オイラーの定数

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

${\displaystyle \Gamma (z)=\lim _{n\to \infty }{\frac {n^{z}n!}{\displaystyle \prod _{k=0}^{n}{(z+k)}}}}$

${\displaystyle {\begin{aligned}\Psi (z)&={\frac {d}{dz}}\log \Gamma (z)={\frac {\Gamma '(z)}{\Gamma (z)}}\\&=\lim _{n\to \infty }\left(\log {n}-\sum _{k=0}^{n}{\frac {1}{z+k}}\right)\end{aligned}}}$

${\displaystyle {\begin{aligned}\Psi (1)=\Gamma '(1)&=\lim _{n\to \infty }\left(\log {n}-\sum _{k=0}^{n}{\frac {1}{1+k}}\right)\\&=\lim _{n\to \infty }\left(\log {n}-\sum _{k=1}^{n}{\frac {1}{k}}\right)\\&=-\gamma \\\end{aligned}}}$

${\displaystyle {\begin{aligned}\gamma &=-\Gamma '(1)\\&=-\int _{0}^{\infty }e^{-t}\log {t}dt\\&=-\int _{0}^{1}\log \log {\frac {1}{u}}du\qquad (u=e^{-t})\\&=-\int _{-\infty }^{\infty }ue^{u-e^{u}}du\qquad (u=\log {t})\\\end{aligned}}}$

${\displaystyle {\begin{aligned}\log {t}&=\int _{1}^{t}{\frac {1}{s}}ds\\&=\int _{1}^{t}\int _{0}^{\infty }e^{-su}duds\\&=\int _{0}^{\infty }\int _{1}^{t}e^{-su}dsdu\\&=\int _{0}^{\infty }{\frac {e^{-u}-e^{-tu}}{u}}du\\\end{aligned}}}$

${\displaystyle {\begin{aligned}\gamma &=-\int _{0}^{\infty }e^{-t}\log {t}dt\\&=-\int _{0}^{\infty }\int _{0}^{\infty }{\frac {e^{-u}-e^{-tu}}{u}}du\;e^{-t}dt\\&=\int _{0}^{\infty }\int _{0}^{\infty }{\frac {e^{-tu}-e^{-u}}{u}}du\;e^{-t}dt\\&=\int _{0}^{\infty }\left(\int _{0}^{\infty }{\frac {e^{-t(u+1)}}{u}}dt-\int _{0}^{\infty }{\frac {e^{-u}e^{-t}}{u}}dt\right)du\\&=\int _{0}^{\infty }\left({\frac {1}{u(u+1)}}-{\frac {e^{-u}}{u}}\right)du\\\end{aligned}}}$

${\displaystyle {\begin{aligned}\left|\int _{\delta }^{e^{\delta }-1}{\frac {1}{u(u+1)}}du\right|&\leq \left|\int _{\delta }^{e^{\delta }-1}{\frac {1}{\delta }}du\right|\\&=O(\delta )\\\end{aligned}}}$

${\displaystyle {\begin{aligned}\gamma &=\lim _{\delta \to +0}\int _{\delta }^{\infty }\left({\frac {1}{u(u+1)}}-{\frac {e^{-u}}{u}}\right)du\\&=\lim _{\delta \to +0}\int _{\delta }^{\infty }{\frac {1}{u(u+1)}}du-\int _{\delta }^{\infty }{\frac {e^{-s}}{s}}ds\\&=\lim _{\delta \to +0}\int _{e^{\delta }-1}^{\infty }{\frac {1}{u(u+1)}}du-\int _{\delta }^{\infty }{\frac {e^{-s}}{s}}ds\\&=\lim _{\delta \to +0}\int _{\delta }^{\infty }{\frac {e^{t}}{(e^{t}-1)e^{t}}}dt-\int _{\delta }^{\infty }{\frac {e^{-s}}{s}}ds\qquad (u=e^{t}-1)\\&=\lim _{\delta \to +0}\int _{\delta }^{\infty }{\frac {e^{-t}}{1-e^{-t}}}dt-\int _{\delta }^{\infty }{\frac {e^{-s}}{s}}ds\\&=\int _{0}^{\infty }\left({\frac {e^{-t}}{1-e^{-t}}}-{\frac {e^{-t}}{t}}\right)dt\end{aligned}}}$

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

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

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

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

${\displaystyle \gamma =\displaystyle \sum _{n=2}^{\infty }{\frac {(-1)^{n}\zeta (n)}{m^{n-1}n}}-m\ln \Gamma \left({\frac {m+1}{m}}\right)}$，ここで ${\displaystyle m\in \mathbb {R} \setminus \left\{0,-1,-{\dfrac {1}{k+1}}\ (k\in \mathbb {Z} \setminus \{-1\})\right\}}$ である。

${\displaystyle \gamma =\lim _{m\rightarrow 0}\left(\displaystyle \sum _{n=2}^{\infty }{\frac {(-1)^{n}\zeta (n)}{m^{n-1}n}}-m\ln \Gamma \left({\frac {m+1}{m}}\right)\right)}$^[3]

${\displaystyle \gamma =\lim _{m\rightarrow -1}\left(\displaystyle \sum _{n=2}^{\infty }{\frac {(-1)^{n}\zeta (n)}{m^{n-1}n}}-m\ln \Gamma \left({\frac {m+1}{m}}\right)\right)}$^[4]

${\displaystyle \gamma =\lim _{m\rightarrow (-1)/(k+1)}\left(\displaystyle \sum _{n=2}^{\infty }{\frac {(-1)^{n}\zeta (n)}{m^{n-1}n}}-m\ln \Gamma \left({\frac {m+1}{m}}\right)\right)}$，ここで ${\displaystyle k\in \mathbb {Z} \setminus \{-1\}}$ である。^[5]

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

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