歐拉-馬斯刻若尼常數

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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
識別
符號	${\displaystyle \gamma }$
位數數列編號	A001620
性質
定義	${\displaystyle \gamma =\lim _{n\rightarrow \infty }\left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]}$ ${\displaystyle \gamma =\int _{1}^{\infty }\left({1 \over \lfloor x\rfloor }-{1 \over x}\right)\,dx}$
連分數	[0; 1, 1, 2, 1, 2, 1, 4, 3, 13, 5, 1, 1, 8, 1, 2, 4, 1, 1, 40, ...]
表示方式
值	${\displaystyle \gamma \approx }$0.57721566490153...
無窮級數	${\displaystyle \gamma =\sum _{k=1}^{\infty }\left[{\frac {1}{k}}-\ln \left(1+{\frac {1}{k}}\right)\right]}$
二进制	0.100100111100010001100111…
十进制	0.577215664901532860606512…
十六进制	0.93C467E37DB0C7A4D1BE3F81…
查 论 编

歐拉-馬斯刻若尼常數是一个数学常数，定义为调和级数与自然对数的差值：

${\displaystyle \gamma =\lim _{n\rightarrow \infty }\left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]=\int _{1}^{\infty }\left({1 \over \lfloor x\rfloor }-{1 \over x}\right)\,dx}$

它的近似值为${\displaystyle \gamma \approx 0.577215664901532860606512090082402431042159335}$^[1]，

歐拉-馬斯刻若尼常數主要应用于数论。

历史

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${\displaystyle C}$作为它的符号，并计算出了它的前6位小数。1761年他又将该值计算到了16位小数。1790年，意大利数学家洛倫佐·馬斯凱羅尼引入了${\displaystyle \gamma }$作为这个常数的符号，并将该常数计算到小数点后32位。但后来的计算显示他在第20位的时候出现了错误。

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]

性质

与伽玛函数的关系

${\displaystyle \ -\gamma =\Gamma '(1)=\Psi (1)}$。

${\displaystyle \gamma =\lim _{x\to \infty }\left[x-\Gamma \left({\frac {1}{x}}\right)\right]}$。

${\displaystyle \gamma =\lim _{n\to \infty }\left[{\frac {\Gamma ({\frac {1}{n}})\Gamma (n+1)\,n^{1+{\frac {1}{n}}}}{\Gamma (2+n+{\frac {1}{n}})}}-{\frac {n^{2}}{n+1}}\right]}$。

与ζ函数的关系

${\displaystyle \gamma =\sum _{m=2}^{\infty }{\frac {(-1)^{m}\zeta (m)}{m}}}$

${\displaystyle =\ln \left({\frac {4}{\pi }}\right)+\sum _{m=1}^{\infty }{\frac {(-1)^{m-1}\zeta (m+1)}{2^{m}(m+1)}}}$。

${\displaystyle \lim _{\varepsilon \to 0}{\frac {\zeta (1+\varepsilon )+\zeta (1-\varepsilon )}{2}}=\gamma }$

${\displaystyle \gamma ={\frac {3}{2}}-\ln 2-\sum _{m=2}^{\infty }(-1)^{m}\,{\frac {m-1}{m}}[\zeta (m)-1]}$

${\displaystyle =\lim _{n\to \infty }\left[{\frac {2\,n-1}{2\,n}}-\ln \,n+\sum _{k=2}^{n}\left({\frac {1}{k}}-{\frac {\zeta (1-k)}{n^{k}}}\right)\right]}$。

${\displaystyle =\lim _{n\to \infty }\left[{\frac {2^{n}}{e^{2^{n}}}}\sum _{m=0}^{\infty }{\frac {2^{m\,n}}{(m+1)!}}\sum _{t=0}^{m}{\frac {1}{t+1}}-n\,\ln 2+O\left({\frac {1}{2^{n}\,e^{2^{n}}}}\right)\right]}$

${\displaystyle \gamma =\lim _{s\to 1^{+}}\sum _{n=1}^{\infty }\left({\frac {1}{n^{s}}}-{\frac {1}{s^{n}}}\right)=\lim _{s\to 1^{+}}\left(\zeta (s)-{\frac {1}{s-1}}\right)}$

${\displaystyle \gamma =\lim _{x\to \infty }\left[x-\Gamma \left({\frac {1}{x}}\right)\right]}$

${\displaystyle =\lim _{n\to \infty }{\frac {1}{n}}\,\sum _{k=1}^{n}\left(\left\lceil {\frac {n}{k}}\right\rceil -{\frac {n}{k}}\right)}$。

${\displaystyle \gamma =\sum _{k=1}^{n}{\frac {1}{k}}-\ln(n)-\sum _{m=2}^{\infty }{\frac {\zeta (m,n+1)}{m}}}$

积分

${\displaystyle \gamma =-\int _{0}^{\infty }{e^{-x}\ln x}\,dx=\int _{\infty }^{0}{e^{-x}\ln x}\,dx}$^{[證明 1]}${\displaystyle =-\int _{0}^{1}{\ln \ln {\frac {1}{x}}}\,dx}$

${\displaystyle =\int _{0}^{\infty }{\left({\frac {1}{1-e^{-x}}}-{\frac {1}{x}}\right)e^{-x}}\,dx}$

${\displaystyle =\int _{0}^{\infty }{{\frac {1}{x}}\left({\frac {1}{1+x}}-e^{-x}\right)}\,dx}$

${\displaystyle \int _{0}^{\infty }{e^{-x^{2}}\ln x}\,dx=-{\tfrac {1}{4}}(\gamma +2\ln 2){\sqrt {\pi }}}$

${\displaystyle \int _{0}^{\infty }{e^{-x}\ln ^{2}x}\,dx=\gamma ^{2}+{\frac {\pi ^{2}}{6}}}$。

${\displaystyle \gamma =\int _{0}^{1}\int _{0}^{1}{\frac {x-1}{(1-x\,y)\ln(x\,y)}}\,dx\,dy=\sum _{n=1}^{\infty }\left({\frac {1}{n}}-\ln {\frac {n+1}{n}}\right)}$

${\displaystyle \sum _{n=1}^{\infty }{\frac {N_{1}(n)+N_{0}(n)}{2n(2n+1)}}=\gamma }$

级数展开式

${\displaystyle \gamma =\sum _{k=1}^{\infty }\left[{\frac {1}{k}}-\ln \left(1+{\frac {1}{k}}\right)\right]}$

${\displaystyle \gamma =1-\sum _{k=2}^{\infty }(-1)^{k}{\frac {\lfloor \log _{2}k\rfloor }{k+1}}}$.

${\displaystyle \gamma =\sum _{k=2}^{\infty }(-1)^{k}{\frac {\left\lfloor \log _{2}k\right\rfloor }{k}}={\tfrac {1}{2}}-{\tfrac {1}{3}}+2\left({\tfrac {1}{4}}-{\tfrac {1}{5}}+{\tfrac {1}{6}}-{\tfrac {1}{7}}\right)+3\left({\tfrac {1}{8}}-\dots -{\tfrac {1}{15}}\right)+\dots }$

${\displaystyle \gamma +\zeta (2)=\sum _{k=1}^{\infty }{\frac {1}{k\lfloor {\sqrt {k}}\rfloor ^{2}}}=1+{\tfrac {1}{2}}+{\tfrac {1}{3}}+{\tfrac {1}{4}}\left({\tfrac {1}{4}}+\dots +{\tfrac {1}{8}}\right)+{\tfrac {1}{9}}\left({\tfrac {1}{9}}+\dots +{\tfrac {1}{15}}\right)+\dots }$

${\displaystyle \gamma =\sum _{k=2}^{\infty }{\frac {k-\lfloor {\sqrt {k}}\rfloor ^{2}}{k^{2}\lfloor {\sqrt {k}}\rfloor ^{2}}}={\tfrac {1}{2^{2}}}+{\tfrac {2}{3^{2}}}+{\tfrac {1}{2^{2}}}\left({\tfrac {1}{5^{2}}}+{\tfrac {2}{6^{2}}}+{\tfrac {3}{7^{2}}}+{\tfrac {4}{8^{2}}}\right)+{\tfrac {1}{3^{2}}}\left({\tfrac {1}{10^{2}}}+\dots +{\tfrac {6}{15^{2}}}\right)+\dots }$

${\displaystyle \gamma =\int _{0}^{1}{\frac {1}{1+x}}\sum _{n=1}^{\infty }x^{2^{n}-1}\,dx}$

${\displaystyle \gamma }$的连分数展开式为：

${\displaystyle \gamma =[0;1,1,2,1,2,1,4,3,13,5,1,1,8,1,2,4,1,1,40,...]\,}$ （OEIS數列A002852）.

渐近展开式

${\displaystyle \gamma \approx H_{n}-\ln \left(n\right)-{\frac {1}{2n}}+{\frac {1}{12n^{2}}}-{\frac {1}{120n^{4}}}+...}$

${\displaystyle \gamma \approx H_{n}-\ln \left({n+{\frac {1}{2}}+{\frac {1}{24n}}-{\frac {1}{48n^{3}}}+...}\right)}$

${\displaystyle \gamma \approx H_{n}-{\frac {\ln \left(n\right)+\ln \left({n+1}\right)}{2}}-{\frac {1}{6n\left({n+1}\right)}}+{\frac {1}{30n^{2}\left({n+1}\right)^{2}}}-...}$

已知位数

${\displaystyle {\boldsymbol {\gamma }}}$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	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

相关证明

1. ${\displaystyle \gamma =-\int _{0}^{\infty }{e^{-x}\ln x}\,dx}$的证明：
   首先根据放缩法（${\displaystyle \int _{k}^{k+1}{\frac {1}{x}}\,dx<{\frac {1}{k}}<\int _{k-1}^{k}{\frac {1}{x}}\,dx}$）容易知道，${\displaystyle \int _{k}^{k-1}{\frac {1}{x}}\,dx-{\frac {1}{k}}<{\frac {1}{k(k-1)}}}$，以及${\displaystyle \ln n<\sum _{k=1}^{n}{\frac {1}{k}}<1+\ln n}$。因此${\displaystyle \gamma }$存在并有限。
   ${\displaystyle \sum _{k=1}^{n}{\frac {1}{k}}}$
   ${\displaystyle =\sum _{k=1}^{n}\int _{0}^{1}t^{k-1}\,dt}$
   ${\displaystyle =\int _{0}^{1}\sum _{k=1}^{n}t^{k-1}\,dt}$
   ${\displaystyle =\int _{0}^{1}{\frac {1-t^{n}}{1-t}}\,dt}$
   ${\displaystyle =\int _{n}^{0}{\frac {1-\left(1-{\frac {x}{n}}\right)^{n}}{1-\left(1-{\frac {x}{n}}\right)}}d\left(1-{\tfrac {x}{n}}\right)}$
   ${\displaystyle =\int _{n}^{0}{\frac {1-\left(1-{\frac {x}{n}}\right)^{n}}{\frac {x}{n}}}\left(-{\frac {1}{n}}\right)dx}$
   ${\displaystyle =\int _{0}^{n}{\frac {1-\left(1-{\frac {x}{n}}\right)^{n}}{x}}dx}$
   而${\displaystyle \ln n=\int _{1}^{n}{\frac {1}{x}}\,dx,}$
   所以${\displaystyle \gamma =\lim _{n\to \infty }\left(\sum _{k=1}^{n}{\frac {1}{k}}-\ln n\right)}$
   ${\displaystyle =\lim _{n\to \infty }\left[\int _{0}^{n}{\frac {1-(1-x/n)^{n}}{x}}\,dx-\int _{1}^{n}{\frac {1}{x}}\,dx\right]}$
   ${\displaystyle =\lim _{n\to \infty }\left[\int _{0}^{1}{\frac {1-(1-x/n)^{n}}{x}}\,dx-\int _{1}^{n}{\frac {(1-x/n)^{n}}{x}}\right]}$
   ${\displaystyle =\int _{0}^{1}{\frac {1-\lim _{n\to \infty }(1-x/n)^{n}}{x}}\,dx-\int _{1}^{\infty }{\frac {\lim _{n\to \infty }(1-x/n)^{n}}{x}}}$ （单调收敛定理）
   ${\displaystyle =\int _{0}^{1}{\frac {1-e^{-x}}{x}}\,dx-\int _{1}^{\infty }{\frac {e^{-x}}{x}}}$
   ${\displaystyle =\left.(1-e^{-x})\ln x\right|_{0}^{1}-\int _{0}^{1}\ln x\,d(1-e^{-x})-\left.e^{-x}\ln x\right|_{1}^{\infty }+\int _{1}^{\infty }\ln x\,de^{-x}}$
   ${\displaystyle =-\int _{0}^{\infty }e^{-x}\ln x\,dx.}$

前面的放缩法主要是证明了

${\displaystyle \left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]}$ 是单调递减并下有界限（0），所有极限存在。放缩法的结论需要使用ln(1+x)和ln(1-x)的泰勒级数展开进行证明。

參考文獻

1. A001620 oeis.org [2014-7-17]
2. Havil 2003 p 97.

1. 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）.  Derives γ as sums over Riemann zeta functions.
2. 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）"
3. Gourdon, Xavier, and Sebah, P. (2004) "The Euler constant: γ. （页面存档备份，存于互联网档案馆）"
4. Donald Knuth (1997) The Art of Computer Programming, Vol. 1, 3rd ed. Addison-Wesley. ISBN 978-0-201-89683-1
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11. Sondow, Jonathan (2005) "New Vacca-type rational series for Euler's constant and its 'alternating' analog ln 4/π."
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19. Karatsuba, E. A. Fast evaluation of transcendental functions. Probl. Inf. Transm. 1991, 27 (44): 339–360. 
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22. 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)

外部連結

• 埃里克·韦斯坦因. Euler-Mascheroni constant. MathWorld. 
• Krämer, Stefan "Euler's Constant γ=0.577... Its Mathematics and History."
• Jonathan Sondow.
• Fast Algorithms and the FEE Method （页面存档备份，存于互联网档案馆）, E.A. Karatsuba (2005)
• Further formulae which make use of the constant: Gourdon and Sebah (2004). （页面存档备份，存于互联网档案馆）