歐拉-馬斯刻若尼常數

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：

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

提示：此條目的主題不是尤拉數。

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	${\displaystyle \gamma =\sum _{k=1}^{\infty }\left[{\frac {1}{k}}-\ln \left(1+{\frac {1}{k}}\right)\right]}$
二進制	0.100100111100010001100111…
十進制	0.577215664901532860606512…
十六進制	0.93C467E37DB0C7A4D1BE3F81…
閱 論 編

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

歐拉-馬斯刻若尼常數主要應用於數論。

歷史

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${\displaystyle C}$作為它的符號，並計算出了它的前6位小數。1761年他又將該值計算到了16位小數。1790年，意大利數學家洛倫佐·馬斯凱羅尼引入了${\displaystyle \gamma }$作為這個常數的符號，並將該常數計算到小數點後32位。但後來的計算顯示他在第20位的時候出現了錯誤。

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]

性質

與伽瑪函數的關係

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

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

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

與ζ函數的關係

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

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

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

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

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

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

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

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

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

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

積分

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

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

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

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

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

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

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

級數展開式

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

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

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

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

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

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

${\displaystyle \gamma }$的連分數展開式為：

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

漸近展開式

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

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

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

已知位數

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

相關證明

1. ${\displaystyle \gamma =-\int _{0}^{\infty }{e^{-x}\ln x}\,dx}$的證明：
   首先根據放縮法（${\displaystyle \int _{k}^{k+1}{\frac {1}{x}}\,dx<{\frac {1}{k}}<\int _{k-1}^{k}{\frac {1}{x}}\,dx}$）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${\displaystyle \gamma }$存在並有限。
   ${\displaystyle \sum _{k=1}^{n}{\frac {1}{k}}}$
   ${\displaystyle =\sum _{k=1}^{n}\int _{0}^{1}t^{k-1}\,dt}$
   ${\displaystyle =\int _{0}^{1}\sum _{k=1}^{n}t^{k-1}\,dt}$
   ${\displaystyle =\int _{0}^{1}{\frac {1-t^{n}}{1-t}}\,dt}$
   ${\displaystyle =\int _{n}^{0}{\frac {1-\left(1-{\frac {x}{n}}\right)^{n}}{1-\left(1-{\frac {x}{n}}\right)}}d\left(1-{\tfrac {x}{n}}\right)}$
   ${\displaystyle =\int _{n}^{0}{\frac {1-\left(1-{\frac {x}{n}}\right)^{n}}{\frac {x}{n}}}\left(-{\frac {1}{n}}\right)dx}$
   ${\displaystyle =\int _{0}^{n}{\frac {1-\left(1-{\frac {x}{n}}\right)^{n}}{x}}dx}$
   而${\displaystyle \ln n=\int _{1}^{n}{\frac {1}{x}}\,dx,}$
   所以${\displaystyle \gamma =\lim _{n\to \infty }\left(\sum _{k=1}^{n}{\frac {1}{k}}-\ln n\right)}$
   ${\displaystyle =\lim _{n\to \infty }\left[\int _{0}^{n}{\frac {1-(1-x/n)^{n}}{x}}\,dx-\int _{1}^{n}{\frac {1}{x}}\,dx\right]}$
   ${\displaystyle =\lim _{n\to \infty }\left[\int _{0}^{1}{\frac {1-(1-x/n)^{n}}{x}}\,dx-\int _{1}^{n}{\frac {(1-x/n)^{n}}{x}}\right]}$
   ${\displaystyle =\int _{0}^{1}{\frac {1-\lim _{n\to \infty }(1-x/n)^{n}}{x}}\,dx-\int _{1}^{\infty }{\frac {\lim _{n\to \infty }(1-x/n)^{n}}{x}}}$ （單調收斂定理）
   ${\displaystyle =\int _{0}^{1}{\frac {1-e^{-x}}{x}}\,dx-\int _{1}^{\infty }{\frac {e^{-x}}{x}}}$
   ${\displaystyle =\left.(1-e^{-x})\ln x\right|_{0}^{1}-\int _{0}^{1}\ln x\,d(1-e^{-x})-\left.e^{-x}\ln x\right|_{1}^{\infty }+\int _{1}^{\infty }\ln x\,de^{-x}}$
   ${\displaystyle =-\int _{0}^{\infty }e^{-x}\ln x\,dx.}$

前面的放縮法主要是證明了

${\displaystyle \left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]}$ 是單調遞減並下有界限（0），所有極限存在。放縮法的結論需要使用ln(1+x)和ln(1-x)的泰勒級數展開進行證明。

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. （原始內容 (PDF)存檔於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. （頁面存檔備份，存於互聯網檔案館）"
3. Gourdon, Xavier, and Sebah, P. (2004) "The Euler constant: γ. （頁面存檔備份，存於互聯網檔案館）"
4. Donald Knuth (1997) The Art of Computer Programming, Vol. 1, 3rd ed. Addison-Wesley. ISBN 978-0-201-89683-1
5. Krämer, Stefan (2005) Die Eulersche Konstante γ und verwandte Zahlen. Diplomarbeit, Universität Göttingen.
6. Sondow, Jonathan (1998) "An antisymmetric formula for Euler's constant," Mathematics Magazine 71: 219-220.
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10. 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.
11. Sondow, Jonathan (2005) "New Vacca-type rational series for Euler's constant and its 'alternating' analog ln 4/π."
12. Sondow, Jonathan; Zudilin, Wadim. Euler's constant, q-logarithms, and formulas of Ramanujan and Gosper. 2006. arXiv:math.NT/0304021 . Ramanujan Journal 12: 225-244.
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外部連結

• 埃里克·韋斯坦因. Euler-Mascheroni constant. MathWorld.
• Krämer, Stefan "Euler's Constant γ=0.577... Its Mathematics and History."
• Jonathan Sondow.
• Fast Algorithms and the FEE Method （頁面存檔備份，存於互聯網檔案館）, E.A. Karatsuba (2005)
• Further formulae which make use of the constant: Gourdon and Sebah (2004). （頁面存檔備份，存於互聯網檔案館）