歐拉-馬斯刻若尼常數

歐拉-馬斯刻若尼常數是一个数学常数，定义为调和级数与自然对数的差值： γ = lim n → ∞ [ ( ∑ k = 1 n 1 k ) − ln ⁡ ( n ) ] = ∫ 1 ∞ ( 1 ⌊ x ⌋ − 1 x ) d x {\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} 它的近似值为 γ ≈ 0.577215664901532860606512090082402431042159335 {\displaystyle \gamma \approx 0.577215664901532860606512090082402431042159335} ， 歐拉-馬斯刻若尼常數主要应用于数论。 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 C {\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，意大利数学家洛倫佐·馬斯凱羅尼引入了 γ {\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。   − γ = Γ ′ ( 1 ) = Ψ ( 1 ) {\displaystyle \ -\gamma =\Gamma '(1)=\Psi (1)} 。 γ = lim x → ∞ [ x − Γ ( 1 x ) ] {\displaystyle \gamma =\lim _{x\to \infty }\left[x-\Gamma \left({\frac {1}{x}}\right)\right]} 。 γ = lim n → ∞ [ Γ ( 1 n ) Γ ( n + 1 ) n 1 + 1 n Γ ( 2 + n + 1 n ) − n 2 n + 1 ] {\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]} 。 γ = ∑ m = 2 ∞ ( − 1 ) m ζ ( m ) m {\displaystyle \gamma =\sum _{m=2}^{\infty }{\frac {(-1)^{m}\zeta (m)}{m}}} = ln ⁡ ( 4 π ) + ∑ m = 1 ∞ ( − 1 ) m − 1 ζ ( m + 1 ) 2 m ( m + 1 ) {\displaystyle =\ln \left({\frac {4}{\pi }}\right)+\sum _{m=1}^{\infty }{\frac {(-1)^{m-1}\zeta (m+1)}{2^{m}(m+1)}}} 。 lim ε → 0 ζ ( 1 + ε ) + ζ ( 1 − ε ) 2 = γ {\displaystyle \lim _{\varepsilon \to 0}{\frac {\zeta (1+\varepsilon )+\zeta (1-\varepsilon )}{2}}=\gamma } γ = 3 2 − ln ⁡ 2 − ∑ m = 2 ∞ ( − 1 ) m m − 1 m [ ζ ( m ) − 1 ] {\displaystyle \gamma ={\frac {3}{2}}-\ln 2-\sum _{m=2}^{\infty }(-1)^{m}\,{\frac {m-1}{m}}[\zeta (m)-1]} = lim n → ∞ [ 2 n − 1 2 n − ln n + ∑ k = 2 n ( 1 k − ζ ( 1 − k ) n k ) ] {\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]} 。 = lim n → ∞ [ 2 n e 2 n ∑ m = 0 ∞ 2 m n ( m + 1 ) ! ∑ t = 0 m 1 t + 1 − n ln ⁡ 2 + O ( 1 2 n e 2 n ) ] {\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]} γ = lim s → 1 + ∑ n = 1 ∞ ( 1 n s − 1 s n ) = lim s → 1 + ( ζ ( s ) − 1 s − 1 ) {\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)} γ = lim x → ∞ [ x − Γ ( 1 x ) ] {\displaystyle \gamma =\lim _{x\to \infty }\left[x-\Gamma \left({\frac {1}{x}}\right)\right]} = lim n → ∞ 1 n ∑ k = 1 n ( ⌈ n k ⌉ − n k ) {\displaystyle =\lim _{n\to \infty }{\frac {1}{n}}\,\sum _{k=1}^{n}\left(\left\lceil {\frac {n}{k}}\right\rceil -{\frac {n}{k}}\right)} 。 γ = ∑ k = 1 n 1 k − ln ⁡ ( n ) − ∑ m = 2 ∞ ζ ( m , n + 1 ) m {\displaystyle \gamma =\sum _{k=1}^{n}{\frac {1}{k}}-\ln(n)-\sum _{m=2}^{\infty }{\frac {\zeta (m,n+1)}{m}}} γ = − ∫ 0 ∞ e − x ln ⁡ x d x = ∫ ∞ 0 e − x ln ⁡ x d x {\displaystyle \gamma =-\int _{0}^{\infty }{e^{-x}\ln x}\,dx=\int _{\infty }^{0}{e^{-x}\ln x}\,dx} = − ∫ 0 1 ln ⁡ ln ⁡ 1 x d x {\displaystyle =-\int _{0}^{1}{\ln \ln {\frac {1}{x}}}\,dx} = ∫ 0 ∞ ( 1 1 − e − x − 1 x ) e − x d x {\displaystyle =\int _{0}^{\infty }{\left({\frac {1}{1-e^{-x}}}-{\frac {1}{x}}\right)e^{-x}}\,dx} = ∫ 0 ∞ 1 x ( 1 1 + x − e − x ) d x {\displaystyle =\int _{0}^{\infty }{{\frac {1}{x}}\left({\frac {1}{1+x}}-e^{-x}\right)}\,dx} ∫ 0 ∞ e − x 2 ln ⁡ x d x = − 1 4 ( γ + 2 ln ⁡ 2 ) π {\displaystyle \int _{0}^{\infty }{e^{-x^{2}}\ln x}\,dx=-{\tfrac {1}{4}}(\gamma +2\ln 2){\sqrt {\pi }}} ∫ 0 ∞ e − x ln 2 ⁡ x d x = γ 2 + π 2 6 {\displaystyle \int _{0}^{\infty }{e^{-x}\ln ^{2}x}\,dx=\gamma ^{2}+{\frac {\pi ^{2}}{6}}} 。 γ = ∫ 0 1 ∫ 0 1 x − 1 ( 1 − x y ) ln ⁡ ( x y ) d x d y = ∑ n = 1 ∞ ( 1 n − ln ⁡ n + 1 n ) {\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)} ∑ n = 1 ∞ N 1 ( n ) + N 0 ( n ) 2 n ( 2 n + 1 ) = γ {\displaystyle \sum _{n=1}^{\infty }{\frac {N_{1}(n)+N_{0}(n)}{2n(2n+1)}}=\gamma } γ = ∑ k = 1 ∞ [ 1 k − ln ⁡ ( 1 + 1 k ) ] {\displaystyle \gamma =\sum _{k=1}^{\infty }\left[{\frac {1}{k}}-\ln \left(1+{\frac {1}{k}}\right)\right]} γ = 1 − ∑ k = 2 ∞ ( − 1 ) k ⌊ log 2 ⁡ k ⌋ k + 1 {\displaystyle \gamma =1-\sum _{k=2}^{\infty }(-1)^{k}{\frac {\lfloor \log _{2}k\rfloor }{k+1}}} . γ = ∑ k = 2 ∞ ( − 1 ) k ⌊ log 2 ⁡ k ⌋ k = 1 2 − 1 3 + 2 ( 1 4 − 1 5 + 1 6 − 1 7 ) + 3 ( 1 8 − ⋯ − 1 15 ) + … {\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 } γ + ζ ( 2 ) = ∑ k = 1 ∞ 1 k ⌊ k ⌋ 2 = 1 + 1 2 + 1 3 + 1 4 ( 1 4 + ⋯ + 1 8 ) + 1 9 ( 1 9 + ⋯ + 1 15 ) + … {\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 } γ = ∑ k = 2 ∞ k − ⌊ k ⌋ 2 k 2 ⌊ k ⌋ 2 = 1 2 2 + 2 3 2 + 1 2 2 ( 1 5 2 + 2 6 2 + 3 7 2 + 4 8 2 ) + 1 3 2 ( 1 10 2 + ⋯ + 6 15 2 ) + … {\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 } γ = ∫ 0 1 1 1 + x ∑ n = 1 ∞ x 2 n − 1 d x {\displaystyle \gamma =\int _{0}^{1}{\frac {1}{1+x}}\sum _{n=1}^{\infty }x^{2^{n}-1}\,dx} γ {\displaystyle \gamma } 的连分数展开式为： γ = [ 0 ; 1 , 1 , 2 , 1 , 2 , 1 , 4 , 3 , 13 , 5 , 1 , 1 , 8 , 1 , 2 , 4 , 1 , 1 , 40 , . . . ] {\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）. γ ≈ H n − ln ⁡ ( n ) − 1 2 n + 1 12 n 2 − 1 120 n 4 + . . . {\displaystyle \gamma \approx H_{n}-\ln \left(n\right)-{\frac {1}{2n}}+{\frac {1}{12n^{2}}}-{\frac {1}{120n^{4}}}+...} γ ≈ H n − ln ⁡ ( n + 1 2 + 1 24 n − 1 48 n 3 + . . . ) {\displaystyle \gamma \approx H_{n}-\ln \left({n+{\frac {1}{2}}+{\frac {1}{24n}}-{\frac {1}{48n^{3}}}+...}\right)} γ ≈ H n − ln ⁡ ( n ) + ln ⁡ ( n + 1 ) 2 − 1 6 n ( n + 1 ) + 1 30 n 2 ( n + 1 ) 2 − . . . {\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}}}-...} γ = − ∫ 0 ∞ e − x ln ⁡ x d x {\displaystyle \gamma =-\int _{0}^{\infty }{e^{-x}\ln x}\,dx} 的证明： 首先根据放缩法（ ∫ k k + 1 1 x d x < 1 k < ∫ k − 1 k 1 x d x {\displaystyle \int _{k}^{k+1}{\frac {1}{x}}\,dx<{\frac {1}{k}}<\int _{k-1}^{k}{\frac {1}{x}}\,dx} ）容易知道， ∫ k k − 1 1 x d x − 1 k < 1 k ( k − 1 ) {\displaystyle \int _{k}^{k-1}{\frac {1}{x}}\,dx-{\frac {1}{k}}<{\frac {1}{k(k-1)}}} ，以及 ln ⁡ n < ∑ k = 1 n 1 k < 1 + ln ⁡ n {\displaystyle \ln n<\sum _{k=1}^{n}{\frac {1}{k}}<1+\ln n} 。因此 γ {\displaystyle \gamma } 存在并有限。 ∑ k = 1 n 1 k {\displaystyle \sum _{k=1}^{n}{\frac {1}{k}}} = ∑ k = 1 n ∫ 0 1 t k − 1 d t {\displaystyle =\sum _{k=1}^{n}\int _{0}^{1}t^{k-1}\,dt} = ∫ 0 1 ∑ k = 1 n t k − 1 d t {\displaystyle =\int _{0}^{1}\sum _{k=1}^{n}t^{k-1}\,dt} = ∫ 0 1 1 − t n 1 − t d t {\displaystyle =\int _{0}^{1}{\frac {1-t^{n}}{1-t}}\,dt} = ∫ n 0 1 − ( 1 − x n ) n 1 − ( 1 − x n ) d ( 1 − x n ) {\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)} = ∫ n 0 1 − ( 1 − x n ) n x n ( − 1 n ) d x {\displaystyle =\int _{n}^{0}{\frac {1-\left(1-{\frac {x}{n}}\right)^{n}}{\frac {x}{n}}}\left(-{\frac {1}{n}}\right)dx} = ∫ 0 n 1 − ( 1 − x n ) n x d x {\displaystyle =\int _{0}^{n}{\frac {1-\left(1-{\frac {x}{n}}\right)^{n}}{x}}dx} 而 ln ⁡ n = ∫ 1 n 1 x d x , {\displaystyle \ln n=\int _{1}^{n}{\frac {1}{x}}\,dx,} 所以 γ = lim n → ∞ ( ∑ k = 1 n 1 k − ln ⁡ n ) {\displaystyle \gamma =\lim _{n\to \infty }\left(\sum _{k=1}^{n}{\frac {1}{k}}-\ln n\right)} = lim n → ∞ [ ∫ 0 n 1 − ( 1 − x / n ) n x d x − ∫ 1 n 1 x d x ] {\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]} = lim n → ∞ [ ∫ 0 1 1 − ( 1 − x / n ) n x d x − ∫ 1 n ( 1 − x / n ) n x ] {\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]} = ∫ 0 1 1 − lim n → ∞ ( 1 − x / n ) n x d x − ∫ 1 ∞ lim n → ∞ ( 1 − x / n ) n x {\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） = ∫ 0 1 1 − e − x x d x − ∫ 1 ∞ e − x x {\displaystyle =\int _{0}^{1}{\frac {1-e^{-x}}{x}}\,dx-\int _{1}^{\infty }{\frac {e^{-x}}{x}}} = ( 1 − e − x ) ln ⁡ x | 0 1 − ∫ 0 1 ln ⁡ x d ( 1 − e − x ) − e − x ln ⁡ x | 1 ∞ + ∫ 1 ∞ ln ⁡ x d e − 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}} = − ∫ 0 ∞ e − x ln ⁡ x d x . {\displaystyle =-\int _{0}^{\infty }e^{-x}\ln x\,dx.} 前面的放缩法主要是证明了 [ ( ∑ k = 1 n 1 k ) − ln ⁡ ( n ) ] {\displaystyle \left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]} 是单调递减并下有界限（0），所有极限存在。放缩法的结论需要使用ln(1+x)和ln(1-x)的泰勒级数展开进行证明。 A001620 oeis.org [2014-7-17] Havil 2003 p 97. 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