为标准正态概率密度函数， Φ ( x ) = ∫ − ∞ x ϕ ( t ) d t = 1 2 ( 1 + erf ( x 2 ) ) {\displaystyle \Phi (x)=\int _{-\infty }^{x}\phi (t)\,dt={\frac {1}{2}}\left(1+\operatorname {erf} \left({\frac {x}{\sqrt {2}}}\right)\right)} 在这些表达式中， ϕ ( x ) = 1 2 π e − 1 2 x 2 {\displaystyle \phi (x)={\frac {1}{\sqrt {2\pi }}}e^{-{\frac {1}{2}}x^{2}}} 为对应的累积分布函数，其中erf为误差函数， 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} 为欧文T函数（英语：Owen's T function）。 ∫ ϕ ( x ) d x = Φ ( x ) + C {\displaystyle \int \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} ∫ 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} 在这些积分中，n!!为双阶乘：若n为偶数，则等于从2到n的所有偶数的乘积；若n为奇数，则等于从1到n的所有奇数的乘积；特别地，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}}\right)+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 \int \phi (x)\phi (a+bx)\,dx={\frac {1}{t}}\phi \left({\frac {a}{t}}\right)\Phi \left(tx+{\frac {ab}{t}}\right)+C,\qquad t={\sqrt {1+b^{2}}}} ∫ 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}}}\left(\phi (a+bx)+a\Phi (a+bx)\right)+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}}}\left((a^{2}+1)\Phi (a+bx)+(a-bx)\phi (a+bx)\right)+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(2\pi )^{n-1}}}}}\Phi \left({\sqrt {n}}(a+bx)\right)+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}}\left((a+bx)\Phi (a+bx)+\phi (a+bx)\right)+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}}}\left((b^{2}x^{2}-a^{2}-1)\Phi (a+bx)+(bx-a)\phi (a+bx)\right)+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={\frac {1}{3b^{3}}}\left((b^{3}x^{3}+a^{3}+3a)\Phi (a+bx)+(b^{2}x^{2}-abx+a^{2}+2)\phi (a+bx)\right)+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}}\left(\left(x^{n+1}-nx^{n-1}\right)\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}{t}}\phi \left({\frac {a}{t}}\right)\Phi \left(xt+{\frac {ab}{t}}\right)-\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)\phi (x)-{\frac {1}{\sqrt {\pi }}}\Phi \left(x{\sqrt {2}}\right)+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} ∫ − ∞ ∞ x 2 ϕ ( x ) n d x = 1 n 3 ( 2 π ) n − 1 {\displaystyle \int _{-\infty }^{\infty }x^{2}\phi (x)^{n}\,dx={\frac {1}{\sqrt {n^{3}(2\pi )^{n-1}}}}} ∫ − ∞ 0 ϕ ( a x ) Φ ( b x ) d x = 1 2 π | a | ( π 2 − arctan ( b | a | ) ) {\displaystyle \int _{-\infty }^{0}\phi (ax)\Phi (bx)dx={\frac {1}{2\pi |a|}}\left({\frac {\pi }{2}}-\arctan \left({\frac {b}{|a|}}\right)\right)} ∫ 0 ∞ ϕ ( a x ) Φ ( b x ) d x = 1 2 π | a | ( π 2 + arctan ( b | a | ) ) {\displaystyle \int _{0}^{\infty }\phi (ax)\Phi (bx)\,dx={\frac {1}{2\pi |a|}}\left({\frac {\pi }{2}}+\arctan \left({\frac {b}{|a|}}\right)\right)} ∫ 0 ∞ x ϕ ( x ) Φ ( b x ) d x = 1 2 2 π ( 1 + b 1 + b 2 ) {\displaystyle \int _{0}^{\infty }x\phi (x)\Phi (bx)\,dx={\frac {1}{2{\sqrt {2\pi }}}}\left(1+{\frac {b}{\sqrt {1+b^{2}}}}\right)} ∫ 0 ∞ x 2 ϕ ( x ) Φ ( b x ) d x = 1 4 + 1 2 π ( b 1 + b 2 + arctan ( b ) ) {\displaystyle \int _{0}^{\infty }x^{2}\phi (x)\Phi (bx)\,dx={\frac {1}{4}}+{\frac {1}{2\pi }}\left({\frac {b}{1+b^{2}}}+\arctan(b)\right)} ∫ 0 ∞ x ϕ ( x ) 2 Φ ( x ) d x = 1 4 π 3 {\displaystyle \int _{0}^{\infty }x\phi (x)^{2}\Phi (x)\,dx={\frac {1}{4\pi {\sqrt {3}}}}} ∫ 0 ∞ Φ ( b x ) 2 ϕ ( x ) d x = 1 2 π ( arctan ( b ) + arctan 1 + 2 b 2 ) {\displaystyle \int _{0}^{\infty }\Phi (bx)^{2}\phi (x)\,dx={\frac {1}{2\pi }}\left(\arctan(b)+\arctan {\sqrt {1+2b^{2}}}\right)} ∫ − ∞ ∞ 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 _{-\infty }^{\infty }\Phi (a+bx)^{2}\phi (x)\,dx=\Phi \left({\frac {a}{\sqrt {1+b^{2}}}}\right)-2T\left({\frac {a}{\sqrt {1+b^{2}}}},{\frac {1}{\sqrt {1+2b^{2}}}}\right)} ∫ − ∞ ∞ 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 _{-\infty }^{\infty }x\Phi (a+bx)^{2}\phi (x)\,dx={\frac {2b}{\sqrt {1+b^{2}}}}\phi \left({\frac {a}{t}}\right)\Phi \left({\frac {a}{{\sqrt {1+b^{2}}}{\sqrt {1+2b^{2}}}}}\right)} ∫ − ∞ ∞ Φ ( b x ) 2 ϕ ( x ) d x = 1 π arctan 1 + 2 b 2 {\displaystyle \int _{-\infty }^{\infty }\Phi (bx)^{2}\phi (x)\,dx={\frac {1}{\pi }}\arctan {\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 _{-\infty }^{\infty }\Phi (a+bx)\phi (x)\,dx=\Phi \left({\frac {a}{\sqrt {1+b^{2}}}}\right)} ∫ − ∞ ∞ x Φ ( a + b x ) ϕ ( x ) d x = b t ϕ ( a t ) , t = 1 + b 2 {\displaystyle \int _{-\infty }^{\infty }x\Phi (a+bx)\phi (x)\,dx={\frac {b}{t}}\phi \left({\frac {a}{t}}\right),\qquad t={\sqrt {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}^{\infty }x\Phi (a+bx)\phi (x)\,dx={\frac {b}{t}}\phi \left({\frac {a}{t}}\right)\Phi \left(-{\frac {ab}{t}}\right)+{\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} Wikiwand in your browser!Seamless Wikipedia browsing. 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