歐拉-馬斯刻若尼常數

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle \gamma }$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）.

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

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

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