Eulerin–Mascheronin vakio

Eulerin–Mascheronin vakion määritteli ensimmäisenä sveitsiläinen matemaatikko Leonhard Euler paperissaan De Progressionibus harmonicis observationes vuonna 1735. Euler käytti vakiolle merkintöjä C ja O ja laski sen arvon viiden desimaalin tarkkuudella. Vuonna 1781 hän oli laskenut vakion 15 desimaalin tarkkuudella.

Vuonna 1790 italialainen matemaatikko Lorenzo Mascheroni esitti vakiolle merkinnän A ja esitti sen arvon 31 desimaalin tarkkuudella, tosin desimaalit 20:nnestä eteenpäin osoittautuivat virheellisiksi. Mascheroni ei koskaan käyttänyt merkintää γ. Vakiolla on yhteys gammafunktioon.^selvennä

Joulukuussa 2006 opiskelija Alexander J. Yee laski Eulerin–Mascheronin vakion 116 580 040 desimaalin tarkkuudella.[1]

${\displaystyle {\begin{aligned}\gamma &=-\int _{0}^{\infty }{e^{-x}\ln x}\,dx=-4\int _{0}^{\infty }{e^{-x^{2}}x\ln x}\,dx\\&=-\int _{0}^{1}\ln \ln \left({\frac {1}{x}}\right)dx\\&=\int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {1}{xe^{x}}}\right)dx=\int _{0}^{1}\left({\frac {1}{\ln x}}+{\frac {1}{1-x}}\right)dx\\&=\int _{0}^{\infty }\left({\frac {1}{1+x^{k}}}-e^{-x}\right){\frac {dx}{x}},\quad k>0\\&=\int _{0}^{\infty }\left({\frac {1}{kx+1}}-e^{-kx}\right){\frac {\mathrm {d} x}{x}},\quad k>0\\&=\int _{0}^{\infty }{\frac {\ln(1+x)}{\ln ^{2}x+\pi ^{2}}}\cdot {\frac {dx}{x^{2}}}\\&={\frac {1}{2}}+2\int _{0}^{\infty }{\frac {\sin(\arctan x)}{(e^{2\pi x}-1){\sqrt {1+x^{2}}}}}\mathrm {d} x=\int _{0}^{1}H_{x}dx=-\int _{0}^{\infty }\left({\frac {\ln x}{e^{x}}}\right)dx\end{aligned}}.}$

Hieman monimutkaisempia integraaleja:

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

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

Kaksoisintegraali gammalle on

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

Vertaa

${\displaystyle \ln \left({\frac {4}{\pi }}\right)=\int _{0}^{1}\int _{0}^{1}{\frac {x-1}{(1+x\,y)\ln(x\,y)}}\,dx\,dy=\sum _{n=1}^{\infty }(-1)^{n-1}\left({\frac {1}{n}}-\ln {\frac {n+1}{n}}\right).}$

Catalan löysi integraalin

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

Euler todisti kaavan

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

Toinen kaava on

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

Vacca on todistanut kaavat

${\displaystyle {\gamma =\sum _{k=2}^{\infty }(-1)^{k}{\frac {\left\lfloor \log _{2}k\right\rfloor }{k}}={\frac {1}{2}}-{\frac {1}{3}}+2\left({\frac {1}{4}}-{\frac {1}{5}}+{\frac {1}{6}}-{\frac {1}{7}}\right)+3\left({\frac {1}{8}}-{\frac {1}{9}}+{\frac {1}{10}}-{\frac {1}{11}}+\dots -{\frac {1}{15}}\right)+\dots }}$

${\displaystyle {\gamma +\zeta (2)=\sum _{k=2}^{\infty }\left({\frac {1}{\lfloor {\sqrt {k}}\rfloor ^{2}}}-{\frac {1}{k}}\right)=\sum _{k=2}^{\infty }{\frac {k-\lfloor {\sqrt {k}}\rfloor ^{2}}{k\lfloor {\sqrt {k}}\rfloor ^{2}}}={\frac {1}{2}}+{\frac {2}{3}}+{\frac {1}{2^{2}}}\sum _{k=1}^{2\times 2}{\frac {k}{k+2^{2}}}+{\frac {1}{3^{2}}}\sum _{k=1}^{3\times 2}{\frac {k}{k+3^{2}}}+\dots }.}$

Toinen kaava on

${\displaystyle \gamma =\ln \pi -4\ln \Gamma ({\tfrac {3}{4}})+{\frac {4}{\pi }}\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {\ln(2k+1)}{2k+1}}.}$

${\displaystyle {\frac {e^{1+\gamma /2}}{\sqrt {2\,\pi }}}=\prod _{n=1}^{\infty }e^{-1+1/(2\,n)}\,\left(1+{\frac {1}{n}}\right)^{n}}$

${\displaystyle {\frac {e^{3+2\gamma }}{2\,\pi }}=\prod _{n=1}^{\infty }e^{-2+2/n}\,\left(1+{\frac {2}{n}}\right)^{n}.}$

${\displaystyle e^{\gamma }=\left({\frac {2}{1}}\right)^{1/2}\left({\frac {2^{2}}{1\cdot 3}}\right)^{1/3}\left({\frac {2^{3}\cdot 4}{1\cdot 3^{3}}}\right)^{1/4}\left({\frac {2^{4}\cdot 4^{4}}{1\cdot 3^{6}\cdot 5}}\right)^{1/5}\cdots .}$

1. http://media.www.dailynorthwestern.com/media/storage/paper853/news/2007/03/02/Campus/Student.Sets.World.Record.For.Math.Constant.Calculation-2754519.shtml (Arkistoitu – Internet Archive)