黎曼澤塔函數 ，寫作ζ(s ) 複數 s Re(s ) > 1
ζ
(
s
)
=
∑
n
=
1
∞
1
n
s
{\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}}
複平面 中一矩形區域之黎曼ζ函數
ζ
(
z
)
{\displaystyle \zeta (z)}
Matplotlib 程式繪圖產生，使用到定義域着色 方法。[ 1] 它亦可以用積分定義：
ζ
(
s
)
=
1
Γ
(
s
)
∫
0
∞
x
s
−
1
e
x
−
1
d
x
{\displaystyle \zeta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }{\frac {x^{s-1}}{e^{x}-1}}\mathrm {d} x}
在區域 {s : Re(s ) > 1} 無窮級數 收斂並為一全純函數 。歐拉 在1740年考慮過 s 柴比雪夫 拓展到 s > 1[ 2] 波恩哈德·黎曼 認識到：ζ函數可以通過解析延拓 ，把定義域 擴展到幾乎整個複數域上的全純函數 ζ(s ) 黎曼猜想 所研究的函數。
雖然黎曼的ζ函數被數學家認為主要和「最純」的數學領域數論 相關，它也出現在應用統計學 （參看齊夫定律 和齊夫-曼德爾布羅特定律 物理 ，以及調音 的數學理論中。
ζ函數最早出現於1350年左右，尼克爾·奧里斯姆 發現了調和級數 發散，即：
ζ
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+
1
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1
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.
.
→
∞
{\displaystyle \zeta (1)=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+...\to \infty }
More information
奧里斯姆對調和級數發散的「證明」
ζ
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1
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{\displaystyle {\begin{aligned}\zeta (1)&=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+...\\&=1+{\frac {1}{2}}+({\frac {1}{3}}+{\frac {1}{4}})+({\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}})+...\\&\geq 1+{\frac {1}{2}}+({\frac {1}{4}}+{\frac {1}{4}})+({\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}})+...\\&=1+{\frac {1}{2}}+{\frac {1}{2}}+{\frac {1}{2}}+...\\&\to \infty \\\end{aligned}}}
Close
第n個調和數（藍點）與Log(n)+γ(紅線)的圖像 之後的一次進展來自萊昂哈德·歐拉 ，他給出了調和級數呈對數發散。
More information
歐拉對調和級數發散速度的證明[ 3]
為了求出調和級數的部分和，使用歐拉-麥克勞林求和公式 （當然，亦可使用阿貝爾求和公式 ）：
∑
y
<
n
≤
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f
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n
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=
∫
y
x
f
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d
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∫
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x
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⌊
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⌋
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f
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⌊
x
⌋
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x
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⌊
y
⌋
−
y
)
{\displaystyle \sum _{y<n\leq x}f(n)=\int _{y}^{x}f(t)\,\mathrm {d} t+\int _{y}^{x}(t-\left\lfloor t\right\rfloor )f'(t)\,\mathrm {d} t+f(x)(\left\lfloor x\right\rfloor -x)-f(y)(\left\lfloor y\right\rfloor -y)}
∑
n
≤
x
1
n
=
1
+
∫
1
x
1
t
d
t
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∫
1
x
(
t
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⌊
t
⌋
)
t
2
d
t
+
⌊
x
⌋
−
x
x
=
1
+
ln
x
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∫
1
x
(
t
−
⌊
t
⌋
)
t
2
d
t
+
O
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1
x
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1
+
ln
x
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∫
1
∞
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⌊
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⌋
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2
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∞
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⌋
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∞
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∞
{
t
}
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O
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x
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{\displaystyle {\begin{aligned}\sum _{n\leq x}{\frac {1}{n}}&=1+\int _{1}^{x}{\frac {1}{t}}\,\mathrm {d} t-\int _{1}^{x}{\frac {(t-\left\lfloor t\right\rfloor )}{t^{2}}}\,\mathrm {d} t+{\frac {\left\lfloor x\right\rfloor -x}{x}}\\&=1+\ln x-\int _{1}^{x}{\frac {(t-\left\lfloor t\right\rfloor )}{t^{2}}}\,\mathrm {d} t+\mathrm {O} \left({\frac {1}{x}}\right)\\&=1+\ln x-\int _{1}^{\infty }{\frac {(t-\left\lfloor t\right\rfloor )}{t^{2}}}\,\mathrm {d} t+\int _{x}^{\infty }{\frac {(t-\left\lfloor t\right\rfloor )}{t^{2}}}\,\mathrm {d} t+\mathrm {O} \left({\frac {1}{x}}\right)\\&=\ln x+1-\int _{1}^{\infty }{\frac {(t-\left\lfloor t\right\rfloor )}{t^{2}}}\,\mathrm {d} t+\int _{x}^{\infty }{\frac {\left\{t\right\}}{t^{2}}}\,\mathrm {d} t+\mathrm {O} \left({\frac {1}{x}}\right)\\\end{aligned}}}
1
−
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1
∞
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t
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⌊
t
⌋
)
t
2
d
t
{\displaystyle 1-\int _{1}^{\infty }{\frac {(t-\left\lfloor t\right\rfloor )}{t^{2}}}\,\mathrm {d} t}
歐拉-馬斯刻若尼常數 γ
再考慮剩下的一個積分，也就是
∫
x
∞
{
t
}
t
2
d
t
{\displaystyle \int _{x}^{\infty }{\frac {\left\{t\right\}}{t^{2}}}\,\mathrm {d} t}
{
t
}
≤
1
{\displaystyle \left\{t\right\}\leq 1}
∫
x
∞
{
t
}
t
2
d
t
≤
∫
x
∞
1
t
2
d
t
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1
x
{\displaystyle \int _{x}^{\infty }{\frac {\left\{t\right\}}{t^{2}}}\,\mathrm {d} t\leq \int _{x}^{\infty }{\frac {1}{t^{2}}}\,\mathrm {d} t={\frac {1}{x}}}
∑
n
≤
x
1
n
=
ln
x
+
γ
+
O
(
1
x
)
{\displaystyle \sum _{n\leq x}{\frac {1}{n}}=\ln x+\gamma +\mathrm {O} ({\frac {1}{x}})}
Close
除此之外，他還在1735年給出了巴塞爾問題 的解答，得到
ζ
(
2
)
=
π
2
6
{\displaystyle \zeta (2)={\frac {\pi ^{2}}{6}}}
巴塞爾問題#歐拉的錯誤證明 中看到，然而那是他的第一個證明，因而廣為人知。[ 4]
More information 歐拉對
歐拉對
ζ
(
2
)
=
π
2
6
{\displaystyle {\begin{smallmatrix}\zeta (2)={\frac {\pi ^{2}}{6}}\end{smallmatrix}}}
下面將寫出歐拉對上式的證明中缺失的嚴格論證的部分，即對連乘積公式的證明部分，而不涉及最終的系數比較
首先考慮當n為奇數時，將
z
n
−
a
n
{\displaystyle z^{n}-a^{n}}
a
,
a
e
2
π
i
1
n
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a
e
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π
i
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n
,
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.
.
,
a
e
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π
i
n
−
1
n
{\displaystyle a,ae^{2\pi i{\frac {1}{n}}},ae^{2\pi i{\frac {2}{n}}},...,ae^{2\pi i{\frac {n-1}{n}}}}
由於n為奇數，所以可以將除了z=a外的其他根及其共軛一一配對，即 將共軛的根一一配對
a
e
2
π
i
k
n
,
a
e
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π
i
n
−
k
n
=
a
e
−
2
π
i
k
n
{\displaystyle ae^{2\pi i{\frac {k}{n}}},ae^{2\pi i{\frac {n-k}{n}}}=ae^{-2\pi i{\frac {k}{n}}}}
韋達定理 可以還原出每對根的最小多項式：
x
1
+
x
2
=
−
a
1
a
0
=
a
e
2
π
i
k
n
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a
e
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π
i
k
n
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cos
(
2
π
k
n
)
+
cos
(
−
2
π
k
n
)
=
2
cos
(
2
π
k
n
)
{\displaystyle x_{1}+x_{2}=-{\frac {a_{1}}{a_{0}}}=ae^{2\pi i{\frac {k}{n}}}+ae^{-2\pi i{\frac {k}{n}}}=\cos \left(2\pi {\frac {k}{n}}\right)+\cos \left(-2\pi {\frac {k}{n}}\right)=2\cos \left({\frac {2\pi k}{n}}\right)}
x
1
x
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=
a
2
a
0
=
a
e
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π
i
k
n
a
e
−
2
π
i
k
n
=
a
2
{\displaystyle x_{1}x_{2}={\frac {a_{2}}{a_{0}}}=ae^{2\pi i{\frac {k}{n}}}ae^{-2\pi i{\frac {k}{n}}}=a^{2}}
a
0
=
1
{\displaystyle a_{0}=1}
a
0
z
2
+
a
1
z
+
a
2
=
z
2
−
2
cos
(
2
π
k
n
)
z
+
a
2
{\displaystyle a_{0}z^{2}+a_{1}z+a_{2}=z^{2}-2\cos \left({\tfrac {2\pi k}{n}}\right)z+a^{2}}
n
−
1
2
{\displaystyle {\tfrac {n-1}{2}}}
z
−
a
{\displaystyle z-a}
z
n
−
a
n
=
(
z
−
a
)
∏
k
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1
n
−
1
2
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z
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a
z
cos
2
k
π
n
+
a
2
)
{\displaystyle z^{n}-a^{n}=(z-a)\prod _{k=1}^{\frac {n-1}{2}}\left(z^{2}-2az\cos {\frac {2k\pi }{n}}+a^{2}\right)}
z
=
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x
N
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a
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N
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n
{\displaystyle z=1+{\frac {x}{N}},a=1-{\frac {x}{N}},N=n}
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x
N
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N
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[
(
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∏
k
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2
[
(
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+
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N
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2
−
2
(
1
+
x
N
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(
1
−
x
N
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cos
(
2
π
k
N
)
+
(
1
−
x
N
)
2
]
=
2
x
N
∏
k
=
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N
−
1
2
[
2
+
2
x
2
N
2
−
2
(
1
−
x
2
N
2
)
cos
(
2
π
k
N
)
]
=
2
x
N
∏
k
=
1
N
−
1
2
[
2
+
2
x
2
N
2
−
2
cos
(
2
π
k
N
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+
2
x
2
N
2
cos
(
2
π
k
N
)
]
=
4
x
N
∏
k
=
1
N
−
1
2
(
(
1
−
cos
(
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π
k
N
)
)
+
(
1
+
cos
(
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π
k
N
)
)
x
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)
=
4
x
N
∏
k
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1
N
−
1
2
{
[
1
−
cos
(
2
π
k
N
)
]
[
1
+
1
+
cos
(
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π
k
N
)
1
−
cos
(
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π
k
N
)
x
2
N
2
]
}
{\displaystyle {\begin{aligned}\left(1+{\frac {x}{N}}\right)^{N}-\left(1-{\frac {x}{N}}\right)^{N}&=\left[\left(1+{\frac {x}{N}}\right)-\left(1-{\frac {x}{N}}\right)\right]\prod _{k=1}^{\frac {N-1}{2}}\left[\left(1+{\frac {x}{N}}\right)^{2}-2\left(1+{\frac {x}{N}}\right)\left(1-{\frac {x}{N}}\right)\cos \left({\frac {2\pi k}{N}}\right)+\left(1-{\frac {x}{N}}\right)^{2}\right]\\&={\frac {2x}{N}}\prod _{k=1}^{\frac {N-1}{2}}\left[2+{\frac {2x^{2}}{N^{2}}}-2\left(1-{\frac {x^{2}}{N^{2}}}\right)\cos \left({\frac {2\pi k}{N}}\right)\right]\\&={\frac {2x}{N}}\prod _{k=1}^{\frac {N-1}{2}}\left[{2+{\frac {2{x^{2}}}{N^{2}}}-2\cos \left({\frac {2\pi k}{N}}\right)+{\frac {2{x^{2}}}{N^{2}}}\cos({\frac {2\pi k}{N}})}\right]\\&={\frac {4x}{N}}\prod _{k=1}^{\frac {N-1}{2}}\left({(1-\cos({\frac {2\pi k}{N}}))+(1+\cos({\frac {2\pi k}{N}})){\frac {x^{2}}{N^{2}}}}\right)\\&={\frac {4x}{N}}\prod _{k=1}^{\frac {N-1}{2}}\left\{\left[1-\cos \left({\frac {2\pi k}{N}}\right)\right]\left[{1+{\frac {1+\cos \left({\frac {2\pi k}{N}}\right)}{1-\cos({\frac {2\pi k}{N}})}}{\frac {x^{2}}{N^{2}}}}\right]\right\}\\\end{aligned}}}
4
N
∏
k
=
1
N
−
1
2
(
1
−
cos
(
2
π
k
N
)
)
{\displaystyle {\frac {4}{N}}\prod _{k=1}^{\frac {N-1}{2}}(1-\cos({\frac {2\pi k}{N}}))}
C
(
N
)
{\displaystyle C(N)}
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1
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N
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N
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x
N
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N
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N
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x
∏
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N
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2
(
1
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+
cos
(
2
π
k
N
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1
−
cos
(
2
π
k
N
)
x
2
N
2
)
{\displaystyle \left(1+{\frac {x}{N}}\right)^{N}-\left(1-{\frac {x}{N}}\right)^{N}={C(N)}x\prod _{k=1}^{\frac {N-1}{2}}\left({1+{\frac {1+\cos({\frac {2\pi k}{N}})}{1-\cos({\frac {2\pi k}{N}})}}{\frac {x^{2}}{N^{2}}}}\right)}
(
1
+
x
N
)
N
=
∑
k
=
0
N
C
N
k
x
k
N
k
{\displaystyle {(1+{\frac {x}{N}})^{N}}=\sum _{k=0}^{N}{C_{N}^{k}{\frac {x^{k}}{N^{k}}}}}
(
1
−
x
N
)
N
=
∑
k
=
0
N
(
−
1
)
k
C
N
k
x
k
N
k
{\displaystyle {(1-{\frac {x}{N}})^{N}}=\sum _{k=0}^{N}{{{(-1)}^{k}}C_{N}^{k}{\frac {x^{k}}{N^{k}}}}}
C
N
1
x
N
−
(
−
1
)
C
N
1
x
N
=
2
C
N
1
x
N
=
2
x
{\displaystyle C_{N}^{1}{\frac {x}{N}}-(-1)C_{N}^{1}{\frac {x}{N}}=2C_{N}^{1}{\frac {x}{N}}=2x}
C
(
N
)
=
2
{\displaystyle C(N)=2}
(
1
+
x
N
)
N
−
(
1
−
x
N
)
N
=
2
x
∏
k
=
1
N
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1
2
(
1
+
1
+
cos
(
2
π
k
N
)
1
−
cos
(
2
π
k
N
)
x
2
N
2
)
{\displaystyle \left(1+{\frac {x}{N}}\right)^{N}-\left(1-{\frac {x}{N}}\right)^{N}=2x\prod _{k=1}^{\frac {N-1}{2}}\left({1+{\frac {1+\cos({\frac {2\pi k}{N}})}{1-\cos({\frac {2\pi k}{N}})}}{\frac {x^{2}}{N^{2}}}}\right)}
θ
=
2
π
k
N
{\displaystyle \theta ={\frac {2\pi k}{N}}}
cos
(
θ
)
=
1
−
θ
2
2
+
O
(
θ
3
)
{\displaystyle \cos(\theta )=1-{\frac {\theta ^{2}}{2}}+\mathrm {O} (\theta ^{3})}
(
1
+
x
N
)
N
−
(
1
−
x
N
)
N
=
2
x
∏
k
=
1
N
−
1
2
[
1
+
1
+
cos
(
2
π
k
N
)
1
−
cos
(
2
π
k
N
)
x
2
N
2
]
=
2
x
∏
k
=
1
N
−
1
2
{
1
+
1
+
[
1
−
θ
2
2
+
O
(
θ
3
)
]
1
−
[
1
−
θ
2
2
+
O
(
θ
3
)
]
x
2
N
2
}
=
2
x
∏
k
=
1
N
−
1
2
[
1
+
2
−
θ
2
2
+
O
(
θ
3
)
θ
2
2
+
O
(
θ
3
)
x
2
N
2
]
=
2
x
∏
k
=
1
N
−
1
2
(
1
+
(
4
−
θ
2
+
O
(
θ
3
)
)
x
2
(
θ
2
+
O
(
θ
3
)
)
N
2
)
=
2
x
∏
k
=
1
N
−
1
2
(
1
+
(
4
−
(
2
k
π
N
)
2
+
O
(
(
2
k
π
N
)
3
)
)
x
2
(
(
2
k
π
N
)
2
+
O
(
(
2
k
π
N
)
3
)
)
N
2
)
=
2
x
∏
k
=
1
N
−
1
2
(
1
+
(
4
−
(
2
k
π
N
)
2
+
O
(
(
2
k
π
N
)
3
)
)
x
2
(
2
k
π
)
2
+
O
(
(
2
k
π
)
3
N
)
)
{\displaystyle {\begin{aligned}\left(1+{\frac {x}{N}}\right)^{N}-\left(1-{\frac {x}{N}}\right)^{N}&=2x\prod _{k=1}^{\frac {N-1}{2}}\left[1+{\frac {1+\cos \left({\frac {2\pi k}{N}}\right)}{1-\cos \left({\frac {2\pi k}{N}}\right)}}{\frac {x^{2}}{N^{2}}}\right]\\&=2x\prod _{k=1}^{\frac {N-1}{2}}\left\{1+{\frac {1+\left[1-{\frac {\theta ^{2}}{2}}+\mathrm {O} \left(\theta ^{3}\right)\right]}{1-\left[1-{\frac {\theta ^{2}}{2}}+\mathrm {O} \left(\theta ^{3}\right)\right]}}{\frac {x^{2}}{N^{2}}}\right\}\\&=2x\prod _{k=1}^{\frac {N-1}{2}}\left[1+{\frac {2-{\frac {\theta ^{2}}{2}}+\mathrm {O} \left(\theta ^{3}\right)}{{\frac {\theta ^{2}}{2}}+\mathrm {O} \left(\theta ^{3}\right)}}{\frac {x^{2}}{N^{2}}}\right]\\&=2x\prod _{k=1}^{\frac {N-1}{2}}\left(1+{\frac {\left(4-\theta ^{2}+\mathrm {O} \left(\theta ^{3}\right)\right)x^{2}}{\left(\theta ^{2}+\mathrm {O} \left(\theta ^{3}\right)\right)N^{2}}}\right)\\&=2x\prod _{k=1}^{\frac {N-1}{2}}\left(1+{\frac {\left(4-\left({\frac {2k\pi }{N}}\right)^{2}+\mathrm {O} \left(\left({\frac {2k\pi }{N}}\right)^{3}\right)\right)x^{2}}{\left(\left({\frac {2k\pi }{N}}\right)^{2}+\mathrm {O} \left(\left({\frac {2k\pi }{N}}\right)^{3}\right)\right)N^{2}}}\right)\\&=2x\prod _{k=1}^{\frac {N-1}{2}}\left(1+{\frac {\left(4-\left({\frac {2k\pi }{N}}\right)^{2}+\mathrm {O} \left(\left({\frac {2k\pi }{N}}\right)^{3}\right)\right)x^{2}}{(2k\pi )^{2}+\mathrm {O} \left({\frac {(2k\pi )^{3}}{N}}\right)}}\right)\end{aligned}}}
lim
N
→
∞
(
1
+
x
N
)
N
−
(
1
−
x
N
)
N
=
e
x
−
e
−
x
{\displaystyle \lim _{N\to \infty }(1+{\frac {x}{N}})^{N}-(1-{\frac {x}{N}})^{N}=e^{x}-e^{-x}}
e
x
−
e
−
x
=
2
x
∏
k
=
1
∞
(
1
+
(
4
+
o
(
1
)
)
x
2
(
2
k
π
)
2
+
o
(
1
)
)
=
2
x
∏
k
=
1
∞
(
1
+
(
1
+
o
(
1
)
)
x
2
k
2
π
2
+
o
(
1
)
)
=
2
x
∏
k
=
1
∞
(
1
+
x
2
k
2
π
2
)
{\displaystyle {\begin{aligned}e^{x}-e^{-x}&=2x\prod _{k=1}^{\infty }\left({1+{\frac {(4+o(1)){x^{2}}}{{{(2k\pi )}^{2}}+o(1)}}}\right)\ \\&=2x\prod _{k=1}^{\infty }\left({1+{\frac {(1+o(1)){x^{2}}}{{k^{2}}{\pi ^{2}}+o(1)}}}\right)\ \\&=2x\prod _{k=1}^{\infty }\left({1+{\frac {x^{2}}{{k^{2}}{\pi ^{2}}}}}\right)\ \\\end{aligned}}}
e
x
−
e
−
x
=
−
∑
k
=
0
∞
(
−
x
)
k
−
x
k
k
!
{\displaystyle e^{x}-e^{-x}=-\sum _{k=0}^{\infty }{\frac {(-x)^{k}-x^{k}}{k!}}}
k
=
3
{\displaystyle k=3}
2
x
3
3
!
{\displaystyle {\frac {2x^{3}}{3!}}}
同時展開右式可得右式的三次項為
(
∑
k
=
1
∞
2
k
2
π
2
)
x
3
{\displaystyle \left(\sum _{k=1}^{\infty }{\frac {2}{k^{2}\pi ^{2}}}\right)x^{3}}
2
3
!
=
∑
k
=
1
∞
2
k
2
π
2
{\displaystyle {\frac {2}{3!}}=\sum _{k=1}^{\infty }{\frac {2}{k^{2}\pi ^{2}}}}
π
2
6
=
∑
k
=
1
∞
1
k
2
=
ζ
(
2
)
{\displaystyle {\frac {\pi ^{2}}{6}}=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}=\zeta \left(2\right)}
Close
歐拉在1737年還發現了歐拉乘積公式 ：
∑
n
=
1
∞
1
n
s
=
∏
p
(
1
−
1
p
s
)
−
1
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=\prod _{p}(1-{\frac {1}{p^{s}}})^{-1}}
證明黎曼ζ函數的歐拉乘積公式 中看到。
Re
(
s
)
>
1
{\displaystyle {\begin{smallmatrix}\operatorname {Re} (s)>1\end{smallmatrix}}}
ζ
(
s
)
>
0
{\displaystyle {\begin{smallmatrix}\zeta (s)>0\end{smallmatrix}}}
1749年，歐拉通過大膽的計算發現了(以下公式當中存在定義域謬誤，後由黎曼透過解析延拓証明以下公式只適用於 Re(s ) > 1 [ 5]
ζ
(
−
1
)
=
1
+
2
+
3
+
4
+
5
+
.
.
.
=
−
1
12
{\displaystyle \zeta (-1)=1+2+3+4+5+...=-{\frac {1}{12}}}
ζ
(
−
2
)
=
1
2
+
2
2
+
3
2
+
4
2
+
5
2
+
.
.
.
=
0
{\displaystyle \zeta (-2)=1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+...=0}
ζ
(
−
3
)
=
1
3
+
2
3
+
3
3
+
4
3
+
5
3
+
.
.
.
=
1
120
{\displaystyle \zeta (-3)=1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+...={\frac {1}{120}}}
1900年，希爾伯特 在巴黎的第二屆國際數學家大會上作了題為《數學問題》的演講，提出了23道最重要的數學問題，黎曼假設在其中作為第8題出現。希爾伯特－波利亞猜想 ，具體時間及場合未知。
虛部介於0與T的零點數量（藍點）與黎曼-馮·曼格爾特公式（紅線）的圖像 1914年，哈那德·玻爾 和愛德蒙·蘭道 證明了玻爾-蘭道定理 ：含有臨界線的任意帶狀區域都幾乎包含了ζ的所有非平凡零點，表明了臨界線為零點匯聚的「中心位置」。
1921年，哈代 和李特爾伍德 證明了存在常數T，使臨界線上虛部位於0與T之間的非平凡零點的數量至少為
K
T
{\displaystyle {\begin{smallmatrix}KT\end{smallmatrix}}}
ζ函數與數論函數存在的聯繫可以通過佩龍公式 轉化為它和數論函數的求和的關係：設
G
(
s
)
=
∑
n
=
1
∞
g
(
n
)
{\displaystyle G(s)={\sum _{n=1}^{\infty }}g(n)}
則由佩龍公式，
A
(
x
)
=
∑
n
≤
x
′
g
(
n
)
=
1
2
π
i
∫
c
−
i
∞
c
+
i
∞
G
(
z
)
x
z
z
d
z
{\displaystyle A(x)={\sum _{n\leq x}}'g(n)={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }G(z){\frac {x^{z}}{z}}\,\mathrm {d} z}
其中右上角的'表示如果x是整數，那麼求和的最後一項要乘以
1
2
{\displaystyle {\begin{smallmatrix}{\frac {1}{2}}\end{smallmatrix}}}
此函數和質數 的關係已由歐拉 所揭示：
ζ
(
s
)
=
∏
p
1
1
−
p
−
s
{\displaystyle \zeta (s)=\prod _{p}{\frac {1}{1-p^{-s}}}}
這是一個延展到所有的質數p 的無窮乘積 ，被稱為歐拉乘積 。這是幾何級數 的公式和算術基本定理 的一個結果。
log
ζ
(
s
)
=
∑
n
=
1
∞
1
n
∑
p
p
−
n
s
{\displaystyle \log \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n}}\sum _{p}p^{-ns}}
More information 首先留意到 Close
lim
ε
→
0
ζ
(
1
+
ε
)
+
ζ
(
1
−
ε
)
2
=
γ
{\displaystyle \lim _{\varepsilon \to 0}{\frac {\zeta (1+\varepsilon )+\zeta (1-\varepsilon )}{2}}=\gamma }
其中γ是歐拉-馬歇羅尼常數 =
0.577215...
{\displaystyle 0.577215...}
同樣由歐拉發現，ζ函數在負整數點的值是有理數，這在模形式中發揮着重要作用，而且ζ函數在負偶整數點的值為零。
事實上
ζ
(
−
n
)
=
−
B
n
+
1
n
+
1
{\displaystyle \zeta (-n)=-{\frac {B_{n+1}}{n+1}}}
B n 白努利數 。
因為 B 2n +1 =0，故ζ函數在負偶整數點的值為零。
ζ
(
x
+
i
y
)
=
∑
k
=
1
∞
cos
(
y
ln
k
)
−
i
sin
(
y
ln
k
)
k
x
,
y
∈
R
{\displaystyle \zeta (x+{\rm {i}}y)=\sum _{k=1}^{\infty }{\frac {\cos(y\ln k)-{\rm {i}}\sin(y\ln k)}{k^{x}}},y\in {\mathbb {R} }}
arg
[
ζ
(
x
+
i
y
)
]
=
−
arctan
∑
k
=
1
∞
sin
(
y
ln
k
)
k
x
∑
k
=
1
∞
cos
(
y
ln
k
)
k
x
{\displaystyle \arg[\zeta (x+{\rm {i}}y)]=-\arctan {\frac {\sum _{k=1}^{\infty }{\frac {\sin(y\ln k)}{k^{x}}}}{\sum _{k=1}^{\infty }{\frac {\cos(y\ln k)}{k^{x}}}}}}
ζ
(
−
2
n
n
∈
Z
+
)
=
0
{\displaystyle \zeta (-2n_{n\in \mathbb {Z} ^{+}})=0}
ζ
(
−
9
)
=
−
1
132
(
R
)
{\displaystyle \zeta (-9)=-{\frac {1}{132}}({\mathfrak {R}})}
ζ
(
−
7
)
=
1
240
(
R
)
{\displaystyle \zeta (-7)={\frac {1}{240}}({\mathfrak {R}})}
ζ
(
−
5
)
=
−
1
252
(
R
)
{\displaystyle \zeta (-5)=-{\frac {1}{252}}({\mathfrak {R}})}
ζ
(
−
3
)
=
1
120
(
R
)
{\displaystyle \zeta (-3)={\frac {1}{120}}({\mathfrak {R}})}
ζ
(
−
1
)
=
−
1
12
(
R
)
{\displaystyle \zeta (-1)=-{\frac {1}{12}}({\mathfrak {R}})}
ζ
(
0
)
=
−
1
2
{\displaystyle \zeta (0)=-{\frac {1}{2}}}
ζ
(
1
−
)
=
−
∞
{\displaystyle \zeta (1^{-})=-\infty }
ζ
(
1
+
)
=
∞
{\displaystyle \zeta (1^{+})=\infty }
ζ
(
2
)
=
π
2
6
{\displaystyle \zeta (2)={\frac {\pi ^{2}}{6}}}
ζ
(
4
)
=
π
4
90
{\displaystyle \zeta (4)={\frac {\pi ^{4}}{90}}}
ζ
(
6
)
=
π
6
945
{\displaystyle \zeta (6)={\frac {\pi ^{6}}{945}}}
ζ
(
8
)
=
π
8
9450
{\displaystyle \zeta (8)={\frac {\pi ^{8}}{9450}}}
ζ
(
10
)
=
π
10
93555
{\displaystyle \zeta (10)={\frac {\pi ^{10}}{93555}}}
臨界線上的數值計算可以通過黎曼-西格爾公式 完成。林德勒夫猜想
ϵ
>
0
{\displaystyle {\begin{smallmatrix}\epsilon >0\end{smallmatrix}}}
ζ
(
1
2
+
i
t
)
=
O
(
t
ϵ
)
{\displaystyle \zeta \left({\frac {1}{2}}+it\right)=\mathrm {O} (t^{\epsilon })}
存档副本 . [2015-01-17 ] . （原始內容 存檔於2015-03-20）. Tom M.Apostol. Introduction to Analytic Number Theory. 北京: 世界圖書出版公司北京公司. 2012年1月: 55 – 56. ISBN 978-7-5100-4062-7（英語） . Theorem 3.2 If x ≥ 1 we have : C. S. Ogilvy & J. T. Anderson Excursions in Number Theory , pp. 29–35, Dover Publications Inc., 1988 ISBN 978-0-486-25778-5 
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![{\displaystyle {\begin{aligned}\left(1+{\frac {x}{N}}\right)^{N}-\left(1-{\frac {x}{N}}\right)^{N}&=\left[\left(1+{\frac {x}{N}}\right)-\left(1-{\frac {x}{N}}\right)\right]\prod _{k=1}^{\frac {N-1}{2}}\left[\left(1+{\frac {x}{N}}\right)^{2}-2\left(1+{\frac {x}{N}}\right)\left(1-{\frac {x}{N}}\right)\cos \left({\frac {2\pi k}{N}}\right)+\left(1-{\frac {x}{N}}\right)^{2}\right]\\&={\frac {2x}{N}}\prod _{k=1}^{\frac {N-1}{2}}\left[2+{\frac {2x^{2}}{N^{2}}}-2\left(1-{\frac {x^{2}}{N^{2}}}\right)\cos \left({\frac {2\pi k}{N}}\right)\right]\\&={\frac {2x}{N}}\prod _{k=1}^{\frac {N-1}{2}}\left[{2+{\frac {2{x^{2}}}{N^{2}}}-2\cos \left({\frac {2\pi k}{N}}\right)+{\frac {2{x^{2}}}{N^{2}}}\cos({\frac {2\pi k}{N}})}\right]\\&={\frac {4x}{N}}\prod _{k=1}^{\frac {N-1}{2}}\left({(1-\cos({\frac {2\pi k}{N}}))+(1+\cos({\frac {2\pi k}{N}})){\frac {x^{2}}{N^{2}}}}\right)\\&={\frac {4x}{N}}\prod _{k=1}^{\frac {N-1}{2}}\left\{\left[1-\cos \left({\frac {2\pi k}{N}}\right)\right]\left[{1+{\frac {1+\cos \left({\frac {2\pi k}{N}}\right)}{1-\cos({\frac {2\pi k}{N}})}}{\frac {x^{2}}{N^{2}}}}\right]\right\}\\\end{aligned}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/23ed868ba0fd3d14af72b827080bb13dcab2a8e8)
![{\displaystyle {\begin{aligned}\left(1+{\frac {x}{N}}\right)^{N}-\left(1-{\frac {x}{N}}\right)^{N}&=2x\prod _{k=1}^{\frac {N-1}{2}}\left[1+{\frac {1+\cos \left({\frac {2\pi k}{N}}\right)}{1-\cos \left({\frac {2\pi k}{N}}\right)}}{\frac {x^{2}}{N^{2}}}\right]\\&=2x\prod _{k=1}^{\frac {N-1}{2}}\left\{1+{\frac {1+\left[1-{\frac {\theta ^{2}}{2}}+\mathrm {O} \left(\theta ^{3}\right)\right]}{1-\left[1-{\frac {\theta ^{2}}{2}}+\mathrm {O} \left(\theta ^{3}\right)\right]}}{\frac {x^{2}}{N^{2}}}\right\}\\&=2x\prod _{k=1}^{\frac {N-1}{2}}\left[1+{\frac {2-{\frac {\theta ^{2}}{2}}+\mathrm {O} \left(\theta ^{3}\right)}{{\frac {\theta ^{2}}{2}}+\mathrm {O} \left(\theta ^{3}\right)}}{\frac {x^{2}}{N^{2}}}\right]\\&=2x\prod _{k=1}^{\frac {N-1}{2}}\left(1+{\frac {\left(4-\theta ^{2}+\mathrm {O} \left(\theta ^{3}\right)\right)x^{2}}{\left(\theta ^{2}+\mathrm {O} \left(\theta ^{3}\right)\right)N^{2}}}\right)\\&=2x\prod _{k=1}^{\frac {N-1}{2}}\left(1+{\frac {\left(4-\left({\frac {2k\pi }{N}}\right)^{2}+\mathrm {O} \left(\left({\frac {2k\pi }{N}}\right)^{3}\right)\right)x^{2}}{\left(\left({\frac {2k\pi }{N}}\right)^{2}+\mathrm {O} \left(\left({\frac {2k\pi }{N}}\right)^{3}\right)\right)N^{2}}}\right)\\&=2x\prod _{k=1}^{\frac {N-1}{2}}\left(1+{\frac {\left(4-\left({\frac {2k\pi }{N}}\right)^{2}+\mathrm {O} \left(\left({\frac {2k\pi }{N}}\right)^{3}\right)\right)x^{2}}{(2k\pi )^{2}+\mathrm {O} \left({\frac {(2k\pi )^{3}}{N}}\right)}}\right)\end{aligned}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/a3463889e69fa6c26e55d810d8a2ef0a8dc7c440)
, 
...
![{\displaystyle \arg[\zeta (x+{\rm {i}}y)]=-\arctan {\frac {\sum _{k=1}^{\infty }{\frac {\sin(y\ln k)}{k^{x}}}}{\sum _{k=1}^{\infty }{\frac {\cos(y\ln k)}{k^{x}}}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/e900016d87e7c6f5d1672b28108042d891a05718)
