다음은 원주율의 미적분식에 관한 내용이다.
- 이용
F ( ω ) = F { f } ( ω ) = 1 2 π ∫ − ∞ ∞ f ( t ) e − i ω t d t {\displaystyle F(\omega )={\mathcal {F}}\{f\}(\omega )={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }f(t)e^{-\mathrm {i} \omega t}\,\mathrm {d} t} .
- 적분형식
f ( z ) = 1 2 π i ∮ γ f ( ζ ) ζ − z d ζ {\displaystyle f(z)={\frac {1}{2\pi \mathrm {i} }}\oint _{\gamma }{\frac {f(\zeta )}{\zeta -z}}\mathrm {d} \zeta } .
∫ − ∞ ∞ e − x 2 d x = π {\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}} =1.7724538509055160272981674833411451827975494561223871282138077898...
∫ − ∞ ∞ e − x 2 d x = π {\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}}
∫ R 2 e − | x | 2 d x = π {\displaystyle \int _{\mathbb {R} ^{2}}e^{-|x|^{2}}\mathrm {d} x=\pi } ∫ − ∞ ∞ d x 1 + x 2 = 2 ⋅ ∫ − 1 1 d x 1 + x 2 = π {\displaystyle \int _{-\infty }^{\infty }{\frac {\mathrm {d} x}{1+x^{2}}}=2\cdot \int _{-1}^{1}{\frac {\mathrm {d} x}{1+x^{2}}}=\pi }
π = 2 ∫ 0 ∞ sin x x d x {\displaystyle \pi =2\int _{0}^{\infty }{\frac {\sin x}{x}}\,\mathrm {d} x}
π = ∫ − 1 1 d t 1 − t 2 {\displaystyle \pi =\int _{-1}^{1}{\frac {dt}{\sqrt {1-t^{2}}}}}
π = 2 ∫ − 1 1 1 − t 2 d t {\displaystyle \pi =2\int _{-1}^{1}{\sqrt {1-t^{2}}}\,dt}
π = 4 ∫ 0 1 d t 1 + t 2 {\displaystyle \pi =4\int _{0}^{1}{\frac {dt}{1+t^{2}}}}
π = 2 ∫ 0 1 d t 1 − t 2 {\displaystyle \pi =2\int _{0}^{1}{\frac {dt}{\sqrt {1-t^{2}}}}} [1]
