2的自然对数

${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=\sum _{n=0}^{\infty }{\frac {1}{(2n+1)(2n+2)}}=\ln 2.}$ 

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

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(4n^{2}-1)}}=2\ln 2-1.}$ 

${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1.}$ 

${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(9n^{2}-1)}}=2\ln 2-{\frac {3}{2}}.}$ 

${\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln 2-{\frac {1}{2}}.}$ 

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

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

${\displaystyle \gamma }$ 是欧拉-马歇罗尼常数，${\displaystyle \zeta }$ 是黎曼ζ函数。

${\displaystyle \ln 2=\sum _{k\geq 1}{\frac {1}{k2^{k}}}.}$ ^[2]:31

${\displaystyle \ln 2=\sum _{k\geq 1}\left({\frac {1}{3^{k}}}+{\frac {1}{4^{k}}}\right){\frac {1}{k}}.}$ 

${\displaystyle \ln 2={\frac {2}{3}}+\sum _{k\geq 1}\left({\frac {1}{2k}}+{\frac {1}{4k+1}}+{\frac {1}{8k+4}}+{\frac {1}{16k+12}}\right){\frac {1}{16^{k}}}.}$ （贝利－波尔温－普劳夫公式）

${\displaystyle \ln 2={\frac {2}{3}}\sum _{k\geq 0}{\frac {1}{(2k+1)9^{k}}}.}$ （基于反双曲函数，可参见计算自然对数的级数。）

${\displaystyle \int _{0}^{1}{\frac {dx}{1+x}}=\ln 2}$ 

${\displaystyle \int _{1}^{\infty }{\frac {dx}{(1+x^{2})(1+x)^{2}}}={\frac {1}{4}}(1-\ln 2)}$ 

${\displaystyle \int _{0}^{\infty }{\frac {dx}{1+e^{nx}}}={\frac {1}{n}}\ln 2;\int _{0}^{\infty }{\frac {dx}{3+e^{nx}}}={\frac {2}{3n}}\ln 2}$ 

${\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {2}{e^{2x}-1}}\right)=\ln 2}$ 

${\displaystyle \int _{0}^{\infty }e^{-x}{\frac {1-e^{-x}}{x}}dx=\ln 2}$ 

${\displaystyle \int _{0}^{1}\ln {\frac {x^{2}-1}{x\ln x}}dx=-1+\ln 2+\gamma }$ 

${\displaystyle \int _{0}^{\frac {\pi }{3}}\tan xdx=2\int _{0}^{\frac {\pi }{4}}\tan xdx=\ln 2}$ 

${\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\ln(\sin x+\cos x)dx=-{\frac {\pi }{4}}\ln 2}$ 

${\displaystyle \int _{0}^{1}x^{2}\ln(1+x)dx={\frac {2}{3}}\ln 2-{\frac {5}{18}}}$ 

${\displaystyle \int _{0}^{1}x\ln(1+x)\ln(1-x)dx={\frac {1}{4}}-\ln 2}$ 

${\displaystyle \int _{0}^{1}x^{3}\ln(1+x)\ln(1-x)dx={\frac {13}{96}}-{\frac {2}{3}}\ln 2}$ 

${\displaystyle \int _{0}^{1}{\frac {\ln x}{(1+x)^{2}}}dx=-\ln 2}$ 

${\displaystyle \int _{0}^{1}{\frac {\ln(1+x)-x}{x^{2}}}dx=1-2\ln 2}$ 

${\displaystyle \int _{0}^{1}{\frac {dx}{x(1-\ln x)(1-2\ln x)}}=\ln 2}$ 

${\displaystyle \int _{1}^{\infty }{\frac {\ln \ln x}{x^{3}}}dx=-{\frac {1}{2}}(\gamma +\ln 2)}$ 

${\displaystyle \gamma }$ 是欧拉-马歇罗尼常数。

用皮尔斯展开式（A091846）表达ln2：

${\displaystyle \log 2={\frac {1}{1}}-{\frac {1}{1\cdot 3}}+{\frac {1}{1\cdot 3\cdot 12}}-\ldots }$ .

用恩格尔展开式A059180表达ln2：

${\displaystyle \log 2={\frac {1}{2}}+{\frac {1}{2\cdot 3}}+{\frac {1}{2\cdot 3\cdot 7}}+{\frac {1}{2\cdot 3\cdot 7\cdot 9}}+\ldots }$ .

用余切展开式A081785表达ln2：

${\displaystyle \log 2=\cot(\operatorname {arccot} 0-\operatorname {arccot} 1+\operatorname {arccot} 5-\operatorname {arccot} 55+\operatorname {arccot} 14187-\ldots )}$ .

范例

• 埃里克·韦斯坦因. Natural logarithm of 2. MathWorld. 
• table of natural logarithms. PlanetMath. 
• Gourdon, Xavier; Sebah, Pascal. The logarithm constant:log 2. [2011-01-08]. （原始内容存档于2020-02-23）. 

• 对数
• 自然对数
• 常用对数
• 超越数