# List of formulae involving π

The following is a list of significant formulae involving the mathematical constant π. The list contains only formulae whose significance is established either in the article on the formula itself, the article Pi, or the article Approximations of π.

## Euclidean geometry

$\pi ={\frac {C}{d}}$

where C is the circumference of a circle, d is the diameter.

$A=\pi r^{2}$

where A is the area of a circle and r is the radius.

$V={4 \over 3}\pi r^{3}$

where V is the volume of a sphere and r is the radius.

$SA=4\pi r^{2}$

where SA is the surface area of a sphere and r is the radius.

## Physics

$\Lambda ={{8\pi G} \over {3c^{2}}}\rho$
$\Delta x\,\Delta p\geq {\frac {h}{4\pi }}$
$R_{\mu \nu }-{\frac {1}{2}}g_{\mu \nu }R+\Lambda g_{\mu \nu }={8\pi G \over c^{4}}T_{\mu \nu }$
$F={\frac {|q_{1}q_{2}|}{4\pi \varepsilon _{0}r^{2}}}$
$\mu _{0}=4\pi \cdot 10^{-7}\,\mathrm {N} /\mathrm {A} ^{2}$
• Period of a simple pendulum with small amplitude:
$T\approx 2\pi {\sqrt {\frac {L}{g}}}$
${\frac {R^{3}}{T^{2}}}={\frac {GM}{4\pi ^{2}}}$
$F={\frac {\pi ^{2}EI}{L^{2}}}$

## Formulae yielding π

### Integrals

$2\int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx=\pi$  (integrating two halfs $y(x)={\sqrt {r^{2}-x^{2}}}$  to obtain the area of a circle of radius $r=1$ )
$\int _{-\infty }^{\infty }\operatorname {sech} (x)\,dx=\pi$
$\int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-1/2t^{2}-x^{2}+xt}\,dx\,dt=\int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-t^{2}-1/2x^{2}+xt}\,dx\,dt=\pi$
$\int _{-1}^{1}{\frac {dx}{\sqrt {1-x^{2}}}}=\pi$
$\int _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi$  (integral form of arctan over its entire domain, giving the period of tan).
$\int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}$  (see Gaussian integral).
$\oint {\frac {dz}{z}}=2\pi i$  (when the path of integration winds once counterclockwise around 0. See also Cauchy's integral formula).
$\int _{-\infty }^{\infty }{\frac {\sin x}{x}}\,dx=\pi$
$\int _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx={22 \over 7}-\pi$  (see also Proof that 22/7 exceeds π).

Note that with symmetric integrands $f(-x)=f(x)$ , formulas of the form $\int _{-a}^{a}f(x)\,dx$  can also be translated to formulas $2\int _{0}^{a}f(x)\,dx$ .

### Efficient infinite series

$\sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\frac {2^{k}k!^{2}}{(2k+1)!}}={\frac {\pi }{2}}$  (see also Double factorial)
$12\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k+3/2}}}={\frac {1}{\pi }}$  (see Chudnovsky algorithm)
${\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}={\frac {1}{\pi }}$  (see Srinivasa Ramanujan, Ramanujan–Sato series)

The following are efficient for calculating arbitrary binary digits of π:

$\sum _{k=0}^{\infty }{\frac {1}{16^{k}}}\left({\frac {4}{8k+1}}-{\frac {2}{8k+4}}-{\frac {1}{8k+5}}-{\frac {1}{8k+6}}\right)=\pi$  (see Bailey–Borwein–Plouffe formula)
${\frac {1}{2^{6}}}\sum _{n=0}^{\infty }{\frac {{(-1)}^{n}}{2^{10n}}}\left(-{\frac {2^{5}}{4n+1}}-{\frac {1}{4n+3}}+{\frac {2^{8}}{10n+1}}-{\frac {2^{6}}{10n+3}}-{\frac {2^{2}}{10n+5}}-{\frac {2^{2}}{10n+7}}+{\frac {1}{10n+9}}\right)=\pi$

### Other infinite series

$\zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}$    (see also Basel problem and Riemann zeta function)
$\zeta (4)={\frac {1}{1^{4}}}+{\frac {1}{2^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{4^{4}}}+\cdots ={\frac {\pi ^{4}}{90}}$
$\zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}\,={\frac {1}{1^{2n}}}+{\frac {1}{2^{2n}}}+{\frac {1}{3^{2n}}}+{\frac {1}{4^{2n}}}+\cdots =(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}$  , where B2n is a Bernoulli number.
$\sum _{n=1}^{\infty }{\frac {3^{n}-1}{4^{n}}}\,\zeta (n+1)=\pi$ 
$\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\arctan {1}={\frac {\pi }{4}}$    (see Leibniz formula for pi)
$\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n^{2}}}={\frac {1}{1^{2}}}-{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}-{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{12}}$
$\sum _{n=1}^{\infty }{\frac {1}{(2n)^{2}}}={\frac {1}{2^{2}}}+{\frac {1}{4^{2}}}+{\frac {1}{6^{2}}}+{\frac {1}{8^{2}}}+\cdots ={\frac {\pi ^{2}}{24}}$
$\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2n+1}}\right)^{2}={\frac {1}{1^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{5^{2}}}+{\frac {1}{7^{2}}}+\cdots ={\frac {\pi ^{2}}{8}}$
$\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2n+1}}\right)^{3}={\frac {1}{1^{3}}}-{\frac {1}{3^{3}}}+{\frac {1}{5^{3}}}-{\frac {1}{7^{3}}}+\cdots ={\frac {\pi ^{3}}{32}}$
$\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2n+1}}\right)^{4}={\frac {1}{1^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{5^{4}}}+{\frac {1}{7^{4}}}+\cdots ={\frac {\pi ^{4}}{96}}$
$\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2n+1}}\right)^{5}={\frac {1}{1^{5}}}-{\frac {1}{3^{5}}}+{\frac {1}{5^{5}}}-{\frac {1}{7^{5}}}+\cdots ={\frac {5\pi ^{5}}{1536}}$
$\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2n+1}}\right)^{6}={\frac {1}{1^{6}}}+{\frac {1}{3^{6}}}+{\frac {1}{5^{6}}}+{\frac {1}{7^{6}}}+\cdots ={\frac {\pi ^{6}}{960}}$
$\sum _{n=0}^{\infty }{\frac {1}{(4n+1)(4n+3)}}={\frac {1}{1\cdot 3}}+{\frac {1}{5\cdot 7}}+{\frac {1}{9\cdot 11}}+\cdots ={\frac {\pi }{8}}$
$\pi =1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}-{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}-{\frac {1}{10}}+{\frac {1}{11}}+{\frac {1}{12}}-{\frac {1}{13}}+\cdots$    (Euler, 1748)

After the first two terms, the signs are determined as follows: If the denominator is a prime of the form 4m − 1, the sign is positive; if the denominator is a prime of the form 4m + 1, the sign is negative; for composite numbers, the sign is equal the product of the signs of its factors.

Also:

$\sum _{n=1}^{\infty }{\frac {F_{2n}}{n^{2}{\binom {2n}{n}}}}={\frac {4\pi ^{2}}{25{\sqrt {5}}}}$

where $F_{n}$  is the n-th Fibonacci number.

Some formulas relating π and harmonic numbers are given here.

### Machin-like formulae

${\frac {\pi }{4}}=\arctan 1$
${\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}}$
${\frac {\pi }{4}}=2\arctan {\frac {1}{2}}-\arctan {\frac {1}{7}}$
${\frac {\pi }{4}}=2\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}$
${\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}$  (the original Machin's formula)
${\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}$
${\frac {\pi }{4}}=6\arctan {\frac {1}{8}}+2\arctan {\frac {1}{57}}+\arctan {\frac {1}{239}}$
${\frac {\pi }{4}}=12\arctan {\frac {1}{49}}+32\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}+12\arctan {\frac {1}{110443}}$
${\frac {\pi }{4}}=44\arctan {\frac {1}{57}}+7\arctan {\frac {1}{239}}-12\arctan {\frac {1}{682}}+24\arctan {\frac {1}{12943}}$
${\frac {\pi }{2}}=\sum _{n=0}^{\infty }\arctan {\frac {1}{F_{2n+1}}}=\arctan {\frac {1}{1}}+\arctan {\frac {1}{2}}+\arctan {\frac {1}{5}}+\arctan {\frac {1}{13}}+\cdots$

where $F_{n}$  is the n-th Fibonacci number.

### Infinite series

Some infinite series involving pi are:

 $\pi ={\frac {1}{Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {((2n)!)^{3}(42n+5)}{(n!)^{6}{16}^{3n+1}}}$ $\pi ={\frac {4}{Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(21460n+1123)}{(n!)^{4}{441}^{2n+1}{2}^{10n+1}}}$ $\pi ={\frac {4}{Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(6n+1)\left({\frac {1}{2}}\right)_{n}^{3}}{{4^{n}}(n!)^{3}}}$ $\pi ={\frac {32}{Z}}$ $Z=\sum _{n=0}^{\infty }\left({\frac {{\sqrt {5}}-1}{2}}\right)^{8n}{\frac {(42n{\sqrt {5}}+30n+5{\sqrt {5}}-1)\left({\frac {1}{2}}\right)_{n}^{3}}{{64^{n}}(n!)^{3}}}$ $\pi ={\frac {27}{4Z}}$ $Z=\sum _{n=0}^{\infty }\left({\frac {2}{27}}\right)^{n}{\frac {(15n+2)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{3}}\right)_{n}\left({\frac {2}{3}}\right)_{n}}{(n!)^{3}}}$ $\pi ={\frac {15{\sqrt {3}}}{2Z}}$ $Z=\sum _{n=0}^{\infty }\left({\frac {4}{125}}\right)^{n}{\frac {(33n+4)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{3}}\right)_{n}\left({\frac {2}{3}}\right)_{n}}{(n!)^{3}}}$ $\pi ={\frac {85{\sqrt {85}}}{18{\sqrt {3}}Z}}$ $Z=\sum _{n=0}^{\infty }\left({\frac {4}{85}}\right)^{n}{\frac {(133n+8)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{6}}\right)_{n}\left({\frac {5}{6}}\right)_{n}}{(n!)^{3}}}$ $\pi ={\frac {5{\sqrt {5}}}{2{\sqrt {3}}Z}}$ $Z=\sum _{n=0}^{\infty }\left({\frac {4}{125}}\right)^{n}{\frac {(11n+1)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{6}}\right)_{n}\left({\frac {5}{6}}\right)_{n}}{(n!)^{3}}}$ $\pi ={\frac {2{\sqrt {3}}}{Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(8n+1)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{9}^{n}}}$ $\pi ={\frac {\sqrt {3}}{9Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(40n+3)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{49}^{2n+1}}}$ $\pi ={\frac {2{\sqrt {11}}}{11Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(280n+19)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{99}^{2n+1}}}$ $\pi ={\frac {\sqrt {2}}{4Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(10n+1)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{9}^{2n+1}}}$ $\pi ={\frac {4{\sqrt {5}}}{5Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(644n+41)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}5^{n}{72}^{2n+1}}}$ $\pi ={\frac {4{\sqrt {3}}}{3Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(28n+3)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{3^{n}}{4}^{n+1}}}$ $\pi ={\frac {4}{Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(20n+3)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{2}^{2n+1}}}$ $\pi ={\frac {72}{Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(260n+23)}{(n!)^{4}4^{4n}18^{2n}}}$ $\pi ={\frac {3528}{Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(21460n+1123)}{(n!)^{4}4^{4n}882^{2n}}}$ where $(x)_{n}$  is the Pochhammer symbol for the rising factorial. See also Ramanujan–Sato series.

### Infinite products

${\frac {\pi }{4}}=\left(\prod _{p\equiv 1{\pmod {4}}}{\frac {p}{p-1}}\right)\cdot \left(\prod _{p\equiv 3{\pmod {4}}}{\frac {p}{p+1}}\right)={\frac {3}{4}}\cdot {\frac {5}{4}}\cdot {\frac {7}{8}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdots ,$  (Euler)
where the numerators are the odd primes; each denominator is the multiple of four nearest to the numerator.
$\prod _{n=1}^{\infty }{\frac {4n^{2}}{4n^{2}-1}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots ={\frac {4}{3}}\cdot {\frac {16}{15}}\cdot {\frac {36}{35}}\cdot {\frac {64}{63}}\cdots ={\frac {\pi }{2}}$  (see also Wallis product)
${\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdot \cdots ={\frac {2}{\pi }}$

### Arctangent formulas

${\frac {\pi }{2^{k+1}}}=\arctan {\frac {\sqrt {2-a_{k-1}}}{a_{k}}},\qquad \qquad k\geq 2$
${\frac {\pi }{4}}=\sum _{k\geq 2}\arctan {\frac {\sqrt {2-a_{k-1}}}{a_{k}}},$

where $a_{k}={\sqrt {2+a_{k-1}}}$  such that $a_{1}={\sqrt {2}}$ .

### Continued fractions

$\pi ={3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+{\cfrac {7^{2}}{6+\ddots \,}}}}}}}}}$
$\pi ={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}$
$\pi ={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+\ddots }}}}}}}}}}$
$2\pi ={6+{\cfrac {2^{2}}{12+{\cfrac {6^{2}}{12+{\cfrac {10^{2}}{12+{\cfrac {14^{2}}{12+{\cfrac {18^{2}}{12+\ddots }}}}}}}}}}}$

For more on the third identity, see Euler's continued fraction formula.

### Miscellaneous

$n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}$  (Stirling's approximation)
$e^{i\pi }+1=0$  (Euler's identity)
$\sum _{k=1}^{n}\varphi (k)\sim {\frac {3n^{2}}{\pi ^{2}}}$  (see Euler's totient function)
$\sum _{k=1}^{n}{\frac {\varphi (k)}{k}}\sim {\frac {6n}{\pi ^{2}}}$  (see Euler's totient function)
$\Gamma \left({1 \over 2}\right)={\sqrt {\pi }}$  (see also Gamma function)
$\pi ={\frac {\Gamma \left({1/4}\right)^{4/3}\operatorname {agm} (1,{\sqrt {2}})^{2/3}}{2}}$  (where agm is the arithmetic–geometric mean)
$\lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}\sum _{k=1}^{n}(n{\bmod {k}})=1-{\frac {\pi ^{2}}{12}}$  (where ${\textstyle n{\bmod {k}}}$  is the remainder upon division of n by k)
$\pi =\lim _{n\rightarrow \infty }{\frac {4}{n^{2}}}\sum _{k=1}^{n}{\sqrt {n^{2}-k^{2}}}$  (Riemann sum to evaluate the area of the unit circle)
$\pi =\lim _{n\rightarrow \infty }{\frac {2^{4n}}{n{2n \choose n}^{2}}}$  (by Stirling's approximation)