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算术级数中素数的解析等价

通过上下文的方式：众所周知，素数定理$\pi（x）\sim x/\log x$（非私人）等同于以下声明$\泽塔$不会在线路上消失$\Re s=1$.

我想让它澄清什么是算术级数中质数的类似等价语句。特别是，我希望能具体提及证明（或至少明确说明）以下等效性的论文或书籍：

Dirichlet定理，在任何可容许的算术级数（modq美元$). 相当于$L（1，\chi）$对于所有Dirichlet字符$\chi\pmod q$编辑：正如已经指出的那样，这种非对称性很可能等价于约化剩余类（mod）中质数的Dirichlet密度相等q美元$). 什么样的分析语句可以等效于所有此类类中质数的无穷大？]
算术级数中的素数定理$\pi（x；q，a）\sim x/（\phi（q）\log x）$等同于以下声明$L（s，\chi）$不会在线路上消失$\Re s=1$对于所有Dirichlet字符$\chi\pmod q$.

当然，如果这些说法本身是错误的，我希望得到纠正，并指出文献。

答案

<trans data-src="Here are two equivalences.  ">这里有两个等价项。</trans><trans data-src="**Theorem 1**.  ">**定理1**。</trans><trans data-src="For each $m \geq 1$, the following are equivalent.">对于每个$m\geq 1$，以下是等价的。</trans><trans data-src="a) For all nontrivial Dirichlet characters $\chi \bmod m$, $L(1,\chi) \not= 0$.">a） 对于所有非平凡的Dirichlet字符$\chi\bmod m$，$L（1，\chi）\not=0$。</trans><trans data-src="b) For all $a \in (\mathbf Z/m\mathbf Z)^\times$, the set of primes $p \equiv a \bmod m$ has Dirichlet density $1/\varphi(m)$.">b） 对于所有$a\in（\mathbfZ/m\mathbf Z）^\乘以$，素数集$p\equiva\bmodm$具有Dirichlet密度$1/\varphi（m）$。</trans><trans data-src="*Proof of Theorem 1*. ">*定理证明1*。</trans><trans data-src="We will will compute the Dirichlet density of $\{p \equiv a \bmod m\}$ without assuming (a) and then see why (a) and (b) are equivalent. ">我们将在不假设（a）的情况下计算${p\equiva\bmodm\}$的Dirichlet密度，然后看看为什么（a）和（b）是等价的。</trans><trans data-src="The trivial Dirichlet character modulo $m$ will be written as $\mathbf 1_m$.">平凡的Dirichlet字符模$m$将被写入$\mathbf 1_m$。</trans><trans data-src="For each Dirichlet character $\chi \bmod m$, $L(s,\chi)$ is analytic for ${\rm Re}(s) > 0$ except for $L(s,{\mathbf 1}_m)$ having a simple pole at $s = 1$.  ">对于每个Dirichlet字符$\chi\bmod m$，$L（s，\chi）$都是${\rm Re}（s）>0$的解析值，但$L（s，{\mathbf 1}_m）$的简单极点在$s=1$处除外。</trans><trans data-src="Set ">设置</trans><trans data-src="$$">$$</trans><trans data-src="n(\chi) := {\rm ord}_{s=1}(L(s,\chi))">n（chi）：={\rm ord}_{s=1}（L（s，chi））</trans><trans data-src="$$ ">$$ </trans><trans data-src="so $n({\mathbf 1}_m) = -1$ and $n(\chi) \geq 0$ for all nontrivial $\chi$.  ">所以对于所有非平凡的$\chi$，$n（{\mathbf1}m）=-1$和$n（\chi）\geq0$。</trans><trans data-src="For ${\rm Re}(s) > 1$ and $(a,m) = 1$, ">对于${\rm Re}>1$且$（a，m）=1$，</trans><trans data-src="$$">$$</trans><trans data-src="\sum_{p \equiv a \bmod m} \frac{1}{p^s} = ">\sum{p\equiva\bmodm}\frac{1}{p^s}=</trans><trans data-src="\frac{1}{\varphi(m)} \sum_{p}\sum_{\chi} \frac{\chi(p)\overline{\chi}(a)}">\frac{1}{\varphi（m）}\sum{p}\sum{\chi}\frac{\chi（p）\overline{\chi{（a）}</trans><trans data-src="{p^s} = ">{p^s}=</trans><trans data-src="\frac{1}{\varphi(m)} \sum_{\chi}\overline{\chi}(a)\left(\sum_{p} \frac{\chi(p)}">\frac{1}{\varphi（m）}\sum{\chi}\上划线{\chi{（a）\左（\sum{p}\frac{\chi（p）}</trans><trans data-src="{p^s}\right)">｛p^s｝\右）</trans><trans data-src="$$">$$</trans><trans data-src="where the sum on the right run over all primes $p$ and all Dirichlet characters $\chi \bmod m$.">其中右边的和覆盖了所有素数$p$和所有Dirichlet字符$\chi\bmod m$。</trans><trans data-src="For a Dirichlet character $\chi \bmod m$ and ${\rm Re}(s) > 1$, ">对于Dirichlet字符$\chi\bmod m$和${\rm Re}>1$，</trans><trans data-src="$$">$$</trans><trans data-src="\log L(s,\chi) = \sum_p \frac{\chi(p)}{p^s} + \sum_{p,k\geq 2} \frac{\chi(p^k)}{kp^{ks}} = \sum_p \frac{\chi(p)}{p^s} + O(1), ">\对数L（s，\chi）=\sum_p\frac{\chi（p）}{p^s}+\sum_{p，k\geq2}\frac}\chi，</trans><trans data-src="$$">$$</trans><trans data-src="where the $O$-constant is $\sum_{p,k \geq 2} 1/(kp^k)$, so ">其中$O$-常量是$\sum_{p，k\geq2}1/（kp^k）$，所以</trans><trans data-src="$$">$$</trans><trans data-src="\sum_{p \equiv a \bmod m} \frac{1}{p^s} = ">\sum{p\equiva\bmodm}\frac{1}{p^s}=</trans><trans data-src="\frac{1}{\varphi(m)}\sum_{\chi \bmod m}  \overline{\chi}(a)\log L(s,\chi) + O(1).">\裂缝{1}{\varphi（m）}\sum_{\chi\bmod-m}\上横线{\chi}（a）\log L（s，\chi）+O（1）。</trans><trans data-src="$$">$$</trans><trans data-src="Now let's bring in the order of vanishing $n(\chi)$ above. ">现在让我们按照上面$n（\chi）$的消失顺序。</trans><trans data-src="For $s$ near $1$, $L(s,\chi) = (s-1)^{n(\chi)}f_\chi(s)$ where $f_\chi(s)$ is an analytic function in a neighborhood of $s = 1$ and $f_\chi(1) \not= 0$. ">对于接近$1$的$s$，$L（s，\chi）=（s-1）^{n（\chi。</trans><trans data-src="Therefore $f_\chi(s)$ has an analytic  logarithm around $s = 1$ (well-defined up to adding an integer multiple of $2\pi i$), so for $s > 1$, ">因此，$f_chi（s）$的解析对数约为$s=1$（定义明确，可以将$2\pi i$的整数倍相加），因此对于$s>1$，</trans><trans data-src="$\log L(s,\chi) = n(\chi)\log(s-1) + \ell_{f_\chi}(s)$, ">$\log L（s，\chi）=n（\chi，\log（s-1）+\ell_{f_chi}$，</trans><trans data-src="where $\ell_{f_\chi}(s)$ is a suitable logarithm of $f_\chi(s)$. ">其中$\ell{f\chi}$是$f\chi$的合适对数。</trans><trans data-src="Thus ">因此</trans><trans data-src="$$  ">$$  </trans><trans data-src="\log L(s,\chi) = n(\chi)\log(s-1) + O_\chi(1)">\log L（s，\chi）=n（\chi）\log（s-1）+O_\chi（1）</trans><trans data-src="$$ ">$$ </trans><trans data-src="for $s$ near $1$ to the right, and plugging this into the above displayed formula, ">对于右侧接近$1$的$s$，并将其插入上面显示的公式中，</trans><trans data-src="\begin{align}">\开始{align}</trans><trans data-src="\sum_{p \equiv a \bmod m} \frac{1}{p^s}  & = \frac{1}{\varphi(m)}\sum_{\chi \bmod m}  \overline{\chi}(a)(n(\chi)\log(s-1)+ O_\chi(1))  + O(1)  \nonumber  \\">\sum{p\equiva\bmodm}\frac{1}{p^s}&=\frac{1}}{\varphi（m）}\sum{\chi\bmodm}\overline{\chi}（a）（n（chi）\log（s-1）+O\chi（1））+O（1）\nonumber\\</trans><trans data-src="& =\frac{1}{\varphi(m)}\left(\sum_{\chi}  \overline{\chi}(a)n(\chi)\right)\log(s-1) + O_m(1).  ">&=\frac{1}{\varphi（m）}\left（\sum_{\chi}\overline（a）n（\chi）\right）\log（s-1）+O_m（1）。</trans><trans data-src="\end{align} ">\结束{对齐}</trans><trans data-src="To compute a Dirichlet density, ">要计算Dirichlet密度，</trans><trans data-src="we want to divide both sides by $\sum_p 1/p^s$ for ">我们想将两边除以$\sum_p1/p^s$</trans><trans data-src="$s$ near $1$ to the right. ">$s$右侧接近$1$。</trans><trans data-src="For such $s$, ">对于此类$s$，</trans><trans data-src="$$">$$</trans><trans data-src="\log \zeta(s) = \sum_p \frac{1}{p^s} + O(1) = -\log(s-1) + O(1). ">\log\zeta（s）=\sum_p\frac{1}{p^s}+O（1）=-\log（s-1）+O（一）。</trans><trans data-src="$$">$$</trans><trans data-src="Therefore $\sum_p 1/p^s \sim -\log(s-1)$ as $s \to 1^+$, so ">因此，$\sum_p 1/p^s\sim-\log（s-1）$作为$s\到1^+$，所以</trans><trans data-src="dividing through by $\sum_p 1/p^s$ and letting $s \to 1^+$ gives us ">除以$\sum_p1/p^s$，然后让$s\到1^+$得到</trans><trans data-src="\begin{equation}">\开始{方程式}</trans><trans data-src="\lim_{s \to 1^+} ">\lim{s\到1^+}</trans><trans data-src="\frac{\sum_{p \equiv a \bmod m} 1/p^s}{\sum_p 1/p^s}  = ">\裂缝{\sum_{p\equiva\bmodm}1/p^s}{\sum_p1/p^s}=</trans><trans data-src="\frac{1}{\varphi(m)}\left(-\sum_{\chi} \overline{\chi}(a)n(\chi)\right), ">\压裂{1}{\varphi（m）}\左（-\sum{\chi}\上横线{\chi}（a）n（\chi）\右），</trans><trans data-src="\end{equation} ">\结束{方程式}</trans><trans data-src="which expresses the Dirichlet density of $\{p \equiv a \bmod m\}$ in terms of the ">它用</trans><trans data-src="orders of vanishing $n(\chi)$ as $\chi$ runs over Dirichlet characters mod $m$. ">$n（\chi）$作为$\chi$消失的顺序遍历Dirichlet字符mod$m$。</trans><trans data-src="If (a) is true then $n(\chi) = 0$ for all nontrivial $\chi$, so the right side of the above limit calculation ">如果（a）为真，则对于所有非平凡的$\chi$，$n（\chi）=0$，因此上述极限计算的右侧</trans><trans data-src="is $(1/\varphi(m))(-n({\mathbf 1}_m)) = 1/\varphi(m)$, which is (b). ">是$（1/\varphi（m））（-n（{\mathbf1}_m））=1/\varphi（m）$，即（b）。</trans><trans data-src="Conversely, if (b) is true then ">相反，如果（b）为真，则</trans><trans data-src="$$">$$</trans><trans data-src="\sum_{\chi}  \overline{\chi}(a)n(\chi) = -1 ">\sum{chi}\上划线{chi}（a）n（chi）=-1</trans><trans data-src="$$ ">$$ </trans><trans data-src="for all $a \in (\mathbf Z/m\mathbf Z)^\times$ by our limit calculation. ">对于所有$a\in（\mathbf Z/m\mathbf Z）^\times$，通过我们的极限计算。</trans><trans data-src="Why does this imply $n(\chi) = 0$ for nontrivial $\chi$? ">为什么这意味着对于非平凡的$\chi$，$n（\chi）=0$？</trans><trans data-src="Using complex vectors indexed by all the Dirichlet characters mod $m$, let ">使用由所有Dirichlet字符mod$m$索引的复数向量，让</trans><trans data-src="${\mathbf n}_m = (n(\chi))_\chi$ and ">${\mathbf n}_m=（n（\chi））_\chi$和</trans><trans data-src="${\mathbf v}_a = (\chi(a))_\chi$ for each $a \in (\mathbf Z/m\mathbf Z)^\times$. ">${\mathbf v}_a=（\chi（a））_\chi$表示每个$a\in（\mathbfZ/m\mathbf-Z）^\乘以$。</trans><trans data-src="The space of all complex vectors $\mathbf z = (z_\chi)_\chi$ has ">所有复向量$\mathbf z=（z_\chi）_\chi$的空间具有</trans><trans data-src="dimension $\varphi(m)$ and it has the Hermitian inner product ">维度$\varphi（m）$，它有Hermitian内积</trans><trans data-src="$\langle \mathbf z, \mathbf w\rangle = \frac{1}{\varphi(m)}\sum_{\chi} z_\chi\overline{w_\chi}$ for which the vectors ${\mathbf v}_a$ are an orthonormal basis by the orthogonality relations for Dirichlet characters mod $m$. ">$\langle\mathbfz，\mathbf w\rangle=\frac{1}{\varphi（m）}\sum_{\chi}z_\chi\上划线{w_\chi}$，其中向量${mathbf v}_a$是通过Dirichlet字符mod$m$的正交关系的正交基。</trans><trans data-src="The above displayed formula ">上面显示的公式</trans><trans data-src="says $\langle {\mathbf n}_m,{\mathbf v}_a\rangle = -1/\varphi(m)$ for all $a$ in $(\mathbf Z/m\mathbf Z)^\times$, so ">表示$\langle{\mathbf n}_m，{\mathbf v}_a\rangle=-1/\varphi（m）$表示$中的所有$a$（\mathbfZ/m\mathbf-Z）^\times$，所以</trans><trans data-src="$$">$$</trans><trans data-src="{\mathbf n}_m = \sum_{a} \langle {\mathbf n}_m,{\mathbf v}_a\rangle{\mathbf v}_a = -\frac{1}{\varphi(m)}\sum_{a}{\mathbf v}_a.">{mathbfn}_m=\sum_{a}\langle{mathbf n}_m，{mathbf-v}_a\rangle{mathbv}_a=-\frac{1}{\varphi（m）}\sum_a}{\mathbfv}_a。</trans><trans data-src="$$">$$</trans><trans data-src="For each nontrivial $\chi \bmod m$, the $\chi$-component of ">对于每个重要的$\chi\bmod m$</trans><trans data-src="$\sum_{a} {\mathbf v}_a$ is $\sum_a \chi(a)$, which is $0$. ">$\sum_{a}{\mathbfv}_a$是$\sum_a\chi（a）$，即$0$。</trans><trans data-src="So the $\chi$-component of ${\mathbf n}_m$, which is $n(\chi)$, is 0. ">因此${\mathbfn}_m$的$\chi$-组件$n（\chi）$为0。</trans><trans data-src="That is (a). ">那就是（a）。</trans><trans data-src="QED Theorem 1.  ">QED定理1。</trans><trans data-src="(I only realized after copying and pasting this that I had already copy and pasted it earlier as an answer to the MO question [here][1].) ">（我只是在复制和粘贴之后才意识到，我之前已经复制并粘贴了它，作为对MO问题的回答[此处][1]。）</trans><trans data-src="**Theorem 2**.  ">**定理2**。</trans><trans data-src="For each $m \geq 1$, the following are equivalent.">对于每个$m\geq 1$，以下是等价的。</trans><trans data-src="a) For all Dirichlet characters $\chi \bmod m$, $L(s,\chi) \not= 0$ when ${\rm Re}(s) = 1$.">a） 对于所有Dirichlet字符$\chi\bmod m$，当${\rm Re}=1$时，$L（s，\chi）\ not=0$。</trans><trans data-src="b) $\sum_{n \leq x} \chi(n)\Lambda(n) = o(x)$ ">b） $\sum_{n\leqx}\chi（n）\Lambda（n）=o（x）$</trans><trans data-src="for nontrivial Dirichlet characters $\chi \bmod m$ ">对于非平凡的Dirichlet字符$\chi\bmod m$</trans><trans data-src="and $\sum_{n \leq x} \chi_{{\mathbf 1}_m}(n)\Lambda(n) \sim x$, ">和$\sum_{n\leqx}\chi_{{mathbf1}_m}（n）\Lambda（n）\sim x$，</trans><trans data-src="c) For all $a \in (\mathbf Z/m\mathbf Z)^\times$, $|\{p \leq x : p \equiv a \bmod m\}| \sim (1/\varphi(m))x/\log x$.">c） 对于所有$a\in（\mathbf Z/m\mathbfZ）^\乘以$，$|\{p\leq x:p\equiv a\bmod m\}|\sim（1/\varphi（m））x/\log x$。</trans><trans data-src="Comparing the proof of Theorem 2 below to the sketch in the answer by 2734364041, we will also be using a Tauberian theroem (to prove (b) implies (c)), but we will not need an explicit formula.">将下面定理2的证明与2734364041答案中的草图进行比较，我们也将使用Tauberian定理（证明（b）隐含（c）），但我们不需要显式公式。</trans><trans data-src="*Proof of Theorem 2*.  ">*定理证明2*。</trans><trans data-src="We will show (a) is equivalent to (b) and (b) is equivalent to (c).">我们将证明（a）等于（b），（b）等于（c）。</trans><trans data-src="First we show (a) implies (b). ">首先，我们显示（a）意味着（b）。</trans><trans data-src="Set $\psi_\chi(x) = \sum_{n \leq x} \chi(n)\Lambda(n)$ for all $\chi$, so ">为所有$\chi$设置$\psi_\chi（x）=\sum_{n\leqx}\chi</trans><trans data-src="(b) says $\psi_\chi(x) = o(x)$ for nontrivial $\chi$ and $\psi_{{\mathbf 1}_m}(x) \sim x$. ">（b） 对于非平凡的$\chi$和$\psi{\mathbf1}_m}（x）\simx$，表示$\psi.chi（x）=o（x）$。</trans><trans data-src="For $\sigma > 1$, $-L'(s,\chi)/L(s,\chi) = \sum \chi(n)\Lambda(n)/n^s$, for all Dirichlet characters $\chi \bmod m$, ">对于$\sigma>1$，$-L'（s，\chi）/L（s，\ chi）=\sum\chi（n）\Lambda（n）/n^s$，对于所有Dirichlet字符$\chi\bmod m$，</trans><trans data-src="so $\psi_\chi(x)$ is a partial sum of coefficients of $-L'(s,\chi)/L(s,\chi)$. ">所以$\psi_\chi（x）$是$-L'（s，\chi）/L（s，\chi）$的系数的偏和。</trans><trans data-src="Since $L(s,{\mathbf 1}_m) \not= 0$ on $\sigma = 1$ by (a), $-L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$ is holomorphic on $\sigma \geq 1$ except for a simple pole at $s = 1$ with residue 1 and it has nonnegative Dirichlet series coefficients with $\psi_{{\mathbf 1}_m}(x) = O(x)$.  ">由于$L（s，{mathbf 1}_m）在$\sigma=1$by（a）上不=0$，$-L'（s，｛mathbf 1}_m。</trans><trans data-src="Therefore $\psi_{{\mathbf 1}_m}(x) \sim x$, which is part of (b), by Newman's Tauberian theorem. ">因此，根据纽曼的Tauberian定理，$\psi_{{\mathbf1}_m}（x）\simx$是（b）的一部分。</trans><trans data-src="To get the rest of (b), namely $\psi_\chi(x) = o(x)$ for nontrivial $\chi$, we have $-L'(s,\chi)/L(s,\chi)$ being holomorphic on $\sigma \geq 1$ by (a) and its Dirichlet series coefficients satisfy $|\chi(n)\Lambda(n)| \leq {\mathbf 1}_m(n)\Lambda(n)$ for all $n$, so $\psi_\chi(x) = o(x)">为了得到（b）的其余部分，即非平凡$\chi$的$\psi_\chi（x）=o（x）$，我们有$-L'（s，\chi）/L（s，\ chi）$由（a）在$\sigma\geq1$上全纯，并且它的Dirichlet级数系数满足$|\chi</trans><trans data-src="$ by a corollary of Newman's Tauberian theorem for $-L'(s,\chi)/L(s,\chi)$ using comparison Dirichlet series $-L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$ . ">通过Newman的Tauberian定理对$-L'（s，\chi）/L（s，\ chi）$的推论，使用比较Dirichlet级数$-L'。</trans><trans data-src="Thus (a) implies (b). ">因此（a）意味着（b）。</trans><trans data-src="To show (b) implies (a), we will use the following fact. ">为了表明（b）意味着（a），我们将使用以下事实。</trans><trans data-src="For a function $a(x)$ on $[1,\infty)$ that is bounded and Riemann integrable on $[1,T]$ for all $T \geq 1$, so $f(s) := \int_1^\infty (a(x)/x^s)dx/x$ is absolutely convergent on $\sigma > 1$, if $a(x) \to 0$ as $x \to \infty$ and $f$ extends to a meromorphic function on $\sigma  = 1$ then $f$ in fact is holomorphic on $\sigma = 1$.  ">对于$[1，\infty）$上的函数$a（x）$是有界的，并且对于所有$T\geq 1$，在$[1，T]$上是Riemann可积的，所以$f（s）：=\int_1^\infty（a（x）/x^s）dx/x$在$\sigma>1$上是绝对收敛的，如果$a（x）\to0$作为$x\to\infty$并且$f$扩展到$\sigma=1$上的亚纯函数，那么$f$实际上在$\sigma=1$上是全纯的。</trans><trans data-src="(This is used to show the condition $\psi(x) \sim x$ implies $\zeta(s) \not= 0$ on $\sigma = 1$ by using $a(x) = \psi(x)/x - 1$.)  ">（这用于通过使用$a（x）=\psi（x）/x-1$在$\sigma=1$上显示条件$\psi（x）\sim x$表示$\zeta（s）\not=0$。）</trans><trans data-src="Because of the integral representations ">由于积分表示</trans><trans data-src="$$">$$</trans><trans data-src="-\frac{L'(s,\chi)}{sL(s,\chi)} = \int_1^\infty \frac{\psi_\chi(x)}{x} \frac{dx}{x^s}">-\压裂{L'（s，\chi）}{sL（s，\ chi）{=\int_1^\infty\frac{\psi_\chi（x）}{x}\frac}{x^s}</trans><trans data-src="$$">$$</trans><trans data-src="and">和</trans><trans data-src="$$">$$</trans><trans data-src="-\frac{L'(s,{\mathbf 1}_m)}{sL(s,{\mathbf 1}_m)}  - \frac{1}{s-1} = \int_1^\infty \left(\frac{\psi_{\mathbf 1}(x)}{x} -1\right)\frac{dx}{x^s},">-\压裂{L'（s，{\mathbf 1}_m）}{sL（s，}\mathbf1}_m）}-\frac{1}{s-1}=\int_1^\infty\left（\frac}\psi{\matHBf1}（x）}{x}-1\right）\frac[dx}{x^s}，</trans><trans data-src="$$">$$</trans><trans data-src="for ${\rm Re}(s) > 1$, ">对于${\rm Re}>1$，</trans><trans data-src="where $\chi$ is nontrivial in the first equation.">其中$\chi$在第一个等式中很重要。</trans><trans data-src="we can use the above fact when $a(x) = \psi_\chi(x)/x$ for nontrivial $\chi$ and ">当$a（x）=\psi_\chi（x）/x$用于非平凡的$\chi$和</trans><trans data-src="$a(x) = \psi_{{\mathbf 1}_m}(x)/x - 1$ to conclude that  $L'(s,\chi)/L(s,\chi)$ is holomorphic on ">$a（x）=\psi_{{\mathbf 1}_m}（x）/x-1$得出$L'（s，\chi）/L（s，\ chi）$在</trans><trans data-src="$\sigma = 1$ for nontrivial $\chi$ and $L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$ is holomorphic on $\sigma = 1$ except for a simple pole at $s = 1$, so ">对于非平凡的$\chi$和$L'（s，{\mathbf 1}_m）/L（s，}\mathbf1}_m$）$，$\sigma=1$是全纯的，除了$s=1$处的一个简单极点，所以</trans><trans data-src="$L(s,\chi)$ is nonvanishing on $\sigma = 1$ and ">$L（s，\chi）$在$\sigma=1$上是无效的，并且</trans><trans data-src="$L(s,{\mathbf 1}_m)$ is nonvanishing on $\sigma = 1$. ">$L（s，{\mathbf 1}_m）$在$\sigma=1$上是非Anishing的。</trans><trans data-src="Thus (b) implies (a).">因此（b）意味着（a）。</trans><trans data-src="That (b) implies (c) follows from the above integral representation of $-L'(s,\chi)/L(s,\chi)$ for all nontrivial $\chi$ by a standard method to prove (c). ">这（b）意味着（c）是由所有非平凡的$\chi$的$-L'（s，\chi）/L（s，\ chi）$的上述积分表示通过标准方法证明（c）的。</trans><trans data-src="Our last step is showing (c) implies (b). ">我们的最后一步是显示（c）暗示（b）。</trans><trans data-src="Set $\pi(x;a \bmod m) = |\{p \leq x : p \equiv a \bmod m\}|$ when $(a,m) = 1$ and $\pi_\chi(x) = \sum_{p \leq x} \chi(p)$, ">当$（a，m）=1$和$\pi_\chi（x）=\sum_{p\leqx}\chi，</trans><trans data-src="where $\chi$ is a Dirichlet character mod $m$.">其中$\chi$是Dirichlet字符mod$m$。</trans><trans data-src="Write $\chi$ as a linear combination of ">将$\chi$写为以下项的线性组合</trans><trans data-src="delta-functions on $(\mathbf Z/m\mathbf Z)^\times$: ">$上的delta-functions（\mathbf Z/m\mathbf-Z）^\乘以$：</trans><trans data-src="$\chi = \sum_{a \in  (\mathbf Z/m\mathbf Z)^\times} \chi(a)\delta_a$. ">$\chi=\sum_{a\in（\mathbf Z/m\mathbf-Z）^\times}\chi（a）\delta_a$。</trans><trans data-src="Then ">然后</trans><trans data-src="\begin{align*}">\开始{align*}</trans><trans data-src="\pi_\chi(x) & = \sum_{p \leq x} \chi(p) \\">\pi_\chi（x）&=\sum{p\leqx}\chi\\</trans><trans data-src="& = \sum_{p \leq x} \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\delta_a(p) \\">&=sum_{p\leqx}\sum_{a\in（\mathbfZ/m\mathbf Z）^\times}\chi（a）\delta_a（p）\\</trans><trans data-src="& = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\left(\sum_{p \leq x} \delta_a(p)\right) \\">&=\sum_{a\in（\mathbf Z/m\mathbf-Z）^\times}\chi（a）\left（\sum_{p\leq x}\delta_a（p）\right）\\</trans><trans data-src="& = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\pi(x; a \bmod m), ">&=\sum_{a\in（\mathbf Z/m\mathbf-Z）^\times}\chi（a）\pi（x；a\bmod-m），</trans><trans data-src="\end{align*}">\结束{align*}</trans><trans data-src="so">所以</trans><trans data-src="$$">$$</trans><trans data-src="\frac{\pi_\chi(x)}{x/\log x} = \sum_{a \in  (\mathbf Z/m\mathbf Z)^\times} \chi(a)\frac{\pi(x;a \bmod m)}{x/\log x}.">\压裂{\pi_\chi（x）}{x/\log x}=\sum_{a\in（\mathbf Z/m\mathbfZ）^\times}\chi[a）\frac{\pi（x；a\bmodm）}{x/\log-x}。</trans><trans data-src="$$">$$</trans><trans data-src="By (c), as $x \to \infty$ the right side tends to ">通过（c），当$x\to\infty$右侧趋于</trans><trans data-src="$\sum_{a \in ({\mathbf Z}/m{\mathbf Z})^\times} \chi(a)/\varphi(m)$, which is 0 if $\chi$ is nontrivial. ">$\sum_{a\in（{\mathbf Z}/m{\mathbf Z}）^\times}\chi（a）/\varphi（m）$，如果$\chi$不平凡，则为0。</trans><trans data-src="Therefore when $\chi$ is nontrivial we have $\pi_\chi(x) = o(x/\log x)$, which implies $\psi_\chi(x) = o(x)$ by ">因此，当$\chi$很重要时，我们有$\pi_\chi（x）=o（x/\log x）$，这意味着$\psi_\chi（x）=o（x）$by</trans><trans data-src="the same argument that $\pi(x) \sim x/\log x$ implies $\psi(x) \sim x$. ">$\pi（x）\sim x/\log x$的相同参数表示$\psi（x）\sim x$。</trans><trans data-src="To show (c) implies ">显示（c）暗示</trans><trans data-src="$\psi_{{\mathbf 1}_m}(x) \sim x$, sum the relation in (c) over all $a$ in $(\mathbf Z/m\mathbf Z)^\times$ to ">$\psi_{{\mathbf 1}_m}（x）\sim x$，将（c）中的关系求和到$中的所有$a$（\mathbf-Z/m\mathbfZ）^\次$到</trans><trans data-src="get $\pi(x) \sim x/\log x$, the Prime Number Theorem, which ">得到$\pi（x）\sim x/\log x$，素数定理，其中</trans><trans data-src="is equivalent to ">等于</trans><trans data-src="$\psi(x) \sim x$, so ">$\psi（x）\sim x$，所以</trans><trans data-src="$\psi_{{\mathbf 1}_m}(x) \sim x$ since  $\psi_{{\mathbf 1}_m}(x)  = \psi(x) + O_m(\log x)$.">$\psi_{{mathbf 1}_m}（x）\sim x$自$\psi.{mathbf1}_m}（x）=\psi（x）+O_m（\log x）$起。</trans><trans data-src="QED Theorem 2. ">QED定理2。</trans><trans data-src="Update (2022): The prime number theorem ">更新（2022）：素数定理</trans><trans data-src="$\pi(x) \sim x/\log x$ is equivalent to two properties involving the Moebius function: $\sum_{n \leq x} \mu(n) = o(x)$ and ">$\pi（x）\sim x/\log x$等价于涉及Moebius函数的两个属性：$\sum_{n\leq x}\mu（n）=o（x）$和</trans><trans data-src="$\sum \mu(n)/n = 0$.  ">$\sum\mu（n）/n=0$。</trans><trans data-src="These each have analogues in Theorem 2.  ">这些都与定理2中的类似。</trans><trans data-src="See [here][2].">请参阅[此处][2]。</trans><trans data-src="[1]:">[1]:</trans><trans data-src="https://mathoverflow.net/questions/403674/how-essential-is-the-vanishing-of-the-dirichlet-l-functions-to-dirichlets-the">https://mathoverflow.net/questions/403674/how-essential-is-the-vanishing-of-the-dirichlet-l-functions-to-irichlets-the</trans><trans data-src="[2]:">[2]:</trans><trans data-src="https://mathoverflow.net/questions/426773/averages-of-m%c3%b6bius">https://mathoverflow.net/questions/426773/averages-of-m%c3%b6bius</trans><trans data-src="-function-in-arithmetic-progressions/426811">-算术级数函数/426811</trans>

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