위키백과 목록 기사
이런 표현들에서는
ϕ ( x ) = 1 2 π e − 1 2 x 2 {\displaystyle \phi (x)={\frac {1}{\sqrt {2\pi }}e^{-{\frac {1}{2}}:} 표준 정규 확률밀도함수,
Φ ( x ) = ∫ − ∞ x ϕ ( t ) d t = 1 2 ( 1 + 에프 ( x 2 ) ) {\displaystyle \Phi(x)=\int_{-\inflit }^{x}\phi(t)\,dt={\frac {1}{1}:{2}}\좌측(1+\operatorname {erf} \{\frac {x}{\sqrt{2}}\오른쪽)}}}} 해당 누적분포함수 (erf 는 오차함수 ) 및
T ( h , a ) = ϕ ( h ) ∫ 0 a ϕ ( h x ) 1 + x 2 d x {\displaystyle T(h,a)=\phi(h)\int _{0}^{a}{\frac {\phi(hx)}{1+x^{2 }}}\,dx} 오웬의 T기능이야 null
오웬은[nb 1] null
무한정집적 ∫ ϕ ( x ) d x = Φ ( x ) + C \phi(x)\,dx=\Phi(x)+ C} ∫ x ϕ ( x ) d x = − ϕ ( x ) + C \displaystyle \int x\phi (x)\,dx=-\phi (x)+ C} ∫ x 2 ϕ ( x ) d x = Φ ( x ) − x ϕ ( x ) + C {\displaystyle \int x^{2}\phi(x)\,dx=\Phi(x)-x\phi(x)++ C} ∫ x 2 k + 1 ϕ ( x ) d x = − ϕ ( x ) ∑ j = 0 k ( 2 k ) ! ! ( 2 j ) ! ! x 2 j + C {\displaystyle \int x^{2k+1}\phi(x)\,dx=-\phi(x)\sum_{j=0}^{k}{\frac {(2k)!! }{{(2j)!! }}}x^{2j}+C} [nb 2] ∫ x 2 k + 2 ϕ ( x ) d x = − ϕ ( x ) ∑ j = 0 k ( 2 k + 1 ) ! ! ( 2 j + 1 ) ! ! x 2 j + 1 + ( 2 k + 1 ) ! ! Φ ( x ) + C {\displaystyle \int x^{2k+2}\phi (x)\,dx=-\phi (x)\sum _{j=0}^{k}{\frac {(2k+1)!! }{{(2j+1)!! }}}x^{2j+1}+(2k+1)!! \,\Phi (x)+ C} 이러한 통합에서, n !!는 이중 요인이다. n 의 경우에도 2 부터 n까지의 짝수 숫자의 산물과 같고, 홀수 n의 경우 1 부터 n까지의 모든 홀수 숫자의 산물이다. 추가적으로 0!! = (−1)!! = 1 .
∫ ϕ ( x ) 2 d x = 1 2 π Φ ( x 2 ) + C {\displaystyle \int \phi(x)^{2}\,dx={\frac {1}{2{\sqrt{\pi }}}}\Phi \left(x{\sqrt {2}}\오른쪽)+ C} ∫ ϕ ( x ) ϕ ( a + b x ) d x = 1 t ϕ ( a t ) Φ ( t x + a b t ) + C , t = 1 + b 2 {\displaystyle \phi(x)\phi(a+bx)\,dx={\frac {1}{1}\phi \leftm\frac {a}{t}\오른쪽)\ Phi \left(tx+{\frac {ab}{t}\오른쪽)+C,\qquad t={\sqrt{1+b^{2} }}}} [nb 3] ∫ x ϕ ( a + b x ) d x = − 1 b 2 ( ϕ ( a + b x ) + a Φ ( a + b x ) ) + C {\displaystyle \int x\phi (a+bx)\,dx=-{\frac {1}{b^{2}}\\좌측(\phi(a+bx)+a)+a \Phi(a+bx)\오른쪽)+ C} ∫ x 2 ϕ ( a + b x ) d x = 1 b 3 ( ( a 2 + 1 ) Φ ( a + b x ) + ( a − b x ) ϕ ( a + b x ) ) + C {\displaystyle \int x^{2}\phi(a+bx)\,dx={\frac {1}{b^{3}}}\좌측(a^{2}+1) \Phi(a+bx)+(a-bx)\phi(a+bx)\오른쪽)+ C} ∫ ϕ ( a + b x ) n d x = 1 b n ( 2 π ) n − 1 Φ ( n ( a + b x ) ) + C \displaystyle \int \phi (a+bx)^ {n}\,dx={\frac {1}{b{\sqrt{n(pi )^{n-1}}\Phi \left({\sqrt{n}})(a+bx)\오른쪽)++ C} ∫ Φ ( a + b x ) d x = 1 b ( ( a + b x ) Φ ( a + b x ) + ϕ ( a + b x ) ) + C {\displaystyle \int \Phi(a+bx)\,dx={\frac {1}{b}\왼쪽(a+bx)\ 피(a+bx)+ \phi (a+bx)\오른쪽)+ C} ∫ x Φ ( a + b x ) d x = 1 2 b 2 ( ( b 2 x 2 − a 2 − 1 ) Φ ( a + b x ) + ( b x − a ) ϕ ( a + b x ) ) + C {\displaystyle \int x\Phi (a+bx)\,dx={\frac {1}{2b^{2}}:}\좌측(b^{2}x^{2}-a-1)\Phi(a+bx)+(a+bx)\우측)+++++++++++++++++++++; C} ∫ x 2 Φ ( a + b x ) d x = 1 3 b 3 ( ( b 3 x 3 + a 3 + 3 a ) Φ ( a + b x ) + ( b 2 x 2 − a b x + a 2 + 2 ) ϕ ( a + b x ) ) + C {\displaystyle \int x^{2}\Phi(a+bx)\,dx={1}{3b^{3}}}}}\(b^{3}x^{3}+a}+3a)\\ Phi (a+bx)+(b^{2}x^{2}-abx+a^{2}+2)\phi (a+bx)\오른쪽)++ C} ∫ x n Φ ( x ) d x = 1 n + 1 ( ( x n + 1 − n x n − 1 ) Φ ( x ) + x n ϕ ( x ) + n ( n − 1 ) ∫ x n − 2 Φ ( x ) d x ) + C {\displaystyle \int x^{n}\Phi(x)\,dx={\frac {1}{n+1}:{n+1}}\좌측(\좌측(x^{n+1}-nx^{n-1}\우측) \Phi (x)+x^{n}\phi (x)+n(n-1)\int x^{n-2}\Phi (x)\dx\right)+++ C} ∫ x ϕ ( x ) Φ ( a + b x ) d x = b t ϕ ( a t ) Φ ( x t + a b t ) − ϕ ( x ) Φ ( a + b x ) + C , t = 1 + b 2 {\displaystyle \int x\phi (x)\Phi (a+bx)\,dx={\frac {b}}}}\phi \좌측({\frac {a}{t}\오른쪽)\\ Phi \left(xt+{\frac {t}}\오른쪽)-\phi(x)\Phi(a+bx)+C,\qquad t={\sqrt {1+b^{2} }}}} ∫ Φ ( x ) 2 d x = x Φ ( x ) 2 + 2 Φ ( x ) ϕ ( x ) − 1 π Φ ( x 2 ) + C {\displaystyle \int \Phi(x)^{2}\,dx=x\Phi(x)^{2}+2\Phi(x)-{\sqrt{\pi }}}\Phi \left(x{\sqrt {2}}\오른쪽)+++ C} ∫ e c x ϕ ( b x ) n d x = e c 2 2 n b 2 b n ( 2 π ) n − 1 Φ ( b 2 x n − c b n ) + C , b ≠ 0 , n > 0 {\displaystyle \int e^{cx}\phi (bx)^{n}\,dx={\frac {e^{\frac {c^{2}}{2nb^{2}}}}{b{\sqrt {n(2\pi )^{n-1}}}}}\Phi \left({\frac {b^{2}xn-c}{b{\sqrt {n}}}}\right)+C,\qquad b\neq 0,n>0} 확정집적 ∫ − ∞ ∞ x 2 ϕ ( x ) n d x = 1 n 3 ( 2 π ) n − 1 {\displaystyle \int_{-\infit _{-\}^{2}\pi (x)^{n}\,dx={\x={1}\sqrt {1}{n^{3}(2\pi )^{n-1}}}}}}} ∫ − ∞ 0 ϕ ( a x ) Φ ( b x ) d x = 1 2 π a ( π 2 − 아크탄의 ( b a ) ) {\displaystyle \int_{\fract \{0}\pi (ax)\Phi (bx)dx={\frac {1}{2\pi a}}}}}{\fractan \left({\frac {b}{}}}}}}}오른쪽)} ∫ 0 ∞ ϕ ( a x ) Φ ( b x ) d x = 1 2 π a ( π 2 + 아크탄의 ( b a ) ) {\displaystyle \int _{0}^{\inflit }\pi (ax)\pi (bx)\,dx={\frac {1}{2\pi a}}}}}{\frac {2}}+\arctan \left({\frac {b}{}{}}}}}오른쪽)\rig)}}}}}}}}}}}} ∫ 0 ∞ x ϕ ( x ) Φ ( b x ) d x = 1 2 2 π ( 1 + b 1 + b 2 ) {\displaystyle \int _{0}^{}\x\phi (x)\phi (bx)\,dx={\frac {1}{2\sqrt{2\pi }}}}}}}\좌측(1+{\frac {b}{1+b^{2)2 }}}} ∫ 0 ∞ x 2 ϕ ( x ) Φ ( b x ) d x = 1 4 + 1 2 π ( b 1 + b 2 + 아크탄의 ( b ) ) {\displaystyle \int_{0}^{0}^{}\x^{2}\pi (x)\Phi (x)\Phi (x)\,dx={\frac{1}{1}{1}{2\pi }}}\frac{1}{1+b^{2}{2}}{2}{2}}}}{2}{2}}}{2}}{2}{2}}}{2}}{2}}{2}}}}{2}}}}}{2}}}} }}}+\arctan(b)\오른쪽)} ∫ − ∞ ∞ x ϕ ( x ) 2 Φ ( x ) d x = 1 4 π 3 {\displaystyle \int _{-\infit _}^{}\x\phi (x)^{2}\Phi (x)\,dx={1}{4\pi {\sqrt{3}}}}}}}} ∫ 0 ∞ Φ ( b x ) 2 ϕ ( x ) d x = 1 2 π ( 아크탄의 ( b ) + 아크탄의 1 + 2 b 2 ) {\displaystyle \int _{0}^{\inful }\Phi (bx)^{2}\pi (x)\,dx={1}{2\pi }}}\좌(\arctan(b)+\sqrt{1+2b^{2} }}}\오른쪽)} ∫ − ∞ ∞ Φ ( a + b x ) 2 ϕ ( x ) d x = Φ ( a 1 + b 2 ) − 2 T ( a 1 + b 2 , 1 1 + 2 b 2 ) {\displaystyle \int _{-\infit }^{\\infit }\Phi (a+bx)^{2}\phi (x)\dx=\\ Phi \left({\frac {a}{\sqrt {1+b^{2) }}}}\{\{\frac {a}{\sqrt{1+b^{2} }}}},{\frac {1}{\sqrt{1+2b^{2} }}}} ∫ − ∞ ∞ x Φ ( a + b x ) 2 ϕ ( x ) d x = 2 b 1 + b 2 ϕ ( a t ) Φ ( a 1 + b 2 1 + 2 b 2 ) {\displaystyle \int _{-\infit _}^{\x\Phi (a+bx)^{2}\phi (x)\,dx={2b}{\sqrt {1+b^{2}{\sqrt {1+b^{2} }}}}\phi \left\frac {a}{t}\오른쪽)\ Phi \left({\frac {a}{{\sqrt {1+b^{2) }}}{\sqrt{1+2b^{2} }}}\오른쪽)} [nb 4] ∫ − ∞ ∞ Φ ( b x ) 2 ϕ ( x ) d x = 1 π 아크탄의 1 + 2 b 2 {\displaystyle \int_{-\infit _{-\infit }^{2}\Phi (bx)^{2}\,dx={\frac{1}{\pi}}}\sqrt{1+2b^{2} }}}} ∫ − ∞ ∞ x ϕ ( x ) Φ ( b x ) d x = ∫ − ∞ ∞ x ϕ ( x ) Φ ( b x ) 2 d x = b 2 π ( 1 + b 2 ) {\displaystyle \int _{-\infty }^{\infty }x\phi (x)\Phi (bx)\,dx=\int _{-\infty }^{\infty }x\phi (x)\Phi (bx)^{2}\,dx={\frac {b}{\sqrt {2\pi (1+b^{2})}}}} ∫ − ∞ ∞ Φ ( a + b x ) ϕ ( x ) d x = Φ ( a 1 + b 2 ) {\displaystyle \int _{-\infit }^{\\infit }\Phi (a+bx)\phi (x)\,dx=\ Phi \left({\frac {a}{\sqrt {1+b^{2) }}}} ∫ − ∞ ∞ x Φ ( a + b x ) ϕ ( x ) d x = b t ϕ ( a t ) , t = 1 + b 2 {\displaystyle \int _{-\infit _}^{}\x\Phi (a+bx)\phi\,dx={\frac {b}{t}}\phi \좌측({\frac {a}{t}\오른쪽)\qqrt={1+b^2 }}}} ∫ 0 ∞ x Φ ( a + b x ) ϕ ( x ) d x = b t ϕ ( a t ) Φ ( − a b t ) + 1 2 π Φ ( a ) , t = 1 + b 2 {\displaystyle \int_{0}^{0}^{\inflt }x\Phi (a+bx)\phi\,dx={\frac {b}}{t}\phi \왼쪽({\frac {a}{t}\오른쪽)\ Phi \left(-{\frac {t}{t}\오른쪽)+{\frac {1}{\sqrt {2\pi }}\Phi(a),\qquad t={\sqrt {1+b^{2} }}}} ∫ − ∞ ∞ ln ( x 2 ) 1 σ ϕ ( x σ ) d x = ln ( σ 2 ) − γ − ln 2 ≈ ln ( σ 2 ) − 1.27036 {\displaystyle \int _{-\infty }^{\infty }\ln(x^{2}){\frac {1}{\sigma }}\phi \left({\frac {x}{\sigma }}\right)\,dx=\ln(\sigma ^{2})-\gamma -\ln 2\approx \ln(\sigma ^{2})-1.27036} 참조 Patel, Jagdish K.; Read, Campbell B. (1996). Handbook of the normal distribution (2nd ed.). CRC Press. ISBN 0-8247-9342-0 Owen, D. (1980). "A table of normal integrals". Communications in Statistics: Simulation and Computation . B9 (4): 389–419. doi :10.1080/03610918008812164 .
