ln 2 （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：（A007524 ）
log
10
2
≈
0.301029995663981195
{\displaystyle \log_{10} 2 \approx 0.301029995663981195}
数学家理查德·施罗培尔 在1972年证明，不寻常数 的自然密度 等于
ln
2
{\displaystyle \ln 2}
u
(
n
)
{\displaystyle u(n)}
n
{\displaystyle n}
a
{\displaystyle a}
a
{\displaystyle \sqrt a}
lim
n
→
∞
u
(
n
)
n
=
ln
(
2
)
=
0.693147
…
.
{\displaystyle \lim_{n \rightarrow \infty} \frac{u(n)}{n} = \ln(2) = 0.693147 \dots\, .}
公式
∑
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}
欧拉-马歇罗尼常数 ，
ζ
{\displaystyle \zeta}
黎曼ζ函数 。
ln
2
=
∑
k
≥
1
1
k
2
k
.
{\displaystyle \ln 2 = \sum_{k\ge 1} \frac{1}{k2^k}.}
ln
2
=
∑
k
≥
1
(
1
3
k
+
1
4
k
)
1
k
.
{\displaystyle \ln 2 = \sum_{k\ge 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\ge 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\ge 0} \frac{1}{(2k+1)9^k}.}
反双曲函数 ，可参见计算自然对数的级数 。）积分公式
∫
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 x dx=2\int_0^{\frac{\pi}{4}} \tan x dx=\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\ln2}
∫
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}
欧拉-马歇罗尼常数 。
其他公式 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\cdot3\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 - ……) 参考文献 Brent, Richard P. Fast multiple-precision evaluation of elementary functions. J. ACM. 1976, 23 (2): 242–251. doi:10.1145/321941.321944 MR 0395314 . Uhler, Horace S. Recalculation and extension of the modulus and of the logarithms of 2, 3, 5, 7 and 17. Proc. Nat. Acac. Sci. U. S. A. 1940, 26 : 205–212. MR 0001523 . Sweeney, Dura W. On the computation of Euler's constant. Mathematics of Computation. 1963, 17 . MR 0160308 . Chamberland, Marc. Binary BBP-formulae for logarithms and generalized Gaussian-Mersenne primes (PDF) . Journal of Integer Sequences. 2003, 6 : 03.3.7 [2011-01-08 ] . MR 2046407 . Gourévitch, Boris; Guillera Goyanes, Jesus. Construction of binomial sums for π and polylogarithmic constants inspired by BBP formulas (PDF) . Applied Math. E-Notes. 2007, 7 : 237–246 [2011-01-08 ] . MR 2346048 . Wu, Qiang. On the linear independence measure of logarithms of rational numbers. Mathematics of Computation. 2003, 72 (242): 901–911. doi:10.1090/S0025-5718-02-01442-4 外部链接 参见 