane $\operatorname{Re}(s)>0$, we inspect the right-hand side. For a fixed prime $p \mid q$, the map \begin{align*} \mathbb{C} &\to \mathbb{C} \\ s &\mapsto 1-p^{-s} \end{align*} is entire, because $p^{-s}=e^{-s\log p}$ and the exponential function is entire. Since there are only finitely many primes dividing $q$, the product \begin{align*} P_q(s)=\prod_{p \mid q}(1-p^{-s}) \end{align*} is also entire. Now use the meromorphic continuation of the Riemann zeta function to the half-plane $\operatorname{Re}(s)>0$, with only a pole at $s=1$ in that region. This prerequisite is not yet resolved to a wiki theorem ID. Multiplying the meromorphic function $\zeta(s)$ by the entire function $P_q(s)$ gives a meromorphic function \begin{align*} s \mapsto \zeta(s)P_q(s) \end{align*} on $\operatorname{Re}(s)>0$. On the smaller open half-plane $\operatorname{Re}(s)>1$, the previous step proved \begin{align*} L(s,\chi_0)=\zeta(s)P_q(s). \end{align*} Therefore the meromorphic function $\zeta(s)P_q(s)$ is the continuation of the original Dirichlet series defining $L(s,\chi_0)$. Consequently, throughout the half-plane $\operatorname{Re}(s)>0$, away from the possible singular point $s=1$, one has \begin{align*} L(s,\chi_0)=\zeta(s)\prod_{p \mid q}(1-p^{-s}). \end{align*} [/guided] [/step] [step:Evaluate the residue at the inherited pole] The Riemann zeta function has a simple pole at $s=1$ with residue $1$ (citing a result not yet in the wiki: simple pole of $\zeta(s)$ at $s=1$). Thus there is a function \begin{align*} h: \{s \in \mathbb{C}:\operatorname{Re}(s)>0\} &\to \mathbb{C} \end{align*} holomorphic near $s=1$ such that \begin{align*} \zeta(s)=\frac{1}{s-1}+h(s) \end{align*} near $s=1$. Since $P_q$ is holomorphic near $s=1$, we obtain \begin{align*} L(s,\chi_0) = \left(\frac{1}{s-1}+h(s)\right)P_q(s). \end{align*} The coefficient of $(s-1)^{-1}$ in this Laurent expansion is $P_q(1)$. Hence \begin{align*} \operatorname{Res}_{s=1}L(s,\chi_0) = P_q(1) = \prod_{p \mid q}(1-p^{-1}). \end{align*} Each factor $1-p^{-1}$ is nonzero, so the pole remains simple. [guided] The only possible singularity in the half-plane $\operatorname{Re}(s)>0$ comes from $\zeta(s)$ at $s=1$, because $P_q$ is entire. The zeta function has a simple pole at $s=1$ with residue $1$; this prerequisite is not yet resolved to a wiki theorem ID. Equivalently, there is a function \begin{align*} h: \{s \in \mathbb{C}:\operatorname{Re}(s)>0\} &\to \mathbb{C} \end{align*} holomorphic in a neighbourhood of $s=1$ such that \begin{align*} \zeta(s)=\frac{1}{s-1}+h(s) \end{align*} near $s=1$. Multiplying by the finite Euler factor gives \begin{align*} L(s,\chi_0) = \zeta(s)P_q(s) = \left(\frac{1}{s-1}+h(s)\right)P_q(s). \end{align*} Since $P_q$ is holomorphic at $s=1$, the coefficient of $(s-1)^{-1}$ in this Laurent expansion is exactly $P_q(1)$. Therefore \begin{align*} \operatorname{Res}_{s=1}L(s,\chi_0) = P_q(1) = \prod_{p \mid q}(1-p^{-1}). \end{align*} No cancellation of the pole occurs, because for every prime $p \mid q$, \begin{align*} 1-p^{-1}>0. \end{align*} Thus $P_q(1)\neq 0$, and the pole inherited from $\zeta(s)$ is still simple. [/guided] [/step] [step:Identify the finite Euler factor with $\varphi(q)/q$] Write the prime factorisation of $q$ as \begin{align*} q=\prod_{j=1}^{r} p_j^{a_j}, \end{align*} where $r \in \mathbb{N}\cup\{0\}$, the $p_j$ are distinct primes, and $a_j \in \mathbb{N}$. The Euler totient product formula gives \begin{align*} \varphi(q)=q\prod_{j=1}^{r}\left(1-\frac{1}{p_j}\right) \end{align*} (citing a result not yet in the wiki: Euler product formula for Euler's totient function). Dividing by $q$ gives \begin{align*} \frac{\varphi(q)}{q} = \prod_{j=1}^{r}\left(1-\frac{1}{p_j}\right) = \prod_{p \mid q}(1-p^{-1}). \end{align*} Combining this with the residue computation proves \begin{align*} \operatorname{Res}_{s=1}L(s,\chi_0) = \prod_{p \mid q}(1-p^{-1}) = \frac{\varphi(q)}{q}. \end{align*} This completes the proof. [/step]