[proofplan] Let $\Lambda: \mathbb{N}\to\mathbb{R}$ denote the von Mangoldt function, and let $\chi_0: \mathbb{Z}\to\mathbb{C}$ denote the principal Dirichlet character modulo $q$. Define the ordinary Chebyshev function $\psi: [1,\infty)\to\mathbb{R}$ by $\psi(x)=\sum_{1\leq n\leq x}\Lambda(n)$, and define the Dirichlet $L$-function attached to $\chi$ on $\operatorname{Re}(s)>1$ by $L(s,\chi)=\sum_{n=1}^{\infty}\chi(n)n^{-s}$. For the principal character, $\chi_0$ removes exactly the prime powers whose prime base divides $q$, so $\psi(x,\chi_0)$ differs from $\psi(x)$ by at most $O_q(\log x)$; the [prime number theorem](/theorems/1742) for $\psi$ then gives $\psi(x,\chi_0)\sim x$. For a nonprincipal character, the Dirichlet series of $\Lambda(n)\chi(n)$ is the logarithmic derivative $-L'(s,\chi)/L(s,\chi)$ on $\operatorname{Re}(s)>1$. The local holomorphic continuation and boundary nonvanishing of nonprincipal Dirichlet $L$-functions, including imprimitive characters, make this logarithmic derivative holomorphic across the line $\operatorname{Re}(s)=1$, and the complex-coefficient Wiener-Ikehara theorem then gives $\sum_{n\leq x}\Lambda(n)\chi(n)=o(x)$. [/proofplan] [step:Compare the principal character sum with the ordinary Chebyshev function] Assume first that $\chi=\chi_0$. Define the ordinary Chebyshev function $\psi: [1,\infty) \to \mathbb{R}$ by \begin{align*} \psi(x) := \sum_{1 \leq n \leq x} \Lambda(n). \end{align*} Since $\chi_0(n)=1$ when $\gcd(n,q)=1$ and $\chi_0(n)=0$ when $\gcd(n,q)>1$, we have \begin{align*} \psi(x)-\psi(x,\chi_0) &= \sum_{\substack{1 \leq n \leq x \\ \gcd(n,q)>1}} \Lambda(n). \end{align*} The von Mangoldt function is supported on prime powers: $\Lambda(n)=\log p$ if $n=p^k$ for some prime $p$ and $k \in \mathbb{N}$, and $\Lambda(n)=0$ otherwise. Therefore only prime powers $p^k \leq x$ with $p \mid q$ contribute to the last sum. Hence \begin{align*} 0 \leq \psi(x)-\psi(x,\chi_0) &= \sum_{p \mid q}\sum_{\substack{k \in \mathbb{N} \\ p^k \leq x}} \log p \\ &= \sum_{p \mid q} \left\lfloor \frac{\log x}{\log p} \right\rfloor \log p \\ &\leq \sum_{p \mid q} \log x. \end{align*} If $\omega(q)$ denotes the number of distinct prime divisors of $q$, this gives \begin{align*} 0 \leq \psi(x)-\psi(x,\chi_0) \leq \omega(q)\log x. \end{align*} Thus \begin{align*} \psi(x,\chi_0)=\psi(x)+O_q(\log x). \end{align*} [guided] We isolate exactly what the principal character changes. The map $\chi_0: \mathbb{Z}\to\mathbb{C}$ is the indicator of the residue classes coprime to $q$: it equals $1$ on integers $n$ with $\gcd(n,q)=1$ and equals $0$ otherwise. Thus $\psi(x,\chi_0)$ is the ordinary Chebyshev sum with the terms not coprime to $q$ removed. Define \begin{align*} \psi: [1,\infty) &\to \mathbb{R} \\ x &\mapsto \sum_{1 \leq n \leq x} \Lambda(n). \end{align*} Then \begin{align*} \psi(x)-\psi(x,\chi_0) &= \sum_{\substack{1 \leq n \leq x \\ \gcd(n,q)>1}} \Lambda(n). \end{align*} The von Mangoldt function vanishes except at prime powers. More precisely, $\Lambda(n)=\log p$ when $n=p^k$ for a prime $p$ and an integer $k \in \mathbb{N}$, and $\Lambda(n)=0$ otherwise. If $\gcd(p^k,q)>1$, then necessarily $p \mid q$. Hence the difference is exactly bounded by summing over prime powers whose prime base divides $q$: \begin{align*} 0 \leq \psi(x)-\psi(x,\chi_0) &= \sum_{p \mid q}\sum_{\substack{k \in \mathbb{N} \\ p^k \leq x}} \log p. \end{align*} For a fixed prime $p$, the condition $p^k \leq x$ is equivalent to $k \leq \log x/\log p$, so the number of such $k$ is $\left\lfloor \log x/\log p \right\rfloor$. Therefore \begin{align*} \sum_{\substack{k \in \mathbb{N} \\ p^k \leq x}} \log p = \left\lfloor \frac{\log x}{\log p} \right\rfloor \log p \leq \log x. \end{align*} Summing over the finitely many prime divisors of $q$ gives \begin{align*} 0 \leq \psi(x)-\psi(x,\chi_0) \leq \omega(q)\log x, \end{align*} where $\omega(q)$ is the number of distinct prime divisors of $q$. This is precisely \begin{align*} \psi(x,\chi_0)=\psi(x)+O_q(\log x). \end{align*} [/guided] [/step] [step:Use the classical prime number theorem for the principal character] By the classical [prime number theorem](/theorems/1692) in Chebyshev form (citing a result not yet in the wiki: Prime Number Theorem for the Chebyshev Function), \begin{align*} \psi(x) \sim x. \end{align*} Since $\log x=o(x)$, the estimate from the preceding step gives \begin{align*} \frac{\psi(x,\chi_0)}{x} = \frac{\psi(x)}{x}+O_q\left(\frac{\log x}{x}\right) \to 1. \end{align*} Thus $\psi(x,\chi_0)\sim x$. [/step] [step:Identify the Dirichlet series for a nonprincipal character] Assume now that $\chi \neq \chi_0$. Define the Dirichlet $L$-function \begin{align*} L(\,\cdot\,,\chi): \{s \in \mathbb{C} : \operatorname{Re}(s)>1\} &\to \mathbb{C} \\ s &\mapsto \sum_{n=1}^{\infty}\frac{\chi(n)}{n^s}. \end{align*} On $\operatorname{Re}(s)>1$, the Euler product is absolutely convergent and gives \begin{align*} L(s,\chi)=\prod_p \left(1-\frac{\chi(p)}{p^s}\right)^{-1}. \end{align*} Taking the logarithmic derivative of the absolutely convergent Euler product gives \begin{align*} -\frac{L'}{L}(s,\chi) &= \sum_p \sum_{k=1}^{\infty} \frac{\chi(p)^k\log p}{p^{ks}}. \end{align*} Because a Dirichlet character is completely multiplicative and $\Lambda(p^k)=\log p$, this double series is exactly \begin{align*} -\frac{L'}{L}(s,\chi) = \sum_{n=1}^{\infty}\frac{\Lambda(n)\chi(n)}{n^s} \end{align*} for every $s \in \mathbb{C}$ with $\operatorname{Re}(s)>1$. [guided] The purpose of this step is to connect the summatory function $\psi(x,\chi)$ with a complex analytic object. For $\operatorname{Re}(s)>1$, define \begin{align*} L(\,\cdot\,,\chi): \{s \in \mathbb{C} : \operatorname{Re}(s)>1\} &\to \mathbb{C} \\ s &\mapsto \sum_{n=1}^{\infty}\frac{\chi(n)}{n^s}. \end{align*} Since $|\chi(n)|\leq 1$ for every $n\in\mathbb{N}$, this series converges absolutely on $\operatorname{Re}(s)>1$. The usual Euler product for Dirichlet characters therefore holds in this half-plane: \begin{align*} L(s,\chi)=\prod_p \left(1-\frac{\chi(p)}{p^s}\right)^{-1}. \end{align*} The product is absolutely convergent there, so taking logarithms and differentiating term by term is justified in $\operatorname{Re}(s)>1$. This gives \begin{align*} -\frac{L'}{L}(s,\chi) &= \sum_p \frac{\chi(p)\log p\,p^{-s}}{1-\chi(p)p^{-s}} \\ &= \sum_p \sum_{k=1}^{\infty} \frac{\chi(p)^k\log p}{p^{ks}}. \end{align*} Now we translate this double sum into a single Dirichlet series. Since $\chi$ is completely multiplicative, $\chi(p^k)=\chi(p)^k$. Since $\Lambda$ is supported on prime powers and satisfies $\Lambda(p^k)=\log p$, the coefficient attached to $p^{-ks}$ is precisely $\Lambda(p^k)\chi(p^k)$. Therefore \begin{align*} -\frac{L'}{L}(s,\chi) = \sum_{n=1}^{\infty}\frac{\Lambda(n)\chi(n)}{n^s} \end{align*} for every $s$ with $\operatorname{Re}(s)>1$. [/guided] [/step] [step:Extend the logarithmic derivative holomorphically across the boundary line] Since $\chi$ is nonprincipal, the standard [analytic continuation](/page/Analytic%20Continuation) theorem for Dirichlet $L$-functions gives the following local statement: for every point $s_0 \in \mathbb{C}$ with $\operatorname{Re}(s_0)=1$, the function $L(\,\cdot\,,\chi)$ extends holomorphically to an open neighbourhood of $s_0$. This applies also when $\chi$ is imprimitive. Indeed, if $\chi_1$ is the primitive character modulo $q_1$ inducing $\chi$, then \begin{align*} L(s,\chi)=L(s,\chi_1)\prod_{\substack{p\mid q \\ p\nmid q_1}}\left(1-\frac{\chi_1(p)}{p^s}\right) \end{align*} on $\operatorname{Re}(s)>1$, and the finite product gives the continuation. For $\operatorname{Re}(s)=1$, each extra factor is nonzero because \begin{align*} \left|\chi_1(p)p^{-s}\right|\leq p^{-1}<1, \end{align*} so $1-\chi_1(p)p^{-s}\neq 0$. The boundary nonvanishing theorem for nonprincipal Dirichlet $L$-functions says that this continuation has no zeros at points with $\operatorname{Re}(s)=1$ (citing a result not yet in the wiki: Nonvanishing of Dirichlet $L$-Functions on the Line $\operatorname{Re}(s)=1$). Hence, for every such $s_0$, there is an open neighbourhood $V_{s_0}\subset\mathbb{C}$ such that $L(s,\chi)\neq 0$ for all $s\in V_{s_0}$ and $L(\,\cdot\,,\chi)$ is holomorphic on $V_{s_0}$. Therefore \begin{align*} s \mapsto -\frac{L'}{L}(s,\chi) \end{align*} is holomorphic on $V_{s_0}$. Thus the Dirichlet series \begin{align*} \sum_{n=1}^{\infty}\frac{\Lambda(n)\chi(n)}{n^s} \end{align*} has a holomorphic continuation to a neighbourhood of every point on the boundary line $\operatorname{Re}(s)=1$. [guided] The Tauberian theorem we will use needs analytic continuation up to and across the boundary line $\operatorname{Re}(s)=1$. The previous step identified the Dirichlet series as $-L'/L$ only in the half-plane $\operatorname{Re}(s)>1$, so we now verify that this quotient has no singularity on the boundary. Because $\chi$ is nonprincipal, the analytic continuation theorem for Dirichlet $L$-functions gives local holomorphic continuation at every point of the line $\operatorname{Re}(s)=1$. If $\chi$ is imprimitive and is induced by a primitive character $\chi_1$ modulo $q_1$, then $L(s,\chi)$ is obtained from $L(s,\chi_1)$ by multiplying by finitely many Euler factors corresponding to primes dividing $q$ but not $q_1$: \begin{align*} L(s,\chi)=L(s,\chi_1)\prod_{\substack{p\mid q \\ p\nmid q_1}}\left(1-\frac{\chi_1(p)}{p^s}\right). \end{align*} These factors are holomorphic near $\operatorname{Re}(s)=1$. They also have no zeros on the line itself: if $\operatorname{Re}(s)=1$, then \begin{align*} \left|\chi_1(p)p^{-s}\right|\leq p^{-1}<1, \end{align*} so $1-\chi_1(p)p^{-s}\neq 0$. Thus the same local continuation and boundary nonvanishing statements apply to imprimitive nonprincipal characters. The boundary nonvanishing theorem for nonprincipal Dirichlet $L$-functions states that this continuation satisfies $L(s,\chi)\neq 0$ whenever $\operatorname{Re}(s)=1$ (citing a result not yet in the wiki: Nonvanishing of Dirichlet $L$-Functions on the Line $\operatorname{Re}(s)=1$). Fix a point $s_0\in\mathbb{C}$ with $\operatorname{Re}(s_0)=1$. Since $L(s_0,\chi)\neq 0$ and $L(\,\cdot\,,\chi)$ is holomorphic near $s_0$, continuity gives an open neighbourhood $V_{s_0}\subset\mathbb{C}$ on which $L(s,\chi)\neq 0$. On this neighbourhood the quotient \begin{align*} s \mapsto -\frac{L'}{L}(s,\chi) \end{align*} is holomorphic, because it is the quotient of two holomorphic functions with nonvanishing denominator. Combining this with the identity \begin{align*} -\frac{L'}{L}(s,\chi) = \sum_{n=1}^{\infty}\frac{\Lambda(n)\chi(n)}{n^s} \end{align*} on $\operatorname{Re}(s)>1$, we obtain the required holomorphic continuation of the Dirichlet series across every point of the line $\operatorname{Re}(s)=1$. [/guided] [/step] [step:Apply the complex-coefficient Tauberian theorem] Define the coefficient sequence $a: \mathbb{N}\to\mathbb{C}$ by \begin{align*} a(n):=\Lambda(n)\chi(n). \end{align*} Then $|a(n)|\leq \Lambda(n)\leq \log n$ for $n\geq 2$, and the Dirichlet series \begin{align*} A(s):=\sum_{n=1}^{\infty}\frac{a(n)}{n^s} \end{align*} converges for $\operatorname{Re}(s)>1$. By the preceding steps, \begin{align*} A(s)=-\frac{L'}{L}(s,\chi) \end{align*} on $\operatorname{Re}(s)>1$, and $A$ extends holomorphically to a neighbourhood of every point of $\operatorname{Re}(s)=1$. We invoke the complex-coefficient Wiener-Ikehara theorem in its logarithmic-growth Dirichlet-series form (citing a result not yet in the wiki: Complex Wiener-Ikehara Theorem). This form states that if $A(s)=\sum_{n=1}^{\infty}a(n)n^{-s}$ converges on $\operatorname{Re}(s)>1$, the coefficients satisfy $|a(n)|\leq C\log n$ for all $n\geq 2$, and $A$ extends holomorphically to a neighbourhood of every point of the line $\operatorname{Re}(s)=1$, then $\sum_{1\leq n\leq x}a(n)=o(x)$. Here these hypotheses hold with $C=1$, and there is no pole at $s=1$. Therefore the theorem gives \begin{align*} \sum_{1 \leq n \leq x} a(n)=o(x). \end{align*} Substituting $a(n)=\Lambda(n)\chi(n)$ yields \begin{align*} \psi(x,\chi)=\sum_{1 \leq n \leq x}\Lambda(n)\chi(n)=o(x). \end{align*} This proves the nonprincipal case and completes the proof. [/step]