含圆周率的公式列表

${\displaystyle \int \limits _{-\infty }^{\infty }{\text{sech}}(x)dx=\pi \!}$

${\displaystyle \int _{0}^{\infty }{\frac {dx}{(x+1){\sqrt {x}}}}=\pi }$

${\displaystyle \int \limits _{-1}^{1}{\sqrt {1-x^{2}}}\,dx={\frac {\pi }{2}}\!}$

${\displaystyle \int \limits _{-1}^{1}{\frac {dx}{\sqrt {1-x^{2}}}}=\pi \!}$

${\displaystyle \int \limits _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi \!}$

${\displaystyle \int \limits _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}\!}$ (参见 正态分布)

${\displaystyle \oint {\frac {dz}{z}}=2\pi i\!}$ (参见 柯西积分公式)

${\displaystyle \int \limits _{-\infty }^{\infty }{\frac {\sin(x)}{x}}\,dx=\pi \!}$

${\displaystyle \int \limits _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx={22 \over 7}-\pi \!}$ (参见 證明22/7大於π)

${\displaystyle {\frac {\pi }{2}}\!=\sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\frac {2^{k}k!^{2}}{(2k+1)!}}}$ (参见 双阶乘)

${\displaystyle {\frac {1}{\pi }}\!=12\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k+{\frac {3}{2}}}}}}$ (参见 楚德诺夫斯基算法)

${\displaystyle {\frac {1}{\pi }}\!={\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}}$ (参见拉马努金)

${\displaystyle \pi \!={\frac {\sqrt {3}}{6^{5}}}\sum _{k=0}^{\infty }{\frac {[(4k)!]^{2}(6k)!}{9^{k+1}(12k)!(2k)!}}\left({\frac {127169}{12k+1}}-{\frac {1070}{12k+5}}-{\frac {131}{12k+7}}+{\frac {2}{12k+11}}\right)}$[1]

以下是任意位的二进制的π计算：:

${\displaystyle \pi \!=\sum _{k=0}^{\infty }{\frac {1}{16^{k}}}\left({\frac {4}{8k+1}}-{\frac {2}{8k+4}}-{\frac {1}{8k+5}}-{\frac {1}{8k+6}}\right)}$ (参见 贝利－波尔温－普劳夫公式)

${\displaystyle \pi ={\frac {1}{2^{6}}}\sum _{n=0}^{\infty }{\frac {{(-1)}^{n}}{2^{10n}}}\left(-{\frac {2^{5}}{4n+1}}-{\frac {1}{4n+3}}+{\frac {2^{8}}{10n+1}}-{\frac {2^{6}}{10n+3}}-{\frac {2^{2}}{10n+5}}-{\frac {2^{2}}{10n+7}}+{\frac {1}{10n+9}}\right)}$

${\displaystyle \zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}\!}$   (参见巴塞尔问题和黎曼ζ函數)

${\displaystyle \zeta (4)={\frac {1}{1^{4}}}+{\frac {1}{2^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{4^{4}}}+\cdots ={\frac {\pi ^{4}}{90}}\!}$

${\displaystyle \zeta (2n)={\frac {1}{1^{2n}}}+{\frac {1}{2^{2n}}}+{\frac {1}{3^{2n}}}+{\frac {1}{4^{2n}}}+\cdots =(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}\!}$

${\displaystyle {\frac {\pi }{4}}\!=\sum _{n=0}^{\infty }{\left[{\frac {(-1)^{n}}{2n+1}}\right]}^{1}={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\arctan {1}=\int _{0}^{1}{\frac {1}{1+x^{2}}}dx}$   (参见Π的莱布尼茨公式)

${\displaystyle {\frac {\pi ^{2}}{8}}\!=\sum _{n=0}^{\infty }{\left[{\frac {(-1)^{n}}{2n+1}}\right]}^{2}={\frac {1}{1^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{5^{2}}}+{\frac {1}{7^{2}}}+\cdots }$

${\displaystyle {\frac {\pi ^{3}}{32}}\!=\sum _{n=0}^{\infty }{\left[{\frac {(-1)^{n}}{2n+1}}\right]}^{3}={\frac {1}{1^{3}}}-{\frac {1}{3^{3}}}+{\frac {1}{5^{3}}}-{\frac {1}{7^{3}}}+\cdots }$

${\displaystyle {\frac {\pi ^{4}}{96}}\!=\sum _{n=0}^{\infty }{\left[{\frac {(-1)^{n}}{2n+1}}\right]}^{4}={\frac {1}{1^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{5^{4}}}+{\frac {1}{7^{4}}}+\cdots }$

${\displaystyle {\frac {5\pi ^{5}}{1536}}\!=\sum _{n=0}^{\infty }{\left[{\frac {(-1)^{n}}{2n+1}}\right]}^{5}={\frac {1}{1^{5}}}-{\frac {1}{3^{5}}}+{\frac {1}{5^{5}}}-{\frac {1}{7^{5}}}+\cdots }$

${\displaystyle {\frac {\pi ^{6}}{960}}\!=\sum _{n=0}^{\infty }{\left[{\frac {(-1)^{n}}{2n+1}}\right]}^{6}={\frac {1}{1^{6}}}+{\frac {1}{3^{6}}}+{\frac {1}{5^{6}}}+{\frac {1}{7^{6}}}+\cdots }$

${\displaystyle {\frac {\pi }{4}}={\frac {3}{4}}\times {\frac {5}{4}}\times {\frac {7}{8}}\times {\frac {11}{12}}\times {\frac {13}{12}}\times {\frac {17}{16}}\times {\frac {19}{20}}\times {\frac {23}{24}}\times {\frac {29}{28}}\times {\frac {31}{32}}\times \cdots \!}$ (欧拉)

${\displaystyle \pi ={1}+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}-{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}-{\frac {1}{10}}+{\frac {1}{11}}+{\frac {1}{12}}-{\frac {1}{13}}+\cdots \!}$ (欧拉, 1748)[2]

参见梅钦公式.

${\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}\!}$ (原始的梅钦公式.)

${\displaystyle {\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}}\!}$

${\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{2}}-\arctan {\frac {1}{7}}\!}$

${\displaystyle {\frac {\pi }{4}}=2\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}\!}$

${\displaystyle {\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}\!}$

${\displaystyle {\frac {\pi }{4}}=12\arctan {\frac {1}{49}}+32\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}+12\arctan {\frac {1}{110443}}\!}$

${\displaystyle {\frac {\pi }{4}}=44\arctan {\frac {1}{57}}+7\arctan {\frac {1}{239}}-12\arctan {\frac {1}{682}}+24\arctan {\frac {1}{12943}}\!}$

一些涉及圆周率的无穷级数:[3]

${\displaystyle \pi ={\frac {1}{Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {[(2n)!]^{3}(42n+5)}{(n!)^{6}{16}^{3n+1}}}\!}$
${\displaystyle \pi ={\frac {4}{Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(21460n+1123)}{(n!)^{4}{441}^{2n+1}{2}^{10n+1}}}}$
${\displaystyle \pi ={\frac {4}{Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(6n+1)\left({\frac {1}{2}}\right)_{n}^{3}}{{4^{n}}(n!)^{3}}}\!}$
${\displaystyle \pi ={\frac {32}{Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }\left({\frac {{\sqrt {5}}-1}{2}}\right)^{8n}{\frac {(42n{\sqrt {5}}+30n+5{\sqrt {5}}-1)\left({\frac {1}{2}}\right)_{n}^{3}}{{64^{n}}(n!)^{3}}}\!}$
${\displaystyle \pi ={\frac {27}{4Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }\left({\frac {2}{27}}\right)^{n}{\frac {(15n+2)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{3}}\right)_{n}\left({\frac {2}{3}}\right)_{n}}{(n!)^{3}}}\!}$
${\displaystyle \pi ={\frac {15{\sqrt {3}}}{2Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }\left({\frac {4}{125}}\right)^{n}{\frac {(33n+4)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{3}}\right)_{n}\left({\frac {2}{3}}\right)_{n}}{(n!)^{3}}}\!}$
${\displaystyle \pi ={\frac {85{\sqrt {85}}}{18{\sqrt {3}}Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }\left({\frac {4}{85}}\right)^{n}{\frac {(133n+8)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{6}}\right)_{n}\left({\frac {5}{6}}\right)_{n}}{(n!)^{3}}}\!}$
${\displaystyle \pi ={\frac {5{\sqrt {5}}}{2{\sqrt {3}}Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }\left({\frac {4}{125}}\right)^{n}{\frac {(11n+1)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{6}}\right)_{n}\left({\frac {5}{6}}\right)_{n}}{(n!)^{3}}}\!}$
${\displaystyle \pi ={\frac {2{\sqrt {3}}}{Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(8n+1)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{9}^{n}}}\!}$
${\displaystyle \pi ={\frac {\sqrt {3}}{9Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(40n+3)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{49}^{2n+1}}}\!}$
${\displaystyle \pi ={\frac {2{\sqrt {11}}}{11Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(280n+19)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{99}^{2n+1}}}\!}$
${\displaystyle \pi ={\frac {\sqrt {2}}{4Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(10n+1)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{9}^{2n+1}}}\!}$
${\displaystyle \pi ={\frac {4{\sqrt {5}}}{5Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(644n+41)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}5^{n}{72}^{2n+1}}}\!}$
${\displaystyle \pi ={\frac {4{\sqrt {3}}}{3Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(28n+3)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{3^{n}}{4}^{n+1}}}\!}$
${\displaystyle \pi ={\frac {4}{Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(20n+3)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{2}^{2n+1}}}\!}$
${\displaystyle \pi ={\frac {72}{Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(260n+23)}{(n!)^{4}4^{4n}18^{2n}}}\!}$
${\displaystyle \pi ={\frac {3528}{Z}}\!}$	${\displaystyle Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(21460n+1123)}{(n!)^{4}4^{4n}882^{2n}}}\!}$

${\displaystyle (x)_{n}\!}$是阶乘幂中下降阶乘幂的符号。

${\displaystyle \prod _{n=1}^{\infty }{\frac {4n^{2}}{4n^{2}-1}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots ={\frac {4}{3}}\cdot {\frac {16}{15}}\cdot {\frac {36}{35}}\cdot {\frac {64}{63}}\cdots ={\frac {\pi }{2}}\!}$ (参见沃利斯乘积)

弗朗索瓦·韦达的公式:

${\displaystyle {\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdot \cdots ={\frac {2}{\pi }}\!}$

${\displaystyle \pi ={3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+{\cfrac {7^{2}}{6+\ddots \,}}}}}}}}}}$

${\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}}$

${\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+\ddots }}}}}}}}}}\,}$

(参见连分数。)

${\displaystyle n!\approx {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}\!}$ (斯特灵公式)

${\displaystyle e^{i\pi }+1=0}$ (歐拉恆等式)

${\displaystyle \sum _{k=1}^{n}\varphi (k)\approx {\frac {3n^{2}}{\pi ^{2}}}\!}$

${\displaystyle \sum _{k=1}^{n}{\frac {\varphi (k)}{k}}\approx {\frac {6n}{\pi ^{2}}}\!}$

${\displaystyle \Gamma \left({1 \over 2}\right)={\sqrt {\pi }}\!}$ (伽玛函数)

${\displaystyle \pi ={\frac {\Gamma \left({\frac {1}{4}}\right)^{\frac {4}{3}}\mathrm {agm} (1,{\sqrt {2}})^{\frac {2}{3}}}{2}}\!}$

${\displaystyle \lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}\sum _{k=1}^{n}(n\;{\bmod {\;}}k)=1-{\frac {\pi ^{2}}{12}}\!}$

${\displaystyle \lim _{n\rightarrow \infty }10^{n+2}\cdot \sin \left({\frac {1^{\circ }}{\underbrace {55\cdots 55^{\circ }} _{\mathrm {n\;digits} }}}\right)=\pi \!}$

${\displaystyle \lim _{n\rightarrow \infty }n\cdot \sin \left({\frac {180^{\circ }}{n}}\right)=\pi }$

${\displaystyle \lim _{n\rightarrow \infty }{\frac {n}{\sqrt {2}}}\cdot {\sqrt {1-\cos \left({\frac {360^{\circ }}{n}}\right)}}=\pi }$