欧拉-马斯刻若尼常数

$\gamma\approx$0.57721566490153...

$\gamma = \lim_{n \rightarrow \infty } \left[ \left( \sum_{k=1}^n \frac{1}{k} \right) - \ln(n) \right]$ $\gamma = \int_1^\infty\left({1\over\lfloor x\rfloor} - {1\over x}\right)\,dx$

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

$$\gamma = \lim_{n \rightarrow \infty } \left[ \left( \sum_{k=1}^n \frac{1}{k} \right) - \ln(n) \right]=\int_1^\infty\left({1\over\lfloor x\rfloor} - {1\over x}\right)\,dx$$

它的近似值为$\gamma \approx 0.57721 56649 01532 86060 65120 90082 40243 10421 59335$。

欧拉-马斯刻若尼常数主要应用于数论。

## 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$C$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，意大利数学家洛伦佐·马斯凯罗尼引入了$\gamma$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。

## 性质

### 与伽玛函数的关系

$\ -\gamma = \Gamma'(1) = \Psi(1)$。

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

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

### 与ζ函数的关系

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

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

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

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

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

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

$$\gamma = \lim_{s \to 1^+} \sum_{n=1}^\infty \left ( \frac{1}{n^s} - \frac{1}{s^n} \right ) = \lim_{s \to 1^+} \left ( \zeta(s) - \frac{1}{s - 1} \right )$$

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

$= \lim_{n \to \infty} \frac{1}{n}\, \sum_{k=1}^n \left ( \left \lceil \frac{n}{k} \right \rceil - \frac{n}{k} \right )$。

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

### 积分

$\gamma = - \int_0^\infty { e^{-x} \ln x }\,dx = \int_\infty^ 0 { e^{-x} \ln x }\,dx$

$\gamma = - \int_0^\infty { e^{-x} \ln x }\,dx$

$\int_k^{k+1} \frac 1x \,dx < \frac 1k < \int_{k-1}^k \frac 1x \,dx$

$\int_{k-1}^k \frac 1x \,dx - \frac 1k < \frac 1{k(k-1)}$

$\ln n < \sum_{k=1}^n \frac 1k < 1 + \ln n$

$\gamma$

$\sum_{k=1}^n \frac{1}{k}$

$= \sum_{k=1}^n \int_0^1 t^{k-1} \,dt$

$= \int_0^1 \sum_{k=1}^n t^{k-1} \,dt$

$= \int_0^1 \frac {1 - t^n}{1 - t} \,dt$

$= \int_n^0 \frac {1 - \left(1-\frac{x}{n}\right)^n}{1 - \left(1-\frac{x}{n}\right)} d\left(1-\tfrac{x}{n}\right)$

$= \int_n^0 \frac {1 - \left(1-\frac{x}{n}\right)^n}{\frac{x}{n}} \left(- \frac 1n\right) dx$

$= \int_0^n \frac {1 - \left(1-\frac{x}{n}\right)^n}{x} dx$

$\ln n = \int_1^n \frac 1x \,dx,$

$\gamma = \lim_{n \to \infty} \left(\sum_{k=1}^n \frac{1}{k} - \ln n\right)$

$= \lim_{n \to \infty} \left[ \int_0^n \frac {1 - (1-x/n)^n}{x} \,dx - \int_1^n \frac 1x \,dx \right]$

```
       <math>= \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]</math><br />
```

$$= \int_0^1 \frac {1 - \lim_{n \to \infty}(1-x/n)^n}{x} \,dx - \int_1^{\infty} \frac {\lim_{n \to \infty}(1-x/n)^n}{x}$$

$= \int_0^1 \frac {1 - e^{-x}}{x} \,dx - \int_1^{\infty} \frac {e^{-x}}{x}$

$= \left. (1 - e^{-x}) \ln x \right|_0^1 - \int_0^1 \ln x \,d(1 - e^{-x}) - \left. e^{-x} \ln x \right|_1^{\infty} + \int_1^{\infty} \ln x \,de^{-x}$

$$= - \int_0^{\infty} e^{-x} \ln x \,dx.$$

$= - \int_0^1 { \ln\ln \frac{1}{x} }\,dx$

$$= \int_0^\infty {\left (\frac{1}{1 - e^{-x}} - \frac{1}{x} \right )e^{ - x} }\,dx$$

$$= \int_0^\infty { \frac{1}{x} \left ( \frac{1}{1+x} - e^{ - x} \right ) }\,dx$$

$$\int_0^\infty { e^{-x^2} \ln x }\,dx = -\tfrac14(\gamma+2 \ln 2) \sqrt{\pi}$$

$\int_0^\infty { e^{-x} \ln^2 x }\,dx = \gamma^2 + \frac{\pi^2}{6}$。

$$\gamma = \int_{0}^{1}\int_{0}^{1} \frac{x - 1}{(1 - x\,y)\ln(x\,y)} \, dx\,dy = \sum_{n=1}^\infty \left ( \frac{1}{n} - \ln\frac{n+1}{n} \right ) <pre>$$

$$\sum_{n=1}^\infty \frac{N_1(n) + N_0(n)}{2n(2n+1)} = \gamma$$

### 级数展开式

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

$\gamma = 1 - \sum_{k=2}^{\infty}(-1)^k\frac{\lfloor\log_2 k\rfloor}{k+1}$.

$$\gamma = \sum_{k=2}^\infty (-1)^k \frac{ \left \lfloor \log_2 k \right \rfloor}{k} <pre> = \tfrac12-\tfrac13 + 2\left(\tfrac14 - \tfrac15 + \tfrac16 - \tfrac17\right) + 3\left(\tfrac18 - \dots - \tfrac1{15}\right) + \dots$$

$$\gamma + \zeta(2) = \sum_{k=1}^{\infty} \frac1{k\lfloor\sqrt{k}\rfloor^2} <pre> = 1 + \tfrac12 + \tfrac13 + \tfrac14\left(\tfrac14 + \dots + \tfrac18\right) + \tfrac19\left(\tfrac19 + \dots + \tfrac1{15}\right) + \dots$$

$$\gamma = \sum_{k=2}^{\infty} \frac{k - \lfloor\sqrt{k}\rfloor^2}{k^2\lfloor\sqrt{k}\rfloor^2} <pre> = \tfrac1{2^2} + \tfrac2{3^2} + \tfrac1{2^2}\left(\tfrac1{5^2} + \tfrac2{6^2} + \tfrac3{7^2} + \tfrac4{8^2}\right) + \tfrac1{3^2}\left(\tfrac1{10^2} + \dots + \tfrac6{15^2}\right) + \dots$$

$$\gamma = \int_0^1 \frac{1}{1+x} \sum_{n=1}^\infty x^{2^n-1} \, dx$$

$\gamma$的连分数展开式为：

$\gamma = [0; 1, 1, 2, 1, 2, 1, 4, 3, 13, 5, 1, 1, 8, 1, 2, 4, 1, 1, 40, ...]\,$ .

### 渐近展开式

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

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

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

## 已知位数

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

## 相关证明

前面的放缩法主要证明了

$\left[ \left( \sum_{k=1}^n \frac{1}{k} \right) - \ln(n) \right]$ 单调递减并下有界限（0），所以极限存在。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。

## 参考文献

- Derives γ as sums over Riemann zeta functions.
- Gourdon, Xavier, and Sebah, P. (2002) " Collection of formulas for Euler's constant, γ. "
- Gourdon, Xavier, and Sebah, P. (2004) " The Euler constant: γ. "
- Donald Knuth (1997) The Art of Computer Programming, Vol. 1, 3rd ed. Addison-Wesley. ISBN 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 71: 219-220.
- Sondow, 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 (2003a) " Criteria for irrationality of Euler's constant," Proceedings of the American Mathematical Society 131: 3335-3344.
- Sondow, Jonathan (2005) " Double integrals for Euler's constant and ln 4/π and an analog of Hadjicostas's formula," American Mathematical Monthly 112: 61-65.
- Sondow, Jonathan (2005) "New Vacca-type rational series for Euler's constant and its 'alternating' analog ln 4/π."
- Ramanujan Journal 12: 225-244.
- 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 03.0130.01
- Carl 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
- 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)
- , Bulletin of the American Mathematical Society 50 (4): 527-628 (2013)

## 外部链接

- Krämer, Stefan " Euler's Constant γ=0.577... Its Mathematics and History."
- Jonathan Sondow.
- Fast Algorithms and the FEE Method , E.A. Karatsuba (2005)
- Further formulae which make use of the constant: Gourdon and Sebah (2004).

## 基本资料

- nav：no
- 基数词：no
- 说明：蓝色区域的面积收敛到欧拉常数
- 货币符号：$\gamma$
- 面值：$\gamma\approx$0.57721566490153...
- OEIS编号：A001620
- 定义：$\gamma = \lim_{n \rightarrow \infty } \left[ \left( \sum_{k=1}^n \frac{1}{k} \right) - \ln(n) \right]$、$\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, ...
- 系列：$\gamma = \sum_{k=1}^\infty \left[ \frac{1}{k} - \ln \left( 1 + \frac{1}{k} \right) \right]$