2的自然对数

ln2（OEIS數列A002162）约为： ln ⁡ 2 ≈ 0.693147 {\displaystyle \ln 2\approx 0.693147} 使用对数公式 log b ⁡ 2 = ln ⁡ 2 ln ⁡ b . {\displaystyle \log _{b}2={\frac {\ln 2}{\ln b}}.} 可以求出log2，它约为：（OEIS數列A007524） log 10 ⁡ 2 ≈ 0.301029995663981195 {\displaystyle \log _{10}2\approx 0.301029995663981195} 。 數學家理查德·施羅培爾（英语：Richard Schroeppel）在1972年證明，不尋常數的自然密度等於 ln ⁡ 2 {\displaystyle \ln 2} 。換言之，若 u ( n ) {\displaystyle u(n)} 表示不大於 n {\displaystyle n} 的自然數之中，有多少個數 a {\displaystyle a} 具有大於 a {\displaystyle {\sqrt {a}}} 的質因數，則有： lim n → ∞ u ( n ) n = ln ⁡ ( 2 ) = 0.693147 … . {\displaystyle \lim _{n\rightarrow \infty }{\frac {u(n)}{n}}=\ln(2)=0.693147\dots \,.} ∑ n = 1 ∞ ( − 1 ) n + 1 n = ∑ n = 0 ∞ 1 ( 2 n + 1 ) ( 2 n + 2 ) = ln ⁡ 2. {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=\sum _{n=0}^{\infty }{\frac {1}{(2n+1)(2n+2)}}=\ln 2.} ∑ n = 1 ∞ ( − 1 ) n ( n + 1 ) ( n + 2 ) = 2 ln ⁡ 2 − 1. {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{(n+1)(n+2)}}=2\ln 2-1.} ∑ n = 1 ∞ 1 n ( 4 n 2 − 1 ) = 2 ln ⁡ 2 − 1. {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(4n^{2}-1)}}=2\ln 2-1.} ∑ n = 1 ∞ ( − 1 ) n n ( 4 n 2 − 1 ) = ln ⁡ 2 − 1. {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1.} ∑ n = 1 ∞ ( − 1 ) n n ( 9 n 2 − 1 ) = 2 ln ⁡ 2 − 3 2 . {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(9n^{2}-1)}}=2\ln 2-{\frac {3}{2}}.} ∑ n = 2 ∞ 1 2 n [ ζ ( n ) − 1 ] = ln ⁡ 2 − 1 2 . {\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln 2-{\frac {1}{2}}.} ∑ n = 1 ∞ 1 2 n + 1 [ ζ ( n ) − 1 ] = 1 − γ − 1 2 ln ⁡ 2. {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2n+1}}[\zeta (n)-1]=1-\gamma -{\frac {1}{2}}\ln 2.} ∑ n = 1 ∞ 1 2 2 n ( 2 n + 1 ) ζ ( 2 n ) = 1 2 ( 1 − ln ⁡ 2 ) . {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2^{2n}(2n+1)}}\zeta (2n)={\frac {1}{2}}(1-\ln 2).} γ {\displaystyle \gamma } 是欧拉-马歇罗尼常数， ζ {\displaystyle \zeta } 是黎曼ζ函數。 ln ⁡ 2 = ∑ k ≥ 1 1 k 2 k . {\displaystyle \ln 2=\sum _{k\geq 1}{\frac {1}{k2^{k}}}.} ln ⁡ 2 = ∑ k ≥ 1 ( 1 3 k + 1 4 k ) 1 k . {\displaystyle \ln 2=\sum _{k\geq 1}\left({\frac {1}{3^{k}}}+{\frac {1}{4^{k}}}\right){\frac {1}{k}}.} ln ⁡ 2 = 2 3 + ∑ k ≥ 1 ( 1 2 k + 1 4 k + 1 + 1 8 k + 4 + 1 16 k + 12 ) 1 16 k . {\displaystyle \ln 2={\frac {2}{3}}+\sum _{k\geq 1}\left({\frac {1}{2k}}+{\frac {1}{4k+1}}+{\frac {1}{8k+4}}+{\frac {1}{16k+12}}\right){\frac {1}{16^{k}}}.} （贝利－波尔温－普劳夫公式） ln ⁡ 2 = 2 3 ∑ k ≥ 0 1 ( 2 k + 1 ) 9 k . {\displaystyle \ln 2={\frac {2}{3}}\sum _{k\geq 0}{\frac {1}{(2k+1)9^{k}}}.} （基於反雙曲函數，可參見計算自然對數的級數。） ∫ 0 1 d x 1 + x = ln ⁡ 2 {\displaystyle \int _{0}^{1}{\frac {dx}{1+x}}=\ln 2} ∫ 1 ∞ d x ( 1 + x 2 ) ( 1 + x ) 2 = 1 4 ( 1 − ln ⁡ 2 ) {\displaystyle \int _{1}^{\infty }{\frac {dx}{(1+x^{2})(1+x)^{2}}}={\frac {1}{4}}(1-\ln 2)} ∫ 0 ∞ d x 1 + e n x = 1 n ln ⁡ 2 ; ∫ 0 ∞ d x 3 + e n x = 2 3 n ln ⁡ 2 {\displaystyle \int _{0}^{\infty }{\frac {dx}{1+e^{nx}}}={\frac {1}{n}}\ln 2;\int _{0}^{\infty }{\frac {dx}{3+e^{nx}}}={\frac {2}{3n}}\ln 2} ∫ 0 ∞ ( 1 e x − 1 − 2 e 2 x − 1 ) = ln ⁡ 2 {\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {2}{e^{2x}-1}}\right)=\ln 2} ∫ 0 ∞ e − x 1 − e − x x d x = ln ⁡ 2 {\displaystyle \int _{0}^{\infty }e^{-x}{\frac {1-e^{-x}}{x}}dx=\ln 2} ∫ 0 1 ln ⁡ x 2 − 1 x ln ⁡ x d x = − 1 + ln ⁡ 2 + γ {\displaystyle \int _{0}^{1}\ln {\frac {x^{2}-1}{x\ln x}}dx=-1+\ln 2+\gamma } ∫ 0 π 3 tan ⁡ x d x = 2 ∫ 0 π 4 tan ⁡ x d x = ln ⁡ 2 {\displaystyle \int _{0}^{\frac {\pi }{3}}\tan xdx=2\int _{0}^{\frac {\pi }{4}}\tan xdx=\ln 2} ∫ − π 4 π 4 ln ⁡ ( sin ⁡ x + cos ⁡ x ) d x = − π 4 ln ⁡ 2 {\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\ln(\sin x+\cos x)dx=-{\frac {\pi }{4}}\ln 2} ∫ 0 1 x 2 ln ⁡ ( 1 + x ) d x = 2 3 ln ⁡ 2 − 5 18 {\displaystyle \int _{0}^{1}x^{2}\ln(1+x)dx={\frac {2}{3}}\ln 2-{\frac {5}{18}}} ∫ 0 1 x ln ⁡ ( 1 + x ) ln ⁡ ( 1 − x ) d x = 1 4 − ln ⁡ 2 {\displaystyle \int _{0}^{1}x\ln(1+x)\ln(1-x)dx={\frac {1}{4}}-\ln 2} ∫ 0 1 x 3 ln ⁡ ( 1 + x ) ln ⁡ ( 1 − x ) d x = 13 96 − 2 3 ln ⁡ 2 {\displaystyle \int _{0}^{1}x^{3}\ln(1+x)\ln(1-x)dx={\frac {13}{96}}-{\frac {2}{3}}\ln 2} ∫ 0 1 ln ⁡ x ( 1 + x ) 2 d x = − ln ⁡ 2 {\displaystyle \int _{0}^{1}{\frac {\ln x}{(1+x)^{2}}}dx=-\ln 2} ∫ 0 1 ln ⁡ ( 1 + x ) − x x 2 d x = 1 − 2 ln ⁡ 2 {\displaystyle \int _{0}^{1}{\frac {\ln(1+x)-x}{x^{2}}}dx=1-2\ln 2} ∫ 0 1 d x x ( 1 − ln ⁡ x ) ( 1 − 2 ln ⁡ x ) = ln ⁡ 2 {\displaystyle \int _{0}^{1}{\frac {dx}{x(1-\ln x)(1-2\ln x)}}=\ln 2} ∫ 1 ∞ ln ⁡ ln ⁡ x x 3 d x = − 1 2 ( γ + ln ⁡ 2 ) {\displaystyle \int _{1}^{\infty }{\frac {\ln \ln x}{x^{3}}}dx=-{\frac {1}{2}}(\gamma +\ln 2)} γ {\displaystyle \gamma } 是欧拉-马歇罗尼常数。 用皮尔斯展开式（A091846）表达ln2： log ⁡ 2 = 1 1 − 1 1 ⋅ 3 + 1 1 ⋅ 3 ⋅ 12 − … {\displaystyle \log 2={\frac {1}{1}}-{\frac {1}{1\cdot 3}}+{\frac {1}{1\cdot 3\cdot 12}}-\ldots } . 用恩格尔展开式A059180表达ln2： log ⁡ 2 = 1 2 + 1 2 ⋅ 3 + 1 2 ⋅ 3 ⋅ 7 + 1 2 ⋅ 3 ⋅ 7 ⋅ 9 + … {\displaystyle \log 2={\frac {1}{2}}+{\frac {1}{2\cdot 3}}+{\frac {1}{2\cdot 3\cdot 7}}+{\frac {1}{2\cdot 3\cdot 7\cdot 9}}+\ldots } . 用余切展开式A081785表达ln2： log ⁡ 2 = cot ⁡ ( arccot ⁡ 0 − arccot ⁡ 1 + arccot ⁡ 5 − arccot ⁡ 55 + arccot ⁡ 14187 − … ) {\displaystyle \log 2=\cot(\operatorname {arccot} 0-\operatorname {arccot} 1+\operatorname {arccot} 5-\operatorname {arccot} 55+\operatorname {arccot} 14187-\ldots )} . 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