فهرست انتگرال‌های توابع گوسی

در این عبارت‌ها، ${\displaystyle \phi (x)={\tfrac {1}{\sqrt {2\pi }}}e^{-{\frac {1}{2}}x^{2}}}$ توزیع نرمال تابع چگالی احتمال است و ${\displaystyle \textstyle \Phi (x)=\int _{-\infty }^{x}\phi (t)dt={\frac {1}{2}}{\big (}1+\operatorname {erf} {\big (}{\frac {x}{\sqrt {2}}}{\big )}{\big )}}$ تابع توزیع تجمعی مربوط به آن است که در آن erf، تابع خطا و ${\displaystyle T(h,a)=\phi (h)\int _{0}^{a}{\frac {\phi (hx)}{1+x^{2}}}\,dx}$ می‌باشد.

در پایین چند نمونه انتگرال آمده‌است:

انتگرال‌های نامعین

${\displaystyle \int \phi (x)\,dx=\Phi (x)+C}$

${\displaystyle \int x\phi (x)\,dx=-\phi (x)+C}$

${\displaystyle \int x^{2}\phi (x)\,dx=\Phi (x)-x\phi (x)+C}$

${\displaystyle \int x^{2k+1}\phi (x)\,dx=-\phi (x)\sum _{j=0}^{k}{\frac {(2k)!!}{(2j)!!}}x^{2j}+C}$[nb 1]

${\displaystyle \int x^{2k+2}\phi (x)\,dx=-\phi (x)\sum _{j=0}^{k}{\frac {(2k+1)!!}{(2j+1)!!}}x^{2j+1}+(2k+1)!!\,\Phi (x)+C}$

(در این انتگرال‌ها، n!! همان دابل فاکتوریل است یعنی: برای nهای زوج برابر است با حاصل ضرب ۲ تا n و برای nهای فرد برابر است با حاصل ضرب ۱ تا n همچنین فرض می‌شود که ۰!! = (−۱)!! = ۱)

${\displaystyle \int \phi (x)^{2}\,dx={\tfrac {1}{2{\sqrt {\pi }}}}\Phi (x{\sqrt {2}})+C}$

${\displaystyle \int \phi (x)\phi (a+bx)\,dx={\tfrac {1}{t}}\phi (a/t)\Phi (tx+ab/t)+C,\quad t={\sqrt {1+b^{2}}}}$[nb 2]

${\displaystyle \int x\phi (a+bx)\,dx=-{\tfrac {1}{b^{2}}}\phi (a+bx)-{\tfrac {a}{b^{2}}}\Phi (a+bx)+C}$

${\displaystyle \int x^{2}\phi (a+bx)\,dx={\tfrac {a^{2}+1}{b^{3}}}\Phi (a+bx)+{\frac {a-bx}{b^{3}}}\phi (a+bx)+C}$

${\displaystyle \int \phi (a+bx)^{n}\,dx={\frac {(2\pi )^{-(n-1)/2}}{b{\sqrt {n}}}}\Phi {\big (}{\sqrt {n}}(a+bx){\big )}+C}$

${\displaystyle \int \Phi (a+bx)\,dx={\tfrac {1}{b}}(a+bx)\Phi (a+bx)+{\tfrac {1}{b}}\phi (a+bx)+C}$

${\displaystyle \int x\Phi (a+bx)\,dx={\tfrac {1}{2b^{2}}}{\big (}(b^{2}x^{2}-a^{2}-1)\Phi (a+bx)+(bx-a)\phi (a+bx){\big )}+C}$

${\displaystyle \int x^{2}\Phi (a+bx)\,dx={\tfrac {1}{3b^{3}}}{\big (}(b^{3}x^{3}+a^{3}+3a)\Phi (a+bx)+(b^{2}x^{2}-abx+a^{2}+2)\phi (a+bx){\big )}+C}$

${\displaystyle \int x^{n}\Phi (x)\,dx={\tfrac {1}{n+1}}{\Big (}(x^{n+1}-nx^{n-1})\Phi (x)+x^{n}\phi (x)+n(n-1)\int x^{n-2}\Phi (x)\,dx{\Big )}+C}$

${\displaystyle \int x\phi (x)\Phi (a+bx)\,dx={\tfrac {b}{t}}\phi (a/t)\Phi (xt+ab/t)-\phi (x)\Phi (a+bx)+C,\quad t={\sqrt {1+b^{2}}}}$

${\displaystyle \int \Phi (x)^{2}\,dx=x\Phi (x)^{2}+2\Phi (x)\phi (x)-{\tfrac {1}{\sqrt {\pi }}}\Phi (x{\sqrt {2}})+C}$

${\displaystyle \int e^{cx}\phi (bx)^{n}\,dx={\frac {1}{b{\sqrt {n(2\pi )^{n-1}}}}}e^{c^{2}/(2nb^{2})}\Phi (bx{\sqrt {n}}-{\tfrac {c}{b{\sqrt {n}}}})+C,\quad b\neq 0,n>0}$

انتگرال‌های معین

${\displaystyle \int _{-\infty }^{\infty }x^{2}\phi (x)^{n}\,\,dx={\Big (}n^{3/2}(2\pi )^{(n-1)/2}{\Big )}^{-1}}$

${\displaystyle \int _{-\infty }^{0}\phi (ax)\Phi (bx)dx=(2\pi a)^{-1}\arctan(a/b)}$

${\displaystyle \int _{0}^{\infty }\phi (ax)\Phi (bx)\,dx=(2\pi a)^{-1}{\big (}{\tfrac {\pi }{2}}-\arctan(b/a){\big )}}$

${\displaystyle \int _{0}^{\infty }x\phi (x)\Phi (bx)\,dx={\frac {1}{2{\sqrt {2\pi }}}}{\bigg (}1+{\frac {b}{\sqrt {1+b^{2}}}}{\bigg )}}$

${\displaystyle \int _{0}^{\infty }x^{2}\phi (x)\Phi (bx)\,dx={\frac {1}{4}}+{\frac {1}{2\pi }}{\bigg (}{\frac {b}{1+b^{2}}}+\arctan b{\bigg )}}$

${\displaystyle \int x\phi (x)^{2}\Phi (x)\,dx={\frac {1}{4\pi {\sqrt {3}}}}}$

${\displaystyle \int _{0}^{\infty }\Phi (bx)^{2}\phi (x)\,dx=(2\pi )^{-1}{\big (}\arctan b+\arctan {\sqrt {1+2b^{2}}}{\big )}}$

${\displaystyle \int _{-\infty }^{\infty }\Phi (a+bx)^{2}\phi (x)\,dx=\Phi {\bigg (}{\frac {a}{\sqrt {1+b^{2}}}}{\bigg )}-2T{\bigg (}{\frac {a}{\sqrt {1+b^{2}}}},{\frac {1}{\sqrt {1+2b^{2}}}}{\bigg )}}$

${\displaystyle \int _{-\infty }^{\infty }x\Phi (a+bx)^{2}\phi (x)\,dx={\frac {2b}{\sqrt {1+b^{2}}}}\phi (a/t)\Phi {\bigg (}{\frac {a}{{\sqrt {1+b^{2}}}{\sqrt {1+2b^{2}}}}}{\bigg )}}$[nb 3]

${\displaystyle \int _{-\infty }^{\infty }\Phi (bx)^{2}\phi (x)\,dx=\pi ^{-1}\arctan {\sqrt {1+2b^{2}}}}$

${\displaystyle \int _{-\infty }^{\infty }x\phi (x)\Phi (bx)\,dx=\int _{-\infty }^{\infty }x\phi (x)\Phi (bx)^{2}\,dx={\frac {b}{\sqrt {2\pi (1+b^{2})}}}}$

${\displaystyle \int _{-\infty }^{\infty }\Phi (a+bx)\phi (x)\,dx=\Phi {\big (}a/{\sqrt {1+b^{2}}}{\big )}}$

${\displaystyle \int _{-\infty }^{\infty }x\Phi (a+bx)\phi (x)\,dx=(b/t)\phi (a/t),\quad t={\sqrt {1+b^{2}}}}$

${\displaystyle \int _{0}^{\infty }x\Phi (a+bx)\phi (x)\,dx=(b/t)\phi (a/t)\Phi (-ab/t)+(2\pi )^{-1/2}\Phi (a),\quad t={\sqrt {1+b^{2}}}}$

${\displaystyle \int _{-\infty }^{\infty }\ln(x^{2}){\tfrac {1}{\sigma }}\phi {\big (}{\tfrac {x}{\sigma }}{\big )}\,dx=\ln(\sigma ^{2})-\gamma -\ln 2\approx \ln(\sigma ^{2})-1.27036}$

منابع

• 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. pp. 389–419.

1. Patel & Read (1996) list this integral without the minus sign, which is an error. See calculation by WolframAlpha
2. Patel & Read (1996) report this integral with error, see WolframAlpha
3. Patel & Read (1996) report this integral incorrectly by omitting x from the integrand