提示 ：此條目的主題不是
尤拉數 。
歐拉-馬斯刻若尼常數 是一個數學常數 ，定義為調和級數 與自然對數 的差值：
歐拉-馬斯刻若尼常數 歐拉-馬斯刻若尼常數 藍色區域的面積收斂到歐拉常數
符號
γ
{\displaystyle \gamma }
位數 數列編號 A001620 定義
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}
連分數 [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 }
無窮級數
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]}
二進制 0.10010011 1100 0100 0110 0111 … 十進制 0.57721566 4901 5328 6060 6512 … 十六進制 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}
它的近似值為
γ
≈
0.577215664901532860606512090082402431042159335
{\displaystyle \gamma \approx 0.577215664901532860606512090082402431042159335}
[ 1]
歐拉-馬斯刻若尼常數主要應用於數論 。
該常數最先由瑞士 數學家萊昂哈德·歐拉 在1735年發表的文章De Progressionibus harmonicus observationes 中定義。歐拉曾經使用
C
{\displaystyle C}
意大利 數學家洛倫佐·馬斯凱羅尼 引入了
γ
{\displaystyle \gamma }
目前尚不知道該常數是否為有理數 ，但是分析表明如果它是一個有理數，那麼它的分母位數將超過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}
[ 證明 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 }
連分數 展開式為：
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 數列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 }}}
的已知位數
日期
位數
計算者
1734年
5
萊昂哈德·歐拉
1736年
15
萊昂哈德·歐拉
1790年
19
洛倫佐·馬斯凱羅尼
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
約翰·柯西·亞當斯
1962年
1,271
高德納
1962年
3,566
D.W. Sweeney
1977年
20,700
Richard P. Brent
1980年
30,100
Richard P. Brent 和埃德溫·麥克米倫
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}}
=
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∫
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∞
e
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x
ln
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d
x
.
{\displaystyle =-\int _{0}^{\infty }e^{-x}\ln x\,dx.}
前面的放縮法主要是證明了
[
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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 原始內容 (PDF) 存檔於2006-09-25）. Gourdon, Xavier, and Sebah, P. (2002) "Collection of formulas for Euler's constant, γ. （頁面存檔備份 ，存於互聯網檔案館 ）"
Gourdon, Xavier, and Sebah, P. (2004) "The Euler constant: γ. （頁面存檔備份 ，存於互聯網檔案館 ）"
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 （頁面存檔備份 ，存於互聯網檔案館 ）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)
.