# 欧拉-马歇罗尼常数

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

## 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

$\gamma = \sum_{m=2}^{\infty} \frac{(-1)^m\zeta(m)}{m}$
$= \ln \left ( \frac{4}{\pi} \right ) + \sum_{m=1}^{\infty} \frac{(-1)^{m-1} \zeta(m+1)}{2^m (m+1)}$
$\gamma = \frac{3}{2} - \ln 2 - \sum_{m=2}^\infty (-1)^m\,\frac{m-1}{m} [\zeta(m) - 1]$
$= \lim_{n \to \infty} \left [ \frac{2\,n-1}{2\,n} - \ln\,n + \sum_{k=2}^n \left ( \frac{1}{k} - \frac{\zeta(1-k)}{n^k} \right ) \right ]$
$= \lim_{n \to \infty} \left [ \frac{2^n}{e^{2^n}} \sum_{m=0}^\infty \frac{2^{m \,n}}{(m+1)!} \sum_{t=0}^m \frac{1}{t+1} - n\, \ln 2+ O \left ( \frac{1}{2^n\,e^{2^n}} \right ) \right ]$
$\gamma = \lim_{s \to 1^+} \sum_{n=1}^\infty \left ( \frac{1}{n^s} - \frac{1}{s^n} \right ) = \lim_{s \to 1^+} \left ( \zeta(s) - \frac{1}{s - 1} \right )$
$\gamma = \lim_{x \to \infty} \left [ x - \Gamma \left ( \frac{1}{x} \right ) \right ]$
$= \lim_{n \to \infty} \frac{1}{n}\, \sum_{k=1}^n \left ( \left \lceil \frac{n}{k} \right \rceil - \frac{n}{k} \right )$
$\gamma = \sum_{k=1}^n \frac{1}{k} - \ln(n) - \sum_{m=2}^\infty \frac{\zeta (m,n+1)}{m}$

### 积分

$\gamma = - \int_0^\infty { e^{-x} \ln x }\,dx = \int_\infty^ 0 { e^{-x} \ln x }\,dx$[證明 1]
$= - \int_0^1 { \ln\ln \frac{1}{x} }\,dx$
$= \int_0^\infty {\left (\frac{1}{1 - e^{-x}} - \frac{1}{x} \right )e^{ - x} }\,dx$
$= \int_0^\infty { \frac{1}{x} \left ( \frac{1}{1+x} - e^{ - x} \right ) }\,dx$
$\int_0^\infty { e^{-x^2} \ln x }\,dx = -\tfrac14(\gamma+2 \ln 2) \sqrt{\pi}$
$\int_0^\infty { e^{-x} \ln^2 x }\,dx = \gamma^2 + \frac{\pi^2}{6}$
$\gamma = \int_{0}^{1}\int_{0}^{1} \frac{x - 1}{(1 - x\,y)\ln(x\,y)} \, dx\,dy = \sum_{n=1}^\infty \left ( \frac{1}{n} - \ln\frac{n+1}{n} \right ).$
$\sum_{n=1}^\infty \frac{N_1(n) + N_0(n)}{2n(2n+1)} = \gamma$

### 级数展开式

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

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

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

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

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

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

$\gamma$连分数展开式为：

$\gamma = [0; 1, 1, 2, 1, 2, 1, 4, 3, 13, 5, 1, 1, 8, 1, 2, 4, 1, 1, 40, ...]\,$OEIS中的数列A002852）.

### 渐近展开式

$\gamma \sim H_n - \ln \left( n \right) - \frac{1}{{2n}} + \frac{1}{{12n^2 }} - \frac{1}{{120n^4 }} + .$
$\gamma \sim H_n - \ln \left( {n + \frac{1}{2} + \frac{1}{{24n}} - \frac{1}{{48n^3 }} + ...} \right)$
$\gamma \sim H_n - \frac{{\ln \left( n \right) + \ln \left( {n + 1} \right)}}{2} - \frac{1}{{6n\left( {n + 1} \right)}} + \frac{1}{{30n^2 \left( {n + 1} \right)^2 }} - .$

## 已知位数

$\boldsymbol{\gamma}$的已知位数

1734年 5 莱昂哈德·欧拉
1736年 15 莱昂哈德·欧拉
1790年 19 Lorenzo Mascheroni
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

## 相关证明

1. ^ $\gamma = - \int_0^\infty { e^{-x} \ln x }\,dx$的证明：
首先根据放缩法（$\int_k^{k+1} \frac 1x \,dx < \frac 1k < \int_{k-1}^k \frac 1x \,dx$）容易知道，$\int_k^{k-1} \frac 1x \,dx - \frac 1k < \frac 1{k(k-1)}$，以及$\ln n < \sum_{k=1}^n \frac 1k < 1 + \ln n$。因此$\gamma$存在并有限。
$\sum_{k=1}^n \frac{1}{k}$
$= \sum_{k=1}^n \int_0^1 t^{k-1} \,dt$
$= \int_0^1 \sum_{k=1}^n t^{k-1} \,dt$
$= \int_0^1 \frac {1 - t^n}{1 - t} \,dt$
$= \int_n^0 \frac {1 - (1-x/n)^n}{1 - (1-x/n)} \,d(1-x/n)$
$= \int_n^0 \frac {1 - (1-x/n)^n}{x/n} (- \frac 1n) \,dx$
$= \int_0^n \frac {1 - (1-x/n)^n}{x} \,dx,$
$\ln n = \int_1^n \frac 1x \,dx,$
所以$\gamma = \lim_{n \to \infty} [\sum_{k=1}^n \frac{1}{k} - \ln n]$
$= \lim_{n \to \infty} \left[ \int_0^n \frac {1 - (1-x/n)^n}{x} \,dx - \int_1^n \frac 1x \,dx \right]$
$= \lim_{n \to \infty} \left[ \int_0^1 \frac {1 - (1-x/n)^n}{x} \,dx - \int_1^n \frac {(1-x/n)^n}{x} \right]$
$= \int_0^1 \frac {1 - \lim_{n \to \infty}(1-x/n)^n}{x} \,dx - \int_1^{\infty} \frac {\lim_{n \to \infty}(1-x/n)^n}{x}$ （单调收敛定理）
$= \int_0^1 \frac {1 - e^{-x}}{x} \,dx - \int_1^{\infty} \frac {e^{-x}}{x}$
$= \left. (1 - e^{-x}) \ln x \right|_0^1 - \int_0^1 \ln x \,d(1 - e^{-x}) - \left. e^{-x} \ln x \right|_1^{\infty} + \int_1^{\infty} \ln x \,de^{-x}$
$= - \int_0^{\infty} e^{-x} \ln x \,dx.$

## 參考文獻

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