欧拉-马斯刻若尼常数

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，意大利数学家洛伦佐·马斯凯罗尼引入了${\displaystyle \gamma }$ 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]

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

${\displaystyle \gamma =-\int _{0}^{\infty }{e^{-x}\ln x}\,dx=\int _{\infty }^{0}{e^{-x}\ln x}\,dx}$ ^{[证明 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数列A002852）.

渐近展开式

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

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

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

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

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• Fast Algorithms and the FEE Method （页面存档备份，存于互联网档案馆）, E.A. Karatsuba (2005)
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