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3 Dirichlet L-functions
3 狄利克雷 L 函数
Let N N N be a natural number. Let Z N \mathbb{Z}{N} ZN denote the ring of integers modulo N N N. Let Z N ∗ \mathbb{Z}{N}^{*} ZN∗ denote the group of units of Z N \mathbb{Z}{N} ZN. The group Z N ∗ \mathbb{Z}{N}^{*} ZN∗ consists of the residues of the numbers coprime to N N N, and has ϕ ( N ) \phi(N) ϕ(N) elements, where ϕ \phi ϕ denotes Euler's totient function.
设 N N N 为自然数， Z N \mathbb{Z}_N ZN 表示模 N N N 整数环， Z N ∗ \mathbb{Z}_N^* ZN∗ 表示 Z N \mathbb{Z}_N ZN 的单位群。 Z N ∗ \mathbb{Z}_N^* ZN∗ 由全体与 N N N 互素的模 N N N 剩余类构成，元素个数为欧拉函数 ϕ ( N ) \phi(N) ϕ(N)。
A Dirichlet character modulo N N N is a character of the group Z N ∗ \mathbb{Z}{N}^{*} ZN∗. Define X N X{N} XN to be the group of Dirichlet characters modulo N N N. We use ε \varepsilon ε to denote the trivial Dirichlet character. We consider a Dirichlet character χ ∈ X N \chi \in X_{N} χ∈XN as a function on Z \mathbb{Z} Z as follows: we set
χ ( a ) = { χ ( a + N Z ) , if a is coprime to N , 0 , if a is not coprime to N . \chi(a)= \begin{cases} \chi(a + N\mathbb{Z}), & \text{if } a \text{ is coprime to } N, \\ 0, & \text{if } a \text{ is not coprime to } N. \end{cases} χ(a)={χ(a+NZ),0,if a is coprime to N,if a is not coprime to N.
模 N N N 狄利克雷特征标定义为群 Z N ∗ \mathbb{Z}_N^* ZN∗ 的特征标；全体模 N N N 狄利克雷特征标构成的群记作 X N X_N XN，用 ε \varepsilon ε 表示平凡狄利克雷特征标。将 χ ∈ X N \chi\in X_N χ∈XN 延拓为定义在全体整数 Z \mathbb{Z} Z 上的函数：
χ ( a ) = { χ ( a + N Z ) , 若 a 与 N 互素 , 0 , 若 a 与 N 不互素 . \chi(a)= \begin{cases} \chi(a + N\mathbb{Z}), & \text{若 } a \text{ 与 } N \text{ 互素}, \\ 0, & \text{若 } a \text{ 与 } N \text{ 不互素}. \end{cases} χ(a)={χ(a+NZ),0,若 a 与 N 互素,若 a 与 N 不互素.
χ ( a ) = { χ ( a m o d N ) , gcd ( a , N ) = 1 , 0 , gcd ( a , N ) > 1. \chi(a)= \begin{cases} \chi(a \bmod N), & \gcd(a,N)=1,\\ 0, & \gcd(a,N)>1 . \end{cases} χ(a)={χ(amodN),0,gcd(a,N)=1,gcd(a,N)>1.
Let χ ∈ X N \chi \in X_{N} χ∈XN, and let s s s be real. We define the Dirichlet L-function L ( s , χ ) L(s, \chi) L(s,χ) as whenever this sum converges.
任取 χ ∈ X N \chi\in X_N χ∈XN， s s s 为实数，当级数收敛时定义狄利克雷 L 函数：
L ( s , χ ) = ∑ n = 1 ∞ χ ( n ) n s L(s, \chi)=\sum_{n=1}^{\infty} \frac{\chi(n)}{n^{s}} L(s,χ)=n=1∑∞nsχ(n)
Before we can determine when the series for the L-function converges, we need a preliminary result.
分析 L 函数级数的收敛域前，先证明一条预备引理。
Theorem 8
定理 8
Let χ ∈ X N \chi \in X_{N} χ∈XN, and suppose that χ ≠ ε \chi \neq\varepsilon χ=ε. Define a n = ∑ j = 1 n χ ( j ) a_{n}=\sum_{j=1}^{n} \chi(j) an=∑j=1nχ(j). Then ∣ a n ∣ ≤ N |a_{n}| \leq N ∣an∣≤N for all nonnegative integers n n n.
设 χ ∈ X N \chi\in X_N χ∈XN 为非平凡特征标，记部分和 a n = ∑ j = 1 n χ ( j ) a_n=\sum_{j=1}^n \chi(j) an=∑j=1nχ(j)，则对任意非负整数 n n n，有 ∣ a n ∣ ≤ N |a_n|\leq N ∣an∣≤N。
Proof
证明
First of all note that by Theorem 6
由定理6，有
∑ j = 1 N χ ( j ) = ∑ a ∈ Z N ∗ χ ( a ) = 0. \sum_{j=1}^N \chi(j)=\sum_{a\in\mathbb{Z}_N^*}\chi(a)=0 . j=1∑Nχ(j)=a∈ZN∗∑χ(a)=0.
Suppose that n > N n>N n>N. Then
若 n > N n>N n>N，则
a n = ∑ j = 1 n χ ( j ) = ∑ j = 1 N χ ( j ) + ∑ j = N + 1 n χ ( j ) = 0 + ∑ j = 1 n − N χ ( j ) = a n − N . a_n=\sum_{j=1}^n\chi(j)=\sum_{j=1}^N\chi(j)+\sum_{j=N+1}^n\chi(j)=0+\sum_{j=1}^{n-N}\chi(j)=a_{n-N} . an=j=1∑nχ(j)=j=1∑Nχ(j)+j=N+1∑nχ(j)=0+j=1∑n−Nχ(j)=an−N.
By iteration, a n = a r a_{n}=a_{r} an=ar where r r r is the least nonnegative residue of n n n modulo N N N. Then
迭代可得 a n = a r a_n=a_r an=ar，其中 r r r 是 n n n 模 N N N 的最小非负剩余。于是
∣ a n ∣ = ∣ a r ∣ ≤ ∑ j = 1 r ∣ χ ( j ) ∣ = r ≤ N . |a_n|=|a_r| \leq \sum_{j=1}^r |\chi(j)| = r \leq N. ∣an∣=∣ar∣≤j=1∑r∣χ(j)∣=r≤N.
We can now determine where the L-functions converge.
下面给出 L 函数级数的收敛域。
Theorem 9
定理 9
Let χ ∈ X N \chi \in X_{N} χ∈XN. If χ = ε \chi=\varepsilon χ=ε then the series for L ( s , χ ) L(s, \chi) L(s,χ) converges for s > 1 s>1 s>1. If χ ≠ ε \chi \neq\varepsilon χ=ε then the series for L ( s , χ ) L(s, \chi) L(s,χ) converges for s > 0 s>0 s>0.
设 χ ∈ X N \chi\in X_N χ∈XN：若 χ = ε \chi=\varepsilon χ=ε（平凡特征标），则 L ( s , χ ) L(s,\chi) L(s,χ) 的级数在 s > 1 s>1 s>1 收敛；若 χ ≠ ε \chi\neq\varepsilon χ=ε（非平凡特征标），则级数在 s > 0 s>0 s>0 收敛。
Proof
证明
When χ = ε \chi=\varepsilon χ=ε, then χ ( n ) ∈ { 0 , 1 } \chi(n) \in\{0,1\} χ(n)∈{0,1} for all n n n. Then
当 χ = ε \chi=\varepsilon χ=ε 时，对所有整数 n n n， χ ( n ) ∈ { 0 , 1 } \chi(n)\in\{0,1\} χ(n)∈{0,1}，故
0 ≤ ∑ n = 1 M χ ( n ) n s ≤ ∑ n = 1 M 1 n s 0\leq \sum_{n=1}^M \frac{\chi(n)}{n^s}\leq \sum_{n=1}^M\frac{1}{n^s} 0≤n=1∑Mnsχ(n)≤n=1∑Mns1
and by comparison with the series for ζ ( s ) \zeta(s) ζ(s) this series converges for s > 1 s>1 s>1.
与 ζ ( s ) \zeta(s) ζ(s) 级数比较，得该级数在 s > 1 s>1 s>1 收敛。
Suppose that χ ≠ ε \chi \neq\varepsilon χ=ε. We use the technique of partial summation. Let
若 χ ≠ ε \chi\neq\varepsilon χ=ε，采用分部求和技巧。记
a n = ∑ j = 1 n χ ( j ) a_n=\sum_{j=1}^n\chi(j) an=j=1∑nχ(j)
Then a 0 = 0 a_{0}=0 a0=0 and χ ( n ) = a n − a n − 1 \chi(n)=a_{n}-a_{n-1} χ(n)=an−an−1 for n ≥ 1 n \geq1 n≥1. Thus
显然 a 0 = 0 a_0=0 a0=0，且对 n ≥ 1 n\geq1 n≥1 有 χ ( n ) = a n − a n − 1 \chi(n)=a_n-a_{n-1} χ(n)=an−an−1，因此
∑ n = 1 M χ ( n ) n s = ∑ n = 1 M a n − a n − 1 n s = ∑ n = 1 M a n n s − ∑ n = 1 M a n − 1 n s = ∑ n = 1 M a n n s − ∑ n = 1 M − 1 a n ( n + 1 ) s = a M M s + ∑ n = 1 M − 1 a n ( 1 n s − 1 ( n + 1 ) s ) . \begin{aligned} \sum_{n=1}^M \frac{\chi(n)}{n^s} &=\sum_{n=1}^M \frac{a_n-a_{n-1}}{n^s} \\ &=\sum_{n=1}^M \frac{a_n}{n^s}-\sum_{n=1}^M \frac{a_{n-1}}{n^s} \\ &=\sum_{n=1}^M \frac{a_n}{n^s}-\sum_{n=1}^{M-1} \frac{a_n}{(n+1)^s} \\ &=\frac{a_M}{M^s}+\sum_{n=1}^{M-1} a_n\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right) . \end{aligned} n=1∑Mnsχ(n)=n=1∑Mnsan−an−1=n=1∑Mnsan−n=1∑Mnsan−1=n=1∑Mnsan−n=1∑M−1(n+1)san=MsaM+n=1∑M−1an(ns1−(n+1)s1).
We apply the mean value theorem to the function x ↦ x − s x \mapsto x^{-s} x↦x−s to get
对函数 f ( x ) = x − s f(x)=x^{-s} f(x)=x−s 使用微分中值定理，
1 n s − 1 ( n + 1 ) s = s b n s + 1 \frac{1}{n^s}-\frac{1}{(n+1)^s}=\frac{s}{b_n^{s+1}} ns1−(n+1)s1=bns+1s
where n < b n < n + 1 n<b_{n}<n+1 n<bn<n+1. In particular
存在 b n ∈ ( n , n + 1 ) b_n\in(n,n+1) bn∈(n,n+1) 使得，于是有上界估计
1 n s − 1 ( n + 1 ) s < s n s + 1 \frac{1}{n^s}-\frac{1}{(n+1)^s}<\frac{s}{n^{s+1}} ns1−(n+1)s1<ns+1s
If s > 0 s>0 s>0 then a M / M s → 0 a_{M} / M^{s} \to 0 aM/Ms→0 as M → ∞ M \to \infty M→∞, as by Theorem 8, the sequence { a n } \{a_{n}\} {an} is bounded. Also
当 s > 0 s>0 s>0 时，由定理8， { a n } \{a_n\} {an} 有界，故 a M / M s → 0 ( M → ∞ ) a_M/M^s \to 0\ (M\to\infty) aM/Ms→0 (M→∞)。设 A A A 为 { ∣ a n ∣ } \{|a_n|\} {∣an∣} 的上界，则
∑ n = 1 M − 1 a n ( 1 n s − 1 ( n + 1 ) s ) ≤ A s ∑ n = 1 ∞ 1 n s + 1 = A s ζ ( s + 1 ) \sum_{n=1}^{M-1} a_n\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right)\leq A s \sum_{n=1}^\infty \frac{1}{n^{s+1}}=A s \zeta(s+1) n=1∑M−1an(ns1−(n+1)s1)≤Asn=1∑∞ns+11=Asζ(s+1)
where A A A is an upper bound for the sequence { a n } \{a_{n}\} {an}. Hence the series for L ( s , χ ) L(s, \chi) L(s,χ) converges for s > 0 s>0 s>0, and indeed
因此 L ( s , χ ) L(s,\chi) L(s,χ) 的级数在 s > 0 s>0 s>0 收敛，且成立表达式
L ( s , χ ) = ∑ n = 1 ∞ a n ( 1 n s − 1 ( n + 1 ) s ) . L(s,\chi)=\sum_{n=1}^\infty a_n\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right) . L(s,χ)=n=1∑∞an(ns1−(n+1)s1).
Dirichlet L-functions also have Euler products when s > 1 s>1 s>1.
当 s > 1 s>1 s>1 时，狄利克雷 L 函数同样存在欧拉乘积展开。
Theorem 10
定理 10
Let χ ∈ X N \chi \in X_{N} χ∈XN and let s > 1 s>1 s>1. Then
设 χ ∈ X N \chi\in X_N χ∈XN， s > 1 s>1 s>1，则
L ( s , χ ) = ∏ p ( 1 − χ ( p ) p s ) − 1 L(s,\chi)=\prod_{p}\left(1-\frac{\chi(p)}{p^s}\right)^{-1} L(s,χ)=p∏(1−psχ(p))−1
where the product is over all prime numbers p p p.
式中乘积遍历全体素数 p p p。
Proof
证明
Let M M M be a natural number. Then
任取自然数 M M M，有
∏ p ≤ M ( 1 − χ ( p ) p s ) − 1 = ∏ p ≤ M ∑ k = 0 ∞ χ ( p ) k p k s = ∑ n ∈ A M χ ( n ) n s \prod_{p\leq M}\left(1-\frac{\chi(p)}{p^s}\right)^{-1}=\prod_{p\leq M}\sum_{k=0}^\infty \frac{\chi(p)^k}{p^{ks}}=\sum_{n\in A_M}\frac{\chi(n)}{n^s} p≤M∏(1−psχ(p))−1=p≤M∏k=0∑∞pksχ(p)k=n∈AM∑nsχ(n)
where A M A_{M} AM is the set of natural numbers none of whose prime factors exceed M M M. This follows from the fact that χ \chi χ is completely multiplicative: χ ( a b ) = χ ( a ) χ ( b ) \chi(ab)=\chi(a) \chi(b) χ(ab)=χ(a)χ(b) for all a a a and b b b. Certainly A M A_{M} AM contains all natural numbers n ≤ M n \leq M n≤M and so
其中 A M A_M AM 是全体素因子不超过 M M M 的自然数集合；该等式由 χ \chi χ 完全积性 χ ( a b ) = χ ( a ) χ ( b ) \chi(ab)=\chi(a)\chi(b) χ(ab)=χ(a)χ(b) 推出。显然所有满足 n ≤ M n\leq M n≤M 的整数属于 A M A_M AM，因此
∣ L ( s , χ ) − ∏ p ≤ M ( 1 − χ ( p ) p s ) − 1 ∣ ≤ ∑ n = M + 1 ∞ 1 n s \left|L(s,\chi)-\prod_{p\leq M}\left(1-\frac{\chi(p)}{p^s}\right)^{-1}\right|\leq \sum_{n=M+1}^\infty \frac{1}{n^s} L(s,χ)−p≤M∏(1−psχ(p))−1 ≤n=M+1∑∞ns1
which is the tail in the convergent series for ζ ( s ) \zeta(s) ζ(s). Letting M M M tend to infinity we conclude that
右侧是收敛级数 ζ ( s ) \zeta(s) ζ(s) 的余项。令 M → ∞ M\to\infty M→∞ 取极限，即得
∏ p ( 1 − χ ( p ) p s ) − 1 = L ( s , χ ) . \prod_{p}\left(1-\frac{\chi(p)}{p^s}\right)^{-1}=L(s,\chi) . p∏(1−psχ(p))−1=L(s,χ).
When χ = ε \chi=\varepsilon χ=ε, the L-function is essentially the same as the ζ \zeta ζ-function.
当 χ = ε \chi=\varepsilon χ=ε（平凡特征标）时，对应的 L 函数与 ζ \zeta ζ 函数紧密关联。
Theorem 11
定理 11
Let χ = ε ∈ X N \chi=\varepsilon \in X_{N} χ=ε∈XN. Then
lim s → 1 + ( s − 1 ) L ( s , χ ) = ϕ ( N ) N \lim _{s \to 1^{+}}(s-1) L(s, \chi)=\frac{\phi(N)}{N} s→1+lim(s−1)L(s,χ)=Nϕ(N)
where ϕ \phi ϕ denotes Euler's ϕ \phi ϕ-function.
其中 ϕ \phi ϕ 为欧拉函数。
Proof
证明
Let s > 1 s>1 s>1. We have
取 s > 1 s>1 s>1，由欧拉乘积公式
L ( s , χ ) = ∏ p ( 1 − χ ( p ) p s ) − 1 L(s, \chi)=\prod_{p}\left(1-\frac{\chi(p)}{p^{s}}\right)^{-1} L(s,χ)=p∏(1−psχ(p))−1
and
ζ ( s ) = ∏ p ( 1 − 1 p s ) − 1 \zeta(s)=\prod_{p}\left(1-\frac{1}{p^{s}}\right)^{-1} ζ(s)=p∏(1−ps1)−1
But χ ( p ) = 0 \chi(p)=0 χ(p)=0 if p ∣ N p \mid N p∣N and χ ( p ) = 1 \chi(p)=1 χ(p)=1 if p ∤ N p \nmid N p∤N. Hence
对素数 p p p，若 p ∣ N p\mid N p∣N 则 χ ( p ) = 0 \chi(p)=0 χ(p)=0；若 p ∤ N p\nmid N p∤N 则 χ ( p ) = 1 \chi(p)=1 χ(p)=1，因此
L ( s , χ ) = ζ ( s ) ∏ p ∣ N ( 1 − 1 p s ) . L(s, \chi)=\zeta(s) \prod_{p \mid N}\left(1-\frac{1}{p^{s}}\right) . L(s,χ)=ζ(s)p∣N∏(1−ps1).
But lim s → 1 + ( s − 1 ) ζ ( s ) = 1 \lim _{s \to 1^{+}}(s-1) \zeta(s)=1 lims→1+(s−1)ζ(s)=1. Hence
已知 lim s → 1 + ( s − 1 ) ζ ( s ) = 1 \lim_{s\to1^+}(s-1)\zeta(s)=1 lims→1+(s−1)ζ(s)=1，于是
lim s → 1 + ( s − 1 ) L ( s , χ ) = lim s → 1 + ∏ p ∣ N ( 1 − 1 p s ) = ∏ p ∣ N p − 1 p = ϕ ( N ) N . \lim {s \to 1^{+}}(s-1) L(s, \chi)=\lim {s \to 1^{+}} \prod{p \mid N}\left(1-\frac{1}{p^{s}}\right)=\prod{p \mid N} \frac{p-1}{p}=\frac{\phi(N)}{N} . s→1+lim(s−1)L(s,χ)=s→1+limp∣N∏(1−ps1)=p∣N∏pp−1=Nϕ(N).
For nontrivial characters χ \chi χ we shall need to consider the value L ( 1 , χ ) L(1, \chi) L(1,χ) and the behaviour of L ( s , χ ) L(s, \chi) L(s,χ) for s s s just above s = 1 s=1 s=1.
对非平凡特征标 χ \chi χ，我们需要考察 L ( 1 , χ ) L(1,\chi) L(1,χ) 的取值，以及 s s s 从右侧趋近于 1 时 L ( s , χ ) L(s,\chi) L(s,χ) 的渐近行为。
Theorem 12
定理 12
Let χ ∈ X N \chi \in X_{N} χ∈XN with χ ≠ ε \chi \neq\varepsilon χ=ε. There exists C > 0 C>0 C>0 such that
设 χ ∈ X N \chi\in X_N χ∈XN 为非平凡特征标，则存在常数 C > 0 C>0 C>0，使得当 s → 1 + s\to1^+ s→1+ 时
L ( s , χ ) = L ( 1 , χ ) + O ( s − 1 ) L(s,\chi )=L(1,\chi )+O(s-1) L(s,χ)=L(1,χ)+O(s−1)
as s → 1 + s \to 1^{+} s→1+. In particular
特别地
lim s → 1 + L ( s , χ ) = L ( 1 , χ ) . \lim _{s \to 1^{+}} L(s, \chi)=L(1, \chi) . s→1+limL(s,χ)=L(1,χ).
Proof
证明
Let 1 < s < 2 1<s<2 1<s<2. From the proof of Theorem 9 we have
取 1 < s < 2 1<s<2 1<s<2。由定理9证明中的级数表达式，
L ( s , χ ) − L ( 1 , χ ) = ∑ n = 1 ∞ a n ( 1 n s − 1 ( n + 1 ) s ) − ( 1 n − 1 n + 1 ) L(s, \chi) - L(1, \chi) = \sum_{n=1}^\infty a_n\left \\left(\\frac{1}{n\^s}-\\frac{1}{(n+1)\^s}\\right)-\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right) \\right L(s,χ)−L(1,χ)=n=1∑∞an(ns1−(n+1)s1)−(n1−n+11)
where the sequence { a n } \{a_{n}\} {an} is bounded. Applying the mean value theorem to the function s ↦ n − s − ( n + 1 ) − s s \mapsto n^{-s}-(n+1)^{-s} s↦n−s−(n+1)−s gives a sequence { s n } \{s_{n}\} {sn} with 1 < s n < s 1<s_{n}<s 1<sn<s and
其中 { a n } \{a_n\} {an} 是有界序列。对关于 s s s 的函数 f ( s ) = n − s − ( n + 1 ) − s f(s)=n^{-s}-(n+1)^{-s} f(s)=n−s−(n+1)−s 使用微分中值定理，存在 s n ∈ ( 1 , s ) s_n\in(1,s) sn∈(1,s) 使得
L ( s , χ ) − L ( 1 , χ ) = ( s − 1 ) ∑ n = 1 ∞ a n ( log ( n + 1 ) ( n + 1 ) s n − log n n s n ) . L(s, \chi) - L(1, \chi) = (s -1)\sum_{n=1}^\infty a_n\left( \frac{\log(n + 1)}{(n + 1)^{s_n}} -\frac{\log n}{n^{s_n}} \right) . L(s,χ)−L(1,χ)=(s−1)n=1∑∞an((n+1)snlog(n+1)−nsnlogn).
A further application of the mean value theorem, to x ↦ log x x s n x \mapsto \frac{\log x}{x^{s_{n}}} x↦xsnlogx, gives a sequence b n b_{n} bn with n < b n < n + 1 n<b_{n}<n+1 n<bn<n+1 and
再对函数 g ( x ) = log x x s n g(x)=\frac{\log x}{x^{s_n}} g(x)=xsnlogx 使用微分中值定理，存在 b n ∈ ( n , n + 1 ) b_n\in(n,n+1) bn∈(n,n+1)，满足
L ( s , χ ) − L ( 1 , χ ) = ( s − 1 ) ∑ n = 1 ∞ a n ⋅ ( 1 − s n ) log b n b n s n + 1 . L(s, \chi) - L(1, \chi) = (s -1)\sum_{n=1}^\infty a_n \cdot \frac{(1-s_n)\log b_n}{b_n^{s_n+1}} . L(s,χ)−L(1,χ)=(s−1)n=1∑∞an⋅bnsn+1(1−sn)logbn.
Let A A A be an upper bound for the sequence { ∣ a n ∣ } \{|a_{n}|\} {∣an∣}. Define
设 A A A 为 { ∣ a n ∣ } \{|a_n|\} {∣an∣} 的上界，定义常数
C = A ∑ n = 1 ∞ 1 + 2 log ( n + 1 ) n 2 . C = A\sum_{n=1}^\infty \frac{1 + 2 \log(n + 1)}{n^2} . C=An=1∑∞n21+2log(n+1).
To see that this series is convergent first note that log ( n + 1 ) ≤ log ( 2 n ) = log 2 + log n \log (n+1) \leq\log (2 n)=\log 2+\log n log(n+1)≤log(2n)=log2+logn for n ≥ 1 n \geq1 n≥1. Now define f ( x ) = log x x 1 / 2 f(x)=\frac{\log x}{x^{1 / 2}} f(x)=x1/2logx for x > 0 x>0 x>0. Then f ( x ) ≥ 0 f(x) \geq0 f(x)≥0 for x ≥ 1 x \geq1 x≥1 and
先验证该级数收敛：对 n ≥ 1 n\geq1 n≥1， log ( n + 1 ) ≤ log ( 2 n ) = log 2 + log n \log(n+1)\leq\log(2n)=\log2+\log n log(n+1)≤log(2n)=log2+logn。定义 x > 0 x>0 x>0 上函数 f ( x ) = log x x f(x)=\frac{\log x}{\sqrt{x}} f(x)=x logx，则 x ≥ 1 x\geq1 x≥1 时 f ( x ) ≥ 0 f(x)\geq0 f(x)≥0，且导数
f ′ ( x ) = 2 − log x 2 x 3 / 2 . f'(x) = \frac{2 -\log x}{2x^{3/2}} . f′(x)=2x3/22−logx.
Then f ′ ( x ) > 0 f'(x)>0 f′(x)>0 for 1 ≤ x < e 2 1 \leq x<e^{2} 1≤x<e2 and f ′ ( x ) < 0 f'(x)<0 f′(x)<0 for x > e 2 x>e^{2} x>e2. Hence f ( x ) ≤ f ( e 2 ) f(x) \leq f(e^{2}) f(x)≤f(e2) for all x ≥ 1 x \geq1 x≥1, that is log x ≤ e x \log x \leq e\sqrt{x} logx≤ex for all x ≥ 1 x \geq1 x≥1. Hence
当 1 ≤ x < e 2 1\leq x<e^2 1≤x<e2 时 f ′ ( x ) > 0 f'(x)>0 f′(x)>0； x > e 2 x>e^2 x>e2 时 f ′ ( x ) < 0 f'(x)<0 f′(x)<0，故 f ( x ) f(x) f(x) 在 x = e 2 x=e^2 x=e2 处取最大值，即对所有 x ≥ 1 x\geq1 x≥1， log x ≤ e x \log x \leq e\sqrt{x} logx≤ex 。于是对任意正整数 n n n，
0 < 1 + 2 log ( n + 1 ) n 2 ≤ 1 + 2 log 2 + 2 log n n 2 ≤ 1 + 2 log 2 n 2 + 4 e ⋅ n 3 / 2 . 0 < \frac{1 + 2 \log(n + 1)}{n^2} \leq \frac{1 + 2 \log 2 + 2 \log n}{n^2} \leq \frac{1 + 2 \log 2}{n^2}+\frac{4}{e\cdot n^{3/2}} . 0<n21+2log(n+1)≤n21+2log2+2logn≤n21+2log2+e⋅n3/24.
It follows that
因此对任意 N N N，
∑ n = 1 N 1 + 2 log ( n + 1 ) n 2 ≤ ( 1 + 2 log 2 ) ζ ( 2 ) + 4 e ζ ( 3 / 2 ) \sum_{n=1}^N \frac{1 + 2 \log(n + 1)}{n^2}\leq (1+2\log2)\zeta(2)+\frac{4}{e}\zeta(3/2) n=1∑Nn21+2log(n+1)≤(1+2log2)ζ(2)+e4ζ(3/2)
and so the sum defining C C C is convergent. For each n n n,
定义 C C C 的级数收敛。对每一项，
∣ ( 1 − s n ) log b n b n s n + 1 ∣ ≤ 1 + 2 log ( n + 1 ) n 2 \left|\frac{(1-s_n)\log b_n}{b_n^{s_n+1}}\right|\leq \frac{1+2\log(n+1)}{n^2} bnsn+1(1−sn)logbn ≤n21+2log(n+1)
and so indeed
故
∣ L ( s , χ ) − L ( 1 , χ ) ∣ ≤ C ( s − 1 ) |L(s, \chi)-L(1, \chi)| \leq C(s-1) ∣L(s,χ)−L(1,χ)∣≤C(s−1)
for 1 < s < 2 1<s<2 1<s<2. It is immediate that
该不等式对区间 1 < s < 2 1<s<2 1<s<2 成立。令 s → 1 + s\to1^+ s→1+ 即得极限等式
lim s → 1 + L ( s , χ ) = L ( 1 , χ ) . \lim _{s \to 1^{+}} L(s, \chi)=L(1, \chi) . s→1+limL(s,χ)=L(1,χ).
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https://blog.csdn.net/u013669912/article/details/162129145