2的自然對數 2的自然對數 種類 無理數 符號
ln
2
{\displaystyle \ln {2}}
連分數 [0; 1, 2, 3, 1, 6, 3, 1, 1, 2, 1, 1, 1, 1, 3, 10] （OEIS 數列A016730 ）
0
+
1
1
+
1
2
+
1
3
+
1
1
+
1
6
+
⋱
{\displaystyle 0+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{3+{\cfrac {1}{1+{\cfrac {1}{6+\ddots }}}}}}}}}}}
以此為根 的多項式或函數
e
x
−
2
=
0
{\displaystyle e^{x}-2=0}
[ 1] 值
ln
2
≈
{\displaystyle \ln {2}\approx }
二進制 0.10110001 0111 0010 0001 0111 … 十進制 0.69314718 0559 9453 0941 7232 … 十六進制 0.B17217F7 D1CF 79AB C9E3 B398 …
ln 2 （OEIS 數列A002162 ）約為：
ln
2
≈
0.693147
{\displaystyle \ln 2\approx 0.693147}
使用對數公式
log
b
2
=
ln
2
ln
b
.
{\displaystyle \log _{b}2={\frac {\ln 2}{\ln b}}.}
可以求出log2，它約為：（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 不尋常數 的自然密度 等於
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 \,.}
∑
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 }
歐拉-馬歇羅尼常數 ，
ζ
{\displaystyle \zeta }
黎曼ζ函數 。
ln
2
=
∑
k
≥
1
1
k
2
k
.
{\displaystyle \ln 2=\sum _{k\geq 1}{\frac {1}{k2^{k}}}.}
[ 2] :31
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}}}.}
貝利－波爾溫－普勞夫公式 ）
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}}}.}
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 }
歐拉-馬歇羅尼常數 。
用皮爾斯展開式（A091846 ）表達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 }
用恩格爾展開式 A059180 表達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 }
用餘切展開式A081785 表達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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![{\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln 2-{\frac {1}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b61c76e20b8727253043baa534c03f6a85808e72)
![{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2n+1}}[\zeta (n)-1]=1-\gamma -{\frac {1}{2}}\ln 2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52102e7a19d5707bdb52e85feebca116766023d3)
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