提示 ：此条目的主题不是
尤拉数 。
欧拉-马斯刻若尼常数 是一个数学常数 ，定义为调和级数 与自然对数 的差值：
欧拉-马斯刻若尼常数 欧拉-马斯刻若尼常数 蓝色区域的面积收敛到欧拉常数
符号
γ
{\displaystyle \gamma }
位数 数列编号 A001620 定义
γ
=
lim
n
→
∞
[
(
∑
k
=
1
n
1
k
)
−
ln
(
n
)
]
{\displaystyle \gamma =\lim _{n\rightarrow \infty }\left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]}
γ
=
∫
1
∞
(
1
⌊
x
⌋
−
1
x
)
d
x
{\displaystyle \gamma =\int _{1}^{\infty }\left({1 \over \lfloor x\rfloor }-{1 \over x}\right)\,dx}
连分数 [0; 1, 1, 2, 1, 2, 1, 4, 3, 13, 5, 1, 1, 8, 1, 2, 4, 1, 1, 40, ...] 值
γ
≈
{\displaystyle \gamma \approx }
无穷级数
γ
=
∑
k
=
1
∞
[
1
k
−
ln
(
1
+
1
k
)
]
{\displaystyle \gamma =\sum _{k=1}^{\infty }\left[{\frac {1}{k}}-\ln \left(1+{\frac {1}{k}}\right)\right]}
二进制 0.10010011 1100 0100 0110 0111 … 十进制 0.57721566 4901 5328 6060 6512 … 十六进制 0.93C467E3 7DB0 C7A4 D1BE 3F81 …
γ
=
lim
n
→
∞
[
(
∑
k
=
1
n
1
k
)
−
ln
(
n
)
]
=
∫
1
∞
(
1
⌊
x
⌋
−
1
x
)
d
x
{\displaystyle \gamma =\lim _{n\rightarrow \infty }\left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]=\int _{1}^{\infty }\left({1 \over \lfloor x\rfloor }-{1 \over x}\right)\,dx}
它的近似值为
γ
≈
0.577215664901532860606512090082402431042159335
{\displaystyle \gamma \approx 0.577215664901532860606512090082402431042159335}
[ 1]
欧拉-马斯刻若尼常数主要应用于数论 。
该常数最先由瑞士 数学家莱昂哈德·欧拉 在1735年发表的文章De Progressionibus harmonicus observationes 中定义。欧拉曾经使用
C
{\displaystyle C}
意大利 数学家洛伦佐·马斯凯罗尼 引入了
γ
{\displaystyle \gamma }
目前尚不知道该常数是否为有理数 ，但是分析表明如果它是一个有理数，那么它的分母位数将超过10242080 。[ 2]
−
γ
=
Γ
′
(
1
)
=
Ψ
(
1
)
{\displaystyle \ -\gamma =\Gamma '(1)=\Psi (1)}
γ
=
lim
x
→
∞
[
x
−
Γ
(
1
x
)
]
{\displaystyle \gamma =\lim _{x\to \infty }\left[x-\Gamma \left({\frac {1}{x}}\right)\right]}
γ
=
lim
n
→
∞
[
Γ
(
1
n
)
Γ
(
n
+
1
)
n
1
+
1
n
Γ
(
2
+
n
+
1
n
)
−
n
2
n
+
1
]
{\displaystyle \gamma =\lim _{n\to \infty }\left[{\frac {\Gamma ({\frac {1}{n}})\Gamma (n+1)\,n^{1+{\frac {1}{n}}}}{\Gamma (2+n+{\frac {1}{n}})}}-{\frac {n^{2}}{n+1}}\right]}
γ
=
∑
m
=
2
∞
(
−
1
)
m
ζ
(
m
)
m
{\displaystyle \gamma =\sum _{m=2}^{\infty }{\frac {(-1)^{m}\zeta (m)}{m}}}
=
ln
(
4
π
)
+
∑
m
=
1
∞
(
−
1
)
m
−
1
ζ
(
m
+
1
)
2
m
(
m
+
1
)
{\displaystyle =\ln \left({\frac {4}{\pi }}\right)+\sum _{m=1}^{\infty }{\frac {(-1)^{m-1}\zeta (m+1)}{2^{m}(m+1)}}}
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 }
γ
=
3
2
−
ln
2
−
∑
m
=
2
∞
(
−
1
)
m
m
−
1
m
[
ζ
(
m
)
−
1
]
{\displaystyle \gamma ={\frac {3}{2}}-\ln 2-\sum _{m=2}^{\infty }(-1)^{m}\,{\frac {m-1}{m}}[\zeta (m)-1]}
=
lim
n
→
∞
[
2
n
−
1
2
n
−
ln
n
+
∑
k
=
2
n
(
1
k
−
ζ
(
1
−
k
)
n
k
)
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {2\,n-1}{2\,n}}-\ln \,n+\sum _{k=2}^{n}\left({\frac {1}{k}}-{\frac {\zeta (1-k)}{n^{k}}}\right)\right]}
=
lim
n
→
∞
[
2
n
e
2
n
∑
m
=
0
∞
2
m
n
(
m
+
1
)
!
∑
t
=
0
m
1
t
+
1
−
n
ln
2
+
O
(
1
2
n
e
2
n
)
]
{\displaystyle =\lim _{n\to \infty }\left[{\frac {2^{n}}{e^{2^{n}}}}\sum _{m=0}^{\infty }{\frac {2^{m\,n}}{(m+1)!}}\sum _{t=0}^{m}{\frac {1}{t+1}}-n\,\ln 2+O\left({\frac {1}{2^{n}\,e^{2^{n}}}}\right)\right]}
γ
=
lim
s
→
1
+
∑
n
=
1
∞
(
1
n
s
−
1
s
n
)
=
lim
s
→
1
+
(
ζ
(
s
)
−
1
s
−
1
)
{\displaystyle \gamma =\lim _{s\to 1^{+}}\sum _{n=1}^{\infty }\left({\frac {1}{n^{s}}}-{\frac {1}{s^{n}}}\right)=\lim _{s\to 1^{+}}\left(\zeta (s)-{\frac {1}{s-1}}\right)}
γ
=
lim
x
→
∞
[
x
−
Γ
(
1
x
)
]
{\displaystyle \gamma =\lim _{x\to \infty }\left[x-\Gamma \left({\frac {1}{x}}\right)\right]}
=
lim
n
→
∞
1
n
∑
k
=
1
n
(
⌈
n
k
⌉
−
n
k
)
{\displaystyle =\lim _{n\to \infty }{\frac {1}{n}}\,\sum _{k=1}^{n}\left(\left\lceil {\frac {n}{k}}\right\rceil -{\frac {n}{k}}\right)}
γ
=
∑
k
=
1
n
1
k
−
ln
(
n
)
−
∑
m
=
2
∞
ζ
(
m
,
n
+
1
)
m
{\displaystyle \gamma =\sum _{k=1}^{n}{\frac {1}{k}}-\ln(n)-\sum _{m=2}^{\infty }{\frac {\zeta (m,n+1)}{m}}}
γ
=
−
∫
0
∞
e
−
x
ln
x
d
x
=
∫
∞
0
e
−
x
ln
x
d
x
{\displaystyle \gamma =-\int _{0}^{\infty }{e^{-x}\ln x}\,dx=\int _{\infty }^{0}{e^{-x}\ln x}\,dx}
[ 证明 1]
=
−
∫
0
1
ln
ln
1
x
d
x
{\displaystyle =-\int _{0}^{1}{\ln \ln {\frac {1}{x}}}\,dx}
=
∫
0
∞
(
1
1
−
e
−
x
−
1
x
)
e
−
x
d
x
{\displaystyle =\int _{0}^{\infty }{\left({\frac {1}{1-e^{-x}}}-{\frac {1}{x}}\right)e^{-x}}\,dx}
=
∫
0
∞
1
x
(
1
1
+
x
−
e
−
x
)
d
x
{\displaystyle =\int _{0}^{\infty }{{\frac {1}{x}}\left({\frac {1}{1+x}}-e^{-x}\right)}\,dx}
∫
0
∞
e
−
x
2
ln
x
d
x
=
−
1
4
(
γ
+
2
ln
2
)
π
{\displaystyle \int _{0}^{\infty }{e^{-x^{2}}\ln x}\,dx=-{\tfrac {1}{4}}(\gamma +2\ln 2){\sqrt {\pi }}}
∫
0
∞
e
−
x
ln
2
x
d
x
=
γ
2
+
π
2
6
{\displaystyle \int _{0}^{\infty }{e^{-x}\ln ^{2}x}\,dx=\gamma ^{2}+{\frac {\pi ^{2}}{6}}}
γ
=
∫
0
1
∫
0
1
x
−
1
(
1
−
x
y
)
ln
(
x
y
)
d
x
d
y
=
∑
n
=
1
∞
(
1
n
−
ln
n
+
1
n
)
{\displaystyle \gamma =\int _{0}^{1}\int _{0}^{1}{\frac {x-1}{(1-x\,y)\ln(x\,y)}}\,dx\,dy=\sum _{n=1}^{\infty }\left({\frac {1}{n}}-\ln {\frac {n+1}{n}}\right)}
∑
n
=
1
∞
N
1
(
n
)
+
N
0
(
n
)
2
n
(
2
n
+
1
)
=
γ
{\displaystyle \sum _{n=1}^{\infty }{\frac {N_{1}(n)+N_{0}(n)}{2n(2n+1)}}=\gamma }
γ
=
∑
k
=
1
∞
[
1
k
−
ln
(
1
+
1
k
)
]
{\displaystyle \gamma =\sum _{k=1}^{\infty }\left[{\frac {1}{k}}-\ln \left(1+{\frac {1}{k}}\right)\right]}
γ
=
1
−
∑
k
=
2
∞
(
−
1
)
k
⌊
log
2
k
⌋
k
+
1
{\displaystyle \gamma =1-\sum _{k=2}^{\infty }(-1)^{k}{\frac {\lfloor \log _{2}k\rfloor }{k+1}}}
γ
=
∑
k
=
2
∞
(
−
1
)
k
⌊
log
2
k
⌋
k
=
1
2
−
1
3
+
2
(
1
4
−
1
5
+
1
6
−
1
7
)
+
3
(
1
8
−
⋯
−
1
15
)
+
…
{\displaystyle \gamma =\sum _{k=2}^{\infty }(-1)^{k}{\frac {\left\lfloor \log _{2}k\right\rfloor }{k}}={\tfrac {1}{2}}-{\tfrac {1}{3}}+2\left({\tfrac {1}{4}}-{\tfrac {1}{5}}+{\tfrac {1}{6}}-{\tfrac {1}{7}}\right)+3\left({\tfrac {1}{8}}-\dots -{\tfrac {1}{15}}\right)+\dots }
γ
+
ζ
(
2
)
=
∑
k
=
1
∞
1
k
⌊
k
⌋
2
=
1
+
1
2
+
1
3
+
1
4
(
1
4
+
⋯
+
1
8
)
+
1
9
(
1
9
+
⋯
+
1
15
)
+
…
{\displaystyle \gamma +\zeta (2)=\sum _{k=1}^{\infty }{\frac {1}{k\lfloor {\sqrt {k}}\rfloor ^{2}}}=1+{\tfrac {1}{2}}+{\tfrac {1}{3}}+{\tfrac {1}{4}}\left({\tfrac {1}{4}}+\dots +{\tfrac {1}{8}}\right)+{\tfrac {1}{9}}\left({\tfrac {1}{9}}+\dots +{\tfrac {1}{15}}\right)+\dots }
γ
=
∑
k
=
2
∞
k
−
⌊
k
⌋
2
k
2
⌊
k
⌋
2
=
1
2
2
+
2
3
2
+
1
2
2
(
1
5
2
+
2
6
2
+
3
7
2
+
4
8
2
)
+
1
3
2
(
1
10
2
+
⋯
+
6
15
2
)
+
…
{\displaystyle \gamma =\sum _{k=2}^{\infty }{\frac {k-\lfloor {\sqrt {k}}\rfloor ^{2}}{k^{2}\lfloor {\sqrt {k}}\rfloor ^{2}}}={\tfrac {1}{2^{2}}}+{\tfrac {2}{3^{2}}}+{\tfrac {1}{2^{2}}}\left({\tfrac {1}{5^{2}}}+{\tfrac {2}{6^{2}}}+{\tfrac {3}{7^{2}}}+{\tfrac {4}{8^{2}}}\right)+{\tfrac {1}{3^{2}}}\left({\tfrac {1}{10^{2}}}+\dots +{\tfrac {6}{15^{2}}}\right)+\dots }
γ
=
∫
0
1
1
1
+
x
∑
n
=
1
∞
x
2
n
−
1
d
x
{\displaystyle \gamma =\int _{0}^{1}{\frac {1}{1+x}}\sum _{n=1}^{\infty }x^{2^{n}-1}\,dx}
γ
{\displaystyle \gamma }
连分数 展开式为：
γ
=
[
0
;
1
,
1
,
2
,
1
,
2
,
1
,
4
,
3
,
13
,
5
,
1
,
1
,
8
,
1
,
2
,
4
,
1
,
1
,
40
,
.
.
.
]
{\displaystyle \gamma =[0;1,1,2,1,2,1,4,3,13,5,1,1,8,1,2,4,1,1,40,...]\,}
OEIS 数列A002852 ）.
γ
≈
H
n
−
ln
(
n
)
−
1
2
n
+
1
12
n
2
−
1
120
n
4
+
.
.
.
{\displaystyle \gamma \approx H_{n}-\ln \left(n\right)-{\frac {1}{2n}}+{\frac {1}{12n^{2}}}-{\frac {1}{120n^{4}}}+...}
γ
≈
H
n
−
ln
(
n
+
1
2
+
1
24
n
−
1
48
n
3
+
.
.
.
)
{\displaystyle \gamma \approx H_{n}-\ln \left({n+{\frac {1}{2}}+{\frac {1}{24n}}-{\frac {1}{48n^{3}}}+...}\right)}
γ
≈
H
n
−
ln
(
n
)
+
ln
(
n
+
1
)
2
−
1
6
n
(
n
+
1
)
+
1
30
n
2
(
n
+
1
)
2
−
.
.
.
{\displaystyle \gamma \approx H_{n}-{\frac {\ln \left(n\right)+\ln \left({n+1}\right)}{2}}-{\frac {1}{6n\left({n+1}\right)}}+{\frac {1}{30n^{2}\left({n+1}\right)^{2}}}-...}
γ
{\displaystyle {\boldsymbol {\gamma }}}
的已知位数
日期
位数
计算者
1734年
5
莱昂哈德·欧拉
1736年
15
莱昂哈德·欧拉
1790年
19
洛伦佐·马斯凯罗尼
1809年
24
Johann G. von Soldner
1812年
40
F.B.G. Nicolai
1861年
41
Oettinger
1869年
59
William Shanks
1871年
110
William Shanks
1878年
263
约翰·柯西·亚当斯
1962年
1,271
高德纳
1962年
3,566
D.W. Sweeney
1977年
20,700
Richard P. Brent
1980年
30,100
Richard P. Brent 和埃德温·麦克米伦
1993年
172,000
Jonathan Borwein
1997年
1,000,000
Thomas Papanikolaou
1998年12月
7,286,255
Xavier Gourdon
1999年10月
108,000,000
Xavier Gourdon和Patrick Demichel
2006年7月16日
2,000,000,000
Shigeru Kondo和Steve Pagliarulo
2006年12月8日
116,580,041
Alexander J. Yee
2007年7月15日
5,000,000,000
Shigeru Kondo和Steve Pagliarulo
2008年1月1日
1,001,262,777
Richard B. Kreckel
2008年1月3日
131,151,000
Nicholas D. Farrer
2008年6月30日
10,000,000,000
Shigeru Kondo和Steve Pagliarulo
2009年1月18日
14,922,244,771
Alexander J. Yee和Raymond Chan
2009年3月13日
29,844,489,545
Alexander J. Yee和Raymond Chan
2013年
119,377,958,182
Alexander J. Yee
2016年
160,000,000,000
Peter Trueb
2016年
250,000,000,000
Ron Watkins
2017年
477,511,832,674
Ron Watkins
2020年
600,000,000,100
Seungmin Kim和Ian Cutress
^
γ
=
−
∫
0
∞
e
−
x
ln
x
d
x
{\displaystyle \gamma =-\int _{0}^{\infty }{e^{-x}\ln x}\,dx}
∫
k
k
+
1
1
x
d
x
<
1
k
<
∫
k
−
1
k
1
x
d
x
{\displaystyle \int _{k}^{k+1}{\frac {1}{x}}\,dx<{\frac {1}{k}}<\int _{k-1}^{k}{\frac {1}{x}}\,dx}
∫
k
k
−
1
1
x
d
x
−
1
k
<
1
k
(
k
−
1
)
{\displaystyle \int _{k}^{k-1}{\frac {1}{x}}\,dx-{\frac {1}{k}}<{\frac {1}{k(k-1)}}}
ln
n
<
∑
k
=
1
n
1
k
<
1
+
ln
n
{\displaystyle \ln n<\sum _{k=1}^{n}{\frac {1}{k}}<1+\ln n}
γ
{\displaystyle \gamma }
∑
k
=
1
n
1
k
{\displaystyle \sum _{k=1}^{n}{\frac {1}{k}}}
=
∑
k
=
1
n
∫
0
1
t
k
−
1
d
t
{\displaystyle =\sum _{k=1}^{n}\int _{0}^{1}t^{k-1}\,dt}
=
∫
0
1
∑
k
=
1
n
t
k
−
1
d
t
{\displaystyle =\int _{0}^{1}\sum _{k=1}^{n}t^{k-1}\,dt}
=
∫
0
1
1
−
t
n
1
−
t
d
t
{\displaystyle =\int _{0}^{1}{\frac {1-t^{n}}{1-t}}\,dt}
=
∫
n
0
1
−
(
1
−
x
n
)
n
1
−
(
1
−
x
n
)
d
(
1
−
x
n
)
{\displaystyle =\int _{n}^{0}{\frac {1-\left(1-{\frac {x}{n}}\right)^{n}}{1-\left(1-{\frac {x}{n}}\right)}}d\left(1-{\tfrac {x}{n}}\right)}
=
∫
n
0
1
−
(
1
−
x
n
)
n
x
n
(
−
1
n
)
d
x
{\displaystyle =\int _{n}^{0}{\frac {1-\left(1-{\frac {x}{n}}\right)^{n}}{\frac {x}{n}}}\left(-{\frac {1}{n}}\right)dx}
=
∫
0
n
1
−
(
1
−
x
n
)
n
x
d
x
{\displaystyle =\int _{0}^{n}{\frac {1-\left(1-{\frac {x}{n}}\right)^{n}}{x}}dx}
ln
n
=
∫
1
n
1
x
d
x
,
{\displaystyle \ln n=\int _{1}^{n}{\frac {1}{x}}\,dx,}
γ
=
lim
n
→
∞
(
∑
k
=
1
n
1
k
−
ln
n
)
{\displaystyle \gamma =\lim _{n\to \infty }\left(\sum _{k=1}^{n}{\frac {1}{k}}-\ln n\right)}
=
lim
n
→
∞
[
∫
0
n
1
−
(
1
−
x
/
n
)
n
x
d
x
−
∫
1
n
1
x
d
x
]
{\displaystyle =\lim _{n\to \infty }\left[\int _{0}^{n}{\frac {1-(1-x/n)^{n}}{x}}\,dx-\int _{1}^{n}{\frac {1}{x}}\,dx\right]}
=
lim
n
→
∞
[
∫
0
1
1
−
(
1
−
x
/
n
)
n
x
d
x
−
∫
1
n
(
1
−
x
/
n
)
n
x
]
{\displaystyle =\lim _{n\to \infty }\left[\int _{0}^{1}{\frac {1-(1-x/n)^{n}}{x}}\,dx-\int _{1}^{n}{\frac {(1-x/n)^{n}}{x}}\right]}
=
∫
0
1
1
−
lim
n
→
∞
(
1
−
x
/
n
)
n
x
d
x
−
∫
1
∞
lim
n
→
∞
(
1
−
x
/
n
)
n
x
{\displaystyle =\int _{0}^{1}{\frac {1-\lim _{n\to \infty }(1-x/n)^{n}}{x}}\,dx-\int _{1}^{\infty }{\frac {\lim _{n\to \infty }(1-x/n)^{n}}{x}}}
=
∫
0
1
1
−
e
−
x
x
d
x
−
∫
1
∞
e
−
x
x
{\displaystyle =\int _{0}^{1}{\frac {1-e^{-x}}{x}}\,dx-\int _{1}^{\infty }{\frac {e^{-x}}{x}}}
=
(
1
−
e
−
x
)
ln
x
|
0
1
−
∫
0
1
ln
x
d
(
1
−
e
−
x
)
−
e
−
x
ln
x
|
1
∞
+
∫
1
∞
ln
x
d
e
−
x
{\displaystyle =\left.(1-e^{-x})\ln x\right|_{0}^{1}-\int _{0}^{1}\ln x\,d(1-e^{-x})-\left.e^{-x}\ln x\right|_{1}^{\infty }+\int _{1}^{\infty }\ln x\,de^{-x}}
=
−
∫
0
∞
e
−
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{\displaystyle =-\int _{0}^{\infty }e^{-x}\ln x\,dx.}
前面的放缩法主要是证明了
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{\displaystyle \left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]}
^ A001620 oeis.org [2014-7-17]^ Havil 2003 p 97.
Borwein, Jonathan M., David M. Bradley, Richard E. Crandall. Computational Strategies for the Riemann Zeta Function (PDF) . Journal of Computational and Applied Mathematics. 2000, 121 : 11 [2014-07-17 ] . doi:10.1016/s0377-0427(00)00336-8 原始内容 (PDF) 存档于2006-09-25）. Gourdon, Xavier, and Sebah, P. (2002) "Collection of formulas for Euler's constant, γ. （页面存档备份 ，存于互联网档案馆 ）"
Gourdon, Xavier, and Sebah, P. (2004) "The Euler constant: γ. （页面存档备份 ，存于互联网档案馆 ）"
Donald Knuth (1997) The Art of Computer Programming , Vol. 1ISBN 978-0-201-89683-1 Krämer, Stefan (2005) Die Eulersche Konstante γ und verwandte Zahlen . Diplomarbeit, Universität Göttingen.
Sondow, Jonathan (1998) "An antisymmetric formula for Euler's constant, " Mathematics Magazine 71Sondow, Jonathan (2002) "A hypergeometric approach, via linear forms involving logarithms, to irrationality criteria for Euler's constant. " With an Appendix by Sergey Zlobin , Mathematica Slovaca 59: 307-314. Sondow, Jonathan. An infinite product for eγ via hypergeometric formulas for Euler's constant, γ. 2003. arXiv:math.CA/0306008 Sondow, Jonathan (2003a) "Criteria for irrationality of Euler's constant, " Proceedings of the American Mathematical Society 131Sondow, Jonathan (2005) "Double integrals for Euler's constant and ln 4/π and an analog of Hadjicostas's formula, " American Mathematical Monthly 112Sondow, Jonathan (2005) "New Vacca-type rational series for Euler's constant and its 'alternating' analog ln 4/π. "Sondow, Jonathan; Zudilin, Wadim. Euler's constant, q-logarithms, and formulas of Ramanujan and Gosper. 2006. arXiv:math.NT/0304021 G. Vacca (1926), "Nuova serie per la costante di Eulero, C = 0,577…". Rendiconti, Accademia Nazionale dei Lincei, Roma, Classe di Scienze Fisiche, Matematiche e Naturali (6) 3, 19–20.
James Whitbread Lee Glaisher (1872), "On the history of Euler's constant". Messenger of Mathematics. New Series, vol.1, p. 25-30, JFM 03.0130.01Carl Anton Bretschneider (1837). "Theoriae logarithmi integralis lineamenta nova". Crelle Journal, vol.17, p. 257-285 (submitted 1835)
Lorenzo Mascheroni (1790). "Adnotationes ad calculum integralem Euleri, in quibus nonnulla problemata ab Eulero proposita resolvuntur". Galeati, Ticini.Lorenzo Mascheroni (1792). "Adnotationes ad calculum integralem Euleri. In quibus nonnullae formulae ab Eulero propositae evolvuntur". Galeati, Ticini. Both online at: http://books.google.de/books?id=XkgDAAAAQAAJ （页面存档备份 ，存于互联网档案馆 ）Havil, Julian. Gamma: Exploring Euler's Constant . Princeton University Press. 2003. ISBN 0-691-09983-9 Karatsuba, E. A. Fast evaluation of transcendental functions. Probl. Inf. Transm. 1991, 27 (44): 339–360. E.A. Karatsuba, On the computation of the Euler constant γ, J. of Numerical Algorithms Vol.24, No.1-2, pp. 83–97 (2000)
M. Lerch, Expressions nouvelles de la constante d'Euler. Sitzungsberichte der Königlich Böhmischen Gesellschaft der Wissenschaften 42, 5 p. (1897)
Lagarias, Jeffrey C. Euler's constant: Euler's work and modern developments. arXiv:1303.1856 Bulletin of the American Mathematical Society 50 (4): 527-628 (2013)
A001620
![{\displaystyle \gamma =\lim _{n\rightarrow \infty }\left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7af12010d95e43a1b424a6c3a76c92c2727c1c06)
![{\displaystyle \gamma =\sum _{k=1}^{\infty }\left[{\frac {1}{k}}-\ln \left(1+{\frac {1}{k}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7176d3dc104a5ef96f1039306f330096dde13ccd)
![{\displaystyle \gamma =\lim _{n\rightarrow \infty }\left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]=\int _{1}^{\infty }\left({1 \over \lfloor x\rfloor }-{1 \over x}\right)\,dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0cf185b6d9ad97389a24d868420ea2379a75f60a)
![{\displaystyle \gamma =\lim _{x\to \infty }\left[x-\Gamma \left({\frac {1}{x}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6598511fab1a65e9b06953c415d96ee8633c8f2)
![{\displaystyle \gamma =\lim _{n\to \infty }\left[{\frac {\Gamma ({\frac {1}{n}})\Gamma (n+1)\,n^{1+{\frac {1}{n}}}}{\Gamma (2+n+{\frac {1}{n}})}}-{\frac {n^{2}}{n+1}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8fc69773463c0fce9d5e859a34842d25a0da6487)
![{\displaystyle \gamma ={\frac {3}{2}}-\ln 2-\sum _{m=2}^{\infty }(-1)^{m}\,{\frac {m-1}{m}}[\zeta (m)-1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/87c6c8e2b81d1f232a24da48013a831eb1e9192d)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {2\,n-1}{2\,n}}-\ln \,n+\sum _{k=2}^{n}\left({\frac {1}{k}}-{\frac {\zeta (1-k)}{n^{k}}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2dc45371ac6aceac57f05f5412ed5e33c4a69d2)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {2^{n}}{e^{2^{n}}}}\sum _{m=0}^{\infty }{\frac {2^{m\,n}}{(m+1)!}}\sum _{t=0}^{m}{\frac {1}{t+1}}-n\,\ln 2+O\left({\frac {1}{2^{n}\,e^{2^{n}}}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68a0383cf4feb0933d76bad33d30d5077a663d96)
![{\displaystyle \gamma =[0;1,1,2,1,2,1,4,3,13,5,1,1,8,1,2,4,1,1,40,...]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4bd27ebd870468dbb92cf355e1d9525467520096)
![{\displaystyle =\lim _{n\to \infty }\left[\int _{0}^{n}{\frac {1-(1-x/n)^{n}}{x}}\,dx-\int _{1}^{n}{\frac {1}{x}}\,dx\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ad51b5b4b416e2d6f007f02e8c37f90b3f3ea93)
![{\displaystyle =\lim _{n\to \infty }\left[\int _{0}^{1}{\frac {1-(1-x/n)^{n}}{x}}\,dx-\int _{1}^{n}{\frac {(1-x/n)^{n}}{x}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f0df6992c0f206f2e4328fca6798cb2fa271768)
![{\displaystyle \left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/54edfd70bc01e72b44b1a850ab7efad0852c82ee)
.